Question 5 of 8: Influence lines

Question 5: 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.

(a) Influence lines for the frame at B, and the horizontal reaction at E

Given. Top beam A–B–C–D with A at 1 m left of B (free tip), C a hinge 2 m right of B, D a pin 6 m further right (8 m from B); a vertical leg drops 2 m from B to pin E. A unit load travels along A–D.

Find. Influence lines for the moment and shear just right of B, and for the horizontal reaction at E; the peak ordinate of each.

Approach. With a hinge at C, split into body 1 (A–B–C–leg–E) and body 2 (C–D); solve each body’s reactions as a function of the unit-load position $$x$$ (measured from B), then cut just right of B for V and M.

1. Load on A–C $$(-1\le x\le2)$$: body 2 (C–D, unloaded, pin at D only) forces zero vertical hinge shear, so $$R_{y,E}=1$$ (constant) and, taking moments of body 1 about E(0,−2): $$R_{x,E}=x/2$$.
2. Load on C–D $$(2\le x\le8)$$: body 2 alone (simple span C–D=6 m) gives the hinge shear $$=(8-x)/6$$, which becomes $$R_{y,E}=R_{x,E}=(8-x)/6$$ from body 1 (no load on it, so its own moment sum about E must vanish, forcing $$R_{x,E}=R_{y,E}$$).
3. Horizontal reaction at E — influence line. Linear from $$-0.5$$ at A $$(x=-1)$$ up to $$\boxed{+1.0}$$ at C $$(x=2)$$, then back down to 0 at D $$(x=8)$$. Maximum $$|R_{x,E}|=\boxed{1.0}$$, at C.
4. Shear/moment just right of B — cutting the A–B overhang + leg free body: with $$R_{y,E}=1$$ whenever the load sits on A–C, $$V_{B^+}=0$$ for $$x<0$$ (load left of B not yet in this free body) and $$V_{B^+}=-1$$ for $$0, tapering linearly to 0 at D. $$M_{B^+}$$ runs linearly from $$\boxed{-2}$$ at A, through 0 at B, up to $$\boxed{+2}$$ at C, back down to 0 at D.

(b) Influence line for L₁–U₂ and the critical vehicle position

Given. A 24 m simply-supported truss, pin at L₁, roller at L₅; panels of 6 m, height 5 m; the diagonal L₁U₂ spans 2 panels (a counter-diagonal). Idealized vehicle: 100 kN, 100 kN, 50 kN, spaced 2 m then 6 m, travelling left to right.

Find. The influence-line ordinates for L₁U₂, and the maximum compression and tension it experiences as the vehicle crosses.

Approach. Cut the panel between U₁/L₂ and U₂, severing U₁U₂, L₂L₃ and L₁U₂. With a unit load anywhere on the span, use whichever free body (left of the cut or right of it) does not contain the load, so only the known reaction appears; the influence line is then a straight line between panel points 0–6–12–24.

1. Unit load at 0≤x≤6 (right free body, unloaded): $$F_{L1U2}=\dfrac{13}{5}R_{L5}=\dfrac{13}{5}\cdot\dfrac{x}{24}=\dfrac{13x}{120}$$ giving $$F(0)=0,\ F(6)=\boxed{+0.65}$$ (tension).
2. Unit load at 12≤x≤24 (left free body, unloaded): $$F_{L1U2}=-\dfrac{13}{5}R_{L1}=-\dfrac{13}{5}\cdot\dfrac{24-x}{24}$$ giving $$F(12)=\boxed{-1.30}$$ (compression), $$F(24)=0$$. Between x=6 and x=12 the influence line is the straight line joining these two panel-point values.
3. Sweep the vehicle (100–100–50 kN, spacings 2 m/6 m) across the span, evaluating $$\Sigma(\text{load}\times\text{ordinate})$$ at every position where a wheel sits on a panel point (the governing positions for a piecewise-linear IL). Placing the 50 kN wheel at x=6 (ordinate +0.65) and the middle 100 kN wheel at x=12 (ordinate −1.30) simultaneously — possible because those panel points are exactly the vehicle’s 6 m axle spacing apart — with the lead 100 kN at x=14 (ordinate $$-13(10)/120=-1.083$$): $$F_{max,C}=100(-1.083)+100(-1.30)+50(0.65)=\boxed{-205.8\text{ kN (compression)}}$$
4. Maximum tension: keep the vehicle’s rear wheel off the (only 6 m wide) positive zone: lead 100 kN at x=6 (+0.65), middle 100 kN at x=4 $$(13(4)/120=+0.433)$$, rear 50 kN at x=−2 (off the span, no contribution): $$F_{max,T}=100(0.65)+100(0.433)=\boxed{+108.3\text{ kN (tension)}}$$