This article presents a worked example on the design of shear wall to EN 1992-1-1-2004 (Eurocode 2). The shear walls to be designed form the major stabilizing system of a 10-storey building. They are positioned in pairs in the two orthogonal directions of the building as shown in the plan below.

 

The four shear walls are identical, they are 6m in length and 200mm in thickness. Each floor of the building is also 200mm thick and supports an additional permanent load and variable load of 1KN/m² and 4KN/m² respectively. The floor-to-floor height of the building is 3m, which is also the height of the shear walls between successive floors. The shear walls are subjected to lateral load from wind, axial load and transverse moment from floors. The materials for construction of the shear wall are concrete of grade 30/37MPa and steel reinforcement of characteristics tensile strength of 500 MPa.

 

Analysis

Since the shear walls are identical and also symmetrically distributed in a square plan, then the actions on each shear wall are the same. A single shear wall shall be analysed and then designed; the result shall be replicated for the other remaining three walls.

The analysis and design shall be done for the section of the shear wall at ground floor.  

Permanent Load

Unit weight of slab = 0.2 x 25 = 5KN/m²

Additional dead load on slab = 1KN/m²

Distributed Permanent load from single slab to a wall = (5 + 1) x 3 x 6 = 108KN

Permanent Load supported by base of shear wall = No of floors supported by base x Distributed Permanent load from single slab to a wall

Permanent Load supported by base of shear wall = 10 x 108 = 1080KN

Variable Load

The variable loads are gravity floor load and lateral wind load

Variable floor load

Variable load on slab = 4KN/m²

Distributed floor load from single slab to a wall = (4) x 3 x 6 = 72KN

Impose Load supported by base of shear wall = No of floors supported by base x Distributed imposed floor load from single slab to a wall

Imposed floor Load supported by base of shear wall = 10 x 72 = 720KN

Wind Load

The wind load on the building has been analysed in a previous article which has been published here. Wind is exerting a load of 0.82KN/m² from the base of the building to a height of 18m, while a load of 1.0KN/m² is exerted on the remaining upper 12m of the building as shown in the figure below.

However, to reduce computational effort, it shall be conservatively assumed that a load of 1.0KN/m2 is uniformly exerted on the building.

Total horizontal load on the building = 1 x 18 x 30 = 540KN

The shear walls are assumed to resist all horizontal loads. And since the shear wall are symmetrically placed, each pair of shear wall along an axis resist half of the lateral wind load parallel to that axis.

The wind load to be resisted by a single shear wall = 540/2 = 270KN

In-plane Moment due to wind at the base of shear wall = (270 x 30)/2 = 4050KNm

Load Combination

Load combination shall be according to EN 1990:2002 + A1:2005. Ideally, the most critical of load combination of 6.10, 6.10a, and 6.10b should be used. However, to reduce computational effort load combination shall only be according to equation 6.10 and. A1.1, and A1.2(B) of the standard.

Case 1)

Dead + Imposed as leading variable + Wind as accompanying variable

γGj, sup = 1.35, γQ, 1 = 1.5, γQ, i = 1.5, ψ0, i = 0.6

Ultimate Axial Load (N_Ed) = 1.35 x 1080 + 1.5 x 720 = 2538KN

Ultimate Moment due to wind Load = 1.5 x 0.5 x 4050 = 3645KNm

Case 2)

Dead + Wind as leading variable + Imposed as accompanying variable

γGj, sup = 1.35, γQ, 1 = 1.5, γQ, i = 1.5, ψ0, i = 0.7

Ultimate Axial Load (N_Ed) = 1.35 x 1080 + 1.5 x 0.7 x 720 = 2214KN

Ultimate Moment due to wind Load = 1.5 x 4050 = 6075KNm

Transverse Moment

The shear wall is analysed for transverse moment using the subframe below:

Case 1:   Assume variable actions is a leading action

n = 1.35 gk + 1.5 qk = 1.35 x 6 + 1.5 x 4 = 14.1KN/m²

Fixed End Moment (M) =   n x l²/ 12

To convert n to KN/m we shall multiply it by the width of the slab supported by the wall

M = 14.1 x 6 x 6² / 12 = 253.8KNm

To know the moment distributed to the lower wall, we shall calculate the stiffness of each member of the subframe

I (for wall and slab) = bh3/12 = (6 x 0.2³)/12 = 0.004

K_slab = 4 E I/L = 4 x E x 0.004/6 = 0.003E

K_wall = 4EI/L = 4 x E x 0.004/3 = 0.005E

Transversed moment in wall = 253.8 x  $\,\,\,\,\frac{0.005EI}{0.005EI x 2\,\,+\,\,0.003EI}\,\,=\,\,0.333
$  = 97.6KNm

Transverse Moment per meter length of wall = 97.6/6 = 16.3KNm/m

 

Case 2:           Assume variable actions is accompanying action

n = 1.35 gk + 1.5 x 0.7 x qk =  = 12.3KN/m²

Fixed End Moment (M) =   n x l²/ 12

To convert n to KN/m we shall multiply it by the width of the slab supported by the wall

M = 12.3 x 6 x 6² / 12 = 221.4KNm

Transversed moment in wall = 221.4 x $\,\,\,\,\frac{0.005EI}{0.005EI x 2\,\,+\,\,0.003EI}\,\,=\,\,0.333
$  = 85.15KNm

Transverse Moment per meter length of wall = 81.15/6 = 13.53KNm/m

 

 

Design

For the design of shear wall, we shall consider the most onerous of each effect of action for both load combination cases considered. These onerous effects are enumerated below.

Axial load = 2538KN (load case 1)

In-plane Moment due to Wind = 6075KNm (Load case 2)

Transverse Moment = 16.3KNm/m (Load case 1)

Determination of extreme fibre stress on the wall

The extreme fibre stress in the wall due to the combination of in-plane moment from wind action and axial force from floor loads can be computed using:

$f_t=\frac{N_{E d}}{L h} ± \frac{6 M_{E d}}{h L^2}$

$f_t=\frac{2538 x 10³}{6000 x  200} ± \frac{6 x 6075 x 10^6}{200 x 6000^2}$

f_t = 2.1 ± 5.0

f_t = 7.1N/mm² or -2.9N/mm²

Force per length of the wall equals:

F = 7.1 x 200 = 1420KN/m

 

Determination of Design Transverse Moment in the wall

The design transverse moment in the wall is computed as though the wall were a column of 1m length. The analytical moment is 16.3KNm. We shall proceed to compute the design moment.

1. Determine the Slenderness of the wall

We check the slenderness of the wall

Effective length, lo (using figure 5.7 of the standard) = 0.7l

l_o = 0.7 x (3000 – 200)

l_o = 1960

slenderness (λ) = l_o/r

where r is the radius of gyration

$ r\,\,=\,\,\sqrt{\frac{I}{A}}\,\,=\,\,\sqrt{\frac{\frac{bh^3}{12}}{b\,\,X\,\,h}}\,\,=\,\,\sqrt{\frac{\frac{\text{1000}X\,\,200^3}{12}}{\text{1000}X\,\,200}}\,\,\,\,=\,\,57.74 $

•  Calculate the slenderness ratio of the wall

λ = l_o/r   = 1960/57.74      =     33.95

 

• Calculate the limiting slenderness

λ_lim = 20 X A X B X C /√n

A = 0.7, B = 1.1, C = 0.7 (All these assumptions are based on alternative suggestion by the code, check clause 5.8.3.1)

n = N_Ed /A_cf_cd 

$n\,\,\,\,=\,\,\frac{1420 x 10^3}{1000 x 200 x 30/1.5}
$   = 0.49

λ_lim = 20 X 0.7 X 1.1 X 0.7 /√0.49     =    15.4

Since λ_lim (15.4) is less than λ (33.95) then the wall is slender and second-order moment required.

          2. Calculate the second-order moment

 Since the column is slender, second-order moment due to slenderness is necessary.

Second-order moment (M₂) = N_Ed x e₂

Calculate deflection due to slenderness (e₂)

e₂ = (1/r ). l_o²/c

c = 10 (see 5:8:8:2 (4) of EC2)

e₂= 1/r x 1960^2/10

 

Calculate the curvature (1/r)

 1/r = K_r· K_ϴ. 1/r_o

Calculate axial load correction factor (Kr)

K_r = (n_u – n)/ (n_u -n_bal) ≤ 1

n_u = 1 + ω

ω = A_s f_yd/A_cf_cd

Assume A_s to be 600mm2

f_yd = 500/1.15 = 434.8

f_cd = 0.85 x 30/1.5 = 17

ω = (600 x 434.8)/ (1000 x 200 x 17) = 0.08

n_u = 1+ 0.08 = 1.08

n_bal = 0.4 (check clause 5.8.8.3. (4))

n = N_Ed /A_c X f_cd   = 0.49

K_r = (1.08 – 0.49)/ (1.63 -0.4) = 0.48

 

Calculate creep factor (K_ϴ)

K_ϴ = 1 + B φ_ef   ≥ 1

B = 0.35 + f_ck/200 – ⅄/150

= 0.35 + 30/200 – 33.95/150 = 0.27

φ_ef  = 2.14  (This conservative value is gotten by making ϕef the subject of formular in Equ 5.13N of Eurocode 2, when A is assumed to be 0.7)

K_ϴ = 1+ 0.27 x 2.14 = 1.58

 

Calculate 1/r_o

1/r_o = f_yd/(E_s x 0.45d)

434.8/ (200 x 10³ x 0.45 x 200) = 0.00002

1/r = 0.48 x 1.58 x 0.00002 = 0.00002

e₂ = 0.00002 x 1960² /10 = 7.68mm

 

Second -order Moment (M₂ ) = 1420 x 7.68/1000 = 10.9KNm

 

• Compute the design moment (M_Ed)

The design moment is the greatest of:

1. max (M₀₂, M₀₁) + N_Ed x max (e)
2. M_0e+ M₂
3. M₁ + M₂/2

Where:

M₀₂ = 16.3KN

M₀₁= 0

M_0e = 0.4 M02 + 0.6 M01 = 6.52KNm

M₂ = 10.9KNm

 

1. max (M₀₂, M₀₁) + N_Ed x max (e)

maximum eccentricity (e) is max (lo/400, l/30, 20mm)

lo/400 = 1960/400 = 4.9mm

l/30 = 1000/30 = 33.3mm

33.3mm is taken as the maximum eccentricity.

= 16.3 + 1420 x 33.3/1000 = 63.6KNm

 

2. Moe+ M2 = 6.52 + 10.9 = 17.42KNm

 

3. M1 + M2/2 = 0 + 10.9/2 = 5.45KNM

The design moment is 63.6KNm

 

 

Determination of Area of required Reinforcement using Design Chart

We can now then proceed to use column design chart to provide the necessary required reinforcement for the wall

 In order to use the design chart, few parameters have to be calculated

Determine N_ED/bhfck

N_ED/bhfck = 1420X 10^3 /1000 X 200 X 30   = 0.24

 

Determine M_ED/bh²fck

M_ED/bh²fck =    63.6 X 10^6/1000 X 200^2 X 30   = 0.05

Calculate d’/h

d’/h determines the exact graph that shall be used to determine the reinforcement area.

d’ = c + d/2 = 25 + 25/2 = 37.5

d’/h   = 37.5/200 = 0.19 = 0.2

This is approximated to 0.2 and the design chart peculiar to d’/h = 0.2 is used. Fig 9d of “How to design concrete Structures to Eurocode 2” matches this. The chart is reproduced below.

 

From the design chart, the intersection of m/bh2fck = 0.05 and N/bhfck = 0.24 falls below where area of reinforcement is required, hence minimum area of reinforcement should be provided as vertical reinforcement

Vertical Reinforcement

Minimum Area of Reinforcement

A_smin = 0.002Ac

A_smin = 0.002 x 1000 x 200 = 400mm²

 

Maximum Spacing

Maximum spacing = Min (3 x wall thickness, 400mm)

3 x 200 = 600 which is lesser than 400.

Take the spacing between vertical bars to be 250mm

Provide 12mm @ 250c/c (452mm²) at each face of the wall as vertical reinforcement

Horizontal Reinforcement

Recommended value = max (25% of the vertical reinforcement, 0,001 Ac)

Max ((25/100) x 905, 0.001 x 1000 x 200)

Max (226mm², 200)

Area of rebar to be provided = 226mm², hence provide 113mm² on each face.

The spacing between two adjacent horizontal bars should not be greater than 400mm.

Provide 10mm@300mm spacing

 

Transverse reinforcement (links)

Check whether the vertical reinforcement in the wall exceeds 2% of the gross section of the wall

Area beyond which links are required = 0.02 x 200 x 1000 = 4000mm2

Vertical reinforcement in wall section = 2 x 452 = 904mm2

As the vertical steel is one layer only and the percentage does not exceed 2%, transverse reinforcement is not required.