Slope-deflection with sidesway: 16-Civ-B1 Advanced Structural Analysis, May 2014, Q8

Question 8: Slope-deflection with sidesway (24 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 single beam carried by two 6 m columns, one hanging below the beam from an encastré base and one rising above it to an encastré head, with free overhangs at both ends and a UDL over the entire beam.

Given data
Item	Value
Beam ⑤–②–③–⑥	2 m + 6 m + 2 m = 10 m, free at ⑤ and ⑥
Column ①–②	6 m below the beam, encastré at ①
Column ③–④	6 m above the beam, encastré at ④
Uniformly distributed load	12 kN/m over the full 10 m
Flexural rigidity	the same $EI$ everywhere, inextensible
Sidesway	not prevented

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

Question 8: sway frame with one column below and one above the beam.

Approach. Replace each overhang by the force and couple it applies to its joint, write slope-deflection equations for the beam and the two columns in terms of $\theta_2$, $\theta_3$ and the sway $\Delta$, and close the system with the storey-shear equation. Exploiting the frame’s point symmetry cuts the work in half.

Convert the overhangs to joint actions. Each 2 m overhang carries $12(2)=24$ kN whose centroid is 1 m outboard, so it applies 24 kN downward and a 24 kN·m couple to its joint — counter-clockwise at ② and clockwise at ③.
Identify the point symmetry. Rotating the frame $180^\circ$ about the beam mid-point maps ② onto ③ and ① onto ④, so the structure is symmetric under that map while the gravity load is anti-symmetric. The response is therefore anti-symmetric: $\theta_3=-\theta_2$ and $M_{34}=-M_{21}$.
Write the slope-deflection equations. With $h=L=6$ m, the beam gives $M_{23}=\tfrac{2EI}{6}(2\theta_2+\theta_3)+36$ and the columns give $M_{21}=\tfrac{2EI}{6}(2\theta_2+3\Delta/6)$ and $M_{34}=\tfrac{2EI}{6}(2\theta_3-3\Delta/6)$, the opposite signs arising because one column hangs down and the other rises up.
Impose the storey-shear equation. No horizontal load acts, so the two column shears must cancel: $M_{12}+M_{21}=M_{34}+M_{43}$. With $\theta_3=-\theta_2$ this collapses to $\theta_2=-2\Delta/h$ — and it also makes each column shear identically zero, so both columns carry a constant moment.
Solve joint ②. Substituting $\theta_2=-2\Delta/h$ into $M_{21}+M_{23}=+24$ gives $2EI\Delta/h^{2}+4EI\Delta/(Lh)=\mathrm{FEM}-M_{\text{stub}}=36-24=12$, so with $L=h=6$ m, $M_{12}=2EI\Delta/h^{2}=4.0$ kN·m and $EI\Delta=72$ kN·m$^{3}$.
Back-substitute. The beam end moments follow as $M_{23}=+28.0$ and $M_{32}=-28.0$ kN·m, and joint equilibrium checks: $-4.0+28.0=24.0$, the stub couple.

The sway and the joint rotations come out as

$$EI\Delta=72\ \text{kN}\cdot\text{m}^{3},\qquad EI\theta_2=-24\ \text{kN}\cdot\text{m}^{2}=-EI\theta_3$$

and the end moments are

$$\boxed{M_{12}=+4.0,\ M_{21}=-4.0,\ M_{23}=+28.0,\ M_{32}=-28.0,\ M_{34}=+4.0,\ M_{43}=-4.0\ \text{kN}\cdot\text{m}}$$

Because $M_{12}+M_{21}=0$ in both columns, neither column carries any shear and both horizontal reactions are zero: the frame sways 72/EI to one side yet transmits no horizontal force to the ground. The columns instead carry a constant 4.0 kN·m and an axial force of 60 kN — compression in the lower column, tension in the upper one, which literally hangs the right-hand end of the beam.

Question 8: bending moment along the beam, sagging positive.

On the beam the shear runs from $+36$ kN just right of ② to $-36$ kN just left of ③, with the 24 kN overhang shears outboard of each. The moment hogs $-24$ kN·m at each joint from the overhang alone, steps by the 4.0 kN·m column moment to $-28.0$ kN·m inside the main span, and reaches $-28.0+12(6)^{2}/8=+26.0$ kN·m at mid-span.

Result	Value
Sway of the beam	$\Delta = 72/EI$ (rightwards)
Joint rotations	$\theta_2=-24/EI$, $\theta_3=+24/EI$
Column moments (constant)	4.0 kN·m in each
Column shears / horizontal reactions	0
Column axial forces	60 kN (lower in compression, upper in tension)
Beam end moments	−28.0 kN·m at ② and ③
Maximum sagging moment	+26.0 kN·m at mid-span
Maximum beam shear	±36.0 kN at ② and ③
Overhang ordinates	−24.0 kN·m and 24.0 kN shear at each stub root

Topic. Slope-deflection with sidesway, and the economy of point symmetry.

Key relations. One extra unknown $\Delta$ needs one extra equation, the storey shear $\sum V_{\text{col}}=\sum H_{\text{ext}}$, with $V_{\text{col}}=(M_{ij}+M_{ji})/h$. Chord rotation is $\psi=\pm\Delta/h$, the sign depending on which way the member points. A determinate overhang of length $c$ under a UDL applies $wc$ and $wc^{2}/2$ to its joint.

Why this works. The frame looks awkward — one column below the beam, one above — but that is exactly what makes it point-symmetric, and point symmetry under gravity load forces an anti-symmetric response. That one observation halves the unknowns and then delivers the striking result that both column shears vanish, so each column bends in single curvature under a constant moment while the frame still sways. Replacing the overhangs by a force and a couple is the other essential step: the force alone satisfies the storey shear but violates joint moment equilibrium, and the error is invisible until the diagram fails to close.

Common pitfall. Dropping the overhang couple, and giving both columns the same sign of $\psi$. The second error is easy to make because both columns are 6 m long and look alike, but one has its fixed end below the beam and the other above it.

Source. Kassimali, Structural Analysis, 6th ed., Ch. 16 (slope-deflection: frames with sidesway); Hibbeler, Structural Analysis, 10th ed., §11.5.

Question 8 of 9: Slope-deflection with sidesway

Question 8: Slope-deflection with sidesway (24 marks)