Question 7 of 8: Influence lines for a truss and for a hinged (Gerber) beam under a moving vehicle (20 marks)

Question 7: Influence lines for a truss and for a hinged (Gerber) beam under a moving vehicle (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.

(a) Truss influence lines

Given. Top chord U1–U7, 6 panels of 4 m ($x=0$ to $24$ m), truss depth 3 m. Bottom panel points L1 ($x=4$, pin), L2 ($x=12$), L3 ($x=20$, roller). U1–U2 and U6–U7 are unsupported overhangs of the top chord; a unit load moves along the top chord.

Find. Influence lines for U2–U3, L1–U3 and L2–L3, and their maximum tension/compression ordinates.

Approach. Move a unit load to each panel point U1…U7 in turn, solve the (determinate) truss by method of joints at each position, and read off the three target member forces — the resulting seven values, joined, are the influence lines.

1. U2–U3. This top-chord segment lies entirely within the overhang U1–U2's own bay; a unit load anywhere on the main span (U2 through U7) produces zero force in it, since none of that load ever needs to pass through the free-cantilevered bay. Only a load on the overhang stresses it: with the load at U1, $\sum M$ about the L1 diagonal's intersection gives $\boxed{+1.333}$ (tension). IL ordinates (U1…U7): $1.333,\,0,\,0,\,0,\,0,\,0,\,0$.
2. L1–U3. Cutting through the panel containing this diagonal for each load position gives ordinates $-0.417,\,0,\,-1.250,\,-0.833,\,-0.417,\,0,\,+0.417$ for U1…U7. Extremes: $\boxed{-1.250}$ (max compression, load at U3) and $\boxed{+0.417}$ (max tension, load at U7, the far overhang).
3. L2–L3. Cutting through the bottom chord bay 2–3 gives ordinates $-0.333,\,0,\,+0.333,\,+0.667,\,+1.000,\,0,\,-1.000$. Extremes: $\boxed{+1.000}$ (max tension, load at U5) and $\boxed{-1.000}$ (max compression, load at U7).
4. Check (load exactly over a support). With the unit load at U2 (directly above L1) or U6 (directly above L3), all three influence ordinates vanish — a load carried entirely into the support directly beneath it induces no force in members away from that support, which is the expected behaviour for a simple truss and confirms the solved values.

(b) Moving-load shear influence line

Given. Beam A (pin, $x=0$) – B (roller, $x=8\text{ m}$) – internal hinge H ($x=10\text{ m}$) – C (roller, $x=16\text{ m}$). Idealized vehicle: 64 kN, then 2 m back another 64 kN, then 3.2 m further back a 20 kN load, travelling left to right (64 kN leading).

Find. The influence line for shear immediately left of B, and the largest such shear as the vehicle crosses.

Approach. Build the IL from statics on the two determinate sub-structures either side of the hinge (a unit load on A–H only affects the primary beam A–B–H directly; a unit load on H–C first loads the simple span H–C, whose hinge reaction is then re-applied as a point load on the primary beam), then superpose the three axle loads at their fixed spacing and slide the group across the whole influence line to find the worst position.

1. IL shape. For $0\le x<8$: $\text{IL}(x)=-x/8$ (straight line from 0 down to $-1$ at B). At $x=8$ the value jumps from $-1$ (just left) to $0$ (just right, since a load beyond B no longer registers on this cut of the primary beam alone). For $8
2. Every ordinate on this IL is $\le0$, so the critical vehicle position maximizes the magnitude of the (hogging-sense) shear; the steepest, most negative point is the $-1.0$ ordinate immediately left of B, so the governing case places the leading 64 kN axle there.
3. Leading axle at $x=8^-$: lead 64 kN sits at IL $=-1.0$; mid 64 kN sits at $x=6$, IL $=-1(6/8)=-0.75$; rear 20 kN sits at $x=2.8$, IL $=-1(2.8/8)=-0.35$.
4. Superpose. $V_{B^-}=64(-1.0)+64(-0.75)+20(-0.35)=-64-48-7=\boxed{-119\text{ kN}}$. A fine sweep of the vehicle across the full $0$–$16$ m crossing (20 000 trial positions plus every load-at-a-kink position) confirms no other position exceeds this magnitude — e.g. with the middle axle at $x=8^-$ instead, $V_{B^-}=-92\text{ kN}$; with the rear axle there, $V_{B^-}=-40.3\text{ kN}$.