Question 5 of 8: Influence lines and moving loads

Question 5: Influence lines and moving loads (6 + 10 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 M at B and V just left of B

Given. A determinate beam 8 m long. Measuring from the left free end: a 1 m overhang to pin A; 3 m to roller B; 1 m to an internal hinge; 2 m to roller C; and a 1 m overhang to the right free end. Restraints $r = 4$ against three equations plus one hinge condition, so the beam is determinate.

Find. The influence line for the bending moment at B and for the shear immediately to the left of B, and the largest absolute influence coefficient of each.

Approach. Place a unit load at a general position $x$, solve the determinate beam by decomposing at the hinge, and read off the required response as a function of $x$. Because the beam is determinate every influence line is piecewise linear, so only the ordinates at the supports, the hinge and the two tips are needed.

1. Split the beam at the hinge. The reach from the hinge (5 m) to the right tip (8 m) is carried by roller C alone plus the force transmitted through the hinge; the reach from the left tip to the hinge is the anchor beam, supported by pin A and roller B. Consequently, when the unit load stands anywhere on the anchor part the drop-in reach carries nothing at all: its own equilibrium forces both the hinge force and $R_C$ to zero.
2. Influence line for the moment at B. Take the free body to the right of B. For a unit load anywhere left of B that free body is completely unloaded (the hinge force and $R_C$ both vanish), so $$\eta_{M_B}(x) = 0 \qquad\text{for } 0 \le x \le 4\ \text{m}.$$ For the load between B and the hinge the same free body carries only the unit load itself, $\eta = -(x - 4)$, which falls linearly to $$\boxed{\eta_{M_B} = -1.00\ \text{m at the hinge}}$$ For the load on the drop-in reach, moments about C give the hinge force $S = (7-x)/2$ and hence $R_C = (x-5)/2$, so $\eta = 3R_C - (x-4) = 0.5x - 3.5$, which climbs through zero at C and reaches $+0.50$ m at the right tip. The maximum absolute influence coefficient is therefore $1.00$ m, with the unit load standing on the hinge.
3. Influence line for the shear just left of B. With the load on the anchor part, $R_B = (x-1)/3$ and $R_A = (4-x)/3$. The shear just left of B is the sum of the upward forces to the left of the section, $$\eta_{V}(x) = R_A - 1 = \frac{1-x}{3} \quad (0 \le x < 4),\qquad \eta_{V}(x) = R_A = \frac{4-x}{3} \quad (4 < x \le 5).$$ The first branch starts at $+0.333$ at the left tip, crosses zero at A and drops to $$\boxed{\eta_{V} = -1.00 \text{ immediately left of B}}$$ then jumps to zero as the load crosses the support. The unit ordinate at the section is the signature of every shear influence line: the jump across the section is exactly 1.
4. Complete the shear line over the drop-in reach. When the load is beyond the hinge the anchor beam feels only the hinge force $S = (7-x)/2$ pressing down at 5 m, which gives $R_A = -S/3$ and therefore $$\eta_{V}(x) = -\frac{7-x}{6},$$ i.e. $-0.333$ at the hinge, zero at C and $+0.167$ at the right tip. The largest absolute coefficient is 1.00, immediately left of B.
5. Sanity-check the shapes. Both lines are zero at A and at C, because a unit load standing on a support is carried entirely by that support and produces nothing anywhere else in a determinate beam; both are exactly zero over the whole 0–4 m reach for $M_B$, because nothing to the right of B is called upon; and both have a break in slope only at supports and at the hinge. Any influence line for a determinate beam that curves, or that fails to vanish at a support, is wrong.

(b) Maximum moment and shear at C under the combined loadings

Given. A 14 m beam: a 2 m left overhang, pin A at 2 m, point C at 4 m, roller B at 12 m and a 2 m right overhang, so the span is 10 m and C lies 2 m into it. Loading (i) a uniform 6 kN/m that may be placed over any portion or portions of the 14 m; loading (ii) two 30 kN concentrated loads held 2 m apart, positioned anywhere on the beam.

Find. The maximum absolute bending moment and the maximum absolute shear at C from the two loadings acting together.

Approach. Draw both influence lines, then apply the two standard rules for a moving load: a uniform load that may be placed at will contributes (intensity) × (area of the influence line of the wanted sign), while a set of concentrated loads contributes (load) × (sum of the ordinates under it), maximised by scanning the axle group across the line.

1. Influence line for the moment at C. With $a = 2$ m from A and $b = 8$ m to B on a 10 m span, the ordinate at C is the familiar $ab/L$, $$\eta_{M_C}(\text{at C}) = \frac{2(8)}{10} = 1.60\ \text{m},$$ and the line is the triangle A(0)–C(1.60)–B(0) extended straight onto the overhangs, reaching $-1.60$ m at the left tip and $-0.40$ m at the right tip. Its positive area is $\tfrac{1}{2}(10)(1.60) = 8.00$ m² and its negative area is $\tfrac{1}{2}(2)(1.60) + \tfrac{1}{2}(2)(0.40) = 2.00$ m².
2. Influence line for the shear at C. The ordinate just right of C is $b/L = 0.80$ and just left of C is $-a/L = -0.20$, with the usual unit jump between them; extended onto the overhangs the line reaches $+0.20$ at the left tip and $-0.20$ at the right tip. Its positive area is $\tfrac{1}{2}(2)(0.20) + \tfrac{1}{2}(8)(0.80) = 3.40$ m and its negative area is $\tfrac{1}{2}(2)(0.20) + \tfrac{1}{2}(2)(0.20) = 0.40$ m.
3. Uniform load: multiply intensity by area. Because the 6 kN/m may be laid over any portions, the worst case simply covers every region of the required sign: $$M_C^{+,\,\text{UDL}} = 6(8.00) = 48.0\ \text{kN}\cdot\text{m},\qquad M_C^{-,\,\text{UDL}} = -6(2.00) = -12.0\ \text{kN}\cdot\text{m}$$ $$V_C^{+,\,\text{UDL}} = 6(3.40) = 20.4\ \text{kN},\qquad V_C^{-,\,\text{UDL}} = -6(0.40) = -2.40\ \text{kN}$$
4. Two-axle group: maximise the sum of the ordinates. The axles are locked 2 m apart, so the search is over one variable. For the moment the best positive placement puts one axle at the apex C (ordinate 1.60) and the other 2 m into the long side at 6 m (ordinate $0.2(12-6) = 1.20$); the alternative, the second axle at A, scores nothing. Hence $$M_C^{+,\,\text{axles}} = 30(1.60 + 1.20) = 30(2.80) = 84.0\ \text{kN}\cdot\text{m}.$$ The worst negative placement stands one axle on the left tip ($-1.60$) with the other at A (zero), giving $-48.0$ kN·m.
5. Repeat for the shear. The largest positive shear ordinate is the $+0.80$ immediately right of C; the companion axle 2 m further along sits at 6 m with ordinate $0.60$, so $$V_C^{+,\,\text{axles}} = 30(0.80 + 0.60) = 42.0\ \text{kN}.$$ For negative shear the ordinates available are only $-0.20$, either just left of C or at the right tip, and the partner axle always lands on a zero, giving $-6.00$ kN.
6. Combine the two loadings. Adding like signs, $$M_C^{\max} = 48.0 + 84.0 = \boxed{132\ \text{kN}\cdot\text{m (sagging)}}$$ $$V_C^{\max} = 20.4 + 42.0 = \boxed{62.4\ \text{kN}}$$ against opposite-sense extremes of $-60.0$ kN·m and $-8.40$ kN. The maximum absolute values asked for are therefore 132 kN·m and 62.4 kN, both governed by the axle pair rather than by the uniform load — the concentrated loads supply about 64 per cent of the design moment and 67 per cent of the design shear.