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Find the support reaction at B & the load P, for beam shown in fig. If the reaction at support A is zero.
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written 8.4 years ago by
teamques10
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Converting triangular load into point load of 36 KN. Also $VA=0 $
Now, applying conditions of equilibrium
$\sum Fx= 0$ (no force acting in x direction)
$\sum FY= VA - 40 - 36 + RB - P = 0 \\ RB - P = 76 \text {_____}(ii) \\ \sum MFA \cong 0 [ +ve] \\ 20 + 40(2) + 36(5) - RB(6) + P(9) = 0 \\ -6RB + 9P = - 280 \text {_____} (iii) \\ RB = 134.67 KN \\ P = 58.67 KN$
The support reaction at B is 134.67 KN & the load P is $58.67KN$
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