Shear Wall Final Design Example

Assoc.Prof.Dr. Emre AKIN

Shear Wall Final Design Example

• Hw=15 m = 15000 mm

Materials: C25 (fctd=1.2 MPa), S420

Base Shear (Vt): 1500 kN

∑Ag=1.6 m2 ∑Ap=800 m2

• Final Design: The minimum reinforcement according to TEC (2019) will be determined and the corresponding N-M interaction (capacity) diagram will be obtained. The (N, M) demand couples will be compared with the N-M interaction diagram to assess safety (capacity-demand check).

According to the regulations of seismic code:

• Since Hw/lw=15000/2000=7.5>2, we have to constitute boundary elements at both ends of the shear wall.

• The critical height: (2lw=4000 mm)≥ Hcr ≥ max[lw=2000 ; Hw/6=2500 mm]. Selected Hcr: 2500 mm=2.5 m.

• Length of boundary element (assumed to be equal at both ends; lu1=lu2): • Within the critical height: lu≥(2bw=500 mm) and lu≥(0.2lw=400 mm). Selected lu=500 mm=0.5 m

• Above the critical height: lu≥(bw=250 mm) and lu≥(0.1lw=200 mm). Selected lu=250 mm=0.25 m

Assoc.Prof.Dr. Emre AKIN

ADU Civil Eng. Dept.

bw

=25

0 m

m

lw=2000 mm

The dimensions determined from preliminary analysis:

lu1 lu2

Boundary element at the left end

Boundary element at the right end

Reinforcement of the Main Body

• Within the critical height: Amain body=0.25m × (2m-2×0.5m)=0.25 m2

• Above the critical height: Amain body=0.25m × (2m-2×0.25m)=0.375 m2

• Check: 𝐴𝑔

𝐴𝑝=

1.6

800= 0.002 ≥ 0.002 satisfied

𝑉𝑡

𝐴𝑔=1500 𝑘𝑁

1.6 𝑚2 = 937.5𝑘𝑁

𝑚2 ≥ 0.5𝑓𝑐𝑡𝑑 = 0.5 × 1.2 𝑀𝑃𝑎 = 0.6 𝑀𝑃𝑎 = 600𝑘𝑁

𝑚2 satisfied

• Within the critical height: Av=0.0020 × 0.25 m2=0.00050 m2=500 mm2 Select 8ϕ14 (A=1232 mm2)

• Above the critical height: Av=0.0020 × 0.375 m2=0.00075 m2=750 mm2 Select minimum:12ϕ14 (A=1847 mm2)

• Ah=750 mm2. If we use ϕ12,750

𝜋×122 4=1000

𝑠→ 𝑠 = 151 𝑚𝑚. Select: ϕ12/150 mm

Assoc.Prof.Dr. Emre AKIN

ADU Civil Eng. Dept.

Then Av (and Ah)=0.0020. Amain body

sv (and sh)≤300 mm

(if at least one of these conditions is not satisfied we should use 0.0025. Amain body).

lu=0.25 m lu=0.25 mlmain body=1.5 m

sv=250 mm<300 mm

250 mm 250 mm 125 mm

125 mmBoundary

Element

Boundary Element

250 mm 250 mm

Note: The cross-sectional areas are selected to be conservative to satisfy the maximum spacing condition. Here, we might have selected s=300 mm (instead of 250 mm) as the upper limit.

lu=0.5 m lu=0.5 mlmain body=1 m

sv=250 mm<300 mm

250 mm250 mm

125 mm

125 mmBoundary

ElementBoundary Element

bw

=25

0 m

m

Reinforcement of the Boundary Elements

Longitudinal Reinforcement:

• Within the critical height: ≥ [(4ϕ14 =616 mm2); (0.002bwlw=0.002×0.25m×2m=0.001 m2=1000 mm2)].

Select 8ϕ14 =1232 mm2

• Above the critical height: ≥ [(4ϕ14 =616 mm2); (0.001bwlw=0.002×0.25m×2m=0.0005 m2=500 mm2)].

Select 6ϕ14 =924 mm2

Transverse (Lateral) Reinforcement: (ϕ12 reinforcement will be used as in the horizontal reinf. of the main body)

• Above the critical height: 50 mm ≤ s ≤ [(200 mm) ; (bw=250 mm)]. Select ϕ12/200 mm

• Within the critical height: 50 mm ≤ s ≤ [(150 mm) ; (6ϕ=6×12=72 mm) ; (bw/3=250/3=83 mm)]. Select ϕ12/70 mm

Assoc.Prof.Dr. Emre AKIN

ADU Civil Eng. Dept.

Reinforcement of the Boundary Elements

• Within the critical height:

• Above the critical height:

Assoc.Prof.Dr. Emre AKIN

ADU Civil Eng. Dept.

lu=0.5 m lu=0.5 mlmain body=1 m

8ϕ14

bw

=25

0 m

m

8ϕ14

ϕ12/70 mm ϕ12/150 mm ϕ14/250 mm<150 mm

Additional ties,

10 pieces/m2

lu=0.25 m lu=0.25 mlmain body=1.5 m

6ϕ14 6ϕ14

ϕ12/150 mm ϕ14/250 mmϕ12/200 mm<150 mm

Additional ties,

4 pieces/m2

N (kN)

M (kN.m)

(Nd1, Md1)Safe! (Nd1, Md1)

Not Safe!

Note: If you find any

such result, you should

change your design!

N-M Interaction

diagram of the

section should be

obtained.

From Analysis:

(Nd1, Md1) : normal force and bending moment couple from load combination#1

(Nd2, Md2) : normal force and bending moment couple from load combination#2