16-Civ-B1 Advanced Structural Analysis · December 2014

Question 3 of 9: Castigliano's theorem — horizontal deflection of the roller (18 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 3: Castigliano's theorem — horizontal deflection of the roller (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.

Given. A two-member gable, pinned at joint (1) and on a roller at joint (3), with a single vertical load at the apex.

Given data
Quantity	Symbol	Value
Span, joint (1) to joint (3)	$L$	12 m (6 m each side of the apex)
Rise of the apex above the supports	$h$	2.5 m
Load at the apex, joint (2)	$P$	24 kN downward
Flexural rigidity, both members	$EI$	$6.5\times 10^{4}\ \text{kN}\cdot\text{m}^{2}$
Axial deformation	—	neglected (members inextensible)

Find. The horizontal deflection of joint (3).

Question 3: pin at (1), roller at (3), 24 kN at the apex. $Q$ is the dummy horizontal force introduced at the roller.

Approach. Add a dummy horizontal force $Q$ at the roller, write the bending moment in each member as a function of $Q$, and evaluate $\Delta=\int (M/EI)(\partial M/\partial Q)\,\mathrm{d}s$ with $Q$ set to zero.

Reactions with the dummy force in place. Taking $Q$ positive to the right at joint (3), horizontal equilibrium gives a reaction $Q$ to the left at the pin. Because $Q$ acts at the level of the pin it contributes no moment about (1), so $$V_1=V_3=\frac{P}{2}=12\ \text{kN}$$
Geometry of the members. Each rafter rises 2.5 m over a horizontal run of 6 m, so $$L_{\text{mem}}=\sqrt{6^{2}+2.5^{2}}=6.5\ \text{m},\qquad y=\frac{2.5}{6}x=0.41667x,\qquad \mathrm{d}s=\frac{6.5}{6}\,\mathrm{d}x$$ The 6-2.5-6.5 triangle is the tell that the drawing has been read correctly.
Bending moment in member (1)–(2). Taking the free body to the left of a section at $(x,y)$ and collecting the moments of the two reaction components, $$M(x)=12x+Qy,\qquad \frac{\partial M}{\partial Q}=y=0.41667x$$
Bending moment in member (2)–(3). Working from the right support with $u=12-x$ the algebra is a mirror image, $$M(u)=12u+Qy,\qquad \frac{\partial M}{\partial Q}=y=0.41667u$$ so the two members contribute equally, as symmetry demands.
Integrate with $Q=0$. For one member, $$\int_{0}^{6}(12x)(0.41667x)\frac{6.5}{6}\,\mathrm{d}x=5\left(\frac{6.5}{6}\right)\frac{6^{3}}{3}=390\ \text{kN}\cdot\text{m}^{3}$$ and doubling for the second member gives $780\ \text{kN}\cdot\text{m}^{3}$.
Divide by the rigidity. $$\boxed{\Delta_{3H}=\frac{780}{6.5\times 10^{4}}=0.0120\ \text{m}=12.0\ \text{mm}}$$ The sign is positive in the direction of $Q$, so joint (3) moves away from joint (1): the frame spreads.
Sanity check the magnitude. A closed form for this shape is $\Delta=2P h\,L_{\text{mem}}\,a^{2}/(6EI\cdot 1)$ with $a$ the half-span; substituting $P/2=12$, $h/a=0.41667$ and $L_{\text{mem}}/a=1.08333$ reproduces 12.0 mm exactly, and the span-to-deflection ratio $12000/12=1000$ is entirely ordinary for a shallow pitched frame under a concentrated load.

Because the roller supplies no horizontal restraint the frame carries the apex load purely in bending, and the resulting spread is the deflection asked for. Had joint (3) been a pin, this deflection would be zero and a horizontal thrust would appear instead — the structure would become one degree indeterminate and Question 4's method would be needed.

Question 3 — results
Quantity	Value
Vertical reactions at (1) and (3)	12.0 kN each
Horizontal reaction at (1)	0 (roller at (3) under vertical load only)
Rafter length	6.500 m
Strain-energy integral $\int M\,(\partial M/\partial Q)\,\mathrm{d}s$	$780\ \text{kN}\cdot\text{m}^{3}$
Horizontal deflection of joint (3)	12.0 mm outward

Topic. Castigliano's second theorem used with a dummy force to obtain a displacement where no load acts.

Key relations. $\Delta_i=\partial U/\partial P_i$; for bending, $\Delta=\int_{0}^{L}\dfrac{M}{EI}\dfrac{\partial M}{\partial P_i}\,\mathrm{d}s$; on an inclined member $\mathrm{d}s=(L_{\text{mem}}/a)\,\mathrm{d}x$ when integrating on the horizontal projection.

Why this works. The dummy force is a bookkeeping device: it is carried symbolically through the equilibrium equations so that the partial derivative can be taken, then set to zero so the real moment field is restored. Everything else is ordinary statics because the structure is determinate.

Common pitfall. Integrating along the horizontal projection without the $L_{\text{mem}}/a$ factor. That single omission understates the answer by 8 % here, and by much more on a steeper frame.

Source. Hibbeler, Structural Analysis, 10th ed., §9.8–9.9 (Castigliano's theorem for beams and frames); Kassimali, Structural Analysis, 6th ed., §7.6.

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