Virtual-work deflection of a trapezoidal frame (20 marks): 24-Bld-A1 Elementary Structural Analysis, December 2016, Q8

Question 8: Virtual-work deflection of a trapezoidal frame (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. Node 1 (0,0) pin; rafter up to node 2 (1.6, 1.2); flat top chord to node 3 (5.6, 1.2); rafter down to node 4 (7.2, 0) roller. Point loads 12 kN downward at nodes 2 and 3. $EI=4000\text{ kN}\cdot\text{m}^2$; consider flexural strain only (ignore axial shortening).

Find. Vertical deflection at the midpoint of span 2–3.

Fig. Q8 — symmetric trapezoidal frame; the requested point is the midspan of the flat top chord.

Approach. Apply a unit downward virtual load at the midspan point, obtain the real and virtual moment diagrams (both piecewise linear/constant since all loads are point loads), and integrate $\displaystyle\int \frac{mM}{EI}\,dx$ member by member.

Reactions (real load). By symmetry, $R_1=R_4=\boxed{12\text{ kN}}$ each (total load $24\text{ kN}$).
Real moment diagram. Rafter 1–2: $M$ rises linearly from 0 at the pin to $M_2=R_1(1.6)\cos\theta\cdot(\text{lever})=19.2\text{ kN}\cdot\text{m}$ at node 2 (matching $R_1$ times the horizontal lever arm, since the vertical reaction alone produces moment about a horizontal offset). Top chord 2–3: between the two equal point loads with no load in between, shear is zero and moment is constant at $\boxed{19.2\text{ kN}\cdot\text{m}}$ — including at the midspan point requested. Rafter 3–4: falls symmetrically back to 0.
Virtual moment diagram (unit load at midspan of 2–3). Virtual reactions are $0.5$ kN each; $m$ rises linearly $0\to0.8$ kN·m along 1–2, continues linearly $0.8\to1.8$ kN·m along 2–mid (peaking at the unit load), then mirrors back down to $0.8$ at 3 and $0$ at 4.
Integrate $mM/EI$ member by member (Simpson's rule, exact for these low-order polynomials): 1–2 contributes $10.24$; 2–mid contributes $49.92$; mid–3 contributes $49.92$; 3–4 contributes $10.24$ (units kN²·m³). Sum $=120.32$.
Deflection. $\Delta_{\text{mid}}=\dfrac{120.32}{4000}=\boxed{0.03008\text{ m} = 30.1\text{ mm, downward}}$ — confirmed independently by a direct stiffness-method nodal displacement at the same point, which returns the identical 30.08 mm (the EA used for that cross-check was set effectively rigid, so no axial strain enters either method, matching the "flexural strain only" instruction).

Q8 – deflection at midspan (2–3)
Quantity	Value
$R_1=R_4$	12 kN ↑ each
$M$ along top chord 2–3	19.2 kN·m (constant)
$\Delta_{\text{mid}}$	30.1 mm, downward

Topic: Virtual work (unit-load method) for frame deflections, flexural strain only.

Key relations: $\Delta = \displaystyle\sum_{\text{members}}\int \frac{m(x)M(x)}{EI}\,dx$; between two equal point loads with nothing in between, shear is zero and moment is constant — the classic "flat top" of a two-point-load moment diagram.

Why this works: The same complementary-energy argument as Q3 extends directly to a frame: each member's local $x$ and local moment are used in its own integral, and the member's own length and orientation only enter through how $M(x)$ and $m(x)$ were built, not through any extra geometric correction, provided only flexural (bending) strain energy is being tracked.

Common pitfall: forgetting to also include the shortening due to axial force in the inclined rafters when a question does not say "flexural strain only" — here the instruction removes that term deliberately, and a solution that adds it back in would not match the intended answer.

Source: Hibbeler §9.1–9.4 (virtual work, frames); Kassimali §7.5–7.6.

Question 8 of 8: Virtual-work deflection of a trapezoidal frame (20 marks)

Question 8: Virtual-work deflection of a trapezoidal frame (20 marks)