Influence Lines and a Moving Load Train: 07-Str-A1, May 2016, Q5

Question 5: Influence Lines and a Moving Load Train (21 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 a four-support Gerber beam

Given. Measuring $x$ from support A: pin A at $x = 0$, internal hinge $h_1$ at $x = 3\ \text{m}$, roller B at $x = 5\ \text{m}$, roller C at $x = 9\ \text{m}$, internal hinge $h_2$ at $x = 11\ \text{m}$, roller D at $x = 14\ \text{m}$. Four vertical reactions plus one horizontal against three equilibrium equations and two hinge conditions, so $r - 3 - c = 5 - 3 - 2 = 0$ and the beam is determinate.

Find. Influence lines for $R_B$, for the bending moment $M_B$ over support B, and for the shear immediately to the left of C, each with its largest absolute ordinate.

Approach. Decompose the beam into its three Gerber segments. The suspended reasoning is the reverse of the usual: here the end segments A–$h_1$ and $h_2$–D are the simple spans, each carried at one end by its own support and at the other by the cantilever tip of the middle segment $h_1$–$h_2$, which itself sits on B and C with 2 m overhangs. A unit load on an end segment is therefore delivered to the middle segment as a single force at the hinge, scaled by the usual lever ratio.

Set up the load transfer at each hinge. With a unit load at $x$ on the left segment ($0 \le x \le 3$), simple-span statics gives the force delivered to the middle segment at $h_1$: $$P_{h_1} = \frac{x}{3}$$ Symmetrically, a unit load at $x$ on the right segment ($11 \le x \le 14$) delivers $P_{h_2} = (14-x)/3$ at $h_2$. Every influence ordinate on an end segment is therefore the middle-segment ordinate at the hinge multiplied by that linear factor.
Influence line for the reaction at B. On the middle segment, moments about C give $R_B = (9-x)/4$ for a unit load at $x$. At the left hinge this is $(9-3)/4 = 1.5$ and at the right hinge $(9-11)/4 = -0.5$. Chaining through the transfer factors, $$\eta_{R_B}(x)=\begin{cases} 0.5\,x, & 0 \le x \le 3\\[2pt] (9-x)/4, & 3 \le x \le 11\\[2pt] -\tfrac{1}{6}(14-x), & 11 \le x \le 14 \end{cases}$$ a triangle peaking at the left hinge, so $$\boxed{\left|\eta_{R_B}\right|_{\max} = 1.5\ \text{at }x = 3\ \text{m} \ (\text{the left hinge})}$$
Influence line for the bending moment over B. Take the free body of the middle segment to the left of B, i.e. the 2 m stretch from $h_1$ to B. It carries no support, so $M_B$ is simply the moment of whatever load stands on it: $$\eta_{M_B}(x)=\begin{cases} -\tfrac{2}{3}x, & 0 \le x \le 3\\[2pt] -(5-x), & 3 \le x \le 5\\[2pt] 0, & 5 \le x \le 14 \end{cases}$$ Any load to the right of B produces no moment on that stub at all, so the influence line is identically zero over more than half the beam: $$\boxed{\left|\eta_{M_B}\right|_{\max} = 2.0\ \text{m, hogging, at } x = 3\ \text{m}}$$
Influence line for the shear just left of C. The shear at $x = 9^-$ is the sum of the upward forces to the left of that section, namely $R_B$ less any load already passed: $$\eta_{V_{C^-}}(x)=\begin{cases} x/6, & 0 \le x \le 3\\[2pt] (9-x)/4 - 1, & 3 \le x < 9\\[2pt] (9-x)/4, & 9 < x \le 11\\[2pt] -\tfrac{1}{6}(14-x), & 11 \le x \le 14 \end{cases}$$ with the characteristic unit jump across the section: the ordinate falls to $-1.0$ just left of C and returns to $0$ just right of it. Hence $$\boxed{\left|\eta_{V_{C^-}}\right|_{\max} = 1.0\ \text{immediately left of C}}$$
Read the diagram back as a check. All three influence lines must vanish at every support that is not the one being measured, and they do: $\eta_{R_B}$ is zero at A, C and D; $\eta_{M_B}$ is zero at A and everywhere right of B; $\eta_{V_{C^-}}$ is zero at A, B and D. Each is also piecewise linear, as it must be for a statically determinate structure.

Q5(a). Each influence line is plotted against the position of a unit downward load. The peak of every one of them occurs at an internal hinge or immediately beside the section, which is typical of Gerber beams.

Influence line	Largest absolute ordinate	Position of unit load
Reaction at B	$1.5$	$x = 3\ \text{m}$ (left hinge)
Bending moment over B	$2.0\ \text{m}$ (hogging)	$x = 3\ \text{m}$
Shear immediately left of C	$1.0$	$x = 9^-\ \text{m}$

(b) Influence lines at an interior point and the governing load train

Given. Measuring $p$ from the free left end: pin A at $p = 4\ \text{m}$, point C at $p = 8\ \text{m}$, roller B at $p = 20\ \text{m}$, free right end at $p = 24\ \text{m}$. The span A–B is $L = 16\ \text{m}$ with C dividing it into $a = 4\ \text{m}$ and $b = 12\ \text{m}$. The vehicle carries axles of 24, 24 and 8 kN at 0, 2 and 6 m measured from its leading-left axle, and travels left to right, so the 8 kN axle leads.

Find. Influence lines for $M_C$ and $V_C$, with maximum and minimum ordinates labelled, and the largest absolute bending moment and shear force at C as the vehicle crosses.

Approach. The reaction $R_A = (20-p)/16$ holds for a unit load anywhere on the beam, overhangs included, which makes both influence lines easy to write in closed form. Then place the three-axle train so that a heavy axle stands on the peak, and evaluate; because both influence lines are piecewise linear, the extreme value must occur with an axle at a vertex.

Write the influence line for the bending moment at C. Inside the span the ordinate is the familiar $ab/L = 4(12)/16 = 3.0\ \text{m}$ peak, and on the overhangs the same expressions continue without a break: $$\eta_{M_C}(p)=\begin{cases} 0.75\,(p-4), & 0 \le p \le 8\\[2pt] 0.25\,(20-p), & 8 \le p \le 24 \end{cases}$$ so the ordinates are $-3.0\ \text{m}$ at the left tip, $0$ at A, $+3.0\ \text{m}$ at C, $0$ at B and $-1.0\ \text{m}$ at the right tip.
Write the influence line for the shear at C. With $R_A = (20-p)/16$, $$\eta_{V_C}(p)=\begin{cases} R_A - 1 = (4-p)/16, & p < 8\\[2pt] R_A = (20-p)/16, & p > 8 \end{cases}$$ giving $+0.25$ at the left tip, $0$ at A, $-0.25$ just left of C, a unit jump to $+0.75$ just right of C, $0$ at B and $-0.25$ at the right tip.
Position the train for the largest bending moment. Let $q$ be the coordinate of the rear (24 kN) axle; the axles then stand at $q$, $q+2$ and $q+6$. Trying each heavy axle at the apex, $$\begin{aligned} q = 6&:\ 24(1.5) + 24(3.0) + 8(2.0) = 124\ \text{kN}\cdot\text{m}\\ q = 8&:\ 24(3.0) + 24(2.5) + 8(1.5) = 144\ \text{kN}\cdot\text{m} \end{aligned}$$ The second placement governs. Confirm it is a true maximum by differencing: moving the train left of $q = 8$ changes the total at $24(0.75) - 24(0.25) - 8(0.25) = +10$ per metre, and moving it right at $-3.5$ per metre, so $q = 8$ is the peak. $$\boxed{M_{C,\max} = +144\ \text{kN}\cdot\text{m}}$$
Check the opposite sign for the moment. The largest hogging value comes with the train on the left overhang, rear axle at the tip: $$q = 0:\ 24(-3.0) + 24(-1.5) + 8(+1.5) = -96\ \text{kN}\cdot\text{m}$$ Running the train out further only lifts axles off the beam and reduces the total, so $M_{C,\min} = -96\ \text{kN}\cdot\text{m}$ and the governing absolute value remains the sagging 144.
Position the train for the largest shear. The shear influence line jumps to $+0.75$ immediately right of C, so put the rear 24 kN axle there: $$q = 8:\ 24(0.75) + 24(0.625) + 8(0.375) = 18 + 15 + 3 = 36\ \text{kN}$$ $$\boxed{V_{C,\max} = +36.0\ \text{kN}}$$ For the negative extreme the train must sit on the right overhang, where the ordinate reaches $-0.25$ at the tip; with the rear axle at $q = 22\ \text{m}$ the leading 8 kN axle has already run off the beam and $$24(-0.125) + 24(-0.25) = -9.0\ \text{kN}$$ which beats every position left of C (the best of those is only $-5.0\ \text{kN}$). So $V_{C,\min} = -9.0\ \text{kN}$ and the governing absolute value is the $+36.0\ \text{kN}$ found above.

Q5(b). Both influence lines run negative on the overhangs, which is why the governing hogging moment at C is produced with the vehicle entirely off the span.

Quantity	Value	Governing vehicle position
$\eta_{M_C}$ peak	$+3.0\ \text{m}$ at C	—
$\eta_{M_C}$ at the left tip	$-3.0\ \text{m}$	—
$\eta_{V_C}$ jump at C	$-0.25 \to +0.75$	—
Maximum bending moment at C	$+144\ \text{kN}\cdot\text{m}$	rear 24 kN axle at C; axles at 8, 10 and 14 m
Minimum bending moment at C	$-96\ \text{kN}\cdot\text{m}$	rear 24 kN axle at the left tip; axles at 0, 2 and 6 m
Maximum shear at C	$+36.0\ \text{kN}$	rear 24 kN axle just right of C
Minimum shear at C	$-9.0\ \text{kN}$	rear axle at 22 m, leading axle off the beam

Topic. Influence lines for determinate beams and the placement of a moving load train.

Key relations. An influence line gives a single response function as a unit load traverses the structure; for a determinate structure it is always piecewise linear. For a simple span, the moment influence line at a section dividing the span into $a$ and $b$ peaks at $ab/L$, and the shear influence line has a unit jump across the section. The response to a group of loads is $\sum P_i\,\eta(p_i)$, and the governing arrangement always places one axle at a vertex of the influence line.

Why this works. The Gerber beam of part (a) makes the point that an influence line is a property of the structure, not of any load: the ordinate at a hinge is fixed by the geometry of the segment beyond it, and the fact that $\eta_{M_B}$ is identically zero to the right of B says something real — no traffic on that half of the bridge can ever bend the beam over support B. In part (b) the piecewise linearity is what turns an infinite search over vehicle positions into a handful of trials: between vertices the total response is linear in the train position, so its extreme must occur with an axle at a kink. Differencing the slopes on either side of a candidate confirms it without needing to test every placement, and it also shows why the middle axle is not the one to put on the apex here.

Common pitfall. Confusing the influence line with the bending moment diagram. They have the same units and often a similar shape, but the abscissa of an influence line is the position of the load, not the position of the section. A second pitfall in part (b) is testing only the heaviest axle at the peak: with the two 24 kN axles only 2 m apart, putting the rear one on the apex keeps both of them high on the influence line and beats the obvious placement by 20 kN·m.

Source. Hibbeler, Structural Analysis, 10th ed., §6.1–6.4 and §6.7; Kassimali, Structural Analysis, 6th ed., §8.1–8.4 and §9.2.

Question 5 of 8: Influence Lines and a Moving Load Train

Question 5: Influence Lines and a Moving Load Train (21 marks)

(a) Influence lines for a four-support Gerber beam

(b) Influence lines at an interior point and the governing load train