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, EC2 models shear behaviour on 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.

**SOLVED EXAMPLE**

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 mm

^{2}.

**Design Data**:

Concrete Strength = 35 N/mm

^{2}

Grade of steel = 460 N/mm

^{2}

Width of beam = 400mm

Depth of beam = 600mm

Effective depth = 543mm

Area of tension steel provided at section A

_{s1}= 4825 mm

^{2}

**SHEAR DESIGN ACCORDING EC2**

**Support A**; V

_{Ed}= 500.46 KN

V

_{Rd,c}= [C

_{Rd,c}.k.(100ρ

_{1}f

_{ck})

^{(1/3)}+ k

_{1}.σ

_{cp}]b

_{w}.d ≥ (V

_{min}+ k

_{1}.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/mm

^{2}

ρ

_{1}= As/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, Ac = 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.v

_{1}.f

_{cd}) / (cotθ + tanθ)

V

_{1}= 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/mm

^{2}

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 BY BS 8110-1:1997**

**Design of support A**

**Ultimate shear force at centerline of support**

V = 500.46 KN

Using the shear force at the centreline of support;

Shear stress

*v*= V/(bd ) = (500.46 × 10

^{3}) / (400 × 543) = 2.304 N/mm

^{2}

*v*(2.304 N/mm

^{2}) < 0.8√Fcu (4.732 N/mm

^{2}). Hence, dimensions of the cross-section is adequate for shear.

Concrete resistance shear stress

v

_{c}= 0.632 × (100As/bd)

^{1/3}(400/d)

^{1/4}

^{}(100As/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 value as 1.0

v

_{c}= 0.632 × (2.221)

^{1/3}× 1.0 = 0.8245 N/mm

^{2}

^{}For Fcu = 35 N/mm

^{2}, v

_{c}= 0.8245 × (35/25)

^{1/3}= 0.922 N/mm

^{2}

Let us check;

(Vc + 0.4)1.322 N/mm

^{2}< v(2.304 N/mm

^{2}) < 0.8√Fcu(4.732 N/mm

^{2})

Therefore, provide shear reinforcement links.

Let us try 2 legs of Y10mm bars (Area of steel provided = 157 mm

^{2}

_{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%.

Good job engineer. Make it more elaborate next time so that one can follow the steps all the way without the need for a textbook on the side. God bless you.

ReplyDeleteThank you. This is more of a solved example. If you need more understanding on basic design theories and procedures, you can contact me directly.

DeleteI may also post on that in the nearest future. God bless.

• Just wanted to let you know and say very great job on your blog. I for one agree with what you are saying and hope to see more of your posts in the near future. BIM Consulting

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