Question

a beam AB of span 12m as shown in figure
it is hinged A and is on rollers at B . determine the reaction at A and as shown in figure ​

Answer 1

Answer:

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Answer 2

Answer:

The magnitude of reaction at A is = 26.6kN

Explanation:

Free body diagram of forces is given below:

As the system is in equilibrium,

moment about point A = 0

$\Sigma M = 4\times 10+(6\times 15cos(30))+(10\times 20cos(45))-(12\times R_R)=0$

$\implies 40+90\frac{\sqrt{3}}{2}+200\frac{1}{\sqrt{2} } =12R_R$

$\implies R_R=\frac{40+45\sqrt{3}+100\sqrt{2} }{12}$

$\implies R_R=21.61kN$

As the system is in equilibrium, force to left side = to the right side

$R_H=15sin(30)+20cos(45)=\frac{15}{2}+\frac{20}{\sqrt{2}}$

$R_H=21.64kN$

As the system is in equilibrium, upward force= downward force

$R_V+R_R=10+15cos(30)+20cos(45)$

$\implies R_V=10+15cos(30)+20cos(45)-R_R$

$\implies R_V=10+15\frac{\sqrt{3}}{2} +20\frac{1}{\sqrt{2} }-21.61$

$\implies R_V=15.52kN$

The reaction at A = vector sum of $R_V$ and $R_H$

The magnitude of reaction at A = $=\sqrt{(R_V)^2+(R_H)^2}$

$=\sqrt{(15.52)^2+(21.64)^2}$

$=\sqrt{709.16}$

$=26.6kN$

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