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16-Civ-A1 Elementary Structural Analysis · May 2014

Question 8 of 8: Deflection 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 8: Deflection by virtual work (22 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.

2 kN/m5 kN7 kN12341 m3 m2 m8: free tip at node 1 (5 kN), pin at node 2 (1 m), 7 kN at node 3 (4 m), roller at node 4 (6 m); UDL 2 kN/m over 0–4 m.

Given. Beam nodes $1(x{=}0,\text{free tip}),2(1,\text{pin}),3(4),4(6,\text{roller})$. Loads: 5 kN down at the tip, UDL 2 kN/m over 0–4 m, 7 kN down at node 3. $EI=1.7\times10^{3}$ kN·m².

Find. Vertical deflection at node 3 ($x=4$).

Approach. This is a determinate overhanging beam, so the real bending moment follows directly from statics: a hogging cantilever action over the left overhang (the 5 kN tip load plus the UDL) and a sagging span between the two supports. The unit-load theorem gives the deflection at point 3 as $\tfrac1{EI}\int_0^L M\,m\,dx$, where $m$ is the moment produced by a single downward 1 kN load placed at node 3. Because $EI$ is constant it factors out of the integral, which is then evaluated over the tip-load region, the UDL region, and the interior span.

  1. Real reactions. Supports at $x=1,6$. $\sum M_{x=1}=0$: $5R_4 = 5(0{-}1)+8(2{-}1)+7(4{-}1)$ … $\Rightarrow R_4=\boxed{4.8\ \text{kN}\uparrow}$, $R_2=20-4.8=\boxed{15.2\ \text{kN}\uparrow}$ (total load $5+8+7=20$ kN).
  2. Unit system. Apply a 1 kN dummy downward load at node 3 ($x=4$); its reactions are $r_2=0.6,\ r_4=0.4$.
  3. Virtual-work integral. $\displaystyle \delta_3 = \int_0^6 \frac{M(x)\,m(x)}{EI}\,dx$, integrating the real moment $M$ (tip cantilever + UDL + interior span) against the unit moment $m$ over the whole beam.
  4. Result. Evaluating the product gives $\displaystyle \delta_3 = \frac{18.30}{1.7\times10^{3}} = \boxed{0.01076\ \text{m}=10.8\ \text{mm}\downarrow}$.
QuantityValue
Vertical deflection at point 310.8 mm downward
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