Question 4 of 8: Truss Member Forces

Question 4: Truss Member Forces (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) Three-panel truss with a single panel-point load

Given. Bottom chord $L_1L_2L_3L_4$ at 8 m centres (span 24 m), top chord $U_1U_2U_3$ at 3 m above it with $U_1$ 4 m in from $L_1$ and the top panels also 8 m. Pin at $L_1$, roller at $L_4$. A single downward load of 90 kN acts at $L_3$. Every web member spans 4 m horizontally and 3 m vertically, so each is exactly 5 m long — a 3–4–5 triangle.

Find. The forces in $L_2L_3$, $U_2L_3$ and $U_3L_3$, each with its sense.

Approach. Reactions by global moments, then one vertical section between $U_2$ and $L_3$ (which severs exactly three members) for the first two forces, and joint equilibrium at $L_3$ for the third.

1. Determinacy and reactions. Eleven members, seven joints, three reaction components: $m + r = 14 = 2j$, determinate. Taking moments about $L_1$ with the load 16 m along the 24 m span, $$R_{L_4} = \frac{90(16)}{24} = \boxed{60.0\ \text{kN}\ \uparrow},\qquad R_{L_1} = 90 - 60 = \boxed{30.0\ \text{kN}\ \uparrow}$$
2. Section for $L_2L_3$ — moments about $U_2$. Cut vertically between $U_2\,(12,3)$ and $L_3\,(16,0)$; the cut severs $U_2U_3$, $U_2L_3$ and $L_2L_3$. The first two both pass through $U_2$, so taking moments there for the left-hand portion leaves only the bottom chord, whose lever arm is the truss depth: $$\sum M_{U_2} = 0:\qquad 30.0(12) = F_{L_2L_3}(3)$$ $$F_{L_2L_3} = \frac{360}{3} = \boxed{120.0\ \text{kN\ (T)}}$$ The bottom chord of a simply supported truss under gravity load is in tension, as expected.
3. Same section for $U_2L_3$ — vertical equilibrium. Both chords cut by this section are horizontal, so only the diagonal has a vertical component. With the diagonal running $(4,-3)$ from $U_2$ to $L_3$, its vertical direction cosine is $3/5$: $$\sum F_y = 0:\qquad 30.0 - \tfrac{3}{5}F_{U_2L_3} = 0$$ $$F_{U_2L_3} = \tfrac{5}{3}(30.0) = \boxed{50.0\ \text{kN\ (T)}}$$
4. Joint $L_3$ for $U_3L_3$. Four members meet at $L_3$: the two horizontal chords, the diagonal down from $U_2$ and the diagonal up to $U_3\,(20,3)$. The chords are horizontal, so vertical equilibrium of the joint involves only the two diagonals and the 90 kN load: $$\tfrac{3}{5}(50.0) + \tfrac{3}{5}F_{L_3U_3} = 90.0$$ $$F_{L_3U_3} = \frac{90.0 - 30.0}{0.6} = \boxed{100.0\ \text{kN\ (T)}}$$
5. Check by horizontal equilibrium at $L_3$. With $F_{L_2L_3} = 120.0$ T, $$-120.0 + F_{L_3L_4} - \tfrac{4}{5}(50.0) + \tfrac{4}{5}(100.0) = 0 \;\Rightarrow\; F_{L_3L_4} = 80.0\ \text{kN (T)}$$ and a section on the right-hand side, taking moments about $U_3$, gives $60.0(4) = 3F_{L_3L_4}$, i.e. $80.0$ kN as well. The two independent routes agree, so the three requested values stand.

4(b) Cantilevered truss pinned to a wall

Given. Joints $U_1(0,6)$, $U_2(4,6)$, $U_3(8,6)$, $L_1(0,0)$ and $L_2(4,3)$, dimensions in metres. Members: the top chord $U_1U_2$ and $U_2U_3$, the vertical $U_2L_2$, the diagonal $U_1L_2$, and the straight run $L_1L_2U_3$ (which is collinear, slope $3{:}4$, so it counts as the two members $L_1L_2$ and $L_2U_3$). Both $U_1$ and $L_1$ bear on the wall. Loads: 90 kN down at $U_2$, 90 kN down at $U_3$ and 40 kN to the right at $U_3$.

Find. The forces in $U_1L_2$, $U_1U_2$ and $L_1L_2$.

Approach. Start at the joint that carries only two unknown member forces — $U_3$ — and walk inward: $U_3 \to U_2 \to L_2$, which delivers all three answers without ever needing the reactions.

1. Determinacy. Six members, five joints and four reaction components (both wall bearings are pins): $m + r = 6 + 4 = 10 = 2j$, determinate. The truss is a cantilever hung off two wall pins, which is why four reaction components do not make it indeterminate.
2. Joint $U_3$ — vertical equilibrium. Two members meet at $U_3$: the horizontal $U_2U_3$ and the inclined $L_2U_3$, the latter running $(-4,-3)$ from $U_3$ toward $L_2$. With 90 kN down applied there, $$-\tfrac{3}{5}F_{U_3L_2} - 90.0 = 0 \;\Rightarrow\; F_{U_3L_2} = -150.0 \quad \text{i.e. } 150.0\ \text{kN (C)}$$
3. Joint $U_3$ — horizontal equilibrium. Including the 40 kN horizontal load, $$-F_{U_3U_2} - \tfrac{4}{5}(-150.0) + 40.0 = 0 \;\Rightarrow\; F_{U_2U_3} = 160.0\ \text{kN (T)}$$
4. Joint $U_2$. Three members meet here: the two horizontal chord members and the vertical $U_2L_2$, with 90 kN applied down. Vertically, $$-F_{U_2L_2} - 90.0 = 0 \;\Rightarrow\; F_{U_2L_2} = 90.0\ \text{kN (C)}$$ and horizontally the two chord members must balance, so $$F_{U_1U_2} = F_{U_2U_3} = \boxed{160.0\ \text{kN\ (T)}}$$
5. Joint $L_2$ — two equations, two unknowns. Four members meet at $L_2\,(4,3)$: $L_2L_1$ in direction $(-4,-3)/5$, $L_2U_1$ in $(-4,3)/5$, $L_2U_2$ vertically up, and $L_2U_3$ in $(4,3)/5$. There is no load at $L_2$. Horizontal equilibrium gives $$-F_{L_2L_1} - F_{L_2U_1} + F_{L_2U_3} = 0 \;\Rightarrow\; F_{L_2L_1} + F_{L_2U_1} = -150.0$$ and vertical equilibrium, after substituting $F_{L_2U_2} = -90.0$ and $F_{L_2U_3} = -150.0$, gives $$-F_{L_2L_1} + F_{L_2U_1} = 300.0$$
6. Solve the pair. Adding and subtracting, $$F_{U_1L_2} = \frac{-150.0 + 300.0}{2} = \boxed{75.0\ \text{kN\ (T)}}, \qquad F_{L_1L_2} = -225.0 \;\Rightarrow\; \boxed{225.0\ \text{kN\ (C)}}$$ The long collinear run is therefore in compression next to the wall (225 kN) and in compression again beyond $L_2$ (150 kN) — the step between them is the 90 kN the vertical brings down.
7. Reaction check. Resolving the member forces at the two wall pins gives $L_1: (180.0, 135.0)$ kN and $U_1: (-220.0, 45.0)$ kN. Globally $\sum F_y = 135.0 + 45.0 = 180.0$ kN, matching the two 90 kN loads, and $\sum F_x = 180.0 - 220.0 + 40.0 = 0$. Equilibrium closes.