16-Civ-B1 Advanced Structural Analysis · December 2013

Question 2 of 9: Influence lines for a two-beam structure

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

Notes on this paper

National Examination, December 2013 — 98-Civ-B1 Advanced Structural Analysis. Closed book, 3 hours, calculator permitted. Nine questions: #1 and #2 are compulsory, two of #3–#5 and two of #6–#9 are selected, and six questions constitute a complete paper (100 marks). Marks are printed in the left margin. All nine questions are solved below, because the set is a study resource rather than an examination script.

Reference texts. R. C. Hibbeler, Structural Analysis, 10th ed. (slope-deflection ch. 11, moment distribution ch. 12, influence lines ch. 6, energy methods ch. 9); A. Kassimali, Structural Analysis, 6th ed. (force/flexibility method ch. 13, matrix stiffness ch. 18); A. Ghali, A. M. Neville & T. G. Brown, Structural Analysis: A Unified Classical and Matrix Approach, 7th ed. (support-movement and lack-of-fit effects, ch. 4 and 6). Canadian design codes (CSA S16, CSA A23.3, NBCC) are not needed here: every question is an elastic-analysis question, not a design question.

Sign convention used throughout (stated once, obeyed everywhere). Global x to the right, y up; moments and rotations are counter-clockwise positive. For a member running from end $i$ to end $j$, let $\mathbf{e}_1$ be the unit vector from $i$ to $j$ and $\mathbf{e}_2$ that vector turned through $+90^\circ$. The chord rotation is $\psi_{ij}=\left[(\boldsymbol{\Delta}_j-\boldsymbol{\Delta}_i)\cdot\mathbf{e}_2\right]/L$, so the slope-deflection equation reads

$$M_{ij}=\frac{2EI}{L}\left(2\theta_i+\theta_j-3\psi_{ij}\right)+\mathrm{FEM}_{ij}$$

with $\mathrm{FEM}=+wL^{2}/12$ and $+PL/8$ at the i end of a member carrying a downward uniform load or a central point load (and the negative of those at the j end). Ordinary sagging-positive bending moments follow from $M_{\text{sag}}(i)=-M_{ij}$ and $M_{\text{sag}}(j)=+M_{ji}$. Every number below was computed twice — once by hand from these equations and once by an independent direct-stiffness solution — and the two agree.

Question 2: Influence lines for a two-beam structure (8 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.

Given. A compound (Gerber) beam: a free overhang of 3 m, a pin support A, a 9 m span to the roller support B, a 3 m cantilever beyond B ending in an internal hinge, and a 9 m suspended span from the hinge to the roller support C. Measuring $x$ from the left-hand tip, A lies at 3 m, B at 12 m, the hinge at 15 m and C at 24 m.

Find. The influence lines for the bending moment over support B and for the shear force immediately to the right of support A, with the largest ordinate on each labelled.

Question 2: the two-beam (Gerber) structure. The suspended span hangs from the internal hinge at the tip of the 3 m cantilever.

Approach. The structure is statically determinate (four reaction components, three equations plus one condition of construction at the hinge), so each influence line follows from a single free body: for the moment over B, the portion to the right of B; for the shear just right of A, the portion to the left of the section.

Confirm determinacy. The pin at A supplies two components and the rollers at B and C one each, so $r=4$; three equations of statics plus the condition $M=0$ at the internal hinge give four equations. The structure is determinate and the influence lines are therefore straight-line segments.
Influence line for $M_B$ — take the free body to the right. Everything to the right of B consists of the 3 m cantilever and, hanging from its tip hinge, the suspended span. Whenever the unit load stands anywhere between the left tip and support B, that free body carries no load at all, so $$M_B=0\quad\text{for }0\le x\le 12\ \text{m}.$$ This flat stretch, covering more than half the beam, is the feature that distinguishes a Gerber influence line from the familiar continuous-beam one.
Load on the cantilever. For $12\le x\le15$ m the unit load acts directly on the free body at a lever arm $(x-12)$, so $M_B=-(x-12)$, falling linearly to $-3$ m at the hinge.
Load on the suspended span. For $15\le x\le 24$ m the suspended span delivers to the hinge only the reaction $R_H=(24-x)/9$, applied at the tip of the 3 m cantilever, so $$M_B=-3\times\frac{24-x}{9}=-\frac{24-x}{3},$$ which returns linearly to zero at support C. The peak therefore sits at the hinge: $$\left|M_B\right|_{\max}=\boxed{3.00\ \text{m}}\ \ \text{at }x=15\ \text{m}.$$
Influence line for $V_{A^{+}}$ — find $R_A$ first. Taking moments about B for the left-hand beam, $-9R_A-(x-12)=0$ when the load stands on that beam, so $R_A=(12-x)/9$; when the load stands on the suspended span the only action reaching the left-hand beam is the hinge force, and $R_A=-(24-x)/27$.
Subtract the load when it is on the overhang. The section lies immediately to the right of A, so $V_{A^{+}}=R_A$ except when the unit load itself is on the overhang, where $V_{A^{+}}=R_A-1=(3-x)/9$. The influence line therefore starts at $+1/3$ at the tip, falls to zero at A, jumps by unity as the load crosses the support, and then decreases linearly through zero at B to $-1/3$ at the hinge before returning to zero at C. The unit jump is the peak: $$\left|V_{A^{+}}\right|_{\max}=\boxed{1.000}\ \ \text{immediately right of A.}$$

Influence line for the bending moment over support B. Ordinates are in metres; the line is identically zero from the left-hand tip to B.

Influence line for the shear immediately right of support A. The unit step across the support is the maximum ordinate.

Question 2 — influence-line ordinates
Position of the unit load	IL for M_B (m)	IL for V just right of A
Left-hand tip, x = 0	0	+0.333
Support A, x = 3 m (right side)	0	+1.000 (peak)
Support B, x = 12 m	0	0
Internal hinge, x = 15 m	−3.00 (peak)	−0.333
Support C, x = 24 m	0	0

Topic. Influence lines for compound (Gerber) beams by direct free-body reasoning.

Key relations. Müller-Breslau: the influence line for an action is the deflected shape produced by releasing that action and imposing a unit relative displacement; for a determinate structure the shape is piecewise linear. A suspended span transmits to its supporting cantilever only the hinge reaction $R_H=(\text{clear span}-x)/(\text{clear span})$.

Why this works. An influence line answers “how much of a unit load reaches this section?”, so the fastest route is always to name the free body the action belongs to and ask what the load does to it. Over the whole of the left-hand beam, the free body to the right of B is completely unloaded, which is why the moment influence line is flat there; and because the peak sits at the hinge, the worst case for hogging over B is a load parked at the hinge, not one parked over the span. For the shear line, the unit jump across a support is a universal feature: moving the load a hair from one side of the section to the other transfers the whole unit into or out of the free body.

Common pitfall. Drawing the $M_B$ line as the textbook triangle peaking under B, which is the answer for a continuous beam and is wrong here by construction — the hinge, not the support, governs. The second slip is forgetting the $-1$ term for load positions on the overhang, which converts the $+1/3$ ordinate at the tip into an incorrect $+4/3$.

Source. Hibbeler, Structural Analysis, 10th ed., §6.1–6.4 (influence lines for beams and girders); Kassimali, 6th ed., ch. 8.

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