07-Str-A4 · December 2015

Question 4 of 9: Horizontal Deflection of a Frame by Castigliano's Theorem

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

Notes on this paper

Paper format. 07-Str-A4 Advanced Structural Analysis, National Examinations, December 2015 — three hours, closed book (an approved Sharp or Casio calculator is the only aid). Nine questions: Questions 1 and 2 are compulsory (8 and 12 marks); the candidate then answers two of Questions 3, 4, 5 (18 marks each) and two of Questions 6, 7, 8, 9 (22 marks each), so six questions constitute a complete 100-mark paper. Marks are printed in the left margin. All nine questions are solved here, because this set is a study resource rather than an examination script.

Reference texts. R. C. Hibbeler, Structural Analysis, 10th ed. (slope-deflection, moment distribution, influence lines, Castigliano); A. Kassimali, Structural Analysis, 6th ed. (force and displacement methods, degrees of freedom); A. Ghali, A. M. Neville & T. G. Brown, Structural Analysis: A Unified Classical and Matrix Approach, 7th ed. (flexibility and stiffness formulations, lack of fit, support movement); W. Weaver Jr. & J. M. Gere, Matrix Analysis of Framed Structures, 3rd ed. (assembly of [K] and {P}); J. C. McCormac, Structural Analysis: Using Classical and Matrix Methods, 4th ed. Canadian design context: CSA S16, CSA A23.3 and the National Building Code of Canada supply the load and resistance rules that these analyses feed, although this paper is purely an analysis paper.

Check: sign convention used throughout. Member end moments $M_{ij}$ are counter-clockwise positive, which is the convention that matches the standard six-degree-of-freedom stiffness matrix and the vector $\{\theta_2,\theta_3,\delta\}$ the paper itself defines (Question 9 states “positive counter clockwise”). With that choice the fixed-end moment of a downward uniformly distributed load is $\mathrm{FEM}_{ij}=+wL^{2}/12$ at the $i$ end, the chord rotation is $\psi_{ij}=[(\mathbf{d}_j-\mathbf{d}_i)\cdot\mathbf{e}_\perp]/L$ with $\mathbf{e}_\perp$ the member axis turned $+90^{\circ}$, and ordinary sagging moments are recovered as $M_{\text{sag}}(i)=-M_{ij}$, $M_{\text{sag}}(j)=+M_{ji}$. Mixing this with Hibbeler’s clockwise-positive convention produces clean-looking integers that are wrong; every diagram below is plotted in the sagging convention.

Check: printed unit in Question 6. The paper prints “EI = 3.2 × 10⁴ kN.mm²”. Taken literally that is a flexural rigidity some ten orders of magnitude too small for an 18 m beam and would make the 36 mm settlement effect vanish. The intended unit is $\text{kN}\cdot\text{m}^2$: with $EI=3.2\times10^{4}\ \text{kN}\cdot\text{m}^{2}$ the settlement term $3EI\Delta/L^{2}=54\ \text{kN}\cdot\text{m}$ is the same order as the load term $wL^{2}/8=48\ \text{kN}\cdot\text{m}$, which is plainly what the question is testing. The solution below uses $\text{kN}\cdot\text{m}^{2}$ and flags the typographical error.

Question 4: Horizontal Deflection of a Frame by Castigliano's Theorem (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.

Item	Value
Horizontal load at joint 1	$P = 100\ \text{kN}$ (acting to the right)
Vertical member 1–2	6.0 m
Inclined members 2–3 and 2–4	3.6 m run, 4.8 m drop ⇒ $L=6.0$ m each
Support spacing (joints 3 to 4)	7.2 m
Height of joint 1 above the supports	$6.0+4.8 = 10.8$ m
Flexural rigidity	$EI = 1.44\times10^{5}\ \text{kN}\cdot\text{m}^{2}$, all members, inextensible
Supports	Joint 3 pin (two reactions), joint 4 roller (one vertical reaction)

Find. The horizontal deflection of joint 1.

Question 4 — a pin at joint 3 and a roller at joint 4 give exactly three reactions, so this frame is statically determinate and Castigliano's first theorem can be applied directly.

Approach. Confirm the frame is determinate, find the reactions in terms of the applied load $P$, write the bending moment in each member as a linear function of $P$, and evaluate $\Delta = \int M\,(\partial M/\partial P)\,\mathrm{d}s/EI$ member by member.

Check determinacy. With $m=3$, $j=4$ and $r_{\text{reac}} = 2+1 = 3$, $$r = 3(3) + 3 - 3(4) = 0$$ The frame is determinate, so the moments can be written straight from statics with no redundant to chase — the whole 18 marks lie in setting up and evaluating the integrals correctly.
Find the reactions in terms of $P$. Taking moments about the pin at joint 3, the load acts 10.8 m above the support line and the roller is 7.2 m away: $$R_{4y} = \frac{P(10.8)}{7.2} = 1.5P = 150\ \text{kN (up)}$$ $$R_{3y} = -1.5P = -150\ \text{kN (i.e. 150 kN down)},\qquad R_{3x} = -P = -100\ \text{kN}$$ The pin is pulled downward: the frame is trying to overturn about the roller, and the pin is the hold-down.
Write the moment in member 1–2. Measuring $s$ downward from joint 1, the only force on the free body above the section is $P$ itself acting at a lever arm $s$: $$M_{12}(s) = P s,\qquad \frac{\partial M_{12}}{\partial P}=s,\qquad 0\le s\le 6.0\ \text{m}$$ At joint 2 this is $M = 100(6.0) = 600\ \text{kN}\cdot\text{m}$.
Write the moments in the two inclined members. Working from each support, with $t$ the fraction of member length measured from the support end, $$M_{23}(t) = (3.6t)(-1.5P) - (4.8t)(-P) = -0.6Pt$$ $$M_{24}(t) = (-3.6t)(1.5P) = -5.4Pt$$ so at joint 2 the two contributions are $60\ \text{kN}\cdot\text{m}$ and $540\ \text{kN}\cdot\text{m}$. Joint equilibrium is satisfied exactly: $60+540 = 600$, matching the column. This check costs one line and catches any reaction sign error before the integrals are attempted.
Evaluate the three integrals. With $\mathrm{d}s = 6.0\,\mathrm{d}t$ on the inclined members, $$\int_{1}^{2} M\frac{\partial M}{\partial P}\mathrm{d}s = \int_{0}^{6}P s^{2}\,\mathrm{d}s = \frac{P(6.0)^{3}}{3} = 72P = 7200$$ $$\int_{2}^{3} = \int_{0}^{1}(-0.6Pt)(-0.6t)(6.0)\,\mathrm{d}t = \frac{6.0(0.36)P}{3} = 0.72P = 72$$ $$\int_{2}^{4} = \int_{0}^{1}(-5.4Pt)(-5.4t)(6.0)\,\mathrm{d}t = \frac{6.0(29.16)P}{3} = 58.32P = 5832$$ Summing, $\sum \int M(\partial M/\partial P)\,\mathrm{d}s = 13\,104\ \text{kN}^{2}\cdot\text{m}^{3}$. The right-hand inclined member alone supplies 45 % of the total and the column another 55 %: the lightly stressed left member contributes barely half a percent.
Divide by the flexural rigidity. $$\Delta_{1x} = \frac{13\,104}{1.44\times10^{5}} = 0.09100\ \text{m}$$ $$\Delta_{1x} = \boxed{91.0\ \text{mm, in the direction of the 100 kN load}}$$ A direct-stiffness solution of the same frame returns 91.0 mm to four figures, confirming both the moment expressions and the arithmetic.

Quantity	Symbol	Value
Horizontal reaction at the pin	$R_{3x}$	100 kN (toward the load)
Vertical reaction at the pin	$R_{3y}$	150 kN downward (hold-down)
Vertical reaction at the roller	$R_{4y}$	150 kN upward
Moment at joint 2 (column)	$M_{12}$	600 kN·m
Moment at joint 2 (member 2–3 / 2–4)	—	60 and 540 kN·m (sum 600 ✓)
Strain-energy integral	$\sum\int M\,\partial M/\partial P\,\mathrm{d}s$	13 104 kN$^2$·m$^3$
Horizontal deflection of joint 1	$\Delta_{1x}$	91.0 mm (in the direction of $P$)

Topic. Castigliano's first theorem for deflection of a determinate plane frame in flexure.

Key relations. $\Delta_i = \partial U/\partial P_i$; for flexure-dominated frames $U = \int M^{2}\mathrm{d}s/2EI$ so $\Delta = \int M(\partial M/\partial P)\,\mathrm{d}s/EI$; for a determinate structure $M$ is linear in $P$, so $\partial M/\partial P = M/P$.

Why this works. The partial derivative of strain energy with respect to a load equals the displacement of that load's point of application in the load's own direction. Because the applied load already sits at the point and in the direction asked for, no dummy load is needed — a real convenience that this question is deliberately built around. The axial and shear terms are omitted because the members are declared inextensible and are slender enough that shear strain energy is a fraction of a percent of the flexural term.

Common pitfall. Mis-reading joint 4 as a pin. Two pins would make the frame indeterminate to one degree and the direct Castigliano evaluation would be invalid; the drawn support has the two small rollers under the triangle, so it supplies one reaction only. The second trap is integrating along the horizontal projection of the inclined members instead of along their true 6.0 m length, which understates the deflection by a fifth.

Source. Hibbeler, Structural Analysis, 10th ed., §9.7–9.8; Kassimali, Structural Analysis, 6th ed., §7.6.

← Question 3 Next: Question 5 →

Item	Value
Horizontal load at joint 1	\(P = 100\ \text{kN}\) (acting to the right)
Vertical member 1–2	6.0 m
Inclined members 2–3 and 2–4	3.6 m run, 4.8 m drop ⇒ \(L=6.0\) m each
Support spacing (joints 3 to 4)	7.2 m
Height of joint 1 above the supports	\(6.0+4.8 = 10.8\) m
Flexural rigidity	\(EI = 1.44\times10^{5}\ \text{kN}\cdot\text{m}^{2}\), all members, inextensible
Supports	Joint 3 pin (two reactions), joint 4 roller (one vertical reaction)

Quantity	Symbol	Value
Horizontal reaction at the pin	\(R_{3x}\)	100 kN (toward the load)
Vertical reaction at the pin	\(R_{3y}\)	150 kN downward (hold-down)
Vertical reaction at the roller	\(R_{4y}\)	150 kN upward
Moment at joint 2 (column)	\(M_{12}\)	600 kN·m
Moment at joint 2 (member 2–3 / 2–4)	—	60 and 540 kN·m (sum 600 ✓)
Strain-energy integral	\(\sum\int M\,\partial M/\partial P\,\mathrm{d}s\)	13 104 kN\(^2\)·m\(^3\)
Horizontal deflection of joint 1	\(\Delta_{1x}\)	91.0 mm (in the direction of \(P\))