written 6.9 years ago by | • modified 2.9 years ago |
Subject: Structural Analysis 1
Topic: Axial Force, Shear Force and Bending Moment
Difficulty: High
written 6.9 years ago by | • modified 2.9 years ago |
Subject: Structural Analysis 1
Topic: Axial Force, Shear Force and Bending Moment
Difficulty: High
written 6.9 years ago by | • modified 6.7 years ago |
$\begin{align} &\sum M_A=0 (\circlearrowright +ve) \\&-V_D\times4+(20\times4)\times+10\times5=0 \\&\boxed{V_D=52.5\ kN} \\ \\& \sum F_Y=0\ (\uparrow +ve) \\& V_A+V_D=80=10=0 \\& \boxed{V_A=37.5\ kN} \\ \\ &\sum F_X=0 \ (\rightarrow+ve) \\&H_A=H_B=0 \end{align}$
1.Consider part AB:
2.Consider part BCE:
$\begin{align} &\textbf{1. SF calculation: } [\uparrow | \downarrow +ve] \\& SF_{BL} =0 \\& SF_{BR} =37.5\ kN \\& SF_{CL} =37.5-(20\times4)=\underline{-42.5\ kN} \\& SF_{CR} =10 \ kN \\& SF_{EL} =10 \ kN \\& SF_{ER} =0 \\ \\&\textbf{2. BM calculation: } [\circlearrowright | \circlearrowleft +ve] \\& BM_{B} =0 \\& BM_{E} =0 \\& BM_{C} =-10\times1=\underline{-10\ kN} \\&\boxed{ BM_x=M_{max}} \\& \text{To find x, } \frac{37.5}{x}=\frac{42.5}{4-x} \\& \boxed{x=1.875 \ m} \\& BM_x=M_x=37.5\times1.875-20\times1.875\times\frac{1.875}{2} \\&\boxed{BM_{max}=35.156\ kNm} \end{align}$
3.Consider part 'CD'