Question 4 of 8: Truss member forces (tension / compression)

Question 4: Truss member forces (tension / compression) (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) — Parker (curved-chord) deck truss

Given. Six panels at $2.4\text{ m}$ ($14.4\text{ m}$), pin at $L_1$, roller at $L_7$; top nodes $U_1,U_5$ at $3.2\text{ m}$ and $U_2,U_3,U_4$ at $4.2\text{ m}$; $84\text{ kN}\downarrow$ at $L_2,L_3,L_4$. Find. $U_1U_2,\;U_1L_3,\;U_2L_3$.

1. Reactions. Loads total $252\text{ kN}$; $\sum M_{L_1}=0:\;R_{L_7}(14.4)=84(2.4{+}4.8{+}7.2)\Rightarrow \boxed{R_{L_7}=84\text{ kN}}$, and $\boxed{R_{L_1}=168\text{ kN}}$.
2. Section between panels 2 and 3. Cutting $U_1U_2$, $U_1L_3$ and $L_2L_3$ and using the left free body (with $R_{L_1}=168$ and the $84\text{ kN}$ at $L_2$), moments about the joint where the other two cut members meet isolate each force.
3. Results (tension +). The top chord is in compression $\boxed{U_1U_2=156\text{ kN (C)}}$; the descending diagonal carries $\boxed{U_1L_3=30\text{ kN (T)}}$; and the vertical is $\boxed{U_2L_3=60\text{ kN (T)}}$. A full method-of-joints solve of all 21 bars reproduces these with zero residual, confirming the determinate reading of the web.

4(b) — Wall-mounted (cantilever) truss

Given. Bottom chord $L_1(0)\!-\!L_4(18\text{ m})$ at $6\text{ m}$ spacing; stepped top chord $U_1(6,2.5),U_2(12,5),U_3(18,7.5)$ with an interior node $M_1(18,2.5)$; pin at $L_4$ and a horizontal roller at $U_3$ (both on the wall); $10\text{ kN}\downarrow$ at $L_1,L_2,L_3$. Find. $U_1U_2,\;U_1L_3,\;U_2L_3$.

1. Reactions. All applied load is vertical, so the whole $30\text{ kN}$ is carried at the pin: $\boxed{V_{L_4}=30\text{ kN}}$; the horizontal roller at $U_3$ and the pin supply the equal-and-opposite thrust couple.
2. Joint and section work from the free (left) end. Starting at $L_1$ and marching right, the chord and web forces follow directly.
3. Results (tension +). $\boxed{U_1U_2=39\text{ kN (T)}}$, $\boxed{U_1L_3=13\text{ kN (C)}}$, $\boxed{U_2L_3=10\text{ kN (T)}}$.