Subject : Structural Analysis 1

Topic : Axial force, Shear force and Bending moment

Difficulty : Medium

$\sum M_A = 0 (\circlearrowright + ve)$

$8 X 3 X \frac{3}{2} - V_B X 6 = 0$

$\boxed{V_B = 6KN \uparrow}$

$B M _B = 0$

Gives,

$6 X 3 - H_B X 4 = 0$

$H_B = 4.5 KN$

$\boxed {H_A = H_B = 4.5 KN}$

Because of symetrical

$\sum F_y = 0 (\uparrow + ve)$

$V_A - 8 x 3 + 6 = 0$

$\boxed{V_A = 18KN \uparrow}$

Consider Part (AB)

$BM_D = 7.2 X 2.5 - 2.88 X 2.5 X \frac {2.5}{2}$

$ B M_D = 9 KN m $

Now consider Part (BC)