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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 4.6 years ago by |
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$\Sigma M_{A} = 0 +ve$
5 x 1.5 + 7 x 3.5 - $R_{B} \times 4 = 0$
$R_{B} = 8kN$
$\Sigma F_{y} +ve,$
$R_{A} + R_{B} - 5 - 7= 0$
$R_{A} = 12 - 8 = 4kN$
S.F. Calculations +ve
$S.F_{A} = 4kN$
$S.F_{C} (left) = 4kN$
$S.F_{C} (right) 4 - 5 = -1kN$
$S.F_{D} (left) = -1kN$
$S.F_{D} (right) = - 1 - 7 = -8kN$
$S.F_{B} = -8kN$
B.M. Calculations +ve
$B.M_{A} = B.M_{B} = 0$
$B.M_{c} = 4 \times 1.5 = 6KN$
$B.M_{D} = 4 \times 3.5 - 5 \times 2$
$= 4 kN-m$