16-Civ-A1 Elementary Structural Analysis · Undated paper

Question 5 of 8: Influence lines and maximum effects from moving loads

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. Engineers Canada / EGBC national examination 16-Civ-A1 Elementary Structural Analysis, 3 hours, CLOSED BOOK. Six questions constitute a complete paper: answer Q1–Q5 and one of Q6/Q7/Q8. For completeness all eight questions are worked here.

Reference texts: R. C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy and stability (Ch. 2), method of joints and sections (Ch. 3), shear and bending-moment diagrams (Ch. 4), influence lines and moving loads (Ch. 6), virtual-work deflections (Ch. 8–9), slope-deflection and moment distribution (Ch. 10–11); A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges, compound (Gerber) beams, three-hinged frames, Maxwell’s law of reciprocal deflections.

Sign convention. Upward reactions positive; sagging bending moment positive (tension on the underside of a beam), hogging negative; member axial force tension positive (T), compression negative (C). Shear is positive when the resultant of the forces on the left of a section acts upward.

Check: figure reading. All geometry, loads and support symbols used below were read directly from the printed figures. Where a support symbol governs the answer it is called out explicitly in the Given.

Question 5: Influence lines and maximum effects from moving loads (16 marks: 6 + 10)

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.

Part (a) — influence lines for a hinged beam

Given. Measuring from the free left tip: A (pin) at 1 m, B (roller) at 4 m, an internal hinge at 5 m, C (roller) at 7 m, and a free right tip at 8 m. Total length 8 m. With $r = 4$ and $c = 1$ the beam is determinate.

Find. The influence lines for the bending moment at B and for the shear just to the left of B, with the maximum absolute ordinate of each.

Q5(a) — the beam and its two influence lines. Ordinates of the moment influence line are in metres; the shear influence line is dimensionless.

Approach. Place a unit load at a running position $z$ and write the required action from statics. The hinge splits the beam into an anchor part (tip–A–B–hinge) and a suspended part (hinge–C–tip), so treat the two load regions separately.

Unit load on the anchor part, $0 \le z \le 5$. With the suspended span unloaded it delivers nothing at the hinge, so the anchor part is simply the beam A–B with overhangs. Its reactions are $$R_B = \frac{z - 1}{3}, \qquad R_A = 1 - \frac{z-1}{3}$$
Influence line for $M_B$, load on the anchor part. Take the free body to the right of B. For $z \le 4$ there is nothing to the right of B at all, so $$\eta_{M_B}(z) = 0 \qquad (0 \le z \le 4)$$ For $4 \le z \le 5$ the unit load itself lies on that short right-hand piece and hogs the beam over B: $$\eta_{M_B}(z) = -(z - 4)$$ reaching $-1.00\text{ m}$ when the load stands on the hinge.
Unit load on the suspended part, $5 \le z \le 8$. The suspended piece is a simple beam on the hinge and the roller C, with a 1 m tip overhang. Moments about C give the force it hands back to the hinge: $$H = \frac{7 - z}{2}$$ which the anchor part receives as a downward load at $x = 5$. The moment at B is then simply that force times its 1 m lever arm: $$\eta_{M_B}(z) = -\frac{7 - z}{2}$$ so the ordinate runs from $-1.00$ at the hinge, through zero over C, to $+0.50\text{ m}$ at the free right tip.
Maximum ordinate of the $M_B$ influence line. $$\left|\eta_{M_B}\right|_{\max} = \boxed{1.00\ \text{m, at the hinge } (z = 5\ \text{m})}$$ The influence line is identically zero over the whole 4 m from the left tip to B, which is the signature of the hinge: no load placed on the anchor side of B can bend the beam over B.
Influence line for the shear just left of B. Summing the upward forces to the left of the section: for $z < 4$ the unit load is included, $$\eta_{V}(z) = R_A - 1 = -\frac{z - 1}{3}$$ which runs from $+0.333$ at the tip through zero at A down to $-1.00$ just left of B. For $4 < z \le 5$ the load has passed the section: $$\eta_{V}(z) = R_A = 1 - \frac{z-1}{3}$$ falling from 0 to $-0.333$ at the hinge. For $z \ge 5$ only the hinge force acts on the anchor part, giving $\eta_V = -(7-z)/6$, which rises from $-0.333$ through zero at C to $+0.167$ at the right tip.
Maximum ordinate of the shear influence line. The unit jump across the section is the whole story: $$\left|\eta_{V}\right|_{\max} = \boxed{1.00,\ \text{immediately left of B}}$$

Influence line	Ordinate at the tip (z = 0)	At A	At B	At the hinge	At C	At the right tip	Max \|η\|
M_B (m)	0	0	0	−1.00	0	+0.50	1.00 m
V just left of B	+0.333	0	−1.00 / 0	−0.333	0	+0.167	1.00

Part (b) — maximum moment and shear at C under a moving load train

Given. Measuring from the free left tip: A (pin) at 2 m, the section of interest C at 4 m, B (roller) at 12 m, free right tip at 14 m. The span A–B is 10 m with 2 m overhangs at each end. Loads: a 6 kN/m uniformly distributed load that may be placed over any portion or portions of the 14 m length, together with two 30 kN point 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 load systems acting together.

Q5(b) — the beam and the influence lines for moment and shear at C.

Approach. Build the two influence lines, then load them for maximum effect: the free-to-place UDL goes over every region of one sign (its contribution is $w$ times the enclosed area), and the two point loads are positioned so that the sum of the two ordinates under them is extreme. Evaluate both the largest positive and the largest negative effect and report the larger magnitude.

Influence line for the moment at C. With the unit load at $z$, $R_A = (12 - z)/10$. Taking the left free body, for $z \le 4$ $$\eta_{M_C}(z) = 2R_A - (4 - z) = -1.6 + 0.8z$$ and for $z \ge 4$ there is nothing between A and C but the reaction: $$\eta_{M_C}(z) = 2R_A = 2.4 - 0.2z$$ The ordinates are $-1.60\text{ m}$ at the left tip, $0$ at A, a peak of $+1.60\text{ m}$ at C, $0$ at B and $-0.40\text{ m}$ at the right tip.
Areas of the moment influence line. The positive lobe is the triangle from A to B with apex 1.60 m at C, and the negative lobes are the two overhang triangles: $$A^{+} = \tfrac{1}{2}(10)(1.6) = 8.00\ \text{m}^{2}, \qquad A^{-} = \tfrac{1}{2}(2)(1.6) + \tfrac{1}{2}(2)(0.4) = 2.00\ \text{m}^{2}$$
Best position for the 30 kN pair (moment). Placing the leading wheel exactly at the apex and the trailing wheel 2 m into the falling limb gives $\eta(4) = 1.60$ and $\eta(6) = 1.20$; every other 2 m straddle of the apex sums to less. Hence $\sum\eta = 2.80\text{ m}$. For the largest negative effect the pair sits on the left overhang at $z = 0$ and $z = 2$, giving $\sum\eta = -1.60\text{ m}$.
Combine for the maximum moment at C. Positive case — UDL over the whole span A–B and the pair at the apex: $$M_C^{+} = 6(8.00) + 30(2.80) = 48 + 84 = \boxed{+132\ \text{kN}\cdot\text{m}}$$ Negative case — UDL over both overhangs only and the pair on the left overhang: $$M_C^{-} = 6(-2.00) + 30(-1.60) = -12 - 48 = \boxed{-60\ \text{kN}\cdot\text{m}}$$ The governing value is therefore $$\left|M_C\right|_{\max} = \boxed{132\ \text{kN}\cdot\text{m (sagging)}}$$
Influence line for the shear at C. For $z < 4$ the unit load is left of the section, so $\eta_{V_C} = R_A - 1 = (2 - z)/10$; for $z > 4$ it is right of the section, so $\eta_{V_C} = R_A = (12 - z)/10$. The line runs $+0.20$ at the tip, $0$ at A, $-0.20$ just left of C, jumps a full unit to $+0.80$ just right of C, then falls to $0$ at B and $-0.20$ at the right tip. Its areas are $$A^{+} = \tfrac{1}{2}(2)(0.2) + \tfrac{1}{2}(8)(0.8) = 3.40\ \text{m}, \qquad A^{-} = \tfrac{1}{2}(2)(0.2) + \tfrac{1}{2}(2)(0.2) = 0.40\ \text{m}$$
Best position for the pair (shear) and the maximum. The leading wheel goes immediately to the right of C, where $\eta = 0.80$, and the trailing wheel lands at $z = 6$ with $\eta = 0.60$, so $\sum\eta = 1.40$. With the UDL over both positive regions, $$V_C^{+} = 6(3.40) + 30(1.40) = 20.4 + 42 = \boxed{+62.4\ \text{kN}}$$ The best negative arrangement puts the pair on the right overhang ($\sum\eta = -0.20$) with the UDL over the two negative strips: $$V_C^{-} = 6(-0.40) + 30(-0.20) = -2.4 - 6 = \boxed{-8.4\ \text{kN}}$$ so the governing value is $$\left|V_C\right|_{\max} = \boxed{62.4\ \text{kN}}$$

Effect at C	UDL contribution	Point-load contribution	Total
Maximum positive moment	6 × 8.00 = 48 kN·m	30 × 2.80 = 84 kN·m	+132 kN·m
Maximum negative moment	6 × (−2.00) = −12 kN·m	30 × (−1.60) = −48 kN·m	−60 kN·m
Maximum positive shear	6 × 3.40 = 20.4 kN	30 × 1.40 = 42 kN	+62.4 kN
Maximum negative shear	6 × (−0.40) = −2.4 kN	30 × (−0.20) = −6 kN	−8.4 kN
Governing \|M\| at C	132 kN·m
Governing \|V\| at C	62.4 kN

Topic. Influence lines for determinate beams and their use in positioning moving loads for maximum effect.

Key relations. An influence line ordinate $\eta(z)$ is the value of the chosen action when a unit load stands at $z$. For a set of point loads $P_i$ and a distributed load $w$ over a region $R$, the action is $$S = \sum_i P_i\,\eta(z_i) + w\!\!\int_R \eta(z)\,dz$$ so a free-to-place UDL is put over every region where $\eta$ has the wanted sign, and a load train is positioned to maximise $\sum\eta$.

Why this works. Superposition makes the response linear in the load, so the influence line is a complete description of how position maps to effect — the integral above is just superposition written continuously. Because the question allows the UDL over “any portion or portions”, the maximising pattern is a checkerboard: fill every lobe of one sign and leave the others bare. The two point loads are handled by trying every 2 m straddle; for a piecewise linear influence line the optimum always has one wheel at a vertex, which is why the leading wheel lands exactly on the apex at C.

Common pitfall. On part (a), sketching the familiar triangular moment influence line centred on B. The hinge at 5 m makes the true line identically zero from the left tip all the way to B, with the peak at the hinge instead — a load on the anchor side of B simply cannot bend the beam there. On part (b), computing only the positive maximum: the question asks for the maximum absolute values, so both signs must be evaluated, and reporting only the sagging case would be lucky rather than correct.

Source. Hibbeler, Structural Analysis, 10th ed., §6.1–6.4 (influence lines for beams, qualitative Müller-Breslau construction) and §6.5–6.7 (maximum effects from series of concentrated loads); Kassimali, Structural Analysis, 6th ed., Ch. 8.

← Question 4 Next: Question 6 →