16-Civ-A1 Elementary Structural Analysis · May 2013
Question 8 of 8: Influence lines and force under a moving distributed load
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams, May 2013 — 98-Civ-A1 Elementary Structural Analysis. Three-hour, closed-book examination (approved Sharp/Casio calculator only). Format: six questions constitute a complete paper — answer all of #1–#5 and any one of #6, #7 or #8. All eight questions are solved below for completeness.
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; tension member forces positive (T), compression negative (C).
Question 8: Influence lines and force under a moving distributed load (22 marks)
Given. A determinate subdivided truss: top chord $U_1\ldots U_5$ at 5 m, panels 6 m (span 24 m); bottom chord $L_1$ (pin), $L_2$, $L_3$ (roller); mid-height sub-joints $M_1(6,2.5)$ and $M_2(18,2.5)$. Unit loads travel across the top chord.
Find. (a) The three influence lines and their peak ordinates (T/C). (b) The $M_1$–$M_2$ force under a full-span 6 kN/m top-chord load.
Approach. Place a unit load successively at each top panel point $U_1\ldots U_5$, solve the truss (method of joints), and record the target member force — these are the influence ordinates, connected by straight lines. For (b), the member force under a distributed load equals $w\times$(area under its influence line).
Q8 truss (interpreted web); highlighted: the three members whose influence lines are required.
Influence line, $L_1$–$L_2$: peak ordinate ==**1.8 (T)**== at U₂.
Influence line, $M_1$–$L_2$: peak |ordinate| ==**0.65**== (0.65 C at U₂, 0.65 T at U₄).
Influence line, $M_1$–$M_2$: peak |ordinate| ==**1.2 (C)**== at U₂ and U₄.
(a) Ordinates. Solving the truss for a unit load at each of $U_1\ldots U_5$: for $L_1L_2$ the ordinates are $0,\,1.8,\,1.2,\,0.6,\,0$; for $M_1L_2$, $0,\,-0.65,\,0,\,0.65,\,0$; for $M_1M_2$, $0,\,-1.2,\,0,\,-1.2,\,0$. Peaks: ==**$L_1L_2=1.8$ (T)**==, ==**$M_1L_2=0.65$**==, ==**$M_1M_2=1.2$ (C)**==.
(b) Full-span UDL on $M_1$–$M_2$. The area under its influence line is $\int_0^{24} \eta\,dx = 4\times\Big(\tfrac{0+(-1.2)}{2}\times6\Big)=-14.4$ (m, four trapezoids). Hence $F_{M_1M_2}=w\!\int\eta\,dx=6(-14.4)=-86.4$ kN $\Rightarrow$ ==**$F_{M_1M_2}=86.4$ kN (compression)**==.
Quantity
Value
Peak IL $L_1$–$L_2$
1.8 (T) at U₂
Peak IL $M_1$–$L_2$
0.65 (C at U₂ / T at U₄)
Peak IL $M_1$–$M_2$
1.2 (C) at U₂, U₄
$M_1$–$M_2$ force under 6 kN/m
86.4 kN (C)
Check: the printed web layout is ambiguous; the truss is analysed with the determinate diamond-panel interpretation (sub-verticals U₂M₁, U₄M₂; central chord M₁M₂ crossing the U₃–L₂ line without connection). All ordinates below are from a solver check of that model; a different web reading would shift the M₁M₂ magnitude.
