Question 4 of 8: Truss member forces by method of sections (20 marks)

Question 4: Truss member forces by method of sections (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) Asymmetric bowstring-type truss

Given. Bottom chord L1(0,0, pin)–L2(3,0)–L3(9,0)–L4(15,0)–L5(18,0, roller); top chord U1(3,4)–U2(9,6.5, peak)–U3(15,4). Verticals U1–L2, U2–L3, U3–L4; diagonals L1–U1, U1–L3, L3–U3, U3–L5. Point loads 36 kN at U1, 36 kN at U2, 24 kN at U3 (all downward).

Find. Forces in L2–L3, U1–U2, L3–U3.

Approach. Find the reactions from global equilibrium, then cut a section through L2–L3, U1–L3 and U1–U2 and take moments about the point where two of the three unknowns intersect to isolate the third.

1. Reactions. $\sum M_{L1}=0$: $R_{L5}(18)=36(3)+36(9)+24(15)=108+324+360=792 \Rightarrow R_{L5}=44\text{ kN}$. $R_{L1}=96-44=\boxed{52\text{ kN}}$ (both vertical).
2. L2–L3 (moment about U1, which the other two cut members pass through). Free body left of the cut carries only $R_{L1}$ at $(0,0)$ and the force in L2–L3 (horizontal, on the line $y=0$). $\sum M_{U1}=0$: $F_{L2L3}(4)=R_{L1}(3)=52(3)=156 \Rightarrow F_{L2L3}=\boxed{39\text{ kN, tension}}$.
3. U1–U2 (moment about L3, through which L2–L3 and U1–L3 both pass). $\sum M_{L3}=0$ for the same free body (now including the 36 kN load at U1): solving gives $F_{U1U2}=\boxed{42\text{ kN, compression}}$.
4. L3–U3 (method of joints at L3, using the now-known chord forces). With $F_{L2L3}=39\text{ kN (T)}$, $F_{L3L4}=33\text{ kN (T)}$, and $F_{U1L3}=-0.277\text{ kN}$ from the same section, joint equilibrium at L3 closes with $F_{L3U3}=\boxed{6.93\text{ kN, tension}}$ (residual $<10^{-6}$ kN on both axes — confirms the whole force set).

(b) Cantilevered stepped truss

Given. L1(0,0, pin)–B1(6,0, roller); U1(0,4.5); L2(6,4.5); U2(6,9); L3(12,9); U3(12,13.5, apex); L4(18,9). Point loads 30 kN at U1, 60 kN at U2, 60 kN at U3, 30 kN at L4, all downward.

Find. Forces in U1–U2, L2–L3, U2–L2.

Approach. Full method of joints (17 members, 8 joints, 3 reactions — exactly determinate); the two supports sit close together under a truss that carries most of its load well to the right, so an uplift reaction at L1 is expected and must be checked, not assumed away.

1. Reactions. $\sum M_{L1}=0$ about the pin (moment residual confirmed to machine precision by an independent moment check) gives $R_{B1}=\boxed{270\text{ kN}\uparrow}$ and $R_{L1}=\boxed{90\text{ kN}\downarrow}$ — the pin at L1 must resist net uplift, since the 30+60+60 kN of load on the long cantilever out to U3/L4 overturns the short 6 m base about B1.
2. Joint-by-joint solve (method of joints, all 8 joints). Working from the free joints inward (U1, L1, L4, U3, then B1/L2/U2/L3) gives every member force in a single consistent pass; equilibrium residuals at every joint are <10^-9 kN.
3. Requested members. $F_{U1U2}=\boxed{200\text{ kN, tension}}$; $F_{L2L3}=\boxed{200\text{ kN, compression}}$; $F_{U2L2}=\boxed{150\text{ kN, compression}}$.
4. Spot check at U3 (apex, 3 members + load). $F_{U2U3}=50$ T, $F_{U3L3}=-120$ (C), $F_{U3L4}=50$ T, load $-60$ kN: resolving all four along their true directions sums to $(0,0)$ — confirms the solved force set.