16-Civ-B1 Advanced Structural Analysis · December 2015

Question 3 of 9: Three-Bar Truss by the Least-Work Theorem

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

Notes on this paper

Paper format. National Examinations, December 2015 — 98-Civ-B1 Advanced Structural Analysis. 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. 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 3: Three-Bar Truss by the Least-Work 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.

Member	From	To	Projections	Length $L$	Axial rigidity
1–3	apex	left pin	4.8 m across, 6.4 m down	8.0 m	$AE$
1–2	apex	right pin	4.8 m across, 6.4 m down	8.0 m	$AE$
1–4	apex	bottom pin	vertical	10.0 m	$1.6\,AE$

Applied load: 48 kN acting vertically downward at joint 1. Joints 2, 3 and 4 are pin supports; joint 1 is the only free joint.

Find. The axial force in each of the three members, stated as tension or compression.

Question 3 — three bars meeting at a single free joint, each anchored to its own pin. Two equilibrium equations, three unknowns: one degree of statical indeterminacy.

Approach. Express the two inclined forces in terms of the vertical bar force using joint equilibrium, write the strain energy of the three bars, and set $\partial U/\partial X = 0$ — the least-work statement that the redundant force takes the value which minimises the strain energy.

Confirm the degree of indeterminacy and the member lengths. All three bars meet at joint 1, whose equilibrium supplies two equations, while there are three unknown bar forces, so $r=1$. The inclined length follows from the 4.8 m and 6.4 m projections: $$L_{13}=L_{12}=\sqrt{4.8^{2}+6.4^{2}}=\sqrt{23.04+40.96}=8.000\ \text{m}$$ with direction cosines $0.6$ horizontally and $0.8$ vertically. The vertical bar spans $6.4+3.6=10.0\ \text{m}$.
Choose the redundant and write joint equilibrium. Take $X = F_{14}$ as the redundant (tension positive). Horizontal equilibrium at joint 1 is $$-0.6F_{13} + 0.6F_{12} = 0 \;\Longrightarrow\; F_{12}=F_{13}=F$$ which is symmetry, obtained for free. Vertical equilibrium is $$-0.8F - 0.8F - X - 48 = 0 \;\Longrightarrow\; F = -\frac{48+X}{1.6} = -30 - 0.625X$$
Write the strain energy. Each bar stores $N^{2}L/2AE$, and the two inclined bars are identical: $$U = 2\cdot\frac{F^{2}(8.0)}{2AE} + \frac{X^{2}(10.0)}{2(1.6AE)} = \frac{8F^{2}}{AE} + \frac{X^{2}(10.0)}{3.2\,AE}$$ Because every term carries the same $1/AE$, the common factor cancels in the minimisation — the answer depends only on the ratio $1.6$ and on the lengths, not on the absolute value of $AE$.
Apply the least-work condition. With $\partial F/\partial X = -0.625$, $$\frac{\partial U}{\partial X} = \frac{16F}{AE}\left(-0.625\right) + \frac{2X(10.0)}{3.2\,AE} = 0$$ $$-10F + 6.25X = 0 \;\Longrightarrow\; F = 0.625X$$
Solve the pair of relations. Substituting the equilibrium expression for $F$, $$0.625X = -30 - 0.625X \;\Longrightarrow\; 1.25X = -30$$ $$X = F_{14} = \boxed{-24.0\ \text{kN}}$$ and back-substituting, $$F_{12}=F_{13}=0.625(-24.0)= \boxed{-15.0\ \text{kN}}$$ The negative signs mean compression in all three bars, which is physically inevitable: a downward load at the apex pushes on every bar radiating downward from it.
Check the answer against equilibrium. Restoring the forces to the vertical equation, $$-1.6(-15.0) - (-24.0) = 24.0 + 24.0 = 48.0\ \text{kN} = P \quad\checkmark$$ The load splits exactly half-and-half between the pair of inclined bars and the single stiffer vertical bar. Each pin then carries the axial force of its own bar: $9.0\ \text{kN}$ horizontal and $12.0\ \text{kN}$ vertical at joints 2 and 3 (opposite in sign horizontally), and $24.0\ \text{kN}$ vertical at joint 4.

Member	Length	Axial rigidity	Force	Sense
1–2	8.0 m	$AE$	15.0 kN	Compression
1–3	8.0 m	$AE$	15.0 kN	Compression
1–4	10.0 m	$1.6\,AE$	24.0 kN	Compression
Vertical load carried	by the two inclined bars		24.0 kN	50 % of $P$
Vertical load carried	by the vertical bar		24.0 kN	50 % of $P$

Topic. The least-work (second Castigliano) theorem applied to a statically indeterminate pin-jointed assembly.

Key relations. $U=\sum N_i^{2}L_i/2A_iE_i$; $\partial U/\partial X = 0$ for a redundant internal force; equivalently $\sum (N_i L_i / A_iE_i)(\partial N_i/\partial X) = 0$.

Why this works. In a linearly elastic structure the true internal force distribution is the one, among all distributions that satisfy equilibrium, that makes the strain energy stationary. Writing every bar force as a linear function of the single redundant $X$ turns the strain energy into a quadratic in $X$, and one differentiation locates the minimum. The result is a compatibility statement in disguise: $\partial U/\partial X=0$ says the relative displacement across the cut redundant is zero, which is exactly what continuity demands. Because the three flexibilities share the factor $1/AE$, only relative stiffness matters, which is why the question gives $1.6AE$ rather than a number.

Common pitfall. Differentiating $U$ while treating $F$ as a constant. The two inclined forces depend on $X$ through equilibrium, so the chain rule term $\partial F/\partial X=-0.625$ is essential; dropping it gives $X=0$, i.e. the absurd conclusion that the stiffest bar carries nothing. A second trap is scaling the strain energy of the vertical bar with $AE$ instead of $1.6AE$, which shifts the load split away from the exact 50/50 that the calibrated data produce.

Source. Hibbeler, Structural Analysis, 10th ed., §9.7 and §10.5 (least work); Kassimali, Structural Analysis, 6th ed., §7.7 and §13.4.

← Question 2 Next: Question 4 →

Member	Length	Axial rigidity	Force	Sense
1–2	8.0 m	\(AE\)	15.0 kN	Compression
1–3	8.0 m	\(AE\)	15.0 kN	Compression
1–4	10.0 m	\(1.6\,AE\)	24.0 kN	Compression
Vertical load carried	by the two inclined bars		24.0 kN	50 % of \(P\)
Vertical load carried	by the vertical bar		24.0 kN	50 % of \(P\)