16-Civ-A1 Elementary Structural Analysis · May 2018
Question 8 of 8: Truss joint deflections by virtual work
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — May 2018, 16-Civ-A1 Elementary Structural Analysis (closed book, 3 h; Casio/Sharp calculator permitted). Six questions constitute a complete paper: answer Q1–Q5 and one of Q6/Q7/Q8. For completeness all eight questions are worked here.
Reference texts: R. C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy (Ch. 2), method of joints & sections (Ch. 3), shear & moment diagrams (Ch. 4), influence lines (Ch. 6), virtual-work deflections (Ch. 8–9), moment distribution / slope–deflection (Ch. 11–12); A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges, compound (Gerber) beams, indeterminate frames.
Sign convention. Upward reactions positive; sagging bending moment positive (tension on the underside), hogging negative; member axial force tension positive (T), compression negative (C).
These A1 papers are defined entirely by their figures, so every structure is redrawn to scale below.
Question 8: Truss joint deflections by virtual work (22 marks)
Given. Cantilever truss anchored to a wall on the right (pin at $U_a$, horizontal roller at $L_3$); $L_1(0,0)\!-\!L_2(4.2,0)\!-\!L_3(7.2,0)$ along the bottom, $U_1(4.2,1.75)$ on the raking top chord to $U_a(7.2,3)$; loads $21\text{ kN}\downarrow$ at $L_1$ and $14\text{ kN}\downarrow$ at $U_1$. Find. $\delta_{L_1},\,\delta_{L_2}$ (vertical).
Q8 — cantilever truss; deflections sought at $L_1$ and $L_2$.
Real bar forces $N$. Method of joints from the free tip gives, e.g., $L_1U_1=+54.6$, $L_1L_2=-50.4$, $L_2L_3=-64.4$, $L_2U_1=-14.0$, $L_2U_a=+19.8\text{ kN}$ (T +). The wall reactions are $U_{a}=(64.4,\,35)\text{ kN}$, $L_{3x}=-64.4\text{ kN}$.
Virtual bar forces $n$. Apply a unit vertical load at $L_1$ (then, separately, at $L_2$) with all real loads removed, and recover the corresponding $n_1$, $n_2$.
Virtual work. $\displaystyle \delta=\sum \frac{N\,n\,L}{AE}$ over the seven members, with $AE=3.9\times10^{4}\text{ kN}$: $$\boxed{\delta_{L_1}=53.3\text{ mm}\ \downarrow,\qquad \delta_{L_2}=8.0\text{ mm}\ \downarrow}$$