Slope-Deflection Analysis of a Gable Frame with a Fabrication Error: 07-Str-A4, December 2016, Q5

Question 5: Slope-Deflection Analysis of a Gable Frame with a Fabrication Error (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. An unloaded two-member gable frame, pinned at joint 1 and built in at joint 3, in which both members have been fabricated (thermally elongated) too long and forced into place.

Given data
Quantity	Symbol	Value
Coordinates of joint 1 (pin) / 2 (apex) / 3 (fixed)	—	$(0,0)$ / $(1.8,\,2.4)$ / $(5.0,\,0)$ m
Length of member 1–2	$L_{12}$	3.0 m
Length of member 2–3	$L_{23}$	4.0 m
Elongation of member 1–2	$e_{12}$	$+2.4$ mm
Elongation of member 2–3	$e_{23}$	$+3.2$ mm
Flexural rigidity of both members	$EI$	$1.8\times10^{5}\ \text{kN}\cdot\text{m}^{2}$
Applied load	—	none

Find. The member-end moments, and hence the shear force and bending moment diagrams with their maximum and minimum ordinates.

Question 5: unloaded gable frame; each member has been forced into place after being fabricated too long by the amount shown.

Approach. Because the members are inextensible under stress, the two prescribed elongations fix the displacement of the apex completely by geometry. Convert that displacement into member chord rotations, feed them into the slope-deflection equations as the only load terms, and solve the two remaining equilibrium conditions.

Find the apex displacement from the fabrication errors. With $\mathbf{n}_{12}=(0.6,\,0.8)$ and $\mathbf{n}_{23}=(0.8,\,-0.6)$ the unit axial vectors, and both supports immovable, the axial conditions $\mathbf{D}_2\cdot\mathbf{n}_{12}=e_{12}$ and $-\mathbf{D}_2\cdot\mathbf{n}_{23}=e_{23}$ give $$0.6u_2+0.8v_2=0.0024,\qquad -0.8u_2+0.6v_2=0.0032,$$ $$\boxed{u_2=-1.12\ \text{mm},\qquad v_2=+3.84\ \text{mm}.}$$ The apex is pushed up and slightly to the left — the only way two over-length members can both fit between fixed feet.
Convert to chord rotations. With $\mathbf{e}_2$ the member axis turned $+90^\circ$, $$\psi_{12}=\frac{(\mathbf{D}_2-\mathbf{D}_1)\cdot\mathbf{e}_{2}}{L_{12}} =\frac{0.0032}{3.0}=+1.0667\times10^{-3}\ \text{rad},\qquad \psi_{23}=\frac{-0.0024}{4.0}=-6.00\times10^{-4}\ \text{rad}.$$
Write the slope-deflection equations. There are no span loads, so every $\mathrm{FEM}$ is zero. With $k_{12}=2EI/3=1.2\times10^{5}$ and $k_{23}=2EI/4=9.0\times10^{4}$ $\text{kN}\cdot\text{m}$, $$M_{12}=k_{12}\!\left(2\theta_1+\theta_2-3\psi_{12}\right),\quad M_{21}=k_{12}\!\left(2\theta_2+\theta_1-3\psi_{12}\right),$$ $$M_{23}=k_{23}\!\left(2\theta_2-3\psi_{23}\right),\quad M_{32}=k_{23}\!\left(\theta_2-3\psi_{23}\right),$$ with $\theta_3=0$ at the built-in end.
Apply the two conditions. The pin at joint 1 requires $M_{12}=0$, which gives $\theta_1=(0.0032-\theta_2)/2$; joint 2 equilibrium requires $M_{21}+M_{23}=0$. Substituting, $$\left(1.8\times10^{5}\theta_2-192\right)+ \left(1.8\times10^{5}\theta_2+162\right)=0 \;\Longrightarrow\; \theta_2=\frac{30}{3.6\times10^{5}}=8.333\times10^{-5}\ \text{rad},$$ and back substitution gives $\theta_1=1.5583\times10^{-3}$ rad.
Member-end moments. $$M_{12}=0,\qquad \boxed{M_{21}=-177.0\ \text{kN}\cdot\text{m}},\qquad M_{23}=+177.0\ \text{kN}\cdot\text{m},\qquad \boxed{M_{32}=+169.5\ \text{kN}\cdot\text{m}.}$$ A frame carrying no load at all is therefore working at nearly $180\ \text{kN}\cdot\text{m}$ — a 3 mm misfit in a stiff frame is a serious load case, which is exactly the lesson the question is built around.
Shears, axial forces and reactions. Neither member carries a span load, so each shear is constant and follows from $V=(M_{ij}+M_{ji})/L$: $$V_{12}=\frac{0-177.0}{3.0}=59.0\ \text{kN},\qquad V_{23}=\frac{177.0+169.5}{4.0}=86.6\ \text{kN}.$$ The members happen to be mutually perpendicular ($\mathbf{n}_{12}\cdot\mathbf{n}_{23}=0$), so joint 2 equilibrium simply exchanges the two values: member 1–2 carries $86.6$ kN of axial compression and member 2–3 carries $59.0$ kN of compression. The pin reaction at joint 1 is $99.2$ kN horizontal and $33.9$ kN vertical, a resultant of $104.8$ kN, and the built-in end at joint 3 returns the same force together with a moment of $169.5\ \text{kN}\cdot\text{m}$.

Question 5: bending moment diagram (kN.m). Zero at the pin, -177.0 at the apex and +169.5 at the built-in end, with one point of contraflexure 2.04 m from the apex along member 2-3.

Question 5: shear force diagram (kN). Constant within each member because neither carries a span load.

Question 5 — results
Quantity	Value
Apex displacement $u_2$, $v_2$	$-1.12$ mm, $+3.84$ mm
Chord rotations $\psi_{12}$, $\psi_{23}$	$+1.0667\times10^{-3}$, $-6.00\times10^{-4}$ rad
Joint rotations $\theta_1$, $\theta_2$	$1.5583\times10^{-3}$, $8.333\times10^{-5}$ rad
$M_{12}$ (pin)	0
$M_{21}$ / $M_{23}$ at the apex	$-177.0$ / $+177.0\ \text{kN}\cdot\text{m}$
$M_{32}$ at the built-in end (maximum)	$+169.5\ \text{kN}\cdot\text{m}$
Shear in 1–2 / in 2–3 (constant)	$59.0$ / $86.6$ kN
Axial force in 1–2 / in 2–3	$86.6$ / $59.0$ kN compression
Reaction at the pin, joint 1	$99.2$ kN horizontal, $33.9$ kN vertical (resultant $104.8$ kN)
Point of contraflexure	2.04 m from the apex along 2–3

Topic. Slope-deflection analysis of lack of fit: a self-straining load case in which the “load” enters only through the chord rotations.

Key relations. Axial compatibility $(\mathbf{D}_j-\mathbf{D}_i)\cdot\mathbf{n}_{ij}=e_{ij}$ for an inextensible member fabricated $e_{ij}$ too long; chord rotation $\psi_{ij}=[(\mathbf{D}_j-\mathbf{D}_i)\cdot\mathbf{e}_2]/L$; slope-deflection $M_{ij}=(2EI/L)(2\theta_i+\theta_j-3\psi_{ij})$ with $\mathrm{FEM}=0$; constant member shear $V=(M_{ij}+M_{ji})/L$.

Why this works. The instruction to neglect axial strain is what makes the problem tractable and also what makes it severe. Because the members cannot shorten elastically, the fabrication errors are absorbed entirely as rigid geometry: the apex must move to the one position that lets both over-length members span between immovable feet, and that single vector $\mathbf{D}_2$ then determines both chord rotations. From there the problem is an ordinary two-unknown slope-deflection exercise. The severity follows from the same reasoning: with $EI$ of order $10^{5}$ $\text{kN}\cdot\text{m}^{2}$, a chord rotation of $10^{-3}$ rad generates moments of order $EI\psi/L\approx 60$ $\text{kN}\cdot\text{m}$ per unit, so millimetres of misfit produce design-level actions.

Common pitfall. The classic blunder is a sign inversion in $\psi$, which flips the whole answer and typically puts an over-length member into tension. Two cheap checks catch it: the imposed apex displacement must satisfy both axial equations simultaneously (here $u_2$ is negative even though both members grew), and members forced into place after being made too long must finish in compression, as they do here. The second pitfall is trusting a penalty-stiffness computer model for the axial forces after prescribing the joint displacement: the element does not know about the fabrication error and reports a meaningless axial force, so the axial forces must come from joint equilibrium.

Source. Hibbeler, Structural Analysis, 10th ed., §11.4–11.5 (slope-deflection for frames with sidesway and support movements); Ghali, Neville & Brown, 7th ed., §4.6 (lack of fit and temperature effects); Kassimali, 6th ed., §16.6.

Question 5 of 9: Slope-Deflection Analysis of a Gable Frame with a Fabrication Error

Question 5: Slope-Deflection Analysis of a Gable Frame with a Fabrication Error (18 marks)