Question 4 of 8: Member forces in two trusses by the method of sections

Question 4: Member forces in two trusses by the 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.

Given. Two determinate trusses.

Find. Six named member forces, each with its sense stated as tension or compression.

Approach. Take reactions from overall equilibrium, then pass a section through no more than three unknown members and choose the moment centre at the intersection of the two members that are not wanted, so each unknown falls out of a single equation.

1. Part (a) — reactions. Moments about L₁ for the whole truss give $$R_{L5}(24) = 90(6) + 60(12) + 30(18) = 1800 \;\Rightarrow\; R_{L5} = 75.0\ \text{kN}\ (\uparrow)$$ $$R_{L1} = 180 - 75.0 = 105.0\ \text{kN}\ (\uparrow)$$
2. Force in L₁L₂, from joint L₁. Only the end rafter and the bottom chord meet at the pin, and the rafter rises 8 m in 6 m so its length is 10 m. Resolving vertically then horizontally, $$105.0 + N_{L1U1}\left(\frac{8}{10}\right) = 0 \;\Rightarrow\; N_{L1U1} = -131.25\ \text{kN}$$ $$N_{L1L2} = -N_{L1U1}\left(\frac{6}{10}\right) = \boxed{78.75\ \text{kN tension}}$$
3. Force in U₁L₃, from cut 1-1. Cutting between L₂ and L₃ severs U₁U₂, U₁L₃ and L₂L₃. Taking moments about L₃, through which both U₁L₃ and L₂L₃ pass, isolates the top chord. The member U₁U₂ rises 4.5 m in 6 m, length 7.5 m, and its perpendicular distance from L₃ is 10 m, so for the left-hand piece $$-105.0(12) + 90(6) - 10\,N_{U1U2} = 0 \;\Rightarrow\; N_{U1U2} = -\frac{1260 - 540}{10} = -72.0\ \text{kN}$$ Vertical equilibrium of the same piece then releases the diagonal: $$105.0 - 90 + \left(\frac{4.5}{7.5}\right)(-72.0) - \left(\frac{8}{10}\right)N_{U1L3} = 0$$ $$N_{U1L3} = \frac{15.0 - 43.2}{0.8} = \boxed{-35.25\ \text{kN, i.e. } 35.25\ \text{kN compression}}$$
4. Force in U₂U₃, from cut 2-2. Cutting between L₃ and L₄ and working with the right-hand piece, moments about L₃ again eliminate the diagonal and the bottom chord: $$-30(6) + 75.0(12) + 10\,N_{U2U3} = 0 \;\Rightarrow\; N_{U2U3} = -\frac{900 - 180}{10} = \boxed{-72.0\ \text{kN, i.e. } 72.0\ \text{kN compression}}$$ That the two top-chord panels carry the same force is a coincidence of this particular load pattern, not a symmetry: the loads are 90, 60 and 30 kN and the truss is not symmetrically loaded. It is nonetheless a useful check, because both were obtained from independent free bodies.
5. Part (b) — recognise why both supports can be pins. Counting the second truss, there are 18 members, 11 joints and 4 reaction components, so $m + r = 18 + 4 = 22 = 2j$: it is determinate. It can only be so because the bottom chord is interrupted between L₂ and L₃. That gap is the whole point of the arrangement — the truss behaves as a two-pinned arch-like frame in which the horizontal thrust is carried by the supports rather than by a tie.
6. The condition equation the gap provides. Pass a vertical section anywhere between $x = 12$ m and $x = 20$ m. Because the bottom chord is missing over that stretch, only two members are cut, U₃U₄ and U₃L₃, and both of them radiate from the joint U₃ at $(12,\ 18)$. Moments about U₃ for the left-hand piece therefore contain neither of them, nor the 72 kN load that sits at U₃ itself: $$\sum M_{U3} = 0:\quad -12\,V_{L1} + 16\,H_{L1} = 0 \;\Rightarrow\; V_{L1} = \tfrac{4}{3}\,H_{L1}$$ which says that the reaction at L₁ is directed along the line L₁–U₃, a rise of 16 m in a run of 12 m.
7. Complete the reactions. Moments about L₅ at $(36,\ 0)$ for the whole truss, with L₁ at $(0,\ 2)$, give $$-36\,V_{L1} - 2\,H_{L1} + 72(24) + 72(16) + 90(8) + 90(0) = 0$$ $$-36\left(\tfrac{4}{3}H_{L1}\right) - 2H_{L1} + 3600 = 0 \;\Rightarrow\; 50\,H_{L1} = 3600$$ $$\begin{aligned} H_{L1} &= \boxed{72.0\ \text{kN} \rightarrow}, \\ V_{L1} &= \tfrac{4}{3}(72.0) = \boxed{96.0\ \text{kN} \uparrow} \end{aligned}$$ $$\begin{aligned} \sum F_x &= 0:\ H_{L5} = 72.0\ \text{kN} \leftarrow; \\ \sum F_y &= 0:\ V_{L5} = 324 - 96.0 = 228.0\ \text{kN} \uparrow \end{aligned}$$ The vertical components sum to 324 kN, matching the applied 72 + 72 + 90 + 90 kN, and the horizontal components are equal and opposite as they must be.
8. Force in L₂U₃. Joints U₁, U₂ and L₁ carry no applied load, and L₁, L₂ and U₃ are collinear — each step is 6 m across and 8 m up — so L₁L₂U₃ acts as one straight strut 10 m per panel. Resolving horizontally at the pin L₁, $$72.0 + N\left(\frac{6}{10}\right) = 0 \;\Rightarrow\; N = -120.0\ \text{kN}$$ and vertical equilibrium at the same joint, $96.0 - 120.0(8/10) = 0$, confirms it while proving that the end post U₁L₁ is unstressed. Hence $$N_{L2U3} = \boxed{120.0\ \text{kN compression}}$$
9. Force in L₃L₄. Cut between L₃ and L₄, severing U₄U₅, L₃U₅ and L₃L₄, and work with the right-hand piece. The first two meet at U₅ $(28,\ 18)$, so moments there leave only the bottom chord, whose perpendicular offset from U₅ is 9.6 m: $$90(8) - \left[228.0(8) - 72.0(18)\right] + 9.6\,N_{L3L4} = 0$$ $$720 - 528 + 9.6\,N_{L3L4} = 0 \;\Rightarrow\; N_{L3L4} = \boxed{-20.0\ \text{kN, i.e. } 20.0\ \text{kN compression}}$$
10. Force in L₃U₅. From the same cut, the top chord U₄U₅ lies on $y = 18$ and the bottom chord L₃L₄ on the line through $(20,\ 12)$ falling 6 m in 8 m; these intersect at $(12,\ 18)$, the joint U₃. Taking moments about that point for the right-hand piece, with the diagonal again 9.6 m off the centre, $$-90(16) - 90(24) + \left[228.0(24) - 72.0(18)\right] - 9.6\,N_{L3U5} = 0$$ $$-3600 + 4176 - 9.6\,N_{L3U5} = 0 \;\Rightarrow\; N_{L3U5} = \boxed{60.0\ \text{kN tension}}$$ Both members run 8 m across and 6 m vertically, so each is 10 m long. Joint L₃ provides the check: the vertical U₄L₃ delivers 72 kN of compression from above, and the vertical components of the three inclined members meeting there — 24.0 kN from U₃L₃, 36.0 kN from L₃U₅ and 12.0 kN from L₃L₄ — add to exactly 72 kN upward.