Question 4 of 8: Truss member forces (method of sections)

Question 4: Truss member forces (method of sections) (18 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

4(a) Warren truss, 90 kN at $L_3$

Given. Bottom chord $L_1(0,0)\,L_2(8,0)\,L_3(16,0)\,L_4(24,0)$; top $U_1(4,3)\,U_2(12,3)\,U_3(20,3)$; pin $L_1$, roller $L_4$; $90\text{ kN}$ down at $L_3$. Find. $L_2L_3,\,U_2L_3,\,U_3L_3$.

1. Reactions. $\sum M_{L_1}=0:\;R_{L_4}(24)=90(16)\Rightarrow R_{L_4}=60\text{ kN}$; $R_{L_1}=30\text{ kN}$.
2. Section cutting $U_2U_3,\,U_2L_3\;(=U_3L_3\text{ region})$ and $L_2L_3$. Taking the left free body ($R_{L_1}=30$) and moments about $U_2\,(12,3)$ eliminate the two members through $U_2$: $L_2L_3(3)=30(12)\Rightarrow L_2L_3=\boxed{+120\text{ kN (T)}}$.
3. Vertical & joint equilibrium. The web diagonals carry the shear: $U_2L_3=\boxed{+50\text{ kN (T)}}$ and the end diagonal $U_3L_3=\boxed{+100\text{ kN (T)}}$ (both diagonals rise toward the loaded panel, so both pull).

4(b) Cantilever truss

Given. $U_1(0,0)\,U_2(4,0)\,U_3(8,0)$ (top), $L_1(0,-6)$, $L_2(4,-3)$; roller at $U_1$ (horizontal reaction), pin at $L_1$; loads $90\text{ kN}\downarrow$ at $U_2$, and $90\text{ kN}\downarrow + 40\text{ kN}\rightarrow$ at $U_3$. Find. $U_1L_2,\,U_1U_2,\,L_1L_2$.

1. Reactions. $\sum M_{L_1}=0:\;-6U_{1x}=90(4)+90(8)+40(6)\Rightarrow U_{1x}=-220\text{ kN}$ (leftward). Then $L_{1x}=180\text{ kN}$, $L_{1y}=180\text{ kN}$.
2. Joint / section. Resolving at the joints gives $U_1U_2=\boxed{+160\text{ kN (T)}}$ (top chord tension of the cantilever), the web $U_1L_2=\boxed{+75\text{ kN (T)}}$, and the bottom chord $L_1L_2=\boxed{-225\text{ kN (C)}}$.