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$Old$ March 19th, 2009, 07:09 PM

Celia $Celia is offline$

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Solve the reactions for the beam shown below. The applied moment acts as show in the clockwise direction.

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#2

$Old$ March 20th, 2009, 04:24 AM

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Hello Celia

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Solve the reactions for the beam shown below. The applied moment acts as show in the clockwise direction.

Suppose that the reactions at A and B are $R_A$ and $R_B$ respectively. Then, take moments about A:

$M(A):14 \times 3 + 50 = R_B \times 10$

$\Rightarrow R_B = \frac{92}{10} = 9.2$

$M(B): R_A \times 10 + 50 = 14\times 7$

$\Rightarrow R_A = \frac{48}{10} = 4.8$

You can check by resolving vertically that $R_A+R_B = 14$

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