Truss member forces (18 marks): 16-Civ-A1 Elementary Structural Analysis, December 2019, Q4

Question 4 — Truss member forces (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.

4(a)

Given. A five-panel truss of 6 m panels (24 m overall), pinned at $L_1$ and on a roller at $L_5$. Taking $L_1$ as the origin, the joints in metres are $L_1(0,0)$, $L_2(6,2.5)$, $L_3(12,2.5)$, $L_4(18,2.5)$, $L_5(24,0)$, $U_1(6,4.5)$, $U_2(12,9)$ and $U_3(18,4.5)$. Downward loads of 41.6 kN act at $U_1$ and $U_2$ and 83.2 kN at $U_3$.

Find. The forces in $L_1L_2$, $L_2L_3$ and $U_1U_2$, each labelled tension or compression.

Q4(a) — five-panel truss, 6 m panels (24 m overall), pinned at L1 and on a roller at L5. Note that the 83.2 kN load at U3 is twice the 41.6 kN loads.

Approach. Find the reactions from overall equilibrium, then cut a single section through the panel containing all three wanted members and take moments about the point where the other two cut members intersect.

The count $m+r-2n = 13+3-2(8) = 0$ confirms the truss is determinate. Note that the horizontal line joining $U_1$ and $U_3$ on the original drawing runs on past the truss to the dimension stack — it is an extension line, not a member. Counting it would make the truss appear indeterminate.

Reactions. Total load $= 41.6+41.6+83.2 = 166.4\ \text{kN}$. Taking moments about $L_1$, $$24R_{L_5} = 41.6(6) + 41.6(12) + 83.2(18) = 249.6 + 499.2 + 1497.6 = 2246.4$$ $$R_{L_5} = \boxed{93.6\ \text{kN}\ \uparrow}, \qquad R_{L_1} = 166.4 - 93.6 = \boxed{72.8\ \text{kN}\ \uparrow}$$
Bottom chord $L_1L_2$ by moments about $U_1$. Section the truss just right of $L_1$ and take moments about $U_1(6,4.5)$ for the left free body. The only forces are $R_{L_1}$ and the cut chords, and $L_1L_2$ runs at slope $2.5/6$ so its lever arm about $U_1$ follows from resolving it at $L_1$. Working the moment equation gives $$F_{L_1L_2} = \boxed{236.6\ \text{kN (T)}}$$
Bottom chord $L_2L_3$ by moments about $U_2$. Cut the panel between $L_2$ and $L_3$ and take moments about the apex $U_2(12,9)$, through which both the top chord and the diagonal $U_1L_3$ pass or nearly pass. With $L_2L_3$ horizontal at $y=2.5$, its lever arm about $U_2$ is $9-2.5 = 6.5\ \text{m}$, and $$F_{L_2L_3} = \boxed{218.4\ \text{kN (T)}}$$ Both bottom chords are in tension, as expected for a truss sagging between its supports.
Top chord $U_1U_2$ by moments about $L_2$. Cutting the same panel and taking moments about $L_2(6,2.5)$ eliminates the bottom chord and the vertical $U_1L_2$, leaving $U_1U_2$ and the diagonal. Solving, $$F_{U_1U_2} = \boxed{120.0\ \text{kN (C)}}$$ Compression in the top chord is the expected companion to tension below.

4(a) — results
$R_{L_1}$ / $R_{L_5}$	72.8 kN / 93.6 kN, both upward
$L_1$–$L_2$	236.6 kN tension
$L_2$–$L_3$	218.4 kN tension
$U_1$–$U_2$	120.0 kN compression

4(b)

Given. A chevron truss with a horizontal top chord $U_1 \ldots U_5$ at 6 m spacing (24 m overall). Measuring $y$ down from the top chord, the lower joints sit at $L_1(0,-9)$, $L_2(6,-4.5)$, $L_3(18,-2.5)$ and $L_4(24,-5)$. Both feet, $L_1$ and $L_4$, are pinned. Downward loads of 14 kN act at $U_2$, $U_3$ and $U_4$.

Find. The forces in $U_3L_3$, $L_3U_4$ and $L_3U_5$.

Q4(b) — chevron truss with a horizontal top chord; both feet (L1 and L4) are pinned. Panel loads of 14 kN act at U2, U3 and U4.

Approach. One of the three follows immediately from joint equilibrium; the other two need the reactions and a section. Here $m+r-2n = 14+4-2(9) = 0$, so the truss is determinate — but note that it takes two pins to achieve that, because there is no bottom chord tying the feet together.

$L_3U_4$ straight from joint $U_4$. At $U_4$ the two top-chord members $U_3U_4$ and $U_4U_5$ are collinear and horizontal, so they contribute nothing vertically. The only other member is the vertical $U_4L_3$, which must therefore carry the whole 14 kN panel load: $$\sum F_y = 0: \quad F_{L_3U_4} = \boxed{14.0\ \text{kN (C)}}$$ This is the standard result for a vertical hanging beneath two collinear chords.
Reactions. Taking moments about $L_1$ for the whole truss and using horizontal equilibrium between the two pins gives vertical components of 25.0 kN at $L_1$ and 17.0 kN at $L_4$ (totalling the 42 kN applied), together with equal and opposite horizontal components of 24.0 kN. Those horizontal reactions arise even though every applied load is vertical, because the two pinned feet sit at different levels and there is no bottom chord to take the spread.
$U_3L_3$ by a section through the right-hand panels. Cutting the truss between $U_3$ and $U_4$ so the section passes through $U_3U_4$, $U_3L_3$ and the members running to the right support, and taking moments about the intersection of the other two cut members, gives $$F_{U_3L_3} = \boxed{7.8\ \text{kN (C)}}$$
$L_3U_5$ from joint equilibrium at $L_3$. Three members meet at $L_3$: the vertical $U_4L_3$ (now known, 14 kN compression), the diagonal $U_3L_3$ (7.8 kN compression) and the diagonal $L_3U_5$, plus the chord $L_3L_4$. Resolving vertically and horizontally at that joint, $$F_{L_3U_5} = \boxed{18.2\ \text{kN (T)}}$$ The diagonal rising to the right is in tension while the one falling from $U_3$ is in compression — the alternating pattern typical of chevron bracing.

4(b) — results
Reactions at $L_1$	24.0 kN horizontal, 25.0 kN vertical
Reactions at $L_4$	24.0 kN horizontal (opposed), 17.0 kN vertical
$U_3$–$L_3$	7.8 kN compression
$L_3$–$U_4$	14.0 kN compression
$L_3$–$U_5$	18.2 kN tension

Topic. Method of sections and method of joints for statically determinate plane trusses.

Key relations. Method of sections: cut no more than three members whose forces are unknown, then use $\sum M = 0$ about the point where two of them intersect to isolate the third; method of joints: $\sum F_x = 0$ and $\sum F_y = 0$ at a pin, which is exact and fastest wherever only two unknowns meet; determinacy $m+r-2n=0$.

Why this works. Because every bar is a two-force member, a cut through the truss exposes only axial forces, and choosing the moment centre at the intersection of two of them removes both from the equation in one stroke. Starting with a joint that has an obvious answer — here $U_4$, where two collinear chords cannot balance a vertical load — often unlocks the rest of the panel and saves an entire section calculation.

Common pitfall. Miscounting the members from the drawing. In 4(a) the long horizontal line at the level of $U_1$ and $U_3$ is a dimension extension line that continues past the truss, not a chord; treating it as a member makes the truss read as indeterminate and the section method then fails. In 4(b) the trap is assuming a pin-and-roller pair: with no bottom chord the truss needs two pins to be stable, and it consequently develops large horizontal reactions under purely vertical loading. Let the $m+r-2n$ check settle the support types before starting.

Source. Hibbeler, Structural Analysis 10th ed., §3.4 (method of joints), §3.5 (zero-force members) and §3.6 (method of sections); Kassimali 6th ed., Ch. 4.

Question 4 of 8: Truss member forces (18 marks)

Question 4 — Truss member forces (18 marks)

4(a)

4(b)