16-Civ-B1 Advanced Structural Analysis · December 2013

Question 3 of 9: Castigliano deflection of a beam with an internal hinge

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 3: Castigliano deflection of a beam with an internal hinge (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. The beam runs A–B–C–D in three 4 m bays. A is encastré, there is an internal hinge at B, and D is a roller. A uniformly distributed load of 2 kN/m acts over the built-in span AB only, and a 18 kN point load acts downward at C.

Given data
Quantity	Symbol	Value
Bay length (three equal bays)	a	4 m
Distributed load on AB	w	2 kN/m
Point load at C	P	18 kN
Flexural rigidity, both members	EI	2.56 × 10⁴ kN·m²
Support at A / at B / at D	—	fixed / internal hinge / roller

Find. The vertical deflection of point C.

Question 3: three 4 m bays, encastré at A, internal hinge at B, roller at D. The 2 kN/m load acts on span AB only.

Approach. The internal hinge makes the structure determinate, so Castigliano's second theorem can be applied directly to the real 18 kN load, $\delta_C=\partial U/\partial P$, without introducing a dummy force.

Check determinacy before differentiating anything. The fixed end supplies three reaction components and the roller one, so $r=4$; three equations of statics plus the hinge condition $M_B=0$ make four. The structure is statically determinate, which is exactly what Castigliano's second theorem requires if it is to be used in its simple form.
Split at the hinge and find the internal force there. Portion BD is a simple beam of span 8 m supported by the hinge at B and the roller at D, carrying $P$ at its mid-point C. Taking moments about B, $$R_D=\frac{P\times 4}{8}=\frac{P}{2}=9\ \text{kN},\qquad V_B=P-R_D=\frac{P}{2}=9\ \text{kN}.$$ The hinge therefore hands exactly $P/2$ to the cantilever AB, and every internal moment in the structure is linear in $P$.
Write the moment field and its derivative. Measuring $u$ back from the hinge along AB and $s$ forward from the hinge along BD, $$M_{AB}(u)=-\frac{wu^{2}}{2}-\frac{P}{2}u,\qquad \frac{\partial M_{AB}}{\partial P}=-\frac{u}{2},$$ $$M_{BC}(s)=\frac{P}{2}s,\qquad M_{CD}(s)=\frac{P}{2}(8-s),$$ with derivatives $s/2$ and $(8-s)/2$ respectively.
Integrate bay by bay. Castigliano's theorem gives $$\delta_C=\frac{1}{EI}\int M\,\frac{\partial M}{\partial P}\,\mathrm{d}x .$$ Over AB the integrand is $\left(\tfrac{w}{4}u^{3}+\tfrac{P}{4}u^{2}\right)$, which integrates to $16w+\tfrac{16P}{3}=32+96=128$; over BC and over CD each integral is $\tfrac{16P}{3}=96$. Summing, $$\int M\,\frac{\partial M}{\partial P}\,\mathrm{d}x=128+96+96=320\ \text{kN}\cdot\text{m}^{3}.$$
Divide by the rigidity. $$\delta_C=\frac{320}{2.56\times10^{4}}=0.01250\ \text{m}=\boxed{12.5\ \text{mm downward}}$$ The positive sign means the deflection is in the direction of the 18 kN load, i.e. downward, as expected.
Check the reactions. With the hinge force known, the cantilever carries $R_A=8+9=17$ kN and a fixing moment $M_A=8(2)+9(4)=52\ \text{kN}\cdot\text{m}$ hogging, and $R_D=9$ kN; the three add to the total applied load of 26 kN. An independent direct-stiffness run of the same beam reproduces $\delta_C=12.5$ mm and these reactions exactly.

Question 3 — results
Quantity	Value
Force transferred through the hinge at B	9.0 kN
Reaction at A / fixing moment at A	17.0 kN / 52.0 kN·m hogging
Reaction at D	9.0 kN
Strain-energy integral ∫M(∂M/∂P)dx	320 kN·m³
Vertical deflection at C	12.5 mm downward

Topic. Castigliano’s second theorem applied to a determinate compound beam.

Key relations. $\delta_i=\dfrac{\partial U}{\partial P_i}=\displaystyle\int \frac{M}{EI}\frac{\partial M}{\partial P_i}\,\mathrm{d}x$ for flexure; an internal hinge imposes $M=0$ and supplies one extra equation of condition; a simply supported sub-beam delivers $P/2$ to each of its supports when the load is central.

Why this works. The theorem needs a real load acting at the point and in the direction of the wanted deflection; here 18 kN sits exactly at C and acts vertically, so no fictitious force is needed and the differentiation is trivial. The essential preliminary is the hinge: it converts what looks like a propped, indeterminate beam into two determinate pieces, so $M$ can be written in closed form everywhere. Note how cheap this makes the calculation — the whole answer is three polynomial integrals over 4 m bays — and how expensive it becomes if the hinge is overlooked.

Common pitfall. Carrying the numerical 18 kN into the moment expressions before differentiating, which loses $\partial M/\partial P$ altogether. Keep $P$ symbolic until after the derivative is taken and substitute at the end. A second slip is integrating the distributed load past the hinge; the 2 kN/m acts on AB only, and extending it to B–D would inflate the answer by about a third.

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

← Question 2 Next: Question 4 →