5 Examples Shear Wall

EXAMPLE NO.1 ( Design of simple shear wall with enlarged ends & opening ) Design a shear wall of length 4.16 m and thickness 250 mm is subjected to following forces (see fig. ). Assume fck = 20 and fy = 415 N/mm2 and the wall is a high wall with following loadings. Design the reinforcement around opening 1200 mm x 1200 mm. Loading 1.

DL + LL

2. Seismic load

Axial force (kN )

Moment (kNm)

Shear kN

1950

600

20

250

4800

700

Reinforcement around opening

Solution :

 STEP 1 : Determine design loads –

Assuming load factor 0.8 when gravity load assists stability and load factor 1.2 when gravity load opposes stability

Factored Axial Force : P1 (case 1) = (0.8 x 1950 )+ (1.2 x 250 ) = 1860 kN

P2 (case 2) = 1.2(1950 + 250 ) = 2640 kN

Factored Moment = 1.2 (4800 + 600 ) = 6480 kN-m

Factored Shear = 1.2 (700 + 20) = 864 kN

 STEP 2 : Check whether boundary elements are required –

Generally boundary elements are provided when extreme stresses exceeds 4 N/mm2 . Assuming uniform thickness t = 250 mm for L = 4160 mm

bd 3 250  4160 3 I   1.5  1012 mm 4 12 12

A  bd  4160  250  1.04 106 mm2 Combined Axial & Bending stresses

P M  y A I (2.64)10 6 (6.48)10 9 4160 fc   2 = +11.52 and 6.45 N/mm 6 12 2 (1.04)10 (1.5)10

fc 

Extreme Fiber stresses +11.52 & - 6.45 N/mm2

> (0.2 fck) = 0.2 x 25 = 5 N/mm2

Hence the boundary elements are needed.

 STEP 3 : Adopt dimension of boundary element – Adopt a bar bell type wall with central 3400 mm portion and two ends 380 x 760 m giving a total length of (3400 + 380 + 380 ) = 4160 mm

3400 mm

STEP 4 : Check whether two layers of steel is required – Two layers are required if – 1. Shear stress is more than 0.25 fck

2. Thickness of section is more than 200 mm

Checking the above 2 conditions 1. Depth of section = c/c of boundary element = 3400 + 380 = 3780 mm

V 864  10 3   0.92 N / mm 2 Nominal Shear stress v  bd 250  3780 0.25 fck  0.25 20  1.11N / mm 2 2. Thickness of 250 mm > 200 mm Therefore, Use two layers of steel with suitable cover.

 STEP 5 :

Determine Steel required –

Consider min. steel ( Asmin ) p = 0.25%) Asmin = 0.0025 x 250 x 1000 per m length = 625 mm2 in two layers. Provide 10 tor @ 250 mm c/c.

( 314 mm2 per meter length )

Spacing 250 < 450 (max.allowed) Provide the same vertical and horizontal steel.  STEP 6 :

Calculation of Shear capacity of steel “Vs” –

Nominal shear stress, v = 0.92 N/mm2 For 0.25 % steel & fck = 20, Critical shear stress

c = 0.36 N/mm2

(as per Table 9 of IS:456-2000)

Shear taken by steel “Vs”, Vs = (0.92 – 0.36) b d = 0.56 x 250 x 3780 = 529200 N = 529.2 kN

STEP 7 :

Calculate steel necessary to take “Vs” –

Vs Asv  0.87 fy d Sv d = 3780

Required

Asv (529.2)(103 )  Sv (0.87)(415)(3780)

Vs = S.F. resisted by horiz. shear r/f Asv = area of horiz. shear r/f Sv = spacing of shear r/f = 0.388

Consider 1m height = Sv Horizontal steel area = 628 mm2 = Asv

Available

STEP 7a) :

Asv 628   0.628 Sv 1000

> 0.388 Hence min. steel provided is okey for shear requirement Find flexural strength of web of shear wall – P = 0.8(1950)+1.2(250) = 1860 kN

Assuming it as UDL over the area. The axial load for central part of beam - Pw

3400  250 (3400  250)  2(380  760) Pw  1860  0.595  1107kN

Pw  1860

STEP 8 :

Calculate parameter ,  and ( x / L) –

Ref. IS:13920-1993, Appendix “A” : Moment of Resistance of rectangular Shear Wall

Pw 1107  103    0.065 fcktL 20  250  3400 0.87 fy (0.87)( 415)(0.0025)    0.045 fck 20   0.516

x    L 0.36  2 x 0.065  0.045   0.24 L 0.36  0.09

x  0 .5 L

Mu  x ( Ref. IS:13920-1993)   [ 1  ][ 0 . 5  0 . 42 ] 2 fcktL  L Mu 0.065 x  0 . 045 [ 1  ][ 0 . 5  0 . 42 ] 2 fcktL 0.045 L Mu  0.041 2 fcktL M.R. of Shear Wall

Mu  (0.041(20)( 250)(3400) 2  2370kNm This is less than Factored Moment 6480 kN-m required.

 STEP 9 : Calculate moment to be carried by boundary element M1= 6480 – 2370 = 4110 kN-m  STEP 10 : Calculate compression and tension in boundary element due to M1 Distance between boundary elements = 3.480 + 0.380= 3.86 m = c Axial load carried by boundary element = M 1  4110  1065kN

c

3.86

( as per Cl. 9.4.2)

This load acts as tension on one end and as compression at the other end.

 STEP 11 : Calculate compression due to the axial loads at these ends 1  0.595  0.2025 Fraction of area at each end  ( Ref. Step 7a) 2 Factored compression at compression end. Taking worst case. P2 = (0.2025) (2640) = 535 kN

( Ref. Step 1)

Factored compression at tension end (taking P1) P1 = (0.2025 ) (1860 ) = 377 kN

Axial load carried by boundary element

Compression at compression end = 1065 + 535 = 1600 kN Tension at the tension end

= -1065 +377 = -688 kN

 STEP 12 : Design the boundary elements compression a. Design one end as column b. Check laterals for confinement c. Check for anchorage and splice length

STEP 13 : Design tension side of shear wall Provide the same steel also on the tension end as in compression end. (Earthquake forces can act in both direction)

 STEP 14 : Design the reinforcement around opening Assuming opening size as 1200 x 1200 Area of reinforcement cut off by opening =

% steel  1200  250  0.0025  750mm2 100

4 Nos. 16 mm bar area = 804 mm2 Provide 2 nos. 16 mm, one on each face of wall, on all the sides of the hole to compensate for steel cut off by hole.

Reinforcement around opening

X

Example 2 :

(Lateral stiffness of shear walls )

A bar bell type shear wall with central part 3600 x 150 mm and two 400 x 400 mm strong bands at each ends is supported on a footing 8 m x 4 m which rests on soil whose module is 30,000 kN/m3 . Assume fck =20 and height of the wall is 14 m. Determine the lateral stiffness of wall. Solution : Lateral stiffness = Force required at for unit deflection



W total

total = 1 bending +

2 shear + 3 foundation rocking

8m

 STEP 1 : Calculate deflection of wall due to unit load at top (bending deflection)

WH 3 1  3EI Ec  5000 5700 fck  25.5kN / mm2  150(3600) 3  400(400) 3 I   (400) 2 (2000) 2  12 12   I  1.87 1012 mm4 Deflection due to unit load, W = 1 kN (14000)3 7 11   1 . 92  10 mm/ N 3 12 (3)( 25.5 10 )(187 10 )  STEP 2 : Calculate deflection due to shear (shear deflection) WH 2  CAG Where C = constant = shape factor (0.8 for rectangle ) A = (150 x 3600 ) + ( 2 x 400 x 400 ) = 8.6 x 105 mm2

G

E 25.5   10.45kN / mm2 2(1  0.22) 2(1  0.22)

(assume  = 0.22 )

H 0.8 AG 14000  (0.8)(8.6 105 )(10.5 103 )

Deflection for unit load 1 2 

 0.194 10 5 mm / N  STEP 3 : Calculate deflection due to rocking of foundation unit load at top (deflection due to rocking of foundation ) γ = Modulus of sub-grade reaction = 30,000 kN / m3, Foundation size = 8 m x 4 m

 3 rocking 

WH R

2

M 

BL3

  R

= Moment due to rotation θ

12 (30,000)( 4)(8) 3 R  512 10 4 kNm / radian 12

Deflection due to W = 1

H2 (14) 2 5 13    3 . 83  10 mm / N 4 R (512)(10 )

moment produce due to unit rotation of foundation

 STEP 4 : Calculate lateral stiffness of the wall

Stiffness 

Load Deflection

Total deflection due to unit load

1 = 1-1 + 1-2 + 1-3 = (1.92 + 0.19 + 3.83 ) x 10-5 = 5.94 x 10-5 mm

1 105 Stiffness    16.8kN / mm  5.95