16-Civ-B1 Advanced Structural Analysis · December 2013

Question 8 of 9: Slope-deflection analysis of a sway frame

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 8: Slope-deflection analysis of a sway frame (22 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 vertical member runs from an encastré base at joint 1, through joint 2 at 10 m, to joint 3 at 20 m, where a roller bearing on a vertical surface provides horizontal restraint only. A horizontal member 10 m long runs from joint 2 to joint 4, which is a vertical roller. A horizontal point load of 63 kN acts to the right at joint 2, and the horizontal member carries 20 kN/m downward over its full length. No numerical rigidity is given.

Given data
Quantity	Symbol	Value
Lower column 1–2 and upper column 2–3	h	10 m each
Horizontal member 2–4	L	10 m
Horizontal load at joint 2	H_f	63 kN
Uniform load on member 2–4	w	20 kN/m
Support at 1 / 3 / 4	—	fixed / horizontal roller / vertical roller
Flexural rigidity	EI	uniform, unspecified

Find. All member end moments, and the shear and bending-moment diagrams with their extreme ordinates.

Question 8: the sway frame. The bearing at joint 3 restrains horizontal movement only; the roller at joint 4 restrains vertical movement only.

Approach. Identify the single sway degree of freedom, write the member equations in terms of the joint rotation $\theta_2$ and the sway $\Delta$, then close the system with one joint-moment equation and one storey-shear equation.

Establish the degrees of freedom. Column 1–2 is inextensible and joint 1 is fixed, so joint 2 cannot move vertically; column 2–3 then holds joint 3 at the same level. Joint 3 is restrained horizontally by its bearing and joint 1 by its encastré base, but joint 2 is free to translate horizontally by $\Delta$, dragging joint 4 with it through the inextensible horizontal member. The unknowns are therefore $\theta_2$ and $\Delta$ once the two pinned ends are condensed out.
Write the chord rotations. With joint 2 displaced $+\Delta$ and joints 1 and 3 held, $$\psi_{12}=-\frac{\Delta}{10},\qquad \psi_{23}=+\frac{\Delta}{10},\qquad \psi_{24}=0,$$ the two columns rotating in opposite senses as the frame racks.
Assemble the member equations. Putting $k=EI$ and $2EI/L=0.2k$, and imposing $M_{32}=0$ and $M_{42}=0$ to condense the pinned ends, $$M_{12}=0.2k\left(\theta_2+0.3\Delta\right),\qquad M_{21}=0.2k\left(2\theta_2+0.3\Delta\right),$$ $$M_{23}=0.2k\left(1.5\theta_2-0.15\Delta\right),\qquad M_{24}=0.3k\,\theta_2+250,$$ the constant 250 being $wL^{2}/8$, the propped-cantilever fixed-end moment of the horizontal member.
Write the two equilibrium equations. Joint 2 must balance, $M_{21}+M_{23}+M_{24}=0$, giving $$k\theta_2+0.03k\Delta=-250.$$ Horizontal equilibrium of joint 2 involves the two column shears and the applied load, $$63-\frac{M_{12}+M_{21}}{10}+\frac{M_{23}+M_{32}}{10}=0 \quad\Longrightarrow\quad 0.3k\theta_2+0.15k\Delta=630.$$ The horizontal member carries no axial force because joint 4 is a free-sliding roller, so it does not appear in the storey-shear equation.
Solve the pair. $$EI\,\theta_2=-400,\qquad EI\,\Delta=+5000$$ (units kN·m² and kN·m³ respectively). The frame sways to the right, in the direction of the 63 kN load, as it must.
Back-substitute for the end moments. $$M_{12}=\boxed{+220\ \text{kN}\cdot\text{m}},\quad M_{21}=\boxed{+140\ \text{kN}\cdot\text{m}},\quad M_{23}=\boxed{-270\ \text{kN}\cdot\text{m}},\quad M_{24}=\boxed{+130\ \text{kN}\cdot\text{m}}$$ with $M_{32}=M_{42}=0$; the three moments at joint 2 sum to zero, and substituting them back into the storey-shear equation returns 630 exactly.
Draw the diagrams. The lower column carries a constant shear $(220+140)/10=36$ kN and its moment runs from 220 kN·m at the base to 140 kN·m of opposite sense at joint 2, with a point of contraflexure $220/36=6.11$ m above the base. The upper column carries $270/10=27$ kN of shear, and $63-36-27=0$ confirms horizontal equilibrium. On the horizontal member the end shear follows from $-130+10V-1000=0$, so $V=113$ kN falling to $-87$ kN at the roller, and $$x=\frac{113}{20}=5.65\ \text{m},\qquad M_{\max}=\boxed{+189.23\ \text{kN}\cdot\text{m}}.$$

Question 8: bending moment developed up the columns 1–2–3. The 130 kN·m step at joint 2 is the horizontal member’s share.

Question 8: bending moment in the horizontal member 2–4.

Question 8: shear force in the horizontal member 2–4.

Question 8 — results
Member	Bending moment (kN·m)	Shear (kN)
Column 1–2	220 at the fixed base, 140 at joint 2 (opposite sense); contraflexure 6.11 m up	36.0 constant
Column 2–3	270 at joint 2 falling to 0 at the bearing	27.0 constant
Member 2–4	−130 at joint 2; max +189.23 at 5.65 m; 0 at the roller	+113.0 falling to −87.0
Sway of joint 2	Δ = 5000 / EI (to the right)
Reactions: joint 1 = 36.0 kN horizontal, 113.0 kN vertical, 220 kN·m fixing moment; joint 3 = 27.0 kN horizontal; joint 4 = 87.0 kN vertical. Checks: 63 − 36 − 27 = 0 and 113 + 87 = 20(10). ✓

Topic. Slope-deflection with sidesway, and the storey-shear equation that supplies the extra condition.

Key relations. Shear in an unloaded member $=(M_{ij}+M_{ji})/L$; storey equilibrium $\sum(\text{column shears})=\sum(\text{applied horizontal forces})$; propped-cantilever fixed-end moment $wL^{2}/8$; a member with a roller at its far end perpendicular to its axis carries no axial force.

Why this works. A sway frame needs exactly one extra equation per independent translation, and free-body horizontal equilibrium of the storey is the cheapest one to write because it involves only column shears, each of which is already expressed through the end moments. The subtlety in this particular frame is the two unusual supports: the bearing at joint 3 supplies a horizontal reaction with no vertical one, while the roller at joint 4 does the reverse. Between them they leave the frame free to rack, and they keep the horizontal member axially unloaded — which is why the storey-shear equation contains only two terms rather than three.

Common pitfall. Missing the sway entirely because the base is fixed and the top is restrained, and then obtaining a set of moments that fails the storey-shear check by the full 63 kN. A useful sanity test on any sway solution is to add up the column shears and compare with the applied lateral force before drawing anything. The second pitfall is applying $wL^{2}/12$ instead of $wL^{2}/8$ to the horizontal member; its far end is a pin, not a fixed end.

Source. Hibbeler, Structural Analysis, 10th ed., §11.5 (slope-deflection for frames with sidesway); Kassimali, 6th ed., §16.5.

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