Appendix A

Hamilton West-Harbour Re-development

Designer:_D.Huang, M.Morco_Date:___Jan16th ,2012

Verify:____J. Xing, Y,Liu__ Date:__Feb 23rd,2012

A-1

Appendix A: Sample Calculations Reference

Design of Slab:

Floor finish 0.08 kPa

Ceiling 0.20 kPa

Ducts/pipes/wiring 0.25 kPa

Non-structural partition 1.10 kPa

----------------------------------------------------------

Dead Load 1.63 kPa

Live Load (Office) 2.40 kPa

Specified load = LL +

DL = 2.40 +

* 1.63 = 3.8 kPa

The selected composite deck is CD75 - 200 (Galvanneal)

Base Steel = 0.914 mm

Depth = 130 mm,

Maximum specified load = 7.2 kPa> 3.8 kPa Okay

Check deflection:

DP = 155 kN*m

DC = 360

L (beam spacing) = 3,200 mm

Wd =

= 13.14 kPa > 3.8 kPa Okay

Self weight of composite slab = 1.98 kPa

Agwaymetal

INC.

Technical

data sheet of

composite

decks

www.agway

metals.com




A-2

Snow load

Assume flat roof without penthouse

S = Is * [Ss(CbCwCsCa) + Sr]

Is = 1.0 (Importance factor for snow load)

Ss = 0.9 (Ground Snow Load)

Sr = 0.4 (1-in-50 year associated rain load)

Cb = 0.8 (basic roof snow load factor)

Cw = 1.0 (wind exposure factor)

Cs = 1.0 (slope factor)

Ca = 1.0 (shape factor)

S = 1 * (0.9*0.8*1*1*1+0.4)

= 1.12kPa

NBCC 2010

– Clause

4.1.6.2

NBCC 2010

Appendix C

Wind uplift

Use dead load factor = 0.85 for wind uplift

Cpi = 0 (For roof)

Design roof height = 12.0 m

Dead load on the roof,

Roof finish 0.15 kPa

Steel concrete composite slab 1.98 kPa

Insulation 0.10 kPa

Steel deck 0.10 kPa

Ceiling 0.20 kPa

Mechnical system 0.20 kPa

Green roof 0.68 kPa

----------------------------------------------------------

Dead Load 3.41 kPa

Iw = 1.0 (important factor for wind load)

q1/50 = 0.46 KPa

Ce = 0.9 * (h/10)0.2

= 0.93 > 0.9 open terrain ( exposure factor)

Cg = 2.0 Building as a whole ( gust effect factor)

H/D = 12/19.8 = 0.606 < 1 Interior Zone : Cp = -0.5 ; End Zone: Cp = -1.0

Wind Pressure

P = Iw (q Ce CgCp)

- 0.46 [ 1.04 * 2 * (-1) ] = -0.9542 kPa Exterior

- 0.46 [ 1.04 * 2 * (-0.5) ] = -0.4771 kPa Interior

NBCC 2010

– Clause

4.1.7.1




A-3

Beam 1 detail design

Location: on Axis A from Axis 3 to Axis 6 on the roof

Beam spacing = 3.2 m

Beam Length = 6.6 m

Elasticity of steel = 2.00 E +08 kPa

Tributary area = 3.2/2*6.6 = 10.56 m2

Live Load = 1*3.2/2 = 1.6 kN/m

Assumed self weight of beam = 0.207 kN/m ( trails done by Excel )

Dead Load = 3.41 * 3.2 /2 +0.207 = 5.66 kN/m

Snow Load = 1.12 * 3.2 /2 = 1.79 kN/m

Load Combination = 1.25 D + 1.5 S + 0.5 L

= 1.25 * 5.66 + 1.5 * 1.79 + 0.5 * 1.6 = 10.56 kN/m

Max shear = 10.56 * 6.6 /2 = 34.9 kN

Max moment = 10.56 * 6.6 2/8 = 57.5 kN.m

Selection (moment governing ): W310 X 21

Mr = 89.1 kN.m

Vr = 303 kN

Self weight = 0.207 kN/m

Ix = 3.7E+07 mm4

Max deflection due to Live Load :

= 5wl4/(384 E Ix)

= 5 * 1.6 * 66004 / (384 * 2.00 E +08 * 3.7E+07) *1000

= 5.34 mm < 22 mm ok

CISC Steel

Handbook

Pg 5-98

CISC Steel

Handbook

Pg 5-146




A-4


Location: on Axis D from Axis 6 to Axis 8 on the roof


Beam Length = 6.6 m


Tributary area = 3.2*6.6 = 21.12 m2

LLRF = 0.3 + √ = 0.981

Live Load = 1*3.2 *0.981= 3.14 kN/m


Dead Load = 3.41 * 3.2 +0.278 = 11.19 kN/m

Snow Load = 1.12 * 3.2 = 3.58 kN/m

Load Combination = 1.25 D + 1.5 S + 0.5 L

= 1.25 * 11.19 + 1.5 * 3.58 + 0.5 * 3.14 = 20.93 kN/m

Max shear = 20.93* 6.6 /2 = 69.1 kN

Max moment = 20.93 * 6.6 2/8 = 114 kN.m


Mr = 126 kN.m

Vr = 380 kN


Ix = 5.43E+07 mm4


= 5wl4/(384 E Ix)

= 5 * 3.14 * 66004 / (384 * 2.00 E +08 *5.43E+07) *1000

= 7.1 mm < 22 mm ok

CISC Steel

Handbook

Pg 5-98

CISC Steel

Handbook

Pg 5-146




A-5


Location: on Axis D from Axis 6 to Axis 8 on the 3rd

floor slab


Beam Length = 6.6 m



LLRF = 0.3 + √ = 0.981

Live Load = 1.9 *3.2 *0.981= 5.97 kN/m


Dead Load = 3.61 * 3.2 +0.321 = 11.87 kN/m

Load Combination = 1.25 D + 1.5 L

= 1.25 * 11.87 + 1.5 * 5.97 = 23.79 kN/m

Max shear = 23.79* 6.6 /2 = 78.5 kN

Max moment = 23.79 * 6.6 2/8 = 129.5 kN.m

Selection (moment governing ): W360 X33

Mr = 168 kN.m

Vr = 396 kN


Ix = 8.27E+07 mm4


= 5wl4/(384 E Ix)

= 5 * 5.97 * 66004 / (384 * 2.00 E +08 *8.27E+07) *1000

= 8.91 mm < 22 mm ok

CISC Steel

Handbook

Pg 5-98

CISC Steel

Handbook

Pg 5-146




A-6


Location: on Axis G from Axis 6 to Axis 8 on the 3rd

floor slab


Beam Length = 6.6 m



LLRF = 0.3 + √ = 0.981

Live Load = (1.9+4.8)/2 *3.2 *0.981= 10.52 kN/m


Dead Load = 3.61 * 3.2 +0.321 = 11.87 kN/m


= 1.25 * 11.87 + 1.5 * 10.52 = 30.62 kN/m

Max shear = 30.62* 6.6 /2 = 101 kN

Max moment = 30.62 * 6.6 2/8 = 167 kN.m


Mr = 168 kN.m

Vr = 396 kN


Ix = 8.27E+07 mm4


= 5wl4/(384 E Ix)

= 5 * 10.52 * 66004 / (384 * 2.00 E +08 *8.27E+07) *1000

= 15.71 mm < 22 mm ok

CISC Steel

Handbook

Pg 5-98

CISC Steel

Handbook

Pg 5-146




A-7


Location: on Axis I from Axis 6 to Axis 8 on the 3rd

floor slab


Beam Length = 6.6 m



LLRF = 0.3 + √ = 0.981

Live Load = 4.8 *3.2 *0.981= 15.07 kN/m


Dead Load = 3.61 * 3.2 +0.385 = 11.94 kN/m


= 1.25 * 11.94 + 1.5 * 15.07 = 37.53 kN/m

Max shear = 37.53* 6.6 /2 = 124 kN

Max moment = 37.53 * 6.6 2/8 = 204 kN.m


Mr = 227 kN.m

Vr = 480 kN


Ix = 1.27E+08 mm4


= 5wl4/(384 E Ix)

= 5 * 15.07 * 66004 / (384 * 2.00 E +08 *1.27E+08) *1000

= 14.66 mm < 22 mm ok

CISC Steel

Handbook

Pg 5-98

CISC Steel

Handbook

Pg 5-146




A-8


Location: on Axis A from Axis 3 to Axis 6 on the 3rd

floor slab


Beam Length = 6.6 m


Tributary area = 3.2*6.6/2 = 10.56 m2

Live Load = 4.8 *3.2 /2= 7.68 kN/m


Dead Load = 3.61 * 3.2/2 +0.278 = 6.05 kN/m


= 1.25 * 6.05 + 1.5 * 7.68 = 19.09 kN/m

Max shear = 19.09* 6.6 /2 = 63 kN

Max moment = 19.09 * 6.6 2/8 = 104 kN.m


Mr = 126 kN.m

Vr = 380 kN


Ix = 5.43E+07 mm4


= 5wl4/(384 E Ix)

= 5 * 7.68 * 66004 / (384 * 2.00 E +08 *5.43E+07) *1000

= 17.47 mm < 22 mm ok

CISC Steel

Handbook

Pg 5-98

CISC Steel

Handbook

Pg 5-146




A-9


Location: on Axis D from Axis 6 to Axis 8 on the 2nd

floor slab


Beam Length = 6.6 m



LLRF = 0.3 + √ = 0.981

Live Load = 2.4 *3.2 *0.981= 7.54 kN/m


Dead Load = 3.61 * 3.2 +0.321 = 11.87 kN/m


= 1.25 * 11.87 + 1.5 *7.54 = 26.14 kN/m

Max shear = 26.14* 6.6 /2 = 86.3 kN

Max moment = 26.14 * 6.6 2/8 = 142.4 kN.m


Mr = 168 kN.m

Vr = 396 kN


Ix = 8.27E+07 mm4


= 5wl4/(384 E Ix)

= 5 * 7.54 * 66004 / (384 * 2.00 E +08 *8.27E+07) *1000

= 11.26 mm < 22 mm ok

CISC Steel

Handbook

Pg 5-98

CISC Steel

Handbook

Pg 5-146




A-10


Location: on Axis G from Axis 6 to Axis 8 on the 2nd

floor slab


Beam Length = 6.6 m



LLRF = 0.3 + √ = 0.981

Live Load = (2.4+4.8)/2 *3.2 *0.981= 11.3 kN/m


Dead Load = 3.61 * 3.2 +0.385 = 11.94 kN/m


= 1.25 * 11.94 + 1.5 * 11.3 =31.88 kN/m

Max shear = 31.88* 6.6 /2 = 105 kN

Max moment = 31.88 * 6.6 2/8 = 174 kN.m


Mr = 227 kN.m

Vr = 480 kN


Ix = 1.27E+08 mm4


= 5wl4/(384 E Ix)

= 5 * 11.3 * 66004 / (384 * 2.00 E +08 *1.27E+08) *1000

= 10.99 mm < 22 mm ok

CISC Steel

Handbook

Pg 5-98

CISC Steel

Handbook

Pg 5-146




A-11

Beam9 detail design

Location: on Axis I from Axis 6 to Axis 8 on the 2nd

floor slab

Tributary area = 6.6 * 1.6 + 6.6 * 1.6 = 21.12 m2

LLRF = 0.3 + √ = 0.981

Live Load = 0.981 * 4.8 * 3.2 = 15.07 kN/m

Assume self-weight = 0.385 kN/m ( trails done by excel)

Dead Load = 3.2 * 3.5 + 0.385 = 11.94 kN/m

Load combination = 1.25 * 11.94 + 1.5 * 15.07 = 37.53 kN/m

Vmax = (wL)/2 = 6.6 * 37.53 / 2 = 124 kN

Mmax = (wL2)/2 = 37.53 * 6.6

2 / 8 = 204 kN*m


Mr = 227 kN.m

Vr = 480 kN


Ix = 1.27E+08 mm4

Deflection:

Acceptable deflection = Span / 300

= 6,600 / 300 = 22 mm

Maximum deflection =

= 5 * 15.07 * 6,6004 / (384 * 200,000 * 1.27 * 10

8)

= 14.66 mm ok

CISC Steel

Handbook

Pg 5-98

CISC Steel

Handbook

Pg 5-146




A-12


Location: on Axis A from Axis 3 to Axis 6 on the 2nd

floor slab


Beam Length = 6.6 m


Tributary area = 3.2*6.6/2 = 10.56 m2

Live Load = 4.8 *3.2 /2= 7.68 kN/m


Dead Load = 3.61 * 3.2/2 +0.278 = 6.05 kN/m


= 1.25 * 6.05 + 1.5 * 7.68 = 19.09 kN/m

Max shear = 19.09* 6.6 /2 = 63 kN

Max moment = 19.09 * 6.6 2/8 = 104 kN.m


Mr = 126 kN.m

Vr = 380 kN


Ix = 5.43E+07 mm4


= 5wl4/(384 E Ix)

= 5 * 7.68 * 66004 / (384 * 2.00 E +08 *5.43E+07) *1000

= 17.47 mm < 22 mm ok

CISC Steel

Handbook

Pg 5-98

CISC Steel

Handbook

Pg 5-146




A-13

Girder 1 detail design

Location: on Axis 8 from Axis G to Axis J on the roof

Girder Length = 3.2 *3 =9.6 m


Tributary area = 9.6 *6.6 /2 = 31.68 m2

Transferred Load from Beam to Girder:

Concentrated Live Load = 1* 3.2 /2 * 6.6 =10.6 kN

Concentrated Dead Load= 3.41 * 3.2 /2 *6.6= 36 kN

Concentrated Snow Load = 1.12 * 3.2 /2 *6.6 = 11.83 kN

Due to Live Load:

Max shear = 10.6 kN

Max moment = 10.6 * 3.2 = 33.8 kN.m

Due to Dead Load:

Assume self weight of girder = 0.454 kN/m ( trails done by excel )

Max shear = ( 0.454 *9.6 + 36 +36)/2 = 38.2 kN

Max moment = 0.454*9.62/8+36 * 3.2 = 120.5 kN.m

Due to Snow Load:

Max shear = 11.83 kN

Max moment = 11.83 * 3.2 = 37.8 kN.m

Load combination :

Max shear = 1.25 D + 1.5 S + 0.5 L

= 1.25 * 38.2 + 1.5 * 11.83 + 0.5 * 10.6

= 70.76 kN

Max moment = 1.25 D + 1.5 S + 0.5 L

= 1.25 * 120.5 + 1.5 * 37.8 + 0.5 * 33.8

= 224.2 kN.m


Mr= 275 kN.m,

Vr= 578 kN,


Ix = 1.56E+08 mm4


+

a= 3.2 m , L= 9.6m

=12.24 mm < 32 mm ok




A-14


Location: on Axis 6 from Axis G to Axis J on the roof



Tributary area = 9.6 *6.6 = 63.36 m2


Concentrated Live Load = 1* 3.2 * 6.6 =21.1 kN

Concentrated Dead Load= 3.41 * 3.2 *6.6= 72 kN

Concentrated Snow Load = 1.12 * 3.2 *6.6 = 23.7 kN

Due to Live Load:

Max shear = 21.1 kN

Max moment = 21.1* 3.2 = 67.6 kN.m

Due to Dead Load:


Max shear = ( 0.645 *9.6 + 72 +72)/2 = 75.1 kN

Max moment = 0.645*9.62/8+72* 3.2 = 237.9 kN.m

Due to Snow Load:

Max shear = 23.7 kN

Max moment = 23.7 * 3.2 =75.7 kN.m

Load combination :

Max shear = 1.25 D + 1.5 S + 0.5 L

= 1.25 * 75.1 + 1.5 * 23.7 + 0.5 * 21.1

= 140 kN

Max moment = 1.25 D + 1.5 S + 0.5 L

= 1.25 * 237.9 + 1.5 * 75.7 + 0.5 * 67.6

= 445 kN.m


Mr= 484 kN.m,

Vr= 928 kN,


Ix = 3.51E+08 mm4


+

a= 3.2 m , L= 9.6m

=10.46 mm < 32 mm ok




A-15


Location: on Axis 6 from Axis G to Axis J on the 3rd

floor slab



Tributary area = 9.6 *6.6 = 63.36 m2


Concentrated Live Load = 4.8* 3.2 * 6.6 =101.4 kN

Concentrated Dead Load= (3.61*3.2+0.385)*6.6= 78.8 kN

Due to Live Load:


Max moment = 101.4* 3.2 = 324.4 kN.m

Due to Dead Load:


Max shear = ( 0.997 *9.6 + 78.8 +78.8)/2 = 83.6 kN

Max moment = 0.997*9.62/8+78.8* 3.2 = 263.6 kN.m

Load combination :

Max shear = 1.25 D + 1.5 L

= 1.25 * 83.6 + 1.5 * 101.4

= 257 kN

Max moment = 1.25 D + 1.5 L

= 1.25 * 263.6 + 1.5 *324.4

= 816 kN.m


Mr= 900 kN.m,

Vr= 1300 kN,


Ix = 7.64E+08 mm4


+

a= 3.2 m , L= 9.6m

=21.56 mm < 32 mm ok




A-16


Location: on Axis 8 from Axis G to Axis J on the 3rd

floor slab



Tributary area = 9.6 *6.6 /2 = 31.68 m2


Concentrated Live Load = (4.8* 3.2 /2 * 6.6/2)+ (1.9* 3.2 /2 * 6.6/2)=35.4 kN

Concentrated Dead Load= (3.61*3.2+0.321) *6.6/2 = 39.2 kN

Due to Live Load:

Max shear = 35.4 kN

Max moment = 35.4 * 3.2 = 113.2 kN.m

Due to Dead Load:


Max shear = ( 0.510 *9.6 + 39.2 +39.2)/2 = 41.6 kN

Max moment = 0.510*9.62/8+39.2 * 3.2 = 131.3kN.m

Load combination :

Max shear = 1.25 D + 1.5 L

= 1.25 * 41.6 + 1.5 * 35.4

= 105 kN


= 1.25 * 131.3 + 1.5 * 113.2

= 334 kN.m


Mr= 338 kN.m,

Vr= 680 kN,


Ix = 2.12E+08 mm4


+

a= 3.2 m , L= 9.6m

=27.53 mm < 32 mm ok




A-17


Location: on Axis 6 from Axis G to Axis J on the 2nd

floor slab



Tributary area = 9.6 *6.6 = 63.36 m2


Concentrated Live Load = 4.8* 3.2 * 6.6 =101.4 kN

Concentrated Dead Load= (3.61*3.2+0.385)*6.6= 78.8 kN

Due to Live Load:


Max moment = 101.4* 3.2 = 324.4 kN.m

Due to Dead Load:


Max shear = ( 0.997 *9.6 + 78.8 +78.8)/2 = 83.6 kN

Max moment = 0.997*9.62/8+78.8* 3.2 = 263.6 kN.m

Load combination :

Max shear = 1.25 D + 1.5 L

= 1.25 * 83.6 + 1.5 * 101.4

= 257 kN


= 1.25 * 263.6 + 1.5 *324.4

= 816 kN.m


Mr= 900 kN.m,

Vr= 1300 kN,


Ix = 7.64E+08 mm4


+

a= 3.2 m , L= 9.6m

=21.56 mm < 32 mm ok

Steel

Handbook

CISC

section 6

Pg6-48

Steel

Handbook

CISC Table

D1

Pg 1-134




A-18


Location: on Axis 8 from Axis G to Axis J on the 2nd

floor slab



Tributary area = 9.6 *6.6 /2 = 31.68 m2


Concentrated Live Load = 2.4* 3.2 * 6.6/2=25.3 kN

Concentrated Dead Load= (3.61*3.2+0.321) *6.6/2 = 39.2 kN

Due to Live Load:

Max shear = 25.3 kN

Max moment = 25.3 * 3.2 = 81.1 kN.m

Due to Dead Load:


Max shear = ( 0.510 *9.6 + 39.2 +39.2)/2 = 41.6 kN

Max moment = 0.510*9.62/8+39.2 * 3.2 = 131.3kN.m

Load combination :

Max shear = 1.25 D + 1.5 L

= 1.25 * 41.6 + 1.5 * 25.3

= 90.1 kN


= 1.25 * 131.3 + 1.5 * 81.1

= 286 kN.m


Mr= 338 kN.m,

Vr= 680 kN,


Ix = 2.12E+08 mm4


+

a= 3.2 m , L= 9.6m

=20.1 mm < 32 mm ok




A-19

Column 1 detail design

Location: at intersection of Axis G and Axis 6

On each floor, two girders and two beams are sitting on , but only half of each member

bearing load are transferred to the column

Tributary Area = 63.36 m2

LLRF = 0.693

Roof:

LL= 43.9 kN

DL = 224.1 kN

SL= 71 kN

Load combination =1.25 D+1.5 S + 0.5 L= 408.5 kN

3rd

Floor:

LL= 189.6 kN

DL= 240.6 kN

Load combination =1.25 D+1.5 L=585.2 kN

2nd

Floor

LL= 184.5 kN

DL= 240.8 kN

Load combination =1.25 D+1.5 L= 577.79 kN

Total Cf = 408.5+ 585.2 + 577.79= 1570 kN

Selection: W250X73

KL = 5000 mm

Cr = 1680 kN > Cf ok

Steel

Handbook

CISC pg 4-

26




A-20

Column 2 detail design

Location: at intersection of Axis G and Axis 8

On each floor, two girders and one beam are sitting on , but only half of each member

bearing load are transferred to the column

Tributary Area = 31.68 m2

LLRF = 0.856

Roof:

LL= 27.1 kN

DL = 113.3 kN

SL= 35.5 kN

Load combination =1.25 D+1.5 S +0.5 L= 208.42 kN

3rd

Floor:

LL= 104 kN

DL= 121 kN

Load combination =1.25 D+1.5 L=307.2 kN

2nd

Floor

LL= 97.6 kN

DL= 121.2 kN

Load combination =1.25 D+1.5 L= 297.94 kN

Total Cf = 208.42+ 307.2 +297.94 = 813 kN

Selection: W200X52

KL = 5000 mm

Cr = 928 kN > Cf ok

Steel

Handbook

CISC pg 4-

26




A-21

Wind Load

Height = 12 m, Width = 48 m

p = Iw×q×Ce×Cg×Cp

where Iw= 1, q1/50= 0.46 kPa, q1/10 = 0.36, Ce = (H/10)0.2

= 1.037

Cpi = -1 (End Zone) or -0.5 (Interior Zone)

CgCp = 1.15 – (-1) = 2.15 (End Zone)

CgCp = 0.75 – (-0.5) = 1.25 (Interior Zone)

Wind Pressure

(1) Ultimate Limit State

End Zone: p = 1×0.46 ×1.037×2.15= 1.026 kPa

Interior Zone: p = 1×0.46×1.037×1.25 = 0.596 kPa

(2) Serviceability Limit State

End Zone: p = 1×0.36 ×1.037×2.15 = 0.803 kPa

Interior Zone: p = 1×0.36 ×1.037×1.25 = 0.467 kPa

Wind Surface

y = End Zone Width = 0.2×48 = 9.6 m

Eccentricity = 48/2 – 9.6/2 = 19.2 m

Factored Wind Load – End Zone

Height of 2nd

floor, 3rd

floor, and roof

2nd

Floor: (5 + 3.5)/2 = 4.25 m

3rd

Floor: (3.5 + 3.5)/2 =3.5 m

Roof: 3.5 / 2 = 1.75 m

Distribution of Lateral Forces

(αw = 1.4, load factor for wind load)

αw×P1/50 = 1.5 × 1.026 = 1.436 kPa (End Zone)

αw×P1/50 = 1.5 × 0.596 = 0.835 kPa (Interior Zone)

Sample Calculation for Table 12 wind load summary:

At Roof level,

At = Total Wind Surface Area = 48 × 1.75 = 84 (m2)

Ae = End Zone Surface Area = 9.6 × 1.75 = 16.8 (m2)

0.895 * At = Force in the entire area = 0.835 × 84 = 70.1 (m2)

(1.539 – 0.895) * Ae = Additional Force in End Zone = 0.601 × 16.8 =10.1 (m2)

Factored Wind Load = 70.1 + 10.1 = 80.2 (kN)

Factored Torsion = 10.1 × 19.2 = 193.9 (kN)

NBCC 2010

Cl4.1.7.1

Table 4.1.7.1

Appendix C

NBCC

Commentary

I Figure I-7

NBCC

Commentary

I

Figure I-7




A-22

Sample Calculation for Table 13 and Table 14:

d = distance between bracing frame to center of rigidity

F = Floor force in x or y-direction

e = Eccentricity

T = Torsion

K = relative frame stiffness

m = number of frames parallel to F

n = number of frames perpendicular to F

i) Bracing frame 3, along grid G3-4, 3rd

floor, Wind in South-North Direction

The bracing frame is taking both lateral force and torsional force since it is in the same

direction as wind load in this case.

P = 160.5×1/(1+1) + 387.8×0.095/4.8 = 87.92 kN

ii) Bracing frame 3, along G3-4, 3rd

floor, Wind in West-East Direction

The bracing frame is taking torsional force since it is perpendicular to the direction of

wind load in this case.

P = 387.8× 0.095/4.8 = 7.69 kN




A-23

Mass of each floor

Ground level(retails)

Interior wall(100 mm hollow block, 5m high)

Total horizontal wall length = 70.2 m

Total vertical wall length = 97 m

Exterior wall(100 mm hollow block, 5m high)

Total horizontal wall length = 47.7 m

Total vertical wall length = 24 m

*Note that half of weights from walls in ground level are counted as mass of 2nd

level.

2nd

level(Office)

Slab dead load kPa Weight(KN)

Floor Finish 0.08 76

concrete cover slab and

steel deck 1.98 1881

drop ceiling 0.2 190

ducts/pipes/wiring 0.25 237.5

Total 2.51 2384.5

Beams:

Grade Quantity(#) Self Weight(kN/m) Length(m) Weight(kN)

W310X28 6 0.278 6.6 11.01

W360X33 24 0.321 6.6 50.84

W410X39 18 0.385 6.6 45.738

Total 48 108

Girders:


W460X52 9 0.510 9.6 44.064

W610X101 11 0.997 9.6 105.28

Total 20 149

Columns:


W250X73 24 0.715 3.5 72.93

Total 24 72.93




A-24

Interior walls:

100mm Hollow block Length(m) Thickness(m) Height(m) weight(

kN)

Horizontal wall 126.6 0.1 3.5 436.75

Vertical wall 173.4 0.1 3.5 600.54

Total 300 1037

Wall density = 1.1 Kpa ( CISC steel handbook pg 7-41)

(126.6m*3.5m*1.1kPa)/2+(70.2m*5m*1.1kPa)/2=436.75 kN

Exterior walls:

100mm Hollow block Length(m) Thickness(m) Height(m) Weight(kN)

Horizontal wall 54 0.1 3.5 235.13

Vertical wall 24.6 0.1 3.5 113.36

Total 78.6 348

Doors:

Quantity(#) Length(m) Height(m) Weight(kN)

Hinged Door 33 0.9 2 17.82

Sliding Door 6 1 2 3.6

Total 39 21.42

Assumed door density =0.3 kPa

Windows:

Quantity(#) Length(m) Height(m) Weight(KN)

North-south side 28 1.5 2 29.4

East-west side 10 1.5 2 10.5

Total 38 39.9

Total Dead load =2384.5 + 108 +149 + 72.93 + 1037 +348 + 21.42+39.9 =4161 kN

Live Load:

kPa Tributary Area（

m2)

kN

2.4 570.24 1368.57

4.8 380.16 1824.76

Total 950 3193

Total Live Load= 2.4*570.24+4.8*380.16= 3193 kN

Seismic weight(D+25%S) = 4161 +0 = 4161kN

Total(1.25D+1.5L+0.5S) = 1.25 *4161+1.5*3191+0= 9992 kN




A-25

Earthquake Load and Effects

Equivalent Static Force Procedure

IEFaS(0.2) < 0.35

IE = 1.0 (Importance Category - Normal)

Fa = 1.0 Sa(0.2) = 0.32

Ta = 0.025hn = 0.30s

Sa(0.2) = 0.32, Sa(0.5) = 0.17

By linear interpolation Sa(0.30) = 0.27

V = S(Ta)MvIEW/(RdRo)

Mv = Higher mode factor (1.0)

IE = 1.0

Rd = 3.0 (Moderately ductile concentrically braced frame)

Ro = 1.3

W = Total lumped mass from each floor

V = 0.27 * 1.0 * 1.0 * 12,997 / (3 * 1.3)

= 900 kN

Distribute the base shear over the height of the structure

Fx = (V-Ft)Wxhx/(∑Wihi)

Ft may be taken as zero when the fundamental period is less than 0.7s

For instance,

Froof = 900 * 4,537 * 12 / 111,788= 438 kN

Accidental Torsion

eX = distance between the centre of mass and the centre of rigidity

DNX = plan dimension of building at level x perpendicular to the direction of

seismic load

i) TX = FX * (eX + 0.10 * DNX), and

ii) TX = FX * (eX + 0.10 * DNX)

For instance, level 3

i) T3 = Froof * (e3 + 0.10 * DN3)

= 438 * ( 0 + 0.10 * 48)

= 2,103 kN

ii) T3 = FX * (eX - 0.10 * DNX)

= 438 * ( 0 - 0.10 * 48)

= -2,103 kN

NBCC 2010

- Division B

Section 4.1.8

Table 4.1.8.5

Table C-2 ---

Hamilton

Table 4.1.8.9

NBCC

4.1.8.11 (10)




A-26

Torsion Shear

A 100kN force was applied at each different level of both the outside and the inside

bracing frame on SAP2000.The deflections resulting from each simulation is presented in

the table below.

BF1 & 4 BF 2 & 3 Ratios Ratios

Level Force

(kN) u1 (mm) u1 (mm) u1/u1 u1/u1

Roof 100 2.0196 2.6888 0.751 1.331

3rd 100 1.2671 1.5762 0.804 1.244

2nd 100 0.7612 0.9417 0.808 1.237

Since deflection is inversely proportional to rigidity, the ratios obtained from these

simulations will be used in the torsion shear calculation.

Sample torsion shear calculation for BF1 at roof level:

Vyir = [xiRyi/Σ(xi

2Ryi+yi

2Rxi)]exVy

Vout = [youtRout/(2xin2Rin+2yout

2Rout)]M

Since Rin = 0.751 Rout

Therfore, VBF1 = -[9.9Rout/(2*4.82*0.751*Rout+2*9.9

2.Rout)]*2103

VBF1 = -90.27 kN

Masonry

Structures

Behaviour

and Design,

P-474

W460X52

W460X52

W410X46

L152X102X19-10 L152X102X19-10

L152X10

2X19-1

0 L152X102X

19-10

L152X102X19-10 L152X102X19-10

W2

50

X7

3

W2

50

X7

3

100.00

X

Z

X

Z




A-27

Beam to Girder Connection

Girder: W610X101 Beam 5: W410X39

w=10.5 mm w=6.4 mm

d - 2t = 573 mm d - 2t = 381 mm

Bolted to beam web and bolted to both sides of supporting member, supported beams not

coped, bearing-type.

Web-Framing Legs,

Maximum shear for beam W410X39 = 123.8 kN

Vertical line with two bolts provides a capacity of:

350 kN> 123.8 kN Okay

Web thickness, based on bearing, required for steels with Fu=450 MPa

= 8.1 * 123.8 / 350 = 2.87 mm < 6.4 mm Okay

Angle thickness required:

= 7.7 * 123.8 / 350 = 2.72 mm

Minimum Angle length required = 150 mm < 381 mm Okay

Outstanding Legs,

Total reaction on girder web is 2 * 123.8 = 247.6 kN

For beams connected to both sides of supporting member, the bolt capacity is double:

2 * 350 = 700 > 247.6 kN Okay

For angle L = 150 mm, W = 76 or 89 mm, two M20 A325M bolts (Bearing-type

connections, threads included) per vertical line, the girder web thickness is:

10.5 mm > 8.1 mm Okay

Use:

89x89x7.9 connection angles, 150 mm long, two M20 A325M bolts per vertical line

in both web-framing and outstanding legs

Table 3-37

CISC S16-09

Table 3-37

CISC S16-09




A-28

Check minimum number of bolts

Beam: 2 shear planes

1) Shear Resistance of a single bolt.

Vr = 0.60ΦbnmAbFu

= 0.6 × 0.8 × 1 × 2 × 261

= 250.56 kN

2) Bearing resistance

Angle thickness 2t = 2 * 7.94 = 15.88 mm

Beam web thickness = 6.4 Governs

Br = 3ΦbrntdFu

= 3 * 0.8 * 1 * 6.4 *20 * 450

= 138.24 kN

3) Number of bolts = Vf / smaller of Brand Vr

= 123.8 / 138.24

= 0.9 < 2 bolts Okay

Girder: 2 shear planes


Vr = 0.60ΦbnmAbFu

= 0.6 × 0.8 × 1 × 2 × 261

= 250.56 kN



Girder web thickness = 10.5 mm Governs

Br = 3ΦbrntdFu

= 3 * 0.8 * 1 * 10.5 *20 * 450

= 226.8 kN

6) Number of bolts = 2* Vf / smaller of Brand Vr

= 247.6 / 226.8

= 1. 09 < 4 bolts Okay

Check tension and shear failure

An = [(89 - 60) - 0.5 * (20 + 2) ] * 7.94 = 143 mm2

Agv = (80+35) * 7.94 = 913 mm2

Cl

13.12.1.2(c),

Table 3-1

Cl

13.12.1.2(a)

Cl

13.12.1.2(c),

Table 3-1

Cl

13.12.1.2(a)




A-29

Tr = Φu[UtAnFu + 0.6Agv(Fy + Fu) / 2]

= 0.75(0.6 * 143 * 450 + 0.6 * 913(300 + 450) / 2)

= 183 kN> 123.8 kN Okay

Agv = (80+35 * 2) * 7.94 = 1191 mm2 For one angle

Vr = Φu[0.6Agv(Fy + Fu) / 2]

= 0.75(0.6 * 1191(300 + 450) / 2)

= 201 kN> 123.8 kN Okay

Cl 13.11

Beam to Column Connection

Column: W250X73 Beam 9: W410X39

w=8.6 mm w=6.4 mm

d - 2t = 225 mm d - 2t = 381 mm



Web-Framing Legs,

Maximum shear for beam W410X39 = 123.8 kN


350 kN> 123.8 kN Okay


= 8.1 * 123.8 / 350 = 2.87 mm < 6.4 mm Okay


= 7.7 * 123.8 / 350 = 2.72 mm


Outstanding Legs,

Total reaction on column web is 2 * 123.8 = 247.6 kN

For beams connected to both sides of supporting member, the bolt capacity is double:

2 * 350 = 700 > 247.6 kN Okay

Table 3-37

CISC S16-09

Table 3-37

CISC S16-09




A-30


connections, threads included) per vertical line, the girder web thickness is:

10.5 mm > 8.1 mm Okay

Use:

89x89x7.9 connection angles, 150 mm long, two M20 A325M bolts per vertical line


88.9 * 2 + 6.4 = 184.2 mm < 225 mm Okay

Checking,

Beam: 2 shear planes


Vr = 0.60ΦbnmAbFu

= 0.6 × 0.8 × 1 × 2 × 261

= 250.56 kN



Beam web thickness = 6.4 Governs

Br = 3ΦbrntdFu

= 3 * 0.8 * 1 * 6.4 *20 * 450

= 138.24 kN


= 123.8 / 138.24


Column: 2 shear planes


Vr = 0.60ΦbnmAbFu

= 0.6 × 0.8 × 1 × 2 × 261

= 250.56 kN



Column web thickness = 8.6 mm Governs

Cl

13.12.1.2(c),

Table 3-1

Cl

13.12.1.2(a)

Cl

13.12.1.2(c),

Table 3-1




A-31

Br = 3ΦbrntdFu

= 3 * 0.8 * 1 * 8.6 *20 * 450

= 185.76 kN

6) Number of bolts = 2* Vf / smaller of Brand Vr

= 247.6 / 185.76



An = [(89 - 60) - 0.5 * (20 + 2) ] * 7.94 = 143 mm2

Agv = (80+35) * 7.94 = 913 mm2


= 0.75(0.6 * 143 * 450 + 0.6 * 913(300 + 450) / 2)

= 183 kN> 123.8 kN Okay


Vr = Φu[0.6Agv(Fy + Fu) / 2]

= 0.75(0.6 * 1191(300 + 450) / 2)

= 201 kN> 123.8 kN Okay

Cl

13.12.1.2(a)

Cl 13.11

Girder to Column Connection

Column: W250X73 Girder 9: W610X101

tf=14.2 mm w=10.5 mm

d - 2t = 225 mm d - 2t = 573 mm



Web-Framing Legs,

Maximum shear for girder W610X101 = 256.5 kN


350 kN> 256.5 kN Okay


= 8.1 * 256.5 / 350 = 5.94 mm < 10.5 mm Okay

Table 3-37

CISC S16-09




A-32


= 7.7 * 256.5 / 350 = 5.64 mm


Outstanding Legs,

For girders connected to column flange, the bolt capacity is:

350 > 256.5 kN Okay


connections, threads included) per vertical line, the column flange thickness is:

14.2 mm > 8.1 mm Okay

Use:

89x89x13 connection angles, 150 mm long, two M20 A325M bolts per vertical line


Check minimum number of bolts

Girder: 2 shear planes


Vr = 0.60ΦbnmAbFu

= 0.6 × 0.8 × 1 × 2 × 261

= 250.56 kN



Girder web thickness = 10.5 Governs

Br = 3ΦbrntdFu

= 3 * 0.8 * 1 * 10.5 *20 * 450

= 226.8 kN


= 256.5 / 226.8


Column: 1 shear plane


Table 3-37

CISC S16-09

Cl

13.12.1.2(c),

Table 3-1

Cl

13.12.1.2(a)




A-33

Vr = 0.60ΦbnmAbFu

= 0.6 × 0.8 × 1 × 1 × 261

= 125.28 kN


Angle thickness t = 12.7 mm Governs

Column flange thickness = 14.2 mm

Br = 3ΦbrntdFu

= 3 * 0.8 * 1 * 12.7 *20 * 450

= 274 kN


= 256.5 / 125.28



An = [(89 - 60) - 0.5 * (20 + 2)] * 12.7 = 228.6 mm2

Agv = (80+35) * 12.7 = 1460.5 mm2


= 0.75(0.6 * 228.6 * 450 + 0.6 * 1460.5(300 + 450) / 2)

= 292.8 kN> 256.5 kN Okay


Vr = Φu[0.6Agv(Fy + Fu) / 2]

= 0.75(0.6 * 1905(300 + 450) / 2)

= 321.5 kN> 256.5 kN Okay

Cl

13.12.1.2(c),

Table 3-1

Cl

13.12.1.2(a)

Cl 13.11




A-34

1Retrieved from Agway Metals Inc.http://www.agwaymetals.com/products_decking.asp

Slab to Beam Connection

Composite Slab Selection: CD75-200(Galvanneal)1

Base steel = 0.914 mm, slab depth= 130mm.

be = 0.25 * 6600 = 1650 mm yc = (фs Asfy) / (0.85 фcf’c be)

фs = 0.85

As = 4990 mm2

fy = 350 MPa

фc = 0.65

f’c = 25 MPa

yc = (0.9 * 4990 * 350) / 0.85 * 0.65 * 25 * 1650

= 69.0 mm

Ycis less than the depth of the concrete 130 mm, therefore the neutral axis lies in concrete

Qr = Nqrs

N = (0.85 фcf’c b yc * 40%) / qrs

qrs = 32.3 kN - ( 12.7 mm)

N = (0.85 * 0.65 * 25 * 1650 * 69 *0.4) / 32300

= 20 studs

Spacing: S = 6600 / (20 / 2) = 660 mm ( Two studs per row )

CISC,

Cl.17.4.1

CISC P6-48

CISC

TABLE 5-2

http://www.agwaymetals.com/products_decking.asp




A-35

Lateral Bracing System Connections

Factored shear resistance of bolt:

Vr = 0.60фbnmAbFu

Since Vf = 522kN

522,000N = 0.60 x 0.8 x 2 x2xAbx830MPa

Ab = 328 mm2

Therefore, structural bolt M22 is chosen with an Ab of 380 mm2

Factored bearing resistance at bolt holes:

Br = 3фbrntdFu

Br = 3x0.8x2x10x22x450MPa = 475kN< 522kN

Therefore, structural bolt M27 is chosen instead of M22.

Br = 3x0.8x2x10x27x450MPa = 583 kN

Factored tensile resistance:

Double angle:

Tr = фuAneFu

Tr = 0.75x[19x(152-(2x27))]x450 = 628kN

2 angles x 628kN = 1256 kN> 428kN

Double angle:

Tr = фu[UtAnFu+0.6Agv(Fy+Fu)/2]

Tr = 0.75[0.6x76x19x450+0.6x150x19(350+450)/2] = 805 kN

2 angles x 805 kN = 1610 kN> 428kN

Plate:

Tr = фu[UtAnFu+0.6Agv(Fy+Fu)/2]

Tr = 0.75[1x75x10x450+0.6x75x10(350+450)/2] = 388 kN> 428*0.707 =300 kN

Therefore, double angle L152x102x16 with two structural bolts M27 at each beam or

column connection is sufficient to resist the factored loads.

CISC CSA

S16-09,

Cl.13.12.1.2

(c)

CISC CSA

S16-09,

Table 3-1

CISC CSA

S16-09,

Cl.13.12.1.2

(a)

CISC CSA

S16-09,

Cl.13.2

CISC CSA

S16-09,

Cl.13.11




A-36

Column to Column Connections

Use 20M A325M bolts

d = 20mm, Ab = 314mm2

Bearing Capacity

Girder W610X101, Column 250X73

Br = 3ΦbntdFu

= 3×0.8×2×14.9×20×450 = 644 kN

Shear Resistance of a single bolt

Vr = 0.6ΦbnmAbFu

= 0.6×0.8×2×1×261 = 250.56 kN (governs)

Max axial load on Column = 585.2 kN ( page A-19 Column 1 design)

# of bolts required = 585.2 / 250.56 = 2.3 bolts

Use 4 bolts ( 2 per row) for construction purpose

Spacing:

Min. pitch (center to center) = 2.7×20 = 54 mm

Min. edge distance = 34 mm

Max. edge distance = 12 × thickness of the outside connected part BUT LESS THAN

150mm

CSA S16-09

Table 3-1

Cl 13.12.1.2

(a)

CL 13.12.1.2

(c)

Cl 22.3.1

Cl 22.3.2

Table 6 –

shared edge)




A-37

Green Roof Load Calculation:

Roof Area:

Calculation uses loose soil, which has approximately 1000 kg/m3

Soil volume for green roof:

623.5 m2(0.3m) = 187.1 m

3

Total soil weight:

187.1m3(1000kg/m

3)

=18710 kg.

Expected yield of bell pepper per acre is approximately 12000 lb/acre, or 5443 kg/acre.

With addition of roots and above soil plant material: 3500 kg/acre.

Three storey total plant biomass: (3500kg/acre)*(623.5m2)/(4046.9m

2/acre)=539.4 kg.

Local Hamilton Solar Panel Supplier:Westinghouse

Weight per panel: 20.7 kg

Total weight of panels = 36 Panels x 20.7kg = 745.2 kg.

Estimation of Roofing membrane, insulation, supporting structures, and filter

membrane: 50 kg/m2

Estimate of uniform distributed load on roof

(Solar panels are incorporated with total area).

Total area = 811.1 m2

Total distributed load: 69.3 kg/m2

=0.68 kPa

Oklahoma

State

University,

J.E. Motes ,

2005, Pepper

Production.




A-38

Green Roof Water Capture:

Assumption: Porosity of soil is 0.4

Height of soil: 30 cm.

Annual rainfall: 764.8 mm

Annual rainfall to storage tank:

Vegetation (interception and evaporation are neglected):

=623.5m2(0.85)(0.7648m)

=405.3m2

Pervious Walk Path:

=137.1m2(0.9)(0.7648m)

=94.4m2

Solar Panels:

Assume all rain dropped from solar panels surface fall into pervious pavement:

=50.5m2(0.9)(0.7648m)

=34.8m2

Total annual rainwater collection = 405.3m2

+ 94.4m2

+ 34.8m2

= 534,500 Litres

Toilet/Urinal Annual Consumption:

Toilet water usage per flush: 6 Litres

Urinal water usage per flush: 3.8 Litres

Sample: Commercial floor – 8 toilets – 4 Urinals

Number of flush per toilet per day: 30

Number of flush per Urinal per day: 200

Annual Water Consumption for toilet: (8 toilets)(6 Litres per flush)(30 Flushes per

day)(365 days per year) = 525,600 Litres

Annual Water Consumption for Urinals: (4 Urinals)(3.8 Litres per flush)(200 Flushes per

day)(365 days per year) = 1,109,600 Litres

Total Water Consumption for commercial floor = 525,600 + 1,109,600

= 1,635,200 Litres




A-39

Porous pavements Calculations- Full Exfiltration

Minimum Depth Method

1. Calculate ΔQc – additional runoff from contributing areas

Assuming no additional runoff from contributing areas: ΔQc = 0 m2

2. Calculate dp (Base Depth)

dp = Δ

P = design storm rainfall depth (m) = 1 m

f = final infiltration rate (m/hr) = 30 mm/hr = 0.03 m/hr

T = effective filling time of base/sub bass (hr)

According to our design, 15 cm sub bass = 5 hrs

Vr = void ratio (typically 0.4)

dp = 1 – 0.03 * 5 / 0.4 = 0.625 m

3. Calculate dmax

dmax = fTs / Vr

Ts = max allowable storage time (hr) (typically 24 - 48 hrs, assuming 24

for our case)

dmax = 0.03 * 24 / 0.4 = 1.8 m > dp

4. If dp fails increase area of permeable pavement or add underdrains

5. Check structural thickness

6. Check the bottom of the subbase is at least 1 m above seasonal high groundwater

table




A-40

Foundation Design:

Column Base Plate Design:

For W250x73, b = 254 mm, d = 253 mm

Maximum factored axial load = Cf = 1571 kN (obtained from previous calculations)

f'c = 25 MPa

Area of plate required = Cf/Br = Cf/(0.85ϕcf’c)

A = 1571/(0.85x0.65x25/103) = 113738 mm

2

B = C = √A =338 mm

B = C = 350 mm will be chosen, A = 122500 mm2

Determine m and n

0.95d = 0.95x253 = 240 mm, therefore, m = (350 - 240)/2 = 55

0.80b = 0.80x254 = 203 mm, therefore, n = (350 - 203)/2 = 73.5

Use n for design

Plate thickness required = √ [(2xCfxn2)/(BCϕFy)]

tp = √[(2x1571x73.52)/(350x350x0.9x300/10

3) = 22.7 mm

n/5 = 73.5/5 = 14.7 mm < 22.7 mm - OK

Therefore, use 25 mm.

CISC CSA S16

P 6-52

CISC CSA S16

P 4-148

Anchor Rod Design:

Anchor rod tensile resistance:

Tr = ϕarAnFu

Tf = 591 kN

4 Anchor rods will be used

591000/4 = 0.67xAnx450

An = 490 mm2

d = 2x√(An/π) = 25 mm

Therefore, use 1 inch anchor rod with a hole diameter of 34 mm.

Anchor rod shear resistance:

Vr = 0.60ϕarAarFu

Vr = 0.60x0.67x π x (25.4/2)2 x450 = 91.6 kN

Vr = 91.7 kN x 4 = 366 kN > 362 kN

CISC CSA S16

Cl.25.3.2.1

CISC CSA S16

P-4-153

CISC CSA S16-

09

Cl.25.3.3.3




A-41

Bearing Capacity:

ϕ = 28.4o, γsat = 18 kN/m

3 (typical values are assumed; site related values are usually

acquired from the geotechnical investigation report)

Bearing capacity numbers, shape and depth factors:

Assume, L/B = L’/B’ = 1; that is, B/L = B’/L’= 1, and B = B’ = 2m.

(ϕ’p)ps = 9/8(ϕ’p)tr = 9/8x28.4o = 32

o

Nq = eπtan32

tan2(45+32

o/2) = 23.2

Nq-1 = 23.2-1 = 22.2

Assume rough footing.

Nγ = 0.1054exp(9.6ϕ`p) = 0.1054exp(9.6x32xπ/180) = 22.5

sq = 1 + B’/L’tanϕ’p = 1 + tan32o = 1.62

sγ = 1 - 0.4B’/L’ = 0.6

dq = 1+2tanϕ’p(1-sinϕ’p)2tan

-1Df/B’ = 1+2tan32

o(1-sin32

o)2[tan

-1(1.25/2xπ/180)] = 1.

dγ = 1.17

qu = γDf(Nq-1)sqdq+0.5γB’Nγsγdγ

=18x1.25x22.2x1.62x1.17+0.5x18x2x22.5x0.6x1=1190 kPa

qa = qu/FS + γDf = 1190/3 + 18x1.25 = 419 kPa

σmax = Applied load/Area = 1571/2x2 = 393 kPa

Allowable Bearing Capacity with an Inclined Load

Calculate the inclination factors and depth factors

B=B’; ω=21o

n=nB=(2+B’/L’)/(1+B’/L’)=(2+1)/(1+1)=1.5

iq = (1-H/Vn)n = (1-tanω)

n = (1-tan21)

1.5 = 0.48

iγ = (1-H/Vn)n-1

= (1-tanω)n+1

= (1-tan21o)1.5+1

= 0.30

Calculate the ultimate net bearing capacity

qu = γDf(Nq-1)iq+0.5γB’Nγiγ

=18x1.25x22.2x0.48+0.5x18x2x22.5x0.30 = 361 kPa

Soil Mechanics

and Foundations

P-356

Concrete

Design

Handbook

P-9-22




A-42

Footing Design:

1. Compute factored reaction:

qsf = 1571/22 = 393kPa

2. Minimum effective shear depth - one way shear:

ab = (2-0.5)/2 = 0.75m

From Table 9.1, dv = 275mm, thus d = dv/0.9 = 297mm

3. Minimum depth - two way shear:

From Table 9.4 establish d/he based on Af/Ac

Af/Ac = 22/0.5

2 = 16

From interpolation d/he = 0.7, resulting in

dave = hc(0.7) = 350mm.

Since 3dave = 3(0.350)= 1.05m > ab = 0.75m, Cl.13.3.4.4 is not applicable.

Thus two-way shear governs.

4. Total footing depth assuming 20M bars in both directions:

tf = 350+75+20 =445mm

Use tf = 450mm (rounding it up to the next 25mm)

Note that the dave will be used for calculating the flexural reinforcement in both directions.

5. Flexural reinforcement:

Check deep beam action in accordance with Cl.10.7.1

ab/d = 0.75/(0.450-0.075-0.02) = 0.75/0.355 = 2.1 > 2, therefore no deep beam action

qsfab2 = 221 kN, from table 9.7a

As = 950mm2/m

Thus the total reinforcement is:

As = 2(950) = 1900 mm2 > As, min = 0.002(2000)(450) = 1800mm

2

6. Find maximum bar size that can be developed in distance ab-0.075 = 0.675m

From equation 9.7

max db = √f’c(ab-0.075)1000/(0.45k1k2k3k4fy) = 20M

Number of bars = 1900/300 = 6.33 - 20M bars, thus

Use: 7-20M bars BEW (bottom each way)

Appendix A

Documents

Transcript of Appendix A