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16-Civ-A1 Elementary Structural Analysis · May 2014

Question 3 of 8: Deflection of a non-prismatic beam 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 and compound structures. Sign convention: sagging bending moment positive; upward reactions positive; tension member forces positive (T), compression negative (C).

The paper directs candidates to answer Q1–Q4, then one of Q5/Q6 and one of Q7/Q8. For completeness all eight questions are worked here.

Question 3: Deflection of a non-prismatic beam by virtual work (15 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.

30 kN30 kN123452 m4 m4 m2 m3: pin at 1, roller at 5; 30 kN at nodes 2 and 4; segments EI (ends) and 2EI (centre). Point 3 is mid-span.

Given. Simply supported span 12 m (pin 1 at $x=0$, roller 5 at $x=12$); 30 kN downward at node 2 ($x=2$) and node 4 ($x=10$). Flexural rigidity $EI$ on the 2 m end segments, $2EI$ on the 8 m centre, $EI=1.6\times10^{4}$ kN·m².

Find. Vertical deflection at mid-span point 3 ($x=6$).

Approach. The structure and its loading are symmetric about mid-span, so the two reactions are equal and the real bending-moment diagram is a trapezoid that is flat between the loads. We apply a unit dummy load at point 3 and evaluate the virtual-work integral $\int Mm/EI(x)\,dx$ piece-by-piece, taking care to divide the central 8 m length by its doubled rigidity $2EI$ while the two end segments use $EI$. Symmetry lets us integrate over the left half only and double the result.

SegmentRange (m)Real $M(x)$ (kN·m)Unit $m(x)$Rigidity
1–20–2$30x$$0.5x$$EI$
2–32–6$60$$0.5x$$2EI$
  1. Real system. By symmetry $R_1=R_5=30$ kN. Between the loads the moment is constant, $M = 30x-30(x-2)=\boxed{60\ \text{kN}\cdot\text{m}}$.
  2. Unit system. A 1 kN dummy load at point 3 gives $r_1=r_5=0.5$, so $m=0.5x$ up to mid-span.
  3. Virtual-work integral (half-beam, doubled by symmetry). $\displaystyle \delta_3 = 2\!\left[\int_0^2\!\frac{(30x)(0.5x)}{EI}dx + \int_2^6\!\frac{(60)(0.5x)}{2EI}dx\right]$.
  4. Evaluate. First integral $= 15\!\int_0^2 x^2dx/EI = 40/EI$; second $= 15\!\int_2^6 x\,dx/EI = 240/EI$. $\displaystyle \delta_3 = \frac{2(40+240)}{EI}=\frac{560}{1.6\times10^{4}} = \boxed{0.0350\ \text{m}=35.0\ \text{mm}\downarrow}$.
QuantityValue
Vertical deflection at point 335.0 mm downward