Question 4 of 8: Member forces in two determinate trusses

Question 4: Member forces in two determinate 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.

Given.

Find. Six member forces with their sense, tension or compression.

Approach. Get the reactions from global equilibrium, then cut truss (a) with a single vertical section whose only sloping member is the one required, and take truss (b) joint by joint, because every vertical cut through it severs four members.

1. Part (a) — reactions. The pin at $L_1$ and the roller at $L_5$ are both 4 m above the bottom chord, 31.2 m apart. Horizontal equilibrium gives the pin 15.6 kN to the left. The 15.6 kN load acts at $U_3$, which is 4 m above the support line, so it contributes a moment as well: $$V_{L_5}\,(31.2) = 62.4\,(9.6) + 62.4\,(15.6) + 15.6\,(8 - 4) = 599.04 + 973.44 + 62.4$$ $$\begin{aligned} V_{L_5} &= \boxed{52.4\ \text{kN}\uparrow} \\ V_{L_1} &= 124.8 - 52.4 = \boxed{72.4\ \text{kN}\uparrow} \end{aligned}$$
2. Part (a) — the two members at the support joint. Only $L_1$–$U_1$ and $L_1$–$L_2$ meet at $L_1$, so joint equilibrium settles both at once. Both have direction cosines $9.6/10.4$ and $4/10.4$, the first rising and the second falling. Horizontal and vertical equilibrium give $$\begin{aligned} \frac{9.6}{10.4}\big(F_{L_1U_1} + F_{L_1L_2}\big) &= 15.6 \\ \frac{4}{10.4}\big(F_{L_1U_1} - F_{L_1L_2}\big) &= -72.4 \end{aligned}$$ Solving the pair, $$\begin{aligned} F_{L_1U_1} &= \boxed{85.7\ \text{kN compression}} \\ F_{L_1L_2} &= 102.6\ \text{kN tension} \end{aligned}$$
3. Part (a) — section 1–1 for the diagonal. Cut the truss through $U_1$–$U_2$, $U_1$–$L_3$ and $L_2$–$L_3$ and keep the left part, which carries the pin reactions and the 62.4 kN load at $U_1$. Taking moments about the point where the two chord members meet is not available here, so use vertical equilibrium of the cut part instead: both chords are horizontal, so the diagonal alone carries the unbalanced vertical force. The diagonal runs from $U_1$ at $(9.6,\,8)$ to $L_3$ at $(15.6,\,0)$, a 6 by 8 by 10 triangle, so its vertical cosine is $0.8$: $$72.4 - 62.4 - 0.8\,F_{U_1L_3} = 0 \;\Rightarrow\; F_{U_1L_3} = \frac{10.0}{0.8} = \boxed{12.5\ \text{kN tension}}$$
4. Part (a) — the vertical by joint equilibrium at $U_1$. Four members meet there — the diagonal to $L_1$, the top chord to $U_2$, the vertical to $L_2$ and the diagonal to $L_3$ — under the 62.4 kN load. Horizontal equilibrium first fixes the top chord, $$-85.67\Big(\!-\frac{9.6}{10.4}\Big) + F_{U_1U_2} + 12.5\,(0.6) = 0 \;\Rightarrow\; F_{U_1U_2} = -86.58\ \text{kN}$$ and vertical equilibrium then gives the vertical member, $$-85.67\Big(\!-\frac{4}{10.4}\Big) - F_{U_1L_2} - 12.5\,(0.8) - 62.4 = 0 \;\Rightarrow\; F_{U_1L_2} = \boxed{39.5\ \text{kN compression}}$$
5. Part (b) — reactions. The pin at $L_1$ takes the whole 48 kN of horizontal load, acting 48 kN to the left. Moments about $L_1$, with the roller at $L_3$ 8 m away and the two horizontal loads acting 6 m and 3 m above the chord, $$V_{L_3}\,(8) = 24\,(8) + 24\,(6) + 24\,(3) = 192 + 144 + 72 = 408$$ $$\begin{aligned} V_{L_3} &= \boxed{51.0\ \text{kN}\uparrow} \\ V_{L_1} &= 24 - 51 = \boxed{27.0\ \text{kN}\downarrow} \end{aligned}$$ A downward reaction at $L_1$ is expected: the horizontal loads at the tall right-hand end overturn the truss about the roller and the pin has to hold it down.
6. Part (b) — no useful section exists, so work joint by joint. Any vertical cut between $L_2$ and $L_3$ severs four members — $L_2$–$L_3$, $M_1$–$L_3$, $M_1$–$M_2$ and $M_1$–$U_1$ — which is one more than three equilibrium equations can settle, so the method of sections fails on this truss and the joints must be taken in an order that never leaves more than two unknowns. Start at $L_1$, where only $L_1$–$L_2$ and $L_1$–$M_1$ meet; the latter rises 3 in 5: $$-27 + \tfrac{3}{5}F_{L_1M_1} = 0 \;\Rightarrow\; F_{L_1M_1} = 45\ \text{kN tension}$$ $$-48 + F_{L_1L_2} + \tfrac{4}{5}\,(45) = 0 \;\Rightarrow\; F_{L_1L_2} = 12\ \text{kN tension}$$
7. Part (b) — the bottom chord and the top chord. At $L_2$ three members meet and nothing is applied, so vertical equilibrium makes the post $M_1$–$L_2$ a zero-force member and horizontal equilibrium passes the chord force straight through: $$\begin{aligned} F_{M_1L_2} &= 0 \\ F_{L_2L_3} &= F_{L_1L_2} = \boxed{12.0\ \text{kN tension}} \end{aligned}$$ At $U_2$ only the horizontal $U_1$–$U_2$ and the vertical $M_2$–$U_2$ meet under the 24 kN horizontal load, so $F_{M_2U_2} = 0$ and $F_{U_1U_2} = 24$ kN tension.
8. Part (b) — joint $U_1$ unlocks the two sloping members. Three members meet at $U_1$: the chord to $U_2$ now known, and the two 3-4-5 diagonals to $M_1$ and to $M_2$, under the 24 kN downward load. Writing both equations, $$\begin{aligned} -\tfrac{4}{5}F_{M_1U_1} + 24 + \tfrac{4}{5}F_{U_1M_2} &= 0 \\ -\tfrac{3}{5}F_{M_1U_1} - \tfrac{3}{5}F_{U_1M_2} - 24 &= 0 \end{aligned}$$ and solving the pair, $$\begin{aligned} F_{M_1U_1} &= -5.0\ \text{kN} \\ F_{U_1M_2} &= -35.0\ \text{kN} \end{aligned}$$ both compressive.
9. Part (b) — joint $M_1$ closes the answer. Five members meet at $M_1$ and three are now known, so its two equations settle the last two. Vertically, with the post carrying nothing, $$-\tfrac{3}{5}(45) + \tfrac{3}{5}(-5) - \tfrac{3}{5}F_{M_1L_3} = 0 \;\Rightarrow\; F_{M_1L_3} = \boxed{50.0\ \text{kN compression}}$$ and horizontally, $$-\tfrac{4}{5}(45) + \tfrac{4}{5}(-5) + \tfrac{4}{5}(-50) + F_{M_1M_2} = 0 \;\Rightarrow\; F_{M_1M_2} = \boxed{80.0\ \text{kN tension}}$$ As a check, joint $M_2$ balances: $24 - 80 - \tfrac{4}{5}(-35) - \tfrac{4}{5}(-35) = 0$ horizontally and $\tfrac{3}{5}(-35) - \tfrac{3}{5}(-35) = 0$ vertically.