Question 6 of 9: Sway frame with load-carrying stubs, by slope-deflection
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2017 — 07-Str-A4 Advanced Structural Analysis. Three hours, CLOSED BOOK (an approved Sharp or Casio calculator is permitted). Answer BOTH Questions 1 and 2, ONLY TWO of Questions 3, 4 or 5, and ONLY TWO of Questions 6, 7, 8 or 9; six questions constitute a complete paper for 100 marks. Marks are printed in the left margin. All nine questions are worked below, because the complete set is the study resource.
Reference texts. R. C. Hibbeler, Structural Analysis, 10th ed. (Pearson) — Ch. 8 influence lines, Ch. 9–10 the force (flexibility) method, Ch. 11 slope-deflection, Ch. 14–16 the stiffness method; A. Kassimali, Structural Analysis, 6th ed. (Cengage) — Ch. 7 deflections by work and energy (Castigliano), Ch. 13 slope-deflection, Ch. 16 sidesway frames; W. McGuire, R. H. Gallagher & R. D. Ziemian, Matrix Structural Analysis, 2nd ed., for the element stiffness matrix used as the independent check. Once the analysis is complete, member design follows CSA A23.3:19, CSA S16:19 and CSA O86:19, with load combinations from NBCC 2020.
Check: sign convention used throughout. End moments $M_{ij}$ are counter-clockwise positive on the member end — the convention that matches the six-degree-of-freedom planar frame element stiffness matrix, so every answer below can be checked against a direct-stiffness solution. Chord rotation is $\psi_{ij}=\left[(\mathbf{D}_{j}-\mathbf{D}_{i})\cdot\mathbf{e}_{2}\right]/L$, where $\mathbf{e}_{2}$ is the member axis turned $+90^\circ$. The fixed-end moment of a downward uniform load is $\mathrm{FEM}_{ij}=+wL^{2}/12$ at the $i$ end and $-wL^{2}/12$ at the $j$ end; for a member released at its far end it becomes $+wL^{2}/8$. Ordinary sagging moments follow as $M_{\text{sag}}(i)=-M_{ij}$ and $M_{\text{sag}}(j)=+M_{ji}$, and all diagrams are plotted sagging positive. Member shear at the $i$ end is $V_{i}=(M_{ij}+M_{ji})/L$ plus the equivalent nodal shear of any span load. Mixing this with Hibbeler's clockwise-positive convention produces clean-looking integers that are wrong, so the convention is stated once and used everywhere.
Question 6: Sway frame with load-carrying stubs, by slope-deflection (24 marks)
Given. A single-bay portal, both bases built in, columns 4 m from base to beam level, beam 4 m, and a 1 m stub above each top joint carrying a horizontal 21 kN load to the right. All members share the same $EI$ and are inextensible; sidesway is free.
Find. The end moments in every member, the shear force and bending moment diagrams with their maximum and minimum ordinates, and the reactions.
Portal with 1 m stubs: each stub delivers a 21 kN horizontal force and a 21 kN.m clockwise couple to its joint.
Approach. Replace each determinate stub by a joint force and a joint couple, use anti-symmetry to collapse the two joint rotations into one, and close the problem with the storey-shear equation.
Replace the stubs by joint actions. Each stub is a determinate cantilever, so it is statically equivalent to a horizontal force of 21 kN applied at its joint together with a couple$$M_{\text{stub}}=P\,s=21(1)=21\ \text{kN}\cdot\text{m}$$acting clockwise, i.e. $-21$ kN·m in the counter-clockwise-positive convention. Dropping this couple is the classic error: the storey-shear equation still balances without it, so the mistake stays hidden until global overturning fails.
Reduce the unknowns by anti-symmetry. The frame is symmetric and the applied set (two equal rightward forces plus two equal clockwise couples at mirror-image joints) is anti-symmetric, so $\theta_{2}=\theta_{3}\equiv\theta$ and the beam has zero chord rotation. Writing $A=EI\theta$ and $B=EI\Delta$ for the sway, the member equations become$$M_{12}=\frac{2}{h}\!\left(A+\frac{3B}{h}\right),\quad M_{21}=\frac{2}{h}\!\left(2A+\frac{3B}{h}\right),\quad M_{23}=M_{32}=\frac{6A}{L},$$with $h=L=4$ m, since $\psi_{\text{col}}=-\Delta/h$.
Write the joint equation. Moment equilibrium at joint 2 reads $M_{21}+M_{23}=-P s$, that is$$\left(\frac{4}{h}+\frac{6}{L}\right)A+\frac{6}{h^{2}}B=-21 .$$With the numbers, $2.5A+0.375B=-21$.
Write the sway equation. Virtual work on the sway pattern (both joints move 1 to the right, rotations held) equates the work of the two column shears to that of the applied forces, giving$$\frac{M_{12}+M_{21}}{h}+\frac{M_{43}+M_{34}}{h}=2P \;\Longrightarrow\;\frac{6}{h}A+\frac{12}{h^{2}}B=P\,h,$$that is $0.75A+0.375B=42$. Subtracting the joint equation gives $1.75A=-63$, hence$$EI\theta=-36\ \text{kN}\cdot\text{m}^{2},\qquad EI\Delta=\boxed{184\ \text{kN}\cdot\text{m}^{3}}.$$The sway is positive, i.e. the frame moves to the right under a rightward load — the cheapest sanity check available on a sway problem.
Recover the end moments. Substituting back,$$M_{12}=M_{43}=51\ \text{kN}\cdot\text{m},\quad M_{21}=M_{34}=33\ \text{kN}\cdot\text{m},\quad M_{23}=M_{32}=-54\ \text{kN}\cdot\text{m}.$$Joint 2 checks: $33-54=-21$, exactly the stub couple. A second, more telling check is that the stub moment and the column moment add to the beam moment, $21+33=54$: the moment in the continuous column therefore steps by 54 kN·m as it passes the beam connection, which is joint equilibrium, not an error.
Assemble the diagrams and reactions. Each column carries a constant shear of$$V_{\text{col}}=\frac{M_{12}+M_{21}}{h}=\frac{51+33}{4}=21\ \text{kN},$$and the two together carry the applied 42 kN. The beam has no span load, so its moment runs linearly from $+54$ kN·m sagging at joint 2 to $-54$ kN·m at joint 3, passing through zero at mid-span, and its shear is constant at $2(54)/4=27$ kN. Vertical equilibrium then gives 27 kN down at base 1 and 27 kN up at base 4, with 21 kN of horizontal reaction and 51 kN·m of fixing moment at each base. Overturning closes: $2(21)(5)=210$ kN·m of applied moment against $51+51+27(4)=210$ kN·m of restoring moment.
