Influence lines: 16-Civ-A1 Elementary Structural Analysis, May 2014, Q6

Question 6: Influence lines (20 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.

6(a) — Truss influence lines (top-chord loading)

Given. Truss $U_1(0,4.5),L_1(6,0),U_2(12,4.5),L_2(18,0),U_3(24,4.5)$ m; pin at $U_1$, roller at $U_3$; a unit downward load moves along the top chord (applied at $U_1,U_2,U_3$).

Find. Influence lines and peak coefficients for $U_1U_2$ and $U_2L_2$.

Method. Place the unit load at each top panel point and solve the truss; the ordinate is linear between panel points. Load at $U_1$ or $U_3$ (over a support) gives zero in both bars.
$U_1U_2$. Ordinate at $U_2$ $= -0.667$; peak $|\eta| = \boxed{0.667\ \text{(C)}}$ at mid-span.
$U_2L_2$. Ordinate at $U_2$ $= -0.833$; peak $|\eta| = \boxed{0.833\ \text{(C)}}$ at mid-span.

6(b) — Shear influence line + moving vehicle

Section A–A is marked at $x=8$ m. Vehicle: 100 kN — 1.5 m — 100 kN — 3 m — 50 kN.

Given. Beam with pin at $x=2$, roller at $x=17$, free 2 m overhangs each end; shear wanted at A–A, $x=8$ m. Idealised vehicle 100, 100, 50 kN at 1.5 m and 3 m spacing crosses the span.

Find. Influence line for $V_{A}$ and the maximum shear at A–A as the vehicle crosses.

Influence line. With span $a=15$ m between supports and A–A 6 m right of the pin, $\eta_V = +0.6$ just right of A and $-0.4$ just left of A, zero at each support, and $\pm0.133$ at the overhang tips.
Position for maximum. Place the two 100 kN axles on the steep positive branch just right of A: at $x=8,\,9.5$ ($\eta=0.600,0.500$) and the 50 kN at $x=12.5$ ($\eta=0.300$).
Maximum shear. $V_A = 100(0.600)+100(0.500)+50(0.300) = \boxed{125\ \text{kN}}$. (The largest negative, loads on the left branch, is only $-50$ kN.)

Quantity	Value
6(a) $U_1U_2$ peak IL	0.667 (C) at mid-span
6(a) $U_2L_2$ peak IL	0.833 (C) at mid-span
6(b) IL$_{V}$ at A–A	$+0.60$ (right) / $-0.40$ (left)
6(b) max shear at A–A	125 kN

Concept & Theory

Topic: influence lines for a determinate truss bar and for beam shear under a moving axle train.

Key relations: IL by unit-load: solve the structure for a unit load at each load position; peak effect from an axle train is $\sum P_i\eta_i$ maximised over train position; for beam shear the IL steps by 1 across the section.

Why this works: each IL ordinate is the response to a unit load there, so a real load set’s effect is the dot product of loads with ordinates; the train is positioned so the heaviest axles sit on the largest-ordinate branch.

Common pitfall: reading the truss ordinate off the wrong chord level (loads run on the top chord here), and forgetting the shear IL discontinuity of $+1$ across A–A when moving from the left to the right branch.

Source: Hibbeler, Structural Analysis (10th ed.), Chs. 2–12; Kassimali, Structural Analysis (6th ed.).

Question 6 of 8: Influence lines

Question 6: Influence lines (20 marks)

6(a) — Truss influence lines (top-chord loading)

6(b) — Shear influence line + moving vehicle