Slope-deflection analysis of a sway portal with free stubs: 16-Civ-B1 Advanced Structural Analysis, December 2019, Q8

Question 8: Slope-deflection analysis of a sway portal with free stubs (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. Column height $h = 4\ \text{m}$, beam span $L = 4\ \text{m}$, free stub $s = 1\ \text{m}$ above each beam joint, a horizontal load $F = 21\ \text{kN}$ at the top of each stub, both bases built in, uniform $EI$, all members inextensible.

Find. The joint rotation and sway, all member end moments, and the shear force and bending moment diagrams.

Question 8: portal frame with built-in bases and 1 m free stubs above each beam joint, each carrying a 21 kN horizontal load.

Approach. Replace each determinate stub by the force and the couple it delivers to its joint; note that the two loads and the symmetric geometry make the response antisymmetric, so the two joint rotations are equal and one sway parameter remains; write one joint equation and one storey-shear equation and solve the pair.

Replace the stubs. Each stub is a free cantilever standing above its joint, so it is determinate. Transferring its load to the joint gives a horizontal force $F = 21\ \text{kN}$ and a couple $$M_{\text{stub}} = F s = 21(1) = 21.0\ \text{kN}\cdot\text{m}$$ applied to the joint. Forgetting that couple is the classic error on this question — the storey-shear equation is still satisfied without it, so nothing looks wrong until the overturning check fails.
Reduce the unknowns. Both columns are the same height and the beam is inextensible, so joints (2) and (3) translate together by $\Delta$; the geometry and the two equal loads are mirror images, so $\theta_2 = \theta_3 = \theta$. Writing $A = EI\theta$ and $B = EI\Delta$, the whole frame carries two unknowns.
Joint equilibrium at (2). Summing the column and beam end moments against the applied stub couple, $$\left(\frac{4}{h} + \frac{6}{L}\right)A + \frac{6}{h^{2}}B = -Fs$$ $$\left(\frac{4}{4} + \frac{6}{4}\right)A + \frac{6}{16}B = -21.0 \quad\Longrightarrow\quad 2.5A + 0.375B = -21.0$$
Storey shear. The two columns together must carry the $2F$ applied above them, which gives $$\frac{6}{h}A + \frac{12}{h^{2}}B = Fh \quad\Longrightarrow\quad 1.5A + 0.75B = 84.0$$
Solve. Doubling the joint equation and subtracting the shear equation eliminates $B$: $$5.0A + 0.75B = -42.0 \;\Longrightarrow\; 3.5A = -126.0 \;\Longrightarrow\; \boxed{EI\theta = -36.0\ \text{kN}\cdot\text{m}^{2}}$$ $$0.75B = 84.0 - 1.5(-36.0) = 138.0 \;\Longrightarrow\; \boxed{EI\Delta = 184.0\ \text{kN}\cdot\text{m}^{3}}$$ The sway is to the right, as it must be under two rightward loads — a negative $B$ would have signalled a sign slip.
Back-substitute for the member end moments. With $M_{\text{base}} = \frac{2EI}{h}\!\left(\theta + \frac{3\Delta}{h}\right)$ and $M_{\text{top}} = \frac{2EI}{h}\!\left(2\theta + \frac{3\Delta}{h}\right)$, $$M_{\text{base}} = \tfrac{2}{4}\left(-36.0 + 138.0\right) = \boxed{51.0\ \text{kN}\cdot\text{m}},\qquad M_{\text{top}} = \tfrac{2}{4}\left(-72.0 + 138.0\right) = \boxed{33.0\ \text{kN}\cdot\text{m}}$$ for both columns, and the beam carries $$M_{23} = -M_{32} = \frac{2EI}{L}\left(3\theta\right) = -54.0 \;\Longrightarrow\; \boxed{M_{\text{beam}} = 54.0\ \text{kN}\cdot\text{m}}$$ hogging at (2) and sagging at (3), zero at mid-span.
The joint check that finds a dropped stub couple. At joint (2) three members meet — column, beam and stub — so the moments must balance: $$M_{\text{top}} + M_{\text{stub}} = 33.0 + 21.0 = 54.0 = M_{\text{beam}} \quad\checkmark$$ The column moment therefore steps by the full beam moment at the connection, and both ordinates should be shown on the diagram. Had the stub couple been omitted, the beam moment would have come out equal to the column-top moment and this line would not close.
Shears, axial forces and the overturning check. Each column shear is $$V_{\text{col}} = \frac{M_{\text{base}} + M_{\text{top}}}{h} = \frac{51.0 + 33.0}{4} = 21.0\ \text{kN},$$ so the two columns carry the two 21 kN loads exactly. The beam shear is $$V_{\text{beam}} = \frac{2 M_{\text{beam}}}{L} = \frac{2(54.0)}{4} = 27.0\ \text{kN},$$ which is also the axial force in each column — tension (uplift) at base (1), compression at base (4). Global overturning about base (1): $$2F(h+s) = 2(21.0)(5) = 210.0 \;=\; 2M_{\text{base}} + V_{\text{beam}}L = 2(51.0) + 27.0(4) = 210.0 \quad\checkmark$$

Question 8: bending moment plotted normal to each member. The column moment steps by the full beam moment at each joint.

The two diagrams in words. Shear: constant 21.0 kN in each stub, constant 21.0 kN in each column, constant 27.0 kN in the beam — every member is unloaded along its length, so no shear varies. Moment: linear from zero at each stub tip to 21.0 kN·m at the joint; linear down each column from 33.0 kN·m at the top to 51.0 kN·m at the base, with no sign change (both columns bend in the same single curvature, because the loading is a pure storey shear applied above the beam); and linear across the beam from 54.0 hogging at (2) to 54.0 sagging at (3), passing through zero at mid-span.

Quantity	Symbol	Result
Joint rotation (both joints)	$EI\theta$	−36.0 kN·m²
Storey sway	$EI\Delta$	+184.0 kN·m³
Moment at each column base	$M_{12},\,M_{43}$	51.0 kN·m
Moment at each column top	$M_{21},\,M_{34}$	33.0 kN·m
Moment at each stub base	$Fs$	21.0 kN·m
Beam end moments	$M_{23},\,M_{32}$	54.0 kN·m (hogging / sagging)
Column shear (each)	$V_{\text{col}}$	21.0 kN
Beam shear	$V_{\text{beam}}$	27.0 kN
Column axial force	$N$	27.0 kN (tension left, compression right)
Overturning check	—	210.0 = 210.0 kN·m

Topic. Slope-deflection for a single-storey sway frame whose lateral load is applied above the beam on determinate cantilever stubs.

Key relations. With $A = EI\theta$, $B = EI\Delta$, stub $s$, load $F$ at each stub top, columns $h$, beam $L$ and fixed bases: joint $\left(4/h + 6/L\right)A + \left(6/h^{2}\right)B = -Fs$; storey shear $\left(6/h\right)A + \left(12/h^{2}\right)B = Fh$; then $M_{\text{base}} = (2EI/h)(\theta + 3\Delta/h)$, $M_{\text{top}} = (2EI/h)(2\theta + 3\Delta/h)$, $M_{\text{beam}} = 6EI\theta/L$, $V_{\text{beam}} = 2M_{\text{beam}}/L$.

Why this works. A free cantilever stub adds no stiffness and no unknown, so the only thing it does is deliver a force and a couple to its joint — which converts a lateral-load problem into an identical portal problem plus two applied joint moments. Symmetry of the frame under two equal same-direction loads makes the response antisymmetric, which is why both joints rotate the same way and both columns end up with identical moments. Two equations then suffice for a frame that is three times statically indeterminate: the displacement method scales with the number of degrees of freedom, not with the redundancy, and here that is two against three.

Common pitfall. Omitting the stub couple $Fs$. That error leaves the storey-shear equation intact and gives column shears that still sum to the applied load, so the arithmetic looks right; it fails only on the joint check ($33 + 21 = 54$) and on global overturning. The second trap is plotting the column moment as continuous through the beam joint: it genuinely steps by the beam moment there, and both ordinates carry marks.

Source. Hibbeler, Structural Analysis, §11.5 (slope-deflection for frames with sidesway); Kassimali §13.4–13.5.

Question 8 of 9: Slope-deflection analysis of a sway portal with free stubs

Question 8: Slope-deflection analysis of a sway portal with free stubs (22 marks)