07-Str-A1 · December 2013

Question 6 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 Examinations, December 2013 — 07-Str-A1 Elementary Structural Analysis. Three hours, CLOSED BOOK, an approved Sharp or Casio calculator permitted. Six questions constitute a complete paper: candidates answer ALL of Questions 1 to 5 and ONE of Questions 6, 7 or 8. Marks are shown in the left margin (6 + 18 + 18 + 18 + 20 + 20 = 100). All eight questions are solved below, because the three alternatives test different topics and the whole set is useful for study.

Reference texts.

R. C. Hibbeler, Structural Analysis, 10th ed. — Ch. 2 (determinacy and stability), Ch. 3 to 5 (trusses, internal loadings, frames), Ch. 6 (influence lines), Ch. 8 to 9 (deflections and virtual work), Ch. 11 to 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 (deflections by virtual work), Ch. 8 (influence lines), Ch. 16 to 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 after which this exam code is named.
Canadian design context: CSA S16 Design of Steel Structures, CSA A23.3 Design of Concrete Structures and the National Building Code of Canada. This is an analysis paper, so no code clause is needed to answer it, but every result below is expressed in the SI units those documents use.

Sign conventions used throughout. For beams and for each individual frame member, shear is positive when the resultant of the forces to the left of (or below) a section acts upward, and bending moment is positive when it sags the member, that is when it puts the underside of a beam or the inside face of a frame member in tension. Member end moments in the slope-deflection work of Question 5 are clockwise-positive on the member, which is the usual convention for that method. Truss forces are quoted as tension or compression rather than by sign. Reactions are drawn in blue and applied loads in red on every figure.

A note on reading this paper. The figures are hand drawn and carry no dimensions that are not on the drawings, so the geometry of every structure below was read from the support symbols one at a time. Two readings materially affect the answers and are flagged where they arise: the plane on which the roller of Question 2(c) bears, and the position of the 50 kN load in Question 8.

Question 6: Vertical deflection of a truss joint 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.

Item	Value
Joint coordinates	$L_1(0,0)$, $L_2(4,0)$, $L_3(8,3)$, $U_1(4,3)$, $U_2(8,6)$, metres
Members	$L_1L_2$, $L_2L_3$, $L_1U_1$, $U_1U_2$, $U_1L_2$, $U_1L_3$, $U_2L_3$ — seven in all
Supports	pin against the vertical face at $U_2$; roller against the vertical face at $L_3$ (horizontal restraint only)
Loads	36 kN downward at $L_1$ and 36 kN downward at $L_2$
Axial rigidity	$EA = 4.0 \times 10^{4}$ kN for every member

Find. The vertical deflection of joint $L_2$.

The real loading (left) and the virtual unit load at $L_2$ (right). The pin at $U_2$ carries the only vertical restraint; the roller at $L_3$ bears on a vertical face and restrains horizontal movement only.

Approach. Solve the truss twice. Once for the real 36 kN loads to get the member forces N, and once for a single 1 kN downward virtual load at $L_2$ to get the forces n. The deflection is then the sum of $nNL/EA$ over the members.

Check the support arrangement first. Seven members, five joints and three reaction components give $m + r = 7 + 3 = 10 = 2j$, so the truss is determinate. The pin at $U_2$ supplies the only vertical restraint, so it must carry the entire applied load, $$V_{U_2} = 36 + 36 = 72\ \text{kN}\uparrow$$ Taking moments about $U_2$ at $(8,\,6)$, the two 36 kN loads give $36(8) + 36(4) = 432$ kN·m and the horizontal roller reaction at $L_3$ acts 3 m below $U_2$, so $$\begin{aligned} H_{L_3} &= -\frac{432}{3} = 144\ \text{kN}\ \text{to the left} \\ H_{U_2} &= 144\ \text{kN}\ \text{to the right} \end{aligned}$$
Find the real member forces. Working from the joints with only two unknowns, $L_1$ carries only $L_1L_2$ and $L_1U_1$, the latter rising 3 in 5, so $L_1U_1 = +60$ kN and $L_1L_2 = -48$ kN. Joint $L_2$ then gives $U_1L_2 = +72$ kN from vertical equilibrium and $L_2L_3 = -60$ kN from horizontal equilibrium. Joint $U_1$ closes the top of the truss, and the results are $$U_1U_2 = +180, \quad U_1L_3 = -96, \quad U_2L_3 = -36 \quad \text{(kN)}$$
Apply the virtual unit load. Remove the 36 kN loads and hang 1 kN downward at $L_2$. The reactions become 1 kN vertically at $U_2$ and $1(4)/3 = 1.333$ kN horizontally at each support, and repeating the joint work gives $$\begin{aligned} n_{U_1U_2} &= +\tfrac{5}{3} \\ n_{U_1L_2} &= +1 \\ n_{U_1L_3} &= -\tfrac{4}{3} \end{aligned}$$ Every other member has $n = 0$. That is worth pausing on: the three bottom-chord and end-diagonal members do not participate at all, because a unit load hung at $L_2$ is carried straight up the post to $U_1$ and then along the top chord to the support.
Sum the products. Only three members contribute, and their lengths are 5 m, 3 m and 4 m respectively: $$\delta_{L_2} = \sum \frac{nNL}{EA} = \frac{1}{4.0 \times 10^{4}}\Big[\tfrac{5}{3}(180)(5) + (1)(72)(3) + \big(\!-\tfrac{4}{3}\big)(-96)(4)\Big]$$ $$= \frac{1500 + 216 + 512}{4.0 \times 10^{4}} = \frac{2228}{4.0 \times 10^{4}}$$
Evaluate and interpret. $$\delta_{L_2} = \boxed{0.0557\ \text{m} = 55.7\ \text{mm downward}}$$ The result is positive, so the movement is in the direction of the unit load. Note how it is made up: the long top chord alone supplies 37.5 mm of the 55.7 mm, because it is the most heavily loaded member and the one whose virtual force is largest. If the top chord were doubled in area the deflection would fall to about 37 mm, which is the sort of information a virtual work calculation gives for free.

Member	L (m)	N (kN)	n	$nNL/EA$ (mm)
$L_1L_2$	4.000	−48.0	0	0.00
$L_2L_3$	5.000	−60.0	0	0.00
$L_1U_1$	5.000	+60.0	0	0.00
$U_1U_2$	5.000	+180.0	+1.6667	37.50
$U_1L_2$	3.000	+72.0	+1.0000	5.40
$U_1L_3$	4.000	−96.0	−1.3333	12.80
$U_2L_3$	3.000	−36.0	0	0.00
Total				55.70 mm downward

Topic. Joint deflection of a pin-jointed truss by the unit-load method.

Key relations. $1 \cdot \delta = \displaystyle\sum \frac{nNL}{EA}$, with $N$ from the real loading and $n$ from a unit load placed at the joint and in the direction wanted.

Why this works. The unit-load method is the principle of virtual forces written for an assembly of bars. The virtual system need only be in equilibrium with the unit load; it does not have to be compatible with anything, which is why the same real truss with a single dummy load is the natural choice. Members with zero virtual force drop out entirely, so the sum is usually much shorter than the member list suggests. Here four of the seven members contribute nothing, and the whole answer comes from the load path that a unit load at the joint actually uses.

Common pitfall. Mixing the sign conventions of the two analyses. N and n must both be positive for tension, so a member in real compression carrying a compressive virtual force gives a positive product — $U_1L_3$ contributes $+12.8$ mm here, not $-12.8$ mm. Reversing that one sign gives 30.1 mm and looks perfectly plausible.

Source. R. C. Hibbeler, Structural Analysis, 10th ed., Sect. 8.1–8.3 and 9.4; A. Kassimali, Structural Analysis, 6th ed., Sect. 7.3.

← Question 5 Next: Question 7 →