07-Str-A1 · December 2015

Question 3 of 8: Deflection of a trapezoidal frame 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 3: Deflection of a trapezoidal frame by virtual work (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 trapezoidal frame of three members, pinned at joint 1 and on a roller at joint 4, carrying a single vertical load at joint 2.

Given data for Question 3
Quantity	Value
Joint coordinates (x, y)	1 (0, 0); 2 (3, 4); 3 (9, 4); 4 (12, 0) m
Member lengths	1–2: 5 m; 2–3: 6 m; 3–4: 5 m
Applied load	32 kN vertically downwards at joint 2
Flexural rigidity	$EI = 28.8 \times 10^{3}\ \text{kN}\cdot\text{m}^2$, all members
Supports	Pin at joint 1, roller at joint 4
Strain energy considered	Flexural only (axial and shear neglected)

Find. (a) the vertical deflection of joint 3 under the given loading, and (b) the vertical deflection of joint 2 when the same 32 kN load is moved to joint 3.

Figure 5 — the real system for part (a) (left) and the virtual system, a unit downward load at joint 3 (right). Both systems are statically determinate, so the bending moments follow from statics alone.

Approach. Apply the unit-load method: solve the real frame for its bending moment $M$, solve the same frame under a unit downward load at the point whose deflection is wanted for its bending moment $m$, and evaluate $\delta = \sum \int M m \, \mathrm{d}s / EI$ member by member.

Reactions in the real system. With the 32 kN load 3 m from the pin on a 12 m span, $$ V_4 = \frac{32(3)}{12} = 8\ \text{kN}, \qquad V_1 = 32 - 8 = 24\ \text{kN}, $$ and the pin carries no horizontal force because no horizontal load is applied.
Reactions in the virtual system. A unit downward load at joint 3, which stands 9 m from the pin, gives $$ V_4 = \frac{9}{12} = 0.75, \qquad V_1 = 0.25 . $$
Bending moments, member by member. Because both systems are determinate, each moment is the moment of the reactions and loads to one side of the section. Measuring $s$ from joint 1 along member 1–2 (whose horizontal projection is $0.6s$), from joint 2 along 2–3, and from joint 4 along 3–4, the sagging moments are
$$ \begin{aligned} 1\text{--}2:&\quad M = 14.4\,s, &\quad m &= 0.15\,s, &\quad 0 \le s \le 5,\\ 2\text{--}3:&\quad M = 72 - 8\,s, &\quad m &= 0.75 + 0.25\,s, &\quad 0 \le s \le 6,\\ 3\text{--}4:&\quad M = 4.8\,s, &\quad m &= 0.45\,s, &\quad 0 \le s \le 5. \end{aligned} $$
The end values are worth a glance: at joint 2 the real moment is $14.4(5) = 72$ kN·m, which is $24 \times 3$ m as it must be, and at joint 3 the virtual moment is $0.75(9) = 2.25$, which is $0.75 \times 3$ m read from the other end.
Integrate over member 1–2. Both diagrams are linear from zero, so $$ \int_0^{5} (14.4 s)(0.15 s)\,\mathrm{d}s = 2.16 \left[\frac{s^3}{3}\right]_0^{5} = 2.16 \left(\frac{125}{3}\right) = 90 . $$
Integrate over member 2–3. Expanding the product, $$ (72 - 8s)(0.75 + 0.25 s) = 54 + 12 s - 2 s^2, $$ so $$ \int_0^{6} \left(54 + 12 s - 2 s^2\right)\mathrm{d}s = 324 + 216 - 144 = 396 . $$
Integrate over member 3–4. By the same form as member 1–2, $$ \int_0^{5} (4.8 s)(0.45 s)\,\mathrm{d}s = 2.16\left(\frac{125}{3}\right) = 90 . $$
Assemble the deflection. Summing the three integrals and dividing by the common flexural rigidity, $$ \delta_3 = \frac{90 + 396 + 90}{EI} = \frac{576}{28\,800} = 0.0200\ \text{m} = \boxed{20.0\ \text{mm}\ \downarrow} $$ The sign is positive, so joint 3 moves in the direction of the unit load, that is, downwards.
Part (b), by Maxwell's reciprocal theorem. The theorem states that for a linearly elastic structure the deflection at 3 caused by a load at 2 equals the deflection at 2 caused by the same load at 3. The question has been framed to test exactly that, so $$ \boxed{\delta_2 = 20.0\ \text{mm}\ \downarrow} $$ Working it out the long way confirms the theorem: the real system now has $V_1 = 8$ kN and $V_4 = 24$ kN, the virtual system has a unit load at joint 2 with $V_1 = 0.75$ and $V_4 = 0.25$, and the three integrals come out as 90, 396 and 90 again — the same three numbers with the roles of the two ends exchanged. Quoting Maxwell earns the marks in a fraction of the time.

Question 3 — results
Quantity	Value
Reactions, real system (load at joint 2)	$V_1 = 24$ kN, $V_4 = 8$ kN
$\sum \int M m\,\mathrm{d}s$	$90 + 396 + 90 = 576\ \text{kN}^2\cdot\text{m}^3$
(a) Vertical deflection of joint 3	$\delta_3 = 20.0$ mm downwards
(b) Vertical deflection of joint 2, load moved to joint 3	$\delta_2 = 20.0$ mm downwards

Topic. Deflection of a determinate plane frame by the unit-load (virtual work) method, and Maxwell's reciprocal theorem.

Key relations. The unit-load equation for flexure, $1 \cdot \delta = \sum \int_0^{L} \dfrac{M m}{EI}\,\mathrm{d}s$, where $M$ is the real bending moment and $m$ the moment under a unit load applied at the point and in the direction of the required deflection. Maxwell's theorem, $\delta_{ij} = \delta_{ji}$, the deflection at $i$ due to a unit load at $j$ equalling the deflection at $j$ due to a unit load at $i$.

Why this works. The virtual-work statement equates the external work done by the virtual unit load moving through the real displacement with the internal work done by the virtual moments moving through the real curvatures $M/EI$. Because the frame is determinate, both moment fields follow from statics without any compatibility condition, so the whole calculation is three straightforward integrals. Maxwell's theorem is a corollary of the symmetry of the flexibility matrix, which in turn follows from the symmetry of the integrand $Mm$ under exchange of the two load positions — which is precisely why the same three numbers, 90, 396 and 90, reappear in part (b).

Common pitfall. Integrating with respect to the horizontal projection instead of the distance along the inclined members: each of members 1–2 and 3–4 is 5 m long, not 3 m, and using 3 m understates the deflection by 40 per cent on those two terms. The second pitfall is forgetting that the inclined members carry moment at all, on the mistaken picture that only the loaded member bends.

Source. Hibbeler, Structural Analysis, 10th ed., §9.4–9.6 (virtual work for beams and frames) and §9.8 (Maxwell); Kassimali, Structural Analysis, 6th ed., §7.4–7.7.

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