Member forces in two trusses: 16-Civ-A1 Elementary Structural Analysis, May 2013, Q4

Question 4: Member forces in two 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. Two determinate, pin-jointed trusses with the geometry and panel-point loads shown. Both are supported by a pin (left) and a roller (right).

Find. The axial force (with T/C sense) in the three listed members of each truss.

Approach. Find the reactions from global equilibrium, then isolate each member with the method of sections — cut a plane through the target member plus two others and take moments about the point where those two intersect, so the target force follows directly.

(a) Triangular (Fink) truss

(a) Loads 90/60/30 kN at bottom joints L₂/L₃/L₄. Highlighted: the three requested members.

Reactions. $\sum M_{L_1}=0$: $24L_{5y}=90(6)+60(12)+30(18)=1800\Rightarrow L_{5y}=75$ kN; $L_{1y}=180-75=105$ kN.
$U_2$–$U_3$ (top chord right of apex). Section through $U_2U_3$, the vertical $U_2L_3$ and $L_3L_4$; taking moments about $L_3$ (where the other two meet) and using the right-hand part, the horizontal top-chord force resolves to ==**$U_2U_3=72.0$ kN (C)**==.
$L_1$–$L_2$ (bottom chord). Method of joints at $L_1$ with the left rafter $L_1U_1$ then $\sum F_x$ along the bottom gives ==**$L_1L_2=78.75$ kN (T)**==.
$U_1$–$L_3$ (diagonal). A vertical section just right of $U_1$ cutting $U_1U_2$, $U_1L_3$ and $L_2L_3$; vertical equilibrium of the left part isolates the diagonal: ==**$U_1L_3=35.25$ kN (C)**==.

(b) Cranked parallel-chord truss

(b) Loads 72/72/90/90 kN down at U₃–U₆. Highlighted: L₂–U₃, L₃–U₅, L₃–L₄.

Reactions. $\sum M_{L_1}=0$ (loads at $x=12,20,28,36$): $36L_{5y}=72(12)+72(20)+90(28)+90(36)=8064\Rightarrow L_{5y}=224$ kN; $L_{1y}=324-224=100$ kN.
$L_2$–$U_3$ (diagonal). Section cutting $U_2U_3$, $L_2U_3$ and $L_2L_3$; resolving the left part gives ==**$L_2U_3=164.9$ kN (C)**==.
$L_3$–$L_4$ (bottom chord). Section through $U_4U_5$, the diagonal $L_3U_5$ and $L_3L_4$; moments about $U_5$ isolate the bottom chord: ==**$L_3L_4=111.7$ kN (T)**==.
$L_3$–$U_5$ (diagonal). Same section, taking vertical equilibrium (the chords are near-horizontal) gives ==**$L_3U_5=185.0$ kN (T)**==.

Member	Force
(a) $L_1$–$L_2$	78.75 kN (T)
(a) $U_2$–$U_3$	72.0 kN (C)
(a) $U_1$–$L_3$	35.25 kN (C)
(b) $L_2$–$U_3$	164.9 kN (C)
(b) $L_3$–$U_5$	185.0 kN (T)
(b) $L_3$–$L_4$	111.7 kN (T)

Topic. Method of sections for truss member forces.

Key relations. Cut the truss into two rigid parts through at most three unknown members; apply $\sum M=0$ about the intersection of two of them to solve the third directly, or $\sum F_y=0$ for a diagonal.

Why this works. A single cut exposes internal member forces as external forces on a free body, and choosing the moment centre at the meeting point of two cut members eliminates them from that one equation — giving the third force in a single step without back-substitution.

Common pitfall. Cutting more than three unknown members (or three that are concurrent/parallel) leaves the section unsolvable; and the sense (T/C) must be read from the sign relative to the assumed tension direction.

Source. Hibbeler §3.4 (method of sections); Kassimali §4.6.

Question 4 of 8: Member forces in two trusses

Question 4: Member forces in two trusses (18 marks)

(a) Triangular (Fink) truss

(b) Cranked parallel-chord truss