07-Str-A1 · May 2018

Question 8 of 8: Truss Deflections by Virtual Work

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 8: Truss Deflections by Virtual Work (22 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 cantilevered truss carried on a wall: joints L₁(0, 0), L₂(4.2, 0), L₃(7.2, 0), U₁(4.2, 1.75) and U₂(7.2, 3.0), all in metres. The six members are the bottom chords L₁–L₂ and L₂–L₃, the straight inclined chord L₁–U₁–U₂ in two lengths, the vertical U₁–L₂ and the diagonal L₂–U₂. Pins at L₃ and U₂ attach the truss to the wall. The loads are 21 kN downward at L₁ and 14 kN downward at U₁, and $AE = 3.9 \times 10^{4}$ kN for every member.

Find. The vertical deflection of joint L₁ and of joint L₂.

Question 8: the truss. Six members, five joints and four reaction components give $m + r = 10 = 2j$, so it is determinate and can be solved joint by joint starting at the free corner L₁.

Approach. Compute the real member forces $N$ by the method of joints starting at L₁, where only two members meet; repeat twice with a unit downward load at L₁ and then at L₂ to obtain the virtual forces $n$; and evaluate $\Delta = \sum N n L / AE$.

Geometry, and why it is kind. Both chord lengths come from 5–12–13 triangles: L₁–U₁ runs 4.2 m across and 1.75 m up, so $L = 4.55$ m, while U₁–U₂ runs 3.0 m across and 1.25 m up, so $L = 3.25$ m. The diagonal L₂–U₂ rises 3.0 m over 3.0 m, so it stands at 45° with $L = 4.2426$ m, and the vertical U₁–L₂ is 1.75 m. Because L₁, U₁ and U₂ are collinear, the two chord segments share the direction cosines $(12/13,\,5/13)$.
Joint L₁. Only the bottom chord and the inclined chord meet here, and only the latter has a vertical component: $$\tfrac{5}{13}F_{L_1U_1} = 21 \;\Longrightarrow\; F_{L_1U_1} = 54.6\ \text{kN (T)}, \qquad F_{L_1L_2} = -\tfrac{12}{13}(54.6) = -50.4\ \text{kN (C)}.$$
Joint U₁. The two chord segments are collinear, so horizontal equilibrium gives $F_{U_1U_2} = F_{L_1U_1} = 54.6$ kN tension, and their vertical components then cancel, leaving the vertical member to carry the 14 kN load alone: $$F_{U_1L_2} = -14.0\ \text{kN (C)}.$$ This is the observation that collapses the question: a straight chord passing through a loaded joint does not change force across it.
Joint L₂. The 45° diagonal must carry the 14 kN pushed down by the vertical: $$\tfrac{1}{\sqrt2}F_{L_2U_2} = 14.0 \;\Longrightarrow\; F_{L_2U_2} = 19.80\ \text{kN (T)}, \qquad F_{L_2L_3} = -50.4 - 14.0 = -64.4\ \text{kN (C)}.$$ The reactions follow as 64.4 kN horizontal at L₃ and $(64.4,\,35.0)$ kN at U₂, and the 35.0 kN vertical component equals the total applied load, as it must.
Virtual system for L₁. A unit downward load at L₁ alone runs through the same joint sequence, and because there is now no load at U₁ the vertical carries nothing and the 45° diagonal is idle: $$n_{L_1U_1} = n_{U_1U_2} = \tfrac{13}{5} = 2.6, \qquad n_{L_1L_2} = n_{L_2L_3} = -2.4, \qquad n_{U_1L_2} = n_{L_2U_2} = 0.$$
Virtual system for L₂. A unit downward load at L₂ is carried entirely by the 45° diagonal and the adjacent bottom chord, because joint L₁ is now unloaded and therefore force-free: $$n_{L_2U_2} = \sqrt{2} = 1.4142, \qquad n_{L_2L_3} = -1.0, \qquad \text{all others } = 0.$$
Assemble the two sums. For L₁ only four members contribute: $$\sum N n L = (-50.4)(-2.4)(4.2) + (-64.4)(-2.4)(3.0) + (54.6)(2.6)(4.55) + (54.6)(2.6)(3.25)$$ $$= 508.03 + 463.68 + 645.92 + 461.37 = 2079.0\ \text{kN}^{2}\!\cdot\!\text{m},$$ and for L₂ only two: $$\sum N n L = (-64.4)(-1.0)(3.0) + (19.799)(1.4142)(4.2426) = 193.20 + 118.79 = 311.99\ \text{kN}^{2}\!\cdot\!\text{m}.$$
Divide by the axial rigidity. $$\Delta_{L_1} = \frac{2079.0}{3.9 \times 10^{4}} = 0.05331\ \text{m} \;\Longrightarrow\; \boxed{\Delta_{L_1} = 53.3\ \text{mm downward}},$$ $$\Delta_{L_2} = \frac{311.99}{3.9 \times 10^{4}} = 0.008000\ \text{m} \;\Longrightarrow\; \boxed{\Delta_{L_2} = 8.00\ \text{mm downward}}.$$ Both sums are positive, so both joints move in the direction of their unit loads, that is downward.

The ratio of the two answers is worth a sentence in an examination script. Joint L₂ is only 8 mm down because it is braced almost directly back to the wall by the 45° diagonal and the short bottom chord, whereas L₁ hangs 4.2 m beyond it on a chord that must stretch and a bottom chord that must shorten — a lever arm that multiplies both strains. Note also that the vertical U₁–L₂, although it carries 14 kN, contributes nothing to either deflection: it is idle in both virtual systems.

Question 8 — member forces and the virtual-work sums
Member	$L$ (m)	$N$ (kN)	$n$ for L₁	$n$ for L₂	$NnL$ for L₁
L₁–L₂	4.20	−50.4	−2.400	0	508.03
L₂–L₃	3.00	−64.4	−2.400	−1.000	463.68
L₁–U₁	4.55	+54.6	+2.600	0	645.92
U₁–U₂	3.25	+54.6	+2.600	0	461.37
U₁–L₂	1.75	−14.0	0	0	0
L₂–U₂	4.243	+19.80	0	+1.414	0
$\sum NnL$ for L₁					2079.0
$\sum NnL$ for L₂					311.99
Vertical deflection at L₁					53.3 mm down
Vertical deflection at L₂					8.00 mm down

Topic. Joint deflections of a determinate pin-jointed truss by the principle of virtual work (the unit-load method).

Key relations. $1 \cdot \Delta = \sum \dfrac{N n L}{AE}$, in which $N$ is the real member force, $n$ the force from a unit load applied at and along the required displacement, and $NL/AE$ the real elongation of the member. Determinacy: $m + r = 2j$.

Why this works. The virtual unit load is in equilibrium with the virtual member forces $n$, so the external virtual work $1 \cdot \Delta$ must equal the internal virtual work $\sum n \, \delta_{\text{real}}$, and for a bar the real extension is simply $NL/AE$. Because the virtual system need only satisfy equilibrium, any convenient statically admissible force set will do, which is why the unit-load system is chosen to be as sparse as possible. Here that sparseness does most of the work: two of the six members are idle in the first virtual system and four in the second, so the second deflection costs two products.

Common pitfall. Reading the thin vertical line between U₂ and L₃ as a member. It touches neither joint circle and is drawn at the weight of the dimension lines, not the members; counting it gives $m + r = 11$ against $2j = 10$, making the truss indeterminate and the question unanswerable by virtual work alone. The second trap is sign discipline — a compressive $N$ multiplied by a compressive $n$ gives a positive, deflection-increasing product, and dropping either minus sign halves the answer.

Source. Hibbeler, Structural Analysis, §8.1–8.3 and §9.4; Kassimali, Structural Analysis, §7.3.

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