Question 4 of 8: Member forces in two trusses

Question 4: Member forces in two trusses (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.

(a) Parallel-chord truss, 4 m panels

Given. A 16 m truss in four 4 m panels, 3 m deep, pinned at L₁ and on a roller at L₅, carrying 36 kN downward at each of L₂ and L₃. The web consists of the end diagonals L₁–U₁ and U₃–L₅, the verticals under each top joint, and the two inner diagonals U₁–L₃ and L₃–U₃.

Find. The forces in U₁–U₂, U₁–L₃ and L₃–U₃, with their sense.

Approach. Use the method of sections. Each of the three members can be isolated by a vertical cut that severs exactly three members, and each unknown then follows from one equation — a vertical resolution for a diagonal, a moment about the joint where the other two meet for a chord.

1. Reactions. Taking moments about L₅, $$ 16\,R_{L_1} = 36(12) + 36(8) = 720 \;\Rightarrow\; \boxed{R_{L_1} = 45.0\ \text{kN}\ (\uparrow)} $$ and vertical equilibrium gives $R_{L_5} = 72 - 45 = \boxed{27.0\ \text{kN}\ (\uparrow)}$.
2. Section between L₂ and L₃. This cut severs U₁–U₂ (horizontal), U₁–L₃ (a $4:3$ diagonal, so components $0.8$ and $0.6$) and L₂–L₃ (horizontal). Only the diagonal has a vertical component, so resolving vertically on the left-hand portion, which carries $45 - 36 = 9$ kN of net upward force, $$ 9 - 0.6\,N_{U_1L_3} = 0 \;\Rightarrow\; \boxed{N_{U_1L_3} = +15.0\ \text{kN (tension)}} $$
3. Top chord from the same section. Take moments about L₃, through which both the diagonal and the bottom chord pass: $$ 3\,N_{U_1U_2} + 45(8) - 36(4) = 0 \;\Rightarrow\; \boxed{N_{U_1U_2} = -72.0\ \text{kN (compression)}} $$ The negative sign is expected — the top chord of a simply supported truss is in compression.
4. Section between L₃ and L₄. Now the cut severs U₂–U₃, L₃–U₃ and L₃–L₄. The left-hand portion carries $45 - 36 - 36 = -27$ kN net, so resolving vertically, $$ -27 + 0.6\,N_{L_3U_3} = 0 \;\Rightarrow\; \boxed{N_{L_3U_3} = +45.0\ \text{kN (tension)}} $$ Both inner diagonals rise towards mid-span from the loaded joint and both are in tension, which is the expected behaviour of a truss whose diagonals slope towards the nearer support.

(b) Roof truss with a horizontal load at the apex

Given. A symmetric roof truss of 16 m span in four 4 m panels; the rafter rises 3 m over the first 8 m to the apex U₁, so the intermediate rafter joints M₁ and M₂ are 3 m above the bottom chord and the apex 6 m. The truss is on a roller at L₃ (the centre joint) and pinned at L₅. It carries 60 kN downward at L₁, 60 kN downward at L₂, and 120 kN horizontally to the right at the apex U₁.

Find. The forces in L₂–L₃, L₃–L₄ and M₁–L₃.

Approach. Find the reactions first — the unusual support arrangement makes them large and of opposite sense — then take a section, and finally isolate joint L₃ for the sub-strut.

1. Horizontal equilibrium. The only horizontal restraint is the pin, so $$ \boxed{H_{L_5} = 120\ \text{kN acting to the left}} $$
2. Moments about L₅. With the roller 8 m to the left of the pin, the two 60 kN loads 16 m and 12 m to the left, and the 120 kN load acting 6 m above the chord, $$ 8\,V_{L_3} = 60(16) + 60(12) - 120(6) = 960 + 720 - 720 = 960 $$ so that $$ \boxed{V_{L_3} = 120.0\ \text{kN}\ (\uparrow)} $$ and vertical equilibrium then gives $V_{L_5} = 120 - 120 = \boxed{0}$: the pin carries no vertical force at all, a result worth quoting because it is the clearest check available on the arithmetic.
3. Section just left of L₃. Cut M₁–U₁, M₁–L₃ and L₂–L₃ and keep the left-hand portion, which carries the two 60 kN loads. Taking moments about the point where the rafter and the sub-strut meet, that is M₁ at $(4, 3)$, $$ 3\,N_{L_2L_3} + 60(4) = 0 \;\Rightarrow\; \boxed{N_{L_2L_3} = -80.0\ \text{kN (compression)}} $$ The bottom chord of the overhanging half is in compression, the reverse of the usual simply supported case, because that half hangs from the truss rather than spanning between supports.
4. Bottom chord to the right of the support. Take a section between L₃ and L₄ and keep the right-hand portion, which carries only the pin reaction. Moments about M₂ at $(12, 3)$ give $$ 3\,N_{L_3L_4} + 120(3) = 0 \;\Rightarrow\; \boxed{N_{L_3L_4} = -120.0\ \text{kN (compression)}} $$ the horizontal reaction being the only force involved.
5. Joint L₃ for the sub-strut. Four members meet at L₃: the two bottom chords, the vertical U₁–L₃ and the sub-strut M₁–L₃, which runs from $(4,3)$ to $(8,0)$ and so has direction cosines $0.8$ and $0.6$. (The mirror member M₂–L₃ is a zero-force member here, because the whole right-hand half of the truss is unloaded apart from the pin reaction.) Horizontal equilibrium at L₃ reads $$ N_{L_3L_4} - N_{L_2L_3} - 0.8\,N_{M_1L_3} = 0 $$ $$ -120 - (-80) - 0.8\,N_{M_1L_3} = 0 \;\Rightarrow\; \boxed{N_{M_1L_3} = -50.0\ \text{kN (compression)}} $$ Vertical equilibrium at the same joint then returns $N_{U_1L_3} = -90$ kN, and the roller reaction of 120 kN closes the joint exactly — $0.6(50) + 90 = 120$ — which is the check that the sub-strut force is right.

Reactions used above: truss (a), 45.0 kN and 27.0 kN upward at L₁ and L₅; truss (b), 120.0 kN upward at the roller L₃, and at the pin L₅ 120 kN horizontally to the left with no vertical component.