07-Str-A1 · December 2015

Question 8 of 8: Horizontal deflection of a truss by virtual work

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

Notes on this paper

Paper format. National Examinations, December 2015, 07-Str-A1 — Elementary Structural Analysis. Three hours, closed book, approved Sharp or Casio calculator only. Six questions constitute a complete paper: answer ALL of Questions 1 to 4, then ONLY TWO of Questions 5, 6, 7 or 8. Marks are shown in the left margin (6 + 18 + 18 + 18 + 20 + 20 = 100). All eight questions are worked below, because the four alternatives exercise four different methods — moment distribution, three-hinged-frame statics, influence lines and virtual work — and the complete set is the more useful study resource.

Reference texts.

R. C. Hibbeler, Structural Analysis, 10th ed. — Ch. 2 (determinacy and stability), Ch. 3–5 (trusses, internal loadings, frames), Ch. 6 (influence lines), Ch. 8–9 (deflections and virtual work), Ch. 11–12 (slope-deflection and moment distribution).
A. Kassimali, Structural Analysis, 6th ed. — Ch. 3 (equilibrium and determinacy), Ch. 4 (plane trusses), Ch. 5 (beams and frames), Ch. 7 (virtual work), Ch. 8–9 (influence lines), Ch. 16–17 (slope-deflection and moment distribution).
K. Leet, C.-M. Uang and A. Gilbert, Fundamentals of Structural Analysis, 5th ed. — a parallel treatment of the same syllabus.
C. H. Norris, J. B. Wilbur and S. Utku, Elementary Structural Analysis, 4th ed. — the classical text this exam code is named after.

Check: the roller in Question 2(c). The roller under the inclined member of Question 2(c) is drawn on a bed ruled parallel to the member, so its reaction is taken normal to the member axis. A vertical-roller reading is also arithmetically self-consistent. The distinction changes only the reaction components and the axial force: the shear and bending-moment diagrams the question asks for are identical under either reading, and both sets of reactions are reported.

Question 8: Horizontal deflection of a truss by virtual work (20 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 two-panel, two-storey vertical truss pinned to the wall at both bottom corners and loaded horizontally at mid-height.

Given data for Question 8
Quantity	Value
Left chord joints	L₁(0, 0), L₂(0, 3), L₃(0, 6) m
Right chord joints	R₁(8, 0), R₂(8, 3), R₃(8, 6) m
Interior joints	M₁(4, 3), M₂(4, 6) m
Members	12 in all; diagonals are 5 m, horizontals 4 m, chord lengths 3 m
Load	80 kN horizontal, acting to the right at R₂
Supports	Pins at L₁ and R₁
Axial rigidity	$EA = 9.45 \times 10^{4}\ \text{kN}$, all members

Find. The horizontal deflection of joint L₃.

[Figure not reproduced: Figure 12 — the real system (left) with the three members that carry any force, and the virtual system (right) under a unit horizontal load at L 3 . There is no vertical member on the line x = 4 m: the thin line through M 1 and M 2 in the examination figure is a dimension witness line. See the official exam paper.]

Approach. Confirm that the truss is determinate despite having two pins, solve it for the real member forces $N$, solve it again for the forces $n$ under a unit horizontal load at L₃, and evaluate $\delta = \sum N n L / EA$.

Determinacy. There are 12 members, 8 joints and 4 reaction components, so $$ m + r = 12 + 4 = 16 = 2j . $$ Two pins would normally mean one external redundant, but the truss is one member short internally — there is no vertical between M₁ and M₂ — and the two deficiencies cancel exactly. Inventing that missing member would make $m + r = 17$ and the question unanswerable by statics.
Real forces: the top storey is idle. Joint L₃ carries only the vertical chord L₂–L₃ and the horizontal L₃–M₂, with no load, so both are zero; joint R₃ gives the same for R₂–R₃ and M₂–R₃. Joint M₂ then has only the two upper diagonals left, whose vertical components must cancel and whose horizontal components must cancel, so both are zero as well. The entire upper storey is unstressed.
Real forces: joint R₂ and joint M₁. With the upper storey idle, joint R₂ carries the applied 80 kN, the horizontal M₁–R₂ and the vertical R₁–R₂. Horizontal and vertical equilibrium give $$ F_{M_1R_2} = 80\ \text{kN tension}, \qquad F_{R_1R_2} = 0 . $$ At M₁ the horizontal L₂–M₁ is zero (joint L₂ has nothing to give it), so the two lower diagonals must balance the 80 kN between them. Their vertical components cancel and their horizontal components add: $$ 80 + 1.6\,F_{M_1R_1} = 0 \quad\Longrightarrow\quad \boxed{F_{L_1M_1} = 50\ \text{kN (T)}, \quad F_{M_1R_1} = 50\ \text{kN (C)}} $$ Every other member is a zero-force member. The reactions follow as 40 kN to the left with 30 kN down at L₁, and 40 kN to the left with 30 kN up at R₁; they close on both force equations and on moments about either pin.
Virtual forces under a unit load at L₃. The same joint walk with 1 kN acting to the right at L₃ gives, in order, $$ n_{L_3M_2} = -1,\quad n_{L_2M_2} = +0.625,\quad n_{M_2R_2} = -0.625, $$ $$ n_{L_1L_2} = +0.375,\quad n_{L_2M_1} = -0.5,\quad n_{R_1R_2} = -0.375, $$ $$ n_{M_1R_2} = +0.5,\quad n_{L_1M_1} = +0.625,\quad n_{M_1R_1} = -0.625, $$ with $n_{L_2L_3} = n_{R_2R_3} = n_{M_2R_3} = 0$. The pattern mirrors the real system one storey higher, which is what one expects when the unit load is applied one storey up.
Form the products. Only members carrying both a real and a virtual force contribute, so nine of the twelve terms vanish and only three survive:
$$ \begin{aligned} M_1R_2 &: \ (80)(0.5)(4) = 160, \\ L_1M_1 &: \ (50)(0.625)(5) = 156.25, \\ M_1R_1 &: \ (-50)(-0.625)(5) = 156.25, \end{aligned} $$
and the sum is $$ \sum N n L = 160 + 156.25 + 156.25 = 472.5\ \text{kN}^2\cdot\text{m}. $$
Divide by the axial rigidity. $$ \delta_{L_3} = \frac{\sum N n L}{EA} = \frac{472.5}{9.45 \times 10^{4}} = 0.00500\ \text{m} = \boxed{5.00\ \text{mm}} $$ The result is positive, so joint L₃ moves in the direction of the unit load — to the right, the same way as the applied 80 kN.
A sanity check on the magnitude. The three stressed members carry stresses of the order of $80/EA$ per unit length, so a deflection of a few millimetres over an 8 m by 6 m frame is the right order. It is also worth noting that the whole upper storey rides along rigidly: it carries no force, so it does not deform, and the 5 mm at L₃ is entirely the result of the lower storey racking.

Question 8 — virtual work summation
Member	$L$ (m)	$N$ (kN)	$n$	$NnL$ (kN²·m)
M₁–R₂	4	+80	+0.500	160.00
L₁–M₁	5	+50	+0.625	156.25
M₁–R₁	5	−50	−0.625	156.25
All other nine members	—	0	—	0
Total				472.50
Horizontal deflection of L₃				5.00 mm to the right

Topic. Truss deflection by the unit-load form of the principle of virtual work.

Key relations. For a pin-jointed truss, $1 \cdot \delta = \sum \dfrac{N n L}{EA}$, where $N$ is the member force in the real system and $n$ the force in the same member under a unit load applied at the point and in the direction of the required displacement. Determinacy check: $m + r = 2j$ with a stable arrangement.

Why this works. The virtual unit load is in equilibrium with its own set of member forces $n$; the real system supplies a compatible set of member extensions $NL/EA$. Equating the external virtual work $1 \times \delta$ with the internal virtual work $\sum n \,(NL/EA)$ gives the formula directly. It requires only that the $n$ system be in equilibrium, not that it be the actual one — which is why it works unchanged for any determinate truss and why a member that is unstressed in either system drops out of the sum entirely. That is what makes this particular question short: the real system stresses three members and the virtual system leaves all three of them stressed too, so the twelve-term summation collapses to three.

Common pitfall. Reading the vertical line through M₁ and M₂ as a member: it is a dimension witness line, and adding it makes the truss indeterminate to the first degree, so virtual work with statically determined forces no longer applies. The second pitfall is dropping the sign on a compression member: here M₁–R₁ has both $N$ and $n$ negative, so its contribution is positive and equal to that of its tension partner; treating it as negative would halve the answer.

Source. Hibbeler, Structural Analysis, 10th ed., §9.3 (virtual work for trusses); Kassimali, Structural Analysis, 6th ed., §7.3.

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