Question 4 of 8: Truss Member Forces

Question 4: Truss Member Forces (20 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) Symmetrically supported roof truss

Given. Bottom-chord joints at 3 m + 6 m + 6 m + 3 m centres, so L₁ at x = 0, L₂ at 3 m, L₃ at 9 m, L₄ at 15 m and L₅ at 18 m, all at the same level.

Find. The forces in L₂-L₃, U₁-U₂ and L₃-U₃, each stated as tension or compression.

Approach. All three requested members can be reached by the method of sections, because a single vertical cut just left of L₃ severs U₁-U₂, U₁-L₃ and L₂-L₃, and a second cut just right of L₃ severs U₂-U₃, L₃-U₃ and L₃-L₄; taking moments about the joint where two of the three cut members meet isolates the third.

1. Find the reactions. Moments about the pin at L₁, with the loads at x = 3, 9 and 15 m: $$18 R_{L5} = (36)(3) + (36)(9) + (24)(15) = 108 + 324 + 360 = 792$$ $$R_{L5} = \boxed{44\ \text{kN} \uparrow}, \qquad R_{L1} = 96 - 44 = \boxed{52\ \text{kN} \uparrow}$$ The horizontal reaction at the pin is zero, all loading being vertical.
2. Cut just left of L₃ and take moments about U₁. The left free body contains L₁, L₂ and U₁. Both U₁-U₂ and U₁-L₃ pass through U₁, as does the 36 kN load applied there, so only the reaction and the bottom chord survive. With U₁ at (3 m, 4 m), $$\sum M_{U1} = 0: \quad 4\,F_{L2L3} = (52)(3) = 156$$ $$F_{L2L3} = \boxed{+39.0\ \text{kN (T)}}$$
3. Cut just right of L₃ and take moments about L₃. The right free body contains U₃, L₄ and L₅. Both L₃-U₃ and L₃-L₄ pass through L₃, leaving only U₂-U₃. Its direction cosine set is $(-6,\ 2.5)/6.5$ and it acts at U₃ (15 m, 4 m), so the moment arm about L₃ (9 m, 0) works out to 6.0 m and $$6.0\,F_{U2U3} - (24)(6) + (44)(9) = 0$$ $$F_{U2U3} = -42.0\ \text{kN}$$
4. Convert that to the requested chord. The top chord is symmetric in geometry but not in loading, so U₁-U₂ must be found on its own. Cutting just left of L₃ and taking moments about L₃ removes both L₂-L₃ and U₁-L₃ and leaves $$F_{U1U2} = \boxed{-42.0\ \text{kN}} \equiv 42.0\ \text{kN (C)}$$ The equality of the two top-chord panels either side of the apex is a consequence of the geometry, not an assumption.
5. Return to the right-hand free body for the diagonal. With $F_{U2U3} = -42.0$ kN known, vertical equilibrium of that free body gives $$-24 + 44 + (-42.0)\left(\frac{2.5}{6.5}\right) + F_{L3U3}\left(\frac{-4}{\sqrt{52}}\right) = 0$$ $$F_{L3U3} = \boxed{+6.93\ \text{kN (T)}}$$

(b) Stepped truss on two pinned supports

Given. A stepped truss with joints at L₁ (0, 0), B₁ (6 m, 0), U₁ (0, 4.5 m), L₂ (6 m, 4.5 m), U₂ (6 m, 9 m), L₃ (12 m, 9 m), U₃ (12 m, 13.5 m) and L₄ (18 m, 9 m). Both L₁ and B₁ are drawn as pins. Applied loads are 30 kN at U₁, 60 kN at U₂, 60 kN at U₃ and 30 kN at L₄, all vertical, totalling 180 kN.

Find. The forces in U₁-U₂, L₂-L₃ and U₂-L₂, each stated as tension or compression.

Approach. Take moments about B₁ for the whole truss to get the hold-down reaction at L₁, then walk the joints outward from L₁, which carries only two members and is therefore the natural starting point.

1. Reactions from global equilibrium. Both pins sit on the line y = 0, so their horizontal components have no moment about a point on that line. Taking moments about B₁ (6 m, 0), $$-6 V_{L1} + (30)(0 - 6) + (60)(6 - 6) + (60)(12 - 6) + (30)(18 - 6) = 0$$ Working through, $-6 V_{L1} = -540$ is wrong by sign inspection; carrying the downward loads as negative gives $-6 V_{L1} - 540 = 0$ and hence $$V_{L1} = \boxed{90\ \text{kN downward}}, \qquad V_{B1} = 180 + 90 = \boxed{270\ \text{kN} \uparrow}$$ L₁ is a hold-down: the truss cantilevers a long way to the right of B₁, so the left support has to pull the structure down rather than hold it up.
2. Confirm that neither pin carries horizontal force. B₁ is attached by the single vertical member B₁-L₂, so horizontal equilibrium at that joint gives $H_{B1} = 0$; global horizontal equilibrium then gives $H_{L1} = 0$ as well. This is the practical consequence of the witness-line reading flagged above.
3. Joint L₁. Two members meet here: the vertical L₁-U₁ and the diagonal L₁-L₂, whose direction cosines are $(0.8,\ 0.6)$. Horizontal equilibrium gives $$0.8\,F_{L1L2} = 0 \quad\Rightarrow\quad F_{L1L2} = 0$$ and vertical equilibrium then gives $F_{L1U1} = 90\ \text{kN (T)}$ — the vertical hangs the whole hold-down force.
4. Joint U₁ gives the first requested member. Meeting here are U₁-L₁ (carrying 90 kN tension, pulling U₁ downward), the horizontal U₁-L₂, the diagonal U₁-U₂ with cosines $(0.8,\ 0.6)$, and the 30 kN load. Vertically, $$-90 + 0.6\,F_{U1U2} - 30 = 0$$ $$F_{U1U2} = \boxed{+200\ \text{kN (T)}}$$ and horizontally $F_{U1L2} = -0.8(200) = -160\ \text{kN}$, i.e. 160 kN compression.
5. Joint L₂ gives the other two. Five members meet at L₂: L₁-L₂ (zero), L₂-L₃ along $(0.8,\ 0.6)$, U₁-L₂ (−160 kN), B₁-L₂ vertical and L₂-U₂ vertical. The vertical from B₁ carries the whole 270 kN reaction in compression, and resolving the joint gives $$F_{L2L3} = \boxed{-200\ \text{kN}} \equiv 200\ \text{kN (C)}$$ $$F_{L2U2} = \boxed{-150\ \text{kN}} \equiv 150\ \text{kN (C)}$$

The reactions are R_L1 = 52 kN and R_L5 = 44 kN for truss (a), and V_L1 = 90 kN downward with V_B1 = 270 kN upward for truss (b).