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Determine the value of shear force at point A in the figure shown below.

image.png

Awl/6

Bwl/2

Cwl/3

DwI

Answer:

C. wl/3

Read Explanation:

Taking moment about B:

RAL=12WL×2L3RA=WL3R_A L=\frac12WL\times \frac {2L} {3} \Rightarrow R_{A} =\frac{WL}{3}

RB=12WLRA=12WLWL3=WL6R_{B} =\frac12 WL - R_{A} = \frac12WL- \frac{W L}{ 3} = \frac{WL}{ 6}

Fx=RB+12wxL×x=WL6+wx22L F_{x} = - R_{B} +\frac 12 \frac{wx}{L} \times x = -\frac {WL}{6} + \frac{w x ^ 2}{2L}

At x=L:FA=WL3 x = L : F_{A} =\frac {WL}{3}


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