Question 3 of 8: Vertical deflection of a pin-jointed truss by virtual work

Question 3: Vertical deflection of a pin-jointed truss by virtual work (16 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 symmetric parallel-panel truss of 8 m span in four 2 m panels, 1.5 m deep, pinned at L₁ and on a roller at L₅, carrying 18 kN downward at each of L₂, L₃ and L₄.

Find. The vertical deflection of joint L₃.

Approach. Use the unit-load form of virtual work for a pin-jointed frame: solve the truss once for the real 18 kN panel loads to get the member forces $N$, once for a unit downward load at L₃ to get the virtual forces $n$, and sum $NnL/EA$ over the members.

The governing expression for an axially loaded assembly is

in which $N$ is the member force under the real loading, $n$ the force in the same member under a unit load applied at and along the required displacement, and $L$ the member length. Symmetry halves the work: only one half of the truss need be analysed, and both load systems are symmetric about U₂–L₃.

1. Real reactions. The loading is symmetric, so each support carries half the total: $$ R_{L_1} = R_{L_5} = \tfrac{1}{2}(3 \times 18) = \boxed{27.0\ \text{kN}} $$
2. Real member forces by joints. At L₁ the two members are the end diagonal L₁–U₁ (direction $2:1.5$, i.e. components $0.8$ and $0.6$) and the bottom chord. Vertical equilibrium gives $0.6\,N_{L_1U_1} + 27 = 0$, so $N_{L_1U_1} = -45.0$ kN (compression), and horizontal equilibrium then gives $N_{L_1L_2} = +36.0$ kN (tension). Continuing joint by joint to the centre and mirroring, the whole set is $$ \boxed{N_{\text{chords}} = +36\ \text{kN},\quad N_{\text{top}} = -48\ \text{kN}} $$ with $N_{U_1L_2} = N_{U_3L_4} = +18$ kN, $N_{U_2L_3} = 0$, $N_{L_1U_1} = N_{U_3L_5} = -45$ kN and $N_{U_1L_3} = N_{L_3U_3} = +15$ kN.
3. Virtual system. Remove the real loads and apply 1 kN downward at L₃. The reactions are 0.5 kN at each support and the same joint sweep gives $$ n_{\text{chords}} = +\tfrac{2}{3},\quad n_{\text{top}} = -\tfrac{4}{3}, \quad n_{L_1U_1} = -\tfrac{5}{6},\quad n_{U_1L_3} = +\tfrac{5}{6} $$ and $n = 0$ in all three verticals, because with the single load applied at L₃ the joints L₂ and L₄ carry no load and U₂ is a zero-force joint of the usual kind.
4. Assemble the sum, group by group. The four bottom chords contribute $4(36)(\tfrac{2}{3})(2) = 192$; the two top chords contribute $2(-48)(-\tfrac{4}{3})(2) = 256$; the two end diagonals contribute $2(-45)(-\tfrac{5}{6})(2.5) = 187.5$; the two inner diagonals contribute $2(15)(\tfrac{5}{6})(2.5) = 62.5$. The verticals contribute nothing, either because $n = 0$ or because $N = 0$. Hence $$ \sum N n L = 192 + 256 + 187.5 + 62.5 = \boxed{698\ \text{kN}^2\!\cdot\!\text{m}} $$
5. Divide by the axial rigidity. $$ \Delta_{L_3} = \frac{698}{69\,800} = 0.0100\ \text{m} \;\Rightarrow\; \boxed{\Delta_{L_3} = 10.0\ \text{mm downward}} $$ The result is positive, so the joint moves in the direction of the unit load, that is, downward.

Two features of the arithmetic are worth noting because they are the marks the examiner is looking for. Every product $Nn$ is positive — the compression members are compressed by both load systems and the tension members are stretched by both — so no cancellation occurs and a sign slip would be obvious. And the vertical U₂–L₃ carries zero force under the real loading even though the 18 kN load hangs directly beneath it, because at L₃ the two 2.5 m diagonals take the whole panel load between them.