07-Str-A1 · May 2018

Question 4 of 8: Truss Member Forces

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 07-Str-A1 Elementary Structural Analysis, National Examinations, May 2018. Three hours, CLOSED BOOK (an approved Casio or Sharp calculator is permitted). Six questions constitute a complete paper: Questions 1 to 5 are compulsory and one of Questions 6, 7 or 8 is chosen. Marks are shown against each question and total 100. Every question is worked here, including all three of the optional Questions 6, 7 and 8, so that the set can be used as a study resource.

Reference texts. R. C. Hibbeler, Structural Analysis (Ch. 2 stability and determinacy; Ch. 4 influence lines; Ch. 6 deflections by virtual work; Ch. 11 displacement method; Ch. 12 moment distribution); A. Kassimali, Structural Analysis (Ch. 3, 8, 13, 16); K. Leet, C.-M. Uang and A. Gilbert, Fundamentals of Structural Analysis; J. C. McCormac, Structural Analysis: Using Classical and Matrix Methods. For Canadian design practice the companion documents are the National Building Code of Canada (Part 4, load combinations) and CSA S16 Design of Steel Structures; this paper is pure analysis, so no design code is invoked in the answers.

Check — readings taken from the printed figures. Every dimension, support symbol and load extent below was read from the printed figures. Four readings carry the answers and are stated explicitly so that a marker can check them: (i) in Question 2(a) the 10 kN/m block covers only the 2 m overhang, ending at the roller; (ii) in Question 4(a) the truss is a Warren truss whose top nodes are offset 3 m from the bottom nodes (the 3 m dimension at the top right confirms this); (iii) in Question 7 the 10.8 kN/m ruling starts directly above joint 1 and runs to joint 4, so it acts over the whole horizontal projection, inclined member included; (iv) in Question 8 the thin vertical line between U₂ and L₃ touches neither joint circle and is the witness line of the 3 m dimension, not a member — taking it as a member would make the truss statically indeterminate and the question unanswerable by virtual work.

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.

Given. (a) A Warren truss of 24 m span in four 6 m panels and 4 m deep, pinned at L₁ and on a roller at L₅; the top joints U₁ to U₄ stand at $x = 3, 9, 15, 21$ m. Loads: 24 kN horizontal (to the right) at U₁, and 32, 40 and 48 kN vertically downward at L₂, L₃ and L₄. (b) A rectangular truss 6 m wide in two 3 m panels and 4 m deep, pinned to a wall at U₁ and on a vertical-face roller at L₁; the web is the long diagonal U₁–L₃ crossing the short diagonals L₁–U₂ and L₂–U₃, with verticals at each end. Loads: 20 kN down at U₂, 36 kN to the right at U₃, and 60 kN down at each of L₂ and L₃.

Find. The three named member forces in each truss, each labelled tension or compression.

(a) Warren truss

Question 4(a): the Warren truss. The three requested members are highlighted. A vertical section between L₂ and U₂ cuts exactly these three.

Approach. Find the reactions, then pass one vertical section between L₂ and U₂: it severs exactly the three requested members, so moments about the two convenient joints and one vertical resolution deliver all three answers without any joint-by-joint march.

Reactions. Horizontal equilibrium gives $H_{L1} = -24$ kN, that is 24 kN acting to the left. Taking moments about L₁, and remembering that the 24 kN horizontal load acts 4 m above the bottom chord, $$24 V_{L5} = 32(6) + 40(12) + 48(18) + 24(4) = 192 + 480 + 864 + 96 = 1632,$$ so $V_{L5} = 68.0$ kN and $V_{L1} = 120 - 68 = 52.0$ kN, both upward.
Cut the truss and take moments about L₂ for the top chord. The left portion carries the two reaction components at L₁, the 24 kN horizontal load at U₁(3, 4) and the 32 kN load at L₂, which passes through the moment centre. The two web members cut also pass through L₂, leaving $$-6(52) - 4(24) - 4\,F_{U_1U_2} = 0 \;\Longrightarrow\; F_{U_1U_2} = -\frac{312 + 96}{4} = -102.0\ \text{kN},$$ so $\boxed{F_{U_1U_2} = 102.0\ \text{kN compression}}$.
Take moments about U₂ for the bottom chord. Relative to U₂(9, 4) the reaction at L₁ contributes $(-9)(52) - (-4)(-24) = -564$, the horizontal load at U₁ contributes nothing because it is collinear with the moment centre in the vertical sense, and the 32 kN load contributes $+96$: $$-564 + 96 + 4\,F_{L_2L_3} = 0 \;\Longrightarrow\; F_{L_2L_3} = \boxed{117.0\ \text{kN tension}}.$$
Resolve vertically for the diagonal. The diagonal L₂–U₂ runs 3 m across and 4 m up, a 3–4–5 triangle, so its vertical direction cosine is $4/5 = 0.8$. Vertical equilibrium of the left portion gives $$52 - 32 + 0.8\,F_{L_2U_2} = 0 \;\Longrightarrow\; F_{L_2U_2} = -25.0\ \text{kN},$$ that is $\boxed{F_{L_2U_2} = 25.0\ \text{kN compression}}$.
Check horizontally. The horizontal components must close: $$-24 + 24 + (-102.0) + 0.6(-25.0) + 117.0 = -102.0 - 15.0 + 117.0 = 0. \checkmark$$

(b) Rectangular truss with crossing diagonals

This truss is supported on a vertical wall: a pin at U₁ and, at L₁, a roller bearing on the vertical face, which can deliver a horizontal reaction only. Three reaction components against nine members and six joints gives $m + r = 9 + 3 = 12 = 2j$, so the truss is determinate and the method of joints runs straight through.

Question 4(b): the rectangular truss. The long diagonal U₁–L₃ crosses both short diagonals without connecting to them; the requested members are highlighted.

Start where only one unknown carries the load. At joint U₂ the two top chords are horizontal, so the only member able to resist the 20 kN downward load is the diagonal to L₁, whose direction cosines are $(-0.6, -0.8)$: $$-0.8\,F_{L_1U_2} - 20 = 0 \;\Longrightarrow\; F_{L_1U_2} = \boxed{25.0\ \text{kN compression}}.$$
Find the horizontal reaction at L₁. Moments about U₁ for the whole truss, with the wall roller 4 m below it: $$3 H_{L1}\ \text{arm}: \quad 4H_{L1} = 20(3) + 60(3) + 60(6) = 60 + 180 + 360 = 600 \;\Longrightarrow\; H_{L1} = 150.0\ \text{kN}.$$ The 36 kN horizontal load acts along the line through U₁ and contributes no moment. Horizontal equilibrium then gives $H_{U1} = -36 - 150 = -186.0$ kN, and vertical equilibrium gives $V_{U1} = 20 + 60 + 60 = 140.0$ kN upward.
Joint L₁ for the bottom chord. The members at L₁ are the bottom chord, the end vertical and the diagonal already found. Horizontally, $$F_{L_1L_2} + 0.6(-25.0) + 150.0 = 0 \;\Longrightarrow\; F_{L_1L_2} = \boxed{135.0\ \text{kN compression}},$$ and vertically $F_{L_1U_1} = -0.8(-25.0) = 20.0$ kN tension.
Work along the bottom chord to reach the long diagonal. At L₂ only the diagonal to U₃ has a vertical component, so $$0.8\,F_{L_2U_3} = 60 \;\Longrightarrow\; F_{L_2U_3} = 75.0\ \text{kN tension},$$ and horizontal equilibrium at the same joint gives $$F_{L_2L_3} = -135.0 - 0.6(75.0) = -180.0\ \text{kN}.$$
Joint L₃ gives the long diagonal. Member U₁–L₃ runs 6 m across and 4 m up, so its length is $\sqrt{52} = 7.2111$ m and its horizontal direction cosine is $6/7.2111 = 0.83205$. Horizontal equilibrium at L₃ reads $$180.0 - 0.83205\,F_{U_1L_3} = 0 \;\Longrightarrow\; F_{U_1L_3} = \boxed{216.3\ \text{kN tension}}.$$ Its vertical component, $216.3 \times 4/7.2111 = 120.0$ kN, together with the 60 kN load, puts 60.0 kN of compression into the end vertical U₃–L₃, and joint U₁ then closes both force equations exactly.

Question 4 — requested member forces
Truss	Member	Force (kN)	Sense
(a) Warren, 24 m span	U₁–U₂	102.0	Compression
	L₂–U₂	25.0	Compression
	L₂–L₃	117.0	Tension
(b) rectangular, wall-supported	L₁–U₂	25.0	Compression
	L₁–L₂	135.0	Compression
	U₁–L₃	216.3	Tension
Reactions (a)		L₁: 52.0 kN ↑ and 24.0 kN ←; L₅: 68.0 kN ↑
Reactions (b)		U₁: 140.0 kN ↑ and 186.0 kN ←; L₁: 150.0 kN →

Topic. Determinate plane trusses: the method of sections and the method of joints, and the screening that decides which to use.

Key relations. Determinacy: $m + r = 2j$. Method of sections: any cut severing three non-concurrent members is solvable, and a moment centre placed at the intersection of two of them isolates the third. Method of joints: two equations per joint, so start where only two unknowns meet.

Why this works. Every member of an idealised truss is a two-force member, so the whole internal force state is one scalar per member. Cutting the structure produces a free body whose three equilibrium equations involve only the severed members plus the known external forces, which is why a section that crosses exactly three unknown members is enough. When no such cut exists — a sub-divided web, or a panel with crossing diagonals — the joint march is the only route, and the practical trick is to look for a joint whose geometry kills one unknown outright, as U₂ does in part (b) where both chords are horizontal and the diagonal alone can carry the vertical load.

Common pitfall. Mis-reading the truss form. the drawing and the 3 m dimension at the top right show a Warren truss whose top nodes are offset by half a panel, and the wrong reading makes the section cut four members instead of three. In (b), reading the wall roller at L₁ as a pin makes the truss appear indeterminate.

Source. Hibbeler, Structural Analysis, §3.4–3.6; Kassimali, Structural Analysis, §4.5–4.7.

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