Fundaciones B-2 Salida Staad Foundation
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Transcript of Fundaciones B-2 Salida Staad Foundation
Isolated Footing Design(ACI 318-05)
Design For Isolated Footing 3
Design For Isolated Footing 4
Design For Isolated Footing 5
Design For Isolated Footing 2351
Design For Isolated Footing 2352
Design For Isolated Footing 2353
Isolated Footing 3
Footing No. Group ID Foundation Geometry
- - Length Width Thickness
3 1 2.692 m 3.099 m 0.600 m
4 2 1.829 m 2.235 m 0.600 m
5 3 1.778 m 2.184 m 0.600 m
2351 4 2.591 m 2.946 m 0.600 m
2352 5 1.575 m 1.930 m 0.600 m
2353 6 1.626 m 2.032 m 0.600 m
Footing No. Footing Reinforcement Pedestal Reinforcement
- Bottom Reinforcement(Mz) Bottom Reinforcement(M
x) Top Reinforcement(M
z) Top Reinforcement(M
x) Main Steel Trans Steel
3 #5 @ 7 in c/c #4 @ 5 in c/c #4 @ 5 in c/c #3 @ 2 in c/c 24 - #4 #3 @ 8 in
4 #3 @ 2 in c/c #3 @ 2 in c/c #3 @ 2 in c/c #5 @ 8 in c/c 24 - #5 #3 @ 10 in
5 #3 @ 2 in c/c #4 @ 4 in c/c #4 @ 4 in c/c #3 @ 2 in c/c 24 - #3 #3 @ 6 in
2351 #3 @ 2 in c/c #4 @ 4 in c/c #4 @ 4 in c/c #3 @ 2 in c/c 24 - #4 #3 @ 8 in
2352 #3 @ 2 in c/c #4 @ 4 in c/c #4 @ 4 in c/c #4 @ 5 in c/c 24 - #5 #3 @ 10 in
2353 #3 @ 2 in c/c #3 @ 2 in c/c #3 @ 2 in c/c #5 @ 8 in c/c 12 - #5 #3 @ 10 in
Page 1 of 79Isolated Footing Design
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Input Values
Footing Geomtery
Column Dimensions
Pedestal
Design Parameters
Concrete and Rebar Properties
Soil Properties
Sliding and Overturning
Design Type : Calculate Dimension
Footing Thickness (Ft) : 23.620 in
Footing Length - X (Fl) : 40.000 in
Footing Width - Z (Fw) : 40.000 in
Eccentricity along X (Oxd) : 0.000 in
Eccentricity along Z (Ozd) : 0.000 in
Column Shape : Rectangular
Column Length - X (Pl) : 0.300 m
Column Width - Z (Pw) : 0.135 m
Include Pedestal? Yes
Pedestal Shape : Rectangular
Pedestal Height (Ph) : 1.650 m
Pedestal Length - X (Pl) : 0.500 m
Pedestal Width - Z (Pw) : 0.500 m
Unit Weight of Concrete : 156.070 lb/ft3
Strength of Concrete : 2.987 ksi
Yield Strength of Steel : 59.738 ksi
Minimum Bar Size : #3
Maximum Bar Size : #5
Minimum Bar Spacing : 2.000 in
Maximum Bar Spacing : 18.000 in
Pedestal Clear Cover (P, CL) : 3.000 in
Footing Clear Cover (F, CL) : 3.000 in
Soil Type : UnDrained
Unit Weight : 112.370 lb/ft3
Soil Bearing Capacity : 15.156 kip/ft2
Soil Surcharge : 0.000 kip/in2
Depth of Soil above Footing : 65.000 in
Undrained Shear Strength : 0.030 kip/in2
Coefficient of Friction : 0.500
Factor of Safety Against Sliding : 1.500
Factor of Safety Against Overturning : 1.500
------------------------------------------------------
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Design Calculations
Footing Size
Initial Length (Lo) = 1.016 m
Initial Width (Wo) = 1.016 m
Load Combination/s- Service Stress Level
Load CombinationNumber
Load Combination Title
101 1.000 x DL
102 1.000 x DL+1.000 x LL
103 1.000 x DL+0.750 x LL
104 1.000 x DL+1.000 x WLX
105 1.000 x DL+1.000 x WLZ
106 1.000 x DL+0.700 x ELX
107 1.000 x DL+0.700 x ELZ
108 1.000 x DL+0.750 x LL+0.750 x WLX
109 1.000 x DL+0.750 x LL+0.750 x WLZ
110 1.000 x DL+0.750 x LL+0.525 x ELX
111 1.000 x DL+0.750 x LL+0.525 x ELZ
Load Combination/s- Strength Level
Load CombinationNumber
Load Combination Title
201 1.400 x DL
202 1.200 x DL+1.600 x LL
203 1.200 x DL+1.000 x LL
204 1.200 x DL+0.800 x WLX
205 1.200 x DL+0.800 x WLZ
206 1.200 x DL+1.000 x LL+1.600 x WLX
207 1.200 x DL+1.000 x LL+1.600 x WLZ
208 1.200 x DL+1.000 x LL+1.000 x ELX
209 1.200 x DL+1.000 x LL+1.000 x ELZ
210 1.200 x DL+1.600 x WLX
211 1.200 x DL+1.600 x WLZ
212 1.200 x DL+1.000 x ELX
213 1.200 x DL+1.000 x ELZ
Applied Loads - Service Stress Level
LCAxial(kgf)
Shear X(kgf)
Shear Z(kgf)
Moment X(kNm)
Moment Z(kNm)
101 5972.532 -14.707 -633.695 0.004 0.658
102 5810.936 -13.781 -610.400 0.007 0.635
103 5851.335 -14.013 -616.224 0.006 0.641
104 4519.771 1248.679 -710.914 -0.010 -53.949
105 -22051.381 -980.979 4946.066 0.784 39.873
106 6530.984 -328.772 -764.653 -0.012 -12.224
107 14882.900 -37.675 -2252.029 -0.229 -0.249
108 4761.765 933.527 -674.138 -0.004 -40.314
109 -15166.599 -738.716 3568.597 0.591 30.052
110 6270.175 -249.561 -714.442 -0.005 -9.020
111 12534.111 -31.238 -1829.974 -0.168 -0.039
Applied Loads - Strength Level
LCAxial(kgf)
Shear X(kgf)
Shear Z(kgf)
Moment X(kNm)
Moment Z(kNm)
201 8361.545 -20.590 -887.174 0.005 0.921
202 6908.486 -16.167 -723.161 0.010 0.753
203 7005.443 -16.723 -737.139 0.008 0.767
204 6004.829 993.061 -822.210 -0.006 -42.896
205 -15252.091 -790.666 3703.375 0.628 32.161
Page 3 of 79Isolated Footing Design
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Final Footing Size
Pressures at Four Corners
If Au
is zero, there is no uplift and no pressure adjustment is necessary. Otherwise, to account for uplift, areas of negative pressure will be set to zero
and the pressure will be redistributed to remaining corners.
Summary of Adjusted Pressures at 4 corners Four Corners
206 4681.025 2004.696 -860.689 -0.013 -86.604
207 -37832.816 -1562.756 8190.479 1.256 63.511
208 7803.232 -465.387 -924.221 -0.014 -17.636
209 19734.540 -49.534 -3049.044 -0.324 -0.529
210 4842.621 2003.770 -883.985 -0.017 -86.582
211 -37671.222 -1563.683 8167.184 1.252 63.533
212 7964.827 -466.313 -947.517 -0.018 -17.613
213 19896.136 -50.460 -3072.340 -0.328 -0.506
Reduction of force due to buoyancy = 0.000 kgf
Effect due to adhesion = 121072.543 kgf
Area from initial length and width, Ao
=Lo
X Wo
= 1.032 m2
Min. area required from bearing pressure, Amin =P / qmax
= 0.267 m2
Note: Amin
is an initial estimation.
P = Critical Factored Axial Load(without self weight/buoyancy/soil).qmax = Respective Factored Bearing Capacity.
Length (L2) = 2.692 m Governing Load Case : # 105
Width (W2) = 3.099 m Governing Load Case : # 105
Depth (D2) = 0.600 m Governing Load Case : # 105
Area (A2) = 8.343 m2
Load Case
Pressure atcorner 1
(q1)
(kgf/m2)
Pressure atcorner 2
(q2)
(kgf/m2)
Pressure atcorner 3
(q3)
(kgf/m2)
Pressure atcorner 4
(q4)
(kgf/m2)
Area of footingin uplift (Au)
(m2)
107 7487.2136 7455.5098 5092.8780 5124.5818 0.000
104 3199.5297 7639.2279 6896.3548 2456.6566 0.000
104 3199.5297 7639.2279 6896.3548 2456.6566 0.000
105 937.6023 -2413.5310 2788.7537 6139.8869 1.388
Load Case
Pressure atcorner 1 (q1)
(kgf/m2)
Pressure atcorner 2 (q2)
(kgf/m2)
Pressure atcorner 3 (q3)
(kgf/m2)
Pressure atcorner 4 (q4)
(kgf/m2)
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Check for stability against overturning and sliding
Critical Load Case And The Governing Factor Of Safety For Overturning And Sliding - X Direction
Critical Load Case And The Governing Factor Of Safety For Overturning And Sliding - Z Direction
107 7487.2136 7455.5098 5092.8780 5124.5818
104 3199.5297 7639.2279 6896.3548 2456.6566
104 3199.5297 7639.2279 6896.3548 2456.6566
105 845.2744 0.0000 2877.0693 7034.6859
-Factor of safety against
slidingFactor of safety against
overturning
Load CaseNo.
Along X-Direction
Along Z-Direction
About X-Direction
About Z-Direction
101 1481.200 34.377 47.359 585.660
102 1574.887 35.556 48.998 610.126
103 1550.304 35.253 48.576 603.802
104 16.864 29.621 40.771 6.822
105 7.923 1.571 2.149 3.336
106 67.109 28.854 39.713 117.212
107 696.472 11.652 15.974 1190.401
108 22.687 31.416 43.259 9.180
109 15.182 3.143 4.296 6.388
110 87.887 30.700 42.269 164.791
111 802.383 13.697 18.786 1018.369
Critical Load Case for Sliding along X-Direction : 105
Governing Disturbing Force : -980.979 kgf
Governing Restoring Force : 7772.441 kgf
Minimum Sliding Ratio for the Critical Load Case : 7.923
Critical Load Case for Overturning about X-Direction : 105
Governing Overturning Moment : 109.914 kNm
Governing Resisting Moment : 236.191 kNm
Minimum Overturning Ratio for the Critical Load Case : 2.149
Critical Load Case for Sliding along Z-Direction : 105
Governing Disturbing Force : 4946.066 kgf
Governing Restoring Force : 7772.441 kgf
Page 5 of 79Isolated Footing Design
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Shear Calculation
Punching Shear Check
Effective depth, deff, increased until 0.75XVc Punching Shear Force
Punching Shear Force, Vu = 50624.227 kgf, Load Case # 213
Along X Direction
(Shear Plane Parallel to Global X Axis)
Minimum Sliding Ratio for the Critical Load Case : 1.571
Critical Load Case for Overturning about Z-Direction : 105
Governing Overturning Moment : 61.517 kNm
Governing Resisting Moment : 205.215 kNm
Minimum Overturning Ratio for the Critical Load Case : 3.336
Total Footing Depth, D = 0.600m
Calculated Effective Depth, deff
= D - Ccover
- 1.0 = 0.498 m 1 inch is deducted from total depth to cater bar dia(US Convention).
For rectangular column, = Bcol / Dcol = 1.000
From ACI Cl.11.12.2.1, bo for column= 3.993 m
Equation 11-33, Vc1
= 458811.711kgf
Equation 11-34, Vc2 = 534647.648kgf
Equation 11-35, Vc3
= 305874.474kgf
Punching shear strength, Vc
= 0.75 X minimum of (Vc1
, Vc2
, Vc3
) = 229405.856kgf
0.75 X Vc > Vu hence, OK
Page 6 of 79Isolated Footing Design
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Check that 0.75 X Vc
> Vux
where Vux
is the shear force for the critical load cases at a distance deff
from the face of the column caused by bending
about the X axis.
One-Way Shear Check
Along Z Direction
(Shear Plane Parallel to Global Z Axis)
Check that 0.75 X Vc
> Vuz
where Vuz
is the shear force for the critical load cases at a distance deff
from the face of the column caused by bending
about the Z axis.
Design for Flexure about Z Axis
(For Reinforcement Parallel to X Axis)
From ACI Cl.11.3.1.1, Vc
= 103112.394kgf
Distance along X to design for shear,D
x= 0.801 m
From above calculations, 0.75 X Vc = 77334.296 kgf
Critical load case for Vux is # 213 17439.932 kgf
0.75 X Vc > Vux hence, OK
From ACI Cl.11.3.1.1, Vc
= 118676.529 kgf
Distance along X to design for shear, Dz = 2.095 m
From above calculations, 0.75 X Vc = 89007.397 kgf
Critical load case for Vuz is # 210 14558.004 kgf
0.75 X Vc > Vuz hence, OK
Page 7 of 79Isolated Footing Design
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Calculate the flexural reinforcement along the X direction of the footing. Find the area of steel required, A, as per Section 3.8 of Reinforced ConcreteDesign (5th ed.) by Salmon and Wang (Ref. 1)
Critical Load Case # 210
The strength values of steel and concrete used in the formulae are in ksi
Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Because the number of bars is rounded up, make sure new reinforcement ratio < ρmax
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl. 7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
Design for flexure about Z axis isperformed at the face of the column at
a distance, Dx =1.096 m
Ultimate moment, 140.305 kNm
Nominal moment capacity, Mn
= 155.894 kNm
Required = 0.00049
Since OK
Area of Steel Required, As = 4.787 in2
Selected bar Size = #4
Minimum spacing allowed (Smin) = = 2.000 in
Selected spacing (S) = 5.022 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
#4 @ 5.000 in o.c.
Required development length for bars = =0.305 m
Available development length for bars, DL
=1.020 m
Try bar size # 4 Area of one bar = 0.200 in2
Number of bars required, Nbar
= 24
Total reinforcement area, As_total
= Nbar
X (Area of one bar) = 4.800 in2
deff
= D - Ccover
- 0.5 X (dia. of one bar) 0.517 m
Page 8 of 79Isolated Footing Design
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Check to see if width is sufficient to accomodate bars
Design for Flexure about X axis
(For Reinforcement Parallel to Z Axis)
Calculate the flexural reinforcement along the Z direction of the footing. Find the area of steel required, A, as per Section 3.8 of Reinforced ConcreteDesign (5th ed.) by Salmon and Wang (Ref. 1)
Critical Load Case # 213
The strength values of steel and concrete used in the formulae are in ksi
Calculate reinforcement ratio for critical load case
=
Reinforcement ratio, = 0.00193
From ACI Cl.7.6.1, minimum req'd cleardistance between bars, C
d=
max (Diameter of one bar, 1.0,Min. User Spacing) =
5.022 in
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl.7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
Design for flexure about X axis isperformed at the face of the column at
a distance, Dz
=1.299 m
Ultimate moment, 179.476 kNm
Nominal moment capacity, Mn = 199.418 kNm
Required = 0.00077
Since OK
Area of Steel Required, As
= 4.665 in2
Selected Bar Size = #5
Minimum spacing allowed (Smin) = 2.000 in
Selected spacing (S) = 7.644 in
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Based on spacing reinforcement increment; provided reinforcement is
Because the number of bars is rounded up, make sure new reinforcement ratio < ρmax
Check to see if width is sufficient to accomodate bars
Bending moment for uplift cases will be calculated based solely on selfweight, soil depth and surcharge loading.
As the footing size has already been determined based on all servicebility load cases, and design moment calculation is based on selfweight, soil depthand surcharge only, top reinforcement value for all pure uplift load cases will be the same.
Design For Top Reinforcement Parallel to Z Axis
Calculate the flexural reinforcement for Mx. Find the area of steel required
The strength values of steel and concrete used in the formulae are in ksi
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
UnSafe for Cracking Aspect.
#5 @ 7.000 in o.c.
Required development length for bars = =0.344 m
Available development length for bars, DL
=1.020 m
Try bar size # 5 Area of one bar = 0.310 in2
Number of bars required, Nbar = 14
Total reinforcement area, As_total = Nbar X (Area of one bar) = 4.340 in2
deff = D - Ccover - 0.5 X (dia. of one bar)
=
0.503 m
Reinforcement ratio, = 0.00207
From ACI Cl.7.6.1, minimum req'd cleardistance between bars, Cd =
max (Diameter of one bar, 1.0, Min.User Spacing) =
7.644 in
0.850
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Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Design For Top Reinforcement Parallel to X Axis
First load case to be in pure uplift #
Calculate the flexural reinforcement for Mz. Find the area of steel required
Factor from ACI Cl.10.2.7.3 =
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl. 7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
Design for flexure about A axis isperformed at the face of the column
at a distance, Dx =1.299 m
Ultimate moment, 99.673 kNm
Nominal moment capacity, Mn
= 110.748 kNm
Required = 0.00043
Since OK
Area of Steel Required, As = 4.053 in2
Selected bar Size = #3
Minimum spacing allowed (Smin) = 2.000 in
Selected spacing (S) = 2.767 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
#3 @ 2 in o.c.
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The strength values of steel and concrete used in the formulae are in ksi
Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Pedestal Design Calculations
Strength and Moment Along Reinforcement in X direction
Strength and Moment from Concrete
Calculate strength and moment from one bar.
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl.7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
Design for flexure about A axis isperformed at the face of the column
at a distance, Dx =1.096 m
Ultimate moment, 81.644 kNm
Nominal moment capacity, Mn = 90.716 kNm
Required = 0.00029
Since OK
Area of Steel Required, As
= 4.787 in2
Selected bar Size = #4
Minimum spacing allowed (Smin
) = 2.000 in
Selected spacing (S) = 5.022 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
#4 @ 5 in o.c.
Critical Load Case: 213
Bar size : # 4
Number of Bars : 24
Steel Area : 4.0250 sq.in
Neutral Axis Depth (Xb): 0.0537 m
Cc = 40720.223 kgf
Mc
= 90.721 kNm
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Strength and Moment Along Reinforcement in Z direction
Strength and Moment from Concrete
Calculate strength and moment from one bar.
Distance between extreme fiber andbar,
db 0.084 m
Strain in bar, = -0.0017
Maximum Strain, = 0.0021
as
-34712010.141
kgf/m2
0.0016
as
0.000 kgf/m2
-6942.388
kgf
-11.292 kNm
Total Bar Capacity, Cs
= -62827.042
kgf
Capacity of Column = Cc + Cs =-
22106.818kgf
Total Bar Moment, Ms = 7.112 kNm
Total Moment = Mc + Ms = 97.833 kNm
Bar size : # 4
Number of Bars : 24
Steel Area : 4.0250 sq.in
Neutral Axis Depth (Xb): 0.0537 m
Cc = 40720.223 kgf
Mc
= 90.721 kNm
Distance between extreme fiberand bar,
db 0.084 m
Strain in bar, = -0.0017
Maximum Strain, = 0.0021
as
-34712010.141
kgf/m2
0.0016
as
kgf/m2
-6942.388
kgf
Page 13 of 79Isolated Footing Design
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Isolated Footing 4
Input Values
Footing Geomtery
-11.292 kNm
Total Bar Capacity, Cs
= -62827.042
kgf
Capacity of Column = Cc + Cs =-
22106.818kgf
Total Bar Moment, Ms = 7.112 kNm
Total Moment = Mc + Ms = 97.833 kNm
Check for bi-axial bending, 0.502
Design Moment Mnx= 1.470 kNm
Design Moment Mnz= 5669.747 kNm
Total Moment Mox= 9976.404 kNm
Total Moment Moz= 9976.404 kNm
if Mnx
or Mnz
= 0, then = 1.0
otherwise, = 1.24
Design Type : Calculate Dimension
Footing Thickness (Ft) : 23.620 in
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Column Dimensions
Pedestal
Design Parameters
Concrete and Rebar Properties
Soil Properties
Sliding and Overturning
Design Calculations
Footing Size
Footing Length - X (Fl) : 40.000 in
Footing Width - Z (Fw) : 40.000 in
Eccentricity along X (Oxd) : 0.000 in
Eccentricity along Z (Ozd) : 0.000 in
Column Shape : Rectangular
Column Length - X (Pl) : 0.300 m
Column Width - Z (Pw) : 0.135 m
Include Pedestal? Yes
Pedestal Shape : Rectangular
Pedestal Height (Ph) : 1.650 m
Pedestal Length - X (Pl) : 0.500 m
Pedestal Width - Z (Pw) : 0.500 m
Unit Weight of Concrete : 156.070 lb/ft3
Strength of Concrete : 2.987 ksi
Yield Strength of Steel : 59.738 ksi
Minimum Bar Size : #3
Maximum Bar Size : #5
Minimum Bar Spacing : 2.000 in
Maximum Bar Spacing : 18.000 in
Pedestal Clear Cover (P, CL) : 3.000 in
Footing Clear Cover (F, CL) : 3.000 in
Soil Type : UnDrained
Unit Weight : 112.370 lb/ft3
Soil Bearing Capacity : 15.156 kip/ft2
Soil Surcharge : 0.000 kip/in2
Depth of Soil above Footing : 65.000 in
Undrained Shear Strength : 0.030 kip/in2
Coefficient of Friction : 0.500
Factor of Safety Against Sliding : 1.500
Factor of Safety Against Overturning : 1.500
------------------------------------------------------
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Initial Length (Lo) = 1.016 m
Initial Width (Wo) = 1.016 m
Load Combination/s- Service Stress Level
Load CombinationNumber
Load Combination Title
101 1.000 x DL
102 1.000 x DL+1.000 x LL
103 1.000 x DL+0.750 x LL
104 1.000 x DL+1.000 x WLX
105 1.000 x DL+1.000 x WLZ
106 1.000 x DL+0.700 x ELX
107 1.000 x DL+0.700 x ELZ
108 1.000 x DL+0.750 x LL+0.750 x WLX
109 1.000 x DL+0.750 x LL+0.750 x WLZ
110 1.000 x DL+0.750 x LL+0.525 x ELX
111 1.000 x DL+0.750 x LL+0.525 x ELZ
Load Combination/s- Strength Level
Load CombinationNumber
Load Combination Title
201 1.400 x DL
202 1.200 x DL+1.600 x LL
203 1.200 x DL+1.000 x LL
204 1.200 x DL+0.800 x WLX
205 1.200 x DL+0.800 x WLZ
206 1.200 x DL+1.000 x LL+1.600 x WLX
207 1.200 x DL+1.000 x LL+1.600 x WLZ
208 1.200 x DL+1.000 x LL+1.000 x ELX
209 1.200 x DL+1.000 x LL+1.000 x ELZ
210 1.200 x DL+1.600 x WLX
211 1.200 x DL+1.600 x WLZ
212 1.200 x DL+1.000 x ELX
213 1.200 x DL+1.000 x ELZ
Applied Loads - Service Stress Level
LCAxial(kgf)
Shear X(kgf)
Shear Z(kgf)
Moment X(kNm)
Moment Z(kNm)
101 8661.264 -21.119 584.358 0.016 0.964
102 8849.825 -20.736 609.412 0.020 0.977
103 8802.685 -20.832 603.148 0.019 0.974
104 5680.307 1383.753 236.955 -0.028 -61.063
105 31299.688 -1031.955 5610.064 0.763 46.325
106 9417.908 -384.107 462.533 -0.005 -14.134
107 17361.076 -44.994 -991.666 -0.217 0.029
108 6566.968 1032.823 342.596 -0.014 -45.546
109 25781.502 -778.958 4372.428 0.579 34.994
110 9370.169 -293.073 511.780 0.003 -10.350
111 15327.545 -38.737 -578.869 -0.156 0.273
Applied Loads - Strength Level
LCAxial(kgf)
Shear X(kgf)
Shear Z(kgf)
Moment X(kNm)
Moment Z(kNm)
201 12125.769 -29.567 818.101 0.023 1.349
202 10695.215 -24.730 741.316 0.026 1.177
203 10582.078 -24.960 726.283 0.023 1.170
204 8008.751 1098.555 423.307 -0.016 -48.464
205 28504.255 -834.012 4721.794 0.616 37.445
206 5812.548 2222.837 170.439 -0.047 -98.073
207 46803.553 -1642.297 8767.414 1.217 73.747
208 11662.999 -543.514 552.249 -0.008 -20.399
209 23010.380 -59.066 -1525.179 -0.310 -0.166
210 5623.986 2222.453 145.385 -0.051 -98.086
211 46614.993 -1642.680 8742.360 1.214 73.734
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Final Footing Size
Pressures at Four Corners
If Au
is zero, there is no uplift and no pressure adjustment is necessary. Otherwise, to account for uplift, areas of negative pressure will be set to zero
and the pressure will be redistributed to remaining corners.
Summary of Adjusted Pressures at 4 corners Four Corners
212 11474.437 -543.897 527.195 -0.011 -20.412
213 22821.819 -59.449 -1550.233 -0.314 -0.179
Reduction of force due to buoyancy = 0.000 kgf
Effect due to adhesion = 0.000 kgf
Area from initial length and width, Ao
=Lo
X Wo
= 1.032 m2
Min. area required from bearing pressure, Amin =P / qmax
= 0.489 m2
Note: Amin
is an initial estimation.
P = Critical Factored Axial Load(without self weight/buoyancy/soil).qmax = Respective Factored Bearing Capacity.
Length (L2) = 1.829 m Governing Load Case : # 104
Width (W2) = 2.235 m Governing Load Case : # 104
Depth (D2) = 0.600 m Governing Load Case : # 104
Area (A2) = 4.088 m2
Load Case
Pressure atcorner 1
(q1)
(kgf/m2)
Pressure atcorner 2
(q2)
(kgf/m2)
Pressure atcorner 3
(q3)
(kgf/m2)
Pressure atcorner 4
(q4)
(kgf/m2)
Area of footingin uplift (A
u)
(m2)
107 10352.6467 10185.3761 7225.9542 7393.2248 0.000
104 -1912.8829 13080.0054 13776.4494 -1216.4389 0.425
105 9514.2209 -1795.6500 14884.1066 26193.9775 0.033
105 9514.2209 -1795.6500 14884.1066 26193.9775 0.033
Load Case
Pressure atcorner 1 (q1)
(kgf/m2)
Pressure atcorner 2 (q2)
(kgf/m2)
Pressure atcorner 3 (q3)
(kgf/m2)
Pressure atcorner 4 (q4)
(kgf/m2)
107 10352.6467 10185.3761 7225.9542 7393.2248
104 0.0000 13145.7473 13953.2170 0.0000
105 9516.4780 0.0000 14887.7510 26226.1155
105 9516.4780 0.0000 14887.7510 26226.1155
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Check for stability against overturning and sliding
Critical Load Case And The Governing Factor Of Safety For Overturning And Sliding - X Direction
Critical Load Case And The Governing Factor Of Safety For Overturning And Sliding - Z Direction
-Factor of safety against
slidingFactor of safety against
overturning
Load CaseNo.
Along X-Direction
Along Z-Direction
About X-Direction
About Z-Direction
101 644.638 23.298 23.116 170.764
102 661.104 22.495 22.314 171.401
103 656.931 22.689 22.508 171.242
104 8.762 51.165 51.103 2.374
105 24.161 4.444 4.388 6.472
106 36.429 30.252 30.070 44.341
107 399.260 18.115 17.820 315.272
108 12.168 36.682 36.509 3.298
109 28.467 5.071 5.008 7.621
110 47.663 27.294 27.109 64.504
111 437.494 29.277 28.734 269.621
Critical Load Case for Sliding along X-Direction : 104
Governing Disturbing Force : 1383.753 kgf
Governing Restoring Force : 12123.775 kgf
Minimum Sliding Ratio for the Critical Load Case : 8.762
Critical Load Case for Overturning about X-Direction : 105
Governing Overturning Moment : 124.543 kNm
Governing Resisting Moment : 546.527 kNm
Minimum Overturning Ratio for the Critical Load Case : 4.388
Critical Load Case for Sliding along Z-Direction : 105
Governing Disturbing Force : 5610.064 kgf
Governing Restoring Force : 24933.465 kgf
Minimum Sliding Ratio for the Critical Load Case : 4.444
Critical Load Case for Overturning about Z-Direction : 104
Governing Overturning Moment : -91.594 kNm
Governing Resisting Moment : 217.429 kNm
Minimum Overturning Ratio for the Critical Load Case : 2.374
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Shear Calculation
Punching Shear Check
Effective depth, deff
, increased until 0.75XVc
Punching Shear Force
Punching Shear Force, Vu = 49431.647 kgf, Load Case # 207
Along X Direction
(Shear Plane Parallel to Global X Axis)
Check that 0.75 X Vc
> Vux
where Vux
is the shear force for the critical load cases at a distance deff
from the face of the column caused by bending
Total Footing Depth, D = 0.600m
Calculated Effective Depth, deff
= D - Ccover
- 1.0 = 0.498 m 1 inch is deducted from total depth to cater bar dia(US Convention).
For rectangular column, = Bcol / Dcol = 1.000
From ACI Cl.11.12.2.1, bo for column= 3.993 m
Equation 11-33, Vc1
= 458811.711kgf
Equation 11-34, Vc2 = 534647.648kgf
Equation 11-35, Vc3
= 305874.474kgf
Punching shear strength, Vc = 0.75 X minimum of (Vc1, Vc2, Vc3) = 229405.856kgf
0.75 X Vc > Vu hence, OK
From ACI Cl.11.3.1.1, Vc
= 70038.607kgf
Distance along X to design for shear,D
x= 1.866 m
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about the X axis.
One-Way Shear Check
Along Z Direction
(Shear Plane Parallel to Global Z Axis)
Check that 0.75 X Vc
> Vuz
where Vuz
is the shear force for the critical load cases at a distance deff
from the face of the column caused by bending
about the Z axis.
Design for Flexure about Z Axis
(For Reinforcement Parallel to X Axis)
Calculate the flexural reinforcement along the X direction of the footing. Find the area of steel required, A, as per Section 3.8 of Reinforced ConcreteDesign (5th ed.) by Salmon and Wang (Ref. 1)
Critical Load Case # 207
The strength values of steel and concrete used in the formulae are in ksi
From above calculations, 0.75 X Vc = 52528.956 kgf
Critical load case for Vux is # 207 17829.941 kgf
0.75 X Vc > Vux hence, OK
From ACI Cl.11.3.1.1, Vc
= 85602.742 kgf
Distance along X to design for shear, Dz = 0.166 m
From above calculations, 0.75 X Vc = 64202.057 kgf
Critical load case for Vuz is # 207 8973.180 kgf
0.75 X Vc > Vuz hence, OK
Page 20 of 79Isolated Footing Design
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Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Because the number of bars is rounded up, make sure new reinforcement ratio < ρmax
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl. 7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
Design for flexure about Z axis isperformed at the face of the column at
a distance, Dx =0.664 m
Ultimate moment, 110.360 kNm
Nominal moment capacity, Mn
= 122.622 kNm
Required = 0.00054
Since OK
Area of Steel Required, As = 3.453 in2
Selected bar Size = #3
Minimum spacing allowed (Smin) = = 2.000 in
Selected spacing (S) = 2.633 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
#3 @ 2.000 in o.c.
Required development length for bars = =0.305 m
Available development length for bars, DL
=0.588 m
Try bar size # 3 Area of one bar = 0.110 in2
Number of bars required, Nbar
= 32
Total reinforcement area, As_total
= Nbar
X (Area of one bar) = 3.520 in2
deff
= D - Ccover
- 0.5 X (dia. of one bar)
=
0.519 m
Reinforcement ratio, = 0.00196
From ACI Cl.7.6.1, minimum req'd cleardistance between bars, C
d=
max (Diameter of one bar, 1.0,Min. User Spacing) =
2.633 in
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Check to see if width is sufficient to accomodate bars
Design for Flexure about X axis
(For Reinforcement Parallel to Z Axis)
Calculate the flexural reinforcement along the Z direction of the footing. Find the area of steel required, A, as per Section 3.8 of Reinforced ConcreteDesign (5th ed.) by Salmon and Wang (Ref. 1)
Critical Load Case # 207
The strength values of steel and concrete used in the formulae are in ksi
Calculate reinforcement ratio for critical load case
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl.7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
Design for flexure about X axis isperformed at the face of the column at
a distance, Dz =0.868 m
Ultimate moment, 177.097 kNm
Nominal moment capacity, Mn
= 196.775 kNm
Required = 0.00111
Since OK
Area of Steel Required, As = 3.387 in2
Selected Bar Size = #3
Minimum spacing allowed (Smin
) = 2.000 in
Selected spacing (S) = 2.431 in
Smin <= S <= Smax and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
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Based on spacing reinforcement increment; provided reinforcement is
Because the number of bars is rounded up, make sure new reinforcement ratio < ρmax
Check to see if width is sufficient to accomodate bars
Bending moment for uplift cases will be calculated based solely on selfweight, soil depth and surcharge loading.
As the footing size has already been determined based on all servicebility load cases, and design moment calculation is based on selfweight, soil depthand surcharge only, top reinforcement value for all pure uplift load cases will be the same.
Design For Top Reinforcement Parallel to Z Axis
Calculate the flexural reinforcement for Mx. Find the area of steel required
The strength values of steel and concrete used in the formulae are in ksi
#3 @ 2.000 in o.c.
Required development length for bars = =0.305 m
Available development length for bars, DL
=0.588 m
Try bar size # 3 Area of one bar = 0.110 in2
Number of bars required, Nbar = 28
Total reinforcement area, As_total = Nbar X (Area of one bar) = 3.080 in2
deff = D - Ccover - 0.5 X (dia. of one bar)
=
0.506 m
Reinforcement ratio, = 0.00215
From ACI Cl.7.6.1, minimum req'd cleardistance between bars, Cd =
max (Diameter of one bar, 1.0, Min.User Spacing) =
2.431 in
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl. 7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
Page 23 of 79Isolated Footing Design
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Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Design For Top Reinforcement Parallel to X Axis
First load case to be in pure uplift #
Calculate the flexural reinforcement for Mz. Find the area of steel required
The strength values of steel and concrete used in the formulae are in ksi
Design for flexure about A axis isperformed at the face of the column
at a distance, Dx =0.868 m
Ultimate moment, 30.183 kNm
Nominal moment capacity, Mn = 33.536 kNm
Required = 0.00019
Since OK
Area of Steel Required, As = 2.771 in2
Selected bar Size = #5
Minimum spacing allowed (Smin
) = 2.000 in
Selected spacing (S) = 8.172 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
#5 @ 8 in o.c.
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
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Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Pedestal Design Calculations
Strength and Moment Along Reinforcement in X direction
Strength and Moment from Concrete
Calculate strength and moment from one bar.
From ACI Cl.7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
Design for flexure about A axis isperformed at the face of the column
at a distance, Dx =0.664 m
Ultimate moment, 21.634 kNm
Nominal moment capacity, Mn = 24.037 kNm
Required = 0.00011
Since OK
Area of Steel Required, As
= 3.453 in2
Selected bar Size = #3
Minimum spacing allowed (Smin
) = 2.000 in
Selected spacing (S) = 2.633 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
UnSafe for Cracking Aspect.
#3 @ 2 in o.c.
Critical Load Case: 207
Bar size : # 5
Number of Bars : 24
Steel Area : 6.8781 sq.in
Neutral Axis Depth (Xb): 0.0519 m
Cc = 39398.844 kgf
Mc
= 88.063 kNm
Distance between extreme fiber andbar,
db 0.084 m
Strain in bar, = -0.0019
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Strength and Moment Along Reinforcement in Z direction
Strength and Moment from Concrete
Calculate strength and moment from one bar.
Maximum Strain, = 0.0021
as
-37927652.494
kgf/m2
0.0016
as
0.000 kgf/m2
-7585.515
kgf
-12.338 kNm
Total Bar Capacity, Cs
= -64756.423
kgf
Capacity of Column = Cc + Cs =-
25357.579kgf
Total Bar Moment, Ms = 3.974 kNm
Total Moment = Mc + Ms = 92.037 kNm
Bar size : # 5
Number of Bars : 24
Steel Area : 6.8781 sq.in
Neutral Axis Depth (Xb): 0.0519 m
Cc = 39398.844 kgf
Mc
= 88.063 kNm
Distance between extreme fiberand bar,
db 0.084 m
Strain in bar, = -0.0019
Maximum Strain, = 0.0021
as
-37927652.494
kgf/m2
0.0016
as
kgf/m2
-7585.515
kgf
-12.338 kNm
Total Bar Capacity, Cs = -64756.423
kgf
Capacity of Column =-
kgf
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Isolated Footing 5
Input Values
Footing Geomtery
Cc + Cs = 25357.579
Total Bar Moment, Ms = 3.974 kNm
Total Moment = Mc + Ms = 92.037 kNm
Check for bi-axial bending, 0.236
Design Moment Mnx
= 1.267 kNm
Design Moment Mnz
= 2877.644 kNm
Total Moment Mox
= 9385.357 kNm
Total Moment Moz
= 9385.357 kNm
if Mnx or Mnz = 0, then = 1.0
otherwise, = 1.24
Design Type : Calculate Dimension
Footing Thickness (Ft) : 23.620 in
Footing Length - X (Fl) : 40.000 in
Footing Width - Z (Fw) : 40.000 in
Eccentricity along X (Oxd) : 0.000 in
Eccentricity along Z (Ozd) : 0.000 in
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Column Dimensions
Pedestal
Design Parameters
Concrete and Rebar Properties
Soil Properties
Sliding and Overturning
Design Calculations
Footing Size
Column Shape : Rectangular
Column Length - X (Pl) : 0.300 m
Column Width - Z (Pw) : 0.135 m
Include Pedestal? Yes
Pedestal Shape : Rectangular
Pedestal Height (Ph) : 1.650 m
Pedestal Length - X (Pl) : 0.500 m
Pedestal Width - Z (Pw) : 0.500 m
Unit Weight of Concrete : 156.070 lb/ft3
Strength of Concrete : 2.987 ksi
Yield Strength of Steel : 59.738 ksi
Minimum Bar Size : #3
Maximum Bar Size : #5
Minimum Bar Spacing : 2.000 in
Maximum Bar Spacing : 18.000 in
Pedestal Clear Cover (P, CL) : 3.000 in
Footing Clear Cover (F, CL) : 3.000 in
Soil Type : UnDrained
Unit Weight : 112.370 lb/ft3
Soil Bearing Capacity : 15.156 kip/ft2
Soil Surcharge : 0.000 kip/in2
Depth of Soil above Footing : 65.000 in
Undrained Shear Strength : 0.030 kip/in2
Coefficient of Friction : 0.500
Factor of Safety Against Sliding : 1.500
Factor of Safety Against Overturning : 1.500
------------------------------------------------------
Initial Length (Lo) = 1.016 m
Initial Width (Wo) = 1.016 m
Load Combination/s- Service Stress Level
Load CombinationNumber
Load Combination Title
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101 1.000 x DL
102 1.000 x DL+1.000 x LL
103 1.000 x DL+0.750 x LL
104 1.000 x DL+1.000 x WLX
105 1.000 x DL+1.000 x WLZ
106 1.000 x DL+0.700 x ELX
107 1.000 x DL+0.700 x ELZ
108 1.000 x DL+0.750 x LL+0.750 x WLX
109 1.000 x DL+0.750 x LL+0.750 x WLZ
110 1.000 x DL+0.750 x LL+0.525 x ELX
111 1.000 x DL+0.750 x LL+0.525 x ELZ
Load Combination/s- Strength Level
Load CombinationNumber
Load Combination Title
201 1.400 x DL
202 1.200 x DL+1.600 x LL
203 1.200 x DL+1.000 x LL
204 1.200 x DL+0.800 x WLX
205 1.200 x DL+0.800 x WLZ
206 1.200 x DL+1.000 x LL+1.600 x WLX
207 1.200 x DL+1.000 x LL+1.600 x WLZ
208 1.200 x DL+1.000 x LL+1.000 x ELX
209 1.200 x DL+1.000 x LL+1.000 x ELZ
210 1.200 x DL+1.600 x WLX
211 1.200 x DL+1.600 x WLZ
212 1.200 x DL+1.000 x ELX
213 1.200 x DL+1.000 x ELZ
Applied Loads - Service Stress Level
LCAxial(kgf)
Shear X(kgf)
Shear Z(kgf)
Moment X(kNm)
Moment Z(kNm)
101 4388.587 -13.808 17.880 0.258 0.594
102 4389.679 -11.631 18.219 0.265 0.529
103 4389.406 -12.175 18.134 0.263 0.545
104 2650.467 1253.655 1.266 0.019 -53.550
105 4525.944 -971.365 36.701 0.930 40.994
106 4639.584 -181.299 17.218 0.243 -7.420
107 4745.298 -28.917 5.922 -0.033 0.028
108 3085.815 938.422 5.674 0.084 -40.062
109 4492.423 -730.343 32.250 0.767 30.846
110 4577.653 -137.794 17.637 0.251 -5.465
111 4656.939 -23.507 9.165 0.045 0.121
Applied Loads - Strength Level
LCAxial(kgf)
Shear X(kgf)
Shear Z(kgf)
Moment X(kNm)
Moment Z(kNm)
201 6144.022 -19.331 25.032 0.361 0.831
202 5268.051 -13.087 21.999 0.320 0.609
203 5267.396 -14.393 21.795 0.316 0.648
204 3875.808 997.401 8.165 0.118 -42.602
205 5376.190 -782.615 36.513 0.847 33.033
206 2486.403 2013.549 -4.787 -0.066 -85.982
207 5487.167 -1546.483 51.908 1.392 65.288
208 5625.962 -253.666 20.849 0.294 -10.799
209 5776.983 -35.976 4.711 -0.100 -0.160
210 2485.311 2011.372 -5.126 -0.073 -85.917
211 5486.076 -1548.660 51.569 1.385 65.353
212 5624.870 -255.843 20.510 0.288 -10.735
213 5775.892 -38.153 4.373 -0.106 -0.096
Reduction of force due to buoyancy = 0.000 kgf
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Final Footing Size
Pressures at Four Corners
If Au
is zero, there is no uplift and no pressure adjustment is necessary. Otherwise, to account for uplift, areas of negative pressure will be set to zero
and the pressure will be redistributed to remaining corners.
Summary of Adjusted Pressures at 4 corners Four Corners
Check for stability against overturning and sliding
Effect due to adhesion = 0.000 kgf
Area from initial length and width, Ao =Lo X Wo = 1.032 m2
Min. area required from bearing pressure, Amin
=P / qmax = 0.130 m2
Note: Amin is an initial estimation.
P = Critical Factored Axial Load(without self weight/buoyancy/soil).q
max= Respective Factored Bearing Capacity.
Length (L2) = 1.778 m Governing Load Case : # 104
Width (W2) = 2.184 m Governing Load Case : # 104
Depth (D2) = 0.600 m Governing Load Case : # 104
Area (A2) = 3.884 m2
Load Case
Pressure atcorner 1
(q1)
(kgf/m2)
Pressure atcorner 2
(q2)
(kgf/m2)
Pressure atcorner 3
(q3)
(kgf/m2)
Pressure atcorner 4
(q4)
(kgf/m2)
Area of footingin uplift (Au)
(m2)
105 11116.8081 54.6521 305.6068 11367.7628 0.000
104 -1970.4713 12420.3215 12427.1067 -1963.6861 0.466
104 -1970.4713 12420.3215 12427.1067 -1963.6861 0.466
105 11116.8081 54.6521 305.6068 11367.7628 0.000
Load Case
Pressure atcorner 1 (q1)
(kgf/m2)
Pressure atcorner 2 (q2)
(kgf/m2)
Pressure atcorner 3 (q3)
(kgf/m2)
Pressure atcorner 4 (q4)
(kgf/m2)
105 11116.8081 54.6521 305.6068 11367.7628
104 0.0000 12645.0738 12654.9993 0.0000
104 0.0000 12645.0738 12654.9993 0.0000
105 11116.8081 54.6521 305.6068 11367.7628
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Critical Load Case And The Governing Factor Of Safety For Overturning And Sliding - X Direction
Critical Load Case And The Governing Factor Of Safety For Overturning And Sliding - Z Direction
Shear Calculation
-Factor of safety against
slidingFactor of safety against
overturning
Load CaseNo.
Along X-Direction
Along Z-Direction
About X-Direction
About Z-Direction
101 798.241 616.438 361.963 213.941
102 947.688 605.003 354.242 244.581
103 905.316 607.821 356.141 236.127
104 8.099 8017.982 4623.300 2.180
105 11.418 302.193 136.548 3.098
106 61.487 647.453 383.671 56.844
107 387.338 1891.443 2462.198 293.217
108 11.051 1827.783 1062.714 2.976
109 15.163 343.381 160.426 4.112
110 80.676 630.286 371.757 79.952
111 474.600 1217.210 968.142 304.070
Critical Load Case for Sliding along X-Direction : 104
Governing Disturbing Force : 1253.655 kgf
Governing Restoring Force : 10153.035 kgf
Minimum Sliding Ratio for the Critical Load Case : 8.099
Critical Load Case for Overturning about X-Direction : 105
Governing Overturning Moment : 1.740 kNm
Governing Resisting Moment : 237.578 kNm
Minimum Overturning Ratio for the Critical Load Case : 136.548
Critical Load Case for Sliding along Z-Direction : 105
Governing Disturbing Force : 36.701 kgf
Governing Restoring Force : 11090.774 kgf
Minimum Sliding Ratio for the Critical Load Case : 302.193
Critical Load Case for Overturning about Z-Direction : 104
Governing Overturning Moment : -81.210 kNm
Governing Resisting Moment : 177.027 kNm
Minimum Overturning Ratio for the Critical Load Case : 2.180
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Punching Shear Check
Effective depth, deff
, increased until 0.75XVc
Punching Shear Force
Punching Shear Force, Vu = 17692.032 kgf, Load Case # 201
Along X Direction
(Shear Plane Parallel to Global X Axis)
Check that 0.75 X Vc > Vux where Vux is the shear force for the critical load cases at a distance deff from the face of the column caused by bending
about the X axis.
Total Footing Depth, D = 0.600m
Calculated Effective Depth, deff
= D - Ccover
- 1.0 = 0.498 m 1 inch is deducted from total depth to cater bar dia(US Convention).
For rectangular column, = Bcol / Dcol = 1.000
From ACI Cl.11.12.2.1, bo
for column= 3.993 m
Equation 11-33, Vc1
= 458811.711kgf
Equation 11-34, Vc2
= 534647.648kgf
Equation 11-35, Vc3 = 305874.474kgf
Punching shear strength, Vc = 0.75 X minimum of (Vc1, Vc2, Vc3) = 229405.856kgf
0.75 X Vc > Vu hence, OK
From ACI Cl.11.3.1.1, Vc
= 68093.091kgf
Distance along X to design for shear,D
x= 1.841 m
From above calculations, 0.75 X Vc
= 51069.818 kgf
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One-Way Shear Check
Along Z Direction
(Shear Plane Parallel to Global Z Axis)
Check that 0.75 X Vc
> Vuz
where Vuz
is the shear force for the critical load cases at a distance deff
from the face of the column caused by bending
about the Z axis.
Design for Flexure about Z Axis
(For Reinforcement Parallel to X Axis)
Calculate the flexural reinforcement along the X direction of the footing. Find the area of steel required, A, as per Section 3.8 of Reinforced ConcreteDesign (5th ed.) by Salmon and Wang (Ref. 1)
Critical Load Case # 206
The strength values of steel and concrete used in the formulae are in ksi
Critical load case for Vux is # 201 3780.286 kgf
0.75 X Vc > Vux hence, OK
From ACI Cl.11.3.1.1, Vc
= 83657.226 kgf
Distance along X to design for shear, Dz = 0.141 m
From above calculations, 0.75 X Vc
= 62742.919 kgf
Critical load case for Vuz
is # 211 4325.558 kgf
0.75 X Vc > Vuz hence, OK
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
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Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Because the number of bars is rounded up, make sure new reinforcement ratio < ρmax
Check to see if width is sufficient to accomodate bars
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl. 7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
Design for flexure about Z axis isperformed at the face of the column at
a distance, Dx
=0.639 m
Ultimate moment, 93.733 kNm
Nominal moment capacity, Mn
= 104.148 kNm
Required = 0.00047
Since OK
Area of Steel Required, As = 3.375 in2
Selected bar Size = #4
Minimum spacing allowed (Smin
) = = 2.000 in
Selected spacing (S) = 4.969 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
#4 @ 4.000 in o.c.
Required development length for bars = =0.305 m
Available development length for bars, DL
=0.563 m
Try bar size # 4 Area of one bar = 0.200 in2
Number of bars required, Nbar
= 17
Total reinforcement area, As_total
= Nbar
X (Area of one bar) = 3.400 in2
deff
= D - Ccover
- 0.5 X (dia. of one bar)
=
0.517 m
Reinforcement ratio, = 0.00194
From ACI Cl.7.6.1, minimum req'd cleardistance between bars, C
d=
max (Diameter of one bar, 1.0,Min. User Spacing) =
4.969 in
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Design for Flexure about X axis
(For Reinforcement Parallel to Z Axis)
Calculate the flexural reinforcement along the Z direction of the footing. Find the area of steel required, A, as per Section 3.8 of Reinforced ConcreteDesign (5th ed.) by Salmon and Wang (Ref. 1)
Critical Load Case # 206
The strength values of steel and concrete used in the formulae are in ksi
Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl.7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
Design for flexure about X axis isperformed at the face of the column at
a distance, Dz
=0.842 m
Ultimate moment, 45.758 kNm
Nominal moment capacity, Mn
= 50.842 kNm
Required = 0.00030
Since OK
Area of Steel Required, As
= 3.289 in2
Selected Bar Size = #3
Minimum spacing allowed (Smin) = 2.000 in
Selected spacing (S) = 2.447 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
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Because the number of bars is rounded up, make sure new reinforcement ratio < ρmax
Check to see if width is sufficient to accomodate bars
Bending moment for uplift cases will be calculated based solely on selfweight, soil depth and surcharge loading.
As the footing size has already been determined based on all servicebility load cases, and design moment calculation is based on selfweight, soil depthand surcharge only, top reinforcement value for all pure uplift load cases will be the same.
Design For Top Reinforcement Parallel to Z Axis
Calculate the flexural reinforcement for Mx. Find the area of steel required
The strength values of steel and concrete used in the formulae are in ksi
#3 @ 2.000 in o.c.
Required development length for bars = =0.305 m
Available development length for bars, DL
=0.563 m
Try bar size # 3 Area of one bar = 0.110 in2
Number of bars required, Nbar
= 27
Total reinforcement area, As_total
= Nbar
X (Area of one bar) = 2.970 in2
deff
= D - Ccover
- 0.5 X (dia. of one bar)
=
0.503 m
Reinforcement ratio, = 0.00214
From ACI Cl.7.6.1, minimum req'd cleardistance between bars, C
d=
max (Diameter of one bar, 1.0, Min.User Spacing) =
2.447 in
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl. 7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
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Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Design For Top Reinforcement Parallel to X Axis
First load case to be in pure uplift #
Calculate the flexural reinforcement for Mz. Find the area of steel required
The strength values of steel and concrete used in the formulae are in ksi
Design for flexure about A axis isperformed at the face of the column
at a distance, Dx =0.842 m
Ultimate moment, 27.651 kNm
Nominal moment capacity, Mn = 30.724 kNm
Required = 0.00018
Since OK
Area of Steel Required, As
= 2.677 in2
Selected bar Size = #3
Minimum spacing allowed (Smin
) = 2.000 in
Selected spacing (S) = 2.651 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
#3 @ 2 in o.c.
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl.7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
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Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Pedestal Design Calculations
Strength and Moment Along Reinforcement in X direction
Strength and Moment from Concrete
Calculate strength and moment from one bar.
Design for flexure about A axis isperformed at the face of the column
at a distance, Dx =0.639 m
Ultimate moment, 19.556 kNm
Nominal moment capacity, Mn = 21.729 kNm
Required = 0.00010
Since OK
Area of Steel Required, As = 3.375 in2
Selected bar Size = #4
Minimum spacing allowed (Smin
) = 2.000 in
Selected spacing (S) = 4.969 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
#4 @ 4 in o.c.
Critical Load Case: 201
Bar size : # 3
Number of Bars : 24
Steel Area : 2.5188 sq.in
Neutral Axis Depth (Xb): 0.0631 m
Cc = 47843.900 kgf
Mc
= 104.719 kNm
Distance between extreme fiber andbar,
db 0.084 m
Strain in bar, = -0.0010
Maximum Strain, = 0.0021
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Strength and Moment Along Reinforcement in Z direction
Strength and Moment from Concrete
Calculate strength and moment from one bar.
as
-20436179.438
kgf/m2
0.0016
as
0.000 kgf/m2
-4087.227
kgf
-6.648 kNm
Total Bar Capacity, Cs = -54261.561
kgf
Capacity of Column = Cc
+ Cs
=-
6417.661kgf
Total Bar Moment, Ms
= 21.044 kNm
Total Moment = Mc
+ Ms
= 125.764 kNm
Bar size : # 3
Number of Bars : 24
Steel Area : 2.5188 sq.in
Neutral Axis Depth (Xb): 0.0631 m
Cc
= 47843.900 kgf
Mc = 104.719 kNm
Distance between extreme fiberand bar,
db 0.084 m
Strain in bar, = -0.0010
Maximum Strain, = 0.0021
as
-20436179.438
kgf/m2
0.0016
as
kgf/m2
-4087.227
kgf
-6.648 kNm
Total Bar Capacity, Cs = -54261.561
kgf
Capacity of Column = Cc + Cs =-
6417.661kgf
Total Bar Moment, Ms = 21.044 kNm
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Isolated Footing 2351
Input Values
Footing Geomtery
Column Dimensions
Total Moment = Mc + Ms = 125.764 kNm
Check for bi-axial bending, 0.002
Design Moment Mnx
= 0.792 kNm
Design Moment Mnz
= 20.068 kNm
Total Moment Mox
= 12824.569 kNm
Total Moment Moz
= 12824.569 kNm
if Mnx or Mnz = 0, then = 1.0
otherwise, = 1.24
Design Type : Calculate Dimension
Footing Thickness (Ft) : 23.620 in
Footing Length - X (Fl) : 40.000 in
Footing Width - Z (Fw) : 40.000 in
Eccentricity along X (Oxd) : 0.000 in
Eccentricity along Z (Ozd) : 0.000 in
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Pedestal
Design Parameters
Concrete and Rebar Properties
Soil Properties
Sliding and Overturning
Design Calculations
Footing Size
Column Shape : Rectangular
Column Length - X (Pl) : 0.300 m
Column Width - Z (Pw) : 0.135 m
Include Pedestal? Yes
Pedestal Shape : Rectangular
Pedestal Height (Ph) : 1.650 m
Pedestal Length - X (Pl) : 0.500 m
Pedestal Width - Z (Pw) : 0.500 m
Unit Weight of Concrete : 156.070 lb/ft3
Strength of Concrete : 2.987 ksi
Yield Strength of Steel : 59.738 ksi
Minimum Bar Size : #3
Maximum Bar Size : #5
Minimum Bar Spacing : 2.000 in
Maximum Bar Spacing : 18.000 in
Pedestal Clear Cover (P, CL) : 3.000 in
Footing Clear Cover (F, CL) : 3.000 in
Soil Type : UnDrained
Unit Weight : 112.370 lb/ft3
Soil Bearing Capacity : 15.156 kip/ft2
Soil Surcharge : 0.000 kip/in2
Depth of Soil above Footing : 65.000 in
Undrained Shear Strength : 0.030 kip/in2
Coefficient of Friction : 0.500
Factor of Safety Against Sliding : 1.500
Factor of Safety Against Overturning : 1.500
------------------------------------------------------
Initial Length (Lo) = 1.016 m
Initial Width (Wo) = 1.016 m
Load Combination/s- Service Stress Level
Load CombinationNumber
Load Combination Title
101 1.000 x DL
102 1.000 x DL+1.000 x LL
103 1.000 x DL+0.750 x LL
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104 1.000 x DL+1.000 x WLX
105 1.000 x DL+1.000 x WLZ
106 1.000 x DL+0.700 x ELX
107 1.000 x DL+0.700 x ELZ
108 1.000 x DL+0.750 x LL+0.750 x WLX
109 1.000 x DL+0.750 x LL+0.750 x WLZ
110 1.000 x DL+0.750 x LL+0.525 x ELX
111 1.000 x DL+0.750 x LL+0.525 x ELZ
Load Combination/s- Strength Level
Load CombinationNumber
Load Combination Title
201 1.400 x DL
202 1.200 x DL+1.600 x LL
203 1.200 x DL+1.000 x LL
204 1.200 x DL+0.800 x WLX
205 1.200 x DL+0.800 x WLZ
206 1.200 x DL+1.000 x LL+1.600 x WLX
207 1.200 x DL+1.000 x LL+1.600 x WLZ
208 1.200 x DL+1.000 x LL+1.000 x ELX
209 1.200 x DL+1.000 x LL+1.000 x ELZ
210 1.200 x DL+1.600 x WLX
211 1.200 x DL+1.600 x WLZ
212 1.200 x DL+1.000 x ELX
213 1.200 x DL+1.000 x ELZ
Applied Loads - Service Stress Level
LCAxial(kgf)
Shear X(kgf)
Shear Z(kgf)
Moment X(kNm)
Moment Z(kNm)
101 8010.615 328.059 -751.474 0.043 -6.156
102 16650.700 617.464 -1293.364 0.152 -12.959
103 14490.679 545.113 -1157.891 0.124 -11.258
104 7252.085 1111.415 -802.095 0.069 -39.690
105 -19436.578 1115.404 4689.678 0.655 -38.790
106 8417.253 -144.567 -830.229 0.033 -24.702
107 16120.994 308.062 -2241.784 -0.128 -6.708
108 13921.781 1132.630 -1195.857 0.144 -36.409
109 -6094.716 1135.622 2922.973 0.584 -35.734
110 14795.658 190.643 -1216.957 0.118 -25.168
111 20573.462 530.115 -2275.624 -0.004 -11.673
Applied Loads - Strength Level
LCAxial(kgf)
Shear X(kgf)
Shear Z(kgf)
Moment X(kNm)
Moment Z(kNm)
201 11214.862 459.283 -1052.064 0.060 -8.618
202 23436.874 856.719 -1768.792 0.226 -18.272
203 18252.823 683.076 -1443.659 0.160 -14.190
204 9005.915 1020.355 -942.266 0.072 -34.214
205 -12345.015 1023.547 3451.153 0.541 -33.494
206 17039.175 1936.445 -1524.652 0.202 -67.846
207 -25662.685 1942.828 7262.186 1.140 -66.405
208 18833.734 7.896 -1556.165 0.147 -40.685
209 29839.079 654.508 -3572.673 -0.084 -14.980
210 8399.091 1647.040 -982.762 0.093 -61.042
211 -34302.768 1653.423 7804.075 1.031 -59.601
212 10193.649 -281.509 -1014.275 0.038 -33.881
213 21198.994 365.103 -3030.783 -0.193 -8.176
Reduction of force due to buoyancy = 0.000 kgf
Effect due to adhesion = 121072.543 kgf
Area from initial length and width, Ao =Lo X Wo = 1.032 m2
Min. area required from bearing pressure, Amin
=P / qmax = 0.344 m2
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Final Footing Size
Pressures at Four Corners
If Au
is zero, there is no uplift and no pressure adjustment is necessary. Otherwise, to account for uplift, areas of negative pressure will be set to zero
and the pressure will be redistributed to remaining corners.
Summary of Adjusted Pressures at 4 corners Four Corners
Check for stability against overturning and sliding
Note: Amin is an initial estimation.
P = Critical Factored Axial Load(without self weight/buoyancy/soil).qmax = Respective Factored Bearing Capacity.
Length (L2) = 2.591 m Governing Load Case : # 105
Width (W2) = 2.946 m Governing Load Case : # 105
Depth (D2) = 0.600 m Governing Load Case : # 105
Area (A2) = 7.634 m2
Load Case
Pressure atcorner 1
(q1)
(kgf/m2)
Pressure atcorner 2
(q2)
(kgf/m2)
Pressure atcorner 3
(q3)
(kgf/m2)
Pressure atcorner 4
(q4)
(kgf/m2)
Area of footingin uplift (A
u)
(m2)
111 7847.5499 9293.5059 6561.5845 5115.6286 0.000
111 7847.5499 9293.5059 6561.5845 5115.6286 0.000
105 -2830.8344 1091.9853 6757.2626 2834.4429 1.368
111 7847.5499 9293.5059 6561.5845 5115.6286 0.000
Load Case
Pressure atcorner 1 (q
1)
(kgf/m2)
Pressure atcorner 2 (q
2)
(kgf/m2)
Pressure atcorner 3 (q
3)
(kgf/m2)
Pressure atcorner 4 (q
4)
(kgf/m2)
111 7847.5499 9293.5059 6561.5845 5115.6286
111 7847.5499 9293.5059 6561.5845 5115.6286
105 0.0000 0.0000 8351.9046 2451.6504
111 7847.5499 9293.5059 6561.5845 5115.6286
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Critical Load Case And The Governing Factor Of Safety For Overturning And Sliding - X Direction
Critical Load Case And The Governing Factor Of Safety For Overturning And Sliding - Z Direction
Shear Calculation
-Factor of safety against
slidingFactor of safety against
overturning
Load CaseNo.
Along X-Direction
Along Z-Direction
About X-Direction
About Z-Direction
101 64.673 28.233 37.069 40.246
102 41.357 19.744 25.994 24.407
103 44.865 21.122 27.795 26.684
104 18.749 25.979 34.153 8.245
105 6.718 1.598 2.079 3.003
106 148.167 25.800 33.848 25.298
107 82.035 11.273 14.725 47.542
108 21.342 20.213 26.615 10.002
109 12.473 4.846 6.289 5.920
110 129.085 20.222 26.598 21.285
111 51.872 12.084 15.823 29.895
Critical Load Case for Sliding along X-Direction : 105
Governing Disturbing Force : 1115.404 kgf
Governing Restoring Force : 7493.130 kgf
Minimum Sliding Ratio for the Critical Load Case : 6.718
Critical Load Case for Overturning about X-Direction : 105
Governing Overturning Moment : 104.129 kNm
Governing Resisting Moment : 216.505 kNm
Minimum Overturning Ratio for the Critical Load Case : 2.079
Critical Load Case for Sliding along Z-Direction : 105
Governing Disturbing Force : 4689.678 kgf
Governing Restoring Force : 7493.130 kgf
Minimum Sliding Ratio for the Critical Load Case : 1.598
Critical Load Case for Overturning about Z-Direction : 105
Governing Overturning Moment : -63.400 kNm
Governing Resisting Moment : 190.375 kNm
Minimum Overturning Ratio for the Critical Load Case : 3.003
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Punching Shear Check
Effective depth, deff
, increased until 0.75XVc
Punching Shear Force
Punching Shear Force, Vu = 55871.336 kgf, Load Case # 209
Along X Direction
(Shear Plane Parallel to Global X Axis)
Check that 0.75 X Vc > Vux where Vux is the shear force for the critical load cases at a distance deff from the face of the column caused by bending
about the X axis.
Total Footing Depth, D = 0.600m
Calculated Effective Depth, deff
= D - Ccover
- 1.0 = 0.498 m 1 inch is deducted from total depth to cater bar dia(US Convention).
For rectangular column, = Bcol / Dcol = 1.000
From ACI Cl.11.12.2.1, bo
for column= 3.993 m
Equation 11-33, Vc1
= 458811.711kgf
Equation 11-34, Vc2
= 534647.648kgf
Equation 11-35, Vc3 = 305874.474kgf
Punching shear strength, Vc = 0.75 X minimum of (Vc1, Vc2, Vc3) = 229405.856kgf
0.75 X Vc > Vu hence, OK
From ACI Cl.11.3.1.1, Vc
= 99221.361kgf
Distance along X to design for shear,D
x= 0.725 m
From above calculations, 0.75 X Vc
= 74416.020 kgf
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One-Way Shear Check
Along Z Direction
(Shear Plane Parallel to Global Z Axis)
Check that 0.75 X Vc
> Vuz
where Vuz
is the shear force for the critical load cases at a distance deff
from the face of the column caused by bending
about the Z axis.
Design for Flexure about Z Axis
(For Reinforcement Parallel to X Axis)
Calculate the flexural reinforcement along the X direction of the footing. Find the area of steel required, A, as per Section 3.8 of Reinforced ConcreteDesign (5th ed.) by Salmon and Wang (Ref. 1)
Critical Load Case # 206
The strength values of steel and concrete used in the formulae are in ksi
Critical load case for Vux is # 209 18848.792 kgf
0.75 X Vc > Vux hence, OK
From ACI Cl.11.3.1.1, Vc
= 112839.979 kgf
Distance along X to design for shear, Dz = 2.044 m
From above calculations, 0.75 X Vc
= 84629.984 kgf
Critical load case for Vuz
is # 206 15215.757 kgf
0.75 X Vc > Vuz hence, OK
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
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Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Because the number of bars is rounded up, make sure new reinforcement ratio < ρmax
Check to see if width is sufficient to accomodate bars
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl. 7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
Design for flexure about Z axis isperformed at the face of the column at
a distance, Dx
=1.045 m
Ultimate moment, 145.957 kNm
Nominal moment capacity, Mn
= 162.174 kNm
Required = 0.00054
Since OK
Area of Steel Required, As = 4.552 in2
Selected bar Size = #4
Minimum spacing allowed (Smin
) = = 2.000 in
Selected spacing (S) = 4.977 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
#4 @ 4.000 in o.c.
Required development length for bars = =0.305 m
Available development length for bars, DL
=0.969 m
Try bar size # 4 Area of one bar = 0.200 in2
Number of bars required, Nbar
= 23
Total reinforcement area, As_total
= Nbar
X (Area of one bar) = 4.600 in2
deff
= D - Ccover
- 0.5 X (dia. of one bar)
=
0.517 m
Reinforcement ratio, = 0.00195
From ACI Cl.7.6.1, minimum req'd cleardistance between bars, C
d=
max (Diameter of one bar, 1.0,Min. User Spacing) =
4.977 in
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Design for Flexure about X axis
(For Reinforcement Parallel to Z Axis)
Calculate the flexural reinforcement along the Z direction of the footing. Find the area of steel required, A, as per Section 3.8 of Reinforced ConcreteDesign (5th ed.) by Salmon and Wang (Ref. 1)
Critical Load Case # 209
The strength values of steel and concrete used in the formulae are in ksi
Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl.7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
Design for flexure about X axis isperformed at the face of the column at
a distance, Dz
=1.223 m
Ultimate moment, 189.507 kNm
Nominal moment capacity, Mn
= 210.563 kNm
Required = 0.00085
Since OK
Area of Steel Required, As
= 4.436 in2
Selected Bar Size = #3
Minimum spacing allowed (Smin) = 2.000 in
Selected spacing (S) = 2.584 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
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Because the number of bars is rounded up, make sure new reinforcement ratio < ρmax
Check to see if width is sufficient to accomodate bars
Bending moment for uplift cases will be calculated based solely on selfweight, soil depth and surcharge loading.
As the footing size has already been determined based on all servicebility load cases, and design moment calculation is based on selfweight, soil depthand surcharge only, top reinforcement value for all pure uplift load cases will be the same.
Design For Top Reinforcement Parallel to Z Axis
Calculate the flexural reinforcement for Mx. Find the area of steel required
The strength values of steel and concrete used in the formulae are in ksi
#3 @ 2.000 in o.c.
Required development length for bars = =0.305 m
Available development length for bars, DL
=0.969 m
Try bar size # 3 Area of one bar = 0.110 in2
Number of bars required, Nbar
= 38
Total reinforcement area, As_total
= Nbar
X (Area of one bar) = 4.180 in2
deff
= D - Ccover
- 0.5 X (dia. of one bar)
=
0.503 m
Reinforcement ratio, = 0.00207
From ACI Cl.7.6.1, minimum req'd cleardistance between bars, C
d=
max (Diameter of one bar, 1.0, Min.User Spacing) =
2.584 in
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl. 7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
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Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Design For Top Reinforcement Parallel to X Axis
First load case to be in pure uplift #
Calculate the flexural reinforcement for Mz. Find the area of steel required
The strength values of steel and concrete used in the formulae are in ksi
Design for flexure about A axis isperformed at the face of the column
at a distance, Dx =1.223 m
Ultimate moment, 84.992 kNm
Nominal moment capacity, Mn = 94.436 kNm
Required = 0.00038
Since OK
Area of Steel Required, As
= 3.900 in2
Selected bar Size = #3
Minimum spacing allowed (Smin
) = 2.000 in
Selected spacing (S) = 2.732 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
#3 @ 2 in o.c.
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl.7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
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Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Pedestal Design Calculations
Strength and Moment Along Reinforcement in X direction
Strength and Moment from Concrete
Calculate strength and moment from one bar.
Design for flexure about A axis isperformed at the face of the column
at a distance, Dx =1.045 m
Ultimate moment, 70.601 kNm
Nominal moment capacity, Mn = 78.445 kNm
Required = 0.00026
Since OK
Area of Steel Required, As = 4.552 in2
Selected bar Size = #4
Minimum spacing allowed (Smin
) = 2.000 in
Selected spacing (S) = 4.977 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
#4 @ 4 in o.c.
Critical Load Case: 209
Bar size : # 4
Number of Bars : 24
Steel Area : 4.2625 sq.in
Neutral Axis Depth (Xb): 0.0529 m
Cc = 40125.398 kgf
Mc
= 89.527 kNm
Distance between extreme fiber andbar,
db 0.084 m
Strain in bar, = -0.0018
Maximum Strain, = 0.0021
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Strength and Moment Along Reinforcement in Z direction
Strength and Moment from Concrete
Calculate strength and moment from one bar.
as
-36133337.630
kgf/m2
0.0016
as
0.000 kgf/m2
-7226.653
kgf
-11.754 kNm
Total Bar Capacity, Cs = -63679.836
kgf
Capacity of Column = Cc
+ Cs
=-
23554.439kgf
Total Bar Moment, Ms
= 5.725 kNm
Total Moment = Mc
+ Ms
= 95.252 kNm
Bar size : # 4
Number of Bars : 24
Steel Area : 4.2625 sq.in
Neutral Axis Depth (Xb): 0.0529 m
Cc
= 40125.398 kgf
Mc = 89.527 kNm
Distance between extreme fiberand bar,
db 0.084 m
Strain in bar, = -0.0018
Maximum Strain, = 0.0021
as
-36133337.630
kgf/m2
0.0016
as
kgf/m2
-7226.653
kgf
-11.754 kNm
Total Bar Capacity, Cs = -63679.836
kgf
Capacity of Column = Cc + Cs =-
23554.439kgf
Total Bar Moment, Ms = 5.725 kNm
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Isolated Footing 2352
Input Values
Footing Geomtery
Column Dimensions
Total Moment = Mc + Ms = 95.252 kNm
Check for bi-axial bending, 0.609
Design Moment Mnx
= 15.649 kNm
Design Moment Mnz
= 5578.264 kNm
Total Moment Mox
= 9713.192 kNm
Total Moment Moz
= 9713.192 kNm
if Mnx or Mnz = 0, then = 1.0
otherwise, = 1.24
Design Type : Calculate Dimension
Footing Thickness (Ft) : 23.620 in
Footing Length - X (Fl) : 40.000 in
Footing Width - Z (Fw) : 40.000 in
Eccentricity along X (Oxd) : 0.000 in
Eccentricity along Z (Ozd) : 0.000 in
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Pedestal
Design Parameters
Concrete and Rebar Properties
Soil Properties
Sliding and Overturning
Design Calculations
Footing Size
Column Shape : Rectangular
Column Length - X (Pl) : 0.300 m
Column Width - Z (Pw) : 0.135 m
Include Pedestal? Yes
Pedestal Shape : Rectangular
Pedestal Height (Ph) : 1.650 m
Pedestal Length - X (Pl) : 0.500 m
Pedestal Width - Z (Pw) : 0.500 m
Unit Weight of Concrete : 156.070 lb/ft3
Strength of Concrete : 2.987 ksi
Yield Strength of Steel : 59.738 ksi
Minimum Bar Size : #3
Maximum Bar Size : #5
Minimum Bar Spacing : 2.000 in
Maximum Bar Spacing : 18.000 in
Pedestal Clear Cover (P, CL) : 3.000 in
Footing Clear Cover (F, CL) : 3.000 in
Soil Type : UnDrained
Unit Weight : 112.370 lb/ft3
Soil Bearing Capacity : 15.156 kip/ft2
Soil Surcharge : 0.000 kip/in2
Depth of Soil above Footing : 65.000 in
Undrained Shear Strength : 0.030 kip/in2
Coefficient of Friction : 0.500
Factor of Safety Against Sliding : 1.500
Factor of Safety Against Overturning : 1.500
------------------------------------------------------
Initial Length (Lo) = 1.016 m
Initial Width (Wo) = 1.016 m
Load Combination/s- Service Stress Level
Load CombinationNumber
Load Combination Title
101 1.000 x DL
102 1.000 x DL+1.000 x LL
103 1.000 x DL+0.750 x LL
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104 1.000 x DL+1.000 x WLX
105 1.000 x DL+1.000 x WLZ
106 1.000 x DL+0.700 x ELX
107 1.000 x DL+0.700 x ELZ
108 1.000 x DL+0.750 x LL+0.750 x WLX
109 1.000 x DL+0.750 x LL+0.750 x WLZ
110 1.000 x DL+0.750 x LL+0.525 x ELX
111 1.000 x DL+0.750 x LL+0.525 x ELZ
Load Combination/s- Strength Level
Load CombinationNumber
Load Combination Title
201 1.400 x DL
202 1.200 x DL+1.600 x LL
203 1.200 x DL+1.000 x LL
204 1.200 x DL+0.800 x WLX
205 1.200 x DL+0.800 x WLZ
206 1.200 x DL+1.000 x LL+1.600 x WLX
207 1.200 x DL+1.000 x LL+1.600 x WLZ
208 1.200 x DL+1.000 x LL+1.000 x ELX
209 1.200 x DL+1.000 x LL+1.000 x ELZ
210 1.200 x DL+1.600 x WLX
211 1.200 x DL+1.600 x WLZ
212 1.200 x DL+1.000 x ELX
213 1.200 x DL+1.000 x ELZ
Applied Loads - Service Stress Level
LCAxial(kgf)
Shear X(kgf)
Shear Z(kgf)
Moment X(kNm)
Moment Z(kNm)
101 11379.871 246.315 783.037 -0.187 -8.106
102 19681.750 617.937 1341.153 -0.476 -23.117
103 17606.279 525.031 1201.624 -0.404 -19.364
104 9031.872 1232.370 433.075 -0.026 -49.967
105 33472.524 1237.935 5769.972 0.496 -50.506
106 12019.358 -284.923 696.469 -0.201 -30.264
107 19325.367 228.594 -692.243 -0.354 -8.656
108 15845.280 1264.573 939.152 -0.283 -50.760
109 34175.766 1268.746 4941.825 0.109 -51.164
110 18085.896 126.603 1136.698 -0.414 -35.983
111 23565.402 511.741 95.164 -0.529 -19.776
Applied Loads - Strength Level
LCAxial(kgf)
Shear X(kgf)
Shear Z(kgf)
Moment X(kNm)
Moment Z(kNm)
201 15931.820 344.841 1096.252 -0.262 -11.348
202 26938.851 890.173 1832.631 -0.687 -33.744
203 21957.725 667.200 1497.761 -0.514 -24.738
204 11777.446 1084.422 659.675 -0.096 -43.216
205 31329.966 1088.874 4929.192 0.322 -43.647
206 18200.925 2244.888 937.820 -0.256 -91.716
207 57305.966 2253.792 9476.856 0.580 -92.578
208 22871.278 -91.711 1374.091 -0.533 -56.392
209 33308.435 641.884 -609.783 -0.752 -25.523
210 9899.047 1873.266 379.704 0.033 -76.705
211 49004.087 1882.171 8918.740 0.869 -77.568
212 14569.400 -463.333 815.976 -0.244 -41.382
213 25006.556 270.263 -1167.899 -0.463 -10.512
Reduction of force due to buoyancy = 0.000 kgf
Effect due to adhesion = 0.000 kgf
Area from initial length and width, Ao =Lo X Wo = 1.032 m2
Min. area required from bearing pressure, Amin
=P / qmax = 0.528 m2
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Final Footing Size
Pressures at Four Corners
If Au
is zero, there is no uplift and no pressure adjustment is necessary. Otherwise, to account for uplift, areas of negative pressure will be set to zero
and the pressure will be redistributed to remaining corners.
Summary of Adjusted Pressures at 4 corners Four Corners
Check for stability against overturning and sliding
Note: Amin is an initial estimation.
P = Critical Factored Axial Load(without self weight/buoyancy/soil).qmax = Respective Factored Bearing Capacity.
Length (L2) = 1.575 m Governing Load Case : # 104
Width (W2) = 1.930 m Governing Load Case : # 104
Depth (D2) = 0.600 m Governing Load Case : # 104
Area (A2) = 3.040 m2
Load Case
Pressure atcorner 1
(q1)
(kgf/m2)
Pressure atcorner 2
(q2)
(kgf/m2)
Pressure atcorner 3
(q3)
(kgf/m2)
Pressure atcorner 4
(q4)
(kgf/m2)
Area of footingin uplift (A
u)
(m2)
107 10802.0732 14303.6747 11044.9960 7543.3945 0.000
108 -2405.3124 17700.9965 21962.8460 1856.5371 0.102
105 -7693.3984 12197.8595 38847.8269 18956.5691 0.170
105 -7693.3984 12197.8595 38847.8269 18956.5691 0.170
Load Case
Pressure atcorner 1 (q
1)
(kgf/m2)
Pressure atcorner 2 (q
2)
(kgf/m2)
Pressure atcorner 3 (q
3)
(kgf/m2)
Pressure atcorner 4 (q
4)
(kgf/m2)
107 10802.0732 14303.6747 11044.9960 7543.3945
108 0.0000 17676.1079 22074.6042 1778.9737
105 0.0000 11914.1094 39570.9607 18734.9028
105 0.0000 11914.1094 39570.9607 18734.9028
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Critical Load Case And The Governing Factor Of Safety For Overturning And Sliding - X Direction
Critical Load Case And The Governing Factor Of Safety For Overturning And Sliding - Z Direction
Shear Calculation
-Factor of safety against
slidingFactor of safety against
overturning
Load CaseNo.
Along X-Direction
Along Z-Direction
About X-Direction
About Z-Direction
101 51.280 16.131 13.991 14.406
102 27.158 12.513 10.911 7.052
103 29.987 13.102 11.415 7.856
104 9.297 26.455 22.760 2.293
105 19.126 4.104 3.507 4.699
106 45.454 18.595 16.165 8.341
107 72.634 23.985 20.113 18.717
108 11.754 15.827 13.767 2.918
109 18.939 4.862 4.168 4.688
110 126.253 14.062 12.267 6.366
111 36.588 196.753 225.654 9.307
Critical Load Case for Sliding along X-Direction : 104
Governing Disturbing Force : 1232.370 kgf
Governing Restoring Force : 11456.991 kgf
Minimum Sliding Ratio for the Critical Load Case : 9.297
Critical Load Case for Overturning about X-Direction : 105
Governing Overturning Moment : 127.805 kNm
Governing Resisting Moment : 448.221 kNm
Minimum Overturning Ratio for the Critical Load Case : 3.507
Critical Load Case for Sliding along Z-Direction : 105
Governing Disturbing Force : 5769.972 kgf
Governing Restoring Force : 23677.317 kgf
Minimum Sliding Ratio for the Critical Load Case : 4.104
Critical Load Case for Overturning about Z-Direction : 104
Governing Overturning Moment : -77.158 kNm
Governing Resisting Moment : 176.933 kNm
Minimum Overturning Ratio for the Critical Load Case : 2.293
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Punching Shear Check
Effective depth, deff
, increased until 0.75XVc
Punching Shear Force
Punching Shear Force, Vu = 47848.204 kgf, Load Case # 207
Along X Direction
(Shear Plane Parallel to Global X Axis)
Check that 0.75 X Vc > Vux where Vux is the shear force for the critical load cases at a distance deff from the face of the column caused by bending
about the X axis.
Total Footing Depth, D = 0.600m
Calculated Effective Depth, deff
= D - Ccover
- 1.0 = 0.498 m 1 inch is deducted from total depth to cater bar dia(US Convention).
For rectangular column, = Bcol / Dcol = 1.000
From ACI Cl.11.12.2.1, bo
for column= 3.993 m
Equation 11-33, Vc1
= 458811.711kgf
Equation 11-34, Vc2
= 534647.648kgf
Equation 11-35, Vc3 = 305874.474kgf
Punching shear strength, Vc = 0.75 X minimum of (Vc1, Vc2, Vc3) = 229405.856kgf
0.75 X Vc > Vu hence, OK
From ACI Cl.11.3.1.1, Vc
= 60311.023kgf
Distance along X to design for shear,D
x= 1.714 m
From above calculations, 0.75 X Vc
= 45233.267 kgf
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One-Way Shear Check
Along Z Direction
(Shear Plane Parallel to Global Z Axis)
Check that 0.75 X Vc
> Vuz
where Vuz
is the shear force for the critical load cases at a distance deff
from the face of the column caused by bending
about the Z axis.
Design for Flexure about Z Axis
(For Reinforcement Parallel to X Axis)
Calculate the flexural reinforcement along the X direction of the footing. Find the area of steel required, A, as per Section 3.8 of Reinforced ConcreteDesign (5th ed.) by Salmon and Wang (Ref. 1)
Critical Load Case # 207
The strength values of steel and concrete used in the formulae are in ksi
Critical load case for Vux is # 207 13165.387 kgf
0.75 X Vc > Vux hence, OK
From ACI Cl.11.3.1.1, Vc
= 73929.641 kgf
Distance along X to design for shear, Dz = 1.536 m
From above calculations, 0.75 X Vc
= 55447.231 kgf
Critical load case for Vuz
is # 207 1643.412 kgf
0.75 X Vc > Vuz hence, OK
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
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Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Because the number of bars is rounded up, make sure new reinforcement ratio < ρmax
Check to see if width is sufficient to accomodate bars
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl. 7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
Design for flexure about Z axis isperformed at the face of the column at
a distance, Dx
=0.537 m
Ultimate moment, 114.760 kNm
Nominal moment capacity, Mn
= 127.511 kNm
Required = 0.00065
Since OK
Area of Steel Required, As = 2.982 in2
Selected bar Size = #4
Minimum spacing allowed (Smin
) = = 2.000 in
Selected spacing (S) = 4.964 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
#4 @ 4.000 in o.c.
Required development length for bars = =0.305 m
Available development length for bars, DL
=0.461 m
Try bar size # 4 Area of one bar = 0.200 in2
Number of bars required, Nbar
= 15
Total reinforcement area, As_total
= Nbar
X (Area of one bar) = 3.000 in2
deff
= D - Ccover
- 0.5 X (dia. of one bar)
=
0.517 m
Reinforcement ratio, = 0.00194
From ACI Cl.7.6.1, minimum req'd cleardistance between bars, C
d=
max (Diameter of one bar, 1.0,Min. User Spacing) =
4.964 in
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Design for Flexure about X axis
(For Reinforcement Parallel to Z Axis)
Calculate the flexural reinforcement along the Z direction of the footing. Find the area of steel required, A, as per Section 3.8 of Reinforced ConcreteDesign (5th ed.) by Salmon and Wang (Ref. 1)
Critical Load Case # 207
The strength values of steel and concrete used in the formulae are in ksi
Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl.7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
Design for flexure about X axis isperformed at the face of the column at
a distance, Dz
=0.715 m
Ultimate moment, 172.369 kNm
Nominal moment capacity, Mn
= 191.521 kNm
Required = 0.00127
Since OK
Area of Steel Required, As
= 2.906 in2
Selected Bar Size = #3
Minimum spacing allowed (Smin) = 2.000 in
Selected spacing (S) = 2.418 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
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Because the number of bars is rounded up, make sure new reinforcement ratio < ρmax
Check to see if width is sufficient to accomodate bars
Bending moment for uplift cases will be calculated based solely on selfweight, soil depth and surcharge loading.
As the footing size has already been determined based on all servicebility load cases, and design moment calculation is based on selfweight, soil depthand surcharge only, top reinforcement value for all pure uplift load cases will be the same.
Design For Top Reinforcement Parallel to Z Axis
Calculate the flexural reinforcement for Mx. Find the area of steel required
The strength values of steel and concrete used in the formulae are in ksi
#3 @ 2.000 in o.c.
Required development length for bars = =0.305 m
Available development length for bars, DL
=0.461 m
Try bar size # 3 Area of one bar = 0.110 in2
Number of bars required, Nbar
= 24
Total reinforcement area, As_total
= Nbar
X (Area of one bar) = 2.640 in2
deff
= D - Ccover
- 0.5 X (dia. of one bar)
=
0.503 m
Reinforcement ratio, = 0.00215
From ACI Cl.7.6.1, minimum req'd cleardistance between bars, C
d=
max (Diameter of one bar, 1.0, Min.User Spacing) =
2.418 in
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl. 7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
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Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Design For Top Reinforcement Parallel to X Axis
First load case to be in pure uplift #
Calculate the flexural reinforcement for Mz. Find the area of steel required
The strength values of steel and concrete used in the formulae are in ksi
Design for flexure about A axis isperformed at the face of the column
at a distance, Dx =0.715 m
Ultimate moment, 17.662 kNm
Nominal moment capacity, Mn = 19.624 kNm
Required = 0.00013
Since OK
Area of Steel Required, As
= 2.371 in2
Selected bar Size = #4
Minimum spacing allowed (Smin
) = 2.000 in
Selected spacing (S) = 5.045 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
#4 @ 5 in o.c.
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl.7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
Page 63 of 79Isolated Footing Design
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Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Pedestal Design Calculations
Strength and Moment Along Reinforcement in X direction
Strength and Moment from Concrete
Calculate strength and moment from one bar.
Design for flexure about A axis isperformed at the face of the column
at a distance, Dx =0.537 m
Ultimate moment, 12.223 kNm
Nominal moment capacity, Mn = 13.582 kNm
Required = 0.00007
Since OK
Area of Steel Required, As = 2.982 in2
Selected bar Size = #4
Minimum spacing allowed (Smin
) = 2.000 in
Selected spacing (S) = 4.964 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
#4 @ 4 in o.c.
Critical Load Case: 207
Bar size : # 5
Number of Bars : 24
Steel Area : 7.1688 sq.in
Neutral Axis Depth (Xb): 0.0507 m
Cc = 38445.584 kgf
Mc
= 86.134 kNm
Distance between extreme fiber andbar,
db 0.084 m
Strain in bar, = -0.0020
Maximum Strain, = 0.0021
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Strength and Moment Along Reinforcement in Z direction
Strength and Moment from Concrete
Calculate strength and moment from one bar.
as
-40384710.295
kgf/m2
0.0016
as
0.000 kgf/m2
-8076.925
kgf
-13.137 kNm
Total Bar Capacity, Cs = -66230.655
kgf
Capacity of Column = Cc
+ Cs
=-
27785.071kgf
Total Bar Moment, Ms
= 1.576 kNm
Total Moment = Mc
+ Ms
= 87.710 kNm
Bar size : # 5
Number of Bars : 24
Steel Area : 7.1688 sq.in
Neutral Axis Depth (Xb): 0.0507 m
Cc
= 38445.584 kgf
Mc = 86.134 kNm
Distance between extreme fiberand bar,
db 0.084 m
Strain in bar, = -0.0020
Maximum Strain, = 0.0021
as
-40384710.295
kgf/m2
0.0016
as
kgf/m2
-8076.925
kgf
-13.137 kNm
Total Bar Capacity, Cs = -66230.655
kgf
Capacity of Column = Cc + Cs =-
27785.071kgf
Total Bar Moment, Ms = 1.576 kNm
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Isolated Footing 2353
Input Values
Footing Geomtery
Column Dimensions
Total Moment = Mc + Ms = 87.710 kNm
Check for bi-axial bending, 0.301
Design Moment Mnx
= 16.539 kNm
Design Moment Mnz
= 2193.588 kNm
Total Moment Mox
= 8944.094 kNm
Total Moment Moz
= 8944.094 kNm
if Mnx or Mnz = 0, then = 1.0
otherwise, = 1.24
Design Type : Calculate Dimension
Footing Thickness (Ft) : 23.620 in
Footing Length - X (Fl) : 40.000 in
Footing Width - Z (Fw) : 40.000 in
Eccentricity along X (Oxd) : 0.000 in
Eccentricity along Z (Ozd) : 0.000 in
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Pedestal
Design Parameters
Concrete and Rebar Properties
Soil Properties
Sliding and Overturning
Design Calculations
Footing Size
Column Shape : Rectangular
Column Length - X (Pl) : 0.300 m
Column Width - Z (Pw) : 0.135 m
Include Pedestal? Yes
Pedestal Shape : Rectangular
Pedestal Height (Ph) : 1.650 m
Pedestal Length - X (Pl) : 0.500 m
Pedestal Width - Z (Pw) : 0.500 m
Unit Weight of Concrete : 156.070 lb/ft3
Strength of Concrete : 2.987 ksi
Yield Strength of Steel : 59.738 ksi
Minimum Bar Size : #3
Maximum Bar Size : #5
Minimum Bar Spacing : 2.000 in
Maximum Bar Spacing : 18.000 in
Pedestal Clear Cover (P, CL) : 3.000 in
Footing Clear Cover (F, CL) : 3.000 in
Soil Type : UnDrained
Unit Weight : 112.370 lb/ft3
Soil Bearing Capacity : 15.156 kip/ft2
Soil Surcharge : 0.000 kip/in2
Depth of Soil above Footing : 65.000 in
Undrained Shear Strength : 0.030 kip/in2
Coefficient of Friction : 0.500
Factor of Safety Against Sliding : 1.500
Factor of Safety Against Overturning : 1.500
------------------------------------------------------
Initial Length (Lo) = 1.016 m
Initial Width (Wo) = 1.016 m
Load Combination/s- Service Stress Level
Load CombinationNumber
Load Combination Title
101 1.000 x DL
102 1.000 x DL+1.000 x LL
103 1.000 x DL+0.750 x LL
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104 1.000 x DL+1.000 x WLX
105 1.000 x DL+1.000 x WLZ
106 1.000 x DL+0.700 x ELX
107 1.000 x DL+0.700 x ELZ
108 1.000 x DL+0.750 x LL+0.750 x WLX
109 1.000 x DL+0.750 x LL+0.750 x WLZ
110 1.000 x DL+0.750 x LL+0.525 x ELX
111 1.000 x DL+0.750 x LL+0.525 x ELZ
Load Combination/s- Strength Level
Load CombinationNumber
Load Combination Title
201 1.400 x DL
202 1.200 x DL+1.600 x LL
203 1.200 x DL+1.000 x LL
204 1.200 x DL+0.800 x WLX
205 1.200 x DL+0.800 x WLZ
206 1.200 x DL+1.000 x LL+1.600 x WLX
207 1.200 x DL+1.000 x LL+1.600 x WLZ
208 1.200 x DL+1.000 x LL+1.000 x ELX
209 1.200 x DL+1.000 x LL+1.000 x ELZ
210 1.200 x DL+1.600 x WLX
211 1.200 x DL+1.600 x WLZ
212 1.200 x DL+1.000 x ELX
213 1.200 x DL+1.000 x ELZ
Applied Loads - Service Stress Level
LCAxial(kgf)
Shear X(kgf)
Shear Z(kgf)
Moment X(kNm)
Moment Z(kNm)
101 5909.234 126.717 24.610 0.274 -5.257
102 12330.327 557.035 15.037 0.124 -24.840
103 10725.053 449.455 17.430 0.162 -19.945
104 5326.912 1174.801 4.592 0.077 -47.359
105 6236.483 1201.035 47.373 0.974 -48.114
106 6223.717 -156.944 22.722 0.252 -18.861
107 6319.274 110.777 13.047 0.024 -5.882
108 10288.312 1235.518 2.417 0.014 -51.521
109 10970.490 1255.194 34.502 0.687 -52.087
110 10960.915 236.709 16.015 0.145 -30.148
111 11032.583 437.500 8.758 -0.026 -20.413
Applied Loads - Strength Level
LCAxial(kgf)
Shear X(kgf)
Shear Z(kgf)
Moment X(kNm)
Moment Z(kNm)
201 8272.927 177.404 34.454 0.384 -7.360
202 17364.829 840.569 14.215 0.089 -37.642
203 13512.174 582.378 19.959 0.179 -25.892
204 6625.223 990.528 13.518 0.172 -39.990
205 7352.880 1011.515 47.742 0.889 -40.594
206 12580.459 2259.312 -12.069 -0.136 -93.255
207 14035.772 2301.287 56.379 1.299 -94.463
208 13961.435 177.147 17.262 0.148 -45.326
209 14097.945 559.606 3.441 -0.178 -26.784
210 6159.365 1828.995 -2.496 0.014 -73.672
211 7614.679 1870.969 65.952 1.449 -74.880
212 7540.342 -253.170 26.835 0.298 -25.743
213 7676.852 129.288 13.014 -0.028 -7.201
Reduction of force due to buoyancy = 0.000 kgf
Effect due to adhesion = 0.000 kgf
Area from initial length and width, Ao =Lo X Wo = 1.032 m2
Min. area required from bearing pressure, Amin
=P / qmax = 0.233 m2
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Final Footing Size
Pressures at Four Corners
If Au
is zero, there is no uplift and no pressure adjustment is necessary. Otherwise, to account for uplift, areas of negative pressure will be set to zero
and the pressure will be redistributed to remaining corners.
Summary of Adjusted Pressures at 4 corners Four Corners
Check for stability against overturning and sliding
Note: Amin is an initial estimation.
P = Critical Factored Axial Load(without self weight/buoyancy/soil).qmax = Respective Factored Bearing Capacity.
Length (L2) = 1.626 m Governing Load Case : # 104
Width (W2) = 2.032 m Governing Load Case : # 104
Depth (D2) = 0.600 m Governing Load Case : # 104
Area (A2) = 3.303 m2
Load Case
Pressure atcorner 1
(q1)
(kgf/m2)
Pressure atcorner 2
(q2)
(kgf/m2)
Pressure atcorner 3
(q3)
(kgf/m2)
Pressure atcorner 4
(q4)
(kgf/m2)
Area of footingin uplift (A
u)
(m2)
107 5494.8206 7392.2378 7449.1811 5551.7639 0.000
109 -1342.4664 16838.6901 17102.6402 -1078.5163 0.259
109 -1342.4664 16838.6901 17102.6402 -1078.5163 0.259
107 5494.8206 7392.2378 7449.1811 5551.7639 0.000
Load Case
Pressure atcorner 1 (q
1)
(kgf/m2)
Pressure atcorner 2 (q
2)
(kgf/m2)
Pressure atcorner 3 (q
3)
(kgf/m2)
Pressure atcorner 4 (q
4)
(kgf/m2)
107 5494.8206 7392.2378 7449.1811 5551.7639
109 0.0000 16874.4277 17164.5923 0.0000
109 0.0000 16874.4277 17164.5923 0.0000
107 5494.8206 7392.2378 7449.1811 5551.7639
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Critical Load Case And The Governing Factor Of Safety For Overturning And Sliding - X Direction
Critical Load Case And The Governing Factor Of Safety For Overturning And Sliding - Z Direction
Shear Calculation
-Factor of safety against
slidingFactor of safety against
overturning
Load CaseNo.
Along X-Direction
Along Z-Direction
About X-Direction
About Z-Direction
101 82.737 426.010 255.627 20.754
102 24.585 910.722 598.522 5.880
103 28.684 739.630 470.273 6.882
104 8.676 2219.549 1136.120 2.217
105 8.866 224.766 105.073 2.275
106 67.804 468.323 281.368 11.017
107 96.494 819.263 681.942 20.466
108 10.258 5243.348 3746.663 2.565
109 10.369 377.215 179.127 2.601
110 54.962 812.384 520.003 5.864
111 29.819 1489.538 1551.264 6.917
Critical Load Case for Sliding along X-Direction : 104
Governing Disturbing Force : 1174.801 kgf
Governing Restoring Force : 10193.037 kgf
Minimum Sliding Ratio for the Critical Load Case : 8.676
Critical Load Case for Overturning about X-Direction : 105
Governing Overturning Moment : 2.019 kNm
Governing Resisting Moment : 212.177 kNm
Minimum Overturning Ratio for the Critical Load Case : 105.073
Critical Load Case for Sliding along Z-Direction : 105
Governing Disturbing Force : 47.373 kgf
Governing Restoring Force : 10647.823 kgf
Minimum Sliding Ratio for the Critical Load Case : 224.766
Critical Load Case for Overturning about Z-Direction : 104
Governing Overturning Moment : -73.280 kNm
Governing Resisting Moment : 162.491 kNm
Minimum Overturning Ratio for the Critical Load Case : 2.217
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Punching Shear Check
Effective depth, deff
, increased until 0.75XVc
Punching Shear Force
Punching Shear Force, Vu = 22640.520 kgf, Load Case # 202
Along X Direction
(Shear Plane Parallel to Global X Axis)
Check that 0.75 X Vc > Vux where Vux is the shear force for the critical load cases at a distance deff from the face of the column caused by bending
about the X axis.
Total Footing Depth, D = 0.600m
Calculated Effective Depth, deff
= D - Ccover
- 1.0 = 0.498 m 1 inch is deducted from total depth to cater bar dia(US Convention).
For rectangular column, = Bcol / Dcol = 1.000
From ACI Cl.11.12.2.1, bo
for column= 3.993 m
Equation 11-33, Vc1
= 458811.711kgf
Equation 11-34, Vc2
= 534647.648kgf
Equation 11-35, Vc3 = 305874.474kgf
Punching shear strength, Vc = 0.75 X minimum of (Vc1, Vc2, Vc3) = 229405.856kgf
0.75 X Vc > Vu hence, OK
From ACI Cl.11.3.1.1, Vc
= 62256.540kgf
Distance along X to design for shear,D
x= 1.764 m
From above calculations, 0.75 X Vc
= 46692.405 kgf
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One-Way Shear Check
Along Z Direction
(Shear Plane Parallel to Global Z Axis)
Check that 0.75 X Vc
> Vuz
where Vuz
is the shear force for the critical load cases at a distance deff
from the face of the column caused by bending
about the Z axis.
Design for Flexure about Z Axis
(For Reinforcement Parallel to X Axis)
Calculate the flexural reinforcement along the X direction of the footing. Find the area of steel required, A, as per Section 3.8 of Reinforced ConcreteDesign (5th ed.) by Salmon and Wang (Ref. 1)
Critical Load Case # 207
The strength values of steel and concrete used in the formulae are in ksi
Critical load case for Vux is # 202 4284.704 kgf
0.75 X Vc > Vux hence, OK
From ACI Cl.11.3.1.1, Vc
= 77820.675 kgf
Distance along X to design for shear, Dz = 1.561 m
From above calculations, 0.75 X Vc
= 58365.506 kgf
Critical load case for Vuz
is # 202 2090.783 kgf
0.75 X Vc > Vuz hence, OK
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
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Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Because the number of bars is rounded up, make sure new reinforcement ratio < ρmax
Check to see if width is sufficient to accomodate bars
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl. 7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
Design for flexure about Z axis isperformed at the face of the column at
a distance, Dx
=0.563 m
Ultimate moment, 95.457 kNm
Nominal moment capacity, Mn
= 106.063 kNm
Required = 0.00051
Since OK
Area of Steel Required, As = 3.139 in2
Selected bar Size = #3
Minimum spacing allowed (Smin
) = = 2.000 in
Selected spacing (S) = 2.629 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
#3 @ 2.000 in o.c.
Required development length for bars = =0.305 m
Available development length for bars, DL
=0.487 m
Try bar size # 3 Area of one bar = 0.110 in2
Number of bars required, Nbar
= 29
Total reinforcement area, As_total
= Nbar
X (Area of one bar) = 3.190 in2
deff
= D - Ccover
- 0.5 X (dia. of one bar)
=
0.519 m
Reinforcement ratio, = 0.00195
From ACI Cl.7.6.1, minimum req'd cleardistance between bars, C
d=
max (Diameter of one bar, 1.0,Min. User Spacing) =
2.629 in
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Design for Flexure about X axis
(For Reinforcement Parallel to Z Axis)
Calculate the flexural reinforcement along the Z direction of the footing. Find the area of steel required, A, as per Section 3.8 of Reinforced ConcreteDesign (5th ed.) by Salmon and Wang (Ref. 1)
Critical Load Case # 207
The strength values of steel and concrete used in the formulae are in ksi
Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl.7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
Design for flexure about X axis isperformed at the face of the column at
a distance, Dz
=0.766 m
Ultimate moment, 53.978 kNm
Nominal moment capacity, Mn
= 59.975 kNm
Required = 0.00038
Since OK
Area of Steel Required, As
= 3.079 in2
Selected Bar Size = #3
Minimum spacing allowed (Smin) = 2.000 in
Selected spacing (S) = 2.401 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
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Because the number of bars is rounded up, make sure new reinforcement ratio < ρmax
Check to see if width is sufficient to accomodate bars
Bending moment for uplift cases will be calculated based solely on selfweight, soil depth and surcharge loading.
As the footing size has already been determined based on all servicebility load cases, and design moment calculation is based on selfweight, soil depthand surcharge only, top reinforcement value for all pure uplift load cases will be the same.
Design For Top Reinforcement Parallel to Z Axis
Calculate the flexural reinforcement for Mx. Find the area of steel required
The strength values of steel and concrete used in the formulae are in ksi
#3 @ 2.000 in o.c.
Required development length for bars = =0.305 m
Available development length for bars, DL
=0.487 m
Try bar size # 3 Area of one bar = 0.110 in2
Number of bars required, Nbar
= 25
Total reinforcement area, As_total
= Nbar
X (Area of one bar) = 2.750 in2
deff
= D - Ccover
- 0.5 X (dia. of one bar)
=
0.506 m
Reinforcement ratio, = 0.00216
From ACI Cl.7.6.1, minimum req'd cleardistance between bars, C
d=
max (Diameter of one bar, 1.0, Min.User Spacing) =
2.401 in
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl. 7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
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Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Design For Top Reinforcement Parallel to X Axis
First load case to be in pure uplift #
Calculate the flexural reinforcement for Mz. Find the area of steel required
The strength values of steel and concrete used in the formulae are in ksi
Design for flexure about A axis isperformed at the face of the column
at a distance, Dx =0.766 m
Ultimate moment, 20.913 kNm
Nominal moment capacity, Mn = 23.237 kNm
Required = 0.00015
Since OK
Area of Steel Required, As
= 2.463 in2
Selected bar Size = #5
Minimum spacing allowed (Smin
) = 2.000 in
Selected spacing (S) = 8.196 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
Safe for Cracking Aspect.
#5 @ 8 in o.c.
Factor from ACI Cl.10.2.7.3 = 0.850
From ACI Cl. 10.3.2, = 0.02142
From ACI Cl. 10.3.3, = 0.01606
From ACI Cl.7.12.2, = 0.00200
From Ref. 1, Eq. 3.8.4a, constant m = 23.529
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Calculate reinforcement ratio for critical load case
Based on spacing reinforcement increment; provided reinforcement is
Pedestal Design Calculations
Strength and Moment Along Reinforcement in X direction
Strength and Moment from Concrete
Calculate strength and moment from one bar.
Design for flexure about A axis isperformed at the face of the column
at a distance, Dx =0.563 m
Ultimate moment, 14.112 kNm
Nominal moment capacity, Mn = 15.680 kNm
Required = 0.00008
Since OK
Area of Steel Required, As = 3.139 in2
Selected bar Size = #3
Minimum spacing allowed (Smin
) = 2.000 in
Selected spacing (S) = 2.629 in
Smin
<= S <= Smax
and selected bar size < selected maximum bar size...
The reinforcement is accepted.
According to ACI 318-05 Clause No- 10.6.4
Max spacing for Cracking Consideration = 7.566 in
UnSafe for Cracking Aspect.
#3 @ 2 in o.c.
Critical Load Case: 202
Bar size : # 5
Number of Bars : 12
Steel Area : 3.3906 sq.in
Neutral Axis Depth (Xb): 0.0617 m
Cc = 46811.497 kgf
Mc
= 102.725 kNm
Distance between extreme fiber andbar,
db 0.084 m
Strain in bar, = -0.0011
Maximum Strain, = 0.0021
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Strength and Moment Along Reinforcement in Z direction
Strength and Moment from Concrete
Calculate strength and moment from one bar.
as
-22235895.929
kgf/m2
0.0016
as
0.000 kgf/m2
-4447.170
kgf
-7.233 kNm
Total Bar Capacity, Cs = -55341.389
kgf
Capacity of Column = Cc
+ Cs
=-
8529.892kgf
Total Bar Moment, Ms
= 19.288 kNm
Total Moment = Mc
+ Ms
= 122.013 kNm
Bar size : # 5
Number of Bars : 12
Steel Area : 3.3906 sq.in
Neutral Axis Depth (Xb): 0.0617 m
Cc
= 46811.497 kgf
Mc = 102.725 kNm
Distance between extreme fiberand bar,
db 0.084 m
Strain in bar, = -0.0011
Maximum Strain, = 0.0021
as
-22235895.929
kgf/m2
0.0016
as
kgf/m2
-4447.170
kgf
-7.233 kNm
Total Bar Capacity, Cs = -55341.389
kgf
Capacity of Column = Cc + Cs =-
8529.892kgf
Total Bar Moment, Ms = 19.288 kNm
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Total Moment = Mc + Ms = 122.013 kNm
Check for bi-axial bending, 0.047
Design Moment Mnx
= 10.326 kNm
Design Moment Mnz
= 27.001 kNm
Total Moment Mox
= 12442.112 kNm
Total Moment Moz
= 12442.112 kNm
if Mnx or Mnz = 0, then = 1.0
otherwise, = 1.24
Print Calculation Sheet
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