Subject: Structural Analysis 1

Topic: Axial Force, Shear Force and Bending Moment

Difficulty: High

$\begin{align} &\sum M_A=0\ \ (\circlearrowright+ve) \\&8\times3\times\frac{3}{2}-V_B\times6=0 \\&\boxed{ \therefore V_B=6\ kN} (\uparrow) \\ \\&\sum F_Y=0\ (\uparrow +ve) \\&V_A-8\times3+6=0 \\& \boxed{\therefore V_A=18\ kN}(\uparrow) \\ \\&\text{Since B is an internal hinge consider part (BC) } \\& BM_B=0\ \ \ [\circlearrowright|\circlearrowleft+ve] \\& 6\times3-H_B\times4=0 \\&H_B=4.5\ kN \\& \boxed{H_A=H_B=4.5\ kN}\ [\text{because symmetrical} ]\end{align}$

1. Consider pat (AB):

$\begin{align} &l=\sqrt{4^2+3^2}=5\ m \\& \tan Q=\frac{4}{3} \ ; \theta=53.13 ^\circ \end{align}$

$\begin{align} &\text{A.F.D sign }[\leftarrow|\rightarrow(+ve)] \\&BM_D=7.2\times2.5-2.88\times2.5\times\frac{2.55}{2} \\&\phantom{BM_D}=\underline{9\ kNm} \end{align}$

2. Now considder part (BC):