Question

Consider the following frame

The magnitude of bending moment at section x-x is ______ kNm.

• A. 80

Answer 1

The correct option is A 80
Since loading as well as frame is symmetrical.

$\therefore V_A=V_D=\frac{40\times 8}{2}=160kN$



Taking moments about E in portion AE

$(\Sigma M_E=0)$

$\therefore 40\times 4\times 2+H_A\times 4=160\times 4$

$\Rightarrow H_A=80kN$

$\therefore M_x=40\times 2\times 1+80\times 4-160\times 2$

= 80 kNm