A uniformly loaded propped cantilever beam and its free body diagram are shown below. The reactions are

R_{1} = \frac{5 q L}{8}, R_{2} = \frac{3 q L}{8}, M = \frac{- q L^{2}}{8}

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R_{1} = \frac{3 q L}{8}, R_{2} = \frac{5 q L}{8}, M = \frac{q L^{2}}{8}

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R_{1} = \frac{5 q L}{8}, R_{2} = \frac{3 q L}{8}, M = 0

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R_{1} = \frac{3 q L}{8}, R_{2} = \frac{5 q L}{8}, M = 0

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Solution

The correct option is A $R_{1} = \frac{5 q L}{8}, R_{2} = \frac{3 q L}{8}, M = \frac{- q L^{2}}{8}$
The given propped cantilever beam can be assumed to be consisting of two type of loads. First part is simple cantilever with $U D L$ and the second part is cantilever beam with point load of $R_{2}$ at end.
Balancing the forces, we get
$R_{1} + R_{2} = q L$ .....(1)

Balancing the deflection at end point as net deflection at the end is zero.
Deflection at $B$ due to the $U D L$ alone,
$δ_{B} = \frac{q L^{4}}{8 E I}$
Deflection at $B$ due to the $R_{2}$ alone,
$δ_{B}^{^{'}} = \frac{R_{2} L^{3}}{3 E I}$
$∴ δ_{B} = δ_{B}^{^{'}}$
$\Rightarrow \frac{q L^{4}}{8 E I} = \frac{R_{2} L^{3}}{3 E I} \Rightarrow R_{2} = \frac{3 q L}{8}$
From eq. (i). $R_{1} = q L - R_{2}$
$= q L - \frac{3 q L}{8} = \frac{5 q L}{8}$
$M o m e n t : M = R_{2} L - q L \times \frac{L}{2}$
$= \frac{3 q L^{2}}{8} - \frac{q L^{2}}{2} = - \frac{q L^{2}}{8}$

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