In the given figure, a UDL load of 0.5 T/m is applied on the beam. Find the maximum positive shear force. 

Explanation:

To find the reactions of support A and B.

\(R_A+R_B=15\times0.5=7.5T/m\)

By equating the moment about B

\(R_A\times50=0.5\times15*(\frac{15}{2}+40) \)

\(R_A=7.125T/m\)

So, \(R_B=0.375T/m\)

Shear force calculation

F_free = 0 T

F_A(L) = -2.5 T

FA(R) = -2.5 + 7.125 = 4.625 T

FX(L) = 4.625 - (0.5 x 10) = -0.375 T

FX(R) = -0.375 T

FB(L) = -0.375 T

FB(R) = -0.375 + 0.375 = 0 T

So the maximum positive shear force is 4.625 T.

Additional Information The shear force diagram for the given problem is as follows.