Tutorial SD 442 ShearWall-1.ppt desidn of concrete structures

Design of shear wall

1. 1
SC 342
WORKED EXAMPLES
DESIGN OF SHEAR WALL
STRUCTURES

2. 2
Example 1: Wind Load Distribution
Determine the wind load distribution for the following shear wall
structure, given that areal wind load is 1.5 kN/m
2
; the wall thickness,
t=0.3 m; and the height of the building L=24.00 m.
Figure 1:Floor plan of a shear wall structure
1
2 3 4
5
6
7

3. 3; 


o,yi,yo
EI/)yEI(y 


o,zii,zo EI/)z,EI(z
In this collection of formulae the distribution ratios have been denoted
by the symbolsuperscript , where “B” means “bending” and “T”
means “torsion”.
E is assumed to be constant and therefore is not considered.
Coordinates of the shear centre of lumped beam using Ô as
reference
The formulae required are:
;
Coordinates of the shear centre of
lumped beamoii yyy

 oii zzz


Coordinates of the shear wall w.r.t. the specific
coordinate axis
i,yo,y
EIEI 
izoz
EIEI
,,
;
Lumped bending stiffness

4. 412/
3
bh )m(y
i
 )(mzi
 )m(yI
5
ii,y
 )m(zI
5
ii,z

Wall
No.
1 5.40 0.00 25.0 6.00 135.0 0
2 2.28 0.00 15.0 2.25 34.2 0
3 2.28 0.00 10.0 2.25 22.8 0
4 2.28 0.00 0.0 2.25 0.0 0
5 2.28 0.00 0.0 9.75 0.0 0
6 0.00 3.13 7.5 12.000.0 37.56
7 0.00 3.13 17.5 12.000.0 37.56
 14.44 6.26 - - 192.0 75.1212/
3
bh )m(yI
5
ii,y
 )m(zI
5
ii,z

Coordinates of the shear center:m00.1226.6/12.75z
m30.1344.14/0.192y
o
o




(0.3 x 6x6x6/)12=5.4
(0.3 x 4.5x 4.5x 4.5/)12=2.28

5. 5
Using S
0as referenceoii yyy


andoii zzz

 62
,
2
,,
577,1)( mzIyII
iiziiyo


Warping relative stiffness

6. 6
Distribution factor for bending and torsion
5.4/14.
44
3.13/6.26

7. 7
Individual wind loads on the different shear walls0.
/00.30)30.135.12)(50.37(
0
/50.37)00.25)(50.1(
,,
,,
,
,




ooyox
oozox
yoy
zoz
epm
mkNmepm
awp
mkNbwp
From equations;ox
o
mmy
mz
ox
o
mmz
my
m
EI
yEI
P
m
EI
zEI
p
,
,
,
,
,
,
,
,





8. 8);()(
0,0,
xLpxV
zz
 )()()(
0,
0,
,
,0,0, xV
I
I
VxLmxM
z
y
iyB
izxx  )()(
0,
0,
,
,
xM
I
yI
xV
x
iiyT
iz

 );()(
2
)(
)(
0,
0,
,
,
2
0,0,
xM
I
I
xM
xL
pxM
y
y
iyB
iyzy


 )()(
0,
0,
,
,
xM
I
yI
xM
iiyT
iy 

 2
)(
)(
2
0,0,
xL
mxM
x



37.5 x 0.374 30 X 0.04006

9. 9
Example 2: Design of an R.C. shear wall
Carry out preliminary
design of shear wall
No. 2 of the skeleton
structure shown in
Figure 2.
Figure 2: Plan of a shear wall structure1
2
3
4
1'
2'
3'
4'
So
3.0
9
.
0
6
.
0
6
.
0
6.06.0
2
4
.
0
1
2
.0
6
.
0
1
2
.0
12.0 12.012.0
Pz,o
36.0 m

10. 10
Size of each shear wall: b/h = 0.20 m/6.00 m
Height of the building:L = 10x4.00 = 40.0 m
Storey height: h
s= 4.00 m
Number of storey's: n
s= 10
Floor construction: R.C. slabs, one-way spanning,
h = 0.15 m
Solutions
(a)Approximate determination of the loads
(1) dead loads: From R.C slabs2
s m/kN75.315.0x24g 
Floor finishes2
/00.1 mkNg
4.75 kN/m
2
For wall it self per storeystorey/kN12000.4x00.6x20.0x24g
w 

11. 11
(2) imposed loads2
/0.3 mkNq
(3) resultant loads: Influence area for the wall for
the vertical load2
I m270.3x0.9A  storey/kN8127x0.3Q
storey/kN24812027x75.4G
1
1


Load per storey
At foundation level loading is obtained to be:kN81010x81Q
kN248010x248G



12. 12
(4) wind loadsheightmkNxp
mkNw
oz /54365.1
/5.1
,
2


At foundation level moment is obtained to be:kNm200,432/40x54M
2
o,y

Due to symmetric each of the 4 shear walls gets the
same amount of momentkNmMM
oyy
800,104/200,434/
,2,


13. 13
(b) Required reinforcement
(i)Step 1 –neglecting slenderness effects, viz. no
additional moments;
(a)Load cases
Safety factors for:
Dead + wind load (Load case I)

g
= 0.9
w
= 1.4
Dead + imposed + wind load (Load case II)

g
= 1.2
q
= 1.2 
w
= 1.2
Load combination for load case IloadwindaskNmxM
kNxN
y


120,154.1800,10
232,29.02480

14. 14
Load combination for load case IIkNmxM
kNxxN
y
960,122.1800,10
39482.18102.12480


(b) First estimation of the reinforcement
Load case I:2
2262
23
800,106000200009.0
90.0/100
/10.26000200/10120,15/
/86.16000200/102232/
mmxxA
chartsthefrombhA
mmNxxbhM
mmNxxbhN
s
s





15. 15
Load case II:2
2262
23
60006000200005.0
5.0/100
/80.16000200/10960,12/
/29.36000200/103948/
mmxxA
bhA
mmNxxbhM
mmNxxbhN
s
s



 2
s mm800,10A
Provide: 40 Y 25 = 19,600 mm
2
> 10,800mm
2
Hence load case I decisive,

16. 16
(ii) Step 2 –taking account of slenderness effects
(a) Load case I –first trial
N = 2,232 kN M = 15,120 kNm
A
s
= 40 Y 25 = 19,600mm
2
Bending about the major axis, h
x
= 6m; l
ex
= 2 x 40 = 80m
Bending about the minor axis, h
y
= 0.20m; l
ey
= 0.9 x 4.0 = 3.60mkN527,1910x)410x600,19x75.025x6000x200x45.0(N
kNm4.77)
20.0
60.3
0035.01()
20.0
60.3
(
1750
20.0x2232
0M
kNm416,16)
0.6
80
0035.01()
0.6
80
(
1750
00.6x2232
120,15M
3
Ux
2
ty
2
tx





17. 17114.0
527,19
2232
N
N
Ux
 00.1
n kNm320,2210x6000x200x1.3M
62
Ux 
 84.010.074.0)
774
4.77
()
320,22
416,16
(
mm/N86.1
6000x200
10x2232
bh
N
kNm74410x200x6000x1.3M
00.100.1
2
3
62
Uy




The reinforcement provided fulfills the requirements; it is,
however, a little bit too uneconomical. Hence, we try it once
more with A
s= 32 Y 25!
The reader may show that now the requirements are
perfectly fulfilled.
Provide: 32 Y 25 = 15,680mm
2

18. 18
(b) Load case II –First trial
N = 3,948 kNM = 12,960 kNm A
s
= 32 Y 25 = 15,680 mm
2
The reader may show that the control equation leads to
0.996 < 1.00 !
General recommendation for a practical design:
For a reinforced concrete wall the reinforcement required is
comparatively high. It is much more economical and from
an engineering point of view better, to increase the
thickness of the R.C. wall on say t = 0.30 m.

19. 19
(iii) Step iii: Check of side reinforcement
Pursuant to Cl. 3.11.8.2 (BS 8110), with s
b
= 250 mmmm8.10
425
200x250
f
bs
)mm(
y
b

Provide: Y 12 –250 , which meets the requirements

20. 206
0
7
x
3
0
2
5
0
2
5
0
Y
1
2
-
2
5
0
c
/
c
Y
1
0
-
3
0
0
c
/
c
1
6
Y
2
5
4
4
Y
1
2
-
2
5
0
c
/
c
6
0
0
0
200 Figure 3:Sketch of reinforcement details to the shear wall

21. Analysis of Irregular shapes
21

22. 22

23. 23

24. 24

25. 25

26. 26

27. 27

28. 28

29. 29
example

30. 30

31. 31

32. 32

33. 33

34. 34

35. 35
Stress x
area x no.
of bars
(3.14 X 19 X 19)/4
AREA

36. 36

37. 37

38. 38

39. 39

40. 40

41. 41

42. 42

43. 43

44. 44