16-Civ-B1 Advanced Structural Analysis · December 2013

Question 5 of 9: Slope-deflection analysis of a frame with an imposed joint movement

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 5: Slope-deflection analysis of a frame with an imposed joint movement (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. A gable frame of two 5 m rafters. Joint 1 is a roller at the left, the apex joint 2 sits 4 m to its right and 3 m above it, and joint 3 is encastré 4 m further to the right at the level of joint 1. There is no applied load whatever; the only action is a prescribed horizontal movement of 12.0 mm to the left at joint 1.

Given data
Quantity	Symbol	Value
Half-span, each side	a	4 m
Rise of the apex above the springings	b	3 m
Length of each rafter	L	5 m
Prescribed movement of joint 1	Δ	12.0 mm to the left
Flexural rigidity, both members	EI	1.75 × 10⁵ kN·m²
Applied loads	—	none

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

Question 5: the gable frame. Joint 1 is a roller free to slide horizontally; the 12 mm movement is imposed on it, not caused by a load.

Approach. Since both rafters are inextensible and joint 3 is fixed, the prescribed movement of joint 1 determines the whole displacement field kinematically; the chord rotations follow at once and the slope-deflection equations then contain only the two unknown joint rotations.

Find the apex movement from inextensibility. Joint 3 is fixed, so joint 2 can only move perpendicular to rafter 2–3, i.e. along $(0.8,\,0.6)$; requiring rafter 1–2 to keep its length as joint 1 slides 12 mm left gives $$\boldsymbol{\Delta}_2=\boxed{(-6.0,\ -8.0)\ \text{mm}},$$ that is, the apex moves 6 mm to the left and drops 8 mm. Physically the span has been lengthened, so the gable flattens.
Convert to chord rotations. Applying $\psi_{ij}=[(\boldsymbol{\Delta}_j-\boldsymbol{\Delta}_i)\cdot\mathbf{e}_2]/L$ member by member, $$\psi_{12}=-0.00200\ \text{rad},\qquad \psi_{23}=+0.00200\ \text{rad}.$$ For a symmetric gable this reduces to the tidy closed form $\psi_{12}=u/(2b)=0.012/6$, where $u$ is the imposed movement and $b$ the rise. Equal magnitudes of opposite sign are the signature of a flattening gable.
Write the four slope-deflection equations. With $2EI/L=7.00\times10^{4}$ and no fixed-end moments, $$M_{12}=7.0\times10^{4}\left(2\theta_1+\theta_2+0.006\right),\qquad M_{21}=7.0\times10^{4}\left(2\theta_2+\theta_1+0.006\right),$$ $$M_{23}=7.0\times10^{4}\left(2\theta_2-0.006\right),\qquad M_{32}=7.0\times10^{4}\left(\theta_2-0.006\right),$$ using $\theta_3=0$ at the encastré joint.
Impose the two conditions. Joint 1 is a roller, so $M_{12}=0$; joint 2 must balance, so $M_{21}+M_{23}=0$. The second gives $\theta_1=-4\theta_2$ and the first then gives $$\theta_2=\frac{0.006}{7}=+8.571\times10^{-4}\ \text{rad},\qquad \theta_1=-3.429\times10^{-3}\ \text{rad}.$$
Back-substitute for the end moments. $$M_{12}=0,\qquad M_{21}=\boxed{+300\ \text{kN}\cdot\text{m}},\qquad M_{23}=\boxed{-300\ \text{kN}\cdot\text{m}},\qquad M_{32}=\boxed{-360\ \text{kN}\cdot\text{m}}$$ For a symmetric gable these collapse to $M_{21}=-15k\psi/7$ and $M_{32}=18k\psi/7$ with $k=2EI/L$, which is a fast independent check.
Get the shears and the reactions. Neither member carries a span load, so each has a constant transverse shear $\left(M_{ij}+M_{ji}\right)/L$: $$V_{12}=\boxed{+60.0\ \text{kN}},\qquad V_{23}=\boxed{-132.0\ \text{kN}}.$$ The rafters also carry axial tensions of 155 kN and 101 kN, and joint 3 develops $H=160$ kN, $V=45$ kN and a fixing moment of 360 kN·m. The force that must be applied at joint 1 to hold it 12 mm out of place is $\boxed{160\ \text{kN}}$ horizontally.

Question 5: bending moment developed along rafter 1–2 then rafter 2–3. Both diagrams are straight lines because there is no span load.

Question 5: transverse shear, constant within each rafter.

Check: the vertical reaction at the roller comes out as 45 kN acting downward, i.e. the support must be able to hold the frame down. A plain roller cannot do that, so in a real structure joint 1 would need a hold-down detail. This is not an arithmetic error: with no gravity load on the frame, the couple generated by jacking joint 1 sideways can only be balanced by an equal and opposite pair of vertical reactions.

Question 5 — results
Quantity	Value
Movement of the apex, joint 2	6.0 mm left, 8.0 mm down
Chord rotations ψ₁₂ / ψ₂₃	−0.00200 / +0.00200 rad
Joint rotations θ₁ / θ₂	−3.429 × 10⁻³ / +8.571 × 10⁻⁴ rad
M₁₂ / M₂₁	0 / +300 kN·m
M₂₃ / M₃₂	−300 / −360 kN·m
Maximum bending moment	360 kN·m at the fixed joint 3
Shear in rafter 1–2 / rafter 2–3	+60.0 / −132.0 kN
Axial tension in rafter 1–2 / 2–3	155 / 101 kN
Force required at joint 1 to impose the movement	160 kN horizontal

Topic. Slope-deflection with a prescribed support movement and no applied load.

Key relations. $M_{ij}=(2EI/L)(2\theta_i+\theta_j-3\psi_{ij})$; chord rotation from the displacement field, $\psi_{ij}=[(\boldsymbol{\Delta}_j-\boldsymbol{\Delta}_i)\cdot\mathbf{e}_2]/L$; for a symmetric gable of rise $b$ pushed by $u$ at one springing, $\psi=u/(2b)$; shear in an unloaded member $=(M_{ij}+M_{ji})/L$.

Why this works. A support movement is a kinematic action, so the analysis runs in the opposite direction to a load case: the displacement field comes first and the forces come last. Inextensibility does all the work — it forces joint 2 onto a line perpendicular to rafter 2–3, and the length condition on rafter 1–2 then pins it to a single point. Note that the moments are proportional to $EI$: unlike a load case, stiffening the frame makes an imposed movement worse, which is exactly why differential settlement is so damaging in stiff structures and why the shallower the gable (smaller $b$), the larger $\psi=u/2b$ and the more severe the effect.

Common pitfall. Getting the sign of $\psi$ wrong, which flips every moment and typically converts the flattening gable into a rising one. The cheap physical check used above is the apex: pushing the springings apart must lower the apex and put the rafters in tension. A second pitfall is assuming the two chord rotations are equal in sign because the frame is symmetric; the movement is antisymmetric, so they are equal and opposite.

Source. Hibbeler, Structural Analysis, 10th ed., §11.2 (slope-deflection with settlement); Ghali, Neville & Brown, Structural Analysis, 7th ed., ch. 4 (effects of support movements and lack of fit).

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