Influence lines and a moving vehicle (18 marks): 16-Civ-A1 Elementary Structural Analysis, December 2019, Q5

Question 5 — Influence lines and a moving vehicle (18 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.

5(a)

Given. A 20 m compound beam with a pin at A $(x=0)$, an internal hinge at C $(x=8)$, and rollers at D $(x=10)$ and E $(x=20)$. Point B is at $x=4$ m. Spacings are A–B 4 m, B–C 4 m, C–D 2 m and D–E 10 m. A unit load travels from A to E.

Find. Influence lines for $M_B$, for shear immediately left of C, and for $M_D$, each with its largest absolute ordinate.

Q5(a) — 20 m compound beam: pin at A, internal hinge at C, rollers at D and E.

Approach. Split the beam at the hinge into two determinate parts and track how each carries a unit load. Segment 1 (A to C) is effectively a simple span propped by the pin at A and by the hinge; segment 2 (C to E) is a beam on rollers at D and E with a 2 m overhang reaching back to the hinge.

With $r = 2+1+1 = 4$ and one hinge, $\mathrm{DSI} = 4-3-1 = 0$, so the beam is determinate and every influence line is made of straight segments.

(i) Influence line for $M_B$. When the unit load stands at $a$ on segment 1, that segment behaves as a simple span of 8 m between A and the hinge, so $$\eta_{M_B}(a) = \frac{a(8-4)}{8} = \frac{a}{2} \ (a \le 4), \qquad \eta_{M_B}(a) = \frac{4(8-a)}{8} \ (4 \le a \le 8)$$ When the load stands beyond the hinge, segment 1 carries nothing at all — the load is taken entirely by the rollers at D and E — so the ordinate is zero from C to E. The peak sits under B: $$\eta_{\max} = \frac{4 \times 4}{8} = \boxed{2.00\ \text{m}}$$

IL for the bending moment at B: a simple-span triangle on A–C only; the whole of C–E is zero because the load is then carried entirely by the D–E segment.

(ii) Influence line for shear immediately left of C. For a load at $a$ on segment 1, the reaction at A is $(8-a)/8$ and the section at $x=8^-$ has the whole unit load to its left, so $$\eta_{V}(a) = \frac{8-a}{8} - 1 = -\frac{a}{8}$$ This falls linearly from 0 at A to $-1$ immediately left of the hinge. Once the load passes the hinge the ordinate drops instantly to zero and stays there. The largest absolute value is $$|\eta|_{\max} = \boxed{1.00}$$ reached with the load standing just to the left of C.

IL for shear just left of C. It falls linearly to −1 at C and is identically zero beyond the hinge.

(iii) Influence line for $M_D$. D is a support, so the only way to bend it is to load the 2 m overhang between the hinge and D, or to load segment 1, which delivers its hinge reaction onto the tip of that overhang. For a unit load at $a$ on segment 1 the hinge force is $a/8$ pressing down at $x=8$, giving $$\eta_{M_D}(a) = -\frac{a}{8}(10-8) = -\frac{a}{4}$$ which reaches $-2$ when $a=8$. Between C and D the load acts directly on the overhang, and $\eta_{M_D} = -(10-a)$, running from $-2$ at C back to 0 at D. Beyond D the ordinate is zero, because the stretch between the hinge and D is then unloaded. The extreme value is $$\eta_{\max} = \boxed{-2.00\ \text{m}}$$ and the influence line is entirely negative — $M_D$ can only ever hog.

IL for the bending moment at D: purely hogging, peaking at −2 m when the load stands at the hinge C.

5(b)

Given. A 16 m compound beam: pin at A $(x=0)$, roller at C $(x=8)$, internal hinge at D $(x=10)$ and roller at E $(x=16)$. Point B lies 6 m from A, that is 2 m to the left of C. The vehicle is three point loads — 40 kN, 40 kN and 20 kN — spaced 2 m between the two 40 kN axles and 4 m between the second 40 kN axle and the 20 kN axle, travelling left to right.

Find. The influence line for shear at B, and the largest shear the vehicle can produce there.

Q5(b) — 16 m compound beam. B (6 m from A) is the section whose shear is required.

Approach. Build the influence line segment by segment, then place the axles so the sum $\sum P_i\eta_i$ is greatest — for a shear influence line with a unit step, that means putting the heaviest axle immediately on the steeper side of the section.

Ordinates for a load on span A–C. Segment 1 spans A to the hinge D with supports at A and C, so with the load at $a$ the reaction $R_A = (8-a)/8$ and $$\eta_{V_B}(a) = \frac{8-a}{8} - 1 = -\frac{a}{8} \ (a < 6), \qquad \eta_{V_B}(a) = \frac{8-a}{8} \ (a > 6)$$ Immediately left of B the ordinate is $-6/8 = -0.75$ and immediately right of it $+2/8 = +0.25$, the unit step across the section. It falls to zero at the roller C.
Ordinates for a load on the overhang and the suspended span. Between C and the hinge D the ordinate continues linearly to $-0.25$ at D. Beyond the hinge, a unit load at $a$ on D–E delivers $(16-a)/6$ down onto the tip of the overhang at $x=10$, which gives $$\eta_{V_B}(a) = -\frac{1}{4}\cdot\frac{16-a}{6}$$ running from $-0.25$ at D back to zero at E. The governing ordinate is $\boxed{-0.75}$ just left of B.

IL for shear at B, showing the unit step across the section and the reversed lobe produced by the suspended span D–E.

Position the vehicle for maximum effect. The influence line is negative almost everywhere and steepest approaching B from the left, so the worst case puts a 40 kN axle exactly at B (just on its left, where $\eta = -0.75$) and the other axles where the ordinates are still negative. Testing every position, the governing arrangement has the leading 20 kN axle at $x = 10$ m, so the two 40 kN axles stand at $x=4$ m and $x=6$ m.
Evaluate the shear. Reading the ordinates $\eta(4) = -0.50$, $\eta(6^-) = -0.75$ and $\eta(10) = -0.25$, $$V_B = \sum P_i \eta_i = 40(-0.50) + 40(-0.75) + 20(-0.25)$$ $$V_B = -20 - 30 - 5 = \boxed{-55.0\ \text{kN}}$$ so the largest shear at B is 55 kN, acting in the negative sense. Sweeping the vehicle numerically across the whole structure confirms no other position beats it.

Question 5 — results
5(a)(i) max $\eta$ for $M_B$	+2.00 m, load at B
5(a)(ii) max $\eta$ for $V$ left of C	−1.00, load just left of C
5(a)(iii) max $\eta$ for $M_D$	−2.00 m, load at the hinge C
5(b) IL ordinates at B	−0.75 (left) and +0.25 (right)
5(b) governing axle positions	40 kN at 4 m, 40 kN at 6 m, 20 kN at 10 m
5(b) largest shear at B	55.0 kN (negative sense)

Topic. Influence lines for determinate compound beams, and the placement of a moving load train for maximum effect.

Key relations. An influence line gives the value of one response function as a unit load moves; for determinate structures every influence line is piecewise linear. A shear influence line has a unit step at the section; a moment influence line is continuous and peaks under the section. For a set of moving loads the response is $\sum P_i \eta_i$, maximised by trial placements with an axle at each peak.

Why this works. Muller-Breslau’s principle explains the shapes: the influence line for a response is the deflected form obtained by releasing that response and imposing a unit displacement. Releasing shear at a section gives a unit relative slide, hence the step; releasing moment gives a unit relative rotation, hence the kink. An internal hinge transforms these shapes decisively — a load beyond the hinge simply cannot reach the segment on the far side, which is why the ordinates go flatly to zero rather than continuing. Because the ordinates are linear, the governing load position is always at a break point, so a handful of trials suffices.

Common pitfall. Drawing the influence lines as if the beam were continuous and forgetting that the hinge cuts the load path — that error puts non-zero ordinates over C–E in part (a)(i), typically doubling the apparent answer. The second trap is straddling the step in part (b): the axle must be placed on the side of B that gives the larger ordinate ($-0.75$, not $+0.25$), and an axle sitting exactly on the section should be taken on the more severe side.

Source. Hibbeler, Structural Analysis 10th ed., Ch. 6 (§6.1 influence lines for beams, §6.2 Muller-Breslau, §6.4 moving load systems); Kassimali 6th ed., Ch. 8.

Question 5 of 8: Influence lines and a moving vehicle (18 marks)

Question 5 — Influence lines and a moving vehicle (18 marks)

5(a)

5(b)