A uniformly distributed load of 50 kN/m and 5 m long crosses a girder of span 15 m from left to right. What will be the maximum bending moment at a section 6 m from left end?

450 kNm

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750 kNm

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900 kNm

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600 kNm

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Solution

The correct option is B 750 kNm

For maximum bending moment, load position should be such that the section divides the load in the same ratio as it divides the spam.
$∴ \frac{x}{6} = \frac{5 - x}{9}$

$\Rightarrow x = 2 m$

$∴$ Maximum bending moment = w × area under ILD under the UDL.

$= 50 \times [\frac{24 + 3.6}{2} \times 2 + \frac{3.6 + 2.4}{2} \times 3] = 750 k N m$

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A uniformly distributed load of 50 kN/m and 5 m long crosses a girder of span 15 m from left to right. What will be the maximum bending moment at a section 6 m from left end?

The correct option is B 750 kNm For maximum bending moment, load position should be such that the section divides the load in the same ratio as it divides the spam. ∴x6=5−x9 ⇒x=2m ∴ Maximum bending moment = w × area under ILD under the UDL. =50×[24+3.62×2+3.6+2.42×3]=750kNm