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16-Civ-A1 Elementary Structural Analysis · May 2017

Question 5 of 8: Influence lines

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

Notes on this paper

Paper: National Exams — May 2017, 16-Civ-A1 Elementary Structural Analysis (closed book, 3 h). 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), 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 beams, indeterminate frames.

Sign convention: upward reactions positive; sagging bending moment positive (tension on the underside); member tension positive (T), compression negative (C).

These A1 papers are defined entirely by their figures, so every structure has been read from the printed figure and redrawn to scale below.

Question 5: Influence lines (9 + 11 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.

5(a) Truss influence lines (span 24 m)

Given. Top chord $U_1\dots U_5$ at $y{=}2.5\text{ m}$, spacing $6\text{ m}$; bottom $L_1(6),L_2(12),L_3(18)$; pin $U_1$, roller $U_5$; unit load travels along $U_1\dots U_5$. The ¼/½/¾ points are $U_2,U_3,U_4$. Find. the three influence lines.

U1U2U3U4U5L1L2L36 m6 m6 m6 m2.5 m
5(a): the sub-divided truss; the moving load runs along the top chord.

Placing a unit load successively at each top node and solving the (determinate) truss gives the ordinates below; each influence line is piecewise-linear between panel points.

−1.0IL U₃L₂0
IL for the vertical $U_3L_2$.
−1.2−2.4−1.2IL U₃U₄0
IL for the top chord $U_3U_4$.
+0.65+1.30−0.65IL L₂U₄0
IL for the diagonal $L_2U_4$.
Member¼ ($U_2$)½ ($U_3$)¾ ($U_4$)
$U_3L_2$ (vertical)0−1.000
$U_3U_4$ (top chord)−1.20−2.40−1.20
$L_2U_4$ (diagonal)+0.65+1.30−0.65

5(b) Moving vehicle on a 25 m simple span

Given. Simple span $L{=}25\text{ m}$; section at $a{=}5\text{ m}$; vehicle = UDL $20\text{ kN/m}$ over $10\text{ m}$. Find. IL for shear at ①–① and the maximum absolute shear.

①–①5 m25 m
5(b): simple beam, section 5 m from the left support.
  1. Influence line. For unit load right of the cut $V=\tfrac{L-p}{L}$; left of the cut $V=\tfrac{L-p}{L}-1=-\tfrac{p}{L}$. Ordinates: $0$ at $A$, $-0.20$ just left of the section, jumping to $\boxed{+0.80}$ just right, then linearly to $0$ at $x{=}25\text{ m}$.
  2. Maximum shear. The positive lobe dominates; place the $10\text{ m}$ UDL from the section rightward ($x=5$ to $15$), where the ordinate runs $0.80\to0.40$. Area $=\tfrac{(0.80+0.40)}{2}(10)=6.0\text{ m}$, so $V_{\max}=20(6.0)=\boxed{120\text{ kN}}$ (the small negative lobe gives only $\approx10\text{ kN}$).
−0.2+0.8IL shear @ ①–①0
5(b) influence line for shear at ①–①.