Comparison of Shear Design to EC2 and BS 8110

Eurocode 2 uses the variable strut inclination method for shear design. It is slightly more complex than the procedure in BS 8110 but can result in savings in the amount of shear reinforcement required.

Like BS 8110, Eurocode 2 models shear behaviour using the truss analogy in which the concrete acts as the diagonal struts, the stirrups act as the vertical ties, the tension reinforcement forms the bottom chord, and the compression steel/ concrete forms the top chord. Whereas in BS 8110 the strut angle has a fixed value of 45°, in EC2 it can vary between 21.8° and 45° and it is this feature which is responsible for possible reductions in the volume of shear reinforcement.

The concrete strut angle changes with respect to the magnitude of the shear force applied. The angle decreases with a reduction in shear stress, with a minimum angle of 21.8° (where cot = 2.5) which happens to be the most frequently used design strut angle.

Shear design in Eurocode 2 often involves comparing the design shear stress (v_Ed) to the concrete shear resistance (v_Rd,c). If v_Ed < v_Rd,c then in the case of one-way spanning slabs no shear reinforcement is required; in the case of beams nominal shear reinforcement is required. v_Rd,c is dependent on the amount of design tensile reinforcement and the design concrete strength.

If v_Ed < v_Rd,max (ie v_{Rd,cot θ = 2.5}), then the shear reinforcement may be calculated as:
A_sw/s ≥ v_Ed/(0.87 × f_yk × z × cot 2.5)

If v_Ed > v_Rd,max (ie v_{Rd,cot θ = 2.5}), then checks against v_{Rd,cot θ = 1.0} and determination of θ are required.
If v_Ed > v_{Rd,cot θ = 1.0} then a larger section or greater concrete strength is required.

Where applied shear stresses are higher, one has to check against the maximum allowable shear stress at the maximum strut angle of θ = 45º (e.g. 5.28 MPa for a C30/37 concrete when cot θ = 1.0) and where necessary determine the actual intermediate strut angle, θ.

Solved Example for Shear Design

For the beam loaded as shown below and disregarding load factors, we are going to obtain the shear reinforcement required for section A. The shear force at the centreline of support has been adopted for this design, and the area of tension steel provided at this section A_s1 = 4825 mm².

Design Data:
Concrete Strength = 35 N/mm²
Grade of steel = 460 N/mm²
Width of beam = 400mm
Depth of beam = 600mm
Effective depth = 543mm
Area of tension steel provided at section A_s1 = 4825 mm²

Shear Design to EC2

Support A; V_Ed = 500.46 kN

V_Rd,c = [C_Rd,c.k.(100ρ₁ f_ck)^(1/3) + k₁.σ_cp]b_w.d ≥ (V_min + k₁.σ_cp) b_w.d

C_Rd,c = 0.18/γ_c = 0.18/1.5 = 0.12
k = 1 + √(200/d) = 1 + √(200/543) = 1.606 > 2.0, therefore, k = 1.606
V_min = 0.035k^(3/2) f_ck^0.5
V_min = 0.035 × (1.606)^1.5 × 35^0.5 = 0.421 N/mm²
ρ₁ = A_s/bd = 4825/(400 × 543) = 0.022 > 0.02; Therefore take 0.02

σ_cp = N_Ed/Ac < 0.2f_cd (Where N_Ed is the axial force at the section, A_c = cross-sectional area of the concrete), f_cd = design compressive strength of the concrete.) Take N_Ed = 0

V_Rd,c = [0.12 × 1.606 (100 × 0.02 × 35 )^(1/3)] 400 × 543 = 172511.992 N = 172.511 kN

Since V_Rd,c (172.511 kN) < V_Ed (500.46 kN), shear reinforcement is required.
The compression capacity of the compression strut (V_Rd,max) assuming θ = 21.8° (cot θ = 2.5)

V_Rd,max = (b_w.z.v₁.f_cd) / (cot⁡θ + tanθ)
V₁ = 0.6(1 – f_ck/250) = 0.6(1 – 35/250) = 0.516
f_cd = (α_cc f_ck) / γ_c = (0.85 × 35) / 1.5 = 19.833 N/mm²
Let z = 0.9d

V_Rd,max = [(400 × 0.9 × 543 × 0.516 × 19.833) / (2.5 + 0.4)] × 10^-3 = 689.83 kN

Since V_Rd,c < V_Ed < V_Rd,max

Hence A_sw / S = V_Ed / (0.87.F_yk.z.cot θ) = 500460 / (0.87 × 460 × 0.9 × 543 × 2.5 ) = 1.0235

Minimum shear reinforcement;
A_sw / S = ρ_w,min × b_w × sinα (α = 90° for vertical links)
ρ_w,min = (0.08 × √(F_ck)) / F_yk = (0.08 × √35) / 460 = 0.001028
A_sw/S_min = 0.001028 × 400 × 1 = 0.411
Maximum spacing of shear links = 0.75d = 0.75 × 543 = 407.25

Provide X10mm @ 150mm c/c as shear links (A_sw/S = 1.0467) Ok!!!!

Shear Design to BS 8110-1:1997

Design of support A
Ultimate shear force at the centerline of support
V = 500.46 kN

Using the shear force at the centreline of support;
Shear stress v = V/(bd ) = (500.46 × 10³) / (400 × 543) = 2.304 N/mm²

v(2.304 N/mm² ) < 0.8√f_cu (4.732 N/mm² ). Hence, the dimensions of the cross-section are adequate for shear.

Concrete resistance shear stress
v_c = 0.632 × (100A_s/bd)^1/3 (400/d)^1/4
(100A_s/bd) = (100 × 4825) / (400 × 543) = 2.221 < 3 (See Table 3.8 BS 8110-1;1997)
(400/d)^1/4 = (400/543)^1/4 = 0.926; But for members with shear reinforcement, this value should not be less than 1. Therefore take the value as 1.0

v_c = 0.632 × (2.221)^1/3 × 1.0 = 0.8245 N/mm²
For f_cu = 35 N/mm², v_c = 0.8245 × (35/25)^1/3 = 0.922 N/mm²

Let us check;
(V_c + 0.4)1.322 N/mm² < v(2.304 N/mm²) < 0.8√f_cu (4.732 N/mm²)

Therefore, provide shear reinforcement links.
Let us try 2 legs of Y10mm bars (Area of steel provided = 157 mm²

A_sv/Sv = (0.95 × F_yv) / (b_v (v – v_c)) = [400 × (2.304 – 0.922)] / (0.95 × 460) = 1.264

Maximum spacing = 0.75d = 0.75 × 543 = 407.25 mm
Provide Y10mm @ 100 mm c/c links as shear reinforcement (A_sv/Sv = 1.57)

We can therefore see that disregarding load factor and flexural design requirements, EC2 is more economical than BS 8110 in shear design, and in this case study by about 19%.