07-Str-A1 · December 2018

Question 7 of 8: Vertical Deflection of a Truss Joint by Virtual Work

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

Notes on this paper

Paper format. National Exams, December 2018 — 07-Str-A1 Elementary Structural Analysis; 3 hours, CLOSED BOOK (an approved Casio or Sharp calculator is permitted). Six questions constitute a complete paper: answer all of Questions 1–5, then one only of Questions 6, 7 or 8. Marks are shown in the left margin (6, 18, 18, 18, 18, then 22 for the optional question). Every question is worked below, including all three alternatives, because the full set is the more useful study resource.

Reference texts. R. C. Hibbeler, Structural Analysis (Ch. 2 determinacy and stability; Ch. 4 influence lines; Ch. 6–9 deflections and virtual work; Ch. 12 moment distribution); A. Kassimali, Structural Analysis (Ch. 3, 6, 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 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 — two figure readings that the printed figures leave open, both stated where they are used. (i) In Question 2(c) the hatched plane under the upper roller is drawn parallel to the inclined member, so the reaction is taken normal to the member. That reading makes every result an integer (R = 28 kN, shears 32 / −4 / −28 kN, axial forces 25 / 10 / 0 kN), whereas a vertical-plane reading gives nothing round; and because the two readings differ only by a force acting along the member axis, the shear and moment diagrams are identical either way. (ii) In Question 7 both wall supports are drawn as triangles on hatching. The truss is analysable only if one of them releases a component; taking the pin at U₂ and the horizontal-only roller at L₃ gives δ = 29.0 mm, and the same value follows if the thin vertical line between the two wall joints is a dimension witness line rather than a member. The alternative (pin at L₃) would give 31.8 mm.

Question 7: Vertical Deflection of a Truss Joint by Virtual Work (22 marks, optional)

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.

Item	Value
Bottom chord	L₁(0, 0), L₂(3, 0), L₃(6, 0) m
Sloping chord	L₁, U₁(3, 1.25), U₂(6, 2.5) — collinear, U₁ at mid-length
Members	L₁-U₁, U₁-U₂ (3.25 m each); U₁-L₂ (1.25 m); L₂-U₂ (3.905 m); L₁-L₂, L₂-L₃ (3 m each); U₂-L₃ (2.5 m)
Loads	10 kN downward at L₁; 30 kN downward at U₁
Axial rigidity	$AE = 36.0 \times 10^3$ kN for every member

Find. The vertical deflection of the free joint L₁.

Question 7: a wall-mounted truss. The sloping chord L₁-U₁-U₂ is straight, so U₁ sits at 1.25 m, half the depth at the wall.

Approach. Solve the truss twice by the method of joints, starting each time from the free end L₁: once for the real loads to get $N$, once for a unit downward load at L₁ to get $n$. Then $\delta = \Sigma\,N n L /(AE)$, summing only over the members that carry both.

Fix the geometry, and notice that the sloping chord is straight. With L₁ at the origin, U₁ at $(3,\ 1.25)$ and U₂ at $(6,\ 2.5)$, the two chord segments have identical slopes, so U₁ is the mid-point of a single straight chord. Each segment is $$L = \sqrt{3^2 + 1.25^2} = 3.25\ \text{m}$$ another 12-5-13 triangle, with direction cosines $(0.9231,\ 0.3846)$. The truss is determinate: $m + r = 7 + 3 = 10 = 2j$.
Real forces — joint L₁ and joint U₁. At L₁ only the chord and the bottom chord meet, so $$0.3846\,N_{L_1U_1} = 10 \ \Rightarrow \ N_{L_1U_1} = +26.0\ \text{kN}, \qquad N_{L_1L_2} = -24.0\ \text{kN}$$ At U₁ the two chord segments are collinear, so horizontal equilibrium forces them equal, $N_{U_1U_2} = +26.0$ kN, and their vertical components then cancel — which leaves the vertical to take the whole 30 kN: $$N_{U_1L_2} = -30.0\ \text{kN (compression)}$$
Real forces — joint L₂ and the wall. The diagonal L₂-U₂ runs $(3,\ 2.5)$ with length $3.905$ m, so its vertical direction cosine is $2.5/3.905 = 0.6402$: $$0.6402\,N_{L_2U_2} = 30.0 \ \Rightarrow \ N_{L_2U_2} = +46.86\ \text{kN}$$ $$N_{L_2L_3} = -(24.0 + 36.0) = -60.0\ \text{kN (compression)}$$ Equilibrium at U₂ then gives wall reactions of 40 kN vertically and 60 kN horizontally, with an equal and opposite 60 kN at L₃, and leaves $$N_{U_2L_3} = 0$$ The short vertical at the wall is a zero-force member: with no vertical restraint at L₃ there is nothing for it to hang from.
Virtual forces for a unit downward load at L₁. The same walk with 10 kN replaced by 1 kN and the 30 kN removed gives $$n_{L_1U_1} = n_{U_1U_2} = +2.60, \qquad n_{L_1L_2} = n_{L_2L_3} = -2.40$$ and — because there is now no load at U₁ — $$n_{U_1L_2} = 0, \qquad n_{L_2U_2} = 0, \qquad n_{U_2L_3} = 0$$ Three of the seven members drop out of the summation entirely, which is worth spotting before any arithmetic: a unit load at L₁ is carried to the wall purely by the straight chord and the bottom chord.
Sum the virtual work. Only four terms survive:

Member	$L$ (m)	$N$ (kN)	$n$	$NnL$ (kN·m)
L₁-U₁	3.25	+26.0	+2.60	+219.70
U₁-U₂	3.25	+26.0	+2.60	+219.70
L₁-L₂	3.00	−24.0	−2.40	+172.80
L₂-L₃	3.00	−60.0	−2.40	+432.00
U₁-L₂	1.25	−30.0	0	0
L₂-U₂	3.905	+46.86	0	0
U₂-L₃	2.50	0	0	0
Sum				+1044.2

Divide by the axial rigidity. Every term is positive, so there is no cancellation to worry about: $$\delta_{L_1} = \frac{\Sigma N n L}{AE} = \frac{1044.2}{36.0 \times 10^3} = 0.02901\ \text{m} = \boxed{29.0\ \text{mm downward}}$$ The positive sign means the deflection is in the direction of the unit load, i.e. downwards, which is what a cantilever tip loaded downwards must do. Note where the deflection comes from: 41 % of the total is the compression of the bottom chord panel next to the wall, and 42 % is the stretching of the two sloping chord segments — the web members contribute nothing at all.

Check: the wall detail, and why the answer is robust to it. Both wall symbols are drawn as triangles on hatching, which would read as two pins and make the truss indeterminate to the first degree, so one of them must release a component. Two readings give the same answer. (i) Pin at U₂ with a horizontal-only roller at L₃, as taken above: U₂-L₃ is then a zero-force member and $\delta_{L_1} = 29.0$ mm. (ii) The thin vertical line between the two wall joints is a dimension witness line rather than a member, in which case there are six bars and two pins, $m + r = 6 + 4 = 10 = 2j$, and the identical member forces and the identical 29.0 mm follow. The only reading that differs is a pin at L₃ with the release at U₂, which puts 40 kN of compression into the short vertical and gives 31.8 mm; it is rejected because the drawing shows the load path arriving at U₂. The 29.0 mm figure is therefore quoted with confidence.

Quantity	Value
Chord forces L₁-U₁ and U₁-U₂	+26.0 kN tension each
Bottom chord L₁-L₂ / L₂-L₃	24.0 / 60.0 kN compression
Vertical U₁-L₂	30.0 kN compression
Diagonal L₂-U₂	46.86 kN tension
Vertical U₂-L₃	zero-force member
Wall reactions	40 kN vertical and 60 kN horizontal at U₂; 60 kN horizontal at L₃
$\Sigma N n L$	1044.2 kN·m
Deflection at L₁	29.0 mm downward

Topic. Virtual work (unit-load method) for pin-jointed trusses

Key relations. $1 \cdot \delta = \sum \dfrac{N n L}{AE}$, with $N$ the real member forces, $n$ the member forces under a unit load applied at and along the required displacement, and tension counted positive in both systems. A positive result means the joint moves in the direction of the unit load.

Why this works. For a truss the internal strain energy is stored purely in axial extension, so the general virtual-work integral collapses to a finite sum — one term per member. The economy of the method comes from the virtual system, which you are free to choose: here the unit load at the free end travels to the wall along the straight chord and the bottom chord only, so three of the seven members have $n = 0$ and their real forces, however large, do no virtual work. Recognising that before computing is worth several minutes in an examination, and it also tells you which members actually control the stiffness of the assembly.

Common pitfall. Mixing sign conventions between the two systems. $N$ and $n$ must both be tension-positive, because the product $Nn$ has to be positive whenever a member elongates in the direction the virtual force pulls it; flipping the sign of one compression member turns a 29.0 mm answer into something smaller and plausible. The other trap is misreading the sloping chord: taking U₁ at 2.5 m instead of 1.25 m (i.e. assuming the top chord is horizontal) changes every member length and every force.

Source. Hibbeler, Structural Analysis, §8.1–8.3 (virtual work for trusses); Kassimali, §7.3; Leet, Uang and Gilbert, §10.4.

← Question 6 Next: Question 8 →