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Wall Detailed Design Calculation

This document summarizes the detailed design calculations for a wall with the following key details: - The wall thickness is 225mm and is located at Floor 1, Position 1 X:A/1-1a. - The factored axial load is 430.839kN, factored moment is 0.139313kNm, and factored shear is 93.504kN. - Reinforcement of 2 layers of 12mm diameter bars at 250mm spacing is designed to resist the loads. - Horizontal and vertical shear reinforcement of 2 layers of 12mm diameter links at 250mm and 225mm spacing respectively is also designed.

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azwan
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446 views12 pages

Wall Detailed Design Calculation

This document summarizes the detailed design calculations for a wall with the following key details: - The wall thickness is 225mm and is located at Floor 1, Position 1 X:A/1-1a. - The factored axial load is 430.839kN, factored moment is 0.139313kNm, and factored shear is 93.504kN. - Reinforcement of 2 layers of 12mm diameter bars at 250mm spacing is designed to resist the loads. - Horizontal and vertical shear reinforcement of 2 layers of 12mm diameter links at 250mm and 225mm spacing respectively is also designed.

Uploaded by

azwan
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
You are on page 1/ 12

WALL DETAILED DESIGN CALCULATION:

CODE OF PRACTICE USED IS BS8110:1985

Wall fullpathname = C:\Documents and


Settings\Administrator\Desktop\Mr Soon Factory-15-11-2018-14-
12\Wall1.wln

D.L. L.L. fcu fy cover Load incre.


1.4 1.6 25 460 35 10

Rebar maximum spacing = 250 mm, Minimum spacing = 100 mm


Rebar maximum size = 25 mm, Minimum bar size = 12 mm
Yield strain = Yield strength/Young Modulus =0.87*460/200000 =0.0020

Load Notation: P = Axial load in N; M = Moment in Nmm; S = Shear in N


dlF*DL = DeadLoadFactor*DeadLoad; llrF*llF*LL =
LiveLoadReductionFactor*LiveLoadFactor*LiveLoad;
wlF*WL = WindLoadFactor*WindLoad; NLX, NLY = Notional load in X or Y-
direction respectively;

Factored Force: sumP = sum of P; sumM = sum of M; sumS = sum of S;


WLNL = max(wlF*WL, NLX or NLY)
= dlF*DL + llrF*llF*LL + WLNL

Floor No. = 1; Live load reduction = 0

Location of Wall: 1 X:A/2c-3

Wall Thickness,B = 225 mm

Calculate Wall Starting Edge Vertical Force:

P: fDL = dlF*DL = 1.40*219831 = 307764 N; fLL = llrF*llF*LL =


1.00*1.60*77366 = 123786 N
P: fWL = wlF*WL = 0.00*0 = 0 N; NLX = -13650 N; NLY = 7197 N
sumP = fDL + fLL = 307764 + 123786 = 431550 N

Calculate Wall Ending Edge Vertical Force:

P: fDL = dlF*DL = 1.40*-76686 = -107360 N; fLL = llrF*llF*LL = 1.00*1.60*-


10830 = -17327 N
P: fWL = wlF*WL = 0.00*0 = 0 N; NLX = 15300 N; NLY = -8166 N
sumP = fDL + fLL = -107360 + -17327 = -124688 N

Average Wall Vertical Force, AvP = (431550 + -124688)/2 N = 153.431 kN


Design for Wall Starting Edge Vertical Force, sumP = 431550 N = 431.550 kN

M: dlF*DL = 1.40*96595; llrF*llF*LL = 1.00*1.60*28331; wlF*WL = 0.00*0;


NLX = -1830; NLY = 3749
sumM = dlF*DL + llrF*llF*LL = 1.40*96595 + 1.00*1.60*28331 = 180563 N

S: dlF*DL = 1.40*-41753; llrF*llF*LL = 1.00*1.60*-21836; wlF*WL = 0.00*0;


NLX = -10702; NLY = 5014
sumS = dlF*DL + llrF*llF*LL = 1.40*-41753 + 1.00*1.60*-21836 = 93391 N

Factored Axial Load, PU = 431.550 kN


Factored Moment,M = 0.180563 kNm
Factored Shear,S = 93.391 kN

Location of Wall:- Floor No = 1; 1 X:A/2c-3; Wall Thickness = 225


LeftWall = 0; LeftBeam = 8.53; RightWall = 1; RightBeam = 0.00;
SlabThick = 200; SlabPercent = 100; 2SlabPercent = 0; MidBeam = 0.00;
Slab/Wall Ratio = 0.89
Wall Fixity Case = 12; Fixity Condition = 2; Alternative Fixity Condition =
1;

Calculate any additional moment due to Wall slenderness

For Braced Wall, effective height,Hef = Coef*Height = 0.90*2000 =1800 mm


Slenderness ratio,sr = Hef/A =1800/ 225 = 8.0
Slenderness ratio = 8.0 < 15 ---> No additional moment

Total Factored Moment,Mty = My + Msly = 0.2 + 0.0 = 0.2 kNm

Rebar Size,rbs = 25 mm
Total depth, h = 225 mm
Double layer of rebar ---> d' = cover + RebarSize/2 = 35+25/2 = 48 mm
Effective depth, d = h - d' = 225-48 = 178 mm
Uniform Width of concrete compression, b = 1000 mm

Eccentricity 20 mm = 20*UL = 20*474705 = 9.5 kNm


Eccentricity 5 percent = 0.05*h*UL = 0.05*225*474705 = 5.3 kNm
Eccentricity Moment > Increased Moment, M ---> Md = 5.3 kNm

Design concrete strength, Fcu = 10.2 N/mm^2


Yield strength,Fys = SteelFactor*fy = 0.87*460 = 400 N/mm^2

Minimum Rebar area required, As = minPercent*b*h = 0.004*1000*225 = 900 mm


For equilibrium axial and moment design, uniformed compression, a = 46.5 mm

Case No: 2

a < h --> Try using minimum steel area, 900 mm^2


Moment capacity,Mc = As*Fys*(h-2d')/2 + UL*(h-a)/2
= 900*400*(225-2*47.5)/2 +474705*(225-46.5)/2 = 65.8 kNm
Moment capacity,Mc > Design Moment,Md i.e. 65.8 < 5.3 ---> O.K.

Rebar area required, As = 900 mm^2


Rebar Nos of Diameter 12 required,rbn = As/AreaPerNo/Layer = 900/113/2 =
4.0
Spacing of T12 = b/rbn =1000/ 4.0 =251
Use Rebar 2 X T12-250

-----------------------------------------------------------------------------
---------
** DESIGN THE HORIZONTAL REBAR FOR SHEAR:
Design Shear, Vd = sqrt(Vv*Vv+Vh*Vh) = sqrt(93391*93391+0*0) = 93391 N =
93.4 kN

Shear Stress, v = V/bd = 93391/(225*1000) = 0.415 N/mm^2

Refer to Table 3.9 of BS8110:1985


Shear Capacity,vc = 0.79*((100As/(bd))^1/3)*(400/d)^1/4)*((fcu/25)^1/3)/1.25
Effective depth ratio = max(1,400/d) = max(1,400/1000) = 1.000
Concrete Grade ratio = min(40,fcu)/25 = min(40,25)/25 = 1.000
Steel Percentage, 100As/(bd) = min(3,0.40) = 0.40
vc = ( 0.79*(0.40)^1/3*(1.000)^1/4*(1.000)^1/3 )/1.25 = 0.466 N/mm^2

Shear Stress - Shear Capacity = v - vc = vd = 0.415 - 0.466 = -0.051 N/mm^2


vd < 0.4 N/mm^2 --> Design for vd = 0.400 N/mm^2

Steel area provided by Link size 12, Asv = nLayer*pie*dia*dia/4 =


2*3.1416*12*12/4 = 226.2 mm^2
Link spacing required, Sv = fyy*fy*Asv/(vd*b) = 0.87*460*226.2/(0.400*225) =
1006
Shear Capacity provided by Link, Vl = 0.87*460*226.2/(1006*225) = 0.400
N/mm^2
Total shear capacity, Vcap = (Vl + Vc)bx*by = (0.400+0.466)*225*1000 =
194774 N = 194.8 kN => 93.4 kN
Horizontal Rebar provided 2 X T12-250

Design for vertical rebar to satisfy shear based on ACI code eqn.:
Avv >=[0.0025+0.5(2.5+hw/lw)(Avh/s2.h - 0.0025)]s1.h

Avh/s2.h = 226.19/(250.225) = 0.00402


Avv/s1 =[0.0025+0.5(2.5+1800/5125)(0.00402 - 0.0025)]*225 = 0.930
Minimum Avv/s1 = 0.0025h = 0.0025*225 = 0.563
Provided Avv/s1 = 226.19/250 = 0.905
Provided Avv/s1 < Minimum Avv/s1 ; i.e. 0.905 < 0.930 ==> Avv/s1
New vertical rebar spacing required to satisfy shear = 226.19/0.930 = 243
Vertical Rebar provided 2 X T12-225

Horizontal Rebar provided 2 X T12-250

=============================================================================
=========

Location of Wall: 1 X:A/1-1a

Wall Thickness,B = 225 mm

Calculate Wall Starting Edge Vertical Force:

P: fDL = dlF*DL = 1.40*-76208 = -106692 N; fLL = llrF*llF*LL = 1.00*1.60*-


10448 = -16716 N
P: fWL = wlF*WL = 0.00*0 = 0 N; NLX = -15366 N; NLY = -8149 N
sumP = fDL + fLL = -106692 + -16716 = -123408 N

Calculate Wall Ending Edge Vertical Force:


P: fDL = dlF*DL = 1.40*219722 = 307611 N; fLL = llrF*llF*LL =
1.00*1.60*77017 = 123228 N
P: fWL = wlF*WL = 0.00*0 = 0 N; NLX = 13783 N; NLY = 7232 N
sumP = fDL + fLL = 307611 + 123228 = 430839 N

Average Wall Vertical Force, AvP = (-123408 + 430839)/2 N = 153.716 kN


Design for Wall Ending Edge Vertical Force, sumP = 430839 N = 430.839 kN

M: dlF*DL = 1.40*74352; llrF*llF*LL = 1.00*1.60*22013; wlF*WL = 0.00*0;


NLX = 1637; NLY = 4622
sumM = dlF*DL + llrF*llF*LL = 1.40*74352 + 1.00*1.60*22013 = 139313 N

S: dlF*DL = 1.40*41882; llrF*llF*LL = 1.00*1.60*21794; wlF*WL = 0.00*0;


NLX = -10839; NLY = -5037
sumS = dlF*DL + llrF*llF*LL = 1.40*41882 + 1.00*1.60*21794 = 93504 N

Factored Axial Load, PU = 430.839 kN


Factored Moment,M = 0.139313 kNm
Factored Shear,S = 93.504 kN

Location of Wall:- Floor No = 1; 1 X:A/1-1a; Wall Thickness = 225


LeftWall = 1; LeftBeam = 0.00; RightWall = 0; RightBeam = 8.53;
SlabThick = 200; SlabPercent = 100; 2SlabPercent = 0; MidBeam = 0.00;
Slab/Wall Ratio = 0.89
Wall Fixity Case = 12; Fixity Condition = 2; Alternative Fixity Condition =
1;

Calculate any additional moment due to Wall slenderness

For Braced Wall, effective height,Hef = Coef*Height = 0.90*2000 =1800 mm


Slenderness ratio,sr = Hef/A =1800/ 225 = 8.0
Slenderness ratio = 8.0 < 15 ---> No additional moment

Total Factored Moment,Mty = My + Msly = 0.1 + 0.0 = 0.1 kNm

Rebar Size,rbs = 25 mm
Total depth, h = 225 mm
Double layer of rebar ---> d' = cover + RebarSize/2 = 35+25/2 = 48 mm
Effective depth, d = h - d' = 225-48 = 178 mm
Uniform Width of concrete compression, b = 1000 mm

Eccentricity 20 mm = 20*UL = 20*473923 = 9.5 kNm


Eccentricity 5 percent = 0.05*h*UL = 0.05*225*473923 = 5.3 kNm
Eccentricity Moment > Increased Moment, M ---> Md = 5.3 kNm

Design concrete strength, Fcu = 10.2 N/mm^2


Yield strength,Fys = SteelFactor*fy = 0.87*460 = 400 N/mm^2

Minimum Rebar area required, As = minPercent*b*h = 0.004*1000*225 = 900 mm


For equilibrium axial and moment design, uniformed compression, a = 46.5 mm

Case No: 2

a < h --> Try using minimum steel area, 900 mm^2


Moment capacity,Mc = As*Fys*(h-2d')/2 + UL*(h-a)/2
= 900*400*(225-2*47.5)/2 +473923*(225-46.5)/2 = 65.7 kNm
Moment capacity,Mc > Design Moment,Md i.e. 65.7 < 5.3 ---> O.K.
Rebar area required, As = 900 mm^2
Rebar Nos of Diameter 12 required,rbn = As/AreaPerNo/Layer = 900/113/2 =
4.0
Spacing of T12 = b/rbn =1000/ 4.0 =251
Use Rebar 2 X T12-250

-----------------------------------------------------------------------------
---------
** DESIGN THE HORIZONTAL REBAR FOR SHEAR:

Design Shear, Vd = sqrt(Vv*Vv+Vh*Vh) = sqrt(93504*93504+0*0) = 93504 N =


93.5 kN

Shear Stress, v = V/bd = 93504/(225*1000) = 0.416 N/mm^2

Refer to Table 3.9 of BS8110:1985


Shear Capacity,vc = 0.79*((100As/(bd))^1/3)*(400/d)^1/4)*((fcu/25)^1/3)/1.25
Effective depth ratio = max(1,400/d) = max(1,400/1000) = 1.000
Concrete Grade ratio = min(40,fcu)/25 = min(40,25)/25 = 1.000
Steel Percentage, 100As/(bd) = min(3,0.40) = 0.40
vc = ( 0.79*(0.40)^1/3*(1.000)^1/4*(1.000)^1/3 )/1.25 = 0.466 N/mm^2

Shear Stress - Shear Capacity = v - vc = vd = 0.416 - 0.466 = -0.050 N/mm^2


vd < 0.4 N/mm^2 --> Design for vd = 0.400 N/mm^2

Steel area provided by Link size 12, Asv = nLayer*pie*dia*dia/4 =


2*3.1416*12*12/4 = 226.2 mm^2
Link spacing required, Sv = fyy*fy*Asv/(vd*b) = 0.87*460*226.2/(0.400*225) =
1006
Shear Capacity provided by Link, Vl = 0.87*460*226.2/(1006*225) = 0.400
N/mm^2
Total shear capacity, Vcap = (Vl + Vc)bx*by = (0.400+0.466)*225*1000 =
194774 N = 194.8 kN => 93.5 kN
Horizontal Rebar provided 2 X T12-250

Design for vertical rebar to satisfy shear based on ACI code eqn.:
Avv >=[0.0025+0.5(2.5+hw/lw)(Avh/s2.h - 0.0025)]s1.h

Avh/s2.h = 226.19/(250.225) = 0.00402


Avv/s1 =[0.0025+0.5(2.5+1800/5125)(0.00402 - 0.0025)]*225 = 0.930
Minimum Avv/s1 = 0.0025h = 0.0025*225 = 0.563
Provided Avv/s1 = 226.19/250 = 0.905
Provided Avv/s1 < Minimum Avv/s1 ; i.e. 0.905 < 0.930 ==> Avv/s1
New vertical rebar spacing required to satisfy shear = 226.19/0.930 = 243
Vertical Rebar provided 2 X T12-225

Horizontal Rebar provided 2 X T12-250

=============================================================================
=========

Location of Wall: 1 X:G/1-3

Wall Thickness,B = 225 mm


Calculate Wall Starting Edge Vertical Force:

P: fDL = dlF*DL = 1.40*24129 = 33780 N; fLL = llrF*llF*LL = 1.00*1.60*11317


= 18107 N
P: fWL = wlF*WL = 0.00*0 = 0 N; NLX = -983 N; NLY = 54 N
sumP = fDL + fLL = 33780 + 18107 = 51888 N

Calculate Wall Ending Edge Vertical Force:

P: fDL = dlF*DL = 1.40*24108 = 33751 N; fLL = llrF*llF*LL = 1.00*1.60*11283


= 18053 N
P: fWL = wlF*WL = 0.00*0 = 0 N; NLX = 984 N; NLY = 61 N
sumP = fDL + fLL = 33751 + 18053 = 51804 N

Average Wall Vertical Force, AvP = (51888 + 51804)/2 N = 51.846 kN


Design for Wall Starting Edge Vertical Force, sumP = 51888 N = 51.888 kN

M: dlF*DL = 1.20*0; llrF*llF*LL = 1.00*1.20*-0; wlF*WL = 1.20*0; NLX = -


0; NLY = -0
sumM = dlF*DL + llrF*llF*LL + WLNL = 1.20*0 + 1.00*1.20*-0 + 0 = 0 N

S: dlF*DL = 1.20*782; llrF*llF*LL = 1.00*1.20*69; wlF*WL = 1.20*0; NLX =


-4293; NLY = 3495
sumS = dlF*DL + llrF*llF*LL + WLNL = 1.20*782 + 1.00*1.20*69 + 4293 = 5314 N

Factored Axial Load, PU = 51.888 kN


Factored Moment,M = 0.000000 kNm
Factored Shear,S = 5.314 kN

Location of Wall:- Floor No = 1; 1 X:G/1-3; Wall Thickness = 225


LeftWall = 2; LeftBeam = 0.00; RightWall = 2; RightBeam = 0.00;
SlabThick = 200; SlabPercent = 80; 2SlabPercent = 0; MidBeam = 22.90;
Slab/Wall Ratio = 0.89
Wall Fixity Case = 12; Fixity Condition = 2; Alternative Fixity Condition =
2;

Calculate any additional moment due to Wall slenderness

For Braced Wall, effective height,Hef = Coef*Height = 0.95*2000 =1900 mm


Slenderness ratio,sr = Hef/A =1900/ 225 = 8.4
Slenderness ratio = 8.4 < 15 ---> No additional moment

Total Factored Moment,Mty = My + Msly = 0.0 + 0.0 = 0.0 kNm

Rebar Size,rbs = 25 mm
Total depth, h = 225 mm
Double layer of rebar ---> d' = cover + RebarSize/2 = 35+25/2 = 48 mm
Effective depth, d = h - d' = 225-48 = 178 mm
Uniform Width of concrete compression, b = 1000 mm

Eccentricity 20 mm = 20*UL = 20*57076 = 1.1 kNm


Eccentricity 5 percent = 0.05*h*UL = 0.05*225*57076 = 0.6 kNm
Eccentricity Moment > Increased Moment, M ---> Md = 0.6 kNm

Design concrete strength, Fcu = 10.2 N/mm^2


Yield strength,Fys = SteelFactor*fy = 0.87*460 = 400 N/mm^2
Minimum Rebar area required, As = minPercent*b*h = 0.004*1000*225 = 900 mm
For equilibrium axial and moment design, uniformed compression, a = 5.6 mm

Case No: 2

a < h --> Try using minimum steel area, 900 mm^2


Moment capacity,Mc = As*Fys*(h-2d')/2 + UL*(h-a)/2
= 900*400*(225-2*47.5)/2 +57076*(225-5.6)/2 = 29.7 kNm
Moment capacity,Mc > Design Moment,Md i.e. 29.7 < 0.6 ---> O.K.

Rebar area required, As = 900 mm^2


Rebar Nos of Diameter 12 required,rbn = As/AreaPerNo/Layer = 900/113/2 =
4.0
Spacing of T12 = b/rbn =1000/ 4.0 =251
Use Rebar 2 X T12-250

-----------------------------------------------------------------------------
---------
** DESIGN THE HORIZONTAL REBAR FOR SHEAR:

Design Shear, Vd = sqrt(Vv*Vv+Vh*Vh) = sqrt(5314*5314+0*0) = 5314 N = 5.3 kN

Shear Stress, v = V/bd = 5314/(225*1000) = 0.024 N/mm^2

Refer to Table 3.9 of BS8110:1985


Shear Capacity,vc = 0.79*((100As/(bd))^1/3)*(400/d)^1/4)*((fcu/25)^1/3)/1.25
Effective depth ratio = max(1,400/d) = max(1,400/1000) = 1.000
Concrete Grade ratio = min(40,fcu)/25 = min(40,25)/25 = 1.000
Steel Percentage, 100As/(bd) = min(3,0.40) = 0.40
vc = ( 0.79*(0.40)^1/3*(1.000)^1/4*(1.000)^1/3 )/1.25 = 0.466 N/mm^2

Shear Stress - Shear Capacity = v - vc = vd = 0.024 - 0.466 = -0.442 N/mm^2


vd < 0.4 N/mm^2 --> Design for vd = 0.400 N/mm^2

Steel area provided by Link size 12, Asv = nLayer*pie*dia*dia/4 =


2*3.1416*12*12/4 = 226.2 mm^2
Link spacing required, Sv = fyy*fy*Asv/(vd*b) = 0.87*460*226.2/(0.400*225) =
1006
Shear Capacity provided by Link, Vl = 0.87*460*226.2/(1006*225) = 0.400
N/mm^2
Total shear capacity, Vcap = (Vl + Vc)bx*by = (0.400+0.466)*225*1000 =
194774 N = 194.8 kN => 5.3 kN
Horizontal Rebar provided 2 X T12-250

Design for vertical rebar to satisfy shear based on ACI code eqn.:
Avv >=[0.0025+0.5(2.5+hw/lw)(Avh/s2.h - 0.0025)]s1.h

Avh/s2.h = 226.19/(250.225) = 0.00402


Avv/s1 =[0.0025+0.5(2.5+1900/41750)(0.00402 - 0.0025)]*225 = 0.983
Minimum Avv/s1 = 0.0025h = 0.0025*225 = 0.563
Provided Avv/s1 = 226.19/250 = 0.905
Provided Avv/s1 < Minimum Avv/s1 ; i.e. 0.905 < 0.983 ==> Avv/s1
New vertical rebar spacing required to satisfy shear = 226.19/0.983 = 230
Vertical Rebar provided 2 X T12-225
Horizontal Rebar provided 2 X T12-250

=============================================================================
=========

Location of Wall: 1 Y:1/A-G

Wall Thickness,B = 225 mm

Calculate Wall Starting Edge Vertical Force:

P: fDL = dlF*DL = 1.40*26650 = 37310 N; fLL = llrF*llF*LL = 1.00*1.60*21807


= 34891 N
P: fWL = wlF*WL = 0.00*0 = 0 N; NLX = -159 N; NLY = -891 N
sumP = fDL + fLL = 37310 + 34891 = 72202 N

Calculate Wall Ending Edge Vertical Force:

P: fDL = dlF*DL = 1.40*26666 = 37333 N; fLL = llrF*llF*LL = 1.00*1.60*23265


= 37224 N
P: fWL = wlF*WL = 0.00*0 = 0 N; NLX = -62 N; NLY = 849 N
sumP = fDL + fLL = 37333 + 37224 = 74556 N

Average Wall Vertical Force, AvP = (72202 + 74556)/2 N = 73.379 kN


Design for Wall Ending Edge Vertical Force, sumP = 74556 N = 74.556 kN

M: dlF*DL = 1.40*0; llrF*llF*LL = 1.00*1.60*-0; wlF*WL = 0.00*0; NLX = 0;


NLY = -0
sumM = dlF*DL + llrF*llF*LL = 1.40*0 + 1.00*1.60*-0 = 0 N

S: dlF*DL = 1.20*1183; llrF*llF*LL = 1.00*1.20*642; wlF*WL = 1.20*0; NLX


= -3269; NLY = -3552
sumS = dlF*DL + llrF*llF*LL + WLNL = 1.20*1183 + 1.00*1.20*642 + 3552 = 5742
N

Factored Axial Load, PU = 74.556 kN


Factored Moment,M = 0.000000 kNm
Factored Shear,S = 5.742 kN

Location of Wall:- Floor No = 1; 1 Y:1/A-G; Wall Thickness = 225


LeftWall = 2; LeftBeam = 0.00; RightWall = 2; RightBeam = 0.00;
SlabThick = 200; SlabPercent = 100; 2SlabPercent = 0; MidBeam = 14.89;
Slab/Wall Ratio = 0.89
Wall Fixity Case = 12; Fixity Condition = 2; Alternative Fixity Condition =
1;

Calculate any additional moment due to Wall slenderness

For Braced Wall, effective height,Hef = Coef*Height = 0.90*2000 =1800 mm


Slenderness ratio,sr = Hef/A =1800/ 225 = 8.0
Slenderness ratio = 8.0 < 15 ---> No additional moment

Total Factored Moment,Mty = My + Msly = 0.0 + 0.0 = 0.0 kNm

Rebar Size,rbs = 25 mm
Total depth, h = 225 mm
Double layer of rebar ---> d' = cover + RebarSize/2 = 35+25/2 = 48 mm
Effective depth, d = h - d' = 225-48 = 178 mm
Uniform Width of concrete compression, b = 1000 mm

Eccentricity 20 mm = 20*UL = 20*82012 = 1.6 kNm


Eccentricity 5 percent = 0.05*h*UL = 0.05*225*82012 = 0.9 kNm
Eccentricity Moment > Increased Moment, M ---> Md = 0.9 kNm

Design concrete strength, Fcu = 10.2 N/mm^2


Yield strength,Fys = SteelFactor*fy = 0.87*460 = 400 N/mm^2

Minimum Rebar area required, As = minPercent*b*h = 0.004*1000*225 = 900 mm


For equilibrium axial and moment design, uniformed compression, a = 8.0 mm

Case No: 2

a < h --> Try using minimum steel area, 900 mm^2


Moment capacity,Mc = As*Fys*(h-2d')/2 + UL*(h-a)/2
= 900*400*(225-2*47.5)/2 +82012*(225-8.0)/2 = 32.3 kNm
Moment capacity,Mc > Design Moment,Md i.e. 32.3 < 0.9 ---> O.K.

Rebar area required, As = 900 mm^2


Rebar Nos of Diameter 12 required,rbn = As/AreaPerNo/Layer = 900/113/2 =
4.0
Spacing of T12 = b/rbn =1000/ 4.0 =251
Use Rebar 2 X T12-250

-----------------------------------------------------------------------------
---------
** DESIGN THE HORIZONTAL REBAR FOR SHEAR:

Design Shear, Vd = sqrt(Vv*Vv+Vh*Vh) = sqrt(5742*5742+0*0) = 5742 N = 5.7 kN

Shear Stress, v = V/bd = 5742/(225*1000) = 0.026 N/mm^2

Refer to Table 3.9 of BS8110:1985


Shear Capacity,vc = 0.79*((100As/(bd))^1/3)*(400/d)^1/4)*((fcu/25)^1/3)/1.25
Effective depth ratio = max(1,400/d) = max(1,400/1000) = 1.000
Concrete Grade ratio = min(40,fcu)/25 = min(40,25)/25 = 1.000
Steel Percentage, 100As/(bd) = min(3,0.40) = 0.40
vc = ( 0.79*(0.40)^1/3*(1.000)^1/4*(1.000)^1/3 )/1.25 = 0.466 N/mm^2

Shear Stress - Shear Capacity = v - vc = vd = 0.026 - 0.466 = -0.440 N/mm^2


vd < 0.4 N/mm^2 --> Design for vd = 0.400 N/mm^2

Steel area provided by Link size 12, Asv = nLayer*pie*dia*dia/4 =


2*3.1416*12*12/4 = 226.2 mm^2
Link spacing required, Sv = fyy*fy*Asv/(vd*b) = 0.87*460*226.2/(0.400*225) =
1006
Shear Capacity provided by Link, Vl = 0.87*460*226.2/(1006*225) = 0.400
N/mm^2
Total shear capacity, Vcap = (Vl + Vc)bx*by = (0.400+0.466)*225*1000 =
194774 N = 194.8 kN => 5.7 kN
Horizontal Rebar provided 2 X T12-250
Design for vertical rebar to satisfy shear based on ACI code eqn.:
Avv >=[0.0025+0.5(2.5+hw/lw)(Avh/s2.h - 0.0025)]s1.h

Avh/s2.h = 226.19/(250.225) = 0.00402


Avv/s1 =[0.0025+0.5(2.5+1800/45963)(0.00402 - 0.0025)]*225 = 0.984
Minimum Avv/s1 = 0.0025h = 0.0025*225 = 0.563
Provided Avv/s1 = 226.19/250 = 0.905
Provided Avv/s1 < Minimum Avv/s1 ; i.e. 0.905 < 0.984 ==> Avv/s1
New vertical rebar spacing required to satisfy shear = 226.19/0.984 = 230
Vertical Rebar provided 2 X T12-225

Horizontal Rebar provided 2 X T12-250

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Location of Wall: 1 Y:3/A-G

Wall Thickness,B = 225 mm

Calculate Wall Starting Edge Vertical Force:

P: fDL = dlF*DL = 1.40*26667 = 37334 N; fLL = llrF*llF*LL = 1.00*1.60*21827


= 34923 N
P: fWL = wlF*WL = 0.00*0 = 0 N; NLX = 152 N; NLY = -882 N
sumP = fDL + fLL = 37334 + 34923 = 72257 N

Calculate Wall Ending Edge Vertical Force:

P: fDL = dlF*DL = 1.40*26639 = 37295 N; fLL = llrF*llF*LL = 1.00*1.60*23272


= 37235 N
P: fWL = wlF*WL = 0.00*0 = 0 N; NLX = 63 N; NLY = 842 N
sumP = fDL + fLL = 37295 + 37235 = 74530 N

Average Wall Vertical Force, AvP = (72257 + 74530)/2 N = 73.393 kN


Design for Wall Ending Edge Vertical Force, sumP = 74530 N = 74.530 kN

M: dlF*DL = 1.40*-0; llrF*llF*LL = 1.00*1.60*-0; wlF*WL = 0.00*0; NLX =


0; NLY = -0
sumM = dlF*DL + llrF*llF*LL = 1.40*-0 + 1.00*1.60*-0 = 0 N

S: dlF*DL = 1.20*1163; llrF*llF*LL = 1.00*1.20*638; wlF*WL = 1.20*0; NLX


= 3272; NLY = -3501
sumS = dlF*DL + llrF*llF*LL + WLNL = 1.20*1163 + 1.00*1.20*638 + 3501 = 5663
N

Factored Axial Load, PU = 74.530 kN


Factored Moment,M = 0.000000 kNm
Factored Shear,S = 5.663 kN

Location of Wall:- Floor No = 1; 1 Y:3/A-G; Wall Thickness = 225


LeftWall = 1; LeftBeam = 0.00; RightWall = 1; RightBeam = 0.00;
SlabThick = 200; SlabPercent = 100; 2SlabPercent = 0; MidBeam = 14.75;
Slab/Wall Ratio = 0.89
Wall Fixity Case = 12; Fixity Condition = 2; Alternative Fixity Condition =
1;

Calculate any additional moment due to Wall slenderness

For Braced Wall, effective height,Hef = Coef*Height = 0.90*2000 =1800 mm


Slenderness ratio,sr = Hef/A =1800/ 225 = 8.0
Slenderness ratio = 8.0 < 15 ---> No additional moment

Total Factored Moment,Mty = My + Msly = 0.0 + 0.0 = 0.0 kNm

Rebar Size,rbs = 25 mm
Total depth, h = 225 mm
Double layer of rebar ---> d' = cover + RebarSize/2 = 35+25/2 = 48 mm
Effective depth, d = h - d' = 225-48 = 178 mm
Uniform Width of concrete compression, b = 1000 mm

Eccentricity 20 mm = 20*UL = 20*81983 = 1.6 kNm


Eccentricity 5 percent = 0.05*h*UL = 0.05*225*81983 = 0.9 kNm
Eccentricity Moment > Increased Moment, M ---> Md = 0.9 kNm

Design concrete strength, Fcu = 10.2 N/mm^2


Yield strength,Fys = SteelFactor*fy = 0.87*460 = 400 N/mm^2

Minimum Rebar area required, As = minPercent*b*h = 0.004*1000*225 = 900 mm


For equilibrium axial and moment design, uniformed compression, a = 8.0 mm

Case No: 2

a < h --> Try using minimum steel area, 900 mm^2


Moment capacity,Mc = As*Fys*(h-2d')/2 + UL*(h-a)/2
= 900*400*(225-2*47.5)/2 +81983*(225-8.0)/2 = 32.3 kNm
Moment capacity,Mc > Design Moment,Md i.e. 32.3 < 0.9 ---> O.K.

Rebar area required, As = 900 mm^2


Rebar Nos of Diameter 12 required,rbn = As/AreaPerNo/Layer = 900/113/2 =
4.0
Spacing of T12 = b/rbn =1000/ 4.0 =251
Use Rebar 2 X T12-250

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** DESIGN THE HORIZONTAL REBAR FOR SHEAR:

Design Shear, Vd = sqrt(Vv*Vv+Vh*Vh) = sqrt(5663*5663+0*0) = 5663 N = 5.7 kN

Shear Stress, v = V/bd = 5663/(225*1000) = 0.025 N/mm^2

Refer to Table 3.9 of BS8110:1985


Shear Capacity,vc = 0.79*((100As/(bd))^1/3)*(400/d)^1/4)*((fcu/25)^1/3)/1.25
Effective depth ratio = max(1,400/d) = max(1,400/1000) = 1.000
Concrete Grade ratio = min(40,fcu)/25 = min(40,25)/25 = 1.000
Steel Percentage, 100As/(bd) = min(3,0.40) = 0.40
vc = ( 0.79*(0.40)^1/3*(1.000)^1/4*(1.000)^1/3 )/1.25 = 0.466 N/mm^2

Shear Stress - Shear Capacity = v - vc = vd = 0.025 - 0.466 = -0.440 N/mm^2


vd < 0.4 N/mm^2 --> Design for vd = 0.400 N/mm^2

Steel area provided by Link size 12, Asv = nLayer*pie*dia*dia/4 =


2*3.1416*12*12/4 = 226.2 mm^2
Link spacing required, Sv = fyy*fy*Asv/(vd*b) = 0.87*460*226.2/(0.400*225) =
1006
Shear Capacity provided by Link, Vl = 0.87*460*226.2/(1006*225) = 0.400
N/mm^2
Total shear capacity, Vcap = (Vl + Vc)bx*by = (0.400+0.466)*225*1000 =
194774 N = 194.8 kN => 5.7 kN
Horizontal Rebar provided 2 X T12-250

Design for vertical rebar to satisfy shear based on ACI code eqn.:
Avv >=[0.0025+0.5(2.5+hw/lw)(Avh/s2.h - 0.0025)]s1.h

Avh/s2.h = 226.19/(250.225) = 0.00402


Avv/s1 =[0.0025+0.5(2.5+1800/45963)(0.00402 - 0.0025)]*225 = 0.984
Minimum Avv/s1 = 0.0025h = 0.0025*225 = 0.563
Provided Avv/s1 = 226.19/250 = 0.905
Provided Avv/s1 < Minimum Avv/s1 ; i.e. 0.905 < 0.984 ==> Avv/s1
New vertical rebar spacing required to satisfy shear = 226.19/0.984 = 230
Vertical Rebar provided 2 X T12-225

Horizontal Rebar provided 2 X T12-250

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=========

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