16-Civ-B1 Advanced Structural Analysis · December 2013

Question 6 of 9: Flexibility method for a non-prismatic beam

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 6: Flexibility method for a non-prismatic beam (22 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 12 m beam, roller-supported at A and encastré at B, built in three segments: 3 m of rigidity $3EI$, then 6 m of rigidity $EI$, then 3 m of $3EI$ again. Two 29 kN point loads act downward, one at each change of section, i.e. 3 m and 9 m from A.

Given data
Quantity	Symbol	Value
Overall span	L	12 m
Stiff end segments (0–3 m and 9–12 m)	—	3EI
Central segment (3–9 m)	—	EI
Point loads at 3 m and at 9 m	P	29 kN each
Support at A / at B	—	roller / fixed

Find. The fixed-end bending moment at B.

Question 6: the non-prismatic beam. The haunched end segments are three times as stiff as the central segment.

Approach. The beam is indeterminate to the first degree; releasing the roller at A leaves a determinate cantilever from B, and one compatibility equation — zero deflection at A — recovers the redundant, after which $M_B$ follows from statics.

Choose the release. One roller plus one encastré end gives $r=4$, so $\mathrm{DSI}=1$. Removing the roller at A leaves a cantilever built in at B, which is the simplest determinate primary structure; the redundant is the vertical reaction $R_A$.
Write the two moment fields. Measuring $x$ from A, the primary structure under the applied loads carries $$M_0(x)=-29\langle x-3\rangle-29\langle x-9\rangle,$$ (Macaulay brackets vanish when negative) and under a unit upward force at A it carries $m(x)=x$. Both are measured with sagging positive.
Integrate the flexibility coefficients segment by segment. Because the rigidity changes, each integral must be split at 3 m and 9 m: $$f_{11}=\int_0^{12}\frac{m^{2}}{EI(x)}\,\mathrm{d}x=\frac{1}{EI}\left(3+234+111\right)=\frac{348}{EI},$$ $$f_{10}=\int_0^{12}\frac{M_0\,m}{EI(x)}\,\mathrm{d}x=-\frac{1}{EI}\left(3654+2784\right)=-\frac{6438}{EI}.$$ The $1/3$ weighting of the haunched segments is what makes the middle 6 m dominate: it supplies 234 of the 348 in $f_{11}$.
Apply compatibility at A. The support does not move, so $$f_{10}+R_A f_{11}=0\quad\Longrightarrow\quad R_A=\frac{6438}{348}=\boxed{18.5\ \text{kN (upward)}}$$ Note that $EI$ cancels completely — only the ratio $3EI:EI$ matters, which is why the question can be set without a numerical rigidity.
Recover the fixed-end moment by statics. Taking the whole beam and summing moments about B, $$M_B=R_A(12)-29(9)-29(3)=222-261-87=\boxed{-126\ \text{kN}\cdot\text{m}}$$ i.e. 126 kN·m hogging. Vertical equilibrium then gives $R_B=58-18.5=39.5$ kN.
Sanity-check against the prismatic case. If the beam were uniform the same calculation returns $R_A=17.4$ kN and $M_B=-139$ kN·m; haunching the ends therefore attracts reaction to A and relieves the built-in end by about 9 %. That is the direction one expects, because stiffening the region near a support draws moment towards it only when the support can rotate — here the stiff segment adjacent to the roller shortens the effective flexible length of the span. An independent direct-stiffness solution with three elements reproduces $R_A=18.5$ kN and $M_B=-126$ kN·m exactly.

Question 6 — results
Quantity	Value
Flexibility coefficient f₁₁	348 / EI
Flexibility coefficient f₁₀	−6438 / EI
Redundant reaction at A	18.5 kN upward
Reaction at B	39.5 kN upward
Bending moment at 3 m / at 9 m	+55.5 / −7.5 kN·m
Fixed-end moment at B	126 kN·m hogging

Topic. The force (flexibility) method applied to a beam of stepped rigidity.

Key relations. $f_{10}+R\,f_{11}=0$ for a rigid support; $f_{ij}=\displaystyle\int \frac{M_i M_j}{EI(x)}\,\mathrm{d}x$; Macaulay notation $\langle x-a\rangle$ for the primary moment field; statics for the recovery of the remaining reactions.

Why this works. The flexibility method is the natural choice when the structure has few redundants but awkward section properties, because the varying $EI$ enters only as a weighting inside two definite integrals — there is no member stiffness matrix to modify. Choosing the roller reaction as the redundant is what keeps those integrals easy: $m(x)=x$ is a single straight line over the whole beam. Had the fixing moment at B been chosen instead, $m(x)$ would be a different straight line and the answer identical, so the choice is one of convenience, not of correctness.

Common pitfall. Integrating across a change of section without splitting the integral, which here would overstate $f_{11}$ by a factor of about two and give $R_A$ around 10 kN. The second pitfall is dropping the $1/3$ weighting on the haunched segments — remember the integrand carries $1/EI(x)$, so a stiffer segment contributes less, not more.

Source. Kassimali, Structural Analysis, 6th ed., §13.1–13.5 (method of consistent deformations, including members of variable section); Hibbeler, 10th ed., ch. 10.

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