A rolled I - section beam is supported on a 70 mm wide bearing plate as shown in the figure. The thickness of flange and web of the I - section are 22 mm and 10 mm, respectively. Root radius of the I - section is 10 mm. Assume material yield stress, $f_{y} = 250 MPa$ and partial safety factor for material $γ_{m 0} = 1.1$ .
As per IS 800 : 2007, the web bearing strength of the beam will be

375 kN

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309.91 kN

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340.91 kN

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231.82 kN

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Solution

The correct option is C 340.91 kN
Web bearing strength $= [b + 2.5 (t_{f} + R)] \times t_{w} \times \frac{f_{y}}{γ_{m 0}}$
$= [70 + 2.5 (22 + 10)] \times 10 \times \frac{210}{1.1} \times 10^{- 3} k N$
$= 340.91 k N$

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The correct option is C 340.91 kNWeb bearing strength =[b+2.5(tf+R)]×tw×fyγm0 =[70+2.5(22+10)]×10×2101.1×10−3kN =340.91 kN

The correct option is C 340.91 kN
Web bearing strength $= [b + 2.5 (t_{f} + R)] \times t_{w} \times \frac{f_{y}}{γ_{m 0}}$
$= [70 + 2.5 (22 + 10)] \times 10 \times \frac{210}{1.1} \times 10^{- 3} k N$
$= 340.91 k N$