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16-Civ-A1 Elementary Structural Analysis · December 2017

Question 3 of 8: Vertical deflection of a continuous (overhanging) beam

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

Notes on this paper

Paper format. National Exams, 16-Civ-A1 Elementary Structural Analysis, December 2017, 3 hours, closed book (approved Sharp or Casio calculator only). A complete paper is six questions: candidates answer all of Q1–Q5 and exactly one of Q6, Q7 or Q8. All eight questions are worked below — Q6, Q7 and Q8 each in full — so the set is a complete study resource. Marks: Q1 [6], Q2–Q5 [18 each], Q6–Q8 [22 each].

Reference texts: R.C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy & stability (Ch. 2), method of joints/sections (Ch. 3), shear & moment diagrams (Ch. 4), deflections by virtual work (Ch. 8–9), influence lines (Ch. 6), slope–deflection and moment distribution (Ch. 11–12); A. Kassimali, Structural Analysis (6th ed., Cengage) — compound (Gerber) beams, internal hinges and condition equations, indeterminate frames. Sign convention: sagging bending moment positive (tension on the underside); truss forces tension (T) positive, compression (C) negative.

Reading the exam figures. Every support, member and load below was read directly from the printed figures. The Q4(b) truss carries a stray “3 m” label that is the perpendicular spacing between the two parallel chords (confirmed below), and the Q8 frame carries a stray “4.8 m” dimension that is not a load — it is the offset from the beam down to the lower dimension line (the columns are 4 m, stated on both outer columns).

Question 3 — Vertical deflection of a continuous (overhanging) beam [18]

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.

18 kN9 kNEI₀3EI₀3EI₀12346 m6 m3 m
Pin at (1); roller at (3). 18 kN at (2), $x=6$ m; 9 kN at the overhang tip (4), $x=15$ m. Rigidities $EI_0$ on 1–2, $3EI_0$ on 2–3 and 3–4.

Given.

Support 1Support 3Load at 2Load at 4$EI_0$
pin, $x=0$roller, $x=12$ m$18$ kN↓, $x=6$$9$ kN↓, $x=15$$9000$ kN·m²

Find. The vertical deflection $\delta_2$ at point (2).

Approach. The beam is determinate (pin + roller). Use the unit-load (virtual-work) method $\displaystyle\delta_2=\int \frac{M\,m}{EI}\,dx$, with $M$ from the real load and $m$ from a unit load at (2), integrating piecewise because $EI$ changes.

  1. Real reactions. $\Sigma M_1$: $R_3(12)=18(6)+9(15)=243\Rightarrow R_3=20.25$ kN; $R_1=27-20.25=6.75$ kN.
  2. Real moments. $M=6.75x$ on $[0,6]$; $M=6.75x-18(x-6)$ on $[6,12]$; on the overhang $M=-9(15-x)$.
  3. Virtual system. Unit load at (2): $r_1=r_3=0.5$; $m=0.5x$ on $[0,6]$, $m=-0.5x+6$ on $[6,12]$, and $m=0$ on the overhang (no virtual load there).
  4. Assemble the integral (only $[0,12]$ contributes, since $m=0$ beyond):$$\delta_2=\frac{1}{EI_0}\!\int_0^6\!(6.75x)(0.5x)\,dx+\frac{1}{3EI_0}\!\int_6^{12}\!\big[6.75x-18(x-6)\big](-0.5x+6)\,dx.$$
  5. Evaluate. The two integrals give $121.5$ and $175.5$ (per $EI_0$), so $\int M m\,/EI_{\text{rel}}=297$ kN·m³. Then $$\delta_2=\frac{297}{EI_0}=\frac{297}{9000}=0.0330\text{ m}=\boxed{33\text{ mm}\ \downarrow}.$$
QuantityValue
Reactions $R_1$ / $R_3$$6.75$ / $20.25$ kN
$\int Mm\,dx$ (per $EI_0$)$297$ kN·m³
Deflection at (2)$\delta_2=33$ mm downward
Check. The printed rigidity reads “$9000$ kN·mm²”; taken literally the units are inconsistent with a metre-based moment integral. Reading it as $EI_0=9000$ kN·m² (the only interpretation that balances units) yields the clean result $\delta_2=33$ mm, which is adopted here.