| written 7.8 years ago by | • modified 3.9 years ago |
Subject: Structural Analysis 1
Topic: Axial Force, Shear Force and Bending Moment
Difficulty: High

| written 7.8 years ago by | • modified 3.9 years ago |
Subject: Structural Analysis 1
Topic: Axial Force, Shear Force and Bending Moment
Difficulty: High

| written 7.8 years ago by | • modified 7.6 years ago |
$\begin{align} &\sum M_A=0\ \ (\circlearrowright+ve) \\&8\times3\times\frac{3}{2}-V_B\times6=0 \\&\boxed{ \therefore V_B=6\ kN} (\uparrow) \\ \\&\sum F_Y=0\ (\uparrow +ve) \\&V_A-8\times3+6=0 \\& \boxed{\therefore V_A=18\ kN}(\uparrow) \\ \\&\text{Since B is an internal hinge consider part (BC) } \\& BM_B=0\ \ \ [\circlearrowright|\circlearrowleft+ve] \\& 6\times3-H_B\times4=0 \\&H_B=4.5\ kN \\& \boxed{H_A=H_B=4.5\ kN}\ [\text{because symmetrical} ]\end{align}$
1. Consider pat …