Verification Manual Staad Fundation v8i

STAAD Foundation Advanced

V8i

Verification ManualDAA039800-1/0001

Last updated: 26 July 2011

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Verification Manual — i

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ii — (Undefined variable: Primary.ProductName)

Chapter 2

Table of Contents

Introduction 1Section 1 Australian Code (AS3600-2001[AMnd2004]) 31.1 General Isolated Foundation 1 3

1.2 General Isolated Foundation 2 6

1.3 General Combined Foundation 1 14


Section 2 British Code (BS8110-1-1997) 232.1 General Isolated Foundation 1 23









2.10 Mat Combined Foundation 83

2.11 General Isolated Foundation with Eccentricity 88

Section 3 Canadian Code (CSA A23.3-2004) 993.1 CSA General Isolated Foundation 1 99

3.2 CSA General Isolated Foundation 2 105

3.3 CSA General Isolated Foundation 3 112

3.5 CSA Pilecap Foundation 1 115

3.4 CSA General Combined Foundation s1 122

Section 4 Indian Code (IS 456 -2000) 1254.1 IS General Isolated Foundation 1 125

4.2 IS General Isolated Foundation 2 129




4.6 IS General Isolated Foundations 6 145


Verification Manual — iii

4.8 IS Toolkit Combined 1 158

4.9 IS Toolkit Combined Foundation 2 164



4.12 IS Pilecap 1 182

4.13 IS Pilecap 2 189

4.14 IS Mat Combined Foundation 1 197

Section 5 United States Code (ACI 318 -2005) 2015.1 US General Isolated Foundation 1 201

5.2 US General Isolated Foundation 2 206






5.8 US General Combined Foundation 1 241




5.12 US Pilecap Foundation 1 264




5.16 US Mat Combined Foundation 1 295

5.17 US General Isolated Foundation with Sliding & Overturning 302

5.18 US General Isolated Foundation with Eccentric Loading 311

Section 6 Deadman Anchors (ACI 318 -2005) 3216.1 Deadman Guy Anchor US 1 321

6.2 Deadman Guy Anchor US 2 330



Section 7 Drilled Pier Foundations 3577.1 Drilled Pier Foundation 1 API 357

7.2 Drilled Pier Foundation 2 API 361

7.3 Drilled Pier Foundation 3 FHWA 366

iv — STAAD Foundation Advanced V8i

Chapter — 3

7.4 Drilled Pier Foundation 4 FHWA 371

7.5 Drilled Pier Foundation 5 VESIC 375

7.6 Drilled Pier Foundation 6 Vesic 380

Section 8 Plant Foundation 3858.1 Vertical Vessel Foundation 1 385

8.2 Vertical Vessel Foundation Design 394

8.3 Vertical Vessel Foundation Design 403

8.4 Vertical Vessel Seismic Load Generation 1 412









8.13 Vertical Vessel Wind Load Generation 1 422




8.17 Horizontal Vessel Applied Loads 1 427

8.18 Horizontal Vessel Applied Loads 2 431

Section 9 Chinese Code (GB50007-2002) 4379.1 Cone Footing Design 437

9.2 PKPM Isolated Footing Design 445

9.3 Stepped Foundation Design 449

9.4 PKPM Stepped Footing Design 457

9.5 Combined Foundation 461

9.6 Pile Foundation Design 470

Section 10 Technical Support 485Index 487List of Figures & Tables 489Figures 489

Tables 493

Verification Manual — v

vi — (Undefined variable: Primary.ProductName)

Chapter 3

IntroductionThis document is intended to use as a hand calculation reference for STAAD FoundationAdvanced V8i (Release 6.0) verification problems. Verification Problems can be found underStart Page > Example > Verification.

Each section in this manual represents either specific design code (e.g., AS3600-2001) orparticular foundation type (e.g., Dead Man Anchor Guy Foundation). Hand calculation title(e.g., AS GEN ISO 1) indicates corresponding STAAD Foundation file name.

At end of each hand calculation a comparison table between hand calculations and programresults is provided for various output parameters like bearing pressure, overturning andsliding factor of safety, shear force, etc.

Verification Manual — 1

2 — (Undefined variable: Primary.ProductName)

Chapter 4

Section 1

Australian Code (AS3600-2001[AMnd 2004])1.1 General Isolated Foundation 1

1.1.1 Reference

1.1.2 ProblemDesign an isolated footing with the given data: Load Fy = 500 KN, fc = 25 MPa, fy = 450MPa, Column Dimension = 300 mm x 300 mm, and Bearing Capacity of Soil = 110 KN/m2.Coefficient of friction =0.5, FOS against sliding =1.5, and FOS against overturning =1.5.Height of soil above footing = 500 mm, GWT is 200 mm from GL.

Surcharge= 10 KN/m2


Figure 1-1: Australian code General isolated foundation

1.1.3 SolutionApproximate area of footing required = 500/110 m2 = 4.545 m2

Assuming 2.4 m x 2.4 m x 0.400 m footing dimension,

Weight of footing = 2.4 x 2.4 x 0.400 x 25 KN = 57.6 KN

Weight of above soil = 2.4 x 2.4 x 0.500 x 18 KN = 51.84 KN

Reduction of Weight due to buoyancy = 2.4 x 2.4 x (0.500+0.400-0.200) x9.81 KN = 39.554 KN

Load due to surcharge = 2.4 x2.4 x 10 KN =57.6 KN

Therefore, total load on the footing = (500+57.6 +51.84+57.6 -39.554 ) KN= 627.486 KN

Maximum pressure = 627.486 /(2.4x2.4) = 108.94 KN/ m2

108.94 KN/m2 <110 KN/m2 (Hence safe)

Critical load case and the governing factor of safety forsliding

Along X Direction

Sliding force =0

4 — STAAD Foundation Advanced V8i

Chapter — 1

1.1 General Isolated Foundation 1

max Resisting force = µ x Total Service load on foundation

Total Service load on foundation = 627.486 KN

Hence Max possible Resisting Sliding force =0.5 x 627.486 = 313.74 KN

Hence OK

Along Z Direction

Sliding force =0



Hence Max possible Resisting Sliding force =0.5 x 627.486 = 313.74 KN

Hence OK

Critical load case and the governing factor of safety foroverturning

WRT Z Direction

Overturning Moment =0

Max Resisting Moment = 0.5 x 2.4 x 627.486 = 752.98 KNm

Hence OK

WRT X Direction


Max Resisting Moment = 0.5 x 2.4 x 627.486 = 752.98 KNm

Hence OK

1.1.4 Comparison

Value of… ReferenceResult

STAAD Foun-dationResult

Percent Dif-ference

Bearing Pressure, KN/m2 108.94 108.63 NegligibleResisting force for sliding (x),KN

313.74 312.87 Negligible

Resisting Moment forOverturning (z), KNm


Resisting force for sliding (z),KN


Resisting Moment forOverturning (x), KNm


Table 1-1: Australian verification example 1 comparison

Section 1 Australian Code (AS3600-2001[AMnd 2004])



1.2 General Isolated Foundation 21.2.1 Reference

1.2.2 ProblemDesign an isolated footing with the given data: Load Fy = 1,500 KN, Mz=50 KNm, Mx=50KNm, fc = 25 N/m2, fy = 450 N/m2m, Column Dimension = 300 mm x 300 mm, andBearing Capacity of Soil = 150 KN/m2. Coefficient of friction =0.5, FOS against sliding=1.5, and FOS against overturning =1.5 (Include SW for Factored Design)

Figure 1-2: Plan and Elevation

1.2.3 SolutionApproximate area of footing required = 1,500/150 m2 = 10.0 m2


Tot Moment wrt Z =50 KNm

Tot Moment wrt X =50 KNm

Stress at four corners ( service condition)

σ1 = V/A – Mx/Zx + Mz/Zz

σ2 = V/A – Mx/Zx - Mz/Zz


Chapter — 1


σ3 = V/A + Mx/Zx - Mz/Zz

σ4 = V/A + Mx/Zx + Mz/Zz

Tot Vertical Load on soil

Self wt of fdn = 4 m (4 m) (0.73 m) (25 KN/m3) = 292.0 KN

Column reaction load = 1,500 KN

Total Vertical load V = 1,792 KN

Zz = Z · X2/6 = 3.935 x 3.9352/6 =10.67 m3

Zx = Z · X2/6 = 3.935 x 3.9352/6 =10.67 m3

Mx= 50 KNm

Mz = 50 KNm

σ1 = V/A – Mx/Zx + Mz/Zz = 112 KN/m2

σ2 = V/A – Mx/Zx - Mz/Zz = 102.6 KN/m2

σ3 = V/A + Mx/Zx - Mz/Zz = 112 KN/m2

σ4 = V/A + Mx/Zx + Mz/Zz = 121.37 KN/m2

Max stress = 122 KN/m2 <150 KN/m2

Hence safe


Along X Direction

Sliding force = 0

max Resisting force = µ x Total Service load on foundation =0.5 (1,792 KN) =896 KN

Hence OK

Along Z Direction

Sliding force =0

max Resisting force = µ x Total Service load on foundation =0.5 (1,792 KN) =896 KN

Hence OK





Along X Direction

Overturning Moment =50 KNm

max resisting Moment = 0.5 · (4 m) · (1,972 KN) = 3,584 KNm

Hence FOS = 3,584/50 = 71.7 > 1.5

Hence OK

Along Z Direction


max resisting Moment = 0.5 · (4 m) · (1,972 KN) = 3,584 KNm

Hence FOS = 3,584/50 = 71.7 > 1.5

Hence OK

Factored Design

Axial Load = 292 KN + 1.4(1,500 KN) = 2,392 KN

MX =1.4 x 50 =70 KNm

MZ =1.4 x 50 =70 KNm

Check For Trial Depth against moment about Z Axis

Average Base Pressure along one edge = 156.07 KN/m2 (left end)

Average Base Pressure along other edge = 142.93 KN/m2 (right end)

Approximate Base Pressure at the left critical section = 150 KN/m2

Approximate Base Pressure at the right critical section = 149.01 KN/m2

Hence, the moment at the left critical section Mu (Left)

F = (156.07 + 150.0)/2 (1.85 m) (4 m) = 1,132.46 KN

LA = (150.0 + 2 · 156.07) (1.85 m) /[3(150.0 + 156.07)] = 0.932 m

Mu(left) = F · LA = 1,132.46 KN (0.932 m) = 1,055.4 KNm

Similarly, the moment at the right critical section Mu (Right):

F = (142.93 + 149.01)/2 (1.85 m) (4 m) = 1,080.2 KN

LA = (142.93 + 2 · 149.01) (1.85 m) /[3(142.93 + 149.01)] = 0.919 m

Mu(right) = F · LA = 1,080.2 KN (0.919 m) = 992.7 KNm

So max moment with respect to the Z axis, Mu(Z) = 1,056 KNm


Chapter — 1


Assuming 50 mm clear cover and 16 mm bar, effective depth

deff = (730 - 50 - 0.5 x 16) mm = 672 mm

m= fc/fy =0.0555

γ = 0.85 - 0.007(fc - 28) = 0.871 (Take γ = 0.85 per Clause 8.1.2.2

Kumax = 0.4 (Clause 8.1.3)

Ku = 0.34 · γ · (1 - 0.2 · γ) = 0.24

Rumax = 0.85 · fc · γ · Kumax · (1 - Kumax /2) = 3.891

Mumax = φ [Rumax · b · d2] = 5,622.7 KNm

Mu < MumaxHence OK

Check For Trial Depth against moment about X Axis

Average Base Pressure along one edge = 142.93 KN/m2(left end)

Average Base Pressure along other edge = 156.07 KN/m2 (right end)

Approximate Base Pressure at the left critical section = 149.01 KN/m2

Approximate Base Pressure at the right critical section = 150.0 KN/m2

Hence, the moment at the critical section Mu (left)

F = (142.93 + 149.01)/2 (1.85 m) (4 m) = 1,080.2 KN

LA = (142.93 + 2 · 149.01) (1.85 m) /[3(142.93 + 149.01)] = 0.919 m

Mu(right) = F · LA = 1,080.2 KN (0.919 m) = 992.7 KNm

Similarly, the moment at the right critical section Mu (Right):

F = (156.07 + 150.0)/2 (1.85 m) (4 m) = 1,132.46 KN

LA = (150.0 + 2 · 156.07) (1.85 m) /[3(150.0 + 156.07)] = 0.932 m




Mu(left) = F · LA = 1,132.46 KN (0.932 m) = 1,055.4 KNm

So max moment with respect to the X axis, Mu(X) = 1,056 KNm


deff = (730 - 50 - 0.5 x 16) mm = 672 mm

m= fc/fy =0.0555

γ = 0.85 - 0.007(fc - 28) = 0.871 (Take γ = 0.85 per Clause 8.1.2.2


Ku = 0.34 · γ · (1 - 0.2 · γ) = 0.24

Rumax = 0.85 · fc · γ · Kumax · (1 - Kumax /2) = 3.891

Mumax = φ [Rumax · b · d2] = 5,622.7 KNm

Mu < MumaxHence OK

Area of Steel Required along X dir

Calculation required steel for balanced section, Astx = 4,427 m2m

Minimum area of steel Astmin = 0.002 · b · d = 5,376 mm2

So, provided area of steel = 5,376 mm2

Area of Steel Required along Z dir

Calculation required steel for balanced section, Astx = 4,427 m2m


So, provided area of steel = 5,376 mm2


Chapter — 1


Check for One-Way Shear

Along X Direction

Critical section for moment is at a distance, d, away from the face of column

Average Base Pressure along one edge = 142.93 kN/m2

Average Base Pressure along other edge = 156.07 kN/m2

Approximate Base Pressure at the left critical section = 156.07 + (142.93 - 156.07) ·1.178/4= 152.2 kN/m2

Approximate Base Pressure at the right critical section = 156.07 + (142.93 - 156.07) · (4 -1.178)/4 = 146.8 kN/m2

Hence, the SF at the left critical section:

F = (156.07 + 152.2)/2 (1.178 m) (4 m) = 726.3 kN

Shear at the right critical section:

F = (142.93 + 146.8)/2 (1.178 m) (4 m) = 682.6 kN

Critical shear is 727 kN

Developed shear stress, τv = 726.3 kN (103)/[4,000 (672)] = 0.44 N/mm2

τcmax = 0.2 · fc = 5 N/mm2

ß1 = 1.1(1.6 - d/1000) = 1.1(1.6 - 672/1,000) = 1.021

ß2 = 1

ß3 = 1

τc = φ · ß1 · ß2 · ß3. · [Ast · fc/(b · d)]1/3 = 0.75{1.021(1)(1)[5,376 · 25/(4,000 · 672)]1/3} = 0.282

N/mm2

Hence OK




Along Z Direction

Critical section for moment is at a distance, d, away from the face of column

Average Base Pressure along one edge = 142.93 kN/m2

Average Base Pressure along other edge = 156.07 kN/m2

Approximate Base Pressure at the left critical section = 156.07 + (142.93 - 156.07) · (4 -1.178)/4 = 146.8 kN/m2

Approximate Base Pressure at the right critical section = 156.07 + (142.93 - 156.07) ·1.178/4= 152.2 kN/m2

Hence, the SF at the left critical section:

F = (142.93 + 146.8)/2 (1.178 m) (4 m) = 682.6 kN

Shear at the right critical section:

F = (156.07 + 152.2)/2 (1.178 m) (4 m) = 726.3 kN

Critical shear is 727 kN

Developed shear stress, τv = 726.3 kN (103)/[4,000 (672)] = 0.27 N/mm2

τcmax = 0.2 · fc = 5 N/mm2

ß1 = 1.1(1.6 - d/1000) = 1.1(1.6 - 672/1,000) = 1.021

ß2 = 1

ß3 = 1

τc = φ · ß1 · ß2 · ß3. · [Ast · fc/(b · d)]1/3 = 0.75{1.021(1)(1)[5,376 · 25/(4,000 · 672)]1/3} =

0.282 N/mm2

Hence OK

Punching Shear

Punching Shear is checked on a perimeter 0.5 · d from the column face.


Chapter — 1


Pm = 3,888 mm

Vmax = 2,251 kN

τv = Vmax/(Pm · d) = 2,251 kN (10)3/(3,888 mm · 672 mm) = 0.862 N/mm2

Punching shear stress capacity

τc = φ · [0.34 · √(fc)] = 0.7 · [0.34 · √(25)] = 1.19 N/mm2

τv < τc

Hence safe




1.2.4 Comparison



Percent Dif-ference

Corner Pressure, KN/m2 112

102.6

112

102.6

111.9

102.6

111.9

121.32

None

Resisting force for sliding,KN

896

896

895.6

895.6

None

Resisting Moment forOverturning, KNm

3,584

3,584

3,582.3

3,582.3

None

Shear Force (One-Way), KN 727

727

726

735

Negligible

Resisting Shear Stress (One-Way), N/mm2

0.284

0.284

0.284

0.284

None

Shear Force (Two-Way), KN 2251 2250 NoneResisting Shear Stress (Two-Way), N/mm2

1.19 1.19 None

Governing Flexural Moment,KNm

1,056

1,056

1,054

1,054

None

Resisting Flexural Moment,KNm

5,622

5,622

5,622

5,622

None

Reinforcement provided indesign, mm2

5,376 ea.way

5,376 ea. way None


1.3 General Combined Foundation 11.3.1 Reference

1.3.2 ProblemDesign a combined footing with the given data: Load Fy = 600 KN each column., fc = 25MPa, fy = 450 MPa, Column Dimension = 300 mm x 300 mm, Pedestal height-500 mm.and C/C column distance = 3,000 mm . Bearing Capacity of Soil = 105 KN/m2. Coefficientof friction =0.5, FOS against sliding =1.5, FOS against overturning =1.5

Ht of soil =450 mm. Depth of GWT=250 mm.


Chapter — 1

1.3 General Combined Foundation 1


1.3.3 SolutionApproximate area of footing required = 2(600)/115 m2 = 10.435 m2

Assuming 5 m x 2.8 m x 0.600 m footing dimension,

( left overhang=right overhang = 1 m)

Weight of footing = 5 (2.8) (0.600) (25) = 210 KN

Weight of pedestal=2(0.3)(0.3)(0.5)(25) = 2.25 KN

Weight of soil above footing = [5(2.8) - 2(0.3)(0.3)] · 0.450 · 18 = 111.9 KN

Reduction of Weight due to buoyancy = 5(2.8) · (0.45 + 0.6 - 0.25) · 9.81 KN = 109.9 KN

Therefore, total load on the footing = (2 · 600 + 210 + 2.25 + 111.9 - 109.9) KN = 1,414.3KN

Maximum pressure= 1,414.3 /(5 · 2.8) = 101.0 KN/ m2

101 KN/ m2 < 105 KN/m2 (Hence safe)


About Z Direction


Total Service load on foundation = 1,414.3 KN




max resisting Moment = 5 m · 1,414.3 KN /2 =3,535.6 KNm

Hence OK

About X Direction

Overturning Moment = 0

max resisting Moment = 2.8 m · 1,414.3 KN /2 = 1,980 KNm

Hence OK

1.3.4 Comparison



Percent Dif-ference

Bearing Pressure, KN/m2 101 101 NoneResisting Moment forOverturning (Z), KNm

3,535.8 3,535 None

Resisting Moment forOverturning (X), KNm

1,980 1,980 None



1.4.2 ProblemDesign a combined footing with the given data: Load Fy = 600 KN and 550 KN on twocol., fc = 25 MPa, fy = 450 MPa, Column Dimension = 300 mm x 300 mm, Pedestalheight-500 mm. and C/C column distance=3000 mm . Bearing Capacity of Soil = 100KN/m2. Coefficient of friction =0.5, FOS against sliding =1.5, FOS against overturning=1.5


Chapter — 1



1.4.3 SolutionApproximate area of footing required = (600+550)/100 m2 = 11.5 m2

Assuming 5 m x 3 m x 0.500 m footing dimension,

( left overhang = right overhang = 1 m)

Weight of footing = 5 m · 3 m · 0.500 m · 25 = 187.5 KN

Weight of pedestal = 2(0.3)(0.3)(0.5)(25) = 2.25 KN

Therefore, total load on the footing = (600 + 550 + 187.5 + 2.25) KN = 1,339.8 KN

Pressure from axial load = 1,339.8 KN/(5 m · 3 m) = 89.3 KN/ m2

CG of foundation raft = 5/2= 2.5 m from left end

CG of load = (1 m · 600 KN + 4 m · 550 KN)/(600 KN + 550 KN) = 2.435 m

Eccentricity= 2.5 - 2.435 = 0.065 m

So Moment Mz = P · e = 1,150 KN (0.065 m) = 75 KNm

Zz = 3 · 52/6 = 12.5 m3

stress due to moment = M/Z = 75 KNm/12.5 m3 = 6 KN/m2

Stress at left end = P/A + M/Z = 89.3 + 6 = 95.3 KN/m2

Stress at right end = P/A - M/Z = 89.3 - 6 = 83.3 KN/m2

So, Maximum stress

95.3 KN/m2 < 100 KN/m2 (Hence safe)





About Z Direction


max resisting Moment = 5 m (1,339.8 KN) /2 = 3,349.5 KNm

Hence OK

Wrt X Direction


max resisting Moment = 3 m (1,339.8 KN) /2 = 2,009.7 KNm

Hence OK

Check For Trial Depth

Moment About Z Axis (sagging)

Bending moment at critical section, Muz = 172 KNm


deff = (500 - 50 - 0.5 · 12) mm = 444 mm

m= fc/fy =0.0556

γ = 0.85 - 0.007 · (fc - 28) = 0.87

take γ = 0.85 ( Clause 8.1.2.2)


Ku = 0.34 ·γ · (1 - 0.2 · λ) = 0.34 · 0.85(1 - 0.2 · 0.85) = 0.24

Rumax = 0.85 · fc · γ · Ku · (1 - γ · Ku /2) = 0.85(25)(0.85)(0.24)(1 - 0.85 · 0.24/2) = 3.891N/mm2

Mumax = φ [Rumax · b · d2] =0.80 [3.891 N/mm2 · 3,000 mm · (444 mm)2]10-6 = 1,840

KNm

Muz < Mumax Hence OK

Moment About Z Axis (hogging)

Bending moment at critical section, Muz = 201 KNm


deff = (500 - 50 - 0.5 · 12) mm = 444 mm

m= fc/fy =0.0556

γ = 0.85 - 0.007 · (fc - 28) = 0.87


Chapter — 1


take γ = 0.85 ( Clause 8.1.2.2)


Ku = 0.34 ·γ · (1 - 0.2 · λ) = 0.34 · 0.85(1 - 0.2 · 0.85) = 0.24



KNm

Muz < Mumax Hence OK

Moment About X Axis

Cantilever length = (3 - 0.3)/2 = 1.35 m

Bending moment at critical section, Mux = 107.34 N/mm2 (5 m) (1.35 m)2/2 =489.1 KNm


deff = (500 - 50 - 0.5 · 12) mm = 444 mm

m= fc/fy =0.0556

γ = 0.85 - 0.007 · (fc - 28) = 0.87

take γ = 0.85 ( Clause 8.1.2.2)


Ku = 0.34 ·γ · (1 - 0.2 · λ) = 0.34 · 0.85(1 - 0.2 · 0.85) = 0.24



KNm

Mu < Mumax Hence OK

Area of Steel Required

Along X Direction (Bottom)

Astx = 1,083 mm2


Provided area = 2,664 mm2

Along X Direction (Top)

Astx = 1,268 mm2






Along Z Direction (Bottom)

Therefore, Astz = 3,096 mm2



Figure 1-5: Graphs of combined strip footing internal forces


Chapter — 1



Developed shear stress V = 299.5(10)3/(3,000 · 444) = 0.225 N/mm2

τcmax = 0.2 · fc = 5 N/mm2

ß1 =1.1(1.6-d/1000) = 1.2716

ß2 = 1

ß3 = 1

τc =ß1.ß2.ß3.(Ast.fc/b.d)1/3 = 0.488 N/mm2

Vumax = 299.5 KN

Developed shear stress, τv = 299.5(10)3/(3,000 · 444) = 0.225 N/mm2

τcmax = 0.2 · fc = 5 N/mm2

ß1 = 1.1(1.6 - d/1000) = 1.1(1.6 - 444/1,000) = 1.272

ß2 = 1

ß3 = 1

τc = φ · ß1 · ß2 · ß3. · [Ast · fc/(b · d)]1/3 = 0.7{1.272(1)(1)[2,664 · 25/(3,000 · 444)]1/3} = 0.328

N/mm2

Hence OK

Punching Shear

For Column One

Punching shear is checked on a perimeter 0.5 · d from the column face.

Two way shear = 777.8 KN

Pm = 4 · (300 mm + 444 mm) = 2,976 mm

τv = Vmax/(Pm · d) = 777.8 KN · 1000/(2,976 mm · 444 mm) = 0.589 N/mm2

τc = φ · [0.34 · √(fc)] = 0.7 · [0.34 · √(25)] = 1.19 N/mm2

τv < τc , Hence safe

For Column Two

Punching shear is checked on a perimeter 0.5d from the column face.

Two way shear= 713.4 KN

Pm = 2,976 mm

τv = Vmax/(Pm · d) = 713.4 KN · 1000/(2,976 mm · 444 mm) = 0.540 N/mm2

τc = φ · [0.34 · √(fc)] = 0.7 · [0.34 · √(25)] = 1.19 N/mm2





1.4.4 Comparison



Percent Dif-ference

Bearing Pressure, KN/m2 95.3

83.3

95.32

83.32

None

Governing Moment, KNm 172

201

489

167

201

512

Negligible

Resisting Moment, KNm 1,840

1,840

3,068

1,840

1,840

3,068

None

Shear Force (One-Way), KN 299.5 299.4 NoneShear capacity (One-Way),N/mm2

0.328 0.328 None

Shear Force (Two-Way), KN 777.8

713.4

777.8

713.4

None

Shear capacity (Two-Way),N/mm2

1.19 1.19 None

Resisting Moment forOverturning (Z), KNm

3,349 3,349 None

Resisting Moment forOverturning (X), KNm

2,010 2,010 None



Chapter — 1


Section 2

British Code (BS8110-1-1997)2.1 General Isolated Foundation 1

2.1.1 Reference‘Reinforced Concrete’ by T.J.Macgingley & B.S.Choo, Page 333 and Example: 11.1.

2.1.2 ProblemA column 400mm X 400mm carries a dead load of 800 kN and an imposed load of 300 kN.The safe bearing pressure is 200 kN/m2. Design a square base to resist the loads. Theconcrete is grade 35 and the reinforcement is grade 460.


2.1.3 SolutionFigure 2-1: Bending section considered

Figure 2-2: One way shear section considered

Figure 2-3: Two way shear section considered


Chapter — 2


Size of base

Self-weight of footing = 2.5 x 2.5 x 0.5 x 25 = 78.125 kN.

Therefore, Service load = Dead load + Imposed load + Self weight

= (800 + 300 + 78.125) kN = 1178.125 kN.

Area required = 1178.125 / 200 m2 = 5.890625 m2.

Make the base 2.5 m x 2.5 m.

Moment Steel

Ultimate load = (1.4 x 800) + (1.6 x 300) = 1600 kN.

Ultimate pressure = 1600 / (2.5 x 2.5) = 256 kNm2

The critical section YY at the column face is shown in Figure 6.1.

MYY= 256 x (2.5 / 2 - 0.4 / 2) x 2.5 x 0.525 = 352.8 kNm.

Try an overall depth of 500 mm with 20 mm bars.

Effective depth = 500 – 40 – 20 – 10 = 430 mm.

Therefore z = 0.95d,

= 1976.31 mm2.

Minimum area of steel = 0.0015 x B x d = 0.0012 x 2500 x 430 = 1625 mm2 < AS (HenceSafe)

Let us provide 10 nos. 16 mm bars, AS = 2010.62 mm2.

One Way Shear

The critical section Y1 Y1 at d = 430 mm from the face of the column is shown in Figure 6.2.

Design shear force, VU = 256 x (2.5 /2 – 0.43 – 0.4 / 2) x 2.5 = 396.8 kN

Design shear stress, v = 396.9(10)3 / (2500 x 430) = 0.369 N/mm2

Now, vC1 = min(0.8 ,5) = 4.7328 N/mm2 > v (Hence Safe)

Section 2 British Code (BS8110-1-1997)



= 0.395 N / mm2 > v (Hence Safe)

Hence no shear reinforcement is required.

Punching Shear

Punching shear is checked on a perimeter 1.5d = 625.5 mm from the column face. Thecritical perimeter is shown in Figure 6.3.

Perimeter = 1690 x 4 = 6760 mm.

Shear = 256 x (2.52 – 1.692) = 868.8 kN.

v = 868 x 103 / (6760 x 430) = 0.3 N / mm2 < VC (Hence Safe).


Spacing

We provided 10 nos. 16 mm bars, AS = 2010.62 mm2.

Spacing = (2500 - 40 x 2 - 16) / (10 -1) = 267.11 mm.

2.1.4 Comparison

Value of Reference Results STAAD FoundationResult Percent Difference

Effective Depth 430 mm 430 mm NoneGoverning Moment 352.8 KN-m 352.8 KN-m NoneArea of Steel 1976.31 1976.31 NoneShear Stress (One-Way) 0.369 N/mm2 0.369 N/mm2 NoneShear Stress (Two-Way) 0.3 N/mm2 0.3 N/mm2 None

Table 2-1: British verification example 1 comparison

2.2 General Isolated Foundation 22.2.1 Reference‘Reinforced Concrete’ by T.J.Macgingley & B.S.Choo, 0 and Example: 11.2.

2.2.2 ProblemThe characteristic loads for an internal column footing in a building are given in thefollowing table. The proposed dimensions for the column and base are shown in Figure6.4. The safe bearing pressure of soil is 150 kN / m2. The materials to be used in thefoundation are grade 35 concrete and grade 460 reinforcement.

Vertical Load (kN) Moment (KN m)Dead Load 770 78Imposed Load 330 34

Table 2-2: Table BS2.1 - Column loads


Chapter — 2



2.2.3 SolutionSelf-weight of footing = 0.5 x 3.6 x 2.8 x 24 = 120.96 kN.

Total axial load = 770 + 330 + 120.96 = 1220.96 kN.

Total moment = 78 + 34 = 112 kN-m.

Base area = 2.8 x 3.6 = 10.08 m2.

Section modulus = (I / y) = (1/12)BD3/(D/2) = 6.048 m3.

Maximum pressure = 1220.96/10.08 + 112/6.048 = 139.65 kN / m2 < 150 kN / m2 (HenceSafe).

Factored axial load = (1.4 x 770) + (1.6 x 330) = 1606 kN.

Factored moment = (1.4 x 78) + (1.6 x 34) = 163.6 kN-m.

Maximum pressure = 1606/10.08 + 163.6/6.048 = 186.38 kN / m2.

Minimum pressure = 1606/10.08 - 163.6/6.048 = 132.28 kN / m2.

Calculation of Reinforcement Along Shorter Span (X1- X1):

Average pressure for section X1X1 (as shown in Figure 6.5) = 159.33 kN / m2.

Moment (MY) = (159.33 x 1.175 x 3.6) x (1.175 / 2) = 395.955 kN-m.

Effective depth (d) = 500 – 40 – 10 = 450 mm.

0.015 < 0.156 (Hence Safe)





= 2119.475 mm2.

The minimum area of steel = 0.13 x 3600 x 500 / 100 = 2340 mm2 > calculated area ofsteel.

Provide minimum steel.

Figure 2-5: Sections considered for bending in both directions

Calculation of Reinforcement Along Longer Span (Y1-Y1):

Pressure at section Y1 Y1 (as shown in Figure 6.5) = 162.7 kN / m2.


Chapter — 2


Moment(MX) = (162.7 x 2.8 x 1.575)x(1.575 / 2)+(0.5 x 1.575 x (186.38 – 162.7) x 2.8)x(2/ 3 x 1.575) = 619.862 kN-m.

Effective depth (d) = 500 – 40 – 20 –10 = 430 mm.

0.034 < 0.156 (Hence Safe)

= 0.96d


= 3472.334 mm2.

The minimum area of steel = 0.13 x 2800 x 500 / 100 = 1820 mm2 <

calculated area of steel. (Hence safe)

One Way Shear Along Section Y2-Y2:

The critical section Y2 Y2 at d = 430 mm from the face of the column is shown in Figure6.6.

Average pressure for the required section = 177.78 kN / m2.

Design shear force, VU = 177.78 x 2.8 x 1.145 = 569.96 kN

Design shear stress, v = =473.388 kN / m2

Now, vC1 = min(0.8 ,5) = 4732.8 kN / m2 > v (Hence Safe)

= 429.6 kN / m2.

Let us consider 1.5 times shear enhancement.

Vce = 1.5 x 429.6 = 644.4 kN/m2 > v (Hence safe)





Figure 2-6: Sections considered for one-way shear in both directions

Along Section X2-X2:

The critical section X2X2 at d = 450 mm from the face of the column is shown in Figure6.6.

Average pressure for the required section = 159.33 kN / m2.

Design shear force, VU = 159.33 x 3.6 x 7.25 = 415.85 kN

Design shear stress, v = =256.698 kN / m2

Now, vC1 = min(0.8 ,5) = 4732.8 kN / m2 > v (Hence Safe)


Chapter — 2


= 409.6 kN / m2.




Punching ShearFigure 2-7: Section considered for punching shear

The punching shear will be calculated for an area outside the area enclosed by the rectangleat a distance 1.5d from the column face as shown in Figure 6.7.

Total pressure under the base = 2.8 x 3.6 x 132.28 + 0.5 x 3.6 x 2.8 x (186.38 – 132.28) =1606.05 kN.

Pressure under enclosed rectangle = (1.74)2 x 146.255 + 0.5 x (1.74)2 x (172.4 – 146.255) =482.38 kN

Punching shear force = 1606.05 – 482.38 = 1123.67 kN.

Critical perimeter = 1.74 x 4 = 6.96 m.

Punching shear stress = 1123.67 / (6.96 x 0.43) = 375.46 kN / m2.




2.2.4 Comparison

Value of ReferenceResult


Percent Dif-ference

Effective Depth (X-X) 430 mm 430 mm NoneEffective Depth (Y-Y) 450 mm 450 mm NoneGoverning Moment (My) 395.955 KN-m 395.943 KN-m NoneGoverning Moment (Mx) 619.862 KN-m 619.909 kN-m NoneArea of Steel (Along X-X) 2340.00 2340.00 NoneArea of Steel (Along Y-Y) 3472.2334 3472.2334 NoneShear Stress (One-Way) (Y1-Y1)

473.388 kN/m2 444.81 kN/m2 Negligible

Shear Stress (One-Way) (X1-X1)

256.698 kN/m2 256.698 kN/m2 None

Shear Stress (Two-Way) 375.46 kN/m2 375.44 kN/m2 None



2.3.2 ProblemDesign an isolated footing with the given data: Load Fy = 1500 KN, fc = 25 MPa, fy = 415MPa, Column Dimension = 300 mm x 300 mm, and Bearing Capacity of Soil = 120KN/m2. Coefficient of friction =0.5, FOS against sliding =1.5, FOS against overturning=1.5


Chapter — 2






Therefore, total load on the footing = (1500+240.865) KN = 1740.865 KN

Maximum pressure =1740.865/(3.85x3.85)=KN/ m2 = 117.45 KN/m2 <120 KN/m2 (Hencesafe)

Ultimate pressure =1500x1.4/(3.85x3.85) KN/m2 = 141.676 KN/m2

Critical load case and the governing factor of safety foroverturning and sliding

Along X Direction

Sliding force =0

max Resisting force = µ x Total Service load on foundation =0.5 x 1740.865 = 870.43 KN




Hence OK


max resisting Moment = 0.5 x 3.85 x 1740.865 = 3351.1 KNm

Hence OK

Along Z Direction

Sliding force =0


Hence OK



Hence OK


Moment About X Axis

Bending moment at critical section, Mux = 141.676 x 3.85 x1.775x1.775/2 = 859.26 KN-m


deff = (650-50-0.5 x 12) mm = 594 mm

K = = =0.0253 < 0.156

Hence safe

Moment About Z Axis

Bending moment at critical section, Muz = 141.676 x 3.85 x1.775x1.775/2 = 859.26 KN-m


deff = (650-50-0.5 x 12) mm = 594 mm

K = = 0.0253 < 0.156

Hence safe


Chapter — 2



Along X Direction

Z = d = 0.971 d>0.95d

So, Z= 0.95d

Therefore, Astx = = 3862.3 m2m

Minimum area of steel Astmin = 0.0013 x B x D = 3253.25 mm2

Along Z Direction

Z = d = 0.971 d>0.95d

So, Z= 0.95d

Therefore, Astz = = 3862.3 m2m



Along X Direction

Percentage of steel pt = = 0.1689

Vumax = 141.676 x 3.85 x = 644.18 KN

Developed shear stress V = = 0.282N/mm2

Now allowable stress= Vc1 = = 4 N/mm2

V < Vc1, Hence Safe

V1 = 0.17 N/mm2

V2 = 1 N/mm2




V3 = 1 N/mm2

Vc = = 0.352 N/mm2

Vce = 0.704 N/mm2

So V < Vce , Hence Safe


Along Z Direction


Vumax = 141.676 x 3.85 x = 644.18 KN



V < Vc1, Hence Safe

V1 = 0.17 N/mm2

V2 = 1 N/mm2

V3 = 1 N/mm2

Vc = = 0.352 N/mm2

Vce = 0.704 N/mm2


Punching Shear

Punching shear is checked on a perimeter 1.5d = 891 mm from the column face.

Area within Critical Perimeter Am = 4.3347 m2

Vmax = 1485.87 KN

Critical perimeter Pm = 2 X ( b + h + 6 x d) = 8.328 m

Vm1 = Vmax/(Pm · d) = 0.3 N/mm2


Chapter — 2


Now allowable stress= Vt1 = = 4 N/mm2

Vm1< Vt1 , Hence safe

V1 = 0.17 N/mm2

V2 = 1 N/mm2

V3 = 1 N/mm2

Vc = = 0.352 N/mm2

Vm1 < Vc, Hence safe

Bar Spacing

In the X Direction

No. of 12 mm bar = = 35

Spacing = = 110 mm

Spacing for 12 mm bar = 110 mm

In the Z Direction

No. of 12 mm bar = = 35

Spacing = = 110 mm


Check For Development Length

Along x & Z direction

Max dia permitted =25 mm

ß = 0.5

Hence Fbu = ß x √fc =0.5 x 5 = 2.5

Ld = 0.95x fy x Ø / 4Fbu = 0.95x 415x25/4x2.5 = 985.6 mm

available length = 1725 mm

Hence OK




2.3.4 Comparison



Percent Dif-ference

Effective Depth 594 mm 594 mm NoneBearing Pressure 117.45

KN/m2117.45 KN/m2 None

Ku 0.0253 0.0252 NegligibleGoverning Moment 859.26 KN-

m856.92 KN-m Negligible

Shear Force(One-Way) 644.18 KN 642.87 KN NegligibleShear Force(Two-Way) 1485.87 KN 1483.64 KN NegligibleSteel required 3862 m2m 3851 m2m 0.28% (

Negligible)Resisting force for sliding 870.43 KN 870.43 KN NoneResisting Moment forOverturning

3351.1 KNm 3351.1 KNm None



2.4.2 ProblemDesign an isolated footing with the given data: Load Fy = 2000 KN, fc = 25 MPa, fy = 415MPa, Column Dimension = 300 mm x 300 mm, and Bearing Capacity of Soil = 100KN/m2. Coefficient of friction =0.5, FOS against sliding =1.5, FOS against overturning=1.5 Soil cover above footing =500 mm

Plan and Elevation


Chapter — 2


2.4.3 SolutionApproximate area of footing required = 2000/100 m2 = 20 m2



Weight of soil = 5.3 x 5.3 x 0.5 x 18 KN = 252.81 KN

Therefore, total load on the footing = (2000+526.687+252.81) KN = 2779.5 KN

Maximum pressure = 2779.5 KN/ (5.3 m x 5.3 m) KN/ m2 = 98.9 KN/m2 <100 KN/m2

(Hence safe)

Ultimate pressure = 1.4(2000 KN) / (5.3 m x 5.3 m)KN/m2 = 99.678 KN/m2


Along X Direction

Sliding force =0

max Resisting force = µ x Total Service load on foundation = 0.5 x 2779.5 = 1389.75 KN

Hence OK





max resisting Moment = 0.5 x 5.3 x 2779.5 = 7365 KNm

Hence OK

Along Z Direction

Sliding force =0

max Resisting force = µ x Total Service load on foundation = 0.5 x 2779.5 = 1389.75 KN

Hence OK



Hence OK

Check for Trial Depth Against Moment

About X Axis

Bending moment at critical section, Mux = 99.68 x 5.3 x 2.5 x 2.5 / 2 = 1651 KN-m


deff = (750-50-0.5 x 16) mm = 692 mm

K = = =0.026 < 0.156

Hence safe

About Z Axis

Bending moment at critical section, Mux = 99.68 x 5.3 x = 1651 KN-m


deff = (750-50-0.5 x 16) mm = 692 mm

K = = =0.026 < 0.156

Hence safe


Along X Direction

Z = d


Chapter — 2


= 0.97 d>0.95d

So, Z= 0.95d



Area of Steel Required along Z dir

Z = d

= 0.97 d>0.95d

So, Z= 0.95d

Therefore, Astz = = 6351.73 m2m



Along X Direction


Vumax = 99.68 x 5.3 x = 955.2 KN

Developed shear stress V = 955.2(103)/(5300 x 692) = 0.26N/mm2


V < Vc1, Hence Safe

V1 = 0.173 N/mm2

V2 = 1 N/mm2

V3 = 1 N/mm2

Vc = = 0.354 N/mm2

Vce = 0.708 N/mm2





Along Z Direction


Vumax = 99.68 x 5.3 x = 955.2 KN



V < Vc1, Hence Safe

V1 = 0.173 N/mm2

V2 = 1 N/mm2

V3 = 1 N/mm2

Vc = = 0.354 N/mm2

Vce = 0.708 N/mm2


Punching Shear



Vmax = 2237.3 KN


Vm1 = Vmax/(Pm · d) = 0.345 N/mm2



V1 = 0.173 N/mm2

V2 = 1 N/mm2

V3 = 1 N/mm2

Vc = = 0.354 N/mm2


Chapter — 2



Bar Spacing

In the X Direction

No. of 16 mm bar = = 32

Spacing = = 165 mm


In the Z Direction

No. of 16 mm bar = = 32

Spacing = = 165 mm



Along X & Z direction


ß = 0.5

Hence Fbu = ß x √fc =0.5 x 5 = 2.5



Hence OK




2.4.4 Comparison



Percent Dif-ference

Effective Depth 692 mm 692 mm NoneBearing Pressure 98.9 KN/m2 98.75 KN/m2 NegligibleKu 0.026 0.026 NegligibleGoverning Moment 1651 KN-m 1647.74 KN-m NegligibleShear Force(One-Way) 955.2 KN 952.85 KN NegligibleShear Force(Two-Way) 2237.3 KN 2232.14 KN NegligibleSteel required 6352 m2m 6357.649 m2m NegligibleResisting force for sliding 1389.75 KN 1374.762 KN NegligibleResisting Moment forOverturning

7365 KNm 7148.63 KN Negligible



2.5.2 ProblemDesign an isolated footing with the given data: Load Fy = 1200 KN,Mz=80 KNm, fc = 25MPa, fy = 415 MPa, Column Dimension = 300 mm x 300 mm, and Bearing Capacity ofSoil = 120 KN/m2. Coefficient of friction =0.5, FOS against sliding =1.5, FOS againstoverturning =1.5


Chapter — 2











Self wt of fdn = 3.65x 3.65x 0.75x25 = 249.8 KN

Dry wt of soil = 0

Col reaction load = 1200 KN

Tot Vertical load V = 1449.8 KN

Zz = Z.X2/6 = 3.65 x 3.652/6 =8.105 m3

Zx = X.Z2/6 = 3.65 x 3.652/6 =8.105 m3




Mx=0

Mz = 80 KNm

σ1 = V/A – Mx/Zx + Mz/Zz = 118.694 KN/m2

σ2 = V/A – Mx/Zx - Mz/Zz = 98.953 KN/m2

σ3 = V/A + Mx/Zx - Mz/Zz = 98.953 KN/m2

σ4 = V/A + Mx/Zx + Mz/Zz = 118.694 KN/m2

= 118.694 KN/m2 <120 KN/m2

Hence safe


Along X Direction

Sliding force =0


Hence OK


max resisting Moment = 0.5 x 3.63x 1449.8 = 2645.885 KNm

Hence OK


Along Z Direction

Sliding force =0


Hence OK



Hence FOS =2645.885/80 = 33.074

Hence OK


Moment About Z Axis

Force creating Moment= (139.922 +127.239) x 0.5 x 1.675 x 3.65 = 816.68 KN

Lever arm =(127.239 + 2 x 139.922) x 1.675/ 3x (139.922 + 127.239) = 0.8508 m

Moment = Fx LA = 694.84 KNm


Chapter — 2



deff = (750-50-0.5 x 12) mm = 692 mm

K = = =0.016 < 0.156

Hence safe

Figure 2-9: Section considered for bending about the Z axis

Moment About X Axis

Force creating Moment= 126.103 x 1.675 x 3.65 = 770.97 KN

Lever arm = 1.675x 0.5 = 0.823 m

Moment = Fx LA =646.08 KNm

K = = =0.01481 < 0.156

Hence safe




Figure 2-10: Section considered for bending about the Z axis


Along X Direction

Z = d = 0.983 d>0.95d

So, Z= 0.95d

Therefore, Astx = = 2594 m2m


So, provided area of steel = 3558.75 mm2

Along Z Direction

Z = d = 0.97 d>0.95d

So, Z= 0.95d

Therefore, Astz = = 2412 m2m


So, provided area of steel = 3558.75 mm2


Chapter — 2



Along X Direction


Vumax =1/2 x (139.922+132.479)x0.983 x 3.65 = 488.69 KN



V < Vc1, Hence Safe

V1 = 0.14 N/mm2

V2 = 1 N/mm2

V3 = 1 N/mm2

Vc = = 0.329 N/mm2

Vce = 0.658 N/mm2





Figure 2-11: Section considered for one-way shear along X direction

Along Z Direction


Vumax 126.103 x 0.983 x 3.65= 452.46 KN

Developed shear stress V = = 0.177 N/mm2


V < Vc1, Hence Safe

V1 = 0.141 N/mm2

V2 = 1 N/mm2

V3 = 1 N/mm2

Vc = = 0.329 N/mm2

Vce = 0.658 N/mm2


Chapter — 2



Figure 2-12: Section considered for one-way shear along z direction

Punching Shear


Area within Critical Perimeter Am = 2.376 x 2.376 =7.68 m2

Vmax = 968.11 KN


Vm1 = Vmax/(Pm · d) = 0.1472 N/mm2



V1 = 0.14 N/mm2

V2 = 1 N/mm2

V3 = 1 N/mm2

Vc = = 0.329 N/mm2





Figure 2-13: Section considered for punching shear

Bar Spacing

In the X Direction

No. of 12 mm bar = = 32

Spacing = = 115 mm


In the Z Direction

No. of 12 mm bar = = 32

Spacing = = 115 mm



Chapter — 2





ß = 0.5

Hence Fbu = ß x √fc =0.5 x 5 = 2.5



Hence OK

2.5.4 Comparison



Percent Dif-ference

Effective Depth (mm) 692 692 NoneCorner Pressure (KN/m2) 118.69

98.95

98.95

118.69

118.69

98.95

98.95

118.69

None

Ku 0.01481

0.0156

0.0146

0.0158

Negligible

Governing Moment (KN-m) 646.08

694.84

643.82

692.8

Negligible

Shear Force ,One-Way (KN) 488.69

452.46

486.65

456.42

Negligible

Shear Force, Two-Way (KN) 968.11 961.62 NegligibleSteel required (mm2) 3558.75

3558.75

3558.75

3558.75

None

Resisting force for sliding(KN)

724.9

724.9

724.898

724.898

Negligible

Resisting Moment forOverturning (KN-m)

2645.8

2645.8

2645.8 Same






2.6.2 ProblemDesign an isolated footing with the given data: Load Fy = 1500 KN, Fx100 KN, fc = 25MPa, fy = 415 MPa, Column Dimension = 300 mm x 300 mm, and Bearing Capacity ofSoil = 90 KN/m2. Coefficient of friction =0.5, FOS against sliding =1.5, FOS againstoverturning =1.5

2.6.3 SolutionApproximate area of footing required = 1500/90 = 16.67 m2


Tot Moment wrt Z =-0.65x100 =-65 KNm









Chapter — 2


Dry wt of soil = 0



Zz = Z.X2/6 = 4.65 x 4.652/6 =16.758 m3

Zx = X.Z2/6 = 4.65 x 4.652/6 =16.758 m3

Mx=0

Mz = -65 KNm

σ1 = V/A – Mx/Zx + Mz/Zz = 81.75 KN/m2

σ2 = V/A – Mx/Zx - Mz/Zz = 89.50 KN/m2

σ3 = V/A + Mx/Zx - Mz/Zz = 89.50 KN/m2

σ4 = V/A + Mx/Zx + Mz/Zz = 81.75 KN/m2

= 89.5 KN/m2 <90 KN/m2

Hence safe


Along X Direction

Sliding force =100 KN


Hence OK



Hence OK

Along Z Direction

Sliding force =0


Hence OK



Hence OK





Moment About Z Axis

Force creating Moment=1/2 x (102.552+97.472) x2.175x4.65 = 1011.5 KN

Lever arm =(97.472 + 2 x 102.552) x 2.175/ 3x (97.472+102.552) = 1.097m



deff = (650-50-0.5 x 12) mm = 594 mm

K = =0.027 < 0.156

Hence safe

Moment About X Axis

Force creating Moment= 97.122 x 2.175 x 4.65 = 982.27 KN

Lever arm =2.175 x 0.5 = 1.088 m

Moment = Fx LA =1068.71 KNm

K = =0.02606 < 0.156

Hence safe


Chapter — 2



Along X Direction

Z = d = 0.969 d > 0.95d

So, Z= 0.95d



So, provided area of steel = 4890 mm2

Along Z Direction

Z = d = 0.97 d>0.95d

So, Z= 0.95d

Therefore, Astz = = 4706 m2m







Along X Direction

Critical section for moment is at d dist from the face of column


Vumax =1/2 x (102.552+98.86)x1.581 x 4.65 = 740.36 KN

Developed shear stress V = 740.36 x1000/4650 x594 = 0.268N/mm2


V < Vc1, Hence Safe

V1 = 0.178 N/mm2

V2 = 1 N/mm2

V3 = 1 N/mm2

Vc = = 0.356 N/mm2

Vce = 0.712 N/mm2



Chapter — 2


Along Z Direction

Critical section for moment is at d dist from the face of column


Vumax = 97.122 x1.58 x4.65= 714 KN

Developed shear stress V = 714 x1000/4650 x 594 =

= 0.258 N/mm2


V < Vc1, Hence Safe

V1 = 0.171 N/mm2

V2 = 1 N/mm2

V3 = 1 N/mm2

Vc = = 0.351 N/mm2

Vce = 0.702 N/mm2





Punching Shear


Area within Critical Perimeter Am = 2.082x2.082 =4.3347 m2

Vmax = 1679 KN


Vm1 = Vmax/(Pm · d) = 0.3394 N/mm2



V1 = 0.171 N/mm2

V2 = 1 N/mm2

V3 = 1 N/mm2

Vc = = 0.351 N/mm2



Chapter — 2


Bar Spacing

In the X Direction

No. of 12 mm bar = = 44

Spacing = = 105 mm


In the Z Direction

No. of 12 mm bar = = 42

Spacing = = 115 mm





ß = 0.5

Hence Fbu = ß x √fc =0.5 x 5 = 2.5



Hence OK




2.6.4 Comparison



Percent Dif-ference

Effective Depth 594 mm 594 mm NoneCorner Pressure 81.75

KN/m2

89.5 KN/m2

89.5 KN/m2

81.75KN/m2

81.74 KN/m2

89.5 KN/m2

89.5 KN/m2

81.74 KN/m2

None

Ku 0.02606

0.027

0.026

0.0269

None

Governing Moment 1068.71KNm

1109.62KNm

1065.83 KNm

1103.31 KNm

Negligible

Shear Force(One-Way) 740.36 KN

714 KN

736.98 KN

712 KN

Negligible

Shear Force(Two-Way) 1679 KN 1677 KN NegligibleSteel required 4890 m2m

4706 m2m

4959.354 m2m

4790.8 m2m

Negligible

Resisting force for sliding 925.68 KN

925.68 KN

925.68 KN

925.68 KN

None

Resisting Moment forOverturning

4304.43KNm

4304.43KNm

4304.43 KNm

4304.43 KNm

None



2.7.2 ProblemDesign an isolated footing with the given data: Load Fy = 1500 KN,Mz=Mx=50 KNm, fc =25 MPa, fy = 460 MPa, Column Dimension = 300 mm x 300 mm, and Bearing Capacity ofSoil = 100 KN/m2. Coefficient of friction =0.5, FOS against sliding =1.5, FOS againstoverturning =1.5


Chapter — 2



2.7.3 SolutionApproximate area of footing required =1500/100 m2 = 15 m2



Tot Moment wrt X =50 KNm








Dry wt of soil = 0






Zz = Z.X2/6 = 4.45 x 4.452/6 =14.686 m3

Zx = X.Z2/6 = 4.45 x 4.452/6 =14.686 m3

Mx= 50 KNm

Mz = 50 KNm

σ1 = V/A – Mx/Zx + Mz/Zz = 92KN/m2

σ2 = V/A – Mx/Zx - Mz/Zz = 85 KN/m2

σ3 = V/A + Mx/Zx - Mz/Zz = 92 KN/m2

σ4 = V/A + Mx/Zx + Mz/Zz = 98.81 KN/m2

= 98.81 KN/m2 <100 KN/m2 (Hence safe)


Along X Direction

Sliding force =0

max Resisting force = µ x Total Service load on foundation =0.5 x 1821.8 = 911 KN

Hence OK



Hence FOS=4053.5/50 = 81> 1.5

Hence OK

Along Z Direction

Sliding force =0

max Resisting force = µ x Total Service load on foundation =0.5 x 1821.8 = 911 KN

Hence OK



Hence FOS=4053.5/50 = 81> 1.5

Hence OK


About Z Axis

Avg Base Pressure at one edge=( 96.5149+106.047)/2 = 101.282 KN/m2

Avg Base Pressure at other edge=( 106.0473+115.5797)/2 = 110.814 KN/m2


Chapter — 2


Force creating Moment= (110.814+106.37)/2 x 2.075x4.45 = 1002.72 KN

Lever arm = (106.37+2x110.814)x2.075/3x(106.37+110.814) m



deff = (650-50-0.5 x 12) mm = 594 mm

K = =0.027 < 0.156

Hence safe

About X Axis

Avg Base Pressure at one edge=( 96.5149+106.047)/2 = 101.282 KN/m2

Avg Base Pressure at other edge=( 106.0473+115.5797)/2 = 110.814 KN/m2

Force creating Moment= (110.814+106.37)/2 x 2.075x4.45 = 1002.72 KN

Lever arm = (106.37+2x110.814)x2.075/3x(106.37+110.814) m



deff = (650-50-0.5 x 12) mm = 594 mm

K = = 0.027 < 0.156

Hence safe





Along X Direction

Z = d =<0.95d

= 0.969 d>0.95d

So, Z= 0.95d


Minimum area of steel Astmin = 0.0013 x B x D = 3760 mm2


Along Z Direction

Z = d

= 0.969 d>0.95d

So, Z= 0.95d

Therefore, Astz = = 4165m2m



Chapter — 2




Along X Direction


Vumax =1/2 x (110.814+107.642)x1.481x4.45 = 719.87 KN

Developed shear stress V = 720x1000/(4450 x594) = 0.272N/mm2


V < Vc1, Hence Safe

V1 = 0.158 N/mm2

V2 = 1 N/mm2

V3 = 1 N/mm2

Vc = = 0.342 N/mm2

Vce = 0.684 N/mm2





Along Z Direction


Vumax =1/2 x (110.814+107.642)x1.481x4.45 = 719.87 KN



V < Vc1, Hence Safe

V1 = 0.158 N/mm2

V2 = 1 N/mm2

V3 = 1 N/mm2

Vc = = 0.342 N/mm2

Vce = 0.684 N/mm2


Punching Shear

Punching shear is checked on a perimeter 1.5d from the column face.


Chapter — 2


Area within Critical Perimeter Am = 2.082x2.082=4.335 m2

Vmax = 1640 KN


Vm1 = Vmax/(Pm · d) = 0.3315 N/mm2



V1 = 0.158 N/mm2

V2 = 1 N/mm2

V3 = 1 N/mm2

Vc = = 0.342 N/mm2






Bar Spacing

In the X Direction

No. of 12 mm bar = = 37

Spacing = = 120 mm


In the Z Direction

No. of 12 mm bar = = 37

Spacing = = 120 mm



Chapter — 2


2.7.4 Comparison



Percent Dif-ference

Effective Depth 594 mm 594 mm NoneCorner Pressure 92KN/m2

85.2 KN/m2

92 KN/m2

98.81KN/m2

92 KN/m2

85.91 KN/m2

92 KN/m2

98.81 KN/m2

None

Ku 0.027

0.027

0.0266

0.0266

Negligible

Governing Moment 1047.85KNm

1047.85KNm

1045 KNm

1045 KNm

Negligible

Shear Force(One-Way) 720 KN

720 KN

718.71 KN

718.71 KN

Negligible

Shear Force(Two-Way) 1640 KN 1638 KN NegligibleSteel required 12 @ 120 c/c 12 @ 118 c/c NegligibleResisting force for sliding 910.9 KN

910.9 KN

910.8 KN

910.8 KN

Negligible


4053.5 KNm

4053.5 KNm

4053.4 KNm

4053.4 KNm

Same



2.8.2 ProblemDesign a combined footing with the given data: Load Fy = 500 KN each column., fc = 25MPa, fy = 450 MPa, Column Dimension = 300 mm x 300 mm, Pedestal height-500 mm.and C/C column distance=3000 mm . Bearing Capacity of Soil = 150 KN/m2. Coefficient offriction =0.5, FOS against sliding =1.5, FOS against overturning =1.5

Ht of soil =400 mm. Depth of GWT=200 mm




2.8.3 SolutionFigure 2-16: Plan and Elevation

Approximate area of footing required = 2x500/150 m2 = 6.67 m2


( left overhang=right overhang=1m)

Weight of footing = 5 m x 1.5 m x 0.600 x25 KN = 112.5 KN

Weight of pedestal=2x0.3x0.3x0.5x25=2.25 KN

Weight of soil above footing = (5 x 1.5-0.3x0.3x2 )x 0.400 x18 KN = 52.704 KN

Reduction of Weight due to buoyancy = 5 x 1.5 x (0.4+0.6-0.2) x9.81 KN = 58.86 KN

Therefore, total load on the footing = (2x500+112.5 +2.25+52.704 -58.86) KN = 1108.6 KN

Maximum pressure= 1108.6 /(5 x1.5) = 147.82 KN/ m2

147.82 KN/ m2 <150 KN/m2

Hence safe

Ultimate pressure = KN/m2 = 186.667 KN/m2


Chapter — 2



About Z Direction


max resisting Moment = 0.5 x 5x 1108.6 = 2771.5 KNm

Hence OK

About X Direction



Hence OK



Bending moment at critical section, Mux = 140 KNm


deff = (600-50-0.5 x 12) mm = 544 mm

K = Mu/fc.b.deff2 = 140x106/(25x1500x544x544) = 0.013 <0.156

Hence OK


Bending moment at critical section, Mux = 174.9 KN-m


deff = (600-50-0.5 x 12) mm = 544 mm

K = Mu/fc.b.deff2 = 0.016= <0.156

Hence OK

Moment About X Axis

Cantilever length=(1.5-0.3)/2 = 0.6 m

Bending moment at critical section, Mux = 186.667 x.5x0.62/2 =168 KNm


deff = (600-50-0.5 x 12) mm = 544 mm

K = Mu/fc.b.deff2 = 168x106/(25x6500x544x544) = 0.005 <0.156




Hence OK



Z = d = 0.998 d>0.95d

So, Z= 0.95d


Minimum area of steel Astmin = 0.0013 x B x D = 1170 mm2 ( as fy>250)

Provided area = 1170 m2m


Z = d = 0.998 d>0.95d

So, Z= 0.95d





Z = d = 0.994 d>0.95d

So, Z= 0.95d







Chapter — 2


Vumax = 225.68 KN



V < Vc1, Hence Safe

V1 = 0.143 N/mm2

V2 = 1 N/mm2

V3 = 1 N/mm2

Vc = = 0.331 N/mm2

Vce = 0.662 N/mm2


Punching Shear

For Column One


2 way shear= 3.24 KN

Pm=300x2+300x2=544x12=7728 mm

τv = Vmax/(Pm · d) = 3.24x1000/(7728)x544 =0.00077 N/mm2


τv < Vt1 , Hence safe

For Column Two



Pm=300x2+300x2=544x12=7728 mm

τv = Vmax/(Pm · d) = 3.24x1000/(7728)x544 =0.00077 N/mm2






Figure 2-17: Shear force and Bending Moment diagrams


Chapter — 2


2.8.4 Comparison



Percent Dif-ference

Bearing Pressure 147.82 KN/m2

147.82 KN/m2 None

Governing Moment 140 KN-m

174.9 KN-m

168 KN-m

135.77 KN-m

174.9 KN-m

168 KN-m

Negligible

Shear Force(One-Way) 225.68 KN 225.69 KN NegligibleShear Force(Two-Way) 3.24 KN

3.24 KN

3.24 KN

3.24 KN

None

Resisting Moment forOverturning (Z)

2771.5 KNm 2771.5 KNm Negligible

Resisting Moment forOverturning (X)




2.9.2 ProblemDesign a combined footing with the given data: Load Fy = 600 KN, Mz=30 KNm, eachcolumn., fc = 25 MPa, fy = 450 MPa, Column Dimension = 300 mm x 300 mm, Pedestalheight-500 mm. and C/C column distance=3500 mm . Bearing Capacity of Soil = 140KN/m2. Coefficient of friction =0.5, FOS against sliding =1.5, FOS against overturning =1.5

No soil above footing and dry condition.




2.9.3 SolutionFigure 2-18: Plan and Elevation

Approximate area of footing required = 2x600/140 m2 = 8.57 m2


( left overhang=right overhang=1.2m)

Weight of footing = 5.9 m x 1.7 m x 0.500 x25 KN = 125.375 KN


Therefore, total load on the footing = (2x600+125.375 +2.25) KN = 1327.63 KN

Maximum pressure from axial load= 1327.63 /(5.9 x1.7) = 132.37 KN/ m2

Total Moment=30+30 = 60 KNm

CG of load= 2.9 m

CG of raft = 2.9 m

Eccentricity =2.95-2.9 =0.05 m

So Moment= P x Eccentricity = 1200x0.05 =60 KNm

Z= 1.7 x 5.92/6= 9.863 m3

So M/Z = 6.09 KN/m2

So stress at left end= P/A + M/Z= 138.46 KN/m2


Chapter — 2


So stress at left end= P/A - M/Z= 126.28 KN/m2

138.46 KN/ m2 <140 KN/m2

Hence safe


About Z Direction



So FOS = 3916.5085/60 = 65.27 >1.5

Hence OK

About X Direction



Hence OK

Check For Trial Depth against




deff = (500-50-0.5 x 12) mm = 444 mm

K = Mu/fc.b.deff2 = 214x106/(25x1700x444x444) = 0.0255 <0.156

Hence OK




deff = (500-50-0.5 x 12) mm = 444 mm

K = Mu/fc.b.deff2 = 0.0276= <0.156

Hence OK

Moment About X Axis


Bending moment at critical section, Mux = (158.98+176.02)/2 x5.9x0.72/2 =242.122 KNm





deff = (500-50-0.5 x 12) mm = 444 mm

K = Mu/fc.b.deff2 = 242x106/(25x6500x544x544) = 0.0083 <0.156

Hence OK



Z = d = 0.971 d>0.95d

So, Z= 0.95d





Z = d = 0.968 d>0.95d

So, Z= 0.95d





Z = d = 0.99 d>0.95d

So, Z= 0.95d


Minimum area of steel Astmin = 0.0013 x B x D =3835 mm2



Chapter — 2




Vumax = 347.24 KN

Developed shear stress V = 347.24x1000/(1700x444) = 0.46 N/mm2


V < Vc1, Hence Safe

V1 = 0.17 N/mm2

V2 = 1 N/mm2

V3 = 1 N/mm2

Vc = = 0.351 N/mm2

Vce = 0.702 N/mm2


Punching Shear

For Column One



Pm=300x2+300x2=444x12=6528 mm

τv = Vmax/(Pm · d) = 380.43x1000/(6528)x544=0.131 N/mm2



For Column Two



Pm=300x2+300x2=444x12=6528 mm

τv = Vmax/(Pm · d) = 407.34x1000/(6528)x544 =0.14 N/mm2






Figure 2-19: Shear force and Bending Moment diagrams


Chapter — 2


2.9.4 Comparison



Percent Dif-ference

Bearing Pressure 138.46KN/m2

126.28KN/m2

138.45 KN/m2

126.28 KN/m2

None


231.62 KNm

242.122KNm

212 KN-m

231.78 KNm

249.42 KNm

Negligible

Shear Force(One-Way) 347.24 KN 347.24 KN NoneShear Force(Two-Way) 380.42 KN

407.34 KN

380.43 KN

407.34 KN

None

Resisting Moment for Over-turning (Z)

3916.5 KNm 3916.5 KNm None

Resisting Moment for Over-turning (X)

1128.48KNm

1128.48 KNm None


2.10 Mat Combined Foundation2.10.1 Reference‘Reinforced Concrete’ by T.J.Macgingley & B.S.Choo, Page 351 and Example: 11.4

2.10.2 ProblemDesign a rectangular base to support two columns carrying the following loads:

Column 1 Dead load 310 kN, imposed load 160 kN

Column 1 Dead load 310 kN, imposed load 160 kN

The columns are each 350 mm square and are spaced at 2.5 m centers. The width of thebase is not to exceed 2.0 m. the safe bearing pressure on the ground is 180 kN/m2. Thematerials are grade 35 concrete and grade 460 reinforcement.


2.10 Mat Combined Foundation



2.10.3 SolutionLet us assume the self weight of the base is 130 kN.

Total vertical load = (310 + 160 + 430 + 220 + 130) = 1250 kN.

Area of base = 1250 / 160 = 7.81 m2 (considering safe base pressure as 160 kN/m2).

Length of base = 7.81/2 = 3.91 m

Let the dimension of the mat is as follows,

Width = 2 m,

Length = 4.5 m,

Depth = 0.6 m.

Hence self-weight of mat = 2 x 4.5 x 0.6 x 24 = 129.6 kN.

Hence, total vertical load = (310 + 160 + 430 + 220 + 129.6) = 1249.6 kN.

Actual base pressure = 1249.6 / (2 x 4.5) = 138.84 kN/m2.

Load Case I (DL & IL on both the column)

The ultimate loads are,

Column 1 load = 1.4 x 310 + 1.6 x 160 = 690 kN,

Column 2 load = 1.4 x 430 + 1.6 x 220 = 954 kN.

The distance of center of gravity from column 1 is checked for service load case 1:

x = (954 x 2.5)/(690 + 954) = 1.45 m.


Chapter — 2


The soil pressure is checked for service loads for case 1:

Base area = 4.5 x 2 =9.0 m2,

Base modulus = 2 x 4.52 / 6 = 6.75 m3.

Direct load = 310 + 160 + 430 + 220 + 129.6 = 1249.6 kN.

The moment about the centerline of the base is,

M = (430 + 220) 1.05 – (310 + 160) 1.45 = 1.0 kN-m.

Maximum pressure = 1249.6 / 9 + 1 / 6.75 = 138.9 kN / m2 < 180 kN / m2 (Hence safe)

Total uniformly distributed upward load = (690 + 954) / (4.5 x 2 / 2) = 365.33 kN / m.

Load Case II (DL & IL on column 1; DL on column 2)


Column 1 load = 1.4 x 310 + 1.6 x 160 = 690 kN,

Column 2 load = 1.0 x 430 = 430 kN.

Direct load = 310 + 160 + 430 + 129.6 = 1029.6 kN.


M = 430 X 1.54 – (310 + 160) 0.96 = 230 kN-m.

Maximum pressure = 1029.6 / 9 + 230 / 6.75 = 148.47 kN / m2 < 180 kN / m2 (Hencesafe)

Load Case III (DL on column 1; DL & IL on column 2)


Column 1 load = 1.0 x 310 = 310 kN,

Column 2 load = 1.4 x 430 + 1.6 x 220 = 954 kN.

Direct load = 310 + 430 + 220 + 129.6 = 1089.6 kN.


M = (430 + 220) 1.05 – 310 X 1.45 = 233 kN-m.

Maximum pressure = 1089.6 / 9 + 233 / 6.75 = 155.59 kN / m2 < 180 kN / m2 (Hence safe)




Design of Longitudinal (Bottom) Steel

The maximum negative BM from Figure 6.18 is 221.7 kN-m

Assuming 40 mm clear cover and 20 mm bar diameter, effective depth = 600 – 40 – 20/2= 550 mm.

0.0104 < 0.156

Hence Safe



Chapter — 2


= 970.95 mm2.


Provide minimum steel.

Provide 16 bars 12 mm in diameter at 125 mm centers to give area of 1808 mm2.

Design of Transverse (Top) Steel

The maximum positive BM from Figure 6.18 is 99.7 kN-m.

Assuming 40 mm clear cover and 20 mm bar diameter, effective depth = 600 – 40 – 20/2= 550 mm.

0.004 < 0.156

Hence Safe


= 436.6 mm2.


Provide minimum steel as above.

Calculation of Vertical Shear

The maximum vertical shear from case 1 is V = 250.8 kN,

V = = 0.228 N/mm2.

= 0.39 N/mm2.





Punching Shear Check

The critical perimeter for punching shear check is at 1.5 d distance from the face of thecolumn. Here the perimeter crosses the boundary of the base on two sides. Hence punchingshear is less critical than the vertical shear in this case.

2.10.4 Comparison



Percent Dif-ference

Max BendingMoment (-)

221.7 kN-m 201.204 kN-m 9.2

Max BendingMoment (+)

99.7 kN-m 108.94 kN-m 9.2

Area of StealRequired

780 mm2/m 780 mm2/m None

Base Pressure 138.84kN/m2

136 kN/m2 2


2.11 General Isolated Foundation withEccentricity

2.11.1 Reference

2.11.2 ProblemDesign an isolated footing with the given data:

Loads:

Fx =-300 KN

Fy =-500 KN

Fz=-200 KN

Mz= 45.89 KNm

Mx=-98.32 KNm

Offset of column in X-direction (Oxd) =300 mm

Offset of column in Z-direction (Ozd) =300 mm

Density of soil =14 KN/m3

Depth of Soil = 0.5m

Density of Concrete = 25 KN/cum

Coefficient of Friction (μ) = 0.5


Chapter — 2

2.11 General Isolated Foundation with Eccentricity

Safe Bearing Capacity of the Soil (σ) = 120 KN/m2

Factor of Safety against Overturning =1.5

Factor of Safety against Sliding = 1.5

Permissible soil pressure =

Column dimension = 0.3m x 0.3m,

Strength of concrete (fc’) = M-30 = 4349.39 Psi

Strength of steel (fy) = Fe-415


Self-wt of footing, wt of soil and surcharge are not included for shear and momentcomputations




2.11.3 SolutionDesign started with trial dimensions of 5.0m X 5.0m X 1.0m

Determination of base area of footing

The base area of footing is determined using service (unfactored) loads with the netpermissible soil pressure.

Net permissible soil pressure = 120 KN/ m2

Required base area of footing = (500 + 0.10 x 500)/ 120= 4.5833 m2

Use a 5 .0m x 5.0m square footing (Af =25 m2).

Using a depth of 1m;

Self-wt. of Footing =(5.0*5.0*1*25) = 625 KN

Wt. of soil = 14*0.5*[(5.0*5.0) –(0.3*0.3)] =174.37

Serviceability Check

The net moments are given by:-

Mz =+45.89 – (-300*1) +(-500*0.3) = 195.89 KNm

Mx= - 98.32 +(-200*1) –(-500*0.3) = -148.32 KNm

The pressure at the four corners are given by:-

σ 1 = ((500+625)/ 25) + (6*195.89/53) - (6*148.32/53) = 47.283 KN/m2

σ 2 = ((500+625)/ 25) - (6*195.89/53) - (6*148.32/53) = 28.478 KN/m2

σ 3 =((500+625)/ 25) - (6*195.89/53) + (6*148.32/53) = 42.717 KN/m2

σ 4 = ((500+625)/ 25) + (6*195.89/53) + (6*148.32/53) =61.522 KN/m2

which is < 120 KN/m2. Hence OK

Stability Check

Calculation for Overturning and Sliding:

For Sliding:

Along X- Direction

Disturbing force = -300 KN

Restoring Force = m*(Wt of Footing + Fy + Wt of Soil) = 649.685 KN

Hence, Factor of Safety against Sliding = (649.685/300) =2.166 > 1.5 Hence Safe


Chapter — 2


Along Z- Direction


Restoring Force = mu*(Wt of Footing + Fy + Wt of Soil) = 649.685 KN

Hence, FacFor Overturning:tor of Safety against Sliding = (649.685/200) =3.2484 > 1.5Hence Safe

About X- Direction

Overturning Moment = Mx + Fz* (Ht of Pedestal + Depth of Footing) = -98.32– 200* (0 +1) = -298.32 KN-m

Restoring Moment = Fy * (Width of Footing *0.5 –Ozd)+ (Wt of Soil + Wt ofFooting) * Width of Footing*0.5 = 3098.425 KN-m

Hence, Factor of Safety against Overturning = (3098.425/298.32) =10.386 > 1.5 HenceSafe

About Z- Direction

Overturning Moment = Mx + Fz* (Ht of Pedestal + Depth of Footing) =45.89+ 300* (0 +1) = 345.89 KN-m


Hence, Factor of Safety against Overturning = (3398.425/345.89) =9.82516 > 1.5 HenceSafe

Base Pressure for Shear and Moment Calculation


σ 1 = (500/ 25) + (6*195.89 /5.03) - (6*148.32 /5.03) = 22.2834KN/m2

σ 2 = (500/25) - (6*195.89 /5.03) - (6*148.32 /5.03) = 3.47792KN/m2

σ 3 =(500/ 25) - (6*195.89 /5.03) + (6*148.32 /5.03) = 17.7167 KN/m2

σ 4 = (500/ 25) + (6*195.89 /5.03) + (6*148.32 /5.03) = 36.52208 KN/m2

Check for Flexure and Calculation for Reinforcement

Factored loads and soil reaction:

To proportion the footing for strength (depth and required reinforcement) factored loads areused. For this problem, the factors used are all 1.0

Figure 2-22: Bending about major axes




Bending About Z-axis Bending About X-axis

Critical section for moment is at the face of column

About X- axis

Average Base Pressure along one edge

=(22.2834+3.47792)/2 =12.8806 KN/m2

Average Base Pressure along other edge

=(17.7167+36.5221)/2 = 27.1194 KN/m2

Approximate Base Pressure at the critical section

=27.1194- {(27.1194 – 12.8806)/5.0*2.05} =21.2815 KN/ m2 [2.05 =5-(5/2+0.3+0.15)]

Hence, the moment at the critical section

Mu =5.0*{21.2815 *2.05*2.05*0.5+0.5*(27.1194-21.2815)* 2.05*2.05*2/3}=264.48 KNm

Effective depth (d) = 1000 – 50 – 20 = 930 mm

Hence safe

Therefore,


Chapter — 2


The minimum area of steel = 0.13 x 5000 x 1000 / 100 = 6500 mm2 > Calculated area ofsteel.

So, provide minimum steel = 6500 mm2

About Z- axis


=(36.5221+22.2834)/2 =29.4027 KN/m2


=(17.7167+3.47792)/2 = 10.5973 KN/m2


=29.4027- {(29.4027-10.5973)/5.0*2.65} =19.4358 KN/ m2 [2.65 =(5/2+0.3-0.15)]


Mu =5.0*{19.4358 *2.65*2.65*0.5+0.5*(29.4027 –19.4358)*2.65*2.65*2/3}=457.874 KNm


Hence safe

Therefore,

The minimum area of steel = 0.13 x 5000 x 1000 / 100 = 6500 mm2 > Calculated area ofsteel.

So, provide minimum steel = 6500 mm2

Check for Shear

Assume overall footing thickness = 1.0m and average effective thickness d = 0.92m (36.22in)

Wide-beam action (One-Way Shear) :




Along Z-Z axis

Vu = qs tributary area

Bw = 5.00m = 196.8504 in

qs is given by:-


=(22.2834+3.47792)/2 =12.8806 KN/m2


=(17.7167+36.5221)/2 = 27.1194 KN/m2


=27.1194- {(27.1194 – 12.8806)/5.0*1.13} =23.9014 KN/ m2 [1.13=5 -(5/2+0.3 +0.92 +0.15)]

Hence, the one- way shear at the critical section

Vux =5.0*{23.9014*1.13+0.5*(27.1194-23.9014)*1.13}= 144.134 KN

Design shear stress, v = 144.134/(5.0 x 0.93) =30.996 kN/ m2

Now, VC1 = min(0.8 √(fcu),5) N/ mm2= 4381.78 kN/ m2 > v (Hence Safe)

=348.5494 kN/ m2




Along X-X axis


Bw = 5.00m = 196.8504 in

qs is given by:-


=(36.5221+22.2834)/2 =29.4027 KN/m2



Chapter — 2


=(17.7167+3.47792)/2 = 10.5973 KN/m2


=29.4027- {(29.4027-10.5973)/5.0*1.73} =22.89603 KN/ m2 [ 1.73=(5/2 +0.3–0.92 –0.15)]

Hence, the Design one-way shear at the critical section

Vuz =5.0*{22.89603*1.73+0.5*(29.4027-22.89603)*1.73}= 226.1924 KN

Design shear stress, v = 226.1924/(5.0 x 0.93) =48.64353 kN/ m2

Now, vC1 = min(0.8 √(fcu),5) N/ mm2= 4381.78 kN/ m2 > v (Hence Safe)

= 348.5494 kN/ m2







Two-way action (Punching Shear)

Along X-X axis

[3090 mm = 300 + 2x(1.5 x 930)]

The punching shear will be calculated for an area outside the area enclosed by therectangle at a distance 1.5d from the column face as shown in figure.

Total pressure under the base = 5.0 x 5.0 x 10.5973 + 0.5 x 5.0 x 5.0 x (29.4027 –10.5973) = 500.00 kN.

Pressure at the critical sections:-

σa = 29.4027 – ((29.4027-10.5973)/5.0*0.955) = 25.810869 KN/m2

σb = 10.5973 + ((29.4027-10.5973)/5.0*0.955) = 14.189131 KN/m2

Pressure under enclosed rectangle = (3.09)2 x 14.189131 + 0.5 x (3.09)2 x (25.810869 –14.189131) = 190.962 kN

Punching shear force = 500.00-190.962 = 309.038 kN.

Critical perimeter = 3.09 x 4 =12.36 m.


Chapter — 2


Punching shear stress = 309.038/(12.36 x 0.93) = 26.885 kN / m2.

Hence, the punching shear stress is less than VC . Hence Safe

Along Z-Z axis

The punching shear will be calculated for an area outside the area enclosed by the rectangleat a distance 1.5d from the column face as shown in figure.

Total pressure under the base = 5.0 x 5.0 x 12.8806 + 0.5 x 5.0 x 5.0 x (27.1194 – 12.8806)= 500.00 kN.

Pressure at the critical sections:-

σa = 27.1194 – ((27.1194-12.8806)/5.0*0.955) = 24.3998 KN/m2

σb = 12.8806 + ((27.1194-12.8806)/5.0*0.955) = 15.60021 KN/m2

Punching shear force = 500.00-190.962 = 309.038 kN.

Critical perimeter = 3.09 x 4 =12.36 m.

Punching shear stress = 309.038/(12.36 x 0.93)= 26.885 kN / m2.




Hence, the punching shear stress is less than VC . Hence Safe


Mu =5.0*{19.4358 *2.65*2.65*0.5+0.5*(29.4027 –19.4358)*2.65*2.65*2/3}=457.874 KNm


2.11.4 Comparison

ReferenceResult

STAADFoundationResult*

Difference (Reasonsthere-of)

Moment about X 264.48KNm

259.98KNm

Error due to approx-imation in base pressureinterpolation

Moment about Z 457.874KNm

448.24KNm


Area of steel aboutX-X

6500.00mm2

6500.00mm2

Negligible

Area of steel aboutZ-Z

6500.00mm2

6500.00mm2

Negligible

Shear Stress

(One way) along X

48.64KN/ m2

46.66 KN/m2


Shear Stress

(One way) along Z

30.996KN/ m2

29.28 KN/m2


Shear Force

(Two way)

309.038KN

305.46 KN Error due to approx-imation in base pressureinterpolation

Factor of Safetyagainst Sliding (X)


Factor of Safetyagainst Sliding (Z)


Factor of Safetyagainst Over-turning (X)


Factor of Safetyagainst Over-turning (Z)


Table 2-12: British verification example 13 comparisons


Chapter — 2


Section 3

Canadian Code (CSAA23.3-2004)3.1 CSA General Isolated Foundation 1

3.1.1 Reference

3.1.2 ProblemDesign of a square Isolated Footing

A tied column, 450 mm square, and reinforced with eight No. 35 bars carries an unfactoreddead load of 1300 kN and an unfactored live load of 1000 kN. Suitable soil with a factoredsoil bearing pressure of 300 kN/m2 is available at a depth of 1.5 m . Design a squarefooting.

The compressive strength f’c is 30 MPa for the column and 25 MPa for the footing. All steelhas fy=400 MPa. Unit weight of concrete and soil is 24 kN/m

2 and 16 kN/m2 respectively.



3.1.3 Solution

Trial Footing Size

Calculate the initial footing size based on soil bearing capacity.

As per CSA A.23.3-04 cls. 8.3 and Annex C. the 2005 National Building Code of Canadaload combination factors must be used:

Factored Load = 1.25 DL + 1.5 LL = (1.25 X 1300 kN) + (1.5 x 1000 kN) =3,125 kN

Required area of footing: 3125/300 = 10.41 m2

Total Axial load = 3125+Self Weight of footing + weight of soil

Self Weight of footing = 3.6 x 3.6 x 0.75 x 24 = 233.28 KN

weight of soil=3.6 x 3.6 x 1.5 x 16 = 311.04 KN

For square footing, the axial force on the footing is:

3125+233.28+311.04 =3669.32 KN

So stress on soil=369.32/(3.6x3.6)=283.12 KN/m2


Chapter — 3

3.1 CSA General Isolated Foundation 1

Calculate Factors of Safety

In this case, we do not have any sliding and overturning forces. The CSA A23.3-04recommends for a footing that experiences horizontal shear, the designer must make surethat this shear is transferred to the subgrade utilizing the passive soil resistance and thefriction between the subgrade and the footing surface. The passive soil resistance will beignored in STAAD Foundation to calculate the factor of safety against sliding check. Factorof safety against overturning must be checked as per the NBCC.

Anyway, max sliding force equals the axial load x coefficient of friction

for coeff. of friction =0.5,

Sliding force= 0.5x 3669.32 =1834.66 KN (same for X & Z dir)

Max resisting moment against overturning = axial force x Dimension/2= 0.5x3669.32x3.6 KNm = 6604.78 KNm (Same wrt both x and z axis).

Stress on soil from Factored load = 3125/(3.6x3.6)=241.126 KN/m2


Along X Direction


deff = (750-50-0.5 x 20) mm = 690 mm

Vumax =

= 768.22 KN

Ø = 0.65, λ = 1

dv = 0.9.deff = 0.9 x 690 = 621 mm

bw=3600 mm

Now allowable shear

Vc = =1030913 N =1031 KN

V < Vc, Hence Safe

Along Z Direction


deff = (750-50-0.5 x 20) mm = 690 mm

Vumax = = 768.22 KN

Ø = 0.65, λ = 1

dv = 0.9.deff=0.9 x 690=621 mm

Section 3 Canadian Code (CSA A23.3-2004)



bw = 3600 mm

Now allowable shear

Vc = = 1030913 N = 1031 KN

V < Vc, Hence Safe

Punching Shear



deff = (750-50-0.5 x 20) mm = 690 mm

Area within Critical Perimeter Am = (450+2x0.5x.69)2=1.2996 m2

Vmax = 241.126x(3.6x3.6-1.2996)=2811.63 KN

Critical perimeter Pm = 2 X ( b + h + 2x d) = 4.56 m

τv = Vmax/(Pm · d) = 0.8936 N/mm2

α=4

ß=L/B =4.5/4.5 =1

1.8525 N/mm2

2.5846 N/mm2

1.235 N/mm2

As effective depth>300 mm so the multiplier=1300/(1000+deff)=0.769

So,

Vr1= 1.424 N/mm2 =1424 KN/m2

Vr2= 1.987N/mm2 = 1987 KN/m2

Vr3= 0.9497 N/mm2 = 949.7 KN/m2

So min {Vr1, Vr2, Vr3} = 949.7 KN/m2

So allowable shear = Vc = 949.7 KN/m2

V < Vc , Hence safe

Development Length

Along Z Axis

ld = 0.45 k1 x k2 x k3 x k4 fy/√(f'c) dbk1 = 1 if clear cover is less than 300 mm or else use 0.45


Chapter — 3


k2 = 1 if coated reinforcement is used

k2 = 1.2 if epoxy coated reinforcement is used

k2 = 1.5 if epoxy coated reinforcement is used and clear cover is less than 3xdbk2 = 1.5 if bar spacing is less than 6x dbk3 = 1 Normal density concrete

k4 = 0.8 for 20M and smaller bar size

k4 = 1 for 20M and larger bar size

ld = 0.45 x 1 x 0.8 x 1 x 0.8 x (400 MPa)/√(25MPa) x 19.5 mm = 449.28 mm

Available Length = (3600-450)/2-50 = 1525 mm

Hence OK

Along X Axis




k2 = 1.5 if epoxy coated reinforcement is used and clear cover is less than 3xdbk2 = 1.5 if bar spacing is less than 6 xdbk3 = 1 Normal density concrete





Hence OK

Check For Trial Depth Against Moment

About X Axis

Bending moment at critical section:

Mux = = 1076.66 KN-m

α1 = 0.85-0.0015.f’c=0.8125

Øs=0.85




= 0.62817


deff = (750-50-0.5 x 20) mm = 690 mm

So, ρ =0.001892

Hence OK

About Z Axis

Bending moment at critical section,

Mux = = 1076.66 KN-m

α1 = 0.85-0.0015.f’c=0.8125

Øs=0.85

= 0.62817


deff = (750-50-0.5 x 20) mm = 690 mm

So,ρ =0.001892

Hence OK



Astx =ρ.B.deff= 4700 m2m

Minimum area of steel Astmin = (0.2x√f’c /fy)xB.D= 6750 mm2



Astz = ρ.L.deff= 4700 m2m







Chapter — 3


(as no uplift is present so min steel is provided)

Along Z Direction (Top)



(as no uplift is present so min steel is provided)

3.1.4 Comparison



Percent Dif-ference

Bearing Pressure 286.45 KN/m2 283.5 KN/m2 NegligibleGoverning Moment 1076.66 KN-m

1076.66 KN-m

1076.64 KN-m

1076.66 KN-m

Negligible


768.22 KN

768.01 KN

768.01 KN

Negligible

Shear Force(Two-Way) 2811.63 KN 2811.49 KN NegligibleResisting Shear Force(One-Way)

1031 KN

1031 KN

1031.14 KN

1031.14 KN

None

Resisting Shear Stress(Two-Way)

949.7 KN/m2 949.86KN/m2

Negligible

Resisting force forsliding

1834.66 KN

1834.66 KN

1837.09 KN

1837.09 KN

Negligible


6604.78 KNm

6604.78 KNm

6613.4 KNm

6613.4 KNm

Negligible

Ast (B) 6750 mm

6750 mm

6750 mm

6750 mm

None

Ast (T) 6750 mm

6750 mm

6750 mm

6750 mm

None

Ld (rqrd) 449.28 mm

449.28 mm

449.28 mm

449.28 mm

None

Ld (available) 1525 mm

1525 mm

1525 mm

1525 mm

None

Table 3-1: CSA verification example 1 comparison

3.2 CSA General Isolated Foundation 23.2.1 Reference




3.2.2 ProblemDesign of a square Isolated Footing

A tied column, 500 mm square, and reinforced with eight No. 35 bars carries an unfactoreddead load of 900 kN and an unfactored live load of 800 kN. Suitable soil with a factoredsoil bearing pressure of 300 kN/m2 is available at a depth of 1.5 m . Design a squarefooting.

The compressive strength f’c is 20 MPa for the column and 20 MPa for the footing. Allsteel has fy=350 MPa. Unit weight of concrete and soil is 24 kN/m

2 and 16 kN/m2

respectively.


3.2.3 Solution

Trial Footing Size

Initial footing size is based on soil bearing capacity.

As per CSA A.23.3-04 cls. 8.3 and Annex C. the 2005 National Building Code of Canadaload combination factors must be used:

Factored Load = 1.25 DL + 1.5 LL = (1.25 X 900 kN) + (1.5 x 800 kN) =2,325 kN

Required area of footing: 2,325 /300 = 7.75 m2


Chapter — 3


Total Axial load = 2,325 +Self Weight of footing + weight of soil

Self Weight of footing = 3 x 3 x 0.6 x 24 = 129.6 KN

weight of soil=3 x 3 x 1.5 x 16 = 216 KN

For square footing, the axial force on the footing is:

2,325 +129.6 + 216 =2670.6 KN

So stress on soil=2670.6 /(3x3)=296.73 KN/m2

Calculate Factors of Safety

In this case, we do not have any sliding and overturning forces. The CSA A23.3-04recommends for a footing that experiences horizontal shear, the designer must make surethat this shear is transferred to the subgrade utilizing the passive soil resistance and thefriction between the subgrade and the footing surface. The passive soil resistance will beignored in STAAD Foundation to calculate the factor of safety against sliding check. Factorof safety against overturning must be checked as per the NBCC.

Anyway, max sliding force =axial load x coeff of friction

for coeff of friction =0.5,

Sliding force= 0.5x 2670.6 =1335.3 KN (same for X & Z dir)

Max resisting moment against overturning = axial force x Dimension/2 = 0.5x2670.6x3 KNm = 4005.45 KNm (Same WRT both x and z axis).

Stress on soil from Factored load=2325/(3x3)=258.33 KN/m2


Along X Direction


deff = (600-50-0.5 x 20) mm = 540 mm

Vumax = = 550.24 KN

Ø = 0.65, λ = 1

dv = 0.9.deff = 0.9 x 490 = 441 mm

bw = 3000 mm

Now allowable shear

Vc = =655986 N =655.99 KN

V < Vc, Hence Safe




Along Z Direction


deff = (600-50-0.5 x 20) mm = 540 mm

Vumax = = 550.24 KN

Ø = 0.65, λ = 1

dv = 0.9.deff = 0.9 x 490 = 441 mm

bw = 3000 mm

Now allowable shear

Vc = = 655986 N = 655.99 KN

V < Vc, Hence Safe

Punching Shear



deff = (600-50-0.5 x 20) mm = 540 mm

Area within Critical Perimeter Am = (500+2x0.5x.54)2 = 1.0816 m2

Vmax = 258.33 x (3x3-1.0816) = 2045.56 KN

Critical perimeter Pm = 2 x ( b + h + 2x d) = 4.16 m

τv = Vmax/(Pm · d) = 0.91 N/mm2

α=4

ß=L/B =5/5 =1

1.657 N/mm2

2.645 N/mm2

1.1046 N/mm2

As effective depth > 300 mm so the multiplier = 1300/(1000+deff) = 0.844

So,


Chapter — 3


Vr1= 1.398 N/mm2 =1398KN/m2

Vr2= 2.232N/mm2 ==2232 KN/m2

Vr3= 0.932 N/mm2 =932 KN/m2

So min{ Vr1,Vr2,Vr3} = 932 KN/m2

So allowable shear = Vc = 932 KN/m2

V < Vc , Hence safe

Development Length

Along Z Axis









Hence OK

Along X Axis









Hence OK





About X Axis

Bending moment at critical section

Mux = = 605.46 KN-m

α1 = 0.85-0.0015.f’c=0.82

Øs=0.85

= 0.0346


deff = (600-50-0.5 x 20) mm = 540 mm

So,ρ =0.002405

Hence OK

About Z Axis


Mux = = 605.46 KN-m

α1 = 0.85-0.0015.f’c=0.82

Øs=0.85

= 0.0346


deff = (600-50-0.5 x 20) mm = 540 mm

So,ρ =0.002405

Hence OK



Astx =ρ.B.deff= 3896 m2m


Chapter — 3


Minimum area of steel Astmin = (0.2x√f’c /fy)xB.D= 4599.9 mm2



Astz = ρ.L.deff= 3896 m2m



Use #20 @ 190 c/c




(as no uplift force is present only min steel is provided)

Use #20 @ 190 c/c

Along Z Direction (Top)



(as no uplift force is present only min steel is provided)

Use #20 @ 190 c/c




3.2.4 Comparison

Value of Reference ResultSTAAD

FoundationResult

PercentDifference

Bearing Pressure 296.73 KN/m2 296.07KN/m2

Negligible

Governing Moment 605.46 KN-m

605.46 KN-m

605.46 KN-m

605.46 KN-m

None

Shear Force(One-Way)

550.24 KN

550.24 KN

550.06 KN

550.06 KN

Negligible

Shear Force(Two-Way)

2045.56 KN 2045.45 KN Negligible

Resisting force forsliding

1335.3KN 1335.3KN 1332.3 KN

1332.3 KN

Negligible

Resisting Momentfor Overturning

4005.45 KNm 4005.45 KNm

3996.827KNm

3996.827KNm

Negligible

Ast (B) #20@190 c/c

#20@190 c/c

#20@190c/c

#20@190c/c

None

Ast (T) #20@190 c/c

#20@190 c/c

#20@190c/c

#20@190c/c

None

Ld (rqrd) 439.52 mm

439.52 mm

439.52 mm

439.52 mm

None

Ld (available) 1200 mm

1200 mm

1200 mm

1200 mm

None


3.3 CSA General Isolated Foundation 3 3.3.1 Reference

3.3.2 ProblemDesign an isolated footing with the given data: Load Fy = 1200 KN, fc = 30 MPa, fy = 400MPa, Column Dimension = 300 mm x 300 mm, and Bearing Capacity of Soil = 90KN/m2.


Chapter — 3


Coefficient of friction =0.5, FOS against sliding =1.5, FOS against overturning =1.5. Heightof soil above footing=450 mm, GWT is 300 mm from GL.

Surcharge= 10 KN/m2





Weight of above soil = 4.3 x 4.3 x 0.450 x 18 KN = 149.77 KN

Reduction of Weight due to buoyancy = 4.3x4.3 x (0.500+0.450-0.300) x 9.81KN = 117.9 KN

Load due to surcharge = 4.3x4.3 x 10 KN =184.9 KN

Therefore, total load on the footing = (1200+231.125 +149.77 +184.9 -117.9)KN = 1647.895 KN

Maximum pressure = 1647.895 /(4.3x4.3) = 89.12 KN/m2

89.12 KN/ m2 <90 KN/m2

Hence safe




Stress for Factor design = 1.25x1200/(4.3x4.3) = 81.12 KN/m2


Along X Direction

Sliding force =0



Max resisting force against sliding = 0.5 x 1647.895 KN = 823.95 KN

Hence OK

Along Z Direction

Sliding force =0



Max resisting force against sliding = 0.5 x 1647.895 KN = 823.95 KN

Hence OK


WRT X Direction



Hence OK

WRT Z Direction



Hence OK


Chapter — 3


3.3.4 Comparison



Percent Dif-ference


89.03 KN/ m2 Negligible

Resisting force for sliding(x)


Resisting Moment forOverturning (z)

3542.97KNm

3539.366KNm

Negligible

Resisting force for sliding(z)


Resisting Moment forOverturning (x)

3542.97KNm

682.862 KNm Negligible


3.5 CSA Pilecap Foundation 13.3.5 Reference

3.3.6 ProblemDesign pilecap foundation with the given data: Load Fy = 800 KN, fc = 25 MPa, fy = 450MPa, Column Dimension = 250 mm x 250 mm. Pedestal ht= 500 mm

Pile Data

Dia of pile= 400 mm.

Vertical capacity = 250 KN,

Horizontal capacity = 100 KN

Uplift capacity = 80 KN


3.5 CSA Pilecap Foundation 1


Figure 3-4: Elevation and Plan, with dimension and loads

3.3.7 Solutiondepth of pilecap= 1.5 x piledia, D=600 mm

Take D=600 mm

c/c pile distance = 3xpile dia =1200 mm. Edge diatance =350 mm

Assuming 4 pile combination,

Coordinates of piles considering pedestal at 0,0,0)

Pile No X Coordinate (mm) Z Coordinate (mm)1 -600 -6002 -600 6003 600 6004 600 -600

pilecap dimension is 1900 mm x1900 mm x 600 mm

Weight of footing = 1.9 x1.9 x 0.60 x 25 KN = 54.15 KN

Weight of pedestal = 0.25 x 0.25 x 0.5 x 25 KN = 0.78 KN

Therefore, total load on the pilecap = (800+54.15 +0.78) KN = 854.93 KN


Chapter — 3


So Pile reaction = 854.93 /4= 213.73 KN < 250 KN, Hence OK

As there is no lateral load, moment or uplift force, so each pile is safe in lateral & upliftcapacity.

Factored Design

Load factor for self wt is taken =1

Load factor for axial load is taken 1.25

SO, Load on pilecap = 1.25x800+54.15 x1+0.78x1= 1054.93 KN

Load on each pile =1054.93 /4=263.73 KN

Figure 3-5: Bending sections considered

Calculation of Moment wrt Z Axis

For moment wrt X1X1

Contribution from pile 1=from pile2=263.73 x 0.475=125.27 KNm

So Total Mz X1X1 = 250.54 KNm

For moment wrt X2X2

Contribution from pile3=from pile4=263.73 x 0.475=125.27 KNm


So Max value of Mz = 250.54 KNm

Check For Trial Depth against moment wrt Z Axis

Bending moment at critical section, Muz = 250.54 KN-m




Assuming 50 mm clear cover, 75 mm pile in pilecap & and 11.3 mm bar ( Bar No 10),effective depth

deff = 463.7 mm

B =1900 mm ,

Øc = Resistance Factor of concrete = 0.6 (Clause No 8.4.2)

α1 = 0.8125 (Clause No 10.1.7.c)

ß1 = 0.8475 (Clause No 10.1.7.c)

c = (700.d)/(700+fy) = 282.25 mm (Clause No 10.5.2)

a = 239.208 mm (Clause No 10.1.7.a)

C = 5539 x 103 KN(Clause No 10.1.7.a)

Resisting Moment=C.(d-a/2)=1906 KNm

Muz < Resisting Moment, Hence OK

Calculation of Moment wrt X Axis

For moment wrt Z1Z1


So Total Mx Z1Z1 = 250.54KNm

For moment wrt Z2Z2


So Total Mx Z2Z2 = 250.54KNm

So Max value of MX = 250.54 KNm

Check For Trial Depth against moment wrt X Axis


deff = 463.7 mm

B =1900 mm ,

C=5539 x 103 KN

Resisting Moment=C.(d-a/2)=1906 KNm

Mux < Resisting Moment, Hence OK


Chapter — 3


Calculation of Shear parallel to X Axis

Figure 3-6: Shear sections considered

For shear wrt X1X1

Contribution from pile 1 =pile2=263.73 x528=139.25 KN

So Total V X1X1 = 274.5 KN

For shear wrt X2X2

Contribution from pile 3=pile4=263.73 x528=139.25 KN


So Max V parallel to X direction = 274.5 KN

Check for One-Way Shear (along X dir)

Shear Strength Vr = 260/(1000+d).Øc.λ.√(f'c ).bw.d=>0.1.λ.Øc.√(f^' c).bw.d

= max(469.497, 264.309) KN = 469.5 KN (Clause no 11.3.5.2)

Where

Øc = Resistance Factor of concrete=0.6 (Clause No 8.4.2)

λ = factor to account Concrete density=1(Clause No 8.6.5)

bw = Section Width=1900 mm

d = effective Depth= 463.7 mm (Clause 11.0)

VX < VResistanceX, Hence Safe

Calculation of Shear parallel to Z Axis

For shear wrt Z1Z1




Contribution from pile 1 =pile 4 =263.73 x528=139.25 KN

So Total V Z1Z1 = 274.5 KN

For shear wrt Z2Z2

Contribution from pile 2=pile 3 =263.73 x528=139.25 KN


So Max V parallel to Z direction = 274.5 KN

Check for One-Way Shear (along Z dir)

Shear Strength Vr=260/(1000+d).Øc.λ.√(f'c ).bw.d=>0.1.λ.Øc.√(f^' c).bw.d

= max(469.497, 264.309) KN = 469.5 KN

VZ < VResistanceZ, Hence Safe

Punching ShearFigure 3-7: Two-way shear sections considered

Punching shear is checked on a perimeter 0.5d =231.85 mm from the column face.

Contribution from pile 1=from pile2=from pile3=from pile 4= 263.73 KN

So total punching shear Vmax= 1054.92 KN

b0=Perimeter of failure line at d/2 distance from column face=2.(Column Length+ColumnWidth+2d)x 2= 4x(250+463.7/2+463.7/2)=2854.88

Therefore Shear stress τc= =Vmax/(b0 x d)= 797 KN/m2

Calculation of punching shear stress capacity

λ = 1 (Clause 8.6.5)

Øc = 0.6 (Clause 8.4.2)

f'c = Strength of concrete (in MPa)

ßc = (Column Length)/(Column Width)=1900/1900=1

αs = 4 (Clause No 13.0 & 13.4.4.a)


Chapter — 3


mod factor=1

Vr1=(1+2/ßc).0.2.λ.Øc.√(f'c).mod factor=1800 KN/m2 (Eqn no 13.5 Clause no 13.4.4a)

Vr2=(0.2+(αs. dv)/bo).λ.Øc.√(f'c).mod factor=2549 KN/m2 (Eqn no 13.6 Clause no13.4.4b)

Vr3=0.4.λ.Øc.√(f'c).mod factor= 1200 KN/m2 (Eqn no 13.7 Clause no 13.4.4c)

Vr=Min (Vr1,Vr2,Vr3)=1200 KN/m2

V < VResistance , Hence safe


Along X Direction

Calculate Kr (neutral axis/depth ratio) for Actual Bending Moment

Where

α1= 0.85- 0.0015.f’c>=0.67 (Clause No 10.1.7)

Øc = 0.6 (clause 8.4.2)

Øs = 0.85 (clause 8.4.3)

Solving the equation ρ (steel area ratio = 0.165 %

Therefore, Astx = ρ.b,d= 1450 m2m

Minimum area of steel Astmin = 0.2/100 x B x D = 2280 mm2

Provided area = 2280 mm2

Along Z Direction

Calculate Kr (neutral axis/depth ratio) for Actual Bending Moment

Solving the previously stated equation ρ (steel area ratio = 0.165 %

Therefore, Astx = ρ.b,d= 1450 m2m

Minimum area of steel Astmin = 0.2/100 x B x D = 2280 mm2





Check for Development Length

Ld (required)=1.15.k1.k2.k3.k4.fy/((dcs+Kr) ).Ab/√(f'c )= 380.7 mm

(CSA A23-3-04 Clause No 12.2.2 & 12.2.3)

Ld (available)Along X=(Length-pedestal length) 1/2 –Cover=775 mm

Ld (required)<Ld (available)Along X

Hence OK

Ld (available)Along Z=(Length-pedestal length) 1/2 -Cover=775 mm

Ld (required)<Ld (available)Along Z

Hence OK

3.3.8 Comparison



Percent Dif-ference

Depth 600 mm 598 mm NegligibleGoverning Moment (wrtX axis)

250.54 KN-m

250.54 KN-m None

Resisting Moment (wrt Zaxis)

1906 KN-m 1933 KN-m Negligible

Governing Moment (wrtX axis)

250.54 KN-m

250.54 KN-m None

Resisting Moment (wrt Zaxis)

1906 KN-m 1933 KN-m Negligible

Shear Force(One-Way) X 274.5 KN 274.3 KN NoneShear Resistance(One-Way) X


Shear Force(One-Way) Z 274.5 KN 274.3 KN NoneShear Resistance(One-Way) Z


Shear stress(Two-Way) 797 KN/m2 787.6 KN/m2 NoneResistance Shear Stress(Two-Way)

1200 KN/m2 1200 KN/m2 None

Ld required 381 mm 381 mm NoneLd Available ( Along X) 775 mm 775 mm NoneLd Available ( Along Z) 775 mm 775 mm NoneAst ( Along X) 2280 mm2 2240 mm2 Negligible


3.4 CSA General Combined Foundation s1 3.4.1 Reference


Chapter — 3

3.4 CSA General Combined Foundation s1


Ht of soil =500 mm. Depth of GWT=200 mm. Surcharge= 5 KN/m2


3.4.3 SolutionApproximate area of footing required = 2x900/120 m2 = 15 m2


left overhang = right overhang = 1.1 m, C/C col dist=3500 mm

Weight of footing = 5.7 x 3.15 x0.500 x25 KN = 224.44 KN


Weight of soil above footing = (5.7 x 3.15-2x0.25x0.25) x 0.500 x18 KN =160.47 KN

Reduction of Weight due to buoyancy = 5.7 x 3.15 x (0.500+0.500-0.200)x9.81 KN = 140.91 KN

Surcharge load = ( 5.7x3.15-2x0.25x0.25)x5= 89.15 KN

Therefore, total load on the footing is




(2x900+224.44 +1.25+160.47 +89.15 -140.91) KN = 2134.4 KN

Maximum pressure= 2134.4 /(5.7 x3.15) = 118.87 KN/m2

118.87 KN/ m2 <120 KN/m2 (Hence safe)


About Z Direction



max resisting Moment = 0.5 x 5.7x 2134.4 = 6083 KNm

Hence OK

About X Direction




Hence OK

3.4.4 Comparison



Percent Dif-ference


118.87 KN/m2 None


6083 KNm 6083 KNm None


3361.7 KNm 3361.7 KNm None



Chapter — 3


Section 4

Indian Code (IS 456 -2000)4.1 IS General Isolated Foundation 1

4.1.1 Reference‘Reinforced Concrete’ by A.K. Jain, Page 539, Example 18.2.

4.1.2 ProblemDesign an isolated footing with the given data: Load Fy = 1000 KN, fc = 15 MPa, fy = 415MPa, Column Dimension = 400 mm X 400 mm, Bearing Capacity of Soil = 100 KN/m2,and Load Factor = 1.5.




Assuming 3.5 m x 3.5 m x 0.6 m footing dimension (I = 12.5 m4)


Therefore, total load on the footing = (1000 +183.75) KN = 1183.75 KN

Maximum pressure = 1183.75/(3.5 x 3.5) KN/ m2 = 96.633 KN/m2 <100KN/m2

Hence safe

Ultimate pressure = 1000 x 1.5/ (3.5 x 3.5) KN/m2 = 122.45 KN/m2


Mu = 122.45 x 3.5 x 1.55 x 1.55/2 = 514.826 KN-m


de = (600-35-0.5 x 10) mm = 560 mm

Ku,max = = 0.479


Chapter — 4

4.1 IS General Isolated Foundation 1

Ru,max = 0.36 x fc x Ku,max x (1-0.42 Ku,max) = 2.066

Mulim = Ru,max x B x de2 = 2267.642 x 106 N-mm = 2267.642 KN-m> Mu

Hence safe


Area of steel required along length,

Ast = 0.5 x x B x de = 2646.4 mm2




Corresponding allowable c = 0.28 N/mm2

Developed shear stress c =

Vumax = 122.45 x 3.5 x = 424.289 KN

Developed shear stress c = = 0.2165 N/mm2 < c,all (Hencesafe)

Check for Two-Way Shear

Vumax = 1500 KN

Developed shear stress c = = 0.698 N/mm2

Ks = min (0.5+1,1) = 1

Allowable shear stress = Ks x c = 1 x 0.25 = 0.968 N/mm2

Note: There is no deduction for the upward force underneath the area enclosed by thecritical perimeter. This approach is conservative.

Section 4 Indian Code (IS 456 -2000)



Spacing

No. of 10 mm bar = = 33.69 (34)

Spacing = = 102.73 mm

Spacing for 10 mm bar = 102.73 mm

Figure 4-2: Plan of Reinforcement


Chapter — 4


Figure 4-3: Cross Section showing Reinforcement

4.1.4 Comparison



Percent Dif-ference

Effective Depth 560 mm 560 mm NoneGoverning Moment 514.826 KN-m 514.821 KN-m NegligibleArea of Steal 2646.40 mm2 2645.01 mm2 0.05Shear Stress (One-Way)

0.216 N/mm2 0.216 N/mm2 None

Shear Stress (Two-Way)

0.698 N/mm2 0.700 N/mm2 0.286

Table 4-1: IS verification example 1 comparison

4.2 IS General Isolated Foundation 24.2.1 Reference‘Reinforced Concrete Structure’ by Punmia-Jain-Jain, Example 25.1.

4.2.2 ProblemDesign an isolated footing with the given data: Load Fy = 600 KN, fc = 15 MPa, fy = 250MPa, Column Dimension = 500 mm x 500 mm, and Bearing Capacity of Soil = 120 KN/m2.





4.2.3 SolutionApproximate area of footing required = 600/ 120 m2 = 5 m2




Maximum pressure = 650.4 / (2.4 x 2.4) KN/ m2 = 112.92 KN/m2 <120KN/m2

Hence safe

Ultimate pressure = 600 x 1.5 /(2.4 x 2.4) KN/m2 = 156.25 KN/m2

Bending moment at critical section, Mu = 56.25 x 2.4 x 0.95 x 0.95 /2=169.21875 KN-m


de = (350-50-0.5 x 12) mm = 294 mm


Chapter — 4


Ku,max = = 0.53

Ru,max = 0.36 x fc x Ku,max x (1- 0.42 Ku,max) = 2.225

Mulim = Ru,max x B x de2 = 461.568 x 106 N-mm = 461.568 KN-m > Mu

Hence safe


Area of steel required along length,

Ast = 0.5 x x B x de = 2837.87 mm2



Percentage of steel pt = 100·Ast/(B·de) = 0.4022

Corresponding allowable c = 0.42 N/mm2

Developed shear stress c =

Vumax = 156.25 x 2.4 x = 246 KN

Developed shear stress c = = 0.3486N/mm2 < c,all

Hence safe

Check for Two-Way Shear

Vumax = 900 KN

Developed shear stress c = = 0.96 N/mm2

Ks = min (0.5+1, 1) = 1

Allowable shear stress = Ks x c = 1 x 0.25 = 0.968 N/mm2

Hence safe





Spacing

No. of 12 mm bar = = 25.09 (26)

Spacing = = 91.52 mm

Spacing for 12 mm bar = 91.52 mm

Figure 4-5: Elevation and Plan showing reinforcement design


Chapter — 4


4.2.4 Comparison



Percent Dif-ference

Effective Depth 294 mm 294 mm NoneGoverning Moment 169.2187 KN-

m169.2187 KN-m None

Area of Steel 2837.87 mm2 2836.34 mm2 0.05Shear Stress (One-Way)

0.3486N/mm2

0.3486 N/mm2 None


0.96 N/mm2 0.96 N/mm2 None


4.3 IS General Isolated Foundation 34.3.1 Reference

4.3.2 ProblemDesign an isolated footing with the given data: Load Fy = 2000 KN, fc = 25 MPa, fy = 415MPa, Column Dimension = 300 mm x 300 mm, and Bearing Capacity of Soil = 100KN/m2. Coefficient of friction =0.5, FOS against sliding =1.5, FOS against overturning =1.5







Weight of footing = 4.95x4.95x0.7 x 25 KN = 428.79 KN


Maximum pressure = 2428.79/(4.95x4.95) = 99.124 KN/ m2 <100 KN/m2

Hence safe



Along X Direction

Sliding force =0


Chapter — 4


max Resisting force = µ x Total Service load on foundation =0.5 x 2428.79 =1214.395 KN

Hence OK



Hence OK

Along Z Direction

Sliding force =0

max Resisting force = µ x Total Service load on foundation =0.5 x 2428.79 =1214.395 KN

Hence OK



Hence OK


About X Axis


Mux = 122.436 x 4.95 x 2.325 x 2.325 / 2 = 1638.06 KN-m


deff = (700-50-0.5 x 16) mm = 642 mm

K=700/(1100+0.87x fy )= 0.479107

Ru= 0.36 .fck. Kumax .(1-0.42Kumax) = 3.4442 N/mm2

Resisting Moment =Ru. B deff2 = 7026 KNm

Hence OK

About Z Axis


Mux = 122.436 x 4.95 x 2.325 x 2.325 / 2 = 1638.06 KN-m


deff = (700-50-0.5 x 16) mm = 642 mm

K=700/(1100+0.87x fy )= 0.479107






Hence OK


Along X Direction

From IS -456-2000 Annex G, G-1, b:

Mu =0.87.fy.Ast.d.(1-Ast.fy/b.d.fck)

So solving equation for Ast,

Astx = 7358 m2m



Along Z Direction




Astx = 7358 m2m




Along X Direction


Vumax = 122.436 x 4.95 x = 1020 KN


Now allowable stress= 0.348 N/mm2

V < τc

Hence Safe


Chapter — 4


Along Z Direction


Vumax = 122.436 x 4.95 x = 1020 KN



V < τc

Hence Safe

Punching Shear



Vmax = 2892 KN


τv = Vmax/(Pm · d) = 1.19 N/mm2

ß=L/B =4.95/4.95 =1

k=0.5 +ß=1.5 , k<=1

Hence, k=1

Now allowable stress

τc =k.0.25.√fck = 1.25 N/mm2





4.3.4 Comparison



Percent Dif-ference


99.124 KN/m2 None

Governing Moment 1638.06KN-m

1638.038 KN-m

None

Shear Force(One-Way) 1020 KN 1019 KN NegligibleShear Force(Two-Way) 2892 KN 2891 KN NegligibleResisting force for sliding(X)

1214.395KN

1214.397 KN None


6011.25KNm

6011.155 KNm None

Resisting force for sliding(Z)

1214.395KN

1214.397 KN None


6011.25KNm

6011.155 KNm None



4.4.2 ProblemDesign an isolated footing with the given data: Load Fy = 1000 KN, fc = 25 MPa, fy = 415MPa, Column Dimension = 300 mm x 300 mm, and Bearing Capacity of Soil = 110KN/m2. Coefficient of friction =0.5, FOS against sliding =1.5, FOS against overturning=1.5 Surcharge=20 KN/m2


Chapter — 4






Surcharge= 20 KN/m2

Surcharge force=20x3.65x3.65= 266.4 KN

Therefore, total load on the footing = (1000+266.4+199.8)= 1466.2 KN

Maximum pressure = 1466.2/(3.65x3.65)=

110 KN/m2 (Hence OK)


Along X Direction

Sliding force =0

Max Resisting force = µ x Total Service load on foundation

Total Service load on foundation =1466.2 KN




Max Resisting force =0.5 x 1466.2 = 733.1 KN

Hence OK



Hence OK

Along Z Direction

Sliding force =0

Max Resisting force = µ x Total Service load on foundation

Total Service load on foundation =1466.2 KN

Max Resisting force =0.5 x 1466.2 = 733.1 KN

Hence OK



Hence OK

4.4.4 Comparison



Percent Dif-ference

Bearing Pressure 110 KN/m2 109.9 KN/m2 NegligibleResisting Moment forOverturning (x)

2675.815KNm





2675.815KNm


Resisting force for sliding(x)




4.5.2 ProblemDesign an isolated footing with the given data: Load Fy = 1200 KN, fc = 25 MPa, fy = 415MPa, Column Dimension = 300 mm x 300 mm, and Bearing Capacity of Soil = 110


Chapter — 4


KN/m2. Coefficient of friction =0.5, FOS against sliding =1.5, FOS against overturning =1.5.Height of soil above footing=450 mm, dry condition.





Weight of above soil = 3.7x3.7x0.45 x 18 KN = 110.889 KN

Therefore, total load on the footing = (1200+188.237+110.889) KN = 1499.126KN

Maximum pressure = 1499.126/(3.7x3.7) = 109.5 KN/m2 <110 KN/m2

Hence safe






Along X Direction

Sliding force =0

max Resisting force = µ x Total Service load on foundation = 0.5 x 1499.126= 749.6 KN

Hence OK



Hence OK

Along Z Direction

Sliding force =0


Hence OK



Hence OK

Check For Trial Depth against moment about X Axis


Mux = 131.4828 x 3.7 x 1.7 x 1.7/2 = 702.97 KN-m


deff = (550-50-0.5 x 16) mm = 492 mm

K=700/(1100+0.87x fy )= 0.479107


Resisting Moment =Ru. B deff2 = 3084.8 KNm

Hence OK

Check For Trial Depth against moment about Z Axis


Mux = 131.4828 x 3.7 x 1.7 x 1.7/2 = 702.97 KN-m



Chapter — 4


deff = (550-50-0.5 x 16) mm = 492 mm

K=700/(1100+0.87x fy )= 0.479107



Hence OK


Along X Direction




Astx = 4120 m2m



Along Z Direction




Astx = 4120 m2m



Check for One-Way Shear ( along X dir)


Vumax = 131.4828 x 3.7 x = 587.7 KN

Developed shear stress τv = 587.7x1000/(3700x492)

= 0.322 N/mm2


τv < τc

Hence Safe




Check for One-Way Shear ( along Z dir)


Vumax = 131.4828 x 3.7 x = 587.7 KN

Developed shear stress τv = 587.7x1000/(3700x492) = 0.322 N/mm2


τv < τc

Hence Safe

Punching Shear



Vmax = 1718 KN


τv = Vmax/(Pm · d) = 1.1 N/mm2

ß=L/B =3.7/3.7 =1

k=0.5 +ß=1.5 , k<=1

Hence, k=1

Now allowable stress= τc =k.0.25.√fck = 1.25 N/mm2



Chapter — 4


4.5.4 Comparison



Percent Dif-ference


109.45 KN/m2 Negligible

Governing Moment (x&Z) 702.97 KN-m

700.1 KN-m Negligible

Shear Force(One-Way) (x&Z)

587.7 KN 586 KN Negligible

Shear Force(Two-Way) 1718 KN 1716.5 KN NegligibleResisting force for sliding(X)




Resisting force for sliding(Z)





4.6 IS General Isolated Foundations 64.6.1 Reference

4.6.2 ProblemDesign an isolated footing with the given data: Load Fy = 2000 KN, fc = 25 MPa, fy = 415MPa, Column Dimension = 300 mm x 300 mm, and Bearing Capacity of Soil = 100KN/m2. Coefficient of friction =0.5, FOS against sliding =1.5, FOS against overturning =1.5.Height of soil above footing=500 mm, depth of GWT =200 mm


4.6 IS General Isolated Foundations 6






Weight of above soil = 4.95x4.95x0.5 x18 KN = 220.522 KN

Reduction of Weight due to buoyancy = 4.95x4.95x0.5 x18 KN = 240.369 KN

Therefore, total load on the footing = (2000+428.79+220.522-240.369) KN =2408.943 KN

Maximum pressure = 2408.943/(4.95x4.95) = 99.12 KN/m2 <100 KN/m2

Hence safe



Chapter — 4



Along X Direction

Sliding force =0


Hence OK



Hence OK

Along Z Direction

Sliding force =0

max Resisting force = µ x Total Service load on foundation=0.5 x 2408.943 =1204.5 KN

Hence OK



Hence OK

Check For Trial Depth against moment

About X Axis


Mux = 122.436 x 4.95 x 2.325 x 2.325 / 2 = 1638.06 KN-m


deff = (700-50-0.5 x 16) mm = 642 mm

K=700/(1100+0.87x fy )= 0.479107



Hence OK

About Z Axis





Mux = 122.436 x 4.95 x 2.325 x 2.325 / 2 = 1638.06 KN-m


deff = (700-50-0.5 x 16) mm = 642 mm

K = 700/(1100+0.87x fy )= 0.479107

Ru = 0.36 .fck. Kumax .(1-0.42Kumax) = 3.4442 N/mm2


Hence OK


Along X Direction




Astx = 7358 m2m



Along Z Direction




Astx = 7358 m2m




Chapter — 4



Along X Direction


Vumax = 122.436 x 4.95 x = 1020 KN



V < τc

Hence Safe

Along Z Direction


Vumax = 122.436 x 4.95 x = 1020 KN



V < τc

Hence Safe

Punching Shear



Vmax = 2892 KN


τv = Vmax/(Pm · d) = 1.19 N/mm2

ß=L/B =4.95/4.95 =1

k=0.5 +ß=1.5 , k<=1

Hence, k=1






4.6.4 Comparison



Percent Dif-ference


98.2832KN/m2

None

Governing Moment (x&Z) 1638.06KN-m

1634 KN-m Negligible

Shear Force(One-Way) (x&Z)

1020 KN 1017 KN Negligible

Shear Force(Two-Way) 2892 KN 2889.7 KN NegligibleResisting force for sliding(x)

1204 KN 1204 KN None


5962 KNm 5960 KNm Negligible


1204 KN 1204 KN None

Resisting Moment forOverturning (x)




4.7.2 ProblemDesign an isolated footing with the given data: Load Fy = 1500 KN, Fz=120 KN fc = 25MPa, fy = 415 MPa, Column Dimension = 300 mm x 300 mm, and Bearing Capacity ofSoil = 120 KN/m2. Coefficient of friction =0.5, FOS against sliding =1.5, FOS againstoverturning =1.5


Chapter — 4






Therefore, total axial load on the footing = (1500+234.04) KN = 1734.04 KN

Maximum axial pressure =P/A= 1734.04 /(3.95x3.95) = 111.138 KN/ m2

Moment due to lateral load= 120x0.6 72 KNm

Total moment (Mx=72 KNm

Zx=3.95x3.952/6 = 10.27 mm3

M/Z=7.01 KN/m2

So,

σ 1=P/A –M/Z=104.129 KN/m2

σ 2=P/A –M/Z=104.129 KN/m2

σ 3=P/A +M/Z=118.148 KN/m2

σ 4=P/A +M/Z=118.148 KN/m2

118.148 KN/m2 <120 KN/m2 (Hence safe)





Along X Direction

Sliding force =0

max Resisting force = µ x Total Service load on foundation = 0.5 x 1734.04 =867.02 KN

Hence OK



FOS = 47.6 > 1.5

Hence OK

Along Z Direction

Sliding force =120

max Resisting force = µ x Total Service load on foundation = 0.5 x 1734.04 =867.02 KN

FOS = 7.22 > 1.5

Hence OK



Hence OK


About X Axis

Critical section for moment is at the face of column; wrt z axis:

Average Base Pressure along one edge = (113.69 + 154.73)x0.5 = 144.21KN/m2 (left end)

Average Base Pressure along other edge = (113.69 + 154.73)x0.5 = 144.21KN/m2 (right end)

Approximate Base Pressure at the left critical section = 144.21 + (144.21-144.21) x 2125/3950 = 144.21 KN/m2

Approximate Base Pressure at the right critical section = 144.21 + (144.21-144.21) x 1825/3950 = 144.21 KN/m2

Hence, the moment at the critical section, Mu = F x LA

Where:


Chapter — 4


F = (144.21 + 144.21) x 0.5 x 1.825 x 3.95 = 1039.57 KN

LA = (144.21 + 2x 144.21) x 1.825/ [3x (144.21 + 144.21)] = 0.913 m

Mu (right) = 949.13 kNm

So, max moment wrt Z axis Mu (z) = 950 KNm


deff = (600-50-0.5 x 16) mm = 542 mm

K = 700/(1100+0.87x fy )= 0.479107



Hence OK

About Z Axis

Critical section for moment is at the face of column; wrt x axis:

Average Base Pressure along one edge = (133.69 + 133.69)x0.5 = 133.69KN/m2 (left end)

Average Base Pressure along other edge = (154.73 + 154.73)x0.5 = 154.73KN/m2 (right end)

Approximate Base Pressure at the left critical section = 154.73 + (133.69-154.731) x 2125/3950 = 143.42 KN/m2




Approximate Base Pressure at the right critical section = 154.73 + (133.69-154.73) x 1825/3950 = 145.01 KN/m2

Hence, the moment at the critical section, Mu = F x LA

Where:

F = (133.69 + 143.42) x 0.5 x 1.825 x 3.95 = 1080.38 KN

LA = (143.42 + 2x 133.69) x 1.825/ [3x (144.21 + 144.21)] = 0.923 m

Mu (right) = 997.19 kNm

So, max moment wrt Z axis Mu (z) = 998 KNm


deff = (600-50-0.5 x 16) mm = 542 mm

K = 700/(1100+0.87x fy )= 0.479107



Hence OK


Along X Direction





Chapter — 4


Astx = 5056 m2m



Along Z Direction




Astx = 5323 m2m




Along X Direction


Average Base Pressure along one edge = (133.69 + 154.73)x0.5 = 144.21KN/m2

Average Base Pressure along other edge = (133.69 + 154.73)x0.5 = 144.21KN/m2

Approximate Base Pressure at the left critical section = 144.21 + (144.21 -144.21) x 2667/3950 = 144.21 KN/m2

Approximate Base Pressure at the right critical section = 144.21 + (144.21 -144.21) x 2667/3950 = 144.21 KN/m2

Hence, the SF at critical section

F = (144.21 + 144.21) x0.5 x 1.283 x 3.95 = 730.84 KN

So max SF along X axis Fux = 731 KN

Developed shear stress τv = 731 x 1000 / (3950 x 542) = 0.341 N/mm2





τv < τc, Hence Safe

Along Z Direction


Average Base Pressure along one edge = (133.69 + 133.69)x0.5 = 133.69KN/m2

Average Base Pressure along other edge = (154.73 + 154.73)x0.5 = 154.73KN/m2

Approximate Base Pressure at the left critical section = 154.73 + (133.69 -154.73) x 2667/3950 = 144.21 KN/m2

Approximate Base Pressure at the right critical section = 154.73 + (133.69 -154.73) x 1283/3950 = 144.21 KN/m2

Hence, the SF at critical section (left)

F = (133.69 + 140.53) x0.5 x 1.283 x 3.95 = 694.84 KN

Hence, the SF at critical section (right)

F = (15473 + 147.90) x0.5 x 1.283 x 3.95 = 766.83 KN

So max SF along X axis Fux = 731 KN


Chapter — 4


So max SF=767 KN

Developed shear stress τv = 767 x 1000 / (3950 x 542) = 0.358 N/mm2


τv < τc

Hence Safe

Punching Shear



Vmax = 2148 KN

Critical perimeter Pm = 2 X ( b + h + 6 x d) = 3368 mm

τv = Vmax/(Pm · d) = 1.177 N/mm2

ß=L/B =3.95/3.95 =1

k=0.5 +ß=1.5 , k<=1

Hence, k=1


τv < τc

Hence safe




Figure 4-11: Final Plan Dimensions

4.7.4 Comparison



Percent Dif-ference

Max BearingPressure

104.129KN/m2

104.129 KN/m2 None

MinBearingPressure

118.148 KN 118.148 KN None

GoverningMoment

999 KN-m

950 KN-m

994 KN-m

946 KN-m

Negligible


767 KN

731 KN

764 KN

728 KN

Negligible


2148 KN 2146 KN Negligible


4.8 IS Toolkit Combined 14.8.1 Reference


Chapter — 4

4.8 IS Toolkit Combined 1



4.8.3 SolutionApproximate area of footing required = 2x400/120 m2 = 6.67 m2


(left overhang = right overhang = 1.375 m)

Weight of footing = 5.75 m x 1.35 m x 0.600 x 25 KN = 116.375 KN

Weight of pedestal=2x 0.3 x 0.3 x 0.5 x 25 = 2.25 KN

Therefore, total load on the footing = (2x400+116.375+2.25) KN = 918.625 KN

Maximum pressure = 918.625 /(5.75x1.35) = 118.34 KN/ m2 < 120 KN/m2

Hence safe

Ultimate pressure = 800 x 1.5/ (5.75 x 1.35) KN/m2 = 154.589 KN/m2





With Respect to Z Direction



Hence OK

With Respect to X Direction



Hence OK


About Z Axis (Sagging)

Bending moment at critical section, Mux = 197.2 KNm


deff = (600-50-0.5 x 12) mm = 544 mm

K = 700/(1100+0.87x fy )= 0.479107



Hence OK

About Z Axis (Hogging)



deff = (600-50-0.5 x 12) mm = 544 mm

K = 700/(1100+0.87x fy )= 0.479107



Hence OK


Chapter — 4


With Respect to X Axis

Cantilever length = (1.35-0.3)/2 = 0.525 m

Bending moment at critical section, Mux = 154.589 x5.75 x0.5252/2 =122.5KNm


deff = (600-50-0.5 x 12) mm = 544 mm

K = 700/(1100+0.87x fy )= 0.479107


Resisting Moment =Ru. B deff2 =5860 KNm

Hence OK






Astx = 1029 m2m







Astx = 192 m2m










Astz = 627 m2m



Figure 4-13: Dimension, Moment, and Shear and diagrams



Vumax = 168.2 KN


Chapter — 4


Developed shear stress V = 168.2 x 1000 / (1350 x 544)= 0.229 N/mm2


V < τc, Hence Safe

Punching Shear

For Column 1



τv = Vmax/(Pm · d) = 489.89 x 1000/(300 x 2 + 300 x 2 + 544 x 4) x 544 =0.2668 N/mm2

ß=L/B = 5.75/1.35 = 4.26

k=0.5 +ß=5.26 , k<=1

Hence, k=1



For Column 2



τv = Vmax/(Pm · d) = 489.89 x 1000 / (300 x 2 + 300 x 2 +544 x 4) x 544= 0.2668 N/mm2

ß=L/B =5.75/1.35 =4.26

k=0.5 +ß=5.26 , k<=1

Hence, k=1






4.8.4 Comparison



Percent Dif-ference


118.35 KN/m2 None


37.49 KN-m

122.5 KN-m

191.4 KN-m

37.4 KN-m

122.5 KN-m

Negligible


489.89 KN

489.88 KN

489.88 KN

Negligible






4.9 IS Toolkit Combined Foundation 24.9.1 Reference


Ht of soil =500 mm. Dry condition


Chapter — 4

4.9 IS Toolkit Combined Foundation 2




( left overhang=right overhang=1m)



Weight of soil above footing = 5 m x 1.65 m x 0.500 x18 KN = 74.25 KN

Therefore, total load on the footing = (2x350+123.75+2.25+74.25) KN =900.25 KN

Maximum pressure = 900.25 /(5 x1.65) = 109.12 KN/ m2

109.12 KN/ m2 <110 KN/m2

Hence safe

Ultimate pressure = 700 x 1.5 / (5.0 x 1.65) = 127.273 KN/m2





About Z Direction


max resisting Moment = 0.5 x 5x 900.25 = 2250 KNm

Hence OK

About X Direction



Hence OK


Moment About Z Axis (Sagging)



deff = (600-50-0.5 x 12) mm = 544 mm

K = 700/(1100+0.87x fy )= 0.479107



Hence OK

Moment About Z Axis (Hogging)



deff = (600-50-0.5 x 12) mm = 544 mm

K = 700/(1100+0.87x fy )= 0.479107



Hence OK

Moment About X Axis



Chapter — 4



Mux = 127.273 x5x0.6752/2 = 144.97 KNm


deff = (600-50-0.5 x 12) mm = 544 mm

K = 700/(1100+0.87x fy )= 0.479107


Resisting Moment =Ru. B deff2 =5096.3 KNm

Hence OK






Astx = 541 m2m







Astx = 677 m2m







Astz = 742 m2m








Vumax = 169.3 KN

Developed shear stress τv = 169.3 x 1000 / (1650 x 544) = 0.1886 N/mm2


τv < τc, Hence Safe

Punching Shear

For Column One



τv = Vmax/(Pm · d) = 434.3 x 1000 / (300 x 2 + 300 x 2 + 544 x 4) x 544= 0.2356 N/mm2

ß=L/B =5.75/1.35 =4.26

k=0.5 +ß=5.26 , k<=1

Hence, k=1



For Column Two




ß=L/B =5.75/1.35 =4.26

k=0.5 +ß=5.26 , k<=1

Hence, k=1




Chapter — 4


Figure 4-15: Shear Force and Bending Moment diagrams




4.9.4 Comparison



Percent Dif-ference


108.92KN/m2

Negligible


131.2 KN-m

144.97 KN-m

101.8 KN-m

131.2 KN-m

144.97 KN-m

Negligible


434.3 KN

434.3 KN

434.3 KN

None




742.7 KNm 741.37KNm None



4.10.2 ProblemDesign a combined footing with the given data: Load Fy = 350 KN each column., fc = 25MPa, fy = 415 MPa, Column Dimension = 300 mm x 300 mm, Pedestal height-500 mm.and C/C column distance=3000 mm . Bearing Capacity of Soil = 130 KN/m2. Coefficientof friction =0.5, FOS against sliding =1.5, FOS against overturning =1.5



Chapter — 4



4.10.3 SolutionApproximate area of footing required = 2 x 500/130 m2 = 7.69 m2


(left overhang = right overhang = 1.2m)

Weight of footing = 5.4 m x 1.65 m x 0.600 x25 KN = 133.65 KN


Weight of soil above footing = 5.4 m x 1.65 m x 0.500 x18 KN = 80.19 KN

Reduction of Weight due to buoyancy = 5.4 m x 1.65 m x (0.5+0.6-0.2) x9.81KN = 78.67 KN

Therefore, total load on the footing = (2x500+133.65+2.25+80.19-78.67) KN =1137.42 KN

Maximum pressure= 1137.42 /(5.4 x1.65) = 127.66 KN/ m2

127.66 KN/ m2 <130 KN/m2

Hence safe

Ultimate pressure = 1000 x 1.5 / (5.4 x 1.65) = 168.35 KN/m2





wrt Z Direction


max resisting Moment = 0.5 x 5x 1137.42 = 3071 KNm

Hence OK

Wrt X Direction



Hence OK




Mux = 200 KNm


deff = (600-50-0.5 x 12) mm = 544 mm

K = 700/(1100+0.87x fy )= 0.479107



Hence OK



Mux = 112.4 KNm


deff = (600-50-0.5 x 12) mm = 544 mm

K = 700/(1100+0.87x fy )= 0.479107



Hence OK


Chapter — 4


About X Axis

Cantilever length = (1.65-0.3)/2 = 0.675 m

Bending moment at critical section, Mux = 168.35 x5.4x0.6752/2 = 207 KNm


deff = (600-50-0.5 x 12) mm = 544 mm

K = 700/(1100+0.87x fy )= 0.479107



Hence OK






Astx = 1040 mm2







Astx = 579 mm2







Astz = 1062 mm2









Vumax = 223.95 KN

Developed shear stress V = 223.95 x 1000 / (1650 x 544) = 0.249 N/mm2


V < τc, Hence Safe


Chapter — 4


Punching Shear

For Column 1




ß=L/B =5.4/1.65 =3.27

k=0.5 +ß=4.27 , k<=1

Hence, k=1



For Column 2




ß=L/B =5.4/1.65 =3.27

k=0.5 +ß=4.27 , k<=1

Hence, k=1






4.10.4 Comparison



Percent Dif-ference


127.66KN/m2

Negligible


112.4 KN-m

207 KN-m

199.9 KN-m

112.4 KN-m

207 KN-m

Negligible

Shear Force(One-Way) 223.95 KN 223.89 KN NegligibleShear Force(Two-Way) 630.07 KN

630.07 KN

630.08 KN

630.08 KN

None




938.4 KNm 937 KNm Negligible



4.11.2 ProblemDesign a combined footing with the given data: Load Fy = 350 KN & Fx=-30 KN eachcolumn., fc = 25 MPa, fy = 415 MPa, Column Dimension = 300 mm x 300 mm, Pedestalheight-500 mm. and C/C column distance=4000 mm . Bearing Capacity of Soil = 90KN/m2. Coefficient of friction =0.5, FOS against sliding =1.5, FOS against overturning=1.5. No soil above footing. Dry condition


Chapter — 4





(left overhang = right overhang = 1 m)


Weight of pedestal = 2 x 0.3 x 0.3 x 0.5 x 25 = 2.25 KN

Therefore, total load on the footing = (2x350+157.5 +2.25) KN = 859.75 KN

Maximum pressure for axial load=P/A = 859.75 /(6x1.75) = 81.88 KN/ m2

Moment on each col = 30 x (0.5 x 0.6) = 33 KNm

So total moment = 33 + 33 = 66 KNm

Z=1.75x62/6 = 10.5 m3

M/Z = 66/10.5 = 6.28 KN/m2

Streaa at left end = 81.88+6.28= 88.16 KN/m2

Streaa at right end = 81.88-6.28= 75.6 KN/m2

88.16 KN/ m2 <90 KN/m2

Hence safe





wrt Z Direction


max resisting Moment = 0.5 x 6 x 859.75 = 2579.25 KNm

Hence OK

Wrt X Direction



Hence OK




Mux = 95 KNm


deff = (600-50-0.5 x 12) mm = 544 mm

K = 700/(1100+0.87x fy )= 0.479107



Hence OK



Mux = 264.25 KNm


deff = (600-50-0.5 x 12) mm = 544 mm

K = 700/(1100+0.87x fy )= 0.479107



Hence OK


Chapter — 4


About X Axis


Bending moment at critical section, Mux = (109.43+90.57) x6x0.7252/2 = 157KNm


deff = (600-50-0.5 x 12) mm = 544 mm

K = 700/(1100+0.87x fy )= 0.479107



Hence OK






Astx = 490 mm2

Minimum area of steel Astmin = 0.0012 x B x D = 1260mm2 ( as fy>250)






Astx = 1380 mm2










Astz = 807 mm2




Percentage of steel pt = 0.133

Vumax = 248.613 KN

Developed shear stress V = 0.261 N/mm2


V < τc, Hence Safe

Punching Shear

For Column One



τv = Vmax/(Pm · d) = 449.289 x 1000 / (300 x 2 + 300 x 2 + 544 x 4) x544 = 0.245 N/mm2

ß=L/B =6/1.75 =3.43

k=0.5 +ß=4.43 , k<=1

Hence, k=1



For Column Two



τv = Vmax/(Pm · d) = 458.245 x 1000 / (300 x 2 + 300 x 2 + 544 x 4) x544 = 0.2496 N/mm2

ß=L/B =6/1.75 =3.43

k=0.5 +ß=4.43 , k<=1

Hence, k=1




Chapter — 4


4.11.4 Comparison


STAADFoundation

Result

Percent Dif-ference


75.59 KN/m2

88.16 KN/m2

75.6 KN/m2

Negligible

Governing Moment 129.7 KN-m

264.25 KN-m

157 KN-m

95 KN-m

264.25 KN-m

167 KN-m

8.5%

None

6%Shear Force(One-Way) 248.61 KN 248.61 KN NoneShear Force(Two-Way) 449.289 KN

458.245 KN449.29 KN

458.24 KN

None


2579.25 KNm 2579.25 KNm None


752.28 KNm 752.28 KNm None


4.12 IS Pilecap 14.12.1 Reference

4.12.2 ProblemDesign pilecap foundation with the given data: Load Fy = 800 KN, fc = 25 MPa, fy = 415MPa, Column Dimension = 250 mm x 250 mm. Pedestal ht= 500 mm

Pile Data- Dia of pile= 400 mm

Vertical capacity = 250 KN




Chapter — 4

4.12 IS Pilecap 1


4.12.3 Solutiondepth of pilecap is equal to 1.5x the pile diameter, D = 600 mm

Take D= 920 mm

c/c pile distance is equal to 3x the pile diameter =1200 mm. Edge distance = 350 mm

Assuming four pile combination, Coordinates of piles (considering pedestal at 0,0,0)

Pile NoX Coor-dinate(mm

Z Coor-dinate(mm)

1 -600 -6002 -600 6003 600 6004 600 -600

Table 4-12: Pile Locations in Plan

pilecap dimension is 1900 mm x1900 mm x 650 mm

Weight of footing = 1.9 x1.9 x 0.92 x 25 KN = 83.03 KN


4.12 IS Pilecap 1



Therefore, total load on the pilecap = (800 + 83.03 + 0.78) KN = 883.81 KN

So Pile reaction = 883.81 /4 = 220.95 KN < 250 KN

Hence OK

As there is no lateral load , moment or uplift force, so each pile is safe in lateral & upliftcapacity.

Factored Design



So, Load on pilecap = 1.5(800) + 1(883.81) + 1(0.78) = 1,283.81 KN

Load on each pile = 1,283.81/4 = 320.95 KN

Calculation of Moment about Z Axis

For moment wrt X1X1

Contribution from pile 1 = from pile 2 = 320.95 x 0.475 = 152.45 KNm


For moment wrt X2X2

Contribution from pile 3 = from pile 4 = 320.95 x 0.475 = 152.45 KNm


So Max value of Mz = 304.9 KNm

Calculation of Moment about X Axis

For moment wrt Z1Z1


Chapter — 4

4.12 IS Pilecap 1

Contribution from pile 1 = from pile 4 = 320.95 6 x 0.475=152.45 KNm

So Total Mx Z1Z1 = 304.9 KNm

For moment wrt Z2Z2

Contribution from pile 2 = from pile 3 = 320.95 6 x 0.475=152.45 KNm


So Max value of MX = 304.9 KNm


About Z Axis


Muz = 304.9 KNm

Assuming 50 mm clear cover, 50 mm pile in pilecap & and 12 mm bar, effective depth

deff = 814 mm

K = 700/(1,100 + 0.87· fy ) = 0.479107

Ru = 0.36·fck·Kumax ·(1-0.42·Kumax) = 3.4442 N/mm2

B = 1,900 mm, deff = 814 mm

Resisting Moment =Ru·B·deff2 = 4,336 KNm

Hence OK

About X Axis


Mux = 304.9 KN-m


deff = 814 mm

K=700/(1,100 + 0.87· fy )= 0.479107

Ru = 0.36·fck·Kumax ·(1-0.42·Kumax) = 3.4442 N/mm2

B =1,900 mm, deff = 814 mm

Resisting Moment =Ru·B·deff2 = 4,336 KNm

Hence OK


Along X Direction



4.12 IS Pilecap 1


M f A d= 0.87u y st

A f

bd f

1 − st y

ck


Astx = 1,050 mm2

Minimum area of steel Astmin = 0.0012 x B x D = 2,098 mm2 ( as fy > 250)


Along Z Direction


M f A d= 0.87u y st

A f

bd f

1 − st y

ck


AstZ = 1,050 mm2



Calculation of Shear

Parallel to X Axis

For shear wrt X1X1

Contribution from pile 1 = pile 2 = 320.95 x 0.67 = 215.03 KN

So Total V X1X1 = 430 KN

For shear wrt X2X2


Chapter — 4

4.12 IS Pilecap 1


So Total V X2X2 = 430 KN

So Maximum V parallel to X direction = 430 KN

Parallel to Z Axis

For shear wrt Z1Z1


So Total V Z1Z1 = 430 KN

For shear wrt Z2Z2



So Max V parallel to Z direction = 430 KN


Along X Direction


Vumax = 430 KN

Developed shear stress V = 430 x 1,000 / 1,900 x 814 = 0.278 N/mm2


V < τc, Hence Safe

Along Z Direction


Vumax = 430 KN

Developed shear stress V = 430 x 1,000 / 1,900 x 814 = 0.278 N/mm2


V < τc, Hence Safe


4.12 IS Pilecap 1


Punching Shear


Contribution from pile 1 = from pile 2 = from pile 3 = from pile 4 = 276.4 KN

So total punching shear Vmax= 1,105.7 KN

Pm = 4 x (250 + 814/2 + 814/2) = 4,256 mm

τv = Vmax/(Pm · d) = 0.319 N/mm2

ß = L/B =1,900/1,900 = 1

k = 0.5 + ß = 1.5 , k ≤ 1

Hence, k = 1

Now allowable stress= τc = k(0.25)√fck = 1.25 N/mm2



Chapter — 4

4.12 IS Pilecap 1

4.12.4 Comparison



Percent Dif-ference

Pile Reaction, Service(KN)

220.95 220.5 None

Pile Reaction, Ultimate(KN)

320.95 320.5 None

Governing Moment, Mx(KNm)

304.9 304.5 None

Governing Moment, Mz(KNm)

304.9 304.5 None

Shear Force, One-Way,X (KN)

430 429.5 None

Shear Force, One-Way,Z (KN)

430 429.5 None

Shear Force, Two-Way(KN)

1,105.7 1,104.3 None


4.13 IS Pilecap 24.13.1 Reference

4.13.2 ProblemDesign pilecap foundation with the given data: Load Fy = 1,100 KN, Mx= 50 KNm, Fz= 50KN,fc = 25 MPa, fy = 415 MPa, Column Dimension = 250 mm x 250 mm. Pedestal ht= 500mm

Diameter of pile= 400 mm.

Vertical capacity =300 KN,



Pedestal dimensions: 250 mm x 250 mm


4.13 IS Pilecap 2


Figure 4-21: Plan, Elevation, and Pedestal dimensions

4.13.3 Solutiondepth of pilecap is equal to 1.5x the pile diameter, D = 600 mm

Take D = 1,255 mm

c/c pile distance = 3x pile diameter =1,200 mm. Edge distance =350 mm

Assuming five pile combination,

Coordinates of piles considering pedestal at 0, 0, 0


Chapter — 4

4.13 IS Pilecap 2

PileNo

X Coordinate(mm)

Z Coordinate(mm)

1 -849 -8492 -849 8493 0 04 849 -8495 849 -849

Table 4-14: Pile Coordinates in Plan

pilecap dimension is 2,400 mm x 2,400 mm x 1,255 mm



Therefore, total load on the pilecap = (1,100 + 180.72 + 0.78) KN = 1,281.5 KN

So Pile reaction from axial load= 1,281.5 /5= 256.3 KN

Moment from lateral load = (1.255 + 0.5) x 50= 87.75 KNm

Moment Mx ( from input) = 50 KNm

So Total moment = 137.75 KNm

Using Rivet theory:

Reaction from moment= ±137.75(0.849)/[4(0.8492)] = ±40.56 KNm

So

Reaction at Pile 2= reaction at pile 5 = 256.3 + 40.56 = 296.86 KN

Reaction at Pile 1= reaction at pile 4 = 256.3 - 40.56 = 215.74 KN

Reaction at Pile 3= 256.3 KN

So Critical vertical reaction= 297 KN< 300 KN

Lateral reaction = 50/5 = 10 KN < 50 KN, Hence OK

As there is no net uplift load, so each pile is safe in uplift capacity.

Factored Design



So, Axial Load on pilecap = 1.5(1,100) + 1(180.72) + 1(0.78) = 1,831.5 KN

Moment on pilecap = 1.5(137.75) = 206.62 KNm

Load on each pile from axial reaction = 1,831.5/5 = 366.3 KN

Reaction from moment= ±206.62(0.849)/[4(0.8492)] = ±60.84 KNm

So

Reaction at Pile 2= reaction at pile 5 = 366.3 + 60.84 = 427.14 KN

Reaction at Pile 1= reaction at pile 4 = 366.3 - 60.84 = 305.46 KN


4.13 IS Pilecap 2


Reaction at Pile 3= 366.3 KN

Calculation of Moment

Moment is calculated at face of column

About Z Axis

For moment wrt X1X1

Contribution from pile 1 = 305.46 x 0.724 = 221.15 KNm

Contribution from pile 2 = 427.14 x 0.724 = 309.25KNm

Contribution from pile 3 = 1.7 KNm


For moment wrt X2X2


Contribution from pile 5 = 427.14 x 0.724 = 309.25KNm



So Max value ofMz = 532.1 KNm

About X Axis

For moment wrt Z1Z1


Chapter — 4

4.13 IS Pilecap 2




So Total Mx Z1Z1 = 444 KNm

For moment wrt Z2Z2





So Max value ofMX = 620.2 KNm


Moment About Z Axis


Muz = 532.1 KN-m


deff = 1,149 mm

K = 700/(1,100 + 0.87x fy ) = 0.479107

Ru = 0.36 (fck) Kumax (1-0.42Kumax) = 3.4442 N/mm2

B =2,400 mm, deff = 1,149 mm

Resisting Moment =Ru. B deff2 = 10,913 KNm

Hence OK

Moment About X Axis


Mux = 620.2 KN-m


deff = 1,149 mm

K = 700/(1,100 + 0.87x fy ) = 0.479107

Ru = 0.36 (fck) Kumax (1-0.42Kumax) = 3.4442 N/mm2

B = 2,400 mm, deff = 1,149 mm

Resisting Moment =Ru. B deff2 = 10,913 KNm

Hence OK


4.13 IS Pilecap 2



Along X Direction


M f A d= 0.87u y st

A f

bd f

1 − st y

ck


Astx = 1,510 mm2



Along Z Direction


M f A d= 0.87u y st

A f

bd f

1 − st y

ck


AstZ = 1,293 mm2




According to Amendment 1shear is checked on a perimeter 0.5d =574.5 mm from thecolumn face.


Chapter — 4

4.13 IS Pilecap 2

Parallel to X Axis

For shear wrt X1X1

Contribution from pile 1 = 305.46 x 0.873 = 266.67 KN



For shear wrt X2X2




So Max V parallel to X direction = 639.56 KN

Parallel to Z Axis

For shear wrt Z1Z1



Contribution from pile 3 = 0 KN


For shear wrt Z2Z2





4.13 IS Pilecap 2



So Max V parallel to Z direction = 745.7 KN


Along X Direction


Vumax = 639.56 KN

Developed shear stress V = 639.56 x 103 / (2,400 x 1,149) = 0.232 N/mm2


V < τc, Hence Safe

Along Z Direction


Vumax = 745.75 KN

Developed shear stress V = 745.75 x 103 / (2,400 x 1,149) = 0.27 N/mm2


V < τc, Hence Safe

Punching Shear

Punching shear is checked on a perimeter 0.5d = 574.5 mm from the column face.

Contribution from pile 1 = from pile 4 = 300.6 KN

Contribution from pile 2 = from pile 5 = 420.3 KN


So total punching shear Vmax= 1,441.8 KN

Pm = 4x(250 + 574.5 + 574.5) = 5,596 mm


Chapter — 4

4.13 IS Pilecap 2

τv = Vmax/(Pm · d) = 0.224 N/mm2

ß = L/B = 2,400/2,400 = s1

k = 0.5 +ß = 1.5 , k ≤ 1

Hence, k = 1

Now allowable stress= τc =k(0.25)√fck = 1.25 N/mm2


4.13.4 Comparison



Percent Dif-ference


215.74

296.86

256.3

215.74

296.86

215.62

296.788

256.203

215.617

296.788

Negligible

Pile Reaction, Ulti-mate (KN)

305.46

427.14

366.3

305.46

427.14

305.32

427.081

366.203

305.324

427.081

Negligible

Governing Moment(KNm)

532

620

531

619.5

Negligible

Shear Force, One-Way(KN)

640

746

641

747

Negligible




4.14 IS Mat Combined Foundation 14.14.1 Reference‘Reinforced Concrete Design’ by Pillai & Menon, Page 652, Example 14.7.

4.14.2 ProblemDesign a combined footing for two columns with the given data: C1 (400 mm x 400 mm)with 4-25 Ø bars and C2 (500 mm x 500mm) with 4-28 Ø bars supporting axial loads P1 =900 KN and P2 = 1600 KN respectively (under service dead and live loads). The column C1is an exterior column whose exterior face is flush with the property line. The center-to-centre distance between C1 and C2 is 4.5 meters. The allowable soil pressure at the base of


4.14 IS Mat Combined Foundation 1


the footing, 1.5 m below ground level, is 240 KN/m2. Assume a steel of grade Fe 415 inthe columns as well as the footing, and a concrete grade of M 20 in the footing.

Figure 4-22: Footing Plan

Figure 4-23: >Loads on Footing

4.14.3 SolutionDimension of Mat (Based on the bearing Capacity given):

Length = 6.16 m

Width = 2 m

Depth = 0.95 m

Calculation for base-pressure

Self-weight of mat = 6.16 x 2 x 0.95 x 25 KN = 292.6 KN

Total load on the mat = (1600+900+200.2) KN = 2792.6 KN

Base pressure = 279.6 / (6.16 x 2) KN/m2 = 226.67 KN/m2 < 240 KN/m2

(Hence Safe)

Ultimate load for C1 = Pu1 = 1.5 x 900 = 1350 KN

Ultimate load for C2 = Pu2 = 1.5 x 1600 = 2400 KN


Chapter — 4


Then uniformly distributed upward load = (Pu1+Pu2)/6.16 KN/m = 608.8KN/m

Developed shear stress,

v = = 0.533 N/mm2 < c,allowable

Hence safe

Maximum shear for C2 = 2400 KN

Developed shear stress,

v = = 0.508 N/mm2 < c,allowable

Hence safe


Calculation of reinforcement

Maximum negative moment Mu(-) = 1227 KN-m

Maximum negative moment/width = 1227/2 KN-m/m = 613.5 KN-m/m

Area of steel required on top face along length,

Ast = 0.5 x x B x deB = 1000 mm

de = 865 mm

Mu= 613.5 x 106 N-mm

Ast = 2067.97 mm2/m

Ast,min = 0.0012 x B x D = 1140mm2/m




Figure 4-24: Shear Force (kN, top) and Bending Moment (kNm, bottom) diagrams

4.14.4 Comparison



Percent Dif-ference

Max BendingMoment(-)

603.201 KN-m/m

613.5 KN-m/m 1.68

Max BendingMoment(+)

219.687 KN-m/m

223 KN-m/m 1.48

Area of StealRequired

2014.835mm2/m

2067.97 mm2/m 2.56

Base Pressure 227 KN/m2 226.67 KN/m2 Negligible



Chapter — 4


Section 5

United States Code (ACI318 -2005)5.1 US General Isolated Foundation 1

5.1.1 Reference‘Notes on ACI318-02 Building Code Requirements for Structural Concrete with DesignApplications’ by B.A. Fanella and B. G. Robert, Page 22-7, Example 22.1-22.3.

5.1.2 ProblemDesign an isolated footing with the given data: Service dead load = 350 kips, Service liveload = 275 kips, Service surcharge = 100 psf, Average weight of soil above footing = 130 pcf,Permissible soil pressure = 4.5 ksf, Column dimension = 30 x 12 in, Strength of concrete(fc’) = 3,000 psi, and Strength of steel (fy) = 60,000 psi.


Figure 5-1: Elevation and loads

5.1.3 SolutionDetermination of base area of footing:


Total weight of surcharge = (0.130 x 5 + 0.1) = 0.75 ksf

Net permissible soil pressure = 4.5 – 0.75 = 3.75 ksf

Required base area of footing = (320 + 275) / 3.75 = 166.667 ft2

[Clause 15.2.2]

Use a 13 x 13 ft square footing (Af = 169 ft2).


To proportion the footing for strength (depth and required reinforcement) factored loadsare used. [Clause 15.2.1]

Pu = 1.2 x 350 + 1.6 x 275 = 860 kips [Eq. 9-2]

qu = (Pu/A) = (860/169) = 5.089 ksf


Chapter — 5

5.1 US General Isolated Foundation 1

Figure 5-2: Considered sections for two-way (bo) and beam (bw) action

Depth requirement for shear usually controls the footing thickness.

Both wide action and two-way action for strength computation need to be investigated todetermine the controlling shear criteria for depth. [Clause 11.2]

Assume overall footing thickness = 33 in. and average effective thickness d = 28 in. = 2.33ft.

Wide-beam action

Vu = qs x tributary area

Bw = 13 ft = 156 in.

Tributary area = 13(6.0 – 2.33) = 47.71 ft2

Vu = 5.089 x 47.71 = 242.796 kips

φV = φ(2√(f'c)bwd) [Eq 11 - 3]

= 0.75(2√(3000) x 156 x 28)/1000 [Clause 9.3.2.3]

= 359 kips > Vu O.K.

Two-way action

Vu = qs x tributary area

Tributary area = = 152.889 ft2

Vu = 5.089 x 152.889 = 778.052 kips

Section 5 United States Code (ACI 318 -2005)



= Minimum of

[Eq 11-33,11-34 & 11-35 respectively]

bo = 2(30 + 28) + 2(12 + 28) = 196 in.

bo/d = 196/28 = 7

= 40 for interior columns

=

φVc = 0.75 x 3.6 √(3000) x 196 x 28/1000 = 812 kips > Vu = 780 kips

O.K.

Calculation of ReinforcementFigure 5-3: Critical section for moment (long projection)

Critical section for moment is at face of column [Clause 15.4.2]

Mu = 5.089 x 13 x 62/2 = 1190.862 ft-kips

Compute required As assuming tension-controlled section ( φ = 0.9)

[Clause 10.3.4, 9.3.2.1]


Chapter — 5


Required Rn = psi

p(gross area) = (d/h) x 0.0022 = (28/33) x 0.0022 = 0.00186

Check minimum As required for footings of uniform thickness; for grade 60 reinforcement:[Clause 10.5.4]

ρmin = 0.00180 < 0.00186 O.K. [Clause 7.12.2]

Required As = ρbd = 0.0022 x 156 x 28 = 9.61 in.2

Try 13-No. 8bars (As = 10.27 in.2) each way

Check for Development Length

Critical section for development length is same as that for moment (at face of column).[Clause 15.6.3]

Ld = [Eq. 12-1]

Clear cover (bottom and side) = 3.0 in.

Center-to-center bar spacing = in.

[Clause 12.2.4]

C = minimum of

Ktr = 0 (no transverse reinforcement)

= 3.5 > 2.5, use 2.5 [Clause 12.2.3]

α = 1.0 (less than 12 in. of concrete below bars) [Clause 12.2.4]

β = 1.0 (uncoated reinforcement)

αβ= 1.0 < 1.7

γ = 1.0 (larger than No.7 bars)




λ = 1.0 (Normal weight concrete)

Ld = = 32.9 in. >12.0 in. O.K. [Clause12.2.1]

Since Ld = 32.9 in. is less than the available embedment length in the short direction(156/2 - 30/2 - 3 = 60 in.), the No.8 bars can be fully developed.

Use 13 – No.8 each way.

5.1.4 Comparison



Percent Dif-ference

Effective Depth 28 in 28 in NoneGoverning Moment 1190.86 ft-

kips1190.77 ft-kips Negligible

Area of Steal 9.61 in2 9.70 in2 0.94Shear Stress (One-Way)

242.79 kips 242.56 kips Negligible


778.05 kips 778.01 kips Negligible

Table 5-1: US verification example 1 comparison

5.2 US General Isolated Foundation 25.2.1 Reference

5.2.2 ProblemDesign an isolated footing with the given data: Load Fy = 200 Kip, fc = 4 Ksi, fy = 60 Ksi,Column Dimension = 12 inch x 12 inch, and Bearing Capacity of Soil = 2.2 Kip/sqft.Coefficient of friction =0.5, FOS against sliding =1.5, FOS against overturning =1.5


Chapter — 5


Figure 5-4: Elevation and Plan

5.2.3 SolutionApproximate area of footing required = 200/2.2 = 90.9 sqft

Assuming 122 inch x122 inch x18 inch footing dimension,

Weight of footing = 122 x122 x18 x 0.159/(123)= 24.66 Kip

Therefore, total load on the footing = (200+24.66) KN = 224.66 Kip

Maximum pressure = 224.66 /(122 x 122) = 0.0151 Ksi=2.17 Kip/sqft <2.2Kip/sqft (Hence safe)

Ultimate pressure =200x1.5/(122x122)= KN/m2 = 0.020156 Ksi


Along X Direction

Sliding force = 0

max Resisting force = µ x Total Service load on foundation = 0.5 x 224.66 =112.33 Kip

Hence OK





max resisting Moment = 0.5x 122 x 224.66 = 13704.26 in·kip

Hence OK

Along Z Direction

Sliding force = 0


Hence OK



Hence OK

Punching Shear

Punching shear is checked on a perimeter 0.5d = 7.5 inch from the column face.

Effective Depth=D-clear cover-1=18-2-1=15 inch

Punching Shear Vm = 0.018813 x 122 x 122 - 0.0189 x 27 x 27 = 266.24 Kip

Critical perimeter Pm = 2 X ( b + h + 2 x d) = 108 inch

Vm1 = Vmax / (Pm*d)= 164.81 Ksi

ßc=Width of col/depth of col=12/12=1

bc=2.(b+d+2.deff)=108 inch

V1 √fc.bc.d = (2+4/1).√(4000) x 108 x 15 = 614746.78 lb

V2 √fc.bc.d = (40x15/108+2). √(4000) x 108 x 15 =774125.57 lb

V2 = 4.√fc.bc.de=4.√(4000) x 108 x 15 = 409831.18 lb

Vc = min {V1, V2, V3} =409831.185 lb

So, 0.75Vc = 307.374 Kip

Vm< 0.75.Vc , Hence safe


Along Z Direction

Vumax = 0.0189 x 122 x = 92.24 Kip


Chapter — 5


Now allowable shear = Vc1 = = 2x√(4000) x 122 x 15 lb =231.48 Kip

0.75.Vc1=173.61 Kip

V < Vc1, Hence Safe

So Vumax < Vc1 , Hence Safe

Along Z Direction

Vumax = 0.0189 x 122 x = 92.24 Kip

Now allowable shear = Vc1 = = 2x√(4000) x 122 x 15 lb =231.48 Kip

0.75.Vc1=173.61 Kip

V < Vc1, Hence Safe



About X Axis

m = fy/0.85.fc=60/(0.85x4) = 17.6471

ß1 = 0.85 ( as fc=4)

ρbal = 0.85.ß1.fc.87/fy.(87+fy) = 0.028507

ρmax = 0.75. ρbal = 0.02138

ρmin = 0.0018 ( as fy=60)


Mux = 0.0189 x 122 x 55 x 55 / 2 = 3487.55 in·kip

So Resisting Moment =Mnz= Muz/Ø =3875.56 in·kip

Rn = Mnz/b.d2 =0.1412 Ksi

2m.Rn/fy <1 , Hence OK

ρ = 1/m. (1-√1-2m.Rn/fy) = 0.002405

Ρmin < ρ < ρmax Hence OK

About Z Axis

m = fy/0.85.fc=60/(0.85x4) = 17.6471

ß1 = 0.85 ( as fc=4)

ρbal = 0.85.ß1.fc.87/fy.(87+fy) = 0.028507




ρmax = 0.75. ρbal = 0.02138

ρmin = 0.0018 ( as fy=60)


Mux = 0.0189 x 122 x 55 x 55 /2 = 3487.55 in·kip

So Resisting Moment = Mnz = Muz/Ø =3875.56 in·kip



ρ = 1/m. (1-√1-2m.Rn/fy) = 0.002405



Along X Direction

ρx = 0.002405

Therefore, Astx = ρx*b*d = 4.4 in2

Along Z Direction

ρz = 0.002405

Therefore, Astz = ρz*b*d = 4.4 in2

5.2.4 Comparison



Percent Dif-ference

Effective Depth 15 inch 15 inch NoneBearing Pressure 0.0151 Ksi 0.0151 Ksi NoneGoverning Moment 3487 in·kip

3487 in·kip

3471 in·kip

3471 in·kip

Negligible

Shear Force(One-Way) 92.24 Kip

92.24 Kip

91.8 Kip

91.8 Kip

Negligible

Shear Force(Two-Way) 266.24 Kip 266.29 Kip NegligibleResisting force for sliding 112.33 Kip 112.33 Kip NoneResisting Moment forOverturning

13704.26in·kip

13703.75 in·kip Negligible



Chapter — 5



5.3.2 ProblemDesign an isolated footing with the given data: Load Fy = 250 Kip, fc = 4 Ksi, fy = 60 Ksi,Column Dimension = 12 inch x 12 inch, and Bearing Capacity of Soil = 2Kip/sqft.Coefficient of friction =0.5, FOS against sliding =1.5, FOS against overturning =1.5


5.3.3 SolutionApproximate area of footing required = 250/2 = 125 sqft


Weight of footing 160 x160 x21 x 0.159/(123)= 49.47 Kip

Weight of soil above footing 160 x160 x32 x 0.159/(123)= 53.57 Kip

Therefore, total load on the footing = (250+49.47 +53.57) KN = 353.04 Kip




Maximum pressure = 353.04/(160x160) = 0.0138 Ksi=1.9872 Kip/sqft <2Kip/sqft

Hence safe

Ultimate pressure = 200 x 1.5/(122 x 122)= KN/m2 = 0.020156 Ksi


Along X Direction

Sliding force =0


Hence OK



Hence OK

Along Z Direction

Sliding force =0

max Resisting force = µ x Total Service load on foundation =0.5 x 353.04 =176.52 Kip

Hence OK



Hence OK

Punching Shear

Punching shear is checked on a perimeter 0.5d = 9 inch from the column face.


Punching Shear Vm= 0.013673x160x160-0.0137x30x30=337.68 Kip




V1 √fc.bc.d = (2+4/1).√(4000) x120x18 =819662.37 lb


Chapter — 5


V2 √fc.bc.d = (40x18/120+2).√(4000) x120x18=1092883.16 lb

V2= 4 .√fc.bc.de=4.√(4000) x120x18=546441.58 lb

Vc =

=546441.58 lb

So, 0.75Vc=409.832 Kip

Vm< 0.75xVc , Hence safe


Along X Direction

Vumax = 0.0137 x 160 x = 122.76 Kip

Now allowable shear = Vc1 = = 2x√(4000) x160x18 lb =364.3 Kip

0.75.Vc1 = 273.2 Kip

So Vumax < 0.75x Vc1, Hence Safe

Along Z Direction

Vumax = 0.0137 x 160 x = 122.76 Kip


0.75.Vc1 = 273.2 Kip

So Vumax <0.75x Vc1 , Hence Safe


About X Axis

m = fy/0.85.fc=60/(0.85x4) = 17.6471

ß1 = 0.85 ( as fc=4)

ρbal = 0.85.ß1.fc.87/fy.(87+fy) = 0.028507

ρmax = 0.75. ρbal = 0.02138

ρmin = 0.0018 ( as fy=60)


Mux = = 0.0137 x 160 x 74 x 74 /2 = 6002 in·kip







ρ = 1/m. (1-√1-2m.Rn/fy) = 0.002188


About Z Axis

m=fy/0.85.fc=60/(0.85x4) = 17.6471

ß1 = 0.85 ( as fc=4)

ρbal = 0.85.ß1.fc.87/fy.(87+fy) = 0.028507

ρmax = 0.75. ρbal = 0.02138

ρmin = 0.0018 ( as fy=60)


Mux = = 0.0137 x 160 x 74 x 74 / 2 = 6002 in·kip




ρ = 1/m. (1-√1-2m.Rn/fy) = 0.002188



Along X Direction

ρx = 0.002188


Along Z Direction

ρz = 0.002188



Chapter — 5


5.3.4 Comparison



Percent Dif-ference

Effective Depth 18 inch 18 inch NoneBearing Pressure 0.0138 Ksi 0.0138 Ksi NoneGoverning Moment 6002 in·kip

6002 in·kip5989 in·kip

5989 in·kip

Negligible


92.24 Kip

91.8 Kip

91.8 Kip

Negligible

Shear Force(Two-Way) 337.68 Kip 337.7 Kip NoneResisting force forsliding

176.52 Kip 176.37 Kip Negligible


28243.2 in·kip 28218.85in·kip

Negligible



5.4.2 ProblemDesign an isolated footing with the given data: Load Fy = 300 Kip, fc = 4 Ksi, fy = 60 Ksi,Column Dimension = 12 inch x 12 inch, and Bearing Capacity of Soil = 2.1Kip/sqft.Coefficient of friction =0.5, FOS against sliding =1.5, FOS against overturning =1.5





5.4.3 SolutionApproximate area of footing required =300/2.1 = 142.86 sqft



Weight of soil above footing 158 x158 x25 x 0.159/(123)= 40.82 Kip

Reduction of Weight due to buoyancy = 158 x158 x(25+21-10) x 3.61398x10-5= 32.479 Kip

Therefore, total load on the footing = (300+48.24 +40.82 -32.479) = 356.581Kip

Maximum pressure = 356.581 /(158x158) = 0.0143 Ksi=2.059 Kip/sqft <2.1Kip/sqft

Hence safe

Ultimate pressure =200x1.5/(122x122)= KN/m2 = 0.020156 Ksi


Chapter — 5



Along X Direction

Sliding force =0

max Resisting force = µ x Total Service load on foundation=0.5 x 356.581 =178.29 Kip

Hence OK



Hence OK

Along Z Direction

Sliding force =0

max Resisting force = µ x Total Service load on foundation=0.5 x 356.581 =178.29 Kip

Hence OK



Hence OK

Punching Shear



Punching Shear Vm= 0.016825 x 158 x 158 - 0.0169 x 30 x 30 = 404.81 Kip


ßc = Width of col/depth of col = 12/12=1

bc = 2.(b+d+2.deff) = 120 inch




Figure 5-7: Section considered for two-way shear

V1 √fc.bc.d = (2+4/1).√(4000) x120x18 =819662.37 lb

V2 √fc.bc.d = (40x18/120+2).√(4000) x120x18=1092883.16lb

V3= 4.√fc.bc.de=4.√(4000) x120x18=546441.58 lb

Vc = min{V1, V2, V3} = 546441.58 lb

So, 0.75Vc=409.832 Kip



Along X Direction

Vumax = 0.0169 x 158 x = 146.87 Kip

Now allowable shear = Vc1 = = 2x√(4000) x158x18 lb = 359.74 Kip

0.75.Vc1=269.81 Kip



Chapter — 5


Along Z Direction

Vumax = 0.0169 x 158 x = 146.87 Kip

Now allowable shear = Vc1 = = 2x√(4000) x158x18 lb = 359.74 Kip

0.75.Vc1=269.81 Kip

So Vumax < 0.75xVc1 , Hence Safe

Check for Trial Depth

Moment About X Axis

m=fy/0.85.fc=60/(0.85x4) = 17.6471

ß1 = 0.85 ( as fc=4)

ρbal = 0.85.ß1.fc.87/fy.(87+fy) = 0.028507

ρmax = 0.75. ρbal = 0.021381

ρmin = 0.0018 ( as fy=60)


Mux = 0.0169 x 158 x 73 x 73 / 2 = 7114.95 in·kip


Rn = Mnx/b.d2 =0.1545 Ksi


ρ = 1/m. (1-√1-2m.Rn/fy) = 0.002637


Moment About Z Axis

m=fy/0.85.fc=60/(0.85x4) = 17.6471

ß1 = 0.85 ( as fc=4)

ρbal = 0.85.ß1.fc.87/fy.(87+fy) = 0.028507

ρmax = 0.75. ρbal = 0.021381

ρmin = 0.0018 ( as fy=60)


Mux = 0.0169 x 158 x 73 x 73 / 2 = 7114.95 in·kip







ρ = 1/m. (1-√1-2m.Rn/fy) = 0.002637



Along X Direction

ρx = 0.002637


Along Z Direction

ρz = 0.002637


5.4.4 Comparison



Percent Dif-ference

Effective Depth 18 inch 18 inch NoneBearing Pressure 0.0143 Ksi 0.0143 Ksi NoneGoverning Moment 7114.95

in·kip

7114.95in·kip

7082.85 in·kip

7082.85 in·kip

Negligible


146.87 Kip

146.2 Kip

146.2 Kip

Negligible

Shear Force(Two-Way) 404.81 Kip 404.86 Kip NoneResisting force for sliding 178.29 Kip 178.18 Kip NegligibleResisting Moment forOverturning

28169.899in·kip

28152.53 in·kip Negligible



5.5.2 ProblemDesign an isolated footing with the given data: Load Fy = 200 Kip,Mx=Mz=240 in·kip. fc= 4 Ksi, fy = 60 Ksi, Column Dimension = 12 inch x 12 inch, and Bearing Capacity of Soil


Chapter — 5


= 2.2 Kip/sqft. Coefficient of friction =0.5, FOS against sliding =1.5, FOS againstoverturning =1.5


5.5.3 SolutionApproximate area of footing required = 200/2.2 = 90.9 sqft



Therefore, total load on the footing = (200+29.55) KN = 229.55 Kip

Zx = 130x1302 = 366166.667 in3

Zz = 130x1302 = 366166.667 in3

Maximum pressure from axial load = 229.55/(130x130) = 0.0136 Ksi

Mx/Zx=240/366166.667 =0.0007 Ksi

Mz/Zz=240/366166.667 =0.0007 Ksi

So stress at four corners –




σ1=P/A-Mx/Zx+Mz/Zz= 0.0136 Ksi

σ2=P/A-Mx/Zx-Mz/Zz= 0.0122 Ksi

σ3=P/A+Mx/Zx-Mz/Zz= 0.0136 Ksi

σ4=P/A+Mx/Zx+Mz/Zz= 0.015 Ksi

Max stress = 0.0136 Ksi = 2.16 Ki/sqft <2.2 Kip/sqft (Hence safe)


Along X Direction

For Sliding

Sliding force =0


Hence OK

Along Z Direction

Sliding force =0

max Resisting force = µ x Total Service load on foundation =0.5 x 229.55 =114.775 Kip

Hence OK

For overturning

About X Direction

Overturning Moment =240 in·kip


FOS= 14920.75/240 = 62.17>1.5

Hence OK

About Z Direction

Overturning Moment =240 in·kip


FOS= 14920.75/240 = 62.17>1.5

Hence OK


Chapter — 5


Punching Shear



Punching Shear Vm= 0.018813x122x122-0.0189x27x27=266.24 Kip


Vm1 = Vmax/(Pm · d) = 164.81 Ksi



V1 √fc.bc.d = (2+4/1).√(4000) x108x15 =614746.78 lb

V2 √fc.bc.d = (40x15/108+2).√(4000) x108x15=774125.57 lb

V2= 4.√fc.bc.de=4.√(4000) x108x15=409831.18 lb

Vc = min{V1, V2, V3} = 409831.185 lb

So, 0.75Vc=307.374 Kip






Along X Direction

Critical section for moment is at distance d from the face of the column.

Shear at Longitudinal Direction (X Axis)

Average Base Pressure along one edge = (0.014733 + 0.016569) x0.5 = 0.157Ksi

Average Base Pressure along other edge = (0.016569 + 0.018405) x 0.5 =0.0175 Ksi

Approximate Base Pressure at the left critical section = 0.0175 + (0.0157 -0.0175) x87/130 = 0.0163 Ksi

Approximate Base Pressure at the right critical section = 0.0175 + (0.0157 -0.0175) x 43/130 = 0.0170 Ksi

Hence, the shear force at the critical section:

F = (0.0175 + 0.0170) x0.5 x 43 x 130 = 96.43 kip

So, max shear force along X axis, Fux = 97 kip


0.75.Vc1 = 197.327 Kip

So Vumax < 0.75x Vc1 , Hence Safe


Chapter — 5


Along Z Direction

Average Base Pressure along one edge = (0.014733 + 0.016569) x0.5 = 0.157Ksi

Average Base Pressure along other edge = (0.016569 + 0.018405) x 0.5 =0.0175 Ksi

Approximate Base Pressure at the left critical section = 0.0175 + (0.0157 -0.0175) x87/130 = 0.0163 Ksi


Hence, the shear force at the critical section:

F = (0.0175 + 0.0170) x0.5 x 43 x 130 = 96.43 kip

So, max shear force along Z axis, Fuz = 97 kip


0.75.Vc1 = 197.327 Kip


About Z Axis

Critical section for moment is at the face of the column.




Average Base Pressure along one edge = (0.014733 + 0.016569) x 0.5 =0.0157 Ksi (left end)

Average Base Pressure along other edge = (0.016569 + 0.018405) x 0.5 =0.0175 Ksi (right end)

Approximate Base Pressure at the left critical section = 0.0175 + (0.0157 -0.0175) x 71/130 = 0.0166 Ksi


Hence, the moment at the critical section (right):

Mu = F x LA

F = (0.0175 + 0.0167) x 0.5 x 59 x 130 = 131.16 Kip

LA = (0.0167 + 2 x 0.0175) x 59/(3x (0.0167 + 0.0175)) = 29.731 inches

Mu (right) = 3899.52 in-kips

So, max moment wrt Z axis, Mu(x) = 3900 in-kips

m=fy/0.85.fc=60/(0.85x4) = 17.6471

ß1 = 0.85 ( as fc=4)

ρbal = 0.85.ß1.fc.87/fy.(87+fy) = 0.028507

ρmax = 0.75. ρbal = 0.021381

ρmin = 0.0018 ( as fy=60)





Chapter — 5


ρ = 1/m. (1-√1-2m.Rn/fy) = 0.002215



About X Axis

Critical section for moment is at the face of the column.

Average Base Pressure along one edge = (0.014733 + 0.016569) x 0.5 = 0.0157Ksi (left end)

Average Base Pressure along other edge = (0.016569 + 0.018405) x 0.5 =0.0175 Ksi (right end)

Approximate Base Pressure at the left critical section = 0.0175 + (0.0157 -0.0175) x 71/130 = 0.0166 Ksi


Hence, the moment at the critical section (right):

Mu = F x LA

F = (0.0175 + 0.0167) x 0.5 x 59 x 130 = 131.16 Kip

LA = (0.0167 + 2 x 0.0175) x 59/(3x (0.0167 + 0.0175)) = 29.731 inches

Mu (right) = 3899.52 in-kips

So, max moment wrt Z axis, Mu(z) = 3900 in-kips

m=fy/0.85.fc=60/(0.85x4) = 17.6471




ß1 = 0.85 ( as fc=4)

ρbal = 0.85.ß1.fc.87/fy.(87+fy) = 0.028507

ρmax = 0.75. ρbal = 0.021381

ρmin = 0.0018 ( as fy=60)




ρ = 1/m. (1-√1-2m.Rn/fy) = 0.002215



Along X Direction

ρx = 0.002215


Along Z Direction

ρz = 0.002215



Chapter — 5


5.5.4 Comparison



Percent Dif-ference

Effective Depth 16 inch 16 inch NoneBearing Pressure 0.0136 Ksi

0.0122 Ksi

0.0136 Ksi

0.015 Ksi

0.0136 Ksi

0.0123 Ksi

0.0136 Ksi

0.0149 Ksi

None

Governing Moment 3899.52in·kip

3899.52in·kip

3893.52 in·kip

3893.52 in·kip

Negligible


96.43 Kip

96.05 Kip

96.05 Kip

Negligible

Shear Force(Two-Way) 266.96 Kip 267.01 Kip NegligibleResisting force for sliding 114.775 Kip

114.775 Kip

114.773 Kip

114.773 Kip

None


14920.75in·kip

14920.75in·kip

14920.466in·kip

14920.466in·kip

Negligible



5.6.2 ProblemDesign an isolated footing with the given data: Load Fy = 1000 KN, Concrete grade M25,Steel Grade Fe 415, Pedestal Dimension = 250 mm x 250 mm x 500 mm.

Bearing Capacity of Soil = 120 KN/m2. Coefficient of friction =0.4, FOS against sliding=1.4, FOS against overturning =1.4

Soil Depth is 500 mm , depth of GWT from GL is 200 mm and surcharge on foundation

is 5 KN/m2. (consider unt wight of soil 18 KN/m2)





5.6.3 SolutionApproximate area of footing required = 1000/120 = 8.33 m2

Assuming 3140 mm x 3140 mm footing dimension, with 490 mm depth

Punching Shear

Weight of footing = 3.14 x 3.14 x 0.49 x 25 = 120.78 KN

Punching Shear Vm = 1400/(3.14 x 3.14) x (3.14 x 3.14-0.665 x 0.665) =1337.2 KN

Critical perimeter Pm = 2 x (b + d + 2 x d) = 2660 mm

Vm1 = Vmax/(Pm .d) = 1211.34 KN/m2

ßc = Width of col/depth of col = 250/250 = 1

bc = 2.(b+d+2.deff) = 2660 mm

V1 = (2+4/ßc)√fc.bc.d = 0.083x(2+4/1)x√25 x 2660 x 415x 1/1000 = 2748.7KN

V2 = (40.d/b+2)√fc.bc.d = 0.083x(40x 415/2660+2)x√25 x 2660 x 415x1/1000 = 3775.17 KN

V3 = 4.√fc.bc.de =0.083 x 4 x √25 x 2660 x 415x 1/1000 =1832.47 KN

Vc =min(V1,V2,V3) =1832.47


Chapter — 5


So, 0.75Vc = 1374.35 KN

Vm < 0.75.Vc , Hence safe


Along X Direction

Vumax = 141.993 x 3.14 x ((3.14-0.25)/2-0.415) = 459.23 KN

Now allowable shear = Vc1 =2√fc.b.d = 0.083x2x√25 x 3140 x 415x 1/1000 =1081.57 KN

0.75.Vc1=811.18 KN

V < Vc1, Hence Safe


Along Z Direction

Vumax = 141.993 x 3.14 x ((3.14-0.25)/2-0.415) = 459.23 KN

Now allowable shear = Vc1 =2√fc.b.d =0.083x2x√25 x 3140 x 415x 1/1000 =1081.57 KN

0.75.Vc1=811.18 KN

V < Vc1, Hence Safe



About X Axis

m = fy/0.85.fc = 415/(0.85x25) = 19.529

ß1 = 0.85 ( as fc < 27.6 N/mm2)

ρbal = 0.85.ß1.fc.87/fy.(87 + fy) = 0.025759

ρmax = 0.75. ρbal = 0.0193

ρmin = 0.0018 ( as fy = 415 N/mm2)

Bending moment at critical section, Mux = 1/2 x 141.993 x 3.14 x [(3.14-0.25)/2)]2 = 465.48 KNm

So Resisting Moment = Mnz = Muz/Ø = 517.2 KNm

Rn = Mnz/b.d2 = 0.956 N/mm2

2m.Rn/fy < 1 , Hence OK

ρ = 1/m. (1-√1-2m.Rn/fy) = 0.002359





About Z Axis

m = fy/0.85.fc = 415/(0.85x25) = 19.529

ß1 = 0.85 ( as fc < 27.6 N/mm2)

ρbal = 0.85.ß1.fc.87/fy.(87 + fy) = 0.025759

ρmax = 0.75. ρbal = 0.0193

ρmin = 0.0018 ( as fy = 415 N/mm2)

Bending moment at critical section, Muz = 1/2 x 141.993 x 3.14 x [(3.14-0.25)/2)]2 = 465.48 KNm

So Resisting Moment = Mnz = Muz/Ø = 517.2 KNm

Rn = Mnz/b.d2 = 0.956 N/mm2


ρ = 1/m. (1-√1-2m.Rn/fy) = 0.002359



Along X Direction

ρz = 0.002359

Therefore, Astx = ρz.b.d = 3074 m2m

Along Z Direction

ρx = 0.002359

Therefore, Astz = ρx.b.d = 3074 m2m


Chapter — 5


5.6.4 Comparison



Percent Dif-ference

Effective Depth 415 mm 415 mm NoneBearing Pressure 119.92 KN/m2 119.96 KN/m2 NoneResisting force for sliding 472.936 KN

472.936 KN

472.945 KN

472.945 KN

None

Resisting Moment for Over-turning

1856.23 KNm

1856.23 KNm

1856.276 KNm

1856.276 KNm

Negligible

Governing Moment 465.48 KNm

465.48 KNm

465.48 KNm

465.48 KNm

None


459.23 KN

459.4 KN

459.4 KN

Negligible

Shear Resistance (One-Way) 811.18 KN

811.18 KN

810.737 KN

810.737 KN

Negligible

Shear Force(Two-Way) 1337.2 KN 1337.28 KN NoneShear Resistance (Two-Way) 1374.35 KN 1372.8 KN NegligibleReinf X 3074 m2m 3076.9 m2m NegligibleReinf Z 3074 m2m 3076.9 m2m Negligible



5.7.2 ProblemDesign an isolated footing with the given data: Load Fy = -1200 KN, Fx=Fz=5 KN,Mx=Mz=10 KNm. fc = 25 N/m2m, fy = 415 N/m2m, Column Dimension = 250 mm x 250mm, and Bearing Capacity of Soil = 120 KN/m2. Assume 500 mm soil depth above footingand depth of GWT 200 mm. Assume unit weight of concrete & soil are 25 & 18 KN/cumrespectively. Coefficient of friction =0.4, FOS against sliding =1.4, FOS against overturning=1.4





5.7.3 SolutionApproximate area of footing required = 1200/120 = 10 m2

Assuming 3500 mm x 3500 mm x 540 mm footing dimension,


Weight of soil = (3.5 x 3.5- 0.25 x 0.25) x 0.5 x 18 = 109.69 KN

Weight reduction for Buoyancy = 3.5 x 3.5 x (0.54 + 0.5 – 0.2) x 9.81 =100.95 KN

Surcharge Load = (3.5 x 3.5 – 0.25 x 0.25) x 5 = 60.94 KN

Therefore, total Axial load on the footing = 1435.063 KN

Total Moment on the footing-

Mx = 10 + 5x 0.54 = 12.7 KNm

Mz = 10 - 5x 0.54 = 7.3 KNm

Zx = 3.5x3.52 /6 = 7.1458 cum

Zz = 3.5x3.52 /6 = 7.1458 cum

Maximum pressure from axial load = 1435.063 /(3.5 x 3.5) = 117.15 KN/m2

Mx/Zx = 12.7 /7.1458 = 1.78 KN/m2

Mz/Zz = 7.3 /7.1458 = 1.03 KN/m2


Chapter — 5


So stress at four corners –

σ1 = P/A-Mx/Zx+Mz/Zz= 116.39 KN/m2

σ2 = P/A-Mx/Zx-Mz/Zz= 114.35 KN/m2

σ3 = P/A+Mx/Zx-Mz/Zz = 117.09 KN/m2

σ4 = P/A+Mx/Zx+Mz/Zz= 119.94 KN/m2

Max stress = 119.94 KN/m2 < 120 KN/m2 (Hence safe)


Along X Direction

For Sliding

Sliding force = 5 KN

max Resisting force = µ x Total Service load on foundation = 0.4 x 1435.063 =574 KN

Hence OK

Along Z Direction

Sliding force = 5 KN

max Resisting force = µ x Total Service load on foundation = 0.4 x 1435.063 =574 KN

Hence OK

For Overturning

About X Direction

Overturning Moment = 12.7 KNm

max resisting Moment = 0.5x 3.5 x 1435.063 = 2511.36 KNm

FOS= 2511.36/12.7 = 197.75 > 1.4

Hence OK

About Z Direction

Overturning Moment = 7.3 KNm

max resisting Moment = 0.5x 3.5 x 1435.063 = 2511.36 KNm

FOS= 2511.36/7.3 = 344 > 1.4

Hence OK




Section Design

Fy = -1.4 x 1200 KN = -1680 KN

Fx = 1.4 x 5 KN = 7 KN

Fz = 1.4 x 5 KN = 7 KN

Mx = 1.4 x 10 KNm = 14 KNm

Mz = 1.4 x 10 KNm = 14 KNm

Modifying Moments due to lateral forces:

Mx= 17.78 KNm & Mz= 10.22 KNm

D= 540 mm, So deff = 650- 50-25 = 465 mm

Punching Shear

Punching shear is checked on a perimeter 0.5d = 232.5 from the column face.

Effective Depth= 465 mm

Punching Shear bc = 1609.9 KN = 1610 KN

Figure 5-11: Corner pressure values on plan for punching shear

Critical perimeter Pm = 2 X ( b + h + 2 x d) = 2860 mm

Vm1 = Vmax/(bc .d) = 1.21 N/m2m

ßc = Width of col/depth of col=250/250=1

V1 = (2+4/ßc)√fc.bc.d = 3311.45 KN

V2 = (40.d/b+2)√fc.bc.d = 4693.15 KN

V3 = 4.√fc.bc.de=2207.63 KN


Chapter — 5


Vc =min(V1,V2,V3) = 2207.63 KN

So, 0.75Vc = 1655.73 KN



Along X Direction

Critical Section for shear is at d distance from face of column

Average Base pressure at left edge = 134.66 KN/m2

Average Base pressure at right edge = 139.64 KN/m2

Approximate Base pressure at left critical section = 136.311 KN/m2

Approximate Base pressure at right critical section = 137.99 KN/m2

Figure 5-12: One-way shear pressure values along x-direction

Hence Maximum Shear Force at critical section = 563.589 KN

Now allowable shear = Vc1 = 2√fc.b.d= 1350.825 KN

0.75.Vc1 = 1013.12 KN


Along Z Direction

Critical Section for shear is at d distance from face of column








Figure 5-13: One-way shear pressure values along z-direction

Hence Maximum Shear Force at critical section = 561 KN

Now allowable shear = Vc1 = 2√fc.b.d = 1350.825 KN

0.75.Vc1 = 1013.12 KN


Check For Flexure

Critical Section for Moment is at the face of column

m = fy/0.85.fc = 19.5295

ß1 = 0.85 (as fc=M 25)

ρbal = 0.85.ß1.fc.87/fy.(87+fy) = 0.025729

ρmax = 0.75. ρbal = 0.019297

ρmin = 0.0018


About Z Axis






Chapter — 5


Figure 5-14: Bending pressure about Z axis

So Maximum Bending Moment at Critical Section = 638.54 KNm

So Resisting Moment = Mnz = Muz/Ø = 710 KNm

Rn = Mnz/b.d2 = 0.93822 N/m2m


ρ = 1/m. (1-√1-2m.Rn/fy) = 0.002313


About X Axis





Figure 5-15: Bending pressure about X axis

So Maximum Bending Moment at Critical Section = 641.93 KNm

So Resisting Moment =Mnz= Muz/Ø = 642 KNm




Rn = Mnz/b.d2 = 0.9426 N/m2m


ρ = 1/m. (1-√1-2m.Rn/fy) = 0.002325



Along X Direction

ρz = 0.002313

Therefore, Astx = ρz.b.d = 3765 m2m

Along Z Direction

ρx = 0.002325

Therefore, Astz = ρx.b.d = 3784 m2m

5.7.4 Comparison

Value of Reference ResultSTAAD Foun-

dationResult

Percent Dif-ference

Effective Depth 540 mm 540 mm NoneBearing Pressure 116.39 KN/m2

114.35 KN/m2

117.9 KN/m2

119.95 KN/m2

116.397 KN/m2

114.35 KN/m2

117.91 KN/m2

119.95 KN/m2

None

Resisting force for sliding 574.05 KN (Xdir)

574.05 KN (Zdir

574.05 KN (Xdir)

574.05 KN (Zdir)

None

Resisting Moment for Over-turning

2511 KNm (WRTZ)

2511 KNm (WRTX)

2511 KNm (WRTZ)

2511 KNm (WRTX)

None

Governing Moment 639 KNm

642 KNm

638 KNm

642 KNm

Negligible

Shear Force(One-Way) 564 KN

561 KN

564 KN

561 KN

Negligible

Shear Force(Two-Way) 1610 KN 1610 KN None

Table 5-7: US Verification problem 9 comparison


Chapter — 5


5.8 US General Combined Foundation 15.8.1 Reference

5.8.2 ProblemDesign a combined footing with the given data: Load Fy = 60 Kip each column., fc = 4 Ksi,fy = 60 Ksi, Column Dimension = 12 inch x 12 inch, Pedestal height-20 inch. and C/Ccolumn distance=195 inch . Bearing Capacity of Soil = 2 Kip/sqft. Coefficient of friction=0.5, FOS against sliding =1.5, FOS against overturning =1.5

Ht of soil =24 inch. Depth of GWT=10 inch

5.8.3 SolutionFigure 5-16: Elevation and Plan

Approximate area of footing required = 2x60/2 sqft = 60 sqft

Assuming 275 inch x40 inch x 24 inch footing dimension,

( left overhang=right overhang = 40 inch)

Weight of footing = 275 x 40 x 24 x 0.159/123 = 24.3Kip


5.8 US General Combined Foundation 1


Weight of pedestal=2 x 12 x 12 x 20 x 0.159 /123 = 0.530 kip

Weight of soil above footing = (275 x 40-2x12x12) x 24 x 0.127/123 = 18.9 Kip

Reduction of Weight due to buoyancy = 275 x 40 x (24 +24-10) x3.61398x10-5 = 15.1Kip

Therefore, total load on the footing = (2x60+24.3+18.9- 15.1) = 148.624 Kip

Maximum pressure= 148.624 /(275x40) = 0.0136 Ksi = 1.9584 Kip/sqft < 2Kip/sqft

Hence safe

Ultimate pressure = 1.4x2x60/(275x40) = 0.015273 Ksi




max resisting Moment = 0.5 x 275x 148.624 = 20435.8 in·kip

Hence OK




Hence OK


Chapter — 5



Punching Shear

For Column 1


deff = (24-2-1) inch = 21 inch

2 way shear Vm = 67.37 Kip






V1 √fc.bc.d = (2+4/1).√(4000) x132x21 = 1051900 lb

V2 √fc.bc.d = (40x21/132+2).√(4000) x132x21 = 1466284.9lb

V3= 4.√fc.bc.de = 4.√(4000) x132x21 = 701266.694 lb

Vc = min{V1, V2, V3} = 701266.694 lb

So, 0.75Vc=525.95 Kip


For Column 2






V1 √fc.bc.d = (2+4/1).√(4000) x132x21 = 1051900 lb

V2 √fc.bc.d = (40x21/132+2).√(4000) x132x21 = 1466284.9lb

V3 = 4.√fc.bc.de = 4.√(4000) x132x21 = 701266.694 lb

Vc = min{V1, V2, V3} = 701266.694 lb

So, 0.75Vc=525.95 Kip



Vumax = 43.072 Kip


0.75.Vc1=79.69 Kip




m=fy/0.85.fc=60/(0.85x4) = 17.6471


Chapter — 5


ß1 = 0.85 (as fc=4)

ρbal = 0.85.ß1.fc.87/fy.(87+fy) = 0.028507

ρmax = 0.75. ρbal = 0.021381

ρmin = 0.0018 ( as fy=60)


Mux = 488.74 in·kip



Rn = Mnx/b.d2 = 0.0308 Ksi


ρ = 1/m. (1-√1-2m.Rn/fy) = 0.000516

Hence OK


m=fy/0.85.fc=60/(0.85x4) = 17.6471

ß1 = 0.85 (as fc=4)

ρbal = 0.85.ß1.fc.87/fy.(87+fy) = 0.028507

ρmax = 0.75. ρbal = 0.021381

ρmin = 0.0018 ( as fy=60


Mux = 2414.897 in·kip



Rn = Mnx/b.d2 = 0.1522 Ksi


ρ = 1/m. (1-√1-2m.Rn/fy) = 0.002597

ρmax Hence OK

With Respect to the X Axis

m=fy/0.85.fc=60/(0.85x4) = 17.6471

ß1 = 0.85 (as fc=4)

ρbal = 0.85.ß1.fc.87/fy.(87+fy) = 0.028507

ρmax = 0.75. ρbal = 0.021381

ρmin = 0.0018 ( as fy=60




Cantilever length=(40-12)/2 = 14 inch


Mux = 411.608 in·kip



Rn = Mnx/b.d2 = 0.0038 Ksi


ρ = 1/m. (1-√1-2m.Rn/fy) = 0.000064

Hence OK

Reinforcement Calculation

ρz (top) = 0.002597

ρz (bot) = 0.0018

ρx = 0.0018

Therefore, Ast =ρ*b*d

Area of Steel Required along X dir ( bot) = 1.512 in2

Area of Steel Required along X dir (top) = 2.18 in2

Area of Steel Required along Z dir ( bot) = 10.395 in2

5.8.4 Comparison



Percent Dif-ference

Bearing Pressure 0.0136 Ksi 0.0135 Ksi NegligibleGoverning Moment 488.736

in·kip

2414.897in·kip

411.608in·kip

461.95 in·kip

2414.41 in·kip

411 in·kip

Negligible

Shear Force(One-Way) 42.73 Kip 43.07 Kip NegligibleShear Force(Two-Way) 66.8 Kip

66.8 Kip

67.37 Kip

67.37 Kip

Negligible


2972.44in·kip

20435.52in·kip

2972.48 in·kip

20435.8 in·kip

Negligible



Chapter — 5



5.9.2 ProblemDesign a combined footing with the given data: Load Fy = 60 Kip, Mz=360 in·kip for eachcolumn., fc = 4 Ksi, fy = 60 Ksi, Column Dimension = 12 inch x 12 inch, Pedestal height-20 inch. and C/C column distance=195 inch . Bearing Capacity of Soil = 2 Kip/sqft.Coefficient of friction =0.5, FOS against sliding =1.5, FOS against overturning =1.5


5.9.3 SolutionApproximate area of footing required = 2x60/2 sqft = 60 sqft

Assuming 275 inch x44 inch x 24 inch footing dimension,

( left overhang=right overhang = 40 inch)

Weight of footing = 275 x 44 x 24 x 0.159/123 = 26.73 Kip

Weight of pedestal=2x12x12x20x0.159/123 = 0.530 kip

Therefore, total load on the footing = (2x60+26.73 +0.530) Kip = 147.26 Kip

Maximum pressure from axial load= P/A = 147.26 /(275x44) = 0.0122 Ksi

Mz=360+360=720 Kip

Mz=0




Zz= 44x2752/6 = 554583.334 inch3

Mz/Zz = 720/554583.334 =0.0013 Ksi

So, stress at left end

σ1 = P/A + Mz/Zz = 0.0135 Ksi

stress at right end

σ1 = P/A - Mz/Zz = 0.0109 Ksi

So maximum stress= 0.0135 Ksi= 1.944 Kip/sqft < 2 Kip/sqft

Hence safe


wrt Z Direction



Hence OK

Wrt X Direction



Hence OK

Punching Shear

For Column One






V1 √fc.bc.d = (2+4/1).√(4000) x132x21 = 1051900 lb

V2 √fc.bc.d = (40x21/132+2).√(4000) x132x21 = 1466284.9lb

V3= 4.√fc.bc.de = 4.√(4000) x132x21 = 701266.694 lb


Chapter — 5


Vc = min{V1, V2, V3} = 701266.694 lb

So, 0.75Vc=525.95 Kip


For Column Two






V1 √fc.bc.d = (2+4/1).√(4000) x132x21 = 1051900 lb

V2 √fc.bc.d = (40x21/132+2).√(4000) x132x21 = 1466284.9 lb

V3 = 4.√fc.bc.de = 4.√(4000) x132x21 = 701266.694 lb

Vc= min{V1, V2, V3} = 701266.694 lb

So, 0.75Vc=525.95 Kip



Vumax = 47.126 Kip


0.75.Vc1=87.66 Kip



About Z Axis (sagging)

m=fy/0.85.fc=60/(0.85x4) = 17.6471

ß1 = 0.85 (as fc=4)

ρbal = 0.85.ß1.fc.87/fy.(87+fy) = 0.028507

ρmax = 0.75. ρbal = 0.0214

ρmin = 0.0018 ( as fy=60)

Bending moment at critical section, Mux = 791 in·kip


So Resisting Moment =Mnz= Muz/Ø =879 in·kip




Rn = Mnx/b.d2 = 0.0453 Ksi


ρ = 1/m. (1-√1-2m.Rn/fy) = 0.000761

Hence OK

About Z Axis (hogging)

m=fy/0.85.fc=60/(0.85x4) = 17.6471

ß1 = 0.85 (as fc=4)

ρbal = 0.85.ß1.fc.87/fy.(87+fy) = 0.028507

ρmax = 0.75. ρbal = 0.0214

ρmin = 0.0018 ( as fy=60

Bending moment at critical section, Mux = 2440 in·kip


Ø = 0.9


Rn = Mnx/b.d2 = 0.1398 Ksi


ρ = 1/m. (1-√1-2m.Rn/fy) = 0.00238

ρmax Hence OK

About X Axis

m=fy/0.85.fc=60/(0.85x4) = 17.6471

ß1 = 0.85 (as fc=4)

ρbal = 0.85.ß1.fc.87/fy.(87+fy) = 0.028507

ρmax = 0.75. ρbal = 0.0214

ρmin = 0.0018 ( as fy=60

Cantilever length=(40-12)/2 = 14 inch

Bending moment at critical section, Mux = 0.013885x275x162/2 =552.82in·kip



Rn = Mnx/b.d2 = 0.0051 Ksi


ρ = 1/m. (1-√1-2m.Rn/fy) = 0.000086

Hence OK


Chapter — 5



ρz (bot) = 0.0018

ρz (top) = 0.00238

ρx(bot) = 0.0018

Therefore, Ast = ρ*b*d

Area of Steel Required along X dir ( bot) = 1.7 in2

Area of Steel Required along X dir (top) = 2.2 in2

Area of Steel Required along Z dir ( bot) = 10.4 in2






Chapter — 5


5.9.4 Comparison

Value of Reference ResultSTAAD

FoundationResult

Percent Dif-ference

Bearing Pressure 0.0135 Ksi

0.0109 Ksi

0.0135 Ksi

0.0109 Ksi

None

Governing Moment 791 in·kip

2440 in·kip

552.82 in·kip

830 in·kip

2439.66in·kip

534 in·kip

Negligible

Shear Force(One-Way) 47.126 Kip 46.8 Kip NegligibleShear Force(Two-Way) 67.48 Kip

70.28 Kip

66.91 Kip

69.81 Kip

Negligible


3239.72 in·kip20248.25 in·kip

3239.518in·kip

20246.985in·kip

Negligible



5.10.2 ProblemDesign a combined footing with the given data: Load Fy = 1500 KN on each column.,Concrete Grade = M 25, Steel Grade Fe 415, Column Dimension = 250 mm x250 mm,Pedestal height-500 mm each. and C/C column distance=5000 mm . Bearing Capacity ofSoil = 120 KN/m2., FOS against overturning =1.4






Approximate area of footing required = 2x1500/120 = 25 m2

Assuming 7000 mm 4500 mm x 840 mm footing dimension,

( left overhang=right overhang = 1000 mm)

Weight of footing = 7 x 4.5 x 0.84 x 25 = 661.5 KN

Weight of pedestal=2 x 0.25 x 0.25 x 0.5 x 25 = 1.5625 KN

Weight of soil above footing = (7 x 4.5- 0.25 x0.25 x 2) x 0.5 x 18 = 282.375 KN

Reduction of Weight due to buoyancy = 7 x 4.5 x (0.84+0.5-0.2) x 9.81 = 352.277 KN

Surcharge = (7 x 4.5 -0.25 x0.25 x2) x 5 = 156.875 KN

Therefore, total load on the footing = (2x1500+661.5 +1.5625 - 352.277 +156.875) = 3750.04KN

Maximum pressure= 3750.04 /(7 x 4.5) = 119.05 KN/m2 =119.05 KN/m2 120 KN/m2

Hence safe


Chapter — 5





max resisting Moment = 0.5 x 7x 3750 = 13125 KNm

Hence OK



max resisting Moment = 0.5 x 4.5 x 3750 = 8437.5 KNm

Hence OK

Ultimate design

Ultimate pressure = 1.4 x 2 x1500/(7 x 4.5) = 133.33 KN/m2

Deff = 784 mm


Punching Shear

For Column 1





2 way shear Vm = 1957.5 KN

ßc = Width of col/depth of col=250/250 = 1

bc = 2·(b + d + 2·deff) = 4136 mm

V1= (2 + 4/βc)√(fc·bc·d) = 8074 KN

V2 = (40·d/b + 2)√fc.bc.d = 12894.7 KN

V3= 4.√fc·bc·de = 5382.76 KN

Vc = min(V1, V2, V3) = 5382.76 KN

So, 0.75·Vc = 4037 KN

Vm < 0.75·Vc , Hence safe

For Column 2


2 way shear Vm = 1957.5 KN


bc = 2·(b + d + 2·deff) = 4136 mm

V1= (2 + 4/βc)√(fc·bc·d) = 8074 KN

V2 = (40·d/b + 2)√fc·bc·d = 12894.7 KN

V3= 4·√fc·bc·de = 5382.76 KN

Vc = min(V1, V2, V3) = 5382.76 KN

So, 0.75·Vc = 4037 KN



Vumax = 954 KN

Now allowable shear = Vc1 = 2 √fc·b·d = 2928 KN

0.75·Vc1 = 2196 KN

So Vumax < 0.75·Vc1 , Hence Safe


m=fy/0.85.fc=415/(0.85x25) = 19.5295

ß1 = 0.85 (as fc=25 N/m2m)

ρbal = 0.85.ß1.fc.87/fy.(87+fy) = 0.025729

ρmax = 0.75. ρbal = 0.019297

ρmin = 0.0018 ( as fy=415 N/m2m)


Chapter — 5




Muz = 300.54 KNm

So Resisting Moment =Mnz= Muz/Ø =333.93 KNm

Rn = Mnz/b.d2 = 0.1208 N/m2m


ρ = 1/m. (1-√1-2m.Rn/fy) = 0.000292 <ρmax Hence OK

Take ρ = ρmin = 0.0018

Hence OK



Muz = 1574.8 KNm

So Resisting Moment =Mnz= Muz/Ø =1749.8 KNm

Rn = Mnz/b.d2 = 0.6327 N/m2m


ρ = 1/m. (1-√1-2m.Rn/fy) = 0.001548 <ρmax Hence OK


Hence OK


Cantilever length = (4500 - 250)/2 = 2125 mm


Mux = 2107.4 KNm

So Resisting Moment =Mnx= Mux/Ø =2341.55 KNm

Rn = Mnx/b.d2 = 0.762 N/m2m


ρ = 1/m. (1-√1-2m.Rn/fy) = 0.001871 < ρmax

Hence OK


ρz (top) = 0.0018

ρz (bot) = 0.0018

ρx = 0.001871




Therefore, Ast =ρ·b·d

Area of Steel Required along X dir ( bot) = 6350 mm2

Area of Steel Required along X dir (top) = 6350 mm2

Area of Steel Required along Z dir ( bot) = 10268 mm2

5.10.4 Comparison



Percent Dif-ference

Bearing Pressure(KN/m2)

119.05 119.05 None


300.54

1574.82

2107.4

293.97

1574.60

2107.4

Negligible

Shear Force, One-Way (KN)

954.8 43.07 Kip Negligible

Shear Force, Two-Way (KN)

1957.4

1957.4

1957.4

1957.4

None



5.11.2 ProblemDesign a combined footing with the given data: Load Fy = -1200 KN & Fx= 10 KN on eachcolumn., Concrete Grade = M 25, Steel Grade Fe 415, Column Dimension = 250 mm x250mm, Pedestal height-500 mm each. and C/C column distance=5000 mm . BearingCapacity of Soil = 120 KN/m2., FOS against overturning =1.4


Chapter — 5



Approximate area of footing required = 2x1200/120 = 20 m2

Assuming 7000 mm 3250 mm x 530 mm footing dimension,

( left overhang=right overhang = 1000 mm)

Weight of footing = 7 x 3.25 x 0.53 x 25 = 301.44 KN

Weight of pedestal=2 x 0.25 x 0.25 x 0.5 x 25 = 1.5625 KN

Therefore, total load on the footing = (2x1200+301.44 +1.5625) = 2703 KN

Mz1 = -10 x (0.5+0.53) = -10.3 KNm

Mz2 = -10 x (0.5+0.53) = -10.3 KNm

Maximum pressure from axial load= 2703/(7 x3.25) = 118.813 KN/ m2

Total Moment = -10.3 + -10.3 = -20.6 KNm

CG of load= 3.5085 m

CG of raft = 3.5 m

Eccentricity =-8.5833 mm




So Moment= P x Eccentricity = -20.6 KNm

Hence OK

Zz= 3.25 x 72/6 = 26.542 m3

So M/Z = -0.78 KN/m2


So stress at right end= P/A + M/Z= 119.6 KN/m2

Maximum pressure =119.6 KN/m2 120 KN/m2

Hence safe



Overturning Moment =20.6 KNm

max resisting Moment = 0.5 x 7x 2703 = 9460 KNm

So FOS= 9460/20.6 = 459.25 >1.4

Hence OK



max resisting Moment = 0.5 x 3.25x 2703 = 4392.4 KNm

Hence OK

Ultimate design

Ultimate pressure from axial load = 1.4 x 2 x1200/(7 x 3.25) = 147.692 KN/m2

Mz1 = 1.4x-10 x90.5+0.53) = -14.42 KNm

Mz2 = 1.4x-10 x90.5+0.53) = -14.42 KNm

Total Moment = -14.42 + -14.42 = -28.84 KNm

CG of load= 3.5085 m

CG of raft = 3.5 m

Eccentricity =-8.5833 mm

So Moment= P x Eccentricity = -28.84 KNm (Hence OK)

Zz= 3.25 x 72/6 = 26.542 m3

So M/Z = -1.09 KN/m2



Chapter — 5


So stress at right end= P/A + M/Z= 148.79 KN/m2

Deff = 474 mm


Punching Shear

For Column 1


Two way shear, Vm = 1602.987 KN





bc = 2·(b + d + 2·deff) = 2896 mm

V1= (2 + 4/βc)√(fc)·bc·d = 3418 KN

V2 = (40·d/b + 2)√(fc)·bc·d = 4868.97 KN

V3= 4.√(fc)·bc·de = 2278.69 KN

Vc = min(V1, V2, V3) = 2278.69 KN

So, 0.75·Vc = 1709 KN

Vm < 0.75·Vc , Hence safe

For Column 2


Two way shear, Vm = 1602.17 KN


bc = 2·(b + d + 2·deff) = 2896 mm

V1= (2 + 4/βc)√(fc)·bc·d = 3418 KN

V2 = (40·d/b + 2)√(fc)·bc·d = 4868.97 KN

V3= 4.√(fc)·bc·de = 2278.69 KN

Vc = min(V1, V2, V3) = 2278.69 KN

So, 0.75·Vc = 1709 KN

Vm< 0.75·Vc , Hence safe


Vumax = 916.99 KN

Now allowable shear = Vc1 = 2 √(fc)·b·d = 1278 KN

0.75·Vc1 = 958.96 KN

So Vumax < 0.75·Vc1 , Hence Safe


m = fy/0.85·fc=415/(0.85x25) = 19.53

ß1 = 0.85 (as fc = 25 N/mm2)

ρbal = 0.85·ß1·fc·87/fy·(87 + fy) = 0.025729

ρmax = 0.75·ρbal = 0.0193

ρmin = 0.0018 ( as fy=415 N/mm2)




Chapter — 5


Muz = 238.94 KNm

So Resisting Moment = Mnz= Muz/φ = 265.5 KNm

Rn = Mnz/b·d2 =0.3636 N/mm2

2·m·Rn/fy <1 , Hence OK

ρ = 1/m· (1-√1-2m·Rn/fy) = 0.000884 <ρmaxHence OK


Hence OK



Muz = 1259.84KNm

So Resisting Moment = Mnz= Muz/φ = 1400 KNm

Rn = Mnz/b·d2 = 1.9171 N/mm2

2·m·Rn/fy < 1 , Hence OK

ρ = 1/m· (1-√1-2m·Rn/fy) = 0.00485 < ρmaxHence OK


Hence OK


Cantilever length = (3250-250)/2 = 1500 mm


Mux = 1163 KNm

So Resisting Moment =Mnx= Mux/φ = 1292 KNm

Rn = Mnz/b·d2 =1.593 N/mm2

2·m·Rn/fy < 1 , Hence OK

ρ = 1/m· (1-√1-2m·Rn/fy) = 0.003995 < ρmaxHence OK


ρz (top) = 0.0018

ρz (bot) = 0.0018

ρx = 0.004

Therefore, Ast =ρ·b·d




Area of Steel Required along X dir ( bot) = 2773 mm2

Area of Steel Required along X dir (top) = 7472 mm2

Area of Steel Required along Z dir ( bot) = 13256 mm2

5.11.4 Comparison



Percent Dif-ference

Bearing Pressure(KN/m2)

119.6

118.04

119.6

118.04

None


239

1260

1163

237

1260

1169

Negligible

Shear Force, One-Way (KN)

917 917 None

Shear Force, Two-Way (KN)

1602.99

1602.17

1602.99

1602.18

None


5.12 US Pilecap Foundation 15.12.1 Reference

5.12.2 ProblemDesign pilecap foundation with the given data: Load Fy = 330 kip, fc = 4 ksi, fy = 60 ksi,Column Dimension = 12 in x 12 in. Pedestal ht= 20 in.

Dia of pile= 18 in

Vertical capacity = 65 kip

Horizontal capacity = 20 kip

Uplift capacity = 15 kip


Chapter — 5

5.12 US Pilecap Foundation 1


5.12.3 SolutionDepth of pilecap= 1.5 x pile diameter, D = 27 in.

Take D = 31 in.

c/c pile distance = 3 x pile diameter =54 mm

Edge distance =18 inch

Assuming 6 pile combination, coordinates of piles (considering pedestal at 0,0,0):




PileNo

X Coor-dinate (in)

Z Coor-dinate (in)

1 -54 -272 -54 273 0 -274 0 275 54 -276 54 27


Pilecap Length = 2(54)+2(18) = 144 inch

Pilecap Width = 54+2(18) = 90 inch

pilecap dimension is 144 inch x90 inch x 31 inch

Weight of footing = 144 x 90 x 31 x 0.16/123 = 37.2 kip

Weight of pedestal = 12 x 12 x 20 x 0.16 /123 = 0.267 kip

Therefore, total load on the pilecap = (330+37.2 +0.267) =367.467 kip

So Pile reaction = 367.467 /6= 61.244 kip < 65 kip

Hence OK


Factored Design

Load factor for self wt is taken = 1

Load factor for axial load is taken = 1.4

So, Load on pilecap = 1.4(330) + 1(37.2)+0 1(.267) = 499.467

Load on each pile =499.467/6=83.245 kip


Chapter — 5



Punching shear is checked on a perimeter 0.5d =13inch from the column face.

Contribution from pile 1 = from pile2 = from pile5 = from pile 6 = 83.245 kip

Contribution from pile 3 = from pile4 = 0.944 x 83.245 = 78.58 kip

So total punching shear Vm= 490.14 kip

Pm = 4·(12+13+13) =152 inch

ß = L/B =12/12 =1

V1= (2 + 4/βc)√(fc)·bc·d = (2+4/1)·√(4000)·152·26 =1499.6·(10)3 lb

V2 = (40·d/b + 2)√(fc)·bc·d = (40·26/152 + 2)·√(4000)·152·26 = 2210·(10)3 lb

V3= 4.√(fc)·bc·de = 4·√(4000)·152·26= 999.78·(10)3 lb

Vc = min(V1, V2, V3) = 999.78 kip

So, 0.75·Vc = 749.8 kip


Punching Shear Check for Corner Piles

Corner piles are 1,2,5,6




Each pile is taking reaction 83.245 kip

d critical = 5.486 inch

Effective depth = 26 inch

d critical< Effective depth, Hence Safe


Parallel to X Axis

For shear wrt X1X1

Contribution from pile 1 =pile2=83.245 Kip

Contribution from pile 3,4,5,6 =0 Kip

So Total V X1X1 = 166.49 Kip

For shear wrt X2X2

Contribution from pile 5 =pile6=83.245 Kip

Contribution from pile 1,2,3,4 =0 Kip


So V parallel to X direction = 166.49 Kip

Now allowable shear = Vc1 =2√(fc)·bc·d =2·√(4000)·90·26 lb =295.99·(10)3 lb

0.75·Vc1=222 Kip

V < Vc1, Hence Safe

Parallel to Z Axis

For shear wrt Z1Z1

Contribution from pile 1 =pile 6 =pile 7 =83.245 x0.222=18.498 Kip

Contribution from pile 2 =pile 4 =pile 6 =0 Kip

So Total V Z1Z1 = 55.5 Kip

For shear wrt Z2Z2

Contribution from pile 2 =pile 4 =pile 6 =83.245 x0.167=18.498 Kip

Contribution from pile 1 =pile 3 =pile 5 =0 Kip


So V parallel to Z direction = 55.5 Kip

Now allowable shear = Vc1 =2√(fc)·bc·d =2·√(4000)·90·26 lb =295.99·(10)3 lb

0.75·Vc1 = 222 Kip

V < Vc1, Hence Safe


Chapter — 5



For either direction:

m = fy/0.85·fc= 60/(0.85·4) = 17.65

ß1 = 0.85 (as fc = 4)

ρbal = 0.85·ß1·fc·87/fy·(87 + fy) = 0.028507

ρmax = 0.75·ρbal = 0.02138

ρmin = 0.0018 ( as fy= 60 Ksi)


Calculation of Mz-

For moment wrt X1X1

Contribution from pile 1=from pile2=83.245 x 48=3995.76 in·kip

Contribution from pile 3=from pile 4=12 in·kip

Contribution from pile 5,6=0 in·kip

So Total Mz X1X1 = 8014 in·kip = 670 ft·kip

For moment wrt X2X2

Contribution from pile 5=from pile6=83.245 x 48=3995.76 in·kip

Contribution from pile 3=from pile 4=12 in·kip





So Total Mz X2X2 = 8014 in·kip = 670 ft·kip

So Resisting Moment = Mnz= Muz/φ = 744.44 ft·kip

Rn = Mnz/b·d2 = 0.147 Ksi

2·m·Rn/fy = 0.0864 <1 , Hence OK

ρ = 1/m· (1-√1-2m·Rn/fy) = 0.0025

ρmin< ρ < ρmax, Hence OK


Calculation of Mx-

For moment wrt Z1Z1

Contribution from pile 1=from pile 3=from pile 5= 83.245 x 21=1748.145in·kip

So Total Mx Z1Z1 = 437 ft·kip

For moment wrt Z2Z2

Contribution from pile 2=from pile 4=from pile 6= 83.245 x 21=1748.145in·kip

So Total Mx Z2Z2 = 437 ft·kip

So Max value of MX = 437 ft·kip

So Resisting Moment = Mnx= Mxux/φ = 485.55 ft·kip

Rn = Mnz/b·d2 = 0.06 Ksi

2·m·Rn/fy = 0.035 <1 , Hence OK

ρ = 1/m· (1-√1-2m·Rn/fy) = 0.001

So ρ= ρmin=0.0018

ρ < ρmax, Hence OK


Along X Direction

ρ = 0.0025

b = 90 in, d=26 in

Therefore, Astx = 5.85 in2

Along Z Direction

ρ = 0.0018


Chapter — 5


b = 144 in , d=26 in

Therefore, Astxz = 6.74 in2

5.12.4 Comparison



Percent Dif-ference

Pile Reaction (kip) 83.245 83.244 NoneGoverning Moment(ft·kip)

670

437

718

502.6

15%

Shear Force, One-Way(kip)

166.49

55.5

166.49

55.5

None

Shear Force, Two-Way(kip)




5.13.2 ProblemDesign pilecap foundation with the given data: Load Fy = 150 kip, Mx= 50 ft·kip. fc = 4ksi, fy = 60 ksi, Column Dimension = 12 inchx12 inch. Pedestal ht= 20 inch

Pile Data- Dia of pile= 18 inch

Vertical capacity = 65 kip,

Horizontal capacity = 20 kip

Uplift capacity = 15 kip






Chapter — 5


5.13.3 Solutiondepth of pilecap= 1.5xpiledia, D=27 inch

Take D=27 inch

c/c pile diatance = 3xpile dia =54 mm. Edge diatance =18 inch

Assume a 4 pile combination.

Coordinates of piles; considering pedestal at 0,0,0

PileNo

X Coor-dinate (in)

Z Coordinate(in)

1 -27 -272 -27 273 27 274 27 -27


Pilecap Length = 54+2x18 = 90 inch

Pilecap Width = 54+2x18 = 90 inch

pilecap dimension is 90 inch x90 inch x 27 inch

Weight of footing = 90 x 90 x 27 x 0.16/123 = 20.25 kip

Weight of pedestal = 12 x 12 x 20 x 0.16 /123 = 0.267 kip

Therefore, total load on the pilecap = (150+20.25 +0.267) =170.517 kip

So Pile reaction from axial load = 170.517 /4= 42.629 kip

Moment Mx ( from input) = 50 ft·kip =600 in·kip

Using Rivet theory-

Reaction from moment= +/- 600x0.27/(4x272) = +/-5.55 Kip

So,

Reaction at Pile 2= reaction at pile 3 = 42.629 +5.55 = 48.185 Kip

Reaction at Pile 1= reaction at pile 4 = 42.629 -5.55 = 37.085Kip

So Critical vertical reaction= 48.185 Kip < 65 kip

As there is no net uplift/lateral load, so each pile is safe in uplift & lateral capacity.

Factored Design



So, Axial Load on pilecap = 1.4x150+20.25 x1+0.267x1= 230.517 Kip

Moment on pilecap = 1.4x50 ft·kip = 840 in·kip

Load on each pile from axial reaction =230.517 /4=57.629 Kip

Reaction from moment= +/- 840x27/(4x272) = +/-7.78 Kip




So,

Reaction at Pile 2= reaction at pile 3 = 57.629 +7.78 = 65.409 Kip

Reaction at Pile 1= reaction at pile 4 = 57.629 -7.78 = 49.849 Kip

Punching ShearFigure 5-27: Section considered for two-way shear

Punching shear is checked on a perimeter 0.5d =11.5 inch from the column face.

Contribution from pile 1=from pile 4= 49.849 Kip

Contribution from pile 2=from pile 3= 65.409 Kip

So total punching shear Vm= 230.516 kip

Pm = 4x(12+11+11) =136 inch

ß=L/B =12/12 =1


Chapter — 5


V1 √fc.bc.d = (2+4/1).√(4000) x136 x 22 =1135.4 x1000 lb

V2 √fc.bc.d = (40x22/136+2).√(4000) x136x22= 1602.9 x 1000 lb

V3 = 4.√fc.bc.de = 4.√(4000) x 136 x 22= 756.9 x 1000 lb

Vc = min{V1, V2, V3} = 756.9 kip

So, 0.75Vc = 567.7 Kip



Corner piles are 1,2,3,4

Pile 1/4 are taking reaction 49.849 Kip & Pile 2/3 are taking reaction 65.409 Kip

d critical = 4.46 inch

Effective depth = 22 inch

d critical< Effective depth, Hence Safe




Calculation of ShearFigure 5-28: Section considered for one-way shear

Parallel to X Axis

For shear wrt X1X1

Contribution from pile 1 =49.849 x 0.444=22.155 Kip


Contribution from pile 3,4 =0 KN


For shear wrt X2X2






Chapter — 5


So V parallel to X direction = 51.195 Kip

Now allowable shear = Vc1 = = 2x√(4000) x 90 x 22 lb = 250.4 x1000 lb

0.75.Vc1=187.84 Kip

V < Vc1, Hence Safe

Parallel to Z Axis

For shear wrt Z1Z1





For shear wrt Z2Z2





So V parallel to Z direction = 58.08 kip

Now allowable shear = Vc1 = = 2x√(4000) x 90 x 22 lb = 250.4 x1000 lb

0.75.Vc1 = 187.8 Kip

V < Vc1, Hence Safe




Calculation of MomentFigure 5-29: Section considered for bending

About Z Axis

Calculation of Mz-

For moment wrt X1X1

Contribution from pile 1=49.849 x21/12=87.24 ft·kip



So Total Mz X1X1 = 207.1 ft·kip

For moment wrt X2X2



Chapter — 5




So Total Mz X2X2 = 207.1 ft·kip

So max Mz = 207.1 ft·kip

So Resisting Moment =Mnz= Muz/Ø =224.11 ft·kip

m=fy/0.85.fc=60/(0.85x4) = 17.6471

ß1 = 0.85 ( as fc=4)

ρbal = 0.85.ß1.fc.87/fy.(87+fy) = 0.028507

ρmax = 0.75. ρbal = 0.02138

ρmin = 0.0018 ( as fy=60)


2m.Rn/fy = 0.0363 <1 , Hence OK

ρ = 1/m. (1-√1-2m.Rn/fy) = 0.001

So, ρ =ρmin = 0.0018

ρ < ρmax Hence OK

About X Axis

Calculation of Mx-

For moment wrt Z1Z1

Contribution from pile 1=from pile 4=49.849 x 21/12=87.24 ft·kip

Contribution from pile 2=from pile 3=0 ft·kip

So Total Mx Z1Z1 = 174.47 ft·kip

For moment wrt Z2Z2

Contribution from pile 2=from pile 3=65.409 x 21/12=114.465 ft·kip

Contribution from pile 1=from pile 4=0 ft·kip

So Total Mx Z2Z2 = 228.93 ft·kip

So Max value of MX = 228.93 ft·kip

So Resisting Moment =Mnz= Muz/Ø =254.37 ft·kip

m=fy/0.85.fc=60/(0.85x4) = 17.6471

ß1 = 0.85 ( as fc=4)

ρbal = 0.85.ß1.fc.87/fy.(87+fy) = 0.028507

ρmax = 0.75. ρbal = 0.02138

ρmin = 0.0018 ( as fy=60


2m.Rn/fy = 0.0412 <1 , Hence OK




ρ = 1/m. (1-√1-2m.Rn/fy) = 0.0012

So ρ= ρmin=0.0018

ρ < ρmax Hence OK


Along X Direction

ρ = 0.0018

b= 90 inch , d=22 inch

Therefore, Astx = 3.6 in2

Along Z Direction

ρ = 0.0018

b= 90 inch , d=22 inch

Therefore, AstZ = 3.6 in2

5.13.4 Comparison

Value of Reference ResultSTAAD Foun-

dationResult

Percent Dif-ference

Pile Reaction 49.849 Kip

65.409 Kip

49.851 Kip

65.407 Kip

None

GoverningMoment

207.1 ft·kip 228.93ft·kip

219.654 ft·kip

294.254 ft·kip

Negligible


51.195 Kip 58.08Kip

51.226 Kip

58.14 Kip

None


230.516 kip 230.517 kip None



5.14.2 ProblemDesign pilecap foundation with the given data: Load Fy = 1000 KN, fc = M 25, fy = Fe 415

Column Dimension = 250 mm x 250 mm, Pedestal ht= 500 mm


Chapter — 5


Dia of pile= 300 mm





5.14.3 SolutionDepth of pilecap= 1.5 x pile diameter, D = 450 mm

Take D = 655 mm

c/c pile distance = 3 x pile diameter =1200 mm

Edge distance =350 mm

Assuming five pile combination, coordinates of piles (considering pedestal at 0,0,0):

PileNo

X Coordinate(mm)

Z Coordinate(mm)

1 -848.528 -848.5282 -848.528 848.5283 0 04 848.528 -848.5285 848.528 848.528





Pilecap Length = 2(850)+2(350) = 2400 mm

Pilecap Width = 2(850)+2(350) = 2400 mm

pilecap dimension is 2400 mm x 2400 mm x 740 mm


Weight of pedestal = 0.25 x 0.25 x 0.5 x 25 = 0.78 KN

Therefore, total load on the pilecap= (1000+106.56 +0.78) =1107.34 KN

So Pile reaction = 1107.34 /5 = 221.47 KN < 250 KN

Hence OK


Factored Design



So, Load on pilecap = 1.4(1000) + 1(106.56)+0 1(0.78) = 1507.34 KN

Load on each pile = 1507.34/5 = 301.468 KN

deff = 740 - 50 - 50 - 25 = 615 mm



Contribution from pile 1 = from pile2 = from pile4 = from pile 5 = 301.468KN


So total punching shear Vm= 1205.9 KN

Pm = 4·(250 + 307.5 + 307.5) =3460 mm

ß = L/B = 250/250 = 1


Chapter — 5


V1= (2 + 4/βc)√(fc)·bc·d = 5298.5 KN

V2 = (40·d/b + 2)√(fc)·bc·d = 5530.9 KN

V3= 4.√(fc)·bc·de = 3532 KN

Vc = min(V1, V2, V3) = 3532 KN

So, 0.75·Vc =2649 KN



Corner piles are 1, 2, 4, & 5

Each pile is taking reaction 301.47 KN

B1 = 1418.27 mm & B2 =2025 mm

d1 = 151 mm

d2 = 211 mm

d critical =212 mm

Effective depth = 615 mm

d critical < Effective depth, Hence Safe


Parallel to X Axis

Shear Plane is at a distance d (615 mm from face of column)

For shear wrt X1X1


Contribution from pile 3,4,5 =0 KN





For shear wrt X2X2

Contribution from pile 4 = pile 5 = 0.8667 x 301.47=261.27 KN

Contribution from pile 3,2,1 =0 KN

So, Total V X2X2 = 522.548 KN

Shear, V, parallel to X direction = 522.548 KN

Now allowable shear = Vc1 =2√(fc)·bc·d = 1225 KN

0.75·Vc1= 918.8 KN

V < Vc1, Hence Safe

Parallel to Z Axis

For shear wrt Z1Z1

Contribution from pile 1 =pile 4 = 0.8667 x 301.47 = 261.27 KN

Contribution from pile 2 = pile 3 = pile 5 = 0 KN


For shear wrt Z2Z2


Contribution from pile 1 = pile 3 = pile 4 =0 KN


Shear, V, parallel to Z direction = 522.548 KN

Now allowable shear = Vc1 =2√(fc)·bc·d = 1225 KN

0.75·Vc1= 918.8 KN

V < Vc1, Hence Safe



m = fy/0.85·fc= 415/(0.85·25) = 19.529

ß1 = 0.85 (as fc = M 25)

ρbal = 0.85·ß1·fc·87/fy·(87 + fy) = 0.025729

ρmax = 0.75·ρbal = 0.019297

ρmin = 0.0018 (as fy= 415 N/mm2)


Chapter — 5


Figure 5-33: Section considered for moment


Calculation of Mz-

For moment wrt X1X1

Contribution from pile 1 = pile 2 = 301.47 x 0.725 = 218.56 KNm

Contribution from pile 3 = 0.083 x 301.47 x 0.0083 = 0.21 KNm

Contribution from pile 4= pile 5 = 0 KNm

So Total Mz X1X1 = 437.34 KNm = 440 KNm

For moment wrt X2X2

Contribution from pile 4 = pile 5 = 301.47 x 0.725 = 218.56 KNm


Contribution from pile 4 = pile 5 =0 KNm


So Resisting Moment = Mnz= Muz/φ = 488.89 KNm = 490 KNm

Rn = Mnz/b·d2 = 0.5398 N/mm2

2·m·Rn/fy = 0.0508 <1 , Hence OK

ρ = 1/m· (1-√1-2m·Rn/fy) =0.00132

ρ= ρmin=0.0018



Calculation of Mx-

For moment wrt Z1Z1

Contribution from pile 1= pile 4 = 301.47 x 0.725 = 218.56 KNm

Contribution from pile 3=0.083 x 301.47 x 0.0083 = 0.21 KNm




Contribution from pile 2 = pile 5 = 0 KNm

So Total Mx Z1Z1 = 437.34 KNm = 440 KNm

For moment wrt Z2Z2

Contribution from pile 2= pile 5 = 301.47 x 0.725 = 218.56 KNm




So Max value of MX = 440 KNm

So Resisting Moment = Mnx= Mxux/φ = 488.89 KNm =490 KNm

Rn = Mnz/b·d2 = 0.5398 N/mm2

2·m·Rn/fy = 0.0508 <1 , Hence OK

ρ = 1/m· (1-√1-2m·Rn/fy) = 0.00132

So ρ= ρmin=0.0018



Along X Direction

ρ = 0.0018

b = 2400 mm, d = 615 mm

Therefore, Astx = 2657 mm2

Along Z Direction

ρ = 0.0018

b = 2400 mm, d = 615 mm

Therefore, Astz = 2657 mm2


Chapter — 5


5.14.4 Comparison



Percent Dif-ference






489

489

485

485

Negligible


523

523

520

520

Negligible


1206 1206 None



5.15.2 ProblemDesign pilecap foundation with the given data: Load Fy = -1200 KN & Mz = 100 KNm, fc =M 25, fy = Fe 415

Column Dimension = 250 mm x 250 mm, Pedestal ht= 500 mm

Dia of pile= 300 mm








5.15.3 SolutionDepth of pilecap= 1.5 x pile diameter, D = 450 mm

Take D = 780 mm

c/c pile distance = 3 x pile diameter = 900 mm

Edge distance =350 mm

Assuming six pile combination, coordinates of piles (considering pedestal at 0,0,0):


Chapter — 5


PileNo

X Coordinate(mm)

Z Coordinate(mm)

1 -900 -4502 -900 4503 0 -4504 0 4505 900 -4506 900 450


Pilecap Length =2(900) + 2(350) = 2500 mm

Pilecap Width = 900 + 2(350) = 1600 mm

pilecap dimension is 2500 mm x 1600 mm x 780 mm

Weight of footing = 2.5 x 1.6 x 0.78 x 25 = 78 KN

Weight of pedestal = 0.25 x 0.25 x 0.5 x 25 = 0.78 KN

Therefore, total load on the pilecap = 1200 + 78 + 0.78 = 1278.78 KN

So Pile reaction from axial load = 1278.78/6= 213.13 KN

So Pile reaction at Pile1 /Pile 2 = 213.13 +100·0.9/(4·0.9·0.9) = 240.91 KN

So Pile reaction at Pile3 /Pile 4 = 213.13 KN

So Pile reaction at Pile5 /Pile 6 = 213.13 - 100·0.9/(4·0.9·0.9) = 185.35 KN

So Critical Pile reaction = 240.91 KN < 250 KN

Hence OK

As there is no lateral load , moment or uplift force, so each pile is safe in lateral & upliftcapacity. Similarly, as there is no tension pile case so each pile is safe again uplift.

Factored Design



So, Load on pilecap = 1.4(1200) + 1(78) + 1(0.78) = 1758.78 KN

So Pile reaction from axial load = 1758.78/6 = 293.13 KN

Moment (Mz) = 1.4(100) = 140 KNm

So Pile reaction at Pile 1 /Pile 2 = 293.13 + 140·0.9/(4·0.9·0.9) = 332.02 KN

So Pile reaction at Pile 3 /Pile 4 = 293.13 KN

So Pile reaction at Pile 5 /Pile 6 = 293.13 -140·0.9/(4·0.9·0.9) = 254.24 KN

deff = 780 - 50 - 50 - 25 = 655 mm

Punching Shear






Contribution from pile 1 = pile 2 = 1(332.02) =332.02 KN

Contribution from pile 3 = pile 4= 0.492(293.13) = 144.22 KN

Contribution from pile 5 = pile 6 = 1(254.24) = 254.24 KN

So total punching shear Vm= 2(333.02 + 144.22 + 254.24) = 1462.96 KN

Pm = 4·(250 + 327.5 + 327.5) = 3620 mm

ß = L/B = 250/250 = 1

V1= (2 + 4/βc)√(fc)·bc·d = 5904 KN

V2 = (40·d/b + 2)√(fc)·bc·d = 9090 KN

V3= 4.√(fc)·bc·de = 3936 KN

Vc = min(V1, V2, V3) = 3936 KN

So, 0.75·Vc = 2952 KN



Corner piles are 1, 2, 5, & 6

Critical Pile 1 & 2 with axial Reaction 333.02 KN

B1 = 1450 mm & B2 =2105 mm

d1 = 163 mm

d2 = 225 mm

d critical =225 mm


Chapter — 5


Effective depth = 655 mm

d critical < Effective depth, Hence Safe


Parallel to X Axis

Shear Plane is at a distance d (655 mm from face of column)

For shear wrt X1X1

Contribution from pile 1 = pile 2 =0.9x 333.03 = 299.72 KN

Contribution from piles 3, 4, 5, & 6 =0 KN

So Total V X1X1 = = 2(299.72) KN= 599.4 KN

For shear wrt X2X2

Contribution from pile 5 = pile 6 = 0.9(254.24) = 228.82 KN

Contribution from piles 1, 2, 3, & 4 = 0 KN

So, Total V X2X2 = 2(228.82) = 457.63 KN

Shear, V, parallel to X direction = 599.4 KN




Now allowable shear = Vc1 =2√(fc)·bc·d = 869.8 KN

0.75·Vc1= 652 KN

V < Vc1, Hence Safe

Parallel to Z Axis

For shear wrt Z1Z1

Contribution from all piles = 0 KN


For shear wrt Z2Z2

Contribution from all piles =0 KN


Shear, V, parallel to Z direction = 0 KN

No shear parallel to the Z axis, Hence Safe



m = fy/0.85·fc= 415/(0.85·25) = 19.529

ß1 = 0.85 (as fc = M 25)

ρbal = 0.85·ß1·fc·87/fy·(87 + fy) = 0.025729

ρmax = 0.75·ρbal = 0.019297

ρmin = 0.0018 (as fy= 415 N/mm2)


Chapter — 5


Figure 5-37: Section considered for moment


Calculation of Mz

For moment wrt X1X1

Contribution from pile 1 = pile 2 = 333.02(0.775) = 258.1 KNm

Contribution from pile 3 = pile 4 = 0.083(293.13·0.0083) = 0.2 KNm



For moment wrt X2X2

Contribution from pile 5 = pile 6 = 254.24(0.775) = 197 KNm

Contribution from pile 3 = pile 4 = 0.083(293.13·0.0083) = 0.2 KNm

Contribution from pile 1 = pile 2 =0 KNm


So Critical value of Mz =520 KNm

So Resisting Moment = Mnz= Muz/φ = 577.78 KNm = 580 KNm

Rn = Mnz/b·d2 = 0.845 N/mm2

2·m·Rn/fy = 0.0795 <1 , Hence OK

ρ = 1/m· (1-√1-2m·Rn/fy) = 0.002

ρ = ρmin=0.0018

ρmin < ρ < ρmax, Hence OK





Calculation of Mx

For moment wrt Z1Z1

Contribution from pile 1 = 332.02(0.325) = 107.9 KNm



Contribution from piles 2, 4, & 6 = 0 KNm

So Total Mx Z1Z1 =285.8 KNm = 286 KNm

For moment wrt Z2Z2




Contribution from piles 1, 3, & 5 = 0 KNm


So Max value of Mx = 286 KNm

So Resisting Moment = Mnx= Mxux/φ = 317.78 KNm =318 KNm

Rn = Mnz/b·d2 = 0.2964 N/mm2

2·m·Rn/fy = 0.0279 <1 , Hence OK

ρ = 1/m· (1-√1-2m·Rn/fy) = 0.00072

So ρ= ρmin=0.0018



Along X Direction

ρ = 0.0020

b = 1600 mm, d = 655 mm

Therefore, Astx = 2096 mm2

Along Z Direction

ρ = 0.0018

b = 2500 mm, d = 655 mm

Therefore, Astz = 2948 mm2


Chapter — 5


5.15.4 Comparison



Percent Dif-ference


241

241

213

213

185

185

241

241

213

213

185

185

Negligible


332

332

293

293

254

254

332

332

293

293

254

254

None


578

318

572

318

Negligible


600

0

599

0

Negligible


1463 1462 Negligible


5.16 US Mat Combined Foundation 15.16.1 Reference‘Foundation Analysis and Design – Fifth Edition’ by J.E. Bowles, Page 475-481, Example9-1.

5.16.2 ProblemDesign a rectangular combined footing using the conventional method given. fc’ = 3000psi(column and footing), fy = 60000 psi, qa = 2 ksf.


5.16 US Mat Combined Foundation 1


Figure 5-38: Elevation and dimensions

Column 1 Column 212 in. x 12 in. 15 in. x 15 in.4 No. 7 bars 4 No. 8 barsDL = 60 kips DL = 110 kipsLL = 60 kips LL = 90 kips

5.16.3 Solution

Step 1- Find out ultimate soil pressure

Pu 1 = 1.2 x 60 + 1.6 x 60 = 168 kips

Pu 2 = 1.2 x 110 + 1.6 x 90 = 276 kips

ΣPw = P1 + P2 = 60 + 60 + 110 + 90 = 320 kips

Ultimate ratio UR= (168 + 276) / 320 = 1.3875

qult = qa x UR= 2 x 1.3875 = 2.775 ksf

This is necessary so that eccentricity is not introduced in finding L using working loadsand then switching to “ultimate” values.

Step 2 - Find footing dimensions L and B; first locate loadresultant from center of column 1

ΣMcol1 = R where R = Pu = 168 + 276 = 444 KN

444 = 15 x 276

= 9.324 ft

To make the resultant 444 kips (factored loads) fall at L/2:

L = (9.324 + 0.5) x 2 = 19.648 ft

We will use 19.75 ft.


Chapter — 5


Step 3 - Find B

BLq = 444

B = = 8.143 ft

Use B = 8.17 ft.

Step 4 - Find out bending moment and shear force

Uniformly distributed upward load = (Pu 1 + Pu 2) / 19.75 = 22.48 kip / ft

Shear between column 1 and 2 using integration (set integration constant by inspection):

dV =

V = 22.48 x – 168

V = 0 at x = = 7.47 ft (locates point of Mmax)

Moment between column 1 and 2 (again set integration constant by inspection):

dM =

M = – 168 (x – 0.5)

At x = 1(right face of column 1),

M = -72.76 ft-kips

Maximum negative M at V = 0 is

M = - 168 (7.47 – 0.5) = -543.76 ft-kips

These values are Mu values and can be directly used to compute steel quantities.




Figure 5-39: Forces on foundation

Maximum positive bending moment is at the face of column 2

= 22.48 x (3.625)2 / 2 = 147.7 ft-kips

Maximum shear force at the face of column 2

= 22.48 x 14.875 – 168 = 166.39 kips

Figure 5-40: Shear and Bending diagrams

Step 5 - Select depth based on analysis for both wide-beam and diagonal tension

a. Critical location for wide-beam is readily obtained.

b. Diagonal tension may have to be investigated for three conditions:i. three-side zone, column 1

ii. four-side zone, column 2

iii. three-side zone, column 2

Check wide beam first (slope of shear diagram = constant) using V diagram:

B vc d = 166.39 – 22.48 d

vc = 11.825 ksf (allowable = )

8.17 x 11.825 d = 166.39 – 22.48 d


Chapter — 5


d = = 1.397 ft (16.765 in)

Checking diagonal tension at column 1 using “d” just obtained for a three-side zone and

vc = = 164.317 psi = 23.66 ksf

Perim. = (12 + 8.38) x 2 + 12 + 16.765 = 69.525 in = 5.794 ft

The net shear is column load – upward soil force in diagonal tension zone:

A = = 4.07 ft2

V = Pcol -Psoil = 168 – 4.07 x = 156.8 kips

Actual v = = 19.37 ksf < 23.66 ksf (O.K.)

At column 2 a four side zone gives

A = = 7.01 ft2

And

V = 276 – 7.01 x = 256.74 kips

Perim. = (15 + 16.765) x = 10.59 ft

v = = 17.354 ksf < 23.66 ksf (O.K.)

By inspection a three-side diagonal tension is not critical at column 2.

Step 6 - Design negative steel (between columns 1 and 2)

For fy = 60 and fc’ = 3 ksi and b = 12 in, obtain a = 1.96As:

As (d - a/2) =

As (16.765 – 0.98 As) = = 14.79

As = 0.94 in2 / ft




p = = 0.00467 > (O.K.) < maximum allowable percent ofsteel

Use 12 No. 8 bars at 8.2 –in spacing across top of footing:

As = 12(0.79) = 9.48 > 7.68 in2

Run 1/3 of bars full length of footing (less 3 in end cover):

Ld is O.K.

Step 7 - Find steel in short direction in column 1

B’ = 12 + 16.765 x 0.75 = 24.574 in = 2.05 ft

L’ = = 3.585 ft

q = = 10.03 ksf (conservative)

M = x 12 = 773.45 in-kips

Take d = (16.765 – 1) in to allow for longitudinal rebars:

As (15.765 – 0.98 As) = = 14.323

As = 0.98 in2 / ft

p is O.K. from previous calculation. Use 4 No. 7 bars at 6 in:

Ld = 0.04 x 0.60 x 60000 x = 26.3 in

or

Ld = 0.004 x 0.875 x 60000 = 21 in

Ld furnished = 3.585 x 12 – 3 = 40 in

Compute short direction steel at column 2; use d= 15 in:

B’ = 15 + 16 x 1.5 = 39 in = 3.25 ft

L’ = = 3.46 ft

q = = 10.394 ksf (conservative)


Chapter — 5


M = x 12 = 746.6 in-kips

As (15.765 – 0.98 As) = = 14.379

As = 0.98 in2 / ft

Use 6 No. 7 bars at 6.5 in

Step 8 - Check dowel requirements of column to footing

At column 1 the supporting area is not on all sides; therefore the bearing stress is limitedto:

fc = 0.75 x 0.7 x fc’ = 1.575 ksi

p = 12 x12 x 1.575 = 226.8 > 168 kips (dowels not required for load transfer)

Use 4 dowels to provide at least 0.05Ag:

As = 0.005 x 144 = 0.72 in2

Use 4 No. 6 for 4 x (0.44) = 1.76 in2 at column 2 with concrete all around

> 2

Use 2

fc = 0.75 x 0.7 fc x 2 = 3.15 ksi

P = 15 x 15 x 3.15 = 708.75 >> 276

Use four dowels same size as column 1.

Step 9 - Conclusion

Steel in cantilever portion is found to be 0.28 for moment and 0.67 in2 / ft for minimumrequirement of 200 / fy (could use 0.28 x 1.33 >= T and S alternatively but would requireincreasing other steel by 1 / 3 also)

Use 10 No. 7 bars

Run five bars full length to use as chairs for short direction steel.




5.16.4 Comparison



Percent Dif-ference

Average Base Pressure 0.01911kip/in2

0.02 kip/in2 4.657

Maximum NegativeBending Moment

543.76 ft-kips

529.064 ft-kips

2.70

Maximum PositiveBending Moment

147.7 ft-kips 149.12 ft-kips 0.96

Required Area ofReinforcement

0.94 in2/ft 0.909 in2/ft 3.3


5.17 US General Isolated Foundation withSliding & Overturning

5.17.1 Reference


Density of soil =14 KN/cum


Density of Concrete = 25 KN/cum

Coefficient of Friction (m) = 0.5

Safe Bearing Capacity of the Soil (s) = 120 KN/m2







Loads:

Fx =-300 KN

Fy =-500 KN

Fz=-200 KN

Mz= -10.32KNm


Chapter — 5

5.17 US General Isolated Foundation with Sliding & Overturning

Mx=45.89 KNm

Note: Self weight of footing, wt of soil and surcharge are not included for shear andmoment computations.



Determination of base area of footing






Required base area of footing = [500 + 0.10(500)]/120 = 4.5833 m2

Use a 3.75m 3.75m square footing (Af =14.0625 m2).


Self-wt. of Footing =(3.75*3.75*1*25) = 351.562 KN

Wt. of soil = 14*0.5*{(3.75*3.75) –(0.3*0.3)} =97.8075 KN


The net moments are given by:-

Mz = -10.32 – (-300*1) = 289.68 KNm

Mx= +45.89 +(-200*1) = -154.11 KNm


s 1 = ((500+351.562)/ 14.0625) + (6*154.11/3.753) - (6*289.68/3.753) = 75.980 KN/m2

s 2 = ((500+351.562)/ 14.0625) - (6*154.11/3.753) - (6*289.68/3.753) = 10.0621 KN/m2

s 3 =((500+351.562)/ 14.0625) - (6*154.11/3.753) + (6*289.68/3.753) = 45.131 KN/m2

s 4 = ((500+351.562)/ 14.0625) + (6*154.11/3.753) + (6*289.68/3.753) = 111.049 KN/m2


Stability Check

Calculation for Overturning and Sliding

For Sliding

Along X- Direction:


Restoring Force = μ*(Wt of Footing + Fy + Wt of Soil) = 474.6848 KN


Along Z- Direction:


Restoring Force = μ*(Wt of Footing + Fy + Wt of Soil) = 474.6848 KN



Chapter — 5


For Overturning

About X- Direction:

Overturning Moment = Mx + Fz* (Ht of Pedestal + Depth of Footing) = 45.89– 200* (0 +1) = -154.11 KN-m

Restoring Moment = (Fy + Wt of Soil + Wt of Footing) *Width of Footing*0.5= 1780.068 KN-m

Hence, Factor of Safety against Overturning = (1780.068/154.11) =11.5506 > 1.5 Hence Safe

About Z- Direction:

Overturning Moment = Mx + Fz* (Ht of Pedestal + Depth of Footing) =-10.32+ 300* (0 +1) = 289.68 KN-m

Restoring Moment = (Fy + Wt of Soil + Wt of Footing) *Width of Footing*0.5= 1780.068 KN-m

Hence, Factor of Safety against Overturning = (1780.068/289.68) =6.1449 > 1.5 Hence Safe

Check For Shear




s 1 = (500/ 14.0625) + (6*154.11/3.753) - (6*289.68/3.753) = 50.9804 KN/m2

s 2 = (500/ 14.0625) - (6*154.11/3.753) - (6*289.68/3.753) = -14.9378 KN/m2

s 3 =(500/ 14.0625) - (6*154.11/3.753) + (6*289.68/3.753) = 20.1307 KN/m2

s 4 = (500/ 14.0625) + (6*154.11/3.753) + (6*289.68/3.753) = 86.049 KN/m2

Average pressure:-

qu = (s 1 +s 2 +s 3+s 4) / 4 = 35.555 KN/ m2

Figure 5-42: Critical section for punching shear is at d/2




Depth requirement for shear usually controls the footing thickness. Both wide action andtwo-way action (punching shear) for strength computation need to be investigated todetermine the controlling shear criteria for depth.


Wide-beam action (One-Way Shear) :

Along Z-Z axis: -


Bw = 3.75m = 147.6378 in

qs is given by:-

Average Base Pressure along one edge = (50.98041-14.9379)/2 =18.02126KN/m2

Average Base Pressure along other edge = (20.1307+86.049)/2 = 53.0899KN/m2

Approximate Base Pressure at the critical section = 53.0899- {(53.0899 -18.02126)/3.75*0.805} =45.56183 KN/ m2 [0.805=3.75-(3.75/2+0.92+0.15)]


Vux =3.75*{45.56183*0.805+0.5*(53.0899-45.56183)*0.805}= 148.9025 KN

Along X-X axis:-


Bw = 3.75m = 147.6378 in

qs is given by:-

Average Base Pressure along one edge = (86.049+50.98041)/2 =68.5147KN/m2

Average Base Pressure along other edge = (20.1307-14.9379)/2 = 2.5964KN/m2

Approximate Base Pressure at the critical section = 68.5147- {(68.5147-2.5964)/3.75*0.805} =54.36424 KN/ m2[0.805=(3.75/2–0.92–0.15)]

Hence, the one-way shear at the critical section

Vuz =3.75*{54.36424*0.805+0.5*(68.5147-54.36424)*0.805}= 185.47 KN

φVu = φ(2√(f'c)bwd) = 0.75(2√(4349.39) 147.6378 36.22)/1000 =529.0018 kips = 2354.06 KN > Vux and Vuz

Hence O.K.

Two-way action (Punching Shear):



Chapter — 5


Tributary area = 14.0625 - (0.3+ 0.92)·(0.3 + 0.92) = 12.5741 m2

Vu = 35.555 12.5741 = 447.0791 KN = 100.467 Kip.

= Minimum of

bo = 2(0.3+0.3+2*0.92) = 4.88m =192.126 in

= = 5.304348


=

Vc = 0.75 4 x 192.126 36.2205/1000 = 1376.82 kips > Vu= 100.467 kips

O.K.

Calculation for reinforcementFigure 5-43: Critical section for moment is at the face of column




Bending about X Axis Bending About Z Axis

About X Axis

Average Base Pressure along one edge = (50.98041-14.9379)/2 =18.02126KN/m2


Approximate Base Pressure at the critical section = 53.0899- {(53.0899 -18.02126)/3.75*1.725} =36.958 KN/ m2 [1.725= 3.75-(3.75/2+0.15)]


Mu =3.75*{36.958*1.725*1.725*0.5+0.5*(53.0899-36.958)*1.725*1.725*2/3}=266.2033 KNm

Required Rn = KN/m2 =13.5105psi

(gross area) = (d/h) 0.00022 = (0.92/1.0) 0.00027 = 0.000207

Check minimum As required for footings of uniform thickness; for grade 415 reinforcement:

min = 0.00180

Since is < min, so = min =0.00180


Chapter — 5


Required As = bd = 0.00180 3.75 0.92 = 0.00621 m2 = 6210 m2m

About Z Axis


Average Base Pressure along other edge = (20.1307-14.9379)/2 = 2.5964KN/m2

Approximate Base Pressure at the critical section = 68.5147- {(68.5147-2.5964)/3.75*1.725} =38.1923 KN/ m2 [1.725=(3.75/2-0.15)]


Mu =3.75*{38.1923*1.725*1.725*0.5+0.5*(68.5147-38.1923)*1.725*1.725*2/3}=325.87126 KNm

Compute required As assuming tension-controlled section ( = 0.9)

Required Rn = KN/m2 =8.44355psi

p(gross area) = (d/h) 0.00014 = (0.92/1.0) 0.00014 = 0.000129


min = 0.00180

Since is < min, so = min =0.00180

Required As = bd = 0.00180 3.75 0.92 = 0.00621 m2 = 6210 m2m

Try with 16mm bars (Af =201.06 m2m)

So, no. of bars reqd = 6210/ 201.06 =31 bars

Check for development length

Critical section for development length is same as that for moment (at face of column).

As per ACI- 12.2.2 and T-1.12.2 of Reinforced Concrete Design by Salmon-Wang; For # 35Mbars and smaller:




Ld = 0.02 Ab = 0.02* 201.06*(415/300.5) = 304.68 mm

Available development length of the bars = 0.5*(L-Dcol) –C cover = 1675 mm> 304.68 mm.

Hence OK

Clear cover (bottom and side) = 50 mm

Center-to-center bar spacing = {3750-2*(50+8)}/30 = 121.13 mm

This is to be checked with the minimum and maximum spacing permissible.


Chapter — 5


5.17.4 Comparison




Moment about X(KNm)

266.2033 269.53 Error due toapproximation in basepressure interpolation

Moment about Z(KNm)

325.87126 328.41 m Error due toapproximation in basepressure interpolation

Area of steel aboutX-X (mm2)

6210 6106.05 Error due toapproximation in basepressure interpolation

Area of steel aboutZ-Z (mm2)

6210 6241.05 Error due toapproximation in basepressure interpolation

Shear force

(One way) along X(KN)


Shear force

(One way) along Z(KN)


Shear force

(Two way) (KN)






Factor of SafetyagainstOverturning (X)


Factor of SafetyagainstOverturning (Z)



5.18 US General Isolated Foundation withEccentric Loading

5.18.1 Reference



5.18 US General Isolated Foundation with Eccentric Loading


Offset of column in X-direction (Oxd) =300 mm

Offset of column in Z-direction (Ozd) =300 mm

Density of soil =14 KN/m3


Density of Concrete = 25 KN/m3

Coefficient of Friction (m) = 0.5

Safe Bearing Capacity of the Soil (s) = 120 KN/m2







Loads:

Fx =-300 KN

Fy =-500 KN

Fz=-200 KN

Mz= 45.89 KNm

Mx=-98.32 KNm

Note: Self weight of footing, wt of soil and surcharge are not included for shear andmoment computations.


Chapter — 5




Determination of base area of footing:



Required base area of footing = [500 + 0.10(500)]/120 = 4.5833 m2

Use a 5 .0m 5.0m square footing (Af =25 m2).





Self-wt. of Footing =(5.0*5.0*1*25) = 625 KN

Wt. of soil = 14*0.5*{(5.0*5.0) –(0.3*0.3)} =174.37 KN


The net moments are given by:

Mz =+45.89 – (-300*1) +(-500*0.3) = 195.89 KNm

Mx= - 98.32 +(-200*1) –(-500*0.3) = -148.32 KNm

The pressure at the four corners are given by:

σ1 = ((500+625)/ 25) + (6*195.89 /5.03) - (6*148.32/5.03) = 47.28336KN/m2

σ2 = ((500+625)/25) - (6*195.89 /5.03) - (6*148.32 /5.03) = 28.478 KN/m2

σ3 =((500+625)/ 25) - (6*195.89 /5.03) + (6*148.32 /5.03) = 42.7167 KN/m2

σ4 = ((500+625)/ 25) + (6*195.89 /5.03) + (6*148.32/5.03) = 61.522 KN/m2


Stability Check

Calculation for Overturning and Sliding

For Sliding

Along X- Direction:



Hence, Factor of Safety against Sliding = (649.685/300) =2.1656 > 1.5

Hence Safe

Along Z- Direction:



Hence, Factor of Safety against Sliding = (649.685/200) =3.2484 > 1.5

Hence Safe

For Overturning

About X- Direction:

Overturning Moment = Mx + Fz* (Ht of Pedestal + Depth of Footing) = -98.32– 200* (0 +1) = -298.32 KN-m


Chapter — 5



Factor of Safety against Overturning = (3098.425/298.32) =10.386 > 1.5

Hence Safe

About Z- Direction:

Overturning Moment = Mx + Fz* (Ht of Pedestal + Depth of Footing) =45.89+ 300* (0 +1) = 345.89 KN-m


Factor of Safety against Overturning = (3398.425/345.89) =9.82516 > 1.5

Hence Safe

Check for Shear

Factored Loads and Soil Reaction


The pressure at the four corners are given by:

σ1 = (500/ 25) + (6*195.89 /5.03) - (6*148.32 /5.03) = 22.2834KN/m2

σ2 = (500/25) - (6*195.89 /5.03) - (6*148.32 /5.03) = 3.47792KN/m2

σ3 =(500/ 25) - (6*195.89 /5.03) + (6*148.32 /5.03) = 17.7167 KN/m2

σ4 = (500/ 25) + (6*195.89 /5.03) + (6*148.32 /5.03) = 36.52208 KN/m2

Average pressure:

qu = (σ1 +σ2 +σ3+σ4) / 4 = 20.000 KN/ m2

Figure 5-45: Critical section for punching shear at d/2

Depth requirement for shear usually controls the footing thickness. Both wide action andtwo-way action (punching shear) for strength computation need to be investigated todetermine the controlling shear criteria for depth.





Wide-beam action (One-Way Shear)

Along Z-Z axis: -


Bw = 5.00m = 196.8504 in

qs is given by:

Average Base Pressure along one edge =(22.2834+3.47792)/2 =12.8806 KN/m2

Average Base Pressure along other edge =(17.7167+36.5221)/2 = 27.1194 KN/m2

Approximate Base Pressure at the critical section = 27.1194- {(27.1194 –12.8806)/5.0*1.13} =23.9014 KN/ m2 [1.13=5 -(5/2 +0.3 +0.92 +0.15)]


Vux =5.0*{23.9014*1.13+0.5*(27.1194-23.9014)*1.13}= 144.134 KN

Along X-X axis:


Bw = 5.00m = 196.8504 in

qs is given by:

Average Base Pressure along one edge =(36.5221+22.2834)/2 =29.4027 KN/m2

Average Base Pressure along other edge =(17.7167+3.47792)/2 = 10.5973 KN/m2

Approximate Base Pressure at the critical section = 29.4027- {(29.4027-10.5973)/5.0*1.73}=22.89603 KN/ m2 [ 1.73=(5/2 +0.3 –0.92 –0.15)]

Hence, the one-way shear at the critical section

Vuz =5.0*{22.89603*1.73+0.5*(29.4027-22.89603)*1.73}= 226.1924 KN

φVu = φ (2√(f'c)bwd) = 0.75(2√(4349.39) 196.8504 36.22)/1000 =705.3265 kips = 3138.703 KN > Vux and Vuz

Hence O.K.

Two-way action (Punching Shear)


Tributary area = [25.000 - (0.3 + 0.92)x(0.3 + 0.92)] = 23.5116 m2

Vu = 20.000 x 23.5116 = 470.232 KN = 105.6701 Kip.


Chapter — 5


= Minimum of

bo = 2(0.3+0.3+2*0.92) = 4.88m =192.126 in

= = 5.304348


=

Vc = 0.75 4 x 192.126 36.2205/1000 = 1376.82 kips > Vu= 100.467 kips O.K.

Calculation for reinforcement

Critical section for moment is at the face of column

Bending about X Axis Bending About Z Axis




About X-Axis



Approximate Base Pressure at the critical section = 27.1194- {(27.1194 –12.8806)/5.0*2.05} =21.2815 KN/ m2 [2.05 =5-(5/2+0.3+0.15)]


Mu = 5.0*{21.2815 *2.05*2.05*0.5+0.5*(27.1194-21.2815)* 2.05*2.05*2/3}=264.48 KNm

Required Rn = KN/m2 =10.067 psi

ρ(gross area) = (d/h) 0.0001675 = (0.92/1.0) 0.0001675 = 0.0001541


ρmin = 0.00180

Since ρ is < ρmin, ρ = ρmin =0.00180

Required As = ρbd = 0.00180 5.0 0.92 = 0.00828 m2 = 8280 mm2

About Z- axis:



Approximate Base Pressure at the critical section = 29.4027- {(29.4027-10.5973)/5.0*2.65} =19.4358 KN/ m2 [2.65 =(5/2+0.3-0.15)]


Mu =5.0*{19.4358 *2.65*2.65*0.5+0.5*(29.4027 –19.4358)*2.65*2.65*2/3}=457.874 KNm

Compute required As assuming tension-controlled section ( = 0.9)


Chapter — 5


Required Rn = KN/m2 =17.429 psi

ρ(gross area) = (d/h) 0.00029 = (0.92/1.0) 0.00029 = 0.000267


ρmin = 0.00180

Since ρ is < ρmin , ρ = ρmin =0.00180

Required As = ρbd = 0.00180 5.0 0.92 = 0.00828 m2 = 8280 mm2

Try with 20 mm bars (Af =314.16 mm2)

So, no. of bars required = 8280/ 314.16 =27 bars

Check for development length

Critical section for development length is same as that for moment (at face of column).

As per ACI- 12.2.2 and T-1.12.2 of Reinforced Concrete Design by Salmon-Wang:

Ld = 0.02 Ab (For # 35M bars and smaller)

= 0.02* 314.16*(415/300.5) = 476.0625m

Available development length of the bars= 0.5*(L-Dcol) –C cover = 2375 mm> 476.0625 mm. Hence OK

Clear cover (bottom and side) = 50mm

Center-to-center bar spacing = {5000-2*(50+10)}/26 = 187.6923 mm

This is to be checked with the minimum and maximum spacing permissible.




5.18.4 Comparison




Moment about X 264.48KNm

264.53KNm

Negligible

Moment about Z 457.874KNm

457.83KNm

Negligible

Area of steel aboutX-X

8280mm2

8906.40mm2

Error due toapproximation in basepressure interpolation

Area of steel aboutZ-Z

8280mm2

8321.40mm2

Error due toapproximation in basepressure interpolation

Shear force

(One way) along X

226.1924KN

225.67 KN Error due toapproximation in basepressure interpolation

Shear force

(One way) along Z

144.134KN

143.58 KN Negligible

Shear force

(Two way)

470.232KN

470.01 KN Negligible





Factor of SafetyagainstOverturning (X)


Factor of SafetyagainstOverturning (Z)




Chapter — 5


Section 6

Deadman Anchors (ACI318 -2005)6.1 Deadman Guy Anchor US 1

6.1.1 Reference380 ft Brenham (Wesley), TX guyed tower Design Project

6.1.2 ProblemDesign an anchor block for a guy rod supporting the following load condition

Axial Tension = 83.815 Kip

Slope with Horizontal = 54.231 degree

Min area required for guy rod =1.677 sq.inch, i.e if single guy rod is used,mindia of rod required = 1.5 inch

Necessary Factors of Safety are as follows

FOS against Uplift = 1.5

FOS against sliding = 2

Ultimate Load Factor = 1.3

Material Specification-

Assume Strength of Concrete = 4 Ksi

Strength of Steel = 60 Ksi


Strength of Gye Rod steel = 50 Ksi

unit weight of Concrete = 150 lb/cu.ft

unit weight of Soil = 62.4 lb/cu.ft

Soil & GWT condition-

LayerIndex No

LayerType

Depth ofLayer

Cohesion(psf)

Angle ofFriction

Dry Density(pcf)

1 Sand (0-2) 0 20 1152 Sand (2-15) 0 30 1153 Sand (15-below) 0 30 115

Table 6-1: Soil Test Report Summary

Assume depth of Ground Water Table from GL = 8 ft

Assume soil cone angle of uplift = 30 degree

6.1.3 SolutionFirst let us calculate the Horizontal & Vertical components of Axial Tension at guy rod

Horizontal component of load (H) = P.cos θ = 83.815 x Cos 54.231 = 48.9656 Kip

Vertical component of load (V)= P.sin θ = 83.815 x Sin 54.231 = 68.0245Kip

Properties of soil (divided into relevant small strips each max 1/2 ft thick)


Chapter — 6

6.1 Deadman Guy Anchor US 1

Kp=tan2 (450+ Ø/2)

Kp=tan2 (450- Ø/2)

Note: Ø is in degrees

Pa= γ.h.KpSo Pa or a particular layer = Pa of previous layer + γ.h.KaWhere:

Ka= Active EP coeff of the present layer

γ = Soil density of Soil at present layer

h= Thickness of present layer)

Section 6 Deadman Anchors (ACI 318 -2005)



Thus

Pp= γ.h.Kp + 2C.√KpSo Pp or a particular layer = Pp of previous layer + γ.h.KpWhere:

Kp= Passive EP coeff of the present layer


h= Thickness of present layer

C= Cohesion of Present layer)

Note: Here, γ= Density of soil which is used when soil layer is above GWT

If Soil layer is below GWT then submerged density of soil (γ-γsoil) is used

Adhesion factor α = 0.31 + 0.34/C

Figure 6-1: Deadman Anchor Guy Tension Block section

α <=1

C= Cohesion in Kip/ft2 unit


Chapter — 6


Figure 6-2: Dispersion of soil against vertical uplift diagram

Soil Depth upper lvl oflayer(ft)

Soil Depth lower lvl oflayer(ft)

avg Pp effective onblock surface (psf)

weightedavg Pp(lb/ft)

0 0.5 0 00.5 1 0 01 1.5 0 01.5 2 0 02 2.5 0 02.5 3 0 03 3.5 0 03.5 4 0 04 4.5 0 04.5 5 0 05 5.5 0 05.5 6 0 06 6.5 0 06.5 7 2167.1918 1083.5967 7.5 2340.028 1170.0147.5 8 2512.8643 1256.4328 8.5 2638.8093 1319.4058.5 9 2717.8631 1358.9329 9.5 2796.9169 1398.4589.5 10 2875.9707 1437.98510 10.5 0 010.5 11 0 011 11.5 0 011.5 12 0 012 12.5 0 0








12.5 13 0 013 13.5 0 013.5 14 0 014 14.5 0 014.5 15 0 0

Figure 6-3: Dispersion line diagram

Check for Safety against Sliding

Tot Passive resistance per unit length = 9024.823 lb/ft

Length = 11.5 ft

Tot Passive resistance = 9024.823 x 11.5 /1000 = 103.786 kip

Allowable Horizontal Load on Anchor = 48.9656 Kip

Safety Factor against Horizontal Load = 103.786 / 48.9656 = 2.12

FOS is greater than min required FOS, Hence OK

Check for Safety against Uplift

given value of dispersion angle = 30 degree

Let us consider wedge at all sides with dimension of 0.5 ft

As wedge is present so dispersion is to be started from bottom of block


Chapter — 6


Height of soil above top level of block = 6.5 ft

Height of water effecting the weight of concrete = 10 - 8 = 2 ft

Weight of Concre Block = LxBxHx unit wt of concrete = 11.5 x 4 x 3.5 x 0.15= 24.15 kip

reduction of concrete weight due to buoyancy = 2 x 4 x 11.5 x 62.4/1000 =5.7408 kip

So, total buoyant weight of concrete = 24.15-5.7408 = 18.4092Kip

Weight of soil over top of anchor in a truncated pyramid = 215.93 Kip

Uplift Resistance due to Soil/Concrete Adhesion = 0 Kip

Therefore, Total resistance against uplift = 18.4092 + 215.93 + 0 = 234.34 Kip

Allowable anchor uplift resistance = V = 68.0245 Kip

Net Safety factor = 234.34 / 68.0245 = 3.445

Safety Factor against Horizontal Load = 1.5





Design checks for the top & front face rebar

Top Rebar Design Check

Figure 6-4: Top rebar force diagram

Vertical Force (V)= 68.0245 Kip

Length (L)= 11.5 Kip

Force/Length = w=V/L = 5.91517391304348 Kip

Bending Moment = w.L2/8 = 97.786 ft·kip

Factored Moment M = 97.786 x 1.3 = 127.1218 ft·kip

Strength of concrete = 4 Ksi


Figure 6-5: Bending moment diagram - top

φ = 0.9

ß = 0.85 - 0.05x (fc-4)

0.85>=ß >=0.65

ß = 0.85

width (B) = 4 ft

Effective Depth = Deff = D-clear cover-Tie bar dia -0.5xtop rebar dia

Hence, deff = 38.19 inch

Area of each top rebar = 0.601015625 sq.inch

Total area of Top Rebar = As = 2.4040625 Ksi

a = As.fy/(ß.fck.B) = 0.883847 inch


Chapter — 6


Resisting Moment = M1 = φ.As.fy.(deff-a/2) =408.37 ft·kip

ratio = 408.37 / 127.1218 = 3.213 >1,

Hence OK

Resisting Moment is greater than Factored Moment, Hence Safe

Front face Rebar Design Check


Length (L) = 11.5 ft

Force/Length = w=V/L = 4.25787826086957 Kip/ft


Factored Moment M = 91.5057 ft·kip


Figure 6-6: Bending moment diagram - front face


φ = 0.9

ß = 0.85 - 0.05x (fc-4)

0.85>=ß >=0.65

ß = 0.85

width (B) = 3.5 ft



Area of each front rebar = 0.601015625 sq.inch

Total area of Front Rebar = As = 1.803046875 sq.inch


Resisting Moment = M1 = φ.As.fy.(deff-a/2) = 355.47 ft·kip

ratio = 355.472 / 91.5057 = 3.885 >1, Hence OK





6.1.4 Comparison



Percent Dif-ference

Sliding Resistance (kips) 103.786 103.71 NegligibleUplift Resistance (kips) 234.34 234.495 NegligibleHorizontal FOS 2.12 2.117 NegligibleVertical FOS 3.445 3.448 NegligibleFactored moment resisted by top rebar(ft-kip)

127.122 127.086 Negligible

Max. moment capacity (vertically hog-ging) (ft-kip)


Factored moment resisted by front rebar(ft-kip)


Max. moment capacity (horizontallyhogging) (ft-kip)

355.472 354.856 Negligible

Table 6-2: Deadman Anchor (US) verification example 1 comparison

6.2 Deadman Guy Anchor US 26.2.1 Reference380 ft Brenham (Wesley), TX guyed tower Design Project

6.2.2 ProblemDesign an anchor block for a guy rod supporting the following load condition-



Min area required for guy rod =1.519 sq.inch, i.e if single rod is used,min dia of rodrequired = 1.4 inch


Chapter — 6



Necessary FOS are as follows-










Soil & GWT condition

LayerIndex No

LayerType

Depth ofLayer

Cohesion(psf)

Angle ofFriction

Dry Density(pcf)

1 clay (0-2) 0 0 1062 clay (2-15) 800 0 1063 clay (15-below) 800 0 106

Table 6-3: Soil test report summary



6.2.3 SolutionFirst let us calculate the Horizontal & Vertical components of Axial Tension at gyerod





Vertical component of load (V)= P.sin θ = 75.926 x Sin 49.816 = 58.0229Kip


Kp=tan2 (450 + Ø/2)

Kp=tan2 (450 - Ø/2)

Note: Ø is in degree

Pa= γ.h.Kp


Chapter — 6


So Pa or a particular layer = Pa of previous layer + γ.h.KaWhere:




Thus





C= Cohesion of Present layer

Note: Here γ= Density of soil which is used when soil layer is above GWT



α <=1


Figure 6-8: Dispersion of soil against vertical uplift








0 0.5 0 00.5 1 0 01 1.5 0 01.5 2 0 02 2.5 0 02.5 3 0 03 3.5 0 03.5 4 0 04 4.5 0 04.5 5 0 05 5.5 0 05.5 6 0 06 6.5 0 06.5 7 2317.4172 1158.7097 7.5 2370.4843 1185.2427.5 8 2423.5513 1211.7768 8.5 2460.9987 1230.4998.5 9 2482.8262 1241.4139 9.5 2504.6538 1252.3279.5 10 2526.4814 1263.24110 10.5 0 010.5 11 0 011 11.5 0 011.5 12 0 012 12.5 0 012.5 13 0 013 13.5 0 013.5 14 0 014 14.5 0 014.5 15 0 0

Table 6-4: Soil layers



Length = 14 ft

Tot Passive resistance = 8543.207 x 14 /1000 = 119.605 kip





Chapter — 6




Let us consider no wedge to support uplift load

As wedge is not present so dispersion is to be started from top of block

Height of soil above top level of block = 6.5 ft


Weight of Concrete Block = LxBxHx unit wt of concrete = 14 x 4 x 3.5 x 0.15= 29.4 kip

reduction of concrete weight due to buoyancy = 2 x 4 x 14 x 62.4/1000 =6.9888 kip



Uplift Resistance due to Soil/Concrete Adhesion = 74.08 Kip

Therefore, Total resistance against uplift = 22.4112 + 38.58 + 74.08 = 135.072Kip











Length (L)= 14 Kip







φ = 0.9

ß = 0.85 - 0.05x (fc-4)

0.85>=ß >=0.65

ß = 0.85

width (B) = 4 ft



Hence, deff =38.19 inch


Total area of Top Rebars = As = 2.4040625 Ksi



Chapter — 6


Resisting Moment = M1 = φ.As.fy.(deff-a/2) = 408.37

ft·kip

ratio = 408.37 / 132.0033 = 3.094 >1, Hence OK




Length (L) = 14 ft







φ = 0.9

ß = 0.85 - 0.05x (fc-4)

0.85>=ß >=0.65

ß = 0.85

width (B) = 3.5 ft







ratio = 355.472 / 111.4087 = 3.191 >1, Hence OK





6.2.4 Comparison


STAADFoundation

Result

Percent Dif-ference

Sliding Resistance (Kip) 119.605 119.542 negligibleUplift Resistance (Kip) 135.072 135.123 negligibleHorizontal FOS 2.443 2.44 negligibleVertical FOS 2.328 2.329 negligiblefactored moment resisted by toprebar ( ft·kip)

132.0033 131.963 negligible

Max moment capacity(vertically hogging) (ft·kip)

408.37 407.66 negligible

factored moment resisted byfront rebar (ft·kip)

111.4087 111.454 negligible

Max moment capacity(horizontally hogging) (ft·kip)

355.472 354.865 negligible


6.3 Deadman Guy Anchor US 36.3.1 Reference380 ft Brenham (Wesley), TX guyed tower Design Project

6.3.2 ProblemDesign an anchor block for a guy rod supporting the following load condition



Min area required for guy rod =1.849 sq.inch, i.e if single rod is used,min diaof rod required = 1.6 inch

Necessary FOS are as follows




Material Specification







Chapter — 6



LayerIndex No

LayerType

Depth ofLayer

Cohesion(psf)

Angle ofFriction

Dry Density(pcf)

1 silt (0-2) 500 20 1052 silt (2-15) 500 20 1053 silt (15-

below)500 20 105

Table 6-6: Soil test report summary



6.3.3 SolutionFirst let us calculate the Horizontal & Vertical components of Axial Tension at gyerod


Vertical component of load (V)= P.sin θ = 92.435 x Sin 31.289 = 48.024 Kip





Kp=tan2 (450+Ø/2)

Kp=tan2 (450-Ø/2)


Pa= γ.h.Kp


Chapter — 6



So Pa or a particular layer = Pa of previous layer + γ.h.KaWhere:












α <=1










0 0.5 0 00.5 1 0 01 1.5 0 01.5 2 0 02 2.5 0 02.5 3 0 03 3.5 0 03.5 4 0 04 4.5 0 04.5 5 0 05 5.5 0 05.5 6 0 06 6.5 0 06.5 7 0 07 7.5 2984.5298 1492.2657.5 8 3091.7854 1545.8938 8.5 3167.1708 1583.5858.5 9 3210.6859 1605.3439 9.5 3254.2011 1627.1019.5 10 3297.7162 1648.85810 10.5 3341.2313 1670.61610.5 11 3384.7465 1692.37311 11.5 3428.2616 1714.13111.5 12 3471.7768 1735.888



Chapter — 6






12 12.5 0 012.5 13 0 013 13.5 0 013.5 14 0 014 14.5 0 014.5 15 0 0



Length = 10 ft







Let us consider no wedge to support uplift load

As wedge is not present so dispersion is to be started from top of block




Figure 6-14: Dispersion line diagram

Height of soil above top level of block = 7 ft


Weight of Concre Block = LxBxHx unit wt of concrete = 10 x 4 x 5 x 0.15 =30 kip


So, total buoyant weight of concrete = 30-9.984 = 20.016Kip




Chapter — 6


Therefore, Total resistance against uplift = 20.016 + 61.98 + 69.3 = 151.296Kip





Now Design check for the top & front face rebar are to bedone




Length (L)= 10 Kip






φ = 0.9

ß = 0.85 - 0.05x (fc-4)

0.85>=ß >=0.65

ß = 0.85

width (B) = 4 ft








Total area of Top Rebars = As = 2.4040625 Ksi



ratio = 603.099 / 78.039 = 7.729 >1, Hence OK




Length (L) = 10 ft







φ = 0.9

ß = 0.85 - 0.05x (fc-4)

0.85>=ß >=0.65

ß = 0.85

width (B) = 5 ft



Chapter — 6







ratio = 356.394 / 128.3438 = 2.777 >1, Hence OK


6.3.4 Comparison


STAADFoundation

Result

Percent Dif-ference

Sliding Resistance (Kip) 163.161 163.306 NegligibleUplift Resistance (Kip) 151.296 151.355 NegligibleHorizontal FOS 2.066 2.064 NegligibleVertical FOS 3.151 3.153 Negligiblefactored moment resisted by toprebar ( ft·kip)


Max moment capacity (verticallyhogging) (ft·kip)



128.3438 128.361 Negligible

Max moment capacity(horizontally hogging) (ft·kip)

356.394 355.775 Negligible


6.4 Deadman Guy Anchor US 46.4.1 Reference

6.4.2 ProblemDesign an anchor block for a guy rod supporting the following load condition-

Axial Tension = 250 Kip

Slope with Horizontal = 50 degree

Min area required for guy rod =4.167 sq.inch, (i.e if single rod is used, thenmin dia of rod required = 2.4 inch).





Necessary FOS are as follows-











LayerIndex No

LayerType

Depth ofLayer

Cohesion(psf)

Angle ofFriction

Dry Density(pcf)

1 sand (0-3) 0 20 1042 silt (3-5) 500 22 1053 silt (5-8) 800 15 1054 silt (8-10) 800 20 1065 silt (10-12) 850 15 1066 silt (12-20) 850 15 1068 silt (20-below) 850 15 106





Chapter — 6


6.4.3 SolutionFirst let us calculate the Horizontal & Vertical components of Axial Tension at guy rod

Horizontal component of load (H) = P.cos θ = 250 x Cos 50 = 160.6297 Kip

Vertical component of load (V)= P.sin θ = 250 x Sin 50 = 191.5676 Kip


Kp=tan2 (450+Ø/2)

Kp=tan2 (450-Ø/2)


Pa= γ.h.KpSo Pa or a particular layer = Pa of previous layer + γ.h.KaWhere:



h= Thickness of present layer)

Pp= γ.h.Kp + 2C.√KpSo Pp or a particular layer = Pp of previous layer + γ.h.KpWhere








α <=1







Chapter — 6




Length = 12 ft










Let us consider wedge at all sides with dimension of 0.5 ft

As wedge is present so dispersion is to be started from bottom of block

Height of soil above top level of block = 8 ft


Weight of Concrete Block = LxBxHx unit wt of concrete = 12 x 6 x 8 x 0.15 =86.4 kip


Chapter — 6






Therefore, Total resistance against uplift = 54.9504 + 491.29 + 172.97 =719.211 Kip









Length (L)= 12 Kip










φ = 0.9

ß = 0.85 - 0.05x (fc-4)

0.85>=ß >=0.65

ß = 0.85

width (B) = 6 ft




Total area of Top Rebar = As = 3.60609375 Ksi



ratio = 1488.835 / 373.5576 = 3.986 >1, Hence OK




Length (L) = 12 ft







φ = 0.9

ß = 0.85 - 0.05x (fc-4)

0.85>=ß >=0.65

ß = 0.85

width (B) = 8 ft


Chapter — 6








ratio = 1101.17 / 313.2285 = 3.516 >1, Hence OK


6.4.4 Comparison


STAADFoundation

Result

Percent Dif-ference

Sliding Resistance (Kip) 413.423 413.174 negligibleUplift Resistance (Kip) 719.211 719.281 negligibleHorizontal FOS 2.574 2.571 negligibleVertical FOS 3.755 3.756 negligiblefactored moment resisted by toprebar ( ft·kip)

373.5576 373.447 negligible

Max moment capacity (verticallyhogging) (ft·kip)

1488.835 1486.29 negligible


313.2285 313.359 negligible

Max moment capacity (hor-izontally hogging) (ft·kip)

1101.17 1099.277 negligible






Chapter 6


Section 7

Drilled Pier Foundations7.1 Drilled Pier Foundation 1 API

7.1.1 ReferenceAPI RP 2A-WSD

7.1.2 ProblemDesign axial capacity for drilled pier with the given data: Design Method: API

Load Fy= 100 kip, fc= 4 ksi, fy= 60 ksi, Straight pier, pier diameter= 2 ft, pier height= 30ft, water level at 40 ft.

Soil Profile by Layer

1. Sand, 8 ft deep, angle of friction 30 deg, Avg Density 108 lb/ft3,

Density- Loose

2. Sand, 13 ft deep, angle of friction 34 deg, Avg Density 110 lb/ft3,

Density- Medium

3. Clay - , 9 ft deep, cohesion 2 kip/ft2, Avg Density 110 lb/ft3,

Density- Dense

Elasticity of Soil 0.3 ksi


Figure 7-1: Pier Elevation

Factor of safety

End bearing - 3

Skin Friction - 3

% of Capacity Used

End bearing - 100%

Skin Friction - 100%

Neglected zone for skin friction

Top - 5ft

Bottom - Pier Dia = 2ft

Concrete Properties

fc= 4ksi

Ec= 3605ksi


Chapter — 7

7.1 Drilled Pier Foundation 1 API

Density= 150lb/ft3

Rebar Properties

fy= 60ksi

Es= 29000ksi

7.1.3 Solution

Critical Depth

Critical depth is set to be calculated by program

Effective Overburden Pressure

Po= (Soil Density of respective layer x Depth to the center of the layer)

Effects of water and critical depth are also considered calculating Po

Layer Effective Overburden Pres-sure (lb/ft2)

Layer1 432Layer2 1579Layer3 2789

Skin Friction

Ψ Factor from API RP 2A-WSD 6.4.2

Ψ = Cohesion / Effective Overburden Pressure

Layer ΨLayer1 0Layer2 0Layer3 0.7171

α Factor from API RP 2A-WSD 6.4.2

α = 0.5 x Ψ -0.5 Ψ <=1.0

α = 0.5 x Ψ -0.25 Ψ >1.0

Layer αLayer1 0Layer2 0Layer3 0.59045

Section 7 Drilled Pier Foundations



K Coefficient of lateral earth pressure from API RP 2A-WSD 6.4.3

K=0.8 (for straight pier)

δ Friction angle between soil and pier

Calculated based on API RP 2A-WSD Table 6.4.3-1

Shaft friction (f), from API RP 2A-WSD 6.4.2 & 6.4.3

For cohesive soil layerf = α x c

For cohesionless soil layerf = K xPo x tanδ

Layer Shaft Friction (psf)Layer1 125.788Layer2 589.040Layer3 1180.889

Skin Friction Resistance (Qf), from API RP 2A-WSD 6.4.1

Qf = f x As

Layer Qf (kip)Layer1 6.3228Layer2 48.1136Layer3 66.7777Total Skin Friction 121.214

Modified Skin Friction Based on Neglected Zones for Skin Friction

Qf (modified) is calculated by considering top and bottom neglect layer

Layer Qf_modified(kip)

Layer1 2.3711Layer2 48.1136Layer3 51.9382Total Modified SkinFriction

102.423

Base Resistance

Unit end bearing (q), from API RP 2A-WSD 6.4.2 & 6.4.3

For cohesive soil layer q = 9 x c

For cohesionless soil layer q =po x Nq

Layer q (kip)Layer1 5.184Layer2 31.58Layer3 18.0

End Bearing for each layer (Qp), from API RP 2A-WSD 6.4.1

Qp = q x Ap


Chapter — 7


Layer Qp (kip)Layer1 16.286Layer2 99.211Layer3 56.549

End Bearing for bottom Layer, Qp_bott = 56.549 kips

Unfactored Capacity

Factored Capacity

7.1.4 Comparison

Value ofSTAAD Foun-

dationResult

ReferenceResult Difference

UnfactoredTip Resistance 56.571 kip 56.549 kip NegligibleUnfactored SkinResistance

102.487 kip 102.423 kip Negligible

Unfactored Total AxialCapacity


FactoredTip Resistance 18.857 kip 18.849 kip NegligibleFactored Skin Resistance 34.162 kip 34.141 kip NegligibleFactored Total AxialCapacity


Table 7-1: Drilled Pier (API) verification example 1 comparison

7.2 Drilled Pier Foundation 2 API7.2.1 ReferenceAPI RP 2A-WSD

7.2.2 ProblemDesign axial capacity for drilled pier with the given data: Design Method:API

Load Fy= 100kip, fc= 4ksi, fy= 60ksi, Straight pier, pier diameter= 4ft, pier height= 39ft,water level at 50ft.




Soil Profile by Layer

1. Clay - , 6ft deep, cohesion 1kip/ft2, Avg Density 105lb/ft3, Density- Very Loose

2. Clay - , 13ft deep, cohesion 1.1kip/ft2, Avg Density 110lb/ft3, Density- Medium

3. Clay - , 8ft deep, cohesion 2kip/ft2, Avg Density 111lb/ft3, Density- Dense

4. Clay - , 12ft deep, cohesion 3kip/ft2, Avg Density 113lb/ft3, Density- Very Dense

Elasticity of Soil 0.3ksi

K = 0.8

Factor of safety

End bearing - 3

Skin Friction - 3

% of Capacity Used

End bearing - 100%



Top - 5ft


Concrete Properties

fc= 4ksi

Ec= 3605ksi

Density= 150lb/ft3

Rebar Properties

fy= 60ksi

Es= 29000ksi


Chapter — 7



7.2.3 Solution

Critical Depth









Layer1 315Layer2 1345Layer3 1874Layer4 1566

Skin Friction

Ψ Factor from API RP 2A-WSD 6.4.2

Ψ = Cohesion / Effective Overburden Pressure

Layer ΨLayer1 3.1746Layer2 0.8178Layer3 1.0672Layer4 1.9157

α Factor

Layer αLayer1 0.55Layer2 0.55Layer3 0.55Layer4 0.55

K Coefficient of lateral earth pressure from API RP 2A-WSD 6.4.3



Calculated based on API RP 2A-WSD Table 6.4.3-1

Shaft friction (f), from API RP 2A-WSD 6.4.2 & 6.4.3



Layer Shaft Friction (psf)Layer1 550Layer2 605Layer3 1100Layer4 1650

Skin Friction Resistance (Qf), from API RP 2A-WSD 6.4.1

Qf = f x As

Layer Qf (psf)Layer1 41.469Layer2 98.835


Chapter — 7


Layer Qf (psf)Layer3 110.584Layer4 248.814Total Skin Friction 499.702



Layer Qf_modified(psf)

Layer1 6.9115Layer2 98.835Layer3 110.584Layer4 165.876Total Skin Fric-tion

382.206

Base Resistance

Unit end bearing (q), from API RP 2A-WSD 6.4.2 & 6.4.3

For cohesive soil layerq = 9 x c

For cohesionless soil layerq =po x Nq

Layer q (psf)Layer1 9000Layer2 9900Layer3 18000Layer4 27000

End Bearing for each layer (Qp), from API RP 2A-WSD 6.4.1

Qp = q x Ap

Layer Qp (psf)Layer1 113.097Layer2 124.407Layer3 226.195Layer4 339.292

End Bearing for bottom Layer

Qp_bott = 339.292 kip

Unfactored Capacity




Factored Capacity

7.2.4 Comparison

Value ofSTAAD Foun-

dationResult








Table 7-2: Drilled Pier (API) verification example 2 comparison

7.3 Drilled Pier Foundation 3 FHWA7.3.1 ReferenceFHWA-IF-99-025

7.3.2 ProblemDesign axial capacity for drilled pier with the given data: Design Method:FHWA

Load Fy= 100kip, fc= 4ksi, fy= 60ksi, Straight pier, pier diameter= 2ft,

pier height= 30ft, water level at 40ft.

Soil Profile

1. Sand, 8ft deep, angle of friction 30deg, Avg Density 108lb/ft3, N60- 11

2. Sand, 13ft deep, angle of friction 34deg, Avg Density 110lb/ft3, N60- 14

3. Clay - , 9ft deep, cohesion 2kip/ft2, Avg Density 110lb/ft3, N60- 16


Factor of safety


Chapter — 7

7.3 Drilled Pier Foundation 3 FHWA

End bearing - 3

Skin Friction - 3

% of Capacity Used

End bearing - 100%



Top - 5ft


Concrete Properties

fc= 4ksi

Ec= 3605ksi

Density= 150lb/ft3

Rebar Properties

fy= 60ksi

Es= 29000ksi





7.3.3 Solution

Critical Depth






Chapter — 7



Layer1 432Layer2 1579Layer3 1925

Skin Friction

α Factor from FHWA-IF-99-025 Eqn 11.6

α = 0.55

Layer αLayer1 0Layer2 0Layer3 0.55

β Dimensionless correlation factor from FHWA-IF-99-025 Eqn 11.18

Layer βLayer1 0.9017Layer2 0.8536Layer3 0


Assumed to be same as soil friction angle

Shaft friction (f), from FHWA-IF-99-025 Eqn 11.16 & 11.17

For cohesive soil layerf = α x su

For cohesionless soil layerf = β x σ vi

Layer Shaft Friction (psf)Layer1 389.50Layer2 1347.86Layer3 1100.0

Skin Friction Resistance (Qf), from FHWA-IF-99-025 Eqn 10.2

Rs = f x As

Layer Rs (kip)Layer1 19.578Layer2 110.095Layer3 62.204Total Skin Friction 191.877






Layer Rs_modified(kip)

Layer1 7.342Layer2 110.095Layer3 48.381Total Modified SkinFriction

165.817

Base Resistance

Unit end bearing (q), from FHWA-IF-99-025 Eqn 11.1, 11.2 & 11.4a



Layer q (kip)Layer1 13.2Layer2 15.6Layer3 5.611

End Bearing for each layer (Qp), from FHWA-IF-99-025 Eqn 10.2

RB = q x Ap

Layer RB (kip)Layer1 41.469Layer2 49.009Layer3 17.628

End Bearing for bottom Layer, RB-bott = 17.628 kips

Un-factored Capacity

Factored Capacity


Chapter — 7


7.3.4 Comparison

Value ofSTAAD Foun-

dationResult








Table 7-3: Drilled Pier (FHWA) verification example 3 comparison

7.4 Drilled Pier Foundation 4 FHWA7.4.1 ReferenceFHWA-IF-99-025

7.4.2 ProblemDesign axial capacity for drilled pier with the given data: Design Method:FHWA


Soil Profile

1. Clay - , 6ft deep, cohesion 1kip/ft2, Avg Density 105lb/ft3, N60 10

2. Clay - , 13ft deep, cohesion 1.1kip/ft2, Avg Density 110lb/ft3, N60 12




Factor of safety

End bearing - 3

Skin Friction - 3

% of Capacity Used

End bearing - 100%


Neglected zone

for skin frictionTop - 5ft





Concrete Properties

fc= 4ksi

Ec= 3605ksi

Density= 150lb/ft3

Rebar Properties

fy= 60ksi

Es= 29000ksi


7.4.3 Solution

Critical Depth



Chapter — 7






1 3152 13453 18744 1380

Skin Friction

α Factor from FHWA-IF-99-025 Eqn 11.6

α = 0.55 (for all layers)

Layer α1 0.552 0.553 0.554 0.55

β Dimensionless correlation factor from FHWA-IF-99-025 Eqn 11.18

β = 0 (for all layers)



Shaft friction (f), from FHWA-IF-99-025 Eqn 11.16 & 11.17

For cohesive soil layerf = α x su

For cohesionless soil layerf = β x σ vi

Layer Shaft Friction (psf)1 5502 6053 11004 1650


Rs = f x As




Layer Rs (kip)1 41.4692 98.8353 110.5844 248.814Σ (Total Skin Friction) 499.702




1 6.9122 98.8353 110.5844 165.876Σ (Total ModifiedSkin Friction)

382.206

Base Resistance




Layer q (kip)1 3.6462 3.8803 5.6114 27


RB = q x Ap

Layer RB (kip)1 45.8152 48.7593 70.5104 339.292

End Bearing for bottom Layer, RB-bott = 339.292 kips



Chapter — 7


Factored Capacity

7.4.4 Comparison

Value ofSTAAD Foun-

dationResult


Unfactored Tip Resistance 339.429 kip 339.292 kip NegligibleUnfactored SkinResistance






Table 7-4: Drilled Pier (FHWA) verification example 4 comparison

7.5 Drilled Pier Foundation 5 VESIC

7.5.1 ReferenceAlternate Vesic Method

7.5.2 ProblemDesign axial capacity for drilled pier with the given data: Design Method: Alternate VesicMethod

Load Fy= 100kip, fc= 4ksi, fy= 60ksi, Straight pier, pier diameter= 2ft,

pier height= 30ft, water level at 40ft.

Soil Profile

1. Sand, 8ft deep, angle of friction 30deg, Avg Density 108lb/ft3,

2. Sand, 13ft deep, angle of friction 34deg, Avg Density 110lb/ft3,

3. Clay, 9ft deep, cohesion 2kip/ft2, Avg Density 110lb/ft3,





Factor of safety

End bearing - 3

Skin Friction - 3

% of Capacity Used

End bearing - 100%



Top - 5ft


Concrete Properties

fc= 4ksi

Ec= 3605ksi

Density= 150lb/ft3

Rebar Properties

fy= 60ksi

Es= 29000ksi


Chapter — 7



7.5.3 Solution

Critical Depth








Layer Effective Overburden Pres-sure (psf)

1 4322 15793 1925

Skin Friction

α Adhesion Factor for Drilled Pier in Cohesive Soil

α = 0.55

Layer α1 02 03 0.55

K Coefficient of lateral earth pressure


β Lateral Earth Pressure and Friction Angle Factor

Layer β1 0.4622 0.5403 0



Shaft friction (f), from Alpha or Beta Method



Layer Shaft Friction (psf)1 199.5322 852.0393 1100

tan(δ1) = 0.675


Rs = f x As

Layer Rs (kip)1 10.0302 69.5963 62.204Σ (Total Skin Friction) 141.829



Chapter — 7




1 3.7612 69.5963 48.381Σ (Total Modified SkinFriction)

121.737

Base Resistance

Cohesive Soil (Bottom Layer)

Factor Fr = 1

Factor Ncp = 9

Cohesionless Soil (Bottom Layer)

Factor Nqp = 0


For cohesive soil layer q = Fr x Nqp x c

For cohesionless soil layer q =po x Nqp

Layer q (psf)1 02 03 18000


RB = q x Ap

Layer RB (kip)1 02 03 56.549

End Bearing for bottom Layer, RB_bott = 56.549 kip





Factored Capacity

7.5.4 Comparison

Value ofSTAAD Foun-

dationResult






FactoredTip Resistance 18.857 kip 5.516 kip NegligibleFactored Skin Resistance 40.608kip 40.579 kip NegligibleFactored Total AxialCapacity


Table 7-5: Drilled Pier (Vesic) verification example 5 comparison

7.6 Drilled Pier Foundation 6 Vesic7.6.1 ReferenceAlternate Vesic Method

7.6.2 ProblemDesign axial capacity for drilled pier with the given data: Design Method: Alternate VesicMethod


Soil Profile

1. Clay - , 6ft deep, cohesion 1kip/ft2, Avg Density 105lb/ft3, Density- Very Loose

2. Clay - , 13ft deep, cohesion 1.1kip/ft2, Avg Density 110lb/ft3, Density- Medium

3. Clay - , 8ft deep, cohesion 2kip/ft2, Avg Density 111lb/ft3, Density- Dense

4. Clay - , 12ft deep, cohesion 3kip/ft2, Avg Density 113lb/ft3, Density- Very Dense


Factor of safety

End bearing - 3

Skin Friction - 3

% of Capacity Used


Chapter — 7

7.6 Drilled Pier Foundation 6 Vesic

End bearing - 100%



Top - 5ft


Concrete Properties

fc= 4ksi

Ec= 3605ksi

Density= 150lb/ft3

Rebar Properties

fy= 60ksi

Es= 29000ksi





7.6.3 Solution

Critical Depth





Layer Effective Overburden Pres-sure (psf)

1 3152 13453 1874

Skin Friction

α Adhesion Factor for Drilled Pier in Cohesive Soil

α = 0.55 (All layers)

K Coefficient of lateral earth pressure


Lateral Earth Pressure and Friction Angle Factor, β = 0 (All layers)



Shaft friction (f), from Alpha or Beta Method



Layer Shaft Friction (psf)1 5502 6053 11004 1650

tan(δ1) = 0


Chapter — 7



Rs = f x As

Layer Rs (kip)1 41.4692 98.8353 110.5844 248.814Σ (Total Skin Friction) 499.702


RB = q x Ap

Layer RB (kip)1 02 03 04 339.292

End Bearing for bottom Layer, RB_bott = 339.292 kip


Factored Capacity




7.6.4 Comparison

Value ofSTAAD Foun-

dationResult








Table 7-6: Drilled Pier (Vesic) verification example 6 comparison


Chapter — 7


Section 8

Plant Foundation8.1 Vertical Vessel Foundation 1

8.1.1 Input Parameters

Geometric Description

Vessel Geometry

Effective Height, Hve = 10ft

Effective Diameter, Dve = 4ft

Center of Gravity, CG = 10 ft

Pedestal Geometry

Height, Tp = 1 ft

Diameter, Dp = 5 ft

Footing Geometry

Minimum Footing Diameter = 10 ft

Maximum Footing Diameter = 10 ft

Minimum Footing Depth = 2 ft

Maximum Footing Depth = 2 ft


Figure 8-1: Tank and foundation elevation

Anchor Bolt Data

Figure 8-2: Anchor bolt plan

Bolt Circle Diameter, BCD = 14.875 ft

Bolt Diameter, BD = 1.5 in.

Sleeve Diameter, SD = 2 in

Number of Anchor Bolts, Nb = 16

Effective Embedment Depth, heff = 1.5 ft


Chapter — 8

8.1 Vertical Vessel Foundation 1

Design Parameters

Soil

Soil Depth, Ts = 0 ft

Soil Density, Vsoil = 110 pcf

Allowable Soil Bearing Pressure, SBC = 3.8 ksf

Concrete

Cover, cc = 0.25 ft

Concrete Density, Vc = 15o pcf

Concrete strength, f'c = 4 ksi

Reinforcement

fy = 60 ksi

Bar Type : Imperial

Minimum Bar Diameter = 4

Maximum Bar Diameter = 11

Stability

Minimum Stability Ratio = 1.5

Primary Load Description

Load Types Axial Force(kip)

Base Moment(ft·kip)

Base Shear(kip)

Empty Load (De) 10 0 0Operating Load(Do)

20 0 0

Test Load (Dt) 0 0 0Erection Load(Dr)

0 0 0

Live Load (Dl) 0 0 0

Table 8-1: Primary load description

8.1.2 Solution

Wind Load

Wind Load Calculation per ASCE 7-05

Partial Wind Case: Percentage of wind = 50%

Section 8 Plant Foundation



Wind Speed, V = 0 mph

Exposure category D, Case 2

Wind Directionality Factor, Kd = 0.95 ................................per ASCE 7-05 Table 6-4

Topographic Factor, Kzt = 1 ................................per ASCE 7-05 Fig. 6-4

Importance factor, IW = 1.15 ................................per ASCE 7-05 Table 6-1

Gust Effect Factor, G = 0.85 ................................per ASCE 7-05 Table 6.5.8

Net Force Coefficient, Cf = 0.9 ................................per ASCE 7-05 Gig. 6-20 & Fig. 6-21

Elevation Kz Pressure Width Area Shear Moment1 1.03023 0 4 4 0 011 1.03023 0 4 40 0 0

Table 8-2: Wind loads

Total Wind Shear = 0 kip

Total Wind Moment = 0 kip-ft

Seismic Load

Importance Factor, I = 1

Fundamental Period, T = 6 s

Long Period, TL = 12 s

Site Class C

Spectral Response Acc. Parameter at Short Period, Ss = 0

Spectral Response Acc. Parameter at 1 Sec, S1 = 0

Short Period Site Coefficient at 0.2s Period, Fa = 1.2

Long Period Site Coefficient at 1.0s Period, Fv = 1.7

Design Spectral Response Acc. Parameter st Short Period, SDS = 0

Design Spectral Response Acc. Parameter at 1 sec, SD1 = 0

Response Modification Factor, R = 2

Calculation Of Seismic Response Coefficient

Cs = 0

Empty Seismic = Cs x De = 0 kip

Operating Seismic = Cs x Do = 0 kip

Test Seismic = Cs x Dt = 0 kip


Chapter — 8


Load Combination Table

Load

CaseEmpty Operating Wind Seismic Test

Erection

LoadLiveLoad

UserLoad1

UserLoad2

UserLoad3

1 0 1 0 0 0 0 1 0 0 02 0 1 1 0 0 0 0 0 0 03 0 1 0 0.7 0 0 0 0 0 04 1 0 1 0 0 0 0 0 0 05 0 0.9 0 0.7 0 0 0 0 0 06 0.9 0 0 0.7 0 0 0 0 0 07 0 0 1 0 0 1 0 0 0 08 0 0 0.83 0 0.83 0 0 0 0 0

Table 8-3: Applied Load Combinations - Allowable Stress Level

Load


Erection

LoadLiveLoad

UserLoad1

UserLoad2

UserLoad3

1 1 0 0 0 0 0 0 0 0 02 0 0 0 0 0 0 0 0 0 03 0 0 0 0 0 0 0 0 0 04 0 0 0 0 0 0 0 0 0 05 0 0 0 0 0 0 0 0 0 06 0 0 0 0 0 0 0 0 0 07 0 0 0 0 0 0 0 0 0 08 0 0 0 0 0 0 0 0 0 0

Table 8-4: Applied Load Combinations - Strength Level

Load

CaseAxial Shear Moment(kips) (kips) (kip-ft)

1 20 0 02 2o 0 03 20 0 04 10 0 05 18 0 06 9 0 07 0 0 08 0 0 0

Table 8-5: Applied Load at Top ofPedestal - Allowable Stress Level

Governing Loads

Axial = 20 kip

Shear = 0 kip




Moment = 0 ft-kip

Load


1 10 0 02 o 0 03 0 0 04 0 0 05 0 0 06 0 0 07 0 0 08 0 0 0

Table 8-6: Applied Load at Top ofPedestal - Strength Level

Governing Loads

Axial = 10 kip

Shear = 0 kip

Moment = 0 kip

Pedestal Design

Fu = 56.64 kip

Check for minimum pedestal dimension is done in accordance with PIP STE 03350 Sect.4.5.1

Minimum Pedestal Dimension = 15.625 ft

Factored O.T.M. At Base Of Pedestal = 0 kip-ft

Seismic Load Governing, hence use Vessel Operating Weight

Nominal Axial Load (Empty/Operating), Du = 56.64 kip

Weight of Pedestal = 3.1066 kip

Dowel Circle Diameter, Dc = BCD

Number of Dowels, Nd = 32

Tensile Force In Each Dowel Per PIP STC03350 4.5.4

Fu = 4·Muped/(Nd·Dc) - 0.9·(Du + Wped)/Nd = -0.369 kip

Area of Dowel Bar Required

As_ped_req = Fu/(φ·fy) = -8.192(10)-3 in2

Minimum Dowel Reinforcement per PIP STC03350 4.5.5 : #5 - 32

Dowel Bar Size Provided = 5

Area of Steel Provided = -0.00819 in2

Area of steel required in pedestal, As, req = (dd)2 x pi/4 = 0.307 in2

Potential Conc. Failure Area per PIP STC03350 Fig. A, An = 5.5086 ft2


Chapter — 8


Compressive Force In Each Dowel Based on PIP STC03350 4.6.2

Pu = Muped/Dc + 0.9· (Du + Wped) = 11.769 kip

db = 0.625 in

Fc = Pu/An = 0.015 ksi

Beta = 1

Weight of Soil = 0 kip

Design Results

Stability Ratio is calculated based on PIP STE03350 Eqn. 15

LoadCase Eccentricity Stability

Ratio1 0 02 03 04 05 06 07 08 0

Table 8-7: Stability Ratio

Soil bearing calculations are per PIP STE03350 4.7.2

Load

CaseMax Soil

Bearing (ksf)Min Soil

Bearing (ksf)1 0.2414 0.24142 0.2414 0.24143 0.2414 0.24144 0.4582 0.45825 0.2173 0.21736 0.4123 0.41237 0 08 0 0

Table 8-8: Soil Bearing Check

Max Diagonal Soil Bearing Pressure, fdia = 0.4582 ksf

Max Diagonal Soil Bearing Pressure, fflat = 0.4233 ksf




Concrete Design

One Way Shear

Figure 8-3: One-way shear dimensions

Location of Pedestal Face from Face of Footing, X1 = 2.7246 ft

Location of Shear check from Face of Footing, X2 = 1.0006 ft

Shear Stress, Vu = 0.266 ksf

Factored Shear Stress Capacity per ACI318-05 Eqn. 11-3, φVC = 13.66 ksf

Two way Shear Check

Figure 8-4: Two-way shear check

Octagonal Perimeter, bo = 21.65 ft

Punching Shear Force, Vu = 6.463 kip

Factored Shear Capacity, φVc = 1019.8 kip

Reinforcement Calculations

Required development length for bars


Chapter — 8


Available development length for bars (From face of Pedestal to face of Footing) = 2.724551ft

m = fy / (0.85 fc) = 17.647

2·m·Rn / fy = 2.597(10)-3


Ast,req = ρ·d·1ft = 0.018 in2

ρmin = 0.0018

Minimum Area of Steel Required

Ast,min = ρmin·d· 1 ft = 0.447 in2

Spacing Required, s = 8 in.

Area of Steel Provided

Ast_prov = (π·db2/4)·1ft/s = 0.205 in2

Bar Size = 5

Final Dimensions

Footing Diameter, Df = 10 ft

Footing Thickness, Tf = 2 ft




8.1.3 Comparison

Hand Cal-culations


Percent Dif-ference

Footing Diagonal 10 10 noneFooting Thickness 2 2 noneFooting Soil Bear-ing

0.458 0.496 8%

Stability Check 0 0 n/aOne Way ShearCheck

0.266 0.266 0%

Punching ShearCheck

6.463 6.564 2%

Reinforcement Pro-vided

0.460 0.447 3%

Table 8-9: Vertical Vessel verification example 1 comparison

Note: Soil Bearing values are generated from graph; %diff also contains human errors.

8.2 Vertical Vessel Foundation Design8.2.1 Input Parameters

Geometrical Description

Vessel Geometry

Effective Height, Hve = 50 ft

Effective Diameter, Dve = 13 ft


Pedestal Geometry

Height, Tp = 4 ft

Diameter, Dp = 14 ft

Footing Geometry






Chapter — 8

8.2 Vertical Vessel Foundation Design


Anchor Bolt Data



Bolt Diameter, BD = 168 in.

Sleeve Diameter, SD = 1.5 in


Effective Embedment Depth, heff = 0.167 ft




Design Parameters

Soil



Allowable Soil Bearing Pressure, SBC = 4 ksf

Concrete

Cover, cc = 0.25 ft



Reinforcement

fy = 60 ksi

Bar Type : Imperial



Stability




Base Moment (ft-kip)

Base Shear(kip)

Empty Load (De) -28 0 0Operating Load(Do)

-66 0 0

Test Load (Dt) -73 0 0Erection Load(Dr)

-70 0 0



Wind Load




Chapter — 8



Exposure category C, Case 2

Wind Directionality Factor, Kd = 0.95 ................................per ASCE 7-05Table 6-4




Net Force Coefficient, Cf = 0.8 ................................per ASCE 7-05 Gig. 6-20& Fig. 6-21

Elevation Kz Pressure Width Area Shear Moment2 0.848884 0.028727 13 26 0.507899 0.50789915 0.848884 0.028727 13 169 3.301344 28.0614320 0.901885 0.030521 13 65 1.349026 23.6079625 0.945265 0.031989 13 65 1.413912 31.8130330 0.982253 0.033241 13 65 1.469238 40.4040540 1.043581 0.035316 13 130 3.121944 109.26850 1.093775 0.037015 13 130 3.272105 147.244752 1.102844 0.037322 13 26 0.659847 33.65219


Total Wind Shear = 15.095 kip

Total Wind Moment = 414.449 ft-kip

Seismic Load


Fundamental Period, T = 4.1 s


Site Class C

Spectral Response Acc. Parameter at Short Period, Ss = 0.105

Spectral Response Acc. Parameter at 1 Sec, S1 = 0.043

Short Period Site Coefficient at 0.2s Period, Fa = 1.2


Design Spectral Response Acc. Parameter st Short Period, SDS = 0.084

Design Spectral Response Acc. Parameter at 1 sec, SD1 = 0.049



Cs = 5.973x10-3




Empty Seismic = Cs x De = 0.167 kip

Operating Seismic = Cs x Do = 0.394 kip

Test Seismic = Cs x Dt = 0.436 kip


Load


Erection

LoadLiveLoad

UserLoad1

UserLoad2

UserLoad3

1 0 1 0 0 0 0 1 0 0 02 0 1 1 0 0 0 0 0 0 03 0 1 0 0.7 0 0 0 0 0 04 1 0 1 0 0 0 0 0 0 05 0 0.9 0 0.7 0 0 0 0 0 06 0.9 0 0 0.7 0 0 0 0 0 07 0 0 1 0 0 1 0 0 0 08 0 0 0.83 0 0.83 0 0 0 0 0

Table 8-12: Applied Load Combination - Allowable Stress Level

Load


Erection

LoadLiveLoad

UserLoad1

UserLoad2

UserLoad3

1 0 1.4 0 0 0 0 0 0 0 02 0 1.2 0 0 0 0 1.6 0 0 03 0 1.2 1.6 0 0 0 0 0 0 04 0 1.2 0 1 0 0 0 0 0 05 0.9 0 1.6 0 0 0 0 0 0 06 0 0.9 0 1 0 0 0 0 0 07 0.9 0 0 1 0 0 0 0 0 08 0 0 1.6 0 0 0.9 0 0 0 09 0 0 0 0 0 0 0 0 0 010 0 0 0 0 0 0 0 0 0 0

Table 8-13: Applied Load Combination - Strength Level

Load


1 66 0 02 66 15.09532 414.55933 66 0.275945 2.2075634 28 15.09532 414.44935 59.4 0.275945 2.2075636 25.2 0.117068 0.9365427 70 7.547658 207.27978 60.59 6.264556 172.0421



Chapter — 8


Governing Loads

Axial = 70 kip

Shear = 15.09532 kip

Moment = 414.5593 ft-kip

Load


1 92.4 0 02 79.2 0 03 79.2 24.15251 663.29494 79.2 0.394208 3.1536615 25.2 24.15251 663.29496 59.4 0.394208 3.1536617 25.2 0.16724 1.3379178 63 12.07625 331.6474

Table 8-15: Applied Load at Top ofPedestal - Strength Level

Governing Loads

Axial = 92.4 kip

Shear = 24.15251 kip

Moment = 663.2949 kip

Pedestal Design

Fu = 56.64 kip



Factored O.T.M. At Base Of Pedestal = 759.9049 kip-ft




Dowel Circle Diameter, Dc = BCD = 1.333 ft



Fu = 4·Muped/(Nd·Dc) - 0.9·(Du + Wped)/Nd = 66.274 kip


As_ped_req = Fu/(φ·fy) = 1.473 in2






Area of Steel Provided = 1.472746 in.2

Potential Conc. Failure Area per PIP STC03350 Fig.

An = 0.060107



Fc = Pu/An = 84.211 ksi

Weight of Soil = 10.935 kip

Design Results



Ratio1 0 02 2.376857 3.365793 0.017400 459.7784 2.913937 2.745435 0.017975 445.0606 0.009203 869.2927 1.165810 6.862198 1.012977 7.89752






Chapter — 8


Concrete Design

One Way Shear


Location of Pedestal Face from Face of Footing (X1) = 1.62742 ft

Location of Shear check from Face of Footing (X2) = 0.904783 ft


Factored Shear Stress Capacity φVC per ACI318-05 Eqn. 11-3 = 13.66104 ksf

Two way Shear Check










Available development length for bars (From face of Pedestal to face of Footing) =1.628742 ft

Rn = Mu / (0.9·1ft·d2) = 0.079 ksi

m = fy / (0.85· fc) = 17.647

2·m·Rn / fy = 0.047


Ast_req = p·d·1ft = 0.139 in2

pmin = 0.0018

Minimum Area of Steel Req

Ast_min = pmin·d·1ft = 0.188 in2

Spacing Required

s = 18 in.



Bar Size = 5

Final Dimensions

Footing Diameter = 16 ft

Footing Thickness = 1 ft


Chapter — 8


8.2.2 Comparison

Hand Cal-culations


Percent Dif-ference

Footing Diagonal 16 16 0Footing Thickness 1 1 0Footing Soil Bear-ing

2.1030 2.102 0

Stability Check 2.7454 2.748 0One Way ShearCheck

4.1857 4.193 0

Punching ShearCheck

17.58 17.951 2


0.20453 0.18765 9


8.3 Vertical Vessel Foundation Design8.3.1 Input Parameters

Geometric Description

Vessel Geometry

Effective Height, Hve = 30 ft

Effective Diameter, Dve - 10 ft


Pedestal Geometry

Height = 2 ft

Diameter = 12 ft

Footing Geometry









Anchor Bolt Data



Bolt Diameter, BD = 132 in.

Sleeve Diameter, SD = 1.5 in.


Effective Embedment Depth, heff = 0.167 fyt


Chapter — 8


Design Parameters

Soil



Allowable Soil Bearing Pressure, SBC = 4 ksf

Concrete

Cover, cc = 0.25 ft



Reinforcement

fy = 60 ksi

Bar Type : Imperial



Stability




Base Moment (ft-kip)

Base Shear(kip)

Empty Load (De) -20 0 0Operating Load(Do)

-40 0 0

Test Load (Dt) -60 0 0Erection Load(Dr)

-30 0 0



8.3.2 Solution

Wind Load







Exposure category B, Case 2

Wind Directionality Factor, Kd = 0.95 ................................per ASCE 7-05 Table 6-4




Net Force Coefficient, Cf = 0.8 ................................per ASCE 7-05 Gig. 6-20 & Fig. 6-21

Elevation Kz Pressure Width Area Shear Moment1 0.57472 0.01302 10 10 0.088534 0.04426715 0.57472 0.01302 10 140 1.23948 9.91583820 0.623954 0.014135 10 50 0.480594 8.41039225 0.66503 0.015066 10 50 0.512232 11.5252230 0.700591 0.015871 10 50 0.539622 14.83962


Total Wind Shear = 2.86 kip

Total Wind Moment = 44.735 ft-kip

Seismic Load


Fundamental Period, T = 4.1 s


Site Class C

Spectral Response Acc. Parameter at Short Period, Ss = 1.997

Spectral Response Acc. Parameter at 1 Sec, S1 = 0.805

Short Period Site Coefficient at 0.2s Period, Fa = 1


Design Spectral Response Acc. Parameter st Short Period, SDS = 1.332

Design Spectral Response Acc. Parameter at 1 sec, SD1 = 0.697



Cs = 0.805

Empty Seismic = Cs x De = 1.701 kip

Operating Seismic = Cs x Do = 3.402 kip

Test Seismic = Cs x Dt = 5.103 kip


Chapter — 8



Load


Erection

LoadLiveLoad

UserLoad1

UserLoad2

UserLoad3

1 0 1 0 0 0 0 1 0 0 02 0 1 1 0 0 0 0 0 0 03 0 1 0 0.7 0 0 0 0 0 04 1 0 1 0 0 0 0 0 0 05 0 0.9 0 0.7 0 0 0 0 0 06 0.9 0 0 0.7 0 0 0 0 0 07 0 0 1 0 0 1 0 0 0 08 0 0 0.83 0 0.83 0 0 0 0 0

Table 8-20: Applied Load Combination - Allowable Stress Level

Load


Erection

LoadLiveLoad

UserLoad1

UserLoad2

UserLoad3

1 0 1.4 0 0 0 0 0 0 0 02 0 1.2 0 0 0 0 1.6 0 0 03 0 1.2 1.6 0 0 0 0 0 0 04 0 1.2 0 1 0 0 0 0 0 05 0.9 0 1.6 0 0 0 0 0 0 06 0 0.9 0 1 0 0 0 0 0 07 0.9 0 0 1 0 0 0 0 0 08 0 0 1.6 0 0 0.9 0 0 0 09 0 0 0 0 0 0 0 0 0 010 0 0 0 0 0 0 0 0 0 0

Table 8-21: Applied Load Combination - Strength Level

Load


1 40 0 02 40 2.8605 44.7353 40 2.3816 19.0534 20 2.8605 44.7355 36 2.3816 19.0536 18 1.1908 9.52647 30 1.4302 22.3688 49.8 1.1871 18.565





Governing Loads

Axial = 49.8 kip

Shear = 2.86046 kip

Moment = 44.73534 kip-ft

Load


1 56 0 02 48 0 03 48 4.5767 71.5774 48 3.4023 27.2185 18 4.5767 71.5776 36 3.4023 27.2187 18 1.7011 13.6098 27 2.2884 35.788

Table 8-23: Applied Load atTop of Pedestal - Strength Level

Governing Loads

Axial = 56 kips

Shear 4.57674 kips

Moment = 71.5765 ft-kip

Pedestal Design

Fu = 56.64 kip



Factored O.T.M. At Base Of Pedestal = 80.730 kip-ft




Dowel Circle Diameter, Dc = BCD = 1.333 ft



Fu = 4·Muped/(Nd·Dc) - 0.9·(Du + Wped)/Nd = 5.212 kip


As_ped_req = Fu/(φ·fy) =0.116 in2


Chapter — 8




Area of Steel Provided = 0.11582 in2

Area of steel required in pedestal, As, req = (dd)2 x pi/4 = 0.307 in2

Potential Conc. Failure Area per PIP STC03350 Fig. A, An = 0.075956ft2



db = 0.625 in

Fc = Pu/An = 12.43 ksi

Beta = 1

Weight of Soil, Wsoil = 9.477 kip

Design Results



Ratio1 0 02 0.4864 14.3923 0.2390 29.2914 0.5949 11.7665 0.2480 28.2226 0.1495 46.8257 0.2676 26.1598 0.1853 37.780



Load

CaseMax Soil

Bearing (ksf)Min Soil

Bearing (ksf)1 0.6751 0.24532 0.8672 0.31643 0.7695 0.28084 0.7440 0.74405 0.7449 0.25396 0.5402 0.54027 0.7100 0.21378 0.8152 0.3399

Table 8-25: Soil Bearing Check






Concrete Design

One Way Shear


Location of Pedestal Face from Face of Footing, X1 = 1.538921 ft

Location of Shear check from Face of Footing, X2 = 0.814963 ft


Factored Shear Stress Capacity per ACI318-05 Eqn. 11-3, φVC = 13.66104 ksf

Two way Shear Check








Chapter — 8


Available development length for bars (From face of Pedestal to face of Footing) = 1.538921ft

Rn = Mu / (0.9·1ft·d2) = 0.014 ksi

m = fy / (0.85· fc) = 17.647

2·m·Rn / fy = 7.944(10)-3


Ast,req = ρ·d·1ft = 0.024 in2

ρmin = 0.0018

Minimum Area of Steel Req

Ast,min = ρmin·d·1 ft = 0.188 in2

Spacing Required, s = 18 in.



Bar Size = 5

Final Dimensions

Footing Diameter = 14 ft

Footing Thickness = 1 ft




8.3.3 Comparison

Hand Cal-culations


Percent Dif-ference

Footing Diagonal 14 14 noneFooting Thickness 1 1 noneFooting Soil Bear-ing

0.8672 0.835 4%

Stability Check 11.766 12.898 9%One Way ShearCheck

0.9052 1.13 20%

Punching ShearCheck

12.085 12.338 2%


0.2045 0.1878 9%


8.4 Vertical Vessel Seismic Load Generation1

Location Santa Ana California

S1 = 0.5312, Spectral Response Acceleration at Short Periods determined in accordancewith ASCE 7 11.4.1

Ss = 1.378, Spectral Response Acceleration at Period of 1 sec determined in accordance withASCE 7 11.4.1

Site Class = A, Based On Soil Prorperties In Accordance With ASCE 7 Chapter 20

R = 2, Response modification Coefficient per ASCE 7 Tables 15.4-1 or 15.4-2

I = 1.25, Importance Factor per ASCE 7 11.5.1

T = 3 sec., Fundamental Period of Vessel

TL = 12 sec., Long-Period Transition Perios per ASCE 7 12.8.2

Empty Weight Of Vessel = 100 kips

Operating Weight of Vessel = 200 kips

Center of Gravity Of Vessel From Top Of Pedestal (CG) = 120 in.

Fa = 0.8 Short-Period Site Coeffiecient per ASCE 7 11.4.3

Fv = 0.8 Long-Period site Coefficient per ASCE 7 11.4.3

SDS = 0.735, Design Spectral Response Acceleration Parameter at short periods per ASCE 711.4.4

SD1 = 0.283, Design Spectral Response Acceleration Parameter at period of 1 sec per ASCE7 11.4.4

Cs = 0.059, Seismic Response Coefficient Per ASCE 7 12.8.1.1

Empty Load Case Base Shear = 5.902 kip


Chapter — 8

8.4 Vertical Vessel Seismic Load Generation 1

Operating Load Case Base Shear = 11.804 kip

Empty Load Case Earthquake Moment = 59.021 kip ft

Operating Load Case Earthquake Moment = 118.043 kip ft

Comparison



Percent Dif-ference

Base Shear Empty Case(kip)

5.902132222 5.902 0.0022403

Base Shear Operating Case(kip)

11.80426444 11.804 0.0022403

Base Moment Empty Case(kip ft)

59.02132222 59.021 0.00054595

Base Moment OperatingCase (kip ft)

118.0426444 118.043 0.00030121



Location Yorba Linda California

Spectral Response Acceleration at Short Periods determined in accordance with ASCE 711.4.1, S1 = 0.857367

Spectral Response Acceleration at Period of 1 sec determined in accordance with ASCE 711.4.1, SS = 2.09511

Site Class = B, Based On Soil Properties In Accordance With ASCE 7 Chapter 20

Response modification Coefficient per ASCE 7 Tables 15.4-1 or 15.4-2, R = 2

Importance Factor per ASCE 7 11.5.1, I = 1.0

Fundamental Period of Vessel, T = 6 sec.

Long-Period Transition Period per ASCE 7 12.8.2, TL = 5 sec.

Empty Weight Of Vessel = 100 kip

Operating Weight of Vessel = 200 kip

Center of Gravity Of Vessel From Top Of Pedestal, CG = 10 ft

Short-Period Site Coefficient per ASCE 7 11.4.3, FA = 1

Long-Period site Coefficient per ASCE 7 11.4.3, FV = 1


SD1 = 0.572, Design Spectral Response Acceleration Parameter at period of 1 sec per ASCE 711.4.4

CS = 0.04, Seismic Response Coefficient Per ASCE 7 12.8.1.1




Empty Load Case Base Shear

Shearempty = CS x Emptywt = 3.969 kips

Operating Load Case Base Shear

Shearoperating = CS x Operatingwt = 7.939 kips

Empty Load Case Earthquake Moment

Momentempty = Shearempty x CG = 39.693 kip*ft

Operating Load Case Earthquake Moment

Momentoperating = Shearoperating x CG = 79.386 kip*ft

Comparison



Percent Dif-ference


3.9693 3.969 0.007


7.9386 7.939 0.005


39.693 39.69 0.007


79.3858 79.39 0.005



Location San Antonio, Texas



Site Class = C, Based On Soil Properties In Accordance With ASCE 7 Chapter 20








Short-Period Site Coefficient per ASCE 7 11.4.3, FA = 1.2


Chapter — 8


Long-Period site Coefficient per ASCE 7 11.4.3, FV = 1.7



CS = 4.91x10-3, Seismic Response Coefficient Per ASCE 7 12.8.1.1






Momentempty = Shearempty x CG =4.91 kip*ft


Momentoperating = Shearoperating x CG = 9.819 kip*ft

Comparison



Percent Dif-ference


0.49096 0.491 0.0081


0.98192 0.982 0.0081


4.9096 4.91 0.0081


9.8192 9.82 0.0081



Location New York, New York



Site Class = D

Site Class Based On Soil Properties In Accordance With ASCE 7 Chapter 20














SD1 = 0.15 Design Spectral Response Acceleration Parameter at period of 1 sec per ASCE 711.4.4









Momentoperating = Shearoperating x CG =24.992 kip*ft

Comparison



Percent Dif-ference


1.249592 1.25 0.03264


2.499184 2.499 0.00736295


12.49592 12.496 0.0006402


24.99184 24.992 0.0006402



Location Nashville, Tennessee


Chapter — 8




Site Class = E, Based On Soil Properties In Accordance With ASCE 7 Chapter 20












CS = 9.039x10-3, Seismic Response Coefficient Per ASCE 7 12.8.1.1








Momentoperating = Shearoperating x CG =18.077 kip*ft




Comparison



Percent Dif-ference


0.903868649 0.904 0.01452995


1.807737298 1.808 0.01452995


9.038686492 9.039 0.00346839


18.07737298 18.077 0.00206331



Location Santa Ana California


Ss = 1.378, Spectral Response Acceleration at Period of 1 sec determined in accordance withASCE 7 11.4.1

Site Class = A; Based On Soil Properties In Accordance With ASCE 7 Chapter 20



TL = 12 sec., Fundamental Period of Vessel

Long-Period Transition Period per ASCE 7 12.8.2

Empty Weight Of Vessel, Ewt = 100 kip

Operating Weight of Vessel, Operatingwt = 200 kip


Fa = 0.9, Short-Period Site Coefficient per ASCE 7 11.4.3

Fv =0.8, Long-Period site Coefficient per ASCE 7 11.4.3




Base Shear Based on Operating Load Condition

V = Cs·Operatingwt = 4.722 kip


Chapter — 8


Comparison



Percent Dif-ference

Seismic Response Coef-ficient (Cs)

0.02361 0.024 1.631

Base Shear OperatingCase (kip)

4.722 4.722 0.006


8.10 Vertical Vessel Seismic LoadGeneration 7

Location Yorba Linda California 92887

S1, 0.8574, Spectral Response Acceleration at Short Periods determined in accordance withASCE 7 11.4.1

SS = 2.0951, Spectral Response Acceleration at Period of 1 sec determined in accordancewith ASCE 7 11.4.1

Site Class = B, Based On Soil Properties In Accordance With ASCE 7 Chapter 20




TL = 12 sec., Long-Period Transition Period per ASCE 7 12.8.2



Center of Gravity Of Vessel From Top Of Pedestal, CG = 2.5 ft

Fa = 1, Short-Period Site Coefficient per ASCE 7 11.4.3

Fv = 1, Long-Period site Coefficient per ASCE 7 11.4.3









Comparison



Percent Dif-ference


0.0715 0.071 0.63


14.289 14.289 0.002



Location Mountain Valley 73062



Site Class = C, Based On Soil Properties In Accordance With ASCE 7 Chapter 20







Center of Gravity Of Vessel From Top Of Pedestal, CG = 2.5 ft

Fa = 1.2, Short-Period Site Coefficient per ASCE 7 11.4.3

Fv = 1.7, Long-Period site Coefficient per ASCE 7 11.4.3

SDS 0.263, Design Spectral Response Acceleration Parameter at short periods per ASCE 711.4.4




V = Cs·Operatingwt =3.339 kip


Chapter — 8


Comparison



Percent Dif-ference


0.0167 0.017 1.787


3.339 3.339 0.007



Location San Bernadino 92411

S1 = 1.415, Spectral Response Acceleration at Short Periods determined in accordance withASCE 7 11.4.1

SS = 2.984, Spectral Response Acceleration at Period of 1 sec determined in accordance withASCE 7 11.4.1

Site Class = D, Based On Soil Properties In Accordance With ASCE 7 Chapter 20

R =2, Response modification Coefficient per ASCE 7 Tables 15.4-1 or 15.4-2


T = 3.5 sec., Fundamental Period of Vessel





Fa = 1, Short-Period Site Coefficient per ASCE 7 11.4.3

Fv = 1.5, Long-Period site Coefficient per ASCE 7 11.4.3



Cs = 0.202, Seismic Response Coefficient Per ASCE 7 12.8.1.1






Comparison



Percent Dif-ference


0.2021 0.202 none


40.429 40.429 none


8.13 Vertical Vessel Wind Load Generation 1Vessel Data

Height of Vessel, H = 60 ft

Diameter of Vessel, D = 15 ft

Vessel Time Period, t = 5 sec.

Pedestal height Above Ground = 1 ft

Vessel Frequency, f = 1/t = 0.2 Hz

Wind Parameters

Wind Speed = 80 mph

Wind Directional Factor, Kd = 0.95, per ASCE 7-05 6.5.6.6

Wind Exposure = C

Exposure Case = 2

Topographic Factor, Kzt = 1.0, per ASCE 7-05 6.5.7.2

Importance factor, I = 1.15, per ASCE 7-05 6.5.5

Gust Wind Effect Factor, G = 0.85, per ASCE 7-05 6.5.8

Net Force Coefficient, Cf = 1, per ASCE 7-05 Fig. 6-20 & 6-21

Velocity Pressure

Exposure Coefficient, Kz, per ASCE 7-05 6.5.6.6

Design Wind Pressure per ASCE 7-05 6.5.10

qz = 0.00256·Kd·Kz·V·I·G·CfShear Force on Top of Pier

F = qz·G·Cf·A

Moment on Top of Pier

M = F x Moment Arm


Chapter — 8

8.13 Vertical Vessel Wind Load Generation 1

Kzqz(psf)

A(ft2) F (kip) M

(ft·kip)0.849 15.195 15 0.194 0.0970.849 15.195 210 2.712 21.6980.902 16.143 75 1.029 18.0100.945 16.920 75 1.079 24.2690.982 17.582 75 1.121 30.8231.044 18.680 150 2.382 83.8231.094 19.578 150 2.496 112.3291.137 20.344 150 2.594 142.6631.141 20.415 15 0.260 15.748

Σ 13.867 448.995

Total Shear Force at Top of Pier = 13.867 kip

Total Moment at Top of Pier = 448.995 ft·kip

Comparison



Percent Dif-ference

Wind Shear on Top of Pier(kip)

13.867 13.867 none

Wind Moment of Top ofPier (ft·kip)

448.995 448.898 negligible


8.14 Vertical Vessel Wind Load Generation 2Vessel DataHeight of Vessel, h = 74 ft

Diameter of Vessel, D = 20 ft




Wind ParametersWind Speed = 90 mph


Wind Exposure = D

Exposure Case = 2







Net Force Coefficient, Cf = 0.8, per ASCE 7-05 Fig. 6-20 & 6-21

Velocity Pressure




F = qz·G·Cf·A


M = F x Moment Arm

Kzqz(psf)

A(ft2)

F(kip) M (ft·kip)

1.030 23.339 20 0.317 0.1591.030 23.339 280 4.444 35.5501.083 24.536 100 1.668 29.1981.126 25.507 100 1.734 39.0261.162 26.329 100 1.790 49.2351.222 27.680 200 3.764 131.7561.270 28.775 200 3.913 176.1031.311 29.702 200 4.039 222.1711.347 30.509 200 4.149 269.7001.369 31.019 140 2.953 217.046

Σ 28.77 1169.945



Comparison



Percent Dif-ference

Wind Shear on Top ofPier (kip)

28.774 28.774 none


1169.945 1168.516 0.12


8.15 Vertical Vessel Wind Load Generation 3Vessel DataHeight of Vessel = 74 ft


Chapter — 8


Diameter of Vessel = 20 ft




Wind ParametersWind Speed = 110 mph


Wind Exposure = B

Exposure Case = 2





Velocity Pressure




F = qz·G·Cf·A


M = F x Moment Arm

Kzqz(psf)

A(ft2) F (kip) M

(ft·kip)0.575 19.449 20 0.265 0.1320.575 19.449 280 3.703 29.6250.624 21.115 100 1.436 25.1270.665 22.505 100 1.530 34.4330.701 23.709 100 1.612 44.3360.761 25.740 200 3.501 122.5220.811 27.434 200 3.731 167.8990.854 28.901 200 3.931 216.1830.892 30.203 200 4.108 266.9930.917 31.037 140 2.955 217.169

Σ 26.771 1124.42

Total Shear Force at Top of Pier 26.771 kip





Comparison



Percent Dif-ference

Wind Shear on Top ofPier (kip)

26.771 26.771 none


1124.42 1123.23 0.11


8.16 Vertical Vessel Wind Load Generation 4Vessel Data

Height of Vessel = 22 ft

Diameter of Vessel = 6 ft




Wind Parameters

Wind Speed = 110 mph


Wind Exposure = B

Exposure Case = 2





Velocity Pressure




F = qz·G·Cf·A


M = F x Moment Arm


Chapter — 8


Kzqz(psf)

A(ft2)

F(kip)

M(ft·kip)

0.575 19.449 6 0.079 0.0400.575 19.449 84 1.111 8.8880.624 21.115 30 0.431 7.5380.657 22.245 24 0.363 7.987

Σ 1.984 24.452



Comparison



Percent Dif-ference

Wind Shear on Top of Pier(kip)

1.984 1.984 none


24.452 24.293 negligible


8.17 Horizontal Vessel Applied Loads 1Pier Design Philosophy

Vessel Loads

You are provided an option of making both piers identical, engineering and installationerrors can be avoided by doing so. If making them identical does not lead to economicalsolution, then check applied load distribution, follow PIP 4.3

Operating Load = 106 kip

Empty Load = 92 kip

Test Load = 118 kip

A major difference between pier loads is caused by longitudinal seismic load (i.e., whenvessel is located in higher seismic area). Thermal load and bundle pull force also contributetowards the difference.

Thermal Load = 0 kip

Live Load = 0 kip

Erection Load = 110 kip

Anchor Bold data

Anchor bold dia., Db = 1 in

BP = 0 kip


8.17 Horizontal Vessel Applied Loads 1


Bolt spacing = 9 in

Bolt edge dist. = 6 in

center-to-center = 2 in

Horizontal vessel Data

Vessel CG from Top of Pier = 7 ft - 5ft/2 = 4.5 ft

Tan to Tan Length, L = 23.5 ft

CL to CL of AB, Lab = 11 ft

Vertical Load DistributionVessel Outer Dim including Insulation = 42 in

% Distribution For Vertical Loads At Fixed End = 40%

% Distribution For Vertical Loads At Sliding End = 60%

CG to Bott of Baseplate = 10 ft

Empty Load at fixed end = 36.8 kip

Operating Load at fixed end = 42.4 kip

Earthquake LoadsTransverse Earthquake Load Operating Condition

Bottom of the Base Plate to HPFS (est) = 10 kip

Longitudinal Earthquake Load Operating Condition = 15 kip

Transverse Earthquake Load Empty Condition = 9 kip

Longitudinal Earthquake Load Empty Condition= 10 kip

X = long

Z = trans

Wind Loadsμ = 0.4

Transverse Wind Load on Vessel = 10 kip

Longitudinal Wind Load on Vessel = 3.5 kip

Longitudinal Wind Load on Pier = 0.5 kip


Chapter — 8


PIP Load combinations

LoadComb. Empty Operating Test Thermal Bundle EQ Trans EQ Long W

TransW

Long Live Erection

1 0 1 0 1 0 0 0 0 0 0 02 0 1 0 1 0 0 0 0 0 1 03 0 1 0 0 0 0 0 1 0 0 04 0 1 0 0 0 0 0 0 1 0 05 0 1 0 0 0 1 0 0 0 0 06 0 1 0 0 0 0 1 0 0 0 07 1 0 0 0 0 0 0 1 0 0 08 1 0 0 0 0 0 0 0 1 0 09 0 0.9 0 0 0 0.7 0 0 0 0 010 0 0.9 0 0 0 0 0.7 0 0 0 011 0.9 0 0 0 0 0.7 0 0 0 0 012 0.9 0 0 0 0 0 0.7 0 0 0 013 0 0 0 0 0 0 0 1 0 0 114 0 0 0 0 0 0 0 0 1 0 115 0 0 0.83 0 0 0 0 0.415 0 0 016 0 0 0.83 0 0 0 0 0 0.415 0 017 1 0 0 0 1 0 0 0 0 0 0

Table 8-40: Service level load combinations per PIP


TransWLong Live Erection

1 0 1.4 0 1.4 0 0 0 0 0 0 02 0 1.2 0 1.2 0 0 0 0 0 1.6 03 0 1.2 0 0 0 0 0 1.6 0 0 04 0 1.2 0 0 0 0 0 0 1.6 0 05 0 1.2 0 0 0 1 0 0 0 0 06 0 1.2 0 0 0 0 1 0 0 0 07 0.9 0 0 0 0 0 0 1.6 0 0 08 0.9 0 0 0 0 0 0 0 1.6 0 09 0 0.9 0 0 0 1 0 0 0 0 010 0 0.9 0 0 0 0 1 0 0 0 011 0.9 0 0 0 0 1 0 0 0 0 012 0.9 0 0 0 0 0 1 0 0 0 013 0 0 0 0 0 0 0 1.6 0 0 0.914 0 0 0 0 0 0 0 0 1.6 0 0.915 0 0 1.2 0 0 0 0 0.8 0 0 016 0 0 1.2 0 0 0 0 0 0.8 0 017 0.9 0 0 0 1.6 0 0 0 0 0 0

Table 8-41: Strength level load combinations per PIP




LoadComb.

AxialLoad(kip)

LongitudinalShear (kip)

TransverseShear (kip)

TransverseMoment(ft-kip)

LongitudinalMoment (ft-

kip)1 42.4 0 0 0 02 42.4 0 0 0 03 42.4 0 5 0 04 42.4 2.25 0 0 05 42.4 0 7 0 06 42.4 10.5 0 0 07 36.8 0 5 0 08 36.8 2.25 0 0 09 38.16 0 4.9 0 010 38.16 7.35 0 0 011 33.12 0 4.41 0 012 33.12 4.9 0 0 013 44.0 0 5 0 014 44.0 2.25 0 0 015 39.176 0 2.075 0 016 39.176 0.9338 0 0 017 36.8 0 0 0 0

Table 8-42: Service level loads applied at the top of the top of the fixedpier

LoadComb.

AxialLoad(kip)





kip)1 59.36 0 0 0 02 50.88 0 0 0 03 50.88 0 4 0 04 50.88 3.6 0 0 05 50.88 0 7 0 06 50.88 10.5 0 0 07 33.12 0 4 0 08 33.12 3.6 0 0 09 38.16 0 7 0 010 38.16 10.5 0 0 011 33.12 0 6.3 0 012 33.12 7 0 0 013 39.60 0 4 0 014 39.60 3.6 0 0 015 56.64 0 2 0 016 56.64 1.8 0 0 017 33.12 0 0 0 0

Table 8-43: Strength level loads applied at the top of the top of thefixed pier


Chapter — 8


ComparisonHand Calculation results exactly match the results of STAAD Foundation analysis.

8.18 Horizontal Vessel Applied Loads 2Pier Design Philosophy

Vessel LoadsYou are provided an option of making both piers identical, engineering and installationerrors can be avoided by doing so. If making them identical does not lead to economicalsolution, then check applied load distribution, follow PIP 4.3

Operating Load = 150 kip

Empty Load = 100 kip

Test Load = 160 kip

A major difference between pier loads is caused by longitudinal seismic load (i.e., whenvessel is located in higher seismic area). Thermal load and bundle pull force also contributetowards the difference.

Thermal Load = 0

Bundle Pull Force = 0

Live Load = 0

Erection Load = 110 kip

Anchor Bold data

Anchor bold dia., Db = 1 in

BP = 0 kip

Bolt spacing = 9 in

Bolt edge dist. = 6 in

center-to-center = 2 in

Horizontal vessel Data

Vessel CG from Top of Pier = 7 ft - 6.5ft/2 = 3.75 ft

Tan to Tan Length, L = 23.5 ft

CL to CL of AB, Lab = 11 ft

Vertical Load DistributionVessel Outer Dim including Insulation = 42 in

% Distribution For Vertical Loads At Fixed End = 40%

% Distribution For Vertical Loads At Sliding End = 60%




CG to Bott of Baseplate = 10 ft

Empty Load at fixed end = 40 kip

Operating Load at fixed end = 60 kip

Earthquake LoadsLocation Corona 92880

S1 = 0.882, Spectral Response Acceleration at Short Periods determined in accordance withASCE 7 11.4.1


Site Class = D, Based On Soil Properties In Accordance With ASCE 7 Chapter 20




TL = 12 sec., Long-Period Transition Periods per ASCE 7 12.8.2

Center of Gravity Of Vessel From Top Of Pedestal = 7 ft + 10.416 ft - 5 ft =

fa = , Short-Period Site Coefficient per ASCE 7 11.4.3

fv = , Long-Period site Coefficient per ASCE 7 11.4.3




Base Shear Based on Operating Load Condition = 11.029 kip

Base Shear Based on Empty Load Condition = 7.352 kip

Transverse Earthquake Load Operating Condition = 11.029 kip

Longitudinal Earthquake Load Operating Condition

Transverse Earthquake Load Empty Condition= 7.352 kip

Longitudinal Earthquake Load Empty Condition= 7.352 kip

Wind LoadsTransverse Wind Load on Vessel = 10 kip

Bottom of the Base Plate to HPFS (est)

μ = 0.4

Longitudinal Wind Load on Vessel = 3.5 kip

Longitudinal Wind Load on Pier = 0.5 kip

Transverse Wind Moment at top of pedestal = 20 ft-kip

X = long


Chapter — 8


Z = trans

Longitudinal Wind Load at top of pedestal = 10 ft-kip

PIP Load combinations


TransW

Long Live Erection

1 0 1 0 1 0 0 0 0 0 0 02 0 1 0 1 0 0 0 0 0 1 03 0 1 0 0 0 0 0 1 0 0 04 0 1 0 0 0 0 0 0 1 0 05 0 1 0 0 0 1 0 0 0 0 06 0 1 0 0 0 0 1 0 0 0 07 1 0 0 0 0 0 0 1 0 0 08 1 0 0 0 0 0 0 0 1 0 09 0 0.9 0 0 0 0.7 0 0 0 0 010 0 0.9 0 0 0 0 0.7 0 0 0 011 0.9 0 0 0 0 0.7 0 0 0 0 012 0.9 0 0 0 0 0 0.7 0 0 0 013 0 0 0 0 0 0 0 1 0 0 114 0 0 0 0 0 0 0 0 1 0 115 0 0 0.83 0 0 0 0 0.415 0 0 016 0 0 0.83 0 0 0 0 0 0.415 0 017 1 0 0 0 1 0 0 0 0 0 0

Table 8-44: Service level load combinations per PIP


TransWLong Live Erection

1 0 1.4 0 1.4 0 0 0 0 0 0 02 0 1.2 0 1.2 0 0 0 0 0 1.6 03 0 1.2 0 0 0 0 0 1.6 0 0 04 0 1.2 0 0 0 0 0 0 1.6 0 05 0 1.2 0 0 0 1 0 0 0 0 06 0 1.2 0 0 0 0 1 0 0 0 07 0.9 0 0 0 0 0 0 1.6 0 0 08 0.9 0 0 0 0 0 0 0 1.6 0 09 0 0.9 0 0 0 1 0 0 0 0 010 0 0.9 0 0 0 0 1 0 0 0 011 0.9 0 0 0 0 1 0 0 0 0 012 0.9 0 0 0 0 0 1 0 0 0 013 0 0 0 0 0 0 0 1.6 0 0 0.914 0 0 0 0 0 0 0 0 1.6 0 0.915 0 0 1.2 0 0 0 0 0.8 0 0 016 0 0 1.2 0 0 0 0 0 0.8 0 017 0.9 0 0 0 1.6 0 0 0 0 0 0

Table 8-45: Strength level load combinations per PIP




LoadComb.

AxialLoad(kip)





kip)1 60 0 0 0 02 60 0 0 0 03 60 0 2.5 5 04 60 1.125 0 0 105 60 0 5.404 20.27 06 60 5.404 0 0 20.277 40 0 2.5 5 08 40 1.125 0 0 109 54 0 3.783 14.19 010 54 3.783 0 0 14.1911 36 0 2.522 9.457 012 36 2.522 0 0 9.45713 44 0 2.5 5 014 44 1.125 0 0 1015 53.12 0 1.038 2.075 016 53.12 0.467 0 0 4.15017 40 0 0 0 0

Table 8-46: Service level loads applied at the top of the top of thefixed pier

LoadComb.

AxialLoad(kip)





kip)1 84 0 0 0 02 72 0 0 0 03 72 0 4 8 04 72 1.8 0 0 165 72 0 5.404 20.27 06 72 5.404 0 0 20.277 36 0 4 8 08 36 1.8 0 0 169 54 0 5.404 20.27 010 54 5.404 0 0 20.2711 36 0 3.602 13.509 012 36 3.602 0 0 13.50913 39.6 0 4 8 014 39.6 1.8 0 0 1615 76.8 0 2 4 016 76.8 0.9 0 0 817 36 0 0 0 0

Table 8-47: Strength level loads applied at the top of the top of thefixed pier


Chapter — 8


ComparisonHand Calculation results exactly match the results of STAAD Foundation analysis.





Chapter 8


Section 9

Chinese Code (GB50007-2002)

The manual test cases in accordance with Chinese standard GB50007-2002 design. PKPMand the Foundation of the input data consistent, easy to check the results

9.1 Cone Footing DesignIndependent bases for design details

This example is included in the \VERIFICATION\CHINESE\ISO_CHN.AFS example file forreference.

Overview of tapered design results based on

NodeNumber

GroupNumber

Basic Geometry (Conical Base)Length(X Dir.)

Width (ZDir.) Thickness Edge height

(should be > 0.2 m)1 1 3.000 m 3.000 m 1.000 m 0.500 m

Table 9-1: Overview of cone footing design results

BaseNumber

Foundation Reinforcement Base Rein-forcement

Bottom Reinf.(Mz)

Bottom Reinf.(Mx)

MainBars Stirrups

1 # 10 @ 60 mm c /c

# 10 @ 60 mm c /c

N / A N / A

Table 9-2: Reinforcement details


9.1.1 Problem

Characteristics of Concrete and Steel

Heavy concrete units: 18.000 kN/m3

Compressive strength of concrete: 11.900 N/mm2

Reinforcement strength: 210.000 N/mm2

Minimum bar size: # 6

Maximum bar size: # 50

Minimum bar spacing: 50.00 mm

Maximum bar spacing: 500.00 mm

Clear Cover

Reinforcement layer thickness (F, CL): 50.00 mm

Soil Characteristics

Unit Weight: 18.00 kN/m3

Foundation bearing capacity: 180.00 kPa

Surcharge: 0.00 kN/m2

Height of soil above footing: 2000.00 mm

Geometry Information

Initial size of base

Thickness (Ft) : 1000.00 mm

Length - X (Fl) : 3000.00 mm

Width - Z (Fw) : 3000.00 mm

Edge height of the cone footing (St) : 500.00 mm

Column Dimension

Column Shape: Rectangular

Column length - X (Pl) : 600.00 mm

Column width - Z (Pw) : 600.00 mm

Column Cap

Column cap length - X : N/A

Column cap width - Z : N/A


Chapter — 9

9.1 Cone Footing Design

9.1.2 Solution

ConditionNo.

VerticalForce (KN)

Shear X(KN)

Shear Z(KN)

Moment X(kN·m)

Moment Z(kN·m)

101 1000.000 0.000 0.000 99.998 99.998

Table 9-3: Loads for foundation base size estimation -For foundationbase (1)

LC Vertical Force(KN)

Shear X(KN)

Shear Z(KN)

Moment X(kN·m)

Moment Z(kN·m)

102 1000.000 0.000 0.000 99.998 99.998

Table 9-4: Loads for Punching shear check and reinforcements- For foun-dation base (1)

Basic dimensions

Initial size (Lo) = 3.00 m

Initial size (Wo) = 3.00 m

Net buoyancy = -0.00 kN

Adhesion = 0.00 kN

The minimum required base area, Amin = P / fa = 7.356 m2

The initial design area , Ao = Lo·Wo = 9.00 m2

Final design size

Length (L2) = 3.00 m

No. of control condition: # 101

Width (W2) = 3.00 m

No. of control condition: # 101

Area (A2) = 9.00 m2

Figure 9-1: Four corners of the calculated stress

Section 9 Chinese Code (GB50007-2002)



LoadCase

Pressure atCorner1(q1)

(KN/m2)


(KN/m2)


(KN/m2)


(KN/m2)

Zero-pres-sure area(Au) (m2)

101 147.1111 102.6667 147.1111 191.5556 0.00101 147.1111 102.6667 147.1111 191.5556 0.00101 147.1111 102.6667 147.1111 191.5556 0.00101 147.1111 102.6667 147.1111 191.5556 0.00

If Au equals zero, that means it is small eccentricity, and do not need to adjust thepressure. Otherwise, the pressure needs to be adjusted. The negative pressure shouldalways set as 0. Keep adjusting if necessary.

Four corners of the stress adjusted data (if any).

No. LoadCondition

Pressure atCorner1 (q1)(KN/m2)




101 147.1111 102.6667 147.1111 191.5556101 147.1111 102.6667 147.1111 191.5556101 147.1111 102.6667 147.1111 191.5556101 147.1111 102.6667 147.1111 191.5556

If necessary, the bottom will be adjusted accordingly based on size.

Zero-pressure area ( if any )

Control the condition number = N / A

Foundation area = 9.00 m2

Zero-pressure area = 0.00 m2

Zero-pressure area percentage = 0.00%

9.1.3 Check Overturning and Sliding StabilityFigure 9-2: Elevation of stability forces


Chapter — 9


Load Case No. Sliding Factor of Safety Overturning Factor of SafetyX Dir. Z Dir. X Dir. Z Dir.

101 N / A N / A 19.763 19.763

Table 9-5: Factor of safety

Critical load cases and governing factor of safety ofoverturning

Along the X Direction

Critical sliding load case along the X direction: 101

Governing sliding force : 0.000 kN

Resisting Force for Sliding: 658.760 kN

Minimum sliding coefficient under critical load case: 0.000

Critical Overturning load case along X direction: 101

Critical overturning moment: 99.998 kN·m

Resisting moment for Overturning: 1976.244 kN·m

Minimum overturning coefficient under critical load case: 19.763

Along the Z Direction

Critical sliding load case along the Z direction : 101

Governing sliding force : 0.000 kN



Critical Overturning load case along Z direction : 101


Resisting Moment for Overturning: 1976.244 kN·m


9.1.4 Check ShearFollowing formulae are used per GB50007 - 2002 code for design of building foundations.

Fl ≤ 0.7·βhpftamh0 (Ref. clause 8.2.7 - 1)

am = (At + ab) / 2 (Ref. clause 8.2.7 - 2)

Fl = Pj·Al (Ref. clause 8.2.7 - 3)

Punching One-way Check

Positive X Side

Control condition = # 102




Punching shear

Fl = Pj·Al = 88.889·724774.976 = 64.424 kN

Punching shear capacity

Fu = 0.7·βhpftamh0 = 0.7·0.92·1.270·(0.600 +2.470) / 2·0.935 = 1169.589 kN

Fl < 0.7·βhpftamh0Hence, Safe

Negative X Side


Punching shear

Fl = Pj·Al = 133.333·724774.976 = 96.637 kN


Fu = 0.7·βhpftamh0 = 0.7·0.92·1.270·(0.600 +2.470) / 2·0.935 = 1169.589 kN


Positive Z Side


Punching shear

Fl = Pj·Al = 133.333·724774.976 = 96.637 kN


Fu = 0.7·βhpftamh0 = 0.7·0.917·1.270·(0.600 +2.470) / 2·0.935 = 1169.589 kN


Negative Z Side


Punching shear

Fl = Pj·Al = 88.889·724774.976 = 64.424 kN


Fu = 0.7·βhpftamh0 = 0.7·0.92·1.270·(0.600 +2.470) / 2·0.935 = 1169.589 kN


Two-way punching test (four sides)



Chapter — 9


Punching shear

Fl = Pj·Al = 111.111·2899099.905 = 322.122 kN


Fu = 0.7·βhpftamh0 = 0.7·0.917·1.270·(2.400 +9.880) / 2·0.935 = 4678.355 kN


9.1.5 Reinforcement

Reinforcement Along the X DirectionFigure 9-3: Reinforcement parallel to the X-direction

A simplified formula for reinforcement is used per GB50010 – 2002.

No control condition = # 102

Minimum reinforcement ratio [per Cl. 9.5.2], ρmin = 0.15%

Cross-sectional area about X-axis, A cross = 2,425,000.005

Minimum reinforcement area

Astmin = Ρmin·A cross- = 0.15 · 2,425,000.005 = 3,637.500

Calculate moment

MI = (A1) 2 [(2·l + a ')·(pmax + p - 2·G / A) + (pmax - p)·l] / 12

= 1,200.0002·[(2·3,000.000 +600.000) (0.169 +0.152 - 2·324,000.000/9,000,000.000) + (0.169 -0.152)·3,000.000] / 12 = 20,3519,997.379 kN·m

Calculate area required

Ast = MI / (0.9·h0·fy) = 20,3519,997.379 / (0.9·935.000·210.000) = 1,139.498

Select rebar size, db = 10.000

Minimum allowable reinforcement spacing, Smin = 50.000 mm




Maximum allowable reinforcement spacing, Smax = 500.000 mm

With actual spacing, S = 60.000 mm

Actual area, Ast (Actual) = 3637.500 mm2

Smin ≤ S ≤ Smax

Selected Reinforcement satisfy the requirements.

Astmin ≤ Ast, with real

Selected Reinforcement satisfy the requirements.

Reinforcement Along the Z DirectionFigure 9-4: Reinforcement parallel to the Z-direction




Cross-sectional area about Z-axis, A cross = 2425000.005


Astmin = Ρmin·A cross = 0.15%·2425000.005 = 3637.500

Calculate moment

MI = (A1) 2 [(2·l + a ')·(pmax + p - 2·G / A) + (pmax - p)·l] / 12

= 1200.0002·[(2·3000.000 +600.000) (0.169 +0.152 - 2·324000.000/9000000.000) + (0.169 -0.152)·3000.000] / 12 = 203519997.379 kN·m

Calculate area required

Ast = MI / (0.9·h0·fy) = 203519997.379 / (0.9·935.000·210.000) = 1151.685

Select rebar size, db = 10.000



Chapter — 9





Smin ≤ S ≤ SmaxReinforced selected to meet the requirements .

Astmin ≤ Ast, with realReinforcement meet the requirements

Reinforcements should be placed at the base bottom.

9.2 PKPM Isolated Footing Design9.2.1 Problem

Elevation and Plan

Foundation type: cast-in-site cone footing

Initial iteration base dimensions:

Length = 3000 mm

Width = 3000 mm


9.2 PKPM Isolated Footing Design


Height = 500 mm

Second iteration

Length = 700 mm

Width = 700 mm

Height = 500 mm

Bottom elevation of the basis: -2.0 m

Shifts of the base: S Direction: 0 mm B direction: 0 mm

Reinforcement at the bottom of the base:

Y direction : 10 @ 200

X direction : 10 @ 200

Weight of the foundation and soil:: 18.0 kPa

Column section information

High column section: 600 mm

Column section width: 600 mm

Eccentric x : 0 mm

Eccentric y : 0 mm

Column angle: 0°

Loading information

Basic values of vertical load: Nk = 1000 kN

Basic value of the moment along X dir.: Mx = 100 kN·M

Basic value of the moment along Y dir.: My = 100 kN·M

9.2.2 Solution

Check Shear

Following formula Per GB5007 - 2002 code for design of building foundation:




Resisting Shear force calculation:

X + direction , height H = 1000

Fl = Pj·Al = 133.33·0.69 = 91.67

Fl ≤ 0.7·βhpft (At + ab)·h0/2 = 0.7·0.98·1270.94·(0.60 +2.50)·0.95 / 2 = 1288.19 KN

Punching Shear check is satisfied in this direction.

X- direction , height H = 1000


Chapter — 9


Fl = Pj·Al = 92.59·0.69 = 63.66

0.7·βhp·ft·(at + ab)·ho / 2 = 0.7·0.98·1270.94·(0.60 +2.50)·0.95 / 2 = 1288.19 KN


Y + direction , height H = 1000

Fl = Pj·Al = 92.59·0.69 = 63.66

0.7·βhp·ft·(at + ab)·ho / 2 = 0.7·0.98·1270.94·(0.60 +2.50)·0.95 / 2 = 1288.19 KN


Y- direction , height H = 1000

Fl = Pj·Al = 133.33·0.69 = 91.67

0.7·βhp·ft·(at + ab)·ho / 2 = 0.7·0.98·1270.94·(0.60 +2.50)·0.95 / 2 = 1288.19 KN


Check Shear Edges

H = 1000.

Fl = N - pk·(bc +2·h0)·(hc +2·h0) = 1000.00 - 111.1·(600.0 + 2·950.0)·(600.0 +2·950.0)·1e-6 =305.56 Kn

Fr = 0.7·βhp·ft·am·h0 = 0.7·0.98·1270.9·(600.0 + 600.0 + 2·950.0)·950.0·1e-6 = 5152.76 Kn

Punching Shear check at edges is satisfied.

X + direction , height H = 1000 mm

Fl = Pj·Al = 133.33·0.56 = 74.67

0.7·βhp·ft·(at + ab)·ho / 2 = 0.7·0.98·1270.94·(0.70 + 2.60)·0.95 / 2 = 1371.30 KN


X- direction , height H = 1000 mm

Fl = Pj·Al = 91.85·0.56 = 51.44

0.7·βhp·ft·(at + ab)·ho / 2 = 0.7·0.98·1270.94·(0.70 + 2.60)·0.95 / 2 = 1371.30 KN


Y + direction , height H = 1000 mm

Fl = Pj·Al = 91.85·0.56 = 51.44

0.7·βhp·ft·(at + ab)·ho / 2 = 0.7·0.98·1270.94·(0.70 + 2.60)·0.95 / 2 = 1371.30 KN


Y- direction , height H = 1000 mm

Fl = Pj·Al = 133.33·0.56 = 74.67

0.7·βhp·ft·(at + ab)·ho / 2 = 0.7·0.98·1270.94·(0.70 + 2.60)·0.95 / 2 = 1371.30 KN





Bending Reinforcement

The following formula are used Per GB5007 - 2002 code for design of building foundation:

M = (1/12)a12·[(2·l + a')(Pjmax + Pj) + (Pjmax + Pj)·l ]

Moment calculations

x direction, h0 = 940 mm

M = (1.20)2·[(2·3.00 + 0.60)·(133333.33 + 115555.55) + (133333.33 - 115555.55)·3.00] / 12 = 203.52kN·m

M =(1.20)2·[(2·3.00 + 0.60)·(88888.89 + 106666.66) + (88888.89 - 106666.66)·3.00] / 12 =148.48 kN·m

y direction , h0 = 940 mm

M = (1.20)2·[(2·3.00 + 0.60)·(88888.89 + 106666.66) + (88888.89 - 106666.66)·3.00] / 12 =148.48 kN·m

M = (1.20)2·[(2·3.00 + 0.60)·(133333.33 + 115555.55) + (133333.33 - 115555.55)·3.00] / 12 = 203.52kN·m

x direction , h0 = 940 mm

M = (1.20)2·[(2·3.00 + 0.60)·(133333.33 + 115555.55) + (133333.33 - 115555.55)·3.00] / 12 = 203.52kN·m

M = (1.20)2·[(2·3.00 + 0.60)·(88888.89 + 106666.66) + (88888.89 - 106666.66)·3.00] / 12 =148.48 kN·m


M = (1.20)2·[(2·3.00 + 0.60)·(88888.89 + 106666.66) + (88888.89 - 106666.66)·3.00] / 12 =148.48 kN·m

M = (1.20)2·[(2·3.00 + 0.60)·(133333.33 + 115555.55) + (133333.33 - 115555.55)·3.00] / 12 = 203.52kN·m

Reinforcement calculation:

M1 = 203.520

AGx = M1 / (0.9·h0·fy) = 203520.016 / (0.9·0.940·210.) = 1145.559 mm2

M2 = 203.520

AGy = M2 / (0.9·h0·fy) = 203520.016 / (0.9·0.940·210.) = 1145.559 mm2

M1 = 203.520

AGx = M1 / (0.9·h0·fy) = 203520.016 / (0.9·0.940·210.) = 1145.559 mm2

M2 = 203.520

AGy = M2 / (0.9·h0·fy) = 203520.016 / (0.9·0.940·210.) = 1145.559 mm2

The area of steel at X direction: 1145.559

The area of steel at Y direction: 1145.559

The initial area of steel along X direction is satisfied.


Chapter — 9


The initial area of steel along Y direction is satisfied.

The area of steel required:

AgX: 10 @ 200

AgY: 10 @ 200

9.3 Stepped Foundation Design

NodeNumber

GroupNumber

Basic Geometry Dimension (Base Level)

Order Length X Dir.(M)

Width Z Dir.(M)

Height(M)

1 1 Total 3.000 m 3.000 m 1.200 m- - Article

(1)Order3.000 m 3.000 m 0.400 m

- - Article (2)order 2.000 m 2.000 m 0.400 m

- - Article (3)order 1.000 m 1.000 m 0.400 m

Table 9-6: Overview of the stepped foundation design

NodeNumber

Foundation Reinforcement Base Rein-forcement

Bottom Reinf.(Mz)

Bottom Reinf.(Mx)

MainBars Stirrups

1 # 10 @ 60 mm c /c

# 10 @ 60 mm c /c

N / A N / A

Table 9-7: Reinforcement details

9.3.1 Problem

Basic Geometry

Height of the base - (Ft): 1200.00 mm

Length of the base - X (Fl): 3000.00 mm

Width of the base - Z (Fw): 3000.00 mm

Column Dimension

Column Shape : Rectangular

Length of the Column section - X (Pl): 600.00 mm

Width of the column section - Z (Pw): 600.00 mm

Base

Base length - X: N / A

Base width - Z: N / A




Concrete and Steel Parameters

Concrete density: 18.000 kN/m3

Concrete strength: 11.900 N/mm2

Reinforcement strength: 210.000 N/mm2

Minimum bar Size: # 6

Maximum bar size: # 40

Minimum bar spacing : 50.00 mm

Maximum bar spacing : 500.00 mm

Clear cover (F, CL): 50.00 mm

Soil Properties

Soil type: Drained

Density: 18.00 kN/m3

Foundation bearing capacity : 180.00 kPa

Surcharge: 0.00 kN/m2

Embedment depth of foundation: 2,000.00 mm

Adhesion: 0.00 kN/m2

Factor of Safety for sliding and overturning

Basal friction coefficient: 0.50

Safety factor of sliding: 1.50

Safety factor of overturning: 1.50

ConditionNo.

VerticalForce (KN)

Shear X(KN)

Shear Z(KN)

Moment X(kN·m)

Moment Z(kN·m)

101 1000.000 0.000 0.000 99.998 99.998

Table 9-8: Critical loads for base size estimation - standard combination

LC Vertical Force(KN)

Shear X(KN)

Shear Z(KN)

Moment X(kN·m)

Moment Z(kN·m)

102 1000.000 0.000 0.000 99.998 99.998

Table 9-9: Loads for foundation design- the basic combination

9.3.2 Solution

Foundation Dimensions

The initial length (Lo) = 76.20 m

The initial width (Wo) = 76.20 m

Buoyancy = -0.00 KN


Chapter — 9


Adhesion = 0.00 kN

Minimum area of steel required

Bearing pressure, Amin = P / qmax = 7.356 m2

Initial foundation area , Ao = Lo x Wo = 5806.44 m2

Final Design Sizes

Length of the Base (L2) = 3.00 m

Number of load case: # 101

Width of the base (W2) = 3.00 m


Height of the base (D2) = 1.20 m


Area (A2) = 9.00 m2

Corner Stresses

Initial pressure at four corners ( before adjustment )

Figure 9-5: Four corners of the calculated stress

LoadCase





Zero-pressurearea (Au) (m2)

101 147.1111 102.6667 147.1111 191.5556 0.00101 147.1111 102.6667 147.1111 191.5556 0.00101 147.1111 102.6667 147.1111 191.5556 0.00101 147.1111 102.6667 147.1111 191.5556 0.00

If Au equals zero, that means it is small eccentricity, and do not need to adjust thepressure. Otherwise, the pressure needs to be adjusted. The negative pressure should alwaysset as 0. Keep adjusting if necessary.




four corners of the stress of adjustment ( if necessary )

No. LoadCondition





101 147.1111 102.6667 147.1111 191.5556101 147.1111 102.6667 147.1111 191.5556101 147.1111 102.6667 147.1111 191.5556101 147.1111 102.6667 147.1111 191.5556

If necessary, the bottom will be adjusted accordingly based on size.

Details of the Zero-pressure zone ( if any )

Design condition number = N / A

Area of Foundation Base = 9.00 sq.m

Zero-pressure area = 0.00 sq.m

Zero-pressure area percentage = 0.00%

Check overturning and sliding stability

Factor of Safety table

Load Case No. Sliding Factor of Safety Overturning Factor of SafetyX Dir. Z Dir. X Dir. Z Dir.

101 N / A N / A 19.782 19.782

Table 9-10: Safety factors

Critical loads and governing factor of safety ofoverturning and sliding

Along the X Direction

Critical sliding load case along X direction: 101

Governing sliding force: 0.000 kN



Critical Overturning load case along X direction: 101


Resisting moment for Overturning: 1978.188 kN·m

Minimum overturning coefficient under critical load case 19.782


Chapter — 9


Along the Z Direction

Critical sliding load case along Z direction: 101

Critical sliding force: 0.000



Critical Overturning load case along Z direction: 101


Resisting Moment for Overturning: 1978.188 kN·m


9.3.3 Check ShearThe following formulae are used per GB50007 - 2002 code for design of buildingfoundations.




Punching One-way Check

Positive X Direction


Punching shear

Fl = Pj·Al = 88.889·190774.972 = 16.958 kN


Fu = 0.7·βhpftamh0 = 0.7 · 0.933 · 1.270 · (0.600 + 2.870) / 2 · 1.135 = 1633.932 kN


Negative X Direction


Punching shear

Fl = Pj·Al = 133.333 · 190774.972 = 25.437 kN


Fu = 0.7·βhpftamh0 =0.7 · 0.933 · 1.270 ·(0.600 +2.870) / 2 · 1.135 = 1633.932 kN





Positive Z Direction


Punching shear

Fl = Pj·Al = 133.333 · 190774.972 = 25.437 kN


Fu = 0.7·βhpftamh0 = 0.7 · 0.933 · 1.270 · (0.600 +2.870) / 2 · 1.135 = 1633.932 kN


Negative Z Direction


Punching shear

Fl = Pj·Al = 88.889 · 190774.972 = 16.958 kN


Fu = 0.7·βhpftamh0 = 0.7 · 0.933 · 1.270 · (0.600 +2.870) / 2 · 1.135 = 1633.932 kN


Two-way punching test (four sides)


Punching shear

Fl = Pj·Al = 111.111 · 763099.890 = 84.789 kN


Fu = 0.7·βhpftamh0 = 0.7 · 0.933 · 1.270 · (2.400 +11.480) / 2 · 1.135 = 6535.727 kN



Chapter — 9


9.3.4 Reinforcement Calculations

Along the X AxisFigure 9-6: Reinforcement parallel to the X-direction


Critical load case number = # 102


Cross-sectional area about X-axis, A cross = 2400000.000


Astmin = 0.15(A cross)- = 0.15(2,400,000.000) = 3600.000

Calculate moment

MI = (A1) 2 [(2·l + a ') · (pmax + p-2·G / A) + (pmax - p)·l] / 12

= 1,200.0002 · [(2 · 3,000.000 + 600.000) (0.169 + 0.152 - 2 · 324,000.000/9,000,000.000) +(0.169 - 0.152) · 3000.000] / 12 = 203,519,997.379 kN·m

Calculate the area required

Ast = MI / (0.9·h0·fy) = 203,519,997.379 / (0.9 · 1,135.000 · 210.000) = 1, 069.687

Select Rebar size, db = 10.000



Actual spacing, S = 60.000 mm

Actual area, Ast (Actual) = 3,600.000 mm2

Smin ≤ S ≤ Smax

Selected Reinforcements satisfy the requirements.






Along the Z AxisFigure 9-7: Reinforcement parallel to the Z-direction


Critical load case number = # 102

Minimal reinforcement ratio [per Cl. 9.5.2], ρmin = 0.15%

Cross-sectional area about Z-axis, A cross = 2400000.000


Astmin = Ρmin · A cross- = 0.15% · 2400000.000 = 3600.000

Calculate moment

MI = (A1) 2 [(2·l + a ') · (pmax + p-2·G / A) + (pmax - p)·l] / 12

= 1,200.0002 · [(2 · 3,000.000 + 600.000) (0.169 + 0.152 - 2 · 324,000.000/9,000,000.000) +(0.169 - 0.152) · 3000.000] / 12 = 203,519,997.379 kN·m

Calculate the area required

Ast = MI / (0.9·h0·fy) = 203,519,997.379 / (0.9 · 1,135.000 · 210.000) = 1,084.241

Reinforced selected size, db = 10.000





Smin ≤ S ≤ Smax



Chapter — 9




Reinforcements should be placed at the base bottom.

9.4 PKPM Stepped Footing Design9.4.1 ProblemFoundation type : Cast-in-place, stepped footing

Initial Single base dimensions:

Length = 3,000 mm

Width = 3,000 mm

Height = 400 mm

Second

Length = 2,000 mm

Width = 2,000 mm

Height = 400 mm

Third

Length = 1,000 mm

Width = 1,000 mm

Height = 400 mm

Bottom elevation of the base: -2.0 m

Shift the basis of the heart: S Direction : 0 mm B direction : 0 mm

Bottom Reinforcement:

Y direction : 10 @ 200

X direction : 10 @ 200

Unit self weight of the soil and footing: 18.0 kPa

Column section information:

Height of the column section: 600 mm

Width of the Column section: 600 mm

Eccentricity x : 0 mm

Eccentricity y : 0 mm

Column angle: 0 °

Loading information

The basic values of vertical load: Nk = 1,000 kN

X direction of the basic value of the moment: Mx = 100 kN·m

Y direction of the basic value of the moment: My = 100 kN·m


9.4 PKPM Stepped Footing Design


Elevation and plan

9.4.2 Solution

Check Shear

The following formulae are used per GB5007 - 2002 code of design of buildingfoundations:




Calculate resisting Shear force:

X + direction , height H = 1000

Fl = Pj·Al = 133.33·0.69 = 91.67

Fl ≤ 0.7·βhpft (At + ab)·h0/2 = 0.7 · 0.97 · 1270.94 · (0.60 +2.90) · 1.15 / 2 = 1,730.76 KN

Punching shear check is satisfied along this direction

X- direction , height H = 1200

Fl = Pj·Al = 89.63 · 0.15 = 13.22


Chapter — 9


0.7·βhp·ft·(at + ab)·ho / 2 = 0.7 · 0.97 · 1270.94 · (0.60 +2.90) · 1.15 / 2 = 1,730.76 KN


Y + direction , height H = 1200

Fl = Pj·Al = 89.63 · 0.15 = 13.22

0.7·βhp·ft·(at + ab)·ho / 2 = 0.7 · 0.97 · 1270.94 · (0.60 +2.90) · 1.15 / 2 = 1,730.76 KN


Y- direction , height H = 1200

Fl = Pj·Al = 133.33 · 0.15 = 19.67

0.7·βhp·ft·(at + ab)·ho / 2 = 0.7 · 0.97 · 1270.94 · (0.60 +2.90) · 1.15 / 2 = 1,730.76 KN


Check Shear Edges

H = 1200.

Fl = N-pk · (bc +2 · h0) · (hc +2 · h0) = 1000.00-111.1 · (600.0 +2 ******)*( 600.0 +2 ******)*1e-6 = 65.56 Kn

Fl = 0.7·βhpftamh0 = 0.7 · 0.97 · 1270.9 · (600.0 +600.0 +2 ******)******* 1e-6 = 6923.04 Kn

Sides punching checking meet


Fl = Pj·Al = 133.33 · 0.69 = 91.67

0.7·βhp·ft·(at + ab)·ho / 2 = 0.7 · 1.00 · 1270.94 · (1.00 +2.50) · 0.75 / 2 = 1167.68 KN



Fl = Pj·Al = 92.59 · 0.69 = 63.66

0.7·βhp·ft·(at + ab)·ho / 2 = 0.7 · 1.00 · 1270.94 · (1.00 +2.50) · 0.75 / 2 = 1167.68 KN



Fl = Pj·Al = 92.59 · 0.69 = 63.66

0.7·βhp·ft·(at + ab)·ho / 2 = 0.7 · 1.00 · 1270.94 · (1.00 +2.50) · 0.75 / 2 = 1167.68 KN



Fl = Pj·Al = 133.33 · 0.69 = 91.67

0.7·βhp·ft·(at + ab)·ho / 2 = 0.7 · 1.00 · 1270.94 · (1.00 +2.50) · 0.75 / 2 = 1167.68 KN



Fl = Pj·Al = 91.11 · 0.43 = 38.95

0.7·βhp·ft·(at + ab)·ho / 2 = 0.7 · 1.00 · 1270.94 · (2.00 +2.70) · 0.35 / 2 = 731.75 KN






Fl = Pj·Al = 133.33 · 0.43 = 57.00

0.7·βhp·ft·(at + ab)·ho / 2 = 0.7 · 1.00 · 1270.94 · (2.00 +2.70) · 0.35 / 2 = 731.75 KN



Fl = Pj·Al = 91.11 · 0.43 = 38.95

0.7·βhp·ft·(at + ab)·ho / 2 = 0.7 · 1.00 · 1270.94 · (2.00 +2.70) · 0.35 / 2 = 731.75 KN



Fl = Pj·Al = 133.33 · 0.43 = 57.00

0.7·βhp·ft·(at + ab)·ho / 2 = 0.7 · 1.00 · 1270.94 · (2.00 +2.70) · 0.35 / 2 = 731.75 KN


Bending Reinforcement

The following formula is used per GB50007 - 2002 code for design of building foundations:

M = (1/12)a12·[(2·l + a')(Pjmax + Pj) + (Pjmax + Pj)·l ]

Moment calculations


M = 0.50 · 0.50 [(2 · 3.00 +0.60) · (133,333.33 + 125,925.92) + (133,333.33 - 125,925.92) · 3.00] /12 = 36.11 kN·m

M = 0.50 · 0.50 [(2 · 3.00 +0.60) · (88,888.89 + 96,296.30) + (88,888.89 - 96,296.30) · 3.00]/ 12 = 25.00 kN·m


M = 0.50 · 0.50 [(2 · 3.00 +0.60) · (88,888.89 + 96,296.30) + (88,888.89 - 96,296.30) · 3.00]/ 12 = 25.00 kN·m

M = 0.50 · 0.50 [(2 · 3.00 +0.60) · (133,333.33 + 125,925.92) + (133,333.33 - 125,925.92) · 3.00] /12 = 36.11 kN·m


M = 1.00 · 1.00 [(2 · 3.00 +0.60) · (133,333.33 + 118,518.52) + (133,333.33 - 118,518.52) · 3.00] /12 = 142.22 kN·m

M = 1.00 · 1.00 [(2 · 3.00 +0.60) · (88,888.89 + 103,703.70) + (88,888.89 - 103,703.70) · 3.00]/ 12 = 102.22 kN·m


M = 1.00 · 1.00 [(2 · 3.00 +0.60) · (88,888.89 + 103,703.70) + (88,888.89 - 103,703.70) · 3.00]/ 12 = 102.22 kN·m


Chapter — 9


M = 1.00 · 1.00 [(2 · 3.00 +0.60) · (133,333.33 + 118,518.52) + (133,333.33 - 118,518.52) · 3.00] / 12= 142.22 kN·m


M = 1.20 · 1.20 [(2 · 3.00 +0.60) · (133,333.33 + 115,555.55) + (133,333.33 - 115,555.55) · 3.00] / 12= 203.52 kN·m

M = 1.20 · 1.20 [(2 · 3.00 +0.60) · (88,888.89 + 106,666.66) + (88,888.89 - 106,666.66) · 3.00]/ 12 = 148.48 kN·m


M = 1.20 · 1.20 [(2 · 3.00 +0.60) · (88,888.89 + 106,666.66) + (88,888.89 - 106,666.66) · 3.00]/ 12 = 148.48 kN·m

M = 1.20 · 1.20 [(2 · 3.00 +0.60) · (133,333.33 + 115,555.55) + (133,333.33 - 115,555.55) · 3.00] / 12= 203.52 kN·m

Reinforcement calculation:

M1 = 36.111

AGx = M1 / (0.9·h0·fy) = 36,111.113 / (0.9 · 0.340 · 210.) = 561.953 mm2

M2 = 36.111

AGy = M2 / (0.9·h0·fy) = 36,111.113 / (0.9 · 0.340 · 210.) = 561.953 mm2

M1 = 142.222

AGx = M1 / (0.9·h0·fy) = 142,222.219 / (0.9 · 0.740 · 210.) = 1,016.890 mm2

M2 = 142.222

AGy = M2 / (0.9·h0·fy) = 142,222.219 / (0.9 · 0.740 · 210.) = 1,016.890 mm2

M1 = 203.520

AGx = M1 / (0.9·h0·fy) = 203,520.016 / (0.9 · 1.140 · 210.) = 944.584 mm2

M2 = 203.520

AGy = M2 / (0.9·h0·fy) = 203,520.016 / (0.9 · 1.140 · 210.) = 944.584 mm2

The area of steel at X direction: 1,016.890

The area of steel at Y direction: 1,016.890

The original area of steel at X direction is satisfied.

The original area of steel at Y direction is satisfied.

Calculated the areas of steel are:

AgX: 10 @ 200

AgY: 10 @ 200

9.5 Combined FoundationPer Chinese standard GB50007-2002.


9.5 Combined Foundation


BaseNumber

Left Can-tilever (M)

Right Can-tilever (M)

Length(M)

Width(M)

Height(M)

1 0.150 2.150 8.300 3.100 0.700

Table 9-11: Overview of the design results

BaseNumber

Top Lon-gitudinal Rein-forcement

Bottom Lon-gitudinal Rein-forcement

Top Trans-verse Rein-forcement

Bottom Trans-verse Rein-forcement

1 #12 @ 55 mmc / c

#12 @ 105 mmc / c

#12 @ 105 mmc / c

#12 @ 105 mmc / c

Table 9-12: Foundation reinforcement details

Elevation and plan


Chapter — 9


9.5.1 Problem

Basic Geometry

Column 1

Column dimensions

Column Shape : Rectangle

Length of the column - X (Pl): 0.30 m

Width of the column - Z (Pw): 0.30 m

No Column caps

Column 2

Column section size

Column Shape : Rectangle

Length of the Column - X (Pl): 0.30 m

Width of the Column - Z (Pw): 0.40 m

No Column cap

Left overhanging length : 0.15 m

Right cantilevered length : 2.15 m

Whether the length of the left cantilever needs design ( or enter a fixed value )? Yes

Whether the length of the right cantilever needs design ( or enter a fixed value )? Yes

The initial input length (Lo) of the foundation: 1500.00 mm

The initial input width (Wo) of the foundation: 3.10 m

The initial input of height (Do) of the foundation: 700.00 mm

Clear Cover and Soil Properties

The thickness of the clear cover for cap : 50.00 mm

The thickness of the clear cover for foundation : 50.00 mm

Density of the Soil: 25.00 kN/m3

Foundation bearing capacity : 200.00 kN/m2

Additional ground pressure : 0.00 kip/in2

Weight of soil about foundation : 1500.00 mm

Groundwater depth : -0.00 KN

Concrete and Steel Properties

Concrete density: 25.000 kN/m3




Compressive strength of concrete : 11.900 N/mm2

Reinforcement strength : 210.000 N/mm2

Minimum bar Size : 12.0 mm

Maximum bar size : 50.0 mm

Minimum bar spacing : 50.00 mm

Maximum bar spacing : 400.00 mm

9.5.2 SolutionBuoyancy generated on the ground = -0.00 kN

Minimum area of steel required

Amin = Pc / qmax: 13.80 m2

Specify the initial cross-sectional area

Ao = L x W: 25.73 m2

Final provided foundation dimensions:

Length of base , L: 8.30 m

Width of base, W: 3.10 m

Height of base, Do: 0.70 m

Area of base, A: 25.73 m2

Left Cantilever length , Llo: 0.15 m

Right Cantilever length , Lro: 2.15 m

ConditionNo.

ColumnNo.

AxialForce (KN)

ShearX (KN)

ShearZ (KN)

MomentX (kN·m)

MomentZ (kN·m)

1 1 105076.125 0.000 0.000 0.000 0.0001 2 210152.249 0.000 0.000 0.000 0.000

Table 9-13: Load cases for base dimensions estimation - standard com-bination

ConditionNo.

ColumnNo.

AxialForce (KN)

ShearX (KN)

ShearZ (KN)

MomentX (kN·m)

MomentZ (kN·m)

1 1 105076.125 0.000 0.000 0.000 0.0001 2 210152.249 0.000 0.000 0.000 0.000

Table 9-14: Load cases for foundation design - basic combinationn

Four corners of the calculated stress

LoadCase

Pressureat Corner1

(q1)(KN/m2)

Pressureat Corner2

(q2)(KN/m2)

Pressureat Corner3

(q3)(KN/m2)

Pressureat Corner4

(q4)(KN/m2)

Zero-pres-sure area(Au) (m2)

1 107.2940 107.2940 107.2940 107.2940 0.001 107.2940 107.2940 107.2940 107.2940 0.001 107.2940 107.2940 107.2940 107.2940 0.001 107.2940 107.2940 107.2940 107.2940 0.00


Chapter — 9


If Au equals zero, that means it is small eccentricity, and do not need to adjust thepressure. Otherwise, the pressure needs to be adjusted. The negative pressure should alwaysset as zero. Keep adjusting if necessary.

Adjusted pressure at corners (if necessary)

ConditionNo.





1 107.2940 107.2940 107.2940 107.29401 107.2940 107.2940 107.2940 107.29401 107.2940 107.2940 107.2940 107.29401 107.2940 107.2940 107.2940 107.2940

Overturning Stability Test

ConditionNo.

MomentX

(kN·m)

MomentZ

(kN·m)

ResistanceMoment X(kN·m)

ResistanceMoment Z(kN·m)

OverturningStabilityFactor X

Overturning Sta-bility Factor Z

1 0.000 0.000 4273.272 11441.341 N / A 633,778,703.608

9.5.3 Check ShearThe following formulae are used per GB50007 - 2002 code for design of buildingfoundations.

Formula is as follows :




One-way Punching Shear Check

Column 1, +X Direction


Punching shear

Fl = Pj·Al = 69.96 · 5,097,642.75 = 356.62 kN


Fu = 0.7·βhpftamh0 = 0.7 · 1.00 · 1270.00 · (0.30 · 1.56) / 2 · 0.64 = 523.64 kN

Fl < 0.7·βhpftamh0Hence, safe.

Column 1, -X Direction


Punching shear

Fl = Pj·Al = 69.96 · 0.00 = 0.0 kN





Fu = 0.7·βhpftamh0 = 0.7 · 1.00 · 1270.00 · (0.30 · 1.56) / 2 · 0.64 = 523.64 kN


Column 1, +Z Direction


Punching shear

Fl = Pj·Al = 69.96 · 1,010,688.00 = 70.70 kN


Fu = 0.7·βhpftamh0 = 0.7 · 1.00 · 1270.00 · (0.30 · 1.56) / 2 · 0.64 = 523.64 kN


Column 1, -Z Direction


Punching shear

Fl = Pj·Al = 69.96 · 1,010,688.00 = 70.70 kN


Fu = 0.7·βhpftamh0 = 0.7 · 1.00 · 1270.00 · (0.30 · 1.56) / 2 · 0.64 = 523.64 kN


Column 2, +X Direction


Punching shear

Fl = Pj·Al = 69.96 · 3,725,275.99 = 260.61 kN


Fu = 0.7·βhpftamh0 = 0.7 · 1.00 · 1270.00 · (0.30 · 1.66) / 2 · 0.64 = 579.83 kN


Column 2, -X Direction


Punching shear

Fl = Pj·Al = 69.96 · 7,548,609.37 = 528.08 kN


Chapter — 9



Fu = 0.7·βhpftamh0 = 0.7 · 1.00 · 1270.00 · (0.30 · 1.66) / 2 · 0.64 = 579.83 kN


Column 2, +Z Direction


Punching shear

Fl = Pj·Al = 69.96 · 163,8475.99 = 114.62 kN


Fu = 0.7·βhpftamh0 = 0.7 · 1.00 · 1270.00 · (0.40 · 1.56) / 2 · 0.64 = 551.73 kN


Column 2, -Z Direction


Punching shear

Fl = Pj·Al = 69.96 · 163,8475.99 = 114.62 kN


Fu = 0.7·βhpftamh0 = 0.7 · 1.00 · 1270.00 · (0.40 · 1.56) / 2 · 0.64 = 551.73 kN


Column 1, Four Edges


Punching shear

Fl = Pj·Al = 139.91 · 3,559,509.38 = 498.03 kN


Fu = 0.7·βhpftamh0 = 0.7 · 1.00 · 1270.00 · (1.20 · 4.99) / 2 · 0.64 = 1,739.48 kN


Column 2, Four Edges


Punching shear

Fl = Pj·Al = 139.91 · 7,275,418.66 = 1,017.94 kN





Fu = 0.7·βhpftamh0 = 0.7 · 1.00 · 1270.00 · (1.20 · 6.46) / 2 · 0.64 = 2,206.94 kN


9.5.4 Reinforcement Design

Top Longitudinal Reinforcement

A simplified formula for reinforcement is used in accordance with GB50010-2002.



Minimum area of steel, Astmin = 3,255.000 mm2

The area of steel required

Ast = MI / (0.9 · h0 · fy) = 740,000,006.857 / (0.9 · 6,44.000 · 210.000) = 6,079.727 mm2

Selected rebar size,db = 12.000 mm




Actual area of steel required, Ast (Actual) = 6220.353 mm2

Smin ≤ S ≤ Smax


Astmin ≤ Ast, with actual


Bottom Longitudinal Reinforcement






Ast = MI / (0.9 · h0 · fy) = -479,467,391.579 / (0.9 · 644.000 · 210.000) = -3,939.231 mm2






Smin ≤ S ≤ Smax


Chapter — 9





Top Transverse Reinforcement



Minimum reinforcement ratio [per Cl. 9.5.2], min = 0.15%

Minimum area of steel, Astmin = 8715.000 mm2


Ast = MI / (0.9 · h0 · fy) = 0.000 / (0.9 · 644.000 · 210.000) = 0.000 mm2






Smin ≤ S ≤ Smax




Bottom Transverse Reinforcement






Ast = MI / (0.9 · h0 · fy) = 569,032,250.186 / (0.9 · 644.000 · 210.000) = 4,675.082 mm2



Maximum allowable reinforcement spacin, Smax = 400.00 mm


Actual area of steel required, Ast (Actual) = 8,821.592 mm2

Smin ≤ S ≤ Smax







9.6 Pile Foundation DesignA typical pile foundation design example is provided here to verify the pile foundationdesign per Chinese codes in the program. The Chinese codes implemented are " GB50007-2002 code for design of building foundations "," GB50009-2001 Load code for design ofbuilding structures "," GB50010-2002 Code for design of Concrete Structures "," TechnicalCode for Building Pile Foundations”

9.6.1 Problem

Basic conditions

Rectangular column foundation. Length of the column is 0.5000 m and the width of thecolumn is 0.5000 m. The height of the base is 0.5000 m , the length of the base is 0.8 m,and the width of the base is 0.8.

Loads: vertical load is 1500 kN and the basic combination distribution factor is 1.4 ( 1500X 1.4 will involve in all members and reinforcements design).

Basic design parameters

Concrete strength: 25 N/mm2

Concrete density: 25 kN/m3

Steel yield strength: 210 N/mm2

Clear cover thickness on sides: 50 mm

Clear cover thickness on bottom: 50 mm

Depth of pile cap: 75 mm

The initial depth of pile cap: 300 mm

Minimum bar diameter: 10

Maximum bar diameter: 45

Pile Parameters

Pile layout: 3 row x 3 row = total 9 piles, spacing 1.5 m , center to edge of pile cap edge 0.5m , pile diameter and 0.5 m , thus base of the cap is 4 m in length and 4 m in width.

Pile bearing capacity: lateral capacity is 100 kN , vertical capacity is 500 kN , upliftcapacity is 300 kN (both single pile). So the reactions of the pile can be calculated as:

Vertical Uplift Lateral-180.888 0.000 0.000-180.888 0.000 0.000-180.888 0.000 0.000-180.888 0.000 0.000

Table 9-15: Pile capacities under load case no. 101


Chapter — 9

9.6 Pile Foundation Design

Vertical Uplift Lateral-180.888 0.000 0.000-180.888 0.000 0.000-180.888 0.000 0.000-180.888 0.000 0.000-180.888 0.000 0.000

Vertical Uplift Lateral-247.555 0.000 0.000-247.555 0.000 0.000-247.555 0.000 0.000-247.555 0.000 0.000-247.555 0.000 0.000-247.555 0.000 0.000-247.555 0.000 0.000-247.555 0.000 0.000-247.555 0.000 0.000

Table 9-16: Pile capacities under load case no. 102

9.6.2 Solution

Pile layout

Column Dimension

Column Shape : Rectangular

Column length - X (Pl): 0.500 m

Column width - Z (Pw): 0.500 m

Base

A base ? Yes

Base Shape : Rectangular

Base height (Ph): 0.500 m

Base length - X (Pl): 0.800 m

Base width - Z (Pw): 0.800 m

Cap geometry

Pile length P CL = 4.000 m

Cap width P CW = 4.000 m

Initial cap height t I = 0.300 m




Pile geometry

Pile spacing, P s = 1.500 m

Distance from the edge of the Pile cap to the center of pile, e = 0.500 m

Pile diameter, d P = 0.500 m

Bearing capacity of pile

Vertical bearing capacity of P P = 500.000 kN

Lateral bearing capacity of P L = 100.000 kN

Pullout capacity of P L = 300.000 kN

Material Properties

Concrete f ' c = 25,000.004 kN/m2

Concrete f ' t = 1,890.000 kN/m2

Steel f y = 210,000.035 kN/m2

Concrete clear cover

Concrete clear cover on the bottom, CC B = 0.050 m

Concrete clear cover on the sides, CC S = 0.050 m

Depth of pile cap depth, PC P = 0.075 m

LoadCase

Fx(kN) Fy (kN) Fz

(kN)Mx (kN-

m)My (kN-

m)Mz (kN-

m)101 0.000 -

1500.000.000 0.000 0.000 0.000

102 0.000 -2100.00

0.000 0.000 0.000 0.000

Table 9-17: Load about the pile cap

Pile Design Calculations

The total number of piles N = 9

Coordinates ofPiles Reactions

Pile No. X (m) Y (m) Vertical(kN)

Lateral(kN)

Uplift(kN)

1 -1.500 -1.500 -247.555 0.000 0.0002 -1.500 0.000 -247.555 0.000 0.0003 -1.500 1.500 -247.555 0.000 0.0004 0.000 -1.500 -247.555 0.000 0.000


Chapter — 9


Coordinates ofPiles Reactions

Pile No. X (m) Y (m) Vertical(kN)

Lateral(kN)

Uplift(kN)

5 0.000 0.000 -247.555 0.000 0.0006 0.000 1.500 -247.555 0.000 0.0007 1.500 -1.500 -247.555 0.000 0.0008 1.500 0.000 -247.555 0.000 0.0009 1.500 1.500 -247.555 0.000 0.000

Check Depth of Pile Cap

One Way Punching Shear Along Length

Critical Load Case #102

Influential factor of sectional height

βhs = (800/h0)1/4 = (800/0)1/4 = 1.000

Shear Span-to-Depth Ratios

λx = ax/h0 = 1.050/0.300 = 3.000

Punching Shear Factor of Pile Cap

α = 1.75/(λx + 1) = 1.75/(3.000 + 1) = 0.438

Shear Capacity of Pile Cap

dVc = βhs · α · ft · b0 · h0 = 1.000 · 0.438 · 1,890.000 · 4.000 · 0.300 = 992.250 kN

Maximum Shear Design Value, V = 742.665 kN

V < dVc

Hence, Safe

One Way Punching Shear Along Width



βhs = (800/h0)1/4 = (800/0)1/4 = 1.000


λy = ay/h0 = 1.050/0.300 = 3.000

Punching Shear Factor of Pile Cap

α = 1.75/(λy + 1) = 1.75/(3.000 + 1) = 0.438


dVc = βhs · α · ft · b0 · h0 = 1.000 · 0.438 · 1,890.000 · 4.000 · 0.300 = 992.250 kN

Maximum Shear Design Value, V = 742.665 kN

V < dVc




Hence, Safe

Punching Shear Check for Column



λ0x = a0x/h0 = (1.050/0.300) = 1.000


λ0y = a0y/h0 = (1.050/0.300) = 1.000


β0x = 0.84/(λ0x + 0.2) = 0.84/(1.000 + 0.2) = 0.700


β0y = 0.84/(λ0y + 0.2) = 0.84/(1.000 + 0.2) = 0.700


dVc = 2 · [β0x · (bc + a0y) + β0y · (hc + a0x)] · βhp · ft · h0 = 2 · [0.700 · (0.500 + 1.050) +0.700 · (0.500 + 1.050)] · 1.000 · 1,890.000 · 0.300 = 2460.780 kN

Maximum Punching Shear Design Value Fl = 1980.440kN

Fl < dVcHence, Safe

Punching Shear Check for Corner Column



λ1x = a1x/h0 = (0.300/0.300) = 1.000


λ1y = a1y/h0 = (0.300/.0.300) = 1.000


β1x = 0.56/(λ1x + 0.2) = 0.56/(1.000 + 0.2) = 0.467


β1y = 0.56/(λ1y + 0.2) = 0.56/(1.000 + 0.2) = 0.467


dVc = [β1x · (c2 + a1y/2) + β1y · (c1 + a1x/2)] · βhp · ft · h0 = [0.467 · (0.700 + 0.300/2) + 0.467· (0.700 + 0.300/2)] · 1.000 · 1,890.000 · 0.300 = 449.820 kN

Maximum Punching Shear Design Value Nl = 247.555kN

Nl < dVcHence, Safe


Chapter — 9



Use the same reinforcement along both top and bottom.

Along X Direction


Critical Moment

My = ∑Nixi = 816.931 kN·m


Asy = My/(0.9 · fy · h0) = 816.931/(0.9 · 210,000.035 · 0.300 = 14,407 mm2


As,min = 0.15 · B · H = 0.15 · 4,000 · 300 = 1,800 mm2

Bar Diameter, ds = 20 mm

Bar Space, S = 80 mm

Min Bar Space, Smin = 50 mm

Max Bar Space, Smax = 500 mm

Actual Bar Area required, As,actual = 49 · π · 10 · 10 = 15,393 mm2

Smin< S < SmaxAs,min < As,actual & Asy < As,actual

Hence, Safe

Along Y Direction


Critical Moment

My = ∑Nixi = 816.931 kN·m


Asy = My/(0.9 · fy · h0) = 816.931/(0.9 · 210,000.035 · 0.300 = 14,407 mm2


As,min = 0.15 · B · H = 0.15 · 4,000 · 300 = 1,800 mm2

Bar Diameter, ds = 20 mm

Bar Space, S = 80 mm

Min Bar Space, Smin = 50 mm

Max Bar Space, Smax = 500 mm

Actual Bar Area required, As,actual = 49 · π · 10 · 10 = 15,393 mm2

Smin< S < SmaxAs,min < As,actual & Asy < As,actual




Hence, Safe

Comparing Results with PKPM***********************************************************-**************************************

Design Iterm: Pile Cap-1

***********************************************************-**************************************

Basic Information

1, cap information

Section shape: Stepped Cast-on-site

Plane Shape: Rectangular

Number of steps: 1-step

Elevation at Cap bottom: 0.0m

Number of cap edges: 4

Pile height: 300mm

Cap eccentric at X direction: 0mm

Cap eccentric at X direction: 0mm

2 pile information

Pile Diameter: 500mm

Pile bearing Capacity: 500.00Kn

Pile Coordinate:

NUm X Y

1 -1500 -1500

2 0 -1500

3 1500 -1500

4 -1500 0

5 0 0

6 1500 0

7 -1500 1500

8 0 1500

9 1500 1500

3, Load information


Chapter — 9


Vertical Load: N = 1500.

Moment at x direction: Mx = 0.00knm

Moment at y direction: My = 0.00knm

Shear force at x direction: Qx = 0.00knm

Shear force at y direction: Qy = 0.00knm

4, column information

Column width: 800mm

Column height: 800mm

5, concrete information

Concrete class: C50

Concrete density: 25.00kn/m3

[Design Results]

6、Pile Reactions Calculation

The following formula is used:




Self weight of pile cap and the soil Gk = B * S *H * γ+ B * S * futu

=( 4000.0* 4000.0* 300.0*25.0*1.E-9+ 4000.0* 4000.0* 0.0*1.E-6)

= 120.0(kn)

∑Xi*Xi = 13500000.0 ∑Yi*Yi =13500000.0

Pile No. X Y Pile ReactionQ(KN) Net Reaction QN(KN)

1 -1500.0 -1500.0 180.00 166.67

2 0.0 -1500.0 180.00 166.67

3 1500.0 -1500.0 180.00 166.67

4 -1500.0 0.0 180.00 166.67

5 0.0 0.0 180.00 166.67

6 1500.0 0.0 180.00 166.67

7 -1500.0 1500.0 180.00 166.67

8 0.0 1500.0 180.00 166.67

9 1500.0 1500.0 180.00 166.67

QP= 1620.0(kN); QAVE= 180.0(kN)

7、Punching Shear Check

Step 1: H = 300.00MM

**Punching Shear Check for the Corner Pile***


Chapter — 9


No.=1

h0= 300. α1x=900.λ1x=1.00 c1=700.

h0= 300. α1y=900.λ1y=1.00 c2=700.

β1x=0.4667 β1y= 0.467βhp=1.00 ft= 1.888

QPC=(β1x*(C2+α1y/2)+β1y*(c1+α1x/2))*βhp*ft*ho

= 449.37KN > QPD = 166.67(*1.35) KN

No.=2

h0= 300. α1x=900.λ1x=1.00 c1=700.

h0= 300. α1y=900.λ1y=1.00 c2=700.

β1x=0.4667 β1y= 0.467βhp=1.00 ft= 1.888


= 449.37KN > QPD = 166.67(*1.35) KN

No.=3

h0= 300. α1x=900.λ1x=1.00 c1=700.

h0= 300. α1y=900.λ1y=1.00 c2=700.

β1x=0.4667 β1y= 0.467βhp=1.00 ft= 1.888


= 449.37KN > QPD = 166.67(*1.35) KN




No.=4

h0= 300. α1x=900.λ1x=1.00 c1=700.

h0= 300. α1y=900.λ1y=1.00 c2=700.

β1x=0.4667 β1y= 0.467βhp=1.00 ft= 1.888


= 449.37KN > QPD = 166.67(*1.35) KN

***Punching Shear Check for Column***

H00= 300.mm

x+ h0= 300. αox= 300. λox=1.000

x- h0= 300. αox= 300. λox=1.000

y+ h0= 300. αoy= 300. λoy=1.000

y+ h0= 300. αoy= 300. λoy=1.000

Step 1 H = 400.00MM

***Punching Shear Check for Column***

H00= 350.mm

x+ h0= 350. αox= 350. λox=1.000

x- h0= 350. αox= 350. λox=1.000

y+ h0= 350. αoy= 350. λoy=1.000

y+ h0= 350. αoy= 350. λoy=1.000

hc= 800. bc= 800. βox= 0.70 βoy= 0.70


Chapter — 9


ft= 1.89 βhp=1.000

QCC = 2*(βox*(bc+αoy)+βoy(hc+αox))βhp*ft*ho

= 2127.89KN

QCC= 2127.89KN > QCD= 1333.33 (* 1.35) KN

***Sheack Check***

VPL = βhs*1.75/(λ+1.0)*b0*h0*ft

=1.000*1.75/(2.571+1.0)*4000.*350.*1.8881*1.e-3

= 1295.2KN

VCI1= 1295.24KN > VDI1= 500.00 (* 1.35)KN

Right h0= 350. αx= 900. λx=2.571

VPL = βhs*1.75/(λ+1.0)*b0*h0*ft

=1.000*1.75/(2.571+1.0)*4000.*350.*1.8881*1.e-3

= 1295.2KN

VCI2= 1295.24KN > VDI2= 500.00 (* 1.35)KN

Down h0= 350. αy= 900. λy=2.571

VPL = βhs*1.75/(λ+1.0)*b0*h0*ft

=1.000*1.75/(2.571+1.0)*4000.*350.*1.8881*1.e-3

= 1295.2KN

VCJ1= 1295.24KN > VDJ1= 500.00 (* 1.35)KN

Top h0= 350. αy= 900. λy=2.571

VPL = βhs*1.75/(λ+1.0)*b0*h0*ft




=1.000*1.75/(2.571+1.0)*4000.*350.*1.8881*1.e-3

= 1295.2KN

VCJ2= 1295.24KN > VDJ2= 500.00 (* 1.35)KN

8、Steel Area Calculation

DMX1 = 742.500

AGX = DMX1/(0.9*h0*fy)/YS = 742.500/(0.9*350.0*210.0)/4.0= 2806.123mm*mm/M

DMX2 = 742.500

AGX = DMX2/(0.9*h0*fy)/YS = 742.500/(0.9*350.0*210.0)/4.0= 2806.123mm*mm/M

DMY1 = 742.500

AGY = DMY1/(0.9*h0*fy)/XS = 742.500/(0.9*350.0*210.0)/4.0= 2806.123mm*mm/M

DMY2 = 742.500

AGY = DMY2/(0.9*h0*fy)/XS = 742.500/(0.9*350.0*210.0)/4.0= 2806.123mm*mm/M

ASX=2806.1mm*mm/M ASY=2806.1mm*mm/M

The Area of Steel at x direction is satisfied,hence, safe.

The Area of Steel at y direction is satisfied,hence, safe.

Actual Areas of Steel required：

AGx: 16@100 AGy: 16@100

Com No ASX ASY H(1)H(2)

1 2806.1 2806.1 400.0

2 2455.4 2455.4 450.0

3 2182.5 2182.5 500.0

4 1964.3 1964.3 550.0


Chapter — 9


5 1785.7 1785.7 600.0

6 1636.9 1636.9 650.0

7 1511.0 1511.0 700.0

9.6.3 Comparison



Percent Dif-ference

Bearing Pressure 108.94 KN/m2 108.63 KN/m2 NegligibleResisting force for sliding (x) 313.74 KN 312.867 KN NegligibleResisting Moment for Overturning(z)


Resisting force for sliding (z) 313.74 KN 312.867 KN NegligibleResisting Moment for Overturning(x)


Table 9-18: Chinese verification example 6 comparison





Chapter 9


Section 10

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Chapter 10

IndexA

ACI 318 201, 321

AS3600 3

Australian 3

General Combined Footing 14,16

General Isolated Footing 3, 6

B

British

Combined Foundation 71, 77

Isolated Foundation 23, 26, 32,38, 44, 53,

62, 88

Mat Combined Foundation 83

British Code 23

BS8110 23

C

Canadian Code 99

Chinese Code 437

CSA A23.3 99

D

Deadman Anchors 321

Drilled Pier 357

E

Eccentricity 88

G

GB50007-2002 437

I

Indian Code 125

IS 456 125

P

Pile Cap

Chinese Code 470

Plant Foundation 385

U

United States Code 201

V

Vertical Vessel Foundation 1 385


List of Figures & TablesFigures

Figure 1-1: Australian code General isolated foundation 4

Figure 1-2: Plan and Elevation 6



Figure 1-5: Graphs of combined strip footing internal forces 20

Figure 2-1: Bending section considered 24

Figure 2-2: One way shear section considered 24

Figure 2-3: Two way shear section considered 24


Figure 2-5: Sections considered for bending in both directions 28

Figure 2-6: Sections considered for one-way shear in both directions 30

Figure 2-7: Section considered for punching shear 31


Figure 2-9: Section considered for bending about the Z axis 47

Figure 2-10: Section considered for bending about the Z axis 48

Figure 2-11: Section considered for one-way shear along X direction 50

Figure 2-12: Section considered for one-way shear along z direction 51






Figure 2-17: Shear force and Bending Moment diagrams 76


Figure 2-19: Shear force and Bending Moment diagrams 82



Figure 2-22: Bending about major axes 91




Figure 3-4: Elevation and Plan, with dimension and loads 116

Figure 3-5: Bending sections considered 117

Figure 3-6: Shear sections considered 119

Figure 3-7: Two-way shear sections considered 120



Figure 4-2: Plan of Reinforcement 128

Figure 4-3: Cross Section showing Reinforcement 129


Figure 4-5: Elevation and Plan showing reinforcement design 132






Figure 4-11: Final Plan Dimensions 158


Figure 4-13: Dimension, Moment, and Shear and diagrams 162


Figure 4-15: Shear Force and Bending Moment diagrams 169







Chapter — 12

Figures

Figure 4-21: Plan, Elevation, and Pedestal dimensions 190

Figure 4-22: Footing Plan 198

Figure 4-23: >Loads on Footing 198

Figure 4-24: Shear Force (kN, top) and Bending Moment (kNm, bottom) dia-grams

200

Figure 5-1: Elevation and loads 202

Figure 5-2: Considered sections for two-way (bo) and beam (bw) action 203

Figure 5-3: Critical section for moment (long projection) 204

Figure 5-4: Elevation and Plan 207



Figure 5-7: Section considered for two-way shear 218




Figure 5-11: Corner pressure values on plan for punching shear 236

Figure 5-12: One-way shear pressure values along x-direction 237

Figure 5-13: One-way shear pressure values along z-direction 238

Figure 5-14: Bending pressure about Z axis 239

Figure 5-15: Bending pressure about X axis 239












Figure 5-27: Section considered for two-way shear 274

Figure 5-28: Section considered for one-way shear 276

Figure 5-29: Section considered for bending 278



List of Figures & Tables

Figures



Figure 5-33: Section considered for moment 285




Figure 5-37: Section considered for moment 293

Figure 5-38: Elevation and dimensions 296

Figure 5-39: Forces on foundation 298

Figure 5-40: Shear and Bending diagrams 298


Figure 5-42: Critical section for punching shear is at d/2 305

Figure 5-43: Critical section for moment is at the face of column 307


Figure 5-45: Critical section for punching shear at d/2 315

Figure 6-1: Deadman Anchor Guy Tension Block section 324

Figure 6-2: Dispersion of soil against vertical uplift diagram 325

Figure 6-3: Dispersion line diagram 326

Figure 6-4: Top rebar force diagram 328

Figure 6-5: Bending moment diagram - top 328

Figure 6-6: Bending moment diagram - front face 329


Figure 6-8: Dispersion of soil against vertical uplift 333






Figure 6-14: Dispersion line diagram 344









Chapter — 12

Figures


Figure 7-1: Pier Elevation 358






Figure 8-1: Tank and foundation elevation 386

Figure 8-2: Anchor bolt plan 386

Figure 8-3: One-way shear dimensions 392

Figure 8-4: Two-way shear check 392









Figure 9-1: Four corners of the calculated stress 439

Figure 9-2: Elevation of stability forces 440

Figure 9-3: Reinforcement parallel to the X-direction 443

Figure 9-4: Reinforcement parallel to the Z-direction 444

Figure 9-5: Four corners of the calculated stress 451

Figure 9-6: Reinforcement parallel to the X-direction 455

Figure 9-7: Reinforcement parallel to the Z-direction 456

Tables

Table 1-1: Australian verification example 1 comparison 5




Table 2-1: British verification example 1 comparison 26

Table 2-2: Table BS2.1 - Column loads 26



Tables










Table 2-12: British verification example 13 comparisons 98

Table 3-1: CSA verification example 1 comparison 105





Table 4-1: IS verification example 1 comparison 129











Table 4-12: Pile Locations in Plan 183


Table 4-14: Pile Coordinates in Plan 191



Table 5-1: US verification example 1 comparison 206






Chapter — 12

Tables


Table 5-7: US Verification problem 9 comparison 240
















Table 6-1: Soil Test Report Summary 322

Table 6-2: Deadman Anchor (US) verification example 1 comparison 330

Table 6-3: Soil test report summary 331

Table 6-4: Soil layers 334


Table 6-6: Soil test report summary 339





Table 7-1: Drilled Pier (API) verification example 1 comparison 361

Table 7-2: Drilled Pier (API) verification example 2 comparison 366

Table 7-3: Drilled Pier (FHWA) verification example 3 comparison 371

Table 7-4: Drilled Pier (FHWA) verification example 4 comparison 375

Table 7-5: Drilled Pier (Vesic) verification example 5 comparison 380

Table 7-6: Drilled Pier (Vesic) verification example 6 comparison 384

Table 8-1: Primary load description 387

Table 8-2: Wind loads 388


Tables


Table 8-3: Applied Load Combinations - Allowable Stress Level 389

Table 8-4: Applied Load Combinations - Strength Level 389

Table 8-5: Applied Load at Top of Pedestal - Allowable Stress Level 389

Table 8-6: Applied Load at Top of Pedestal - Strength Level 390

Table 8-7: Stability Ratio 391

Table 8-8: Soil Bearing Check 391

Table 8-9: Vertical Vessel verification example 1 comparison 394



Table 8-12: Applied Load Combination - Allowable Stress Level 398

Table 8-13: Applied Load Combination - Strength Level 398







Table 8-20: Applied Load Combination - Allowable Stress Level 407

Table 8-21: Applied Load Combination - Strength Level 407




Table 8-25: Soil Bearing Check 409














Chapter — 12

Tables



Table 8-40: Service level load combinations per PIP 429

Table 8-41: Strength level load combinations per PIP 429

Table 8-42: Service level loads applied at the top of the top of the fixed pier 430

Table 8-43: Strength level loads applied at the top of the top of the fixed pier 430

Table 8-44: Service level load combinations per PIP 433

Table 8-45: Strength level load combinations per PIP 433

Table 8-46: Service level loads applied at the top of the top of the fixed pier 434

Table 8-47: Strength level loads applied at the top of the top of the fixed pier 434

Table 9-1: Overview of cone footing design results 437

Table 9-2: Reinforcement details 437

Table 9-3: Loads for foundation base size estimation -For foundation base (1) 439

Table 9-4: Loads for Punching shear check and reinforcements- For foundationbase (1)

439

Table 9-5: Factor of safety 441

Table 9-6: Overview of the stepped foundation design 449

Table 9-7: Reinforcement details 449

Table 9-8: Critical loads for base size estimation - standard combination 450

Table 9-9: Loads for foundation design- the basic combination 450

Table 9-10: Safety factors 452

Table 9-11: Overview of the design results 462

Table 9-12: Foundation reinforcement details 462

Table 9-13: Load cases for base dimensions estimation - standard combination 464

Table 9-14: Load cases for foundation design - basic combinationn 464

Table 9-15: Pile capacities under load case no. 101 470

Table 9-16: Pile capacities under load case no. 102 471

Table 9-17: Load about the pile cap 472

Table 9-18: Chinese verification example 6 comparison 483


Tables



Chapter 12

Tables

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