16-Civ-A1 Elementary Structural Analysis · December 2013
Question 7 of 8: Influence lines
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams, December 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; upward reactions positive; tension member forces positive (T), compression negative (C).
Warren truss 7(a): 24 m span, 4 m deep, supports at L₂ (pin) and L₄ (roller) with 6 m overhangs each side. Unit load travels along the bottom chord.
Given. Warren truss, four 6 m panels, 4 m deep; supports at $L_2$ (pin) and $L_4$ (roller); 6 m overhangs to $L_1$ and $L_5$. A unit downward load travels along the bottom chord $L_1\!\to\!L_5$.
Find. Influence lines and peak (max-|value|) influence coefficients for $U_1U_2$, $L_2L_3$, $U_2L_3$.
Approach. Place a unit load at each bottom panel point $L_1\ldots L_5$, solve the truss, and record the three member forces; because the load path is along the bottom chord the influence line is straight between panel points.
Unit load at
$U_1U_2$
$L_2L_3$
$U_2L_3$
L₁
+1.50
−1.125
+0.625
L₂
0
0
0
L₃
0
+0.375
+0.625
L₄
0
0
0
L₅
0
−0.375
−0.625
$U_1U_2$: non-zero only while the load is on the left ($L_1$) overhang; peak influence coefficient $\boxed{1.50}$ (at $L_1$).
$L_2L_3$ (bottom chord): peak magnitude $\boxed{-1.125}$ (compression, load at $L_1$); $+0.375$ within the span.
$U_2L_3$ (diagonal): peak magnitude $\boxed{0.625}$ (at $L_1$ or $L_3$, sign as tabulated).
7(b) — Influence line for bending moment left of joint C
Structure 7(b): Gerber beam A–B–C–D–E with internal hinges at B and D and a 5 m column fixed under C; end supports at A and E.
Influence line for M immediately left of C: a triangle peaking at −2.0 m over hinge B, zero for load at/right of C. Critical 4 m UDL window shaded (a = 3–7 m).
Given. Compound (Gerber) beam A(0)–B(6)–C(8)–D(10)–E(16) m, internal hinges at B and D, a 5 m column fixed under joint C, end supports at A and E. Vehicle = 10 kN/m over a 4 m length, crossing left to right.
Find. Influence line for bending moment immediately left of C, and the largest negative moment there.
Build the influence line. Using a left free body up to the section at C⁻, and noting the moment is zero at hinge B, the ordinate is $-a/3$ for the load on span A–B ($0\le a\le6$), $-(8-a)$ for the load between B and C ($6\le a\le8$), and zero for any load at or right of C. The IL is a triangle with peak $\boxed{-2.0\ \text{m}}$ under hinge B.
Place the 4 m UDL for the largest negative effect. The most negative area under the IL over a 4 m window occurs with the load spanning $a = 3$ to $7$ m (equal ordinates −1.0 at both ends), giving $\int \text{IL}\,da = -6.0\ \text{m}^2$.
