SHEAR WALL DESIGN

Grade of concrete M 20

Characteristic strength of steel Fy 415

Thickness of web tw 115 mm

Overall length of wall Lw = 1800 mm

Dimension of wall at one end d1 = 230 mm

Dimension of wall at another end d2 = 115 mmEffective length of wall Le = Lw-d1/2-d2/2 Le = 1627.5 mmAxial load (DL+LL) P = 2850 kN

(EQ) P' = 0Shear force (DL+LL) V = 275 kN

(EQ) V' = 0Bending Moment (DL+LL) M = 250 kNm

(EQ) M' = 0

N/mm2

N/mm2

SHEAR DESIGN

Thickness of web tw = 115 mmFactor 1.2

Factored Shear Vu = V * factor Vu = 330 kN

Overall length of wall Lw Lw = 1800 mm

Effective length of wall Le = Lw-d1/2-d2/2 Le = 1627.5 mm

dw = 0.8*Lw dw = 1440 mm

is replaced by Tc

Tc from IS : 456 Tc = 0.36

Tv = Vu Vu = 1.99tw * dw

Condition is Tv > Tc

Tv < Tc

Min % of steel to be provided is 0.25% Ag= 414Diameter of bar used db = 8 mm

Spacing = 122 mm

Reinforcement shall be provided in two curtains

If Tw > 200mm Ag'=Ag/2 Ag' = 207Diameter of bar used db = 8 mm

Spacing = 243 mm

Provide 8 # @ 122 mm c/c

Tv > Tc

Vus = Vu - Tc * tw * dw Vus = 271 kN

Vus = 0.87 * fy * Ah * dw Ah/Sv = 0.52 % Sv

Diameter of bar used db = 8spacing = 90 Trial and errorAh/Sv = 0.56 %

Provide 8 # @ 90 mm c/c

tc

N/mm2

mm2

If thickness of wall is > 200mm or factored sheat stress in thewall exceeds 0.25(fck)1/2,

mm2

11.5 mm

1] lw/5 3602] 3*tw 3453] 450 450

Trial and error

Diameter should be less than tw/10

Spacing not be greater than lesser of

FLEXURAL STRENGTH OF WEB

Axial Compression will increase the moment capacity of wall

Hence the factored axial force Pu= P*0.8+P' *factorfactor 1.2Pu = 3420 kN

Moment of resistance provided by distributed vertical reinforcement across wall section

Considering 0.25% vertical reinforcement in web p = 0.008

0.87*fy*p 0.144 fck

Pu 0.83fck*tw*lw

xu = f+l xu = 1.50 lw lw

xu' = 0.0035 Es = xu' = 0.66 lw 0.0035+0.87*fy/Es lw

0.87*fy 0.516Es*0.0035

Condition is xu'/lw < xu/lw <1

xu/lw < xu'/lw

Muv =

lw lw 3

Muv = 2115 kNm

xu'/lw < xu/lw < 1

0.327

0.129

-0.1088

[ f/b -l ] -0.546

0.140

X = 1.45-0.7497

Muv :-

f = f =

l = l =

2f+0.36

2x105

b = b =

f [ 1+l) (1-0.416xu ) - xu 2 (0.168+b2)

fck*tw*lw2 f

a1 = [0.36+f ( 1-b/2 - 1/2b)] a1 =

a2 = [ 0.15+ f ( 1-b-b2/2-1/3b)] a2 =

a3 = [ f / 6b *(1/(xu/lw) - 3) ] a3 =

a4 = a4 =

a5 = [ f/2b ] a5 =

Substituting xu/lw as X in following equation to derive new value of xu/lw

a1 (X)2 + a4 (X) - a5 =0

Muv = Muv = 748 kNm

-0.10026

[ a1 (X) -a2 (X2) - a3 - l/2 ] * fck * tw * lw2

-747.16

BONDING ELEMENT

Factored Moment Mu Factor 1.2Mu = 300 kNm

Cv - c/c distance b/w bonding element Cv = 1627.5 mmAssume dimension of element b = 230 mm

d = 450 mm

The moment to be resisted by bonding element is Mu-Muv Mu-Muv -448 kNm

C = Mu-Muv C = 275.2688 kNm Cv

Max factored compression on boundry elementCu = Cu = 1004 kN

Max factored tension on boundry elementTu = Tu = 454 kN

Assuming short column action, the axial load capacity of the boundry elementwith minimum reinforcement of 0.8% is P*% of reinforcement > 0.8% but < 6% preferably <4%

p = 1.8P* = [ 0.4 fck +p/100 (0.67 fy - 0.4fck) ] Ag p* = 1332 kN

Ast = 1863Diameter of bar used db = 16

No = 10

Provide 10 No 16 #

Special confining reinforcement shall be provided as in the case of columnIf entire wall is provided by special confinement, boundry element is not required to be provided

C+(0.213*factor*( P+P' )

(0.213*factor*( P+P' ) -C

mm2

If entire wall is provided by special confinement, boundry element is not required to be provided