Determine the value of shear force at point A in the figure shown below. Awl/6Bwl/2Cwl/3DwIAnswer: C. wl/3 Read Explanation: Taking moment about B: RAL=12WL×2L3⇒RA=WL3R_A L=\frac12WL\times \frac {2L} {3} \Rightarrow R_{A} =\frac{WL}{3}RAL=21WL×32L⇒RA=3WL RB=12WL−RA=12WL−WL3=WL6R_{B} =\frac12 WL - R_{A} = \frac12WL- \frac{W L}{ 3} = \frac{WL}{ 6}RB=21WL−RA=21WL−3WL=6WL 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}Fx=−RB+21Lwx×x=−6WL+2Lwx2 At x=L:FA=WL3 x = L : F_{A} =\frac {WL}{3}x=L:FA=3WL Read more in App