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16-Civ-A1 Elementary Structural Analysis · December 2015

Question 8 of 8: Horizontal deflection of a truss by virtual work

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Reference texts: R.C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy, method of joints/sections, virtual-work deflections, moment distribution / slope–deflection, influence lines; A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges, compound (Gerber) beams and three-hinged frames. Sign convention: sagging bending moment positive, upward reactions positive, member tension positive (T) / compression negative (C).

The paper requires Q1–Q4 in full plus two of Q5/Q6/Q7/Q8. For completeness all eight questions are worked here.

Question 8: Horizontal deflection of a truss 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.

Left column $L_1(0,0),L_2(0,3),L_3(0,6)$; right column $R_1(8,0),R_2(8,3),R_3(8,6)$; mid $M_1(4,3),M_2(4,6)$Pins at $L_1,R_1$; $80$ kN → at $R_2$; $EA=9.45\times10^{4}$ kN

Find. Horizontal deflection at $L_3$.

80 kN L3L2L1M2M1R3R2R1
8: truss with pins at $L_1,R_1$; horizontal 80 kN at $R_2$; deflection sought at $L_3$.

Approach. Virtual work: $\delta_{L_3}=\displaystyle\sum \frac{N\,n\,L}{EA}$, where $N$ are the real bar forces (80 kN system) and $n$ those from a unit horizontal load at $L_3$.

  1. Real forces $N$. The 80 kN at $R_2$ travels $R_2\to M_1$ ($+80$ T) then splits into $M_1L_1$ ($+50$ T) and $M_1R_1$ ($-50$ C); all other bars are zero-force. Support reactions: $L_1(-40,-30)$, $R_1(-40,+30)$ kN.
  2. Virtual forces $n$ (unit → at $L_3$). Load path $L_3\to M_1$ diagonal ($-1.25$) and $M_1R_1$ ($-1.25$), with $L_1L_2,L_2L_3=+0.75$; others zero.
  3. Products. Only bar $M_1R_1$ carries both a real and virtual force: $N n L = (-50)(-1.25)(5)=+312.5$. Hence $\sum NnL=\boxed{312.5\ \text{kN}^2\text{m}}$.
  4. Deflection. $\displaystyle\delta_{L_3}=\frac{312.5}{9.45\times10^{4}}=3.31\times10^{-3}\text{ m}=\boxed{3.31\text{ mm}\rightarrow}$ (a direct stiffness solve reproduces $3.31$ mm).
QuantityValue
$\sum NnL$312.5 kN²·m
Horizontal deflection at $L_3$3.31 mm (in the direction of the 80 kN load)

The striking feature of this result is that a single diagonal governs the whole deflection: the 80 kN load and the virtual unit load only share a common force in bar $M_1R_1$, so every other member drops out of the summation. This is a good reminder that virtual work rewards a smart choice of the virtual system — here, a unit load applied exactly where (and in the direction) the deflection is wanted.

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