A simply supported beam of 4 m length carries two point loads of 5 kN and 7 kN at 1.5 m and 3.5 m from the left hand support respectively. Draw SFD and BMD.

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written 5.6 years ago by

teamques10 ★ 69k

modified 2.9 years ago by

RakeshBhuse • 3.2k

strength of materials

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written 5.6 years ago by

teamques10 ★ 69k

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$\Sigma M_{A} = 0 +ve$

5 x 1.5 + 7 x 3.5 - $R_{B} \times 4 = 0$

$R_{B} = 8kN$

$\Sigma F_{y} +ve,$

$R_{A} + R_{B} - 5 - 7= 0$

$R_{A} = 12 - 8 = 4kN$

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S.F. Calculations +ve

$S.F_{A} = 4kN$

$S.F_{C} (left) = 4kN$ …

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