16-Civ-B1 Advanced Structural Analysis · December 2014

Question 6 of 9: Flexibility method — fixed-ended non-prismatic beam (22 marks)

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

Notes on this paper

Paper format. National Examination, December 2014, 98-Civ-B1 Advanced Structural Analysis — 3 hours, closed book (approved Sharp or Casio calculator permitted). Nine questions over six pages: Questions 1 and 2 are compulsory (12 and 8 marks), then any two of Questions 3, 4, 5 (18 marks each) and any two of Questions 6, 7, 8, 9 (22 marks each). Six questions constitute a complete paper and total 100 marks; marks are printed in the left margin. All nine questions are solved here, because the set is a study resource rather than a three-hour sitting.

Reference texts.

R. C. Hibbeler, Structural Analysis, 10th ed. — Ch. 6 (influence lines), Ch. 9 (deflections by energy methods), Ch. 10 (force/flexibility method), Ch. 11 (slope-deflection).
A. Kassimali, Structural Analysis, 6th ed. — Ch. 3 (determinacy and stability), Ch. 8 (influence lines for floor-beam systems), Ch. 13 (least work), Ch. 15–16 (slope-deflection with and without sidesway).
A. Ghali, A. M. Neville and T. G. Brown, Structural Analysis: A Unified Classical and Matrix Approach, 7th ed. — Ch. 4 (force method), Ch. 5 (lack of fit and support movement).
W. McGuire, R. H. Gallagher and R. D. Ziemian, Matrix Structural Analysis, 2nd ed. — Ch. 4–5 (member stiffness, assembly, kinematic constraints).
CSA S6:19 Canadian Highway Bridge Design Code and NBCC 2020 — the Canadian codes that these analysis methods feed; no code check is required by this paper.

Sign convention used throughout. End moments are counter-clockwise positive, the convention that matches the standard six-degree-of-freedom stiffness element. In it the slope-deflection equation is

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

the chord rotation is $\psi_{ij}=\bigl[(\mathbf{d}_j-\mathbf{d}_i)\cdot\mathbf{e}_2\bigr]/L$ with $\mathbf{e}_2$ the member axis turned $+90^\circ$, the fixed-end moment of a downward UDL is $\mathrm{FEM}_{i}=+wL^{2}/12$ and $\mathrm{FEM}_{j}=-wL^{2}/12$, and the ordinary sagging bending moment is recovered as $M_{\text{sag}}(i)=-M_{ij}$ and $M_{\text{sag}}(j)=+M_{ji}$. Every translation degree of freedom is closed with the virtual-work statement

$$\sum_{\text{members}}\bigl(M_{ij}+M_{ji}\bigr)\,\psi^{*}_{ij}\;+\;\sum \mathbf{F}\cdot\mathbf{d}^{*}=0$$

in which $\mathbf{F}$ are the equivalent nodal loads and $\mathbf{d}^{*}$, $\psi^{*}$ belong to the unit sway pattern. Mixing this with the clockwise-positive convention of some textbooks produces clean-looking but wrong answers, which is the single most common failure on this subject.

All work below uses the drawing.

Question 6: Flexibility method — fixed-ended 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 10 m beam built in at both ends, stiffened over its outer 2 m at each end, carrying three equal point loads.

Given data
Station (m from the left support)	Segment	Rigidity	Load
0–2	outer	$2EI$	—
2–5	inner	$EI$	24 kN at 2 m
5–8	inner	$EI$	24 kN at 5 m
8–10	outer	$2EI$	24 kN at 8 m

Find. The bending moments at the two fixed ends.

Question 6: symmetric non-prismatic beam. The stiffened outer segments are where the hogging moment is largest, so the variation of $EI$ genuinely matters.

Approach. Release both fixed ends to a simply supported beam, take the two end moments as redundants, and impose zero end rotation using flexibility coefficients evaluated with the actual variable $EI$.

Choose the released structure and the redundants. Removing the two rotational restraints leaves a simply supported beam — determinate and easy to integrate. The redundants are $M_A$ and $M_B$, and by symmetry of both the beam and the loading $$M_A=M_B=M$$
Primary moment diagram. The released beam carries $R=\tfrac{3(24)}{2}=36$ kN at each end, so $$M_0(x)=36x-\sum 24\,\langle x-x_i\rangle,\qquad M_0(2)=72,\quad M_0(5)=108,\quad M_0(8)=72\ \text{kN}\cdot\text{m}$$
Unit-redundant diagrams. A unit moment applied at A alone gives $m_1=1-x/10$; one applied at B gives $m_2=x/10$. Note that $m_1+m_2=1$ everywhere, which is about to simplify the algebra decisively.
Compatibility at end A. The end rotation must vanish: $$f_{10}+M\,(f_{11}+f_{12})=0,\qquad f_{10}=\int_{0}^{10}\frac{M_0m_1}{EI(x)}\mathrm{d}x,\quad f_{11}+f_{12}=\int_{0}^{10}\frac{m_1(m_1+m_2)}{EI(x)}\mathrm{d}x$$
Evaluate the stiffness integral exactly. Because $m_1+m_2=1$, the second integral is simply $\int m_1/EI(x)\,\mathrm{d}x$, which splits at the rigidity changes: $$\frac{1}{2}\Bigl[x-\frac{x^{2}}{20}\Bigr]_{0}^{2}+\Bigl[x-\frac{x^{2}}{20}\Bigr]_{2}^{8}+\frac{1}{2}\Bigl[x-\frac{x^{2}}{20}\Bigr]_{8}^{10}=0.9+3.0+0.1=\frac{4.0}{EI}$$
Evaluate the load integral. Integrating $M_0(x)m_1(x)$ over the same three ranges with the appropriate rigidity gives $$f_{10}=\frac{306}{EI}\ \text{(units kN}\cdot\text{m}^{2}\text{)}$$
Solve for the redundant. The reference rigidity cancels, as it always does when only one material is involved: $$M=-\frac{306}{4.0}=\boxed{-76.5\ \text{kN}\cdot\text{m}}$$ so both fixed ends hog at 76.5 kN·m.
Complete and check the diagram. Superposing $M(x)=M_0(x)-76.5$, $$M(2)=72-76.5=-4.5,\qquad M(5)=108-76.5=+31.5\ \text{kN}\cdot\text{m}$$ The reactions are unchanged at 36 kN by symmetry, and the two points of contraflexure sit just inside the stiffened zones — consistent with 4.5 kN·m of hogging still present at the 2 m station.

A prismatic beam under the same three loads hogs only 68.4 kN·m at the ends and sags 39.6 kN·m at midspan. Doubling the rigidity over the outer fifths therefore draws an extra 8.1 kN·m towards the supports and sheds 8.1 kN·m from midspan — a 12 % shift, and exactly the reason haunches are used on continuous bridge girders.

Question 6: final bending moment diagram, sagging positive. Hogging of 76.5 kN·m at each fixed end, 31.5 kN·m sagging at midspan.

Question 6 — results
Quantity	Value
End reactions (symmetry)	36.0 kN each
Flexibility coefficient $f_{11}+f_{12}$	$4.0/EI$
Load term $f_{10}$	$306/EI$
Moment at each fixed end	−76.5 kN·m (hogging)
Moment at the outer load points (2 m, 8 m)	−4.5 kN·m
Moment at midspan	+31.5 kN·m

Topic. The force (flexibility) method applied to a beam whose flexural rigidity varies along its length.

Key relations. $[f]\{X\}+\{\Delta_0\}=0$; $f_{ij}=\int m_im_j/EI(x)\,\mathrm{d}x$; $\Delta_{i0}=\int M_0m_i/EI(x)\,\mathrm{d}x$; superposition $M(x)=M_0(x)+\sum X_i m_i(x)$.

Why this works. Non-prismatic members are no harder in the force method than prismatic ones, because $EI(x)$ enters only inside the integrals. Symmetry then halves the work by making the two redundants equal, and the identity $m_1+m_2=1$ turns the stiffness term into a single easy integral.

Common pitfall. Assuming the rigidity variation can be ignored. Any single constant $EI$ cancels from the compatibility equation and returns the prismatic answer of 68.4 kN·m, which understates the true end moment by 10.6 %; the variation matters precisely because the stiff regions coincide with the largest ordinates of $m_1$.

Source. Hibbeler, Structural Analysis, 10th ed., Ch. 10 (force method for beams and frames); Ghali, Neville and Brown, 7th ed., Ch. 4.

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