16-Civ-A1 Elementary Structural Analysis · December 2017
Question 5 of 8: Influence lines and moving loads
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 5 — Influence lines and moving loads [18]
Bottom chord $L_1..L_5$ at 6 m spacing; pin at $L_2$, roller at $L_4$ (so $L_1,L_5$ are overhangs); top chord $U_1,U_2,U_3$ at 4.5 m. Load travels along the bottom chord.
Given. Panel length $6$ m, truss height $4.5$ m; supports at $L_2$ (pin) and $L_4$ (roller). A unit downward load is placed successively at $L_1,\dots,L_5$; between panel points the ordinate varies linearly.
Find. Influence coefficients (with max T / max C) for $U_1\!-\!U_2$, $U_2\!-\!L_3$, $U_2\!-\!L_4$.
Approach. Place the unit load at each bottom node and solve the determinate truss; plot the resulting bar force versus load position.
$U_1\!-\!U_2$ (top chord over the left overhang). Non-zero only while the load is on the $L_1$ overhang; joint analysis at $L_1$–$U_1$ gives coefficient $+\tfrac{4}{3}$ at $L_1$, zero at $L_2$ and everywhere in the main span. $\boxed{\text{max }T=+1.333\ (\text{load at }L_1),\ \text{no }C}$.
$U_2\!-\!L_3$ (central vertical). Carries the panel load only when it sits at $L_3$: triangular IL, $+1.0$ at $L_3$, zero at $L_2$ and $L_4$. $\boxed{\text{max }T=+1.0}$.
$U_2\!-\!L_4$ (diagonal). Coefficients $+\tfrac56$ at $L_1$, $0$ at $L_2$, $-\tfrac56$ at $L_3$, $0$ at $L_4$, $-\tfrac56$ at $L_5$. $\boxed{\text{max }T=+0.833,\ \text{max }C=-0.833}$.
IL for $U_1\!-\!U_2$: $+1.333$ at $L_1$, zero across the span.
IL for $U_2\!-\!L_3$: triangle, peak $+1.0$ at $L_3$.
IL for $U_2\!-\!L_4$: $+0.833$ (T) at $L_1$; $-0.833$ (C) at $L_3$ and $L_5$.
Part (b) — Moving vehicle, shear at Section ①–①
Simple beam: pin at $x=0$, roller at $x=20$, $4$ m overhang to $x=24$. Section ① at $x=6$ m.
Given. Span $20$ m (pin–roller) with a $4$ m overhang; section at $a=6$ m. Vehicle loads $40,40,20$ kN at spacings $2$ m then $4$ m, travelling right.
Find. The IL ordinates at the section, and the maximum shear as the vehicle crosses.
Approach. For a shear IL at a section between supports the ordinate is $-x/L$ for a load left of the section and $(L-x)/L$ for a load right of it; on the overhang it continues linearly. Slide the three-axle group and evaluate $V=\sum P_i\,\eta_i$.
Ordinates. $\eta(0)=0$; just left of the section $\eta(6^-)=-6/20=-0.30$; just right $\eta(6^+)=14/20=+0.70$; $\eta(20)=0$; on the overhang $\eta(24)=(20-24)/20=-0.20$.
Maximise positive shear. Place the rear $40$ kN just right of the section ($\eta=0.70$), the next $40$ at $x=8$ ($\eta=0.60$) and the $20$ at $x=12$ ($\eta=0.40$): $$V_{\max}=40(0.70)+40(0.60)+20(0.40)=28+24+8=\boxed{60\text{ kN}}.$$
Governing negative shear. With the group on the overhang the largest reversal is $V=40(-0.10)+40(-0.20)=-12$ kN.
Shear IL at ①: $-0.30$ just left, $+0.70$ just right of the section, $0$ at the roller, $-0.20$ at the tip.
