EXAMPLE OF SHEAR WALL DESIGN TO EC2 (2)

EXAMPLE OF SHEAR WALL DESIGN TO EC2
Ref
Calculation
Output
The accompanying figure shows a 6 storey rc structure with 250mm thk shear walls
located along Grid 1, 10, A & D. Apart from providing vertical support to suspended
slabs from 1st floor to roof, these shear walls provide lateral stability against wind
forces in both the North-South & East-West direction. The walls are supported on
independent foundation and the ground floor is of non-suspended slab.
Given:
Floor Slab
Roof
Qk (kN/m2)
3.0
0.50
Gk (kN/m2)
1.25
1.5
Note:
i. Gk from finishes & mech. services, not incl. self wt of slab
ii. To include removable partition load of 0.75 kN/m2 in addition to the 2.5 kN/m2
as Qk on slabs
iii. Characteristic wind pressure taken as 1.2 kN/m2 uniformly distributed over the
entire height of building acting in the N-S direction
iv. Characteristic strength of concrete and steel reinforcement taken as 30 N/mm2
and 500 N/mm2 respectively
Ref
Calculation
Output
N
1
2
3
4
5
6
Z
7
8
9
10
10000
10000
A
Ref. Pt A
Cm
C
6000
Y
7000
8000
B
S
7000
6000
Ref. Pt B
15000
D
Shear Wall
9@5000 = 45000
Note:
All slabs 175 mm thk
All Shear Walls 250 mm thk
All Beams 250 x 600 mm
All Columns 400 x 400 mm
Cm= centre of mass
Wind
Ref
Calculation
Output
1.0 Shear Centre Location Y:
Typical Cross Section
6@3500 = 21000
By observation the location of the 4 shear
walls are non symmetrical about both the Z &
Y axis. The second moment of area values
(inertia in m4) of walls Grid/1 and Grid/10
about Z direction (discounting the stiffness of
walls Grid/A and Grid/D) are:
I1y =(0.25 x 203)/12 = 166.7 m4
K1y =(166.7)/(21+1) = 7.58 m3
I10y=(0.25 x 143)/12 = 57.2 m4
K10y=(57.2)/(21+1) = 2.60 m3
Thus the distance of the shear center Cs from
the reference point A:Yc =
∑𝑘𝑘𝑘𝑘𝑘𝑘
∑𝑘𝑘𝑘𝑘
Yc = {(7.58x0.0) + (2.60x45)}/(7.58 + 2.60)
= 11.5 m
Ecc e = (45/2)-11.5 = 11 m
GL
A
B
C
D
Ref
Calculation
Shear Centre Location Z:
The second moment of area values (inertia in m4) of walls Grid/A and Grid/D about Z direction
(discounting the stiffness of walls Grid/1 and Grid/10) are:
IAz =(0.25 x 203)/12 = 166.7 m4
KAz =(166.7)/(21+1) = 7.58 m3
IDz=(0.25 x 153)/12 = 70.3 m4
KDz=(70.3)/(21+1) = 3.20 m3
Thus the distance of the shear center Cs from the reference point B:Zc =
∑𝑘𝑘𝑘𝑘𝑘𝑘
∑𝑘𝑘𝑘𝑘
Zc = {(7.58x0.0) + (3.20x20)}/(7.58 + 3.20)
= 5.93 m
Ecc e = (20/2)-5.93 = 4.07 m
Output
Ref
Calculation
Output
N
4
5
Z
7
6
A
6000
Ref. Pt B
B
Ref. Pt
S
Y
7000
8000
Cm
Ecc e =4070
Cs
Mt
10
10000
10000
Yc =11500
9
8
7000
3
Zc =5930
2
1
6000
C
Ecc e =11000
D
Shear Wall
9@5000 = 45000
+Z
Cs= shear centre
Cm= centre of mass
+Y
Wind
Ref
Calculation
Output
2.0 Lateral Load Distribution
Lateral Force on wall =
𝑘𝑘
𝑘𝑘 𝑘𝑘
M
𝐹𝐹 +
∑𝑘𝑘
∑𝑘𝑘𝑘𝑘2 t
Where:
• F is the total wind load (kN) =1.2x(45x21)= 1134 kN
• Mt is the Torsional Moment about respective axis (kNm) = (1134x11.0) = 12 474 kNm
• y is the perpendicular distance between the axis of each wall and the centre of
rotation about their respective axis (m)
Wall
ky
kz
y
ky
Ky2
Fd
Ft
F
Grid 1
7.58
0
11.5
87.17
1002.46
844.37
-225.57
618.80
Grid 10
2.60
0
33.5
87.10
2917.85
289.63
+225.39
515.02
Grid A
0
7.58
5.93
44.95
266.55
0
-116.32
-116.32
Grid D
0
3.20
14.07
45.02
633.49
0
+116.50
+116.50
∑10.18
∑10.78
∑4820.4 ∑1134
∑ 0.00
∑1134
•
• Fd =
• Ft =
𝑘𝑘
𝐹𝐹 : (Wall @ Grid 1) Fd = (7.58/10.18)x1134 = 844.37 kN
∑𝑘𝑘
𝑘𝑘 𝑘𝑘
M : (Wall @ Grid 1) Ft = (87.17/4820.4)x12474 = 225.57 kN
∑𝑘𝑘𝑘𝑘2 t
Ref
Calculation
Output
SUMMARY OF LATERAL LOAD DISTRIBUTION
N
3
4
7
8
9
10
Zc =5930
6000
6
20000
Yc =11500
A
5
116.5 kN
Cs
Cm
6000
C
116.5 kN
D
15000
9@5000 = 45000
+Z
+Y
1134 kN
515 kN
619 kN
8000
B
14000
2
1
Ref
Calculation
Output
3.0 Gravity Load Distribution (Wall Grid 1)
Actions
Gk kN/m2
Qk kN/m2
Roof:
Variable
0.5
Metal cladding & trusses
0.75
Suspended ceiling & Services
0.50
∑ 1.25
∑ 0.5
Floor Slab:
Variable
2.5
Finishes
0.50
Suspended ceiling & Services
0.50
Removable partition
Self wt 175mm thk slab
0.75
4.375
∑ 5.375
∑ 3.25
Ref
Calculation
Actions
Output
Gk kN/m2
Qk kN/m2
Ground Floor Slab: (Ground
Bearing)
Variable
2.5
Finishes
0.50
Services
0.35
Removable partition
Self wt 175mm thk slab
Actions
0.75
4.375
∑ 5.225
∑ 3.25
Gk kN/m2
Qk kN/m2
Self wt Wall @250mm thk
6.25
Self wt Brick Wall @ 115mm thk
2.60
Wind Load Wk
1.2
Ref
Calculation
Output
4.0 Load Take Down (Wall Grid 1)
Roof
Gk:(5/2x20)(1.25)
Wall
Gk:(3.50x20)(6.25)
Qk:(20x0.5)(5/2)
Qk kN
Gk kN
62.50
25.00
437.50
∑ 500
∑500
@above 5th floor
5th FL
Gk:(5/2x20)(5.375)
Wall
Gk:(3.50x20)(6.25)
∑ 25
Qk:(5/2x20)(3.25)
268.75
162.5
437.50
∑ 706.25
Gk:(5/2x20)(5.375)
Wall
Gk:(3.50x20)(6.25)
∑ 162.5
∑1206.25
@above 4th floor
4th FL
Qk:(5/2x20)(3.25)
268.75
∑ 187.5
162.5
437.50
∑ 706.25
@above 3rd floor
∑ 25
∑ 162.5
∑1912.5
∑ 350
Ref
Calculation
Output
Qk kN
Gk kN
3rd FL
Gk:(5/2x20)(1.25)
Wall
Gk:(3.50x20)(6.25)
Qk:(20x0.5)(5/2)
62.50
162.5
437.50
∑ 706.25
∑2618.75
@above 2nd floor
2nd FL
Gk:(5/2x20)(5.375)
Wall
Gk:(3.50x20)(6.25)
∑ 162.5
Qk:(5/2x20)(3.25)
268.75
162.5
437.50
∑ 706.25
Gk:(5/2x20)(5.375)
Wall
Gk:(3.50x20)(6.25)
∑ 162.5
∑3325
@above 1st floor
1st FL
Qk:(5/2x20)(3.25)
268.75
∑ 675
162.5
437.50
∑ 706.25
@above Ground floor
∑ 512.5
∑ 162.5
∑4031.25
∑ 837.5
Ref
Calculation
Output
Qk kN
Gk kN
Grd FL
Gk:(1x20)(5.375)
Wall
Gk:(1.00x20)(6.25)
125.00
Assume 1m width slab to wall as suspended
∑232.50
@above Foundation
Qk:(1x20)(3.25)
107.50
65
∑ 65
∑4263.75
∑ 902.50
Ref
Calculation
5.0 Summary of Gravity Loads Wall Grid 1 (Ground Floor to 1st Floor)
Gk = 4031.5 kN
Gk/m = 4031.5/20 = 202 kN/m
Qk = an x 837.5
= 0.85 x 837.5
= 712 kN
where an = {2+(4-2)0.7}/4 = 0.85
Qk = 712 kN
Qk/m = 712/20 = 36 kN/m
Output
Ref
Calculation
Output
6@3.5=21m
30 kN/m
6.0 Vertical Loads from Wind Actions: Moments In-Plane
From earlier calculation, Wall Grid 1 takes aprox. 55% (619/1134=0.546) of wind load
Wk = 55% x 1.2 x 45 = 30 kN/m height
@ just above Ground Floor level In Plane
Moments := (30 x 21) x {(21/2) + 1}
= 7245 kNm/m
Using
630 kN
Wk = 6M/L2
= 6 (7245)/202
1st Fl
= 109 kN/m
Grd Fl
1m
1m
20 m
Ref
Calculation
Output
7.0 Effects of Global Imperfections (Notional Horizontal Load)
Global imperfections can be represented by forces Hi at floor level where
Hi =ɸi(Nb – Na) kN
Roof
HR
5th Fl
ɸ
H5
4th Fl
H4
3rd Fl
H3
2nd Fl
1st Fl
Grd Fl
20 m
H2
H1
Hi =ɸi(Nb – Na)
where ɸi = 1/410 (Table 3.1 pg 17)
(Nb – Na) = axial load from each level
Ref
Calculation
Output
(Nb – Na) @ each floor
Roof
Gk:(45x20)(1.25)
B/Wall
Gk:(45+20)x2(2.6)x0
Qk kN
Gk kN
Qk:(45x20)(0.5)
1125
450
0
∑ 1125
∑1125
(Nb – Na) @Roof
5th FL
Gk:(45x20)(5.375)
B/Wall
Gk:(45+20)x2(2.6)x3.075
∑ 450
Qk:(45x20)(3.25)
4837.50
2925
1039.50
∑ 5877
Gk:(45x20)(5.375)
B/Wall
Gk:(45+20)x2(2.6)x3.075
∑ 2925
∑5877
(Nb – Na) @5th floor
4th FL
Qk:(45x20)(3.25)
4837.50
∑ 2925
2925
1039.50
∑ 5877
(Nb – Na) @4th floor
∑ 450
∑ 2925
∑5877
∑ 2925
Ref
Calculation
(Nb – Na) @ each floor
3rd FL
Gk:(45x20)(5.375)
Wall
Gk:(45+20)x2(2.6)x3.075
Output
Qk kN
Gk kN
Qk:(45x20)(3.25)
4837.50
2925
1039.50
∑ 5877
∑5877
(Nb – Na) @3rd floor
2nd FL
Gk:(45x20)(5.375)
Wall
Gk:(45+20)x2(2.6)x3.075
∑ 2925
Qk:(45x20)(3.25)
4837.50
2925
1039.50
∑ 5877
Gk:(45x20)(5.375)
Wall
Gk:(45+20)x2(2.6)x3.075
∑ 2925
∑5877
(Nb – Na) @2nd floor
1st FL
Qk:(45x20)(3.25)
4837.50
∑ 2925
2925
1039.50
∑ 5877
(Nb – Na) @1st floor
∑ 2925
∑ 2925
∑5877
∑ 2925
Ref
Calculation
Output
Summary of Effects of Global Imperfections (Notional Horizontal Load)
Roof
6@3.5=21m
ɸ
5th Fl
HRGk 14.3 kN
HRQk 7.1 kN
4th Fl
HRGk 14.3 kN
HRQk 7.1 kN
3rd Fl
2nd Fl
1st Fl
Grd Fl
20 m
HRGk 2.7 kN
HRQk 1.1 kN
HRGk 14.3 kN
HRQk 7.1 kN
HRGk 14.3 kN
HRQk 7.1 kN
Hi =ɸi(Nb – Na)
where ɸi = 1/410 (Table 3.1 pg 17)
(Nb – Na) = axial load from each level
Ex:
HRGk = {1/410}{1125} = 2.74 kN
HRQk = {1/410}{450} = 1.10 kN
Characteristic design moment @ ground floor:
MkGk=(2.7x21)+{14.3x(17.5+14+10.5+7+3.5)}
= 807.5 kNm
MkQk=(1.1x21)+{7.10x(17.5+14+10.5+7+3.5)}
= 396.0 kNm
HRGk 14.3 kN
HRQk 7.1 kN
Gk = 0.55x(6x807.5/202) = ±6.7 kN/m
Qk = 0.55x(6x396/202) = ±3.3 kN/m
Ref
Calculation
Output
8.0 Design Moments Perpendicular to Wall @ 1st Floor (Transverse Moments)
For the purpose of evaluating the transverse moment, slabs of 1000mm width is assumed
framing into Wall 1. One Free Joint sub frame is used to evaluate transverse moments.
3500
Kwall upper
Stiffness K: EI/L
3.72 x105
Kwall upper= (1000x2503)/12x3500
= 3.72 x105 mm3
Mtupper
Gk & Qk
1st Floor
0.5Kslab 0.45 x105
3500+1000
Mtlower
Kwall lower 2.89 x105
5000
As Ground floor is non suspended, total height
of lower wall taken up to foundation level ie
3500+1000=4500 mm
Ref
Calculation
LC1: Qk as lead action Wk as accompanying action
N Ed = 1.35Gk +1.5Qk + 1.05Wk
Output
LC2: Wk as lead action Qk as accompanying action
N Ed = 1.35Gk +1.5Wk + 1.05Qk
LC#1=1.35Gk +1.5Qk
=1.35(5.375)+1.5(3.25)
=12.15 kN/m2
LC#1=1.35Gk +1.05Qk
=1.35(5.375)+1.05(3.25)
=10.75 kN/m2
FEM = 12.15x52/12
= 25.5 kNm/m width of slab
FEM = 10.75x52/12
= 22.5 kNm/m width of slab
LC2: 1.35Gk +1.05Qk
LC1: 1.35Gk +1.5Qk
10.5 kNm
1st Floor
13.5 kNm
Mt upper:
=(Kupper /∑K)xFEM
=(3.72/7.06)x25.5
=(0.53)x25.5 = 13.5 kNm/m
Mt lower:
=(Klower /∑K)xFEM
=(2.89/7.06)x25.5
=(0.41)x25.5 = 10.5 kNm/m
9.25 kNm
1st Floor
12 kNm
Mt upper:
=(Kupper /∑K)xFEM
=(3.72/7.06)x22.5
=(0.53)x22.5 = 12 kNm/m
Mt lower:
=(Klower /∑K)xFEM
=(2.89/7.06)x22.5
=(0.41)x22.5 = 9.25 kNm/m
Ref
Calculation
LC1: Qk as lead action Wk as accompanying action
N Ed = 1.35Gk +1.5Qk + 1.05Wk
Output
LC2: Wk as lead action Qk as accompanying action
N Ed = 1.35Gk +1.5Wk + 1.05Qk
LC2: 1.35Gk +1.05Qk
LC2: 1.35Gk +1.5Qk
10.5 kNm
1st Floor
9.25 kNm
12 kNm
13.5 kNm
1.05Wk
N Ed = 1.35Gk +1.5Qk + 1.05Wk
= 1.35(202+6.7)+1.5(36+3.3)+1.05(109)
= 455.15 kN/m @ 1st FL to Grd FL
1st Floor
1.5Wk
N Ed = 1.35Gk +1.5Wk + 1.05Qk
= 1.35(202+6.7)+1.5(109)+1.05(36+3.3)
= 486.50 kN/m @ 1st FL to Grd FL
Ref
Calculation
Output
LC3: Wk as lead action favourable vertical loads
N Ed = 1.0Gk + 1.5Wk
T
C
N Ed = 1.0Gk ± 1.5Wk
= 1.0(202+6.7) ± 1.5(109)
= 372.2 or 45.20 kN/m…………No Tension
Ref
Calculation
9.0 Slenderness Check of Wall 1 @Ground Floor to 1st Floor
Effective height Lo= 0.85 (4500-175) = 3676.25 mm
Thus λ = Lo / (I/A)0.5 = 3676.25/{(1000x2503/12)/(250x1000)} 0.5
λ = 3676.25/72.2 = 51
Output
Ref
Calculation
Output
Momen due to imperfections M imp = Ned x (Lo/400) = 486.50 x (3.676/400) = 4.5 kNm/m
λ l i m = 20 A.B.C/√n where A = 0.7, B = 1.1 & C = 1.7 – rm
rm = M01/M02 = -7.15/15 = -0.48
Therefore C – 1.7 – (-0.48) = 2.18
n = Ned/Ac x fcd = 485.50 x 103/{1000x250x(0.85x30/1.5)} = 0.1142
λ l i m = 20 A.B.C/√n = (20x0.7x1.1x2.18)/√0.1142 = 99
λ < λ l i m ………..Non Slender !!
M02 = 10.5 + 4.5 = 15kNm/m
Mmin = Ned x 0.02 = 472.25 x 0.02 = 9.75 kNm/m
Med = 15 kNm/m………Ned = 486.50 kN/m
0.25x10.5 (partially fixed)
Note that Ned obtained from LC2 where as
Med is derived from LC1
2.63 kNm/m
M01 = 2.63 + 4.5 = 7.15 kNm/m
Ref
Calculation
Output
In Summary:
Ned = 486.50 kN/m
(Gk, Qk, Wk)
M02 = 15 kNm/m
(Gk, Qk)
Transverse Moment
1st Fl - Foundation
M01 = 7.15 kNm/m
(Gk, Qk)
Ref
Calculation
Output
10.0 Section Design
10.1 Column Design Procedure (Ned 486.5 kN/m & Med 15 kNm/m)
Adopt cover to horizontal bars as 30mm, thus cover to main vertical bars = 30 +10+(16/2)
where lacer bars taken as 10mm and main vertical bars as 16mm.
Effective depth d = 250 - 48 = 202 mm
d2 = 48 mm
d2 / h = 48 / 250 = 0.192…….use Column Design Chart (d2/h = 0.2)
Ned / bhfck = 486.5 x 103 /(1000 x 250 x 30) = 0.065
Med / bh2fck = 15 x 106 / (1000 x 2502 x 30) = 0.008
}
As fyk /bh fck = 0
Provide minimum reinf……….. As = 0.002bh = 0.002x100 x250 = 500 mm2/m or
250 mm2/m @ each face
cover=48mm
Ref
Calculation
Output
10.2 Walls Subjected to Insignificant Transverse Moment (Axial Load Control)
Thus,
Asc = {(486.5x103) – (0.43 x 30 x 250)} / (0.67 x 500)
= 1443 mm2/m or 722 mm2/m each face….>>250 mm2/m @ each face
(The distance between two adjacent vertical bars should not exceed the lesser of either
three times the wall thickness or 400 mm…………..ok)
Provide:
H16-250
(804 mm2/m
@ each face)
Ref
Calculation
Output
10.3 Horizontal Reinforcement
The minimum area of horizontal reinforcement in walls is the greater of either 25% of
vertical reinforcement or 0.001 Ac. However, where crack control is important, early age
thermal and shrinkage effects should be considered explicitly.
Ashmin = 0.001Ac or 0.25 Asvert
= 0.001(1000 x 250)
or
(0.25 x 1443)
= 250 mm2/m (125 mm2/m each face) or 361 mm2/m (181 mm2/m each face)
Provide:
H10-350
(224 mm2/m
@ each face)
Ref
Calculation
Output
11.0 Horizontal Shear
Max. design ultimate horizontal shear, VEd = 1.5 x 630 = 945 kN
Therefore shear stress
vEd = 945x103/(20x103x250) = 0.19 N/mm2
Normal stress due to min permanent action, Ϭcp = Ned / A = (202+6.7)x103 / 250x1000
= 0.83 N/mm2
vrdc = vmin + k1Ϭcp
= (0.035k3/2 √fck) + k1Ϭcp
where k = 1+√(200/d) =1.99 & k1=0.15
= (0.035x1.993/2 x √30) + (0.15x0.83)
= 0.66 N/mm2 > 0.19 N/mm2 ……….OK!!
vrdc >> vEd
Ref
Calculation
Output
12.0 Check Wall Stability
30 kN/m
6@3.5=21m
Self Wt Component
Say base extends 900 mm beyond either end of wall, thus
foundation dim taken as 21.80 x 2 x 1 m (LxWxH)
630 kN
Grd Fl
1m
H
A
20 m
L= 21.80 m
ii) Global Imperfections (Gk):
= (2.7x23)+(14.3x(19.5+16+12.5+9+5.5)) = 956 kNm
Restoring Moments @ A:
={4263.75 + (0x902.5) + (21.8x2x1x25)} x (21.8/2)
= 58,356 kNm
1st Fl
1m
Overturning Moments @ A:
i) Wind:
= 30 x 21 x (21/2 + 2) = 7875 kNm
@ULS
Overturning Moment:
=(1.5x7875) + (1.1x956) = 12,864.5 kNm
FoS= 4.0
Restoring Moment:
= (0.9x58,356) = 52,520 kNm…>>12,864.5 kNm
OK!!
Ref
Calculation
Output
H10-350 EF
1st Fl
9H16-175
9H16-175
EF
EF
H16-250 EF
Ground Fl
*EF=Each Face