07-Str-A4 · December 2015

Question 5 of 9: Symmetric Gable Frame Forced into Place — Lack of Fit

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 5: Symmetric Gable Frame Forced into Place — Lack of Fit (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
Legs 1–2 and 3–4	3.0 m run, 4.0 m rise ⇒ $L_s = 5.0$ m
Top member 2–3	10.0 m
Fabricated span, joint 1 to joint 4	16.00 m
As-built support spacing	16.03 m (30 mm too far apart; 15 mm each end)
Applied loads	None
Flexural rigidity	$EI = 5.0\times10^{5}\ \text{kN}\cdot\text{m}^{2}$, all members, inextensible
Supports	Pins at joints 1 and 4

Find. The moments, shears and axial forces induced purely by forcing the frame into the over-wide supports, and the corresponding diagrams.

Question 5 — a symmetric two-pinned gable frame stretched 30 mm to reach its supports. No external load acts; the entire moment field is a lack-of-fit effect.

Approach. Impose the support movement as a prescribed displacement, use symmetry to kill the sway mechanism, obtain the knee drop and the chord rotations from inextensibility alone, then run one slope-deflection equation at a knee.

Establish the kinematics before any statics. Displacement unknowns are $u_2,v_2,u_3,v_3$; the three inextensibility conditions (one per member) leave a single free mechanism. That mechanism is anti-symmetric (both knees translate the same way), whereas the imposed spread is symmetric, so the mechanism amplitude is zero and $$u_2=u_3=0$$ This single observation is what “take advantage of symmetry” means here, and it removes the sway equation entirely.
Find the knee drop. Support 1 moves 15 mm outward, i.e. $\Delta u_1=-0.015$ m. The leg 1–2 cannot change length, so its relative end displacement must be perpendicular to its own axis $(0.6,\,0.8)$: $$0.6\,(u_2 + 0.015) + 0.8\,v_2 = 0 \;\Longrightarrow\; v_2 = -\frac{0.6(0.015)}{0.8} = -0.01125\ \text{m}$$ $$\text{knee drop} = \boxed{11.25\ \text{mm downward at both knees}}$$ Pulling the feet apart forces the legs to lie down, and the whole top member simply translates down with them.
Compute the chord rotations. Using $\psi_{ij}=[(\mathbf{d}_j-\mathbf{d}_i)\cdot\mathbf{e}_\perp]/L$, the leg carries $$\psi_{12} = \frac{(0.015,\,-0.01125)\cdot(-0.8,\,0.6)}{5.0} = \frac{-0.01875}{5.0} = -0.003750\ \text{rad}$$ and by symmetry $\psi_{34}=+0.003750$, while $\psi_{23}=0$ because both knees translate identically. A useful closed form falls out: for this geometry $\psi_{12} = -e/h$ with $e$ the half-spread and $h$ the leg rise, i.e. $-0.015/4.0$, which is a one-line check on the whole kinematic step.
Write the slope-deflection equations. With pins at joints 1 and 4 the modified stiffness applies to the legs, and symmetry gives $\theta_3=-\theta_2$: $$M_{21} = \frac{3EI}{L_s}\left(\theta_2-\psi_{12}\right) = 0.6EI\left(\theta_2 + 0.003750\right)$$ $$M_{23} = \frac{2EI}{10}\left(2\theta_2+\theta_3\right) = \frac{EI}{5}\,\theta_2$$ There are no loads, so every fixed-end moment is zero.
Enforce moment equilibrium at knee 2 and solve. $$M_{21}+M_{23}=0 \;\Longrightarrow\; 3\left(\theta_2+0.003750\right)+\theta_2 = 0$$ $$4\theta_2 = -0.011250 \;\Longrightarrow\; \theta_2 = -0.0028125\ \text{rad}\quad(\theta_3=+0.0028125)$$
Back-substitute for the moments. $$M_{21} = 0.6\left(5.0\times10^{5}\right)\left(-0.0028125+0.003750\right) = \boxed{281.25\ \text{kN}\cdot\text{m}}$$ and $M_{23}=-281.25$, $M_{32}=+281.25$, $M_{34}=-281.25$, with $M_{12}=M_{43}=0$ at the pins. In sagging terms the top member carries a constant sagging moment of 281.25 kN·m and each leg a moment rising linearly from zero at its pin to 281.25 kN·m at its knee.
Recover the reactions and the internal force set. With no external load the whole frame is a two-force body between its two pins, so the reactions must be equal, opposite and collinear along the line joining them — that is, purely horizontal: $$H = \frac{M_{\text{knee}}}{h} = \frac{281.25}{4.0} = \boxed{70.3125\ \text{kN}},\qquad V = 0$$ The legs carry a transverse shear $281.25/5.0 = 56.25\ \text{kN}$ and an axial tension $0.6H = 42.19\ \text{kN}$; the top member carries zero shear (its end moments are equal and opposite in the sagging sense) and an axial tension of $70.31\ \text{kN}$. Everything is in tension, as it must be for a frame that has been stretched.
Sanity-check the magnitude. A 30 mm misfit on a 16 m frame is a strain of under two parts in a thousand, yet it generates a moment of 281 kN·m — comparable to a serious service load. That is the engineering message of the question: with stiff members and no load path to relieve the mismatch, fabrication tolerance translates directly into locked-in stress, which is why erection procedures for such frames call for jacking or slotted holes rather than brute force.

Quantity	Value
Knee translation, horizontal	0 (killed by symmetry)
Knee translation, vertical	11.25 mm downward
Knee rotation $\theta_2 = -\theta_3$	$2.8125\times10^{-3}$ rad
Chord rotation of each leg	$3.750\times10^{-3}$ rad
Moment at each knee	281.25 kN·m
Moment at each pin	0
Top member	Constant sagging 281.25 kN·m, zero shear, 70.31 kN tension
Each leg	0 to 281.25 kN·m linear, shear 56.25 kN, tension 42.19 kN
Reactions	$H = 70.3125$ kN outward at each pin, $V = 0$

Topic. Lack of fit (fabrication error) analysed by slope-deflection with prescribed support movement.

Key relations. $M_{ij} = \frac{2EI}{L}\left(2\theta_i+\theta_j-3\psi_{ij}\right)+\mathrm{FEM}_{ij}$; modified form at a pinned far end $M_{ij}=\frac{3EI}{L}\left(\theta_i-\psi_{ij}\right)$; inextensibility $(\mathbf{d}_j-\mathbf{d}_i)\cdot\mathbf{e}_{ij}=0$; for this gable $\psi_{\text{leg}} = -e/h$.

Why this works. A lack-of-fit problem has no load vector at all: the entire right-hand side of the equilibrium equations comes from the chord rotations that the imposed displacement generates. Because the structure is redundant, the misfit cannot be absorbed by rigid-body motion and must be taken out in bending — and because the response scales linearly with $EI$, a stiffer frame is punished harder by the same tolerance. Symmetry does two jobs here: it forces $\theta_3=-\theta_2$, halving the unknowns, and it zeroes the anti-symmetric sway mechanism, removing the translation equation.

Common pitfall. Getting the sign of $\psi$ backwards, which puts the frame into compression instead of tension. The physical check is unambiguous: a frame made too short and stretched into place pulls its supports inward, so the reactions on the frame act outward and every member ends up in tension. If the arithmetic says the top member is in compression, $\psi$ has been inverted. A second trap is applying $4EI/L$ rather than $3EI/L$ to the legs; the supports are pins, and the drawn triangle at joints 1 and 4 carries no rollers, so no moment can develop there.

Source. Ghali, Neville & Brown, Structural Analysis: A Unified Classical and Matrix Approach, 7th ed., §4.5 and §10.4 (support movement and lack of fit); Hibbeler, Structural Analysis, 10th ed., §11.3–11.4.

← Question 4 Next: Question 6 →

Item	Value
Legs 1–2 and 3–4	3.0 m run, 4.0 m rise ⇒ \(L_s = 5.0\) m
Top member 2–3	10.0 m
Fabricated span, joint 1 to joint 4	16.00 m
As-built support spacing	16.03 m (30 mm too far apart; 15 mm each end)
Applied loads	None
Flexural rigidity	\(EI = 5.0\times10^{5}\ \text{kN}\cdot\text{m}^{2}\), all members, inextensible
Supports	Pins at joints 1 and 4