Question 8 of 9: Slope-deflection analysis of a sway frame (22 marks)

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 three-member frame built in at joints (1) and (4), with two inclined members and one vertical member, loaded only at the joints.

Find. Shear force and bending moment diagrams for all three members, with extreme ordinates labelled.

Approach. Use inextensibility to reduce six joint displacements to one sway parameter, write the three chord rotations in terms of it, and close the system with two joint equations and one virtual-work sway equation.

1. Reduce the kinematics. Both inclined members are parallel (each rises 2.5 m over 6 m), so each forces its joint to move perpendicular to itself: $v_2=2.4u_2$ and $v_3=2.4u_3$. The vertical member is inextensible, so $v_2=v_3$ and hence $u_2=u_3=\Delta$. Three unknowns remain: $$\boxed{\Delta,\ \theta_2,\ \theta_3}$$
2. Chord rotations. Resolving the joint displacements onto each member's transverse axis, $$\psi_{12}=\psi_{43}=\frac{2.6\Delta}{6.5}=0.4\Delta,\qquad \psi_{23}=\frac{u_3-u_2}{6.5}=0$$ The vertical member has no chord rotation at all — a useful simplification that follows from $u_2=u_3$.
3. Slope-deflection equations. Writing $k=2EI/6.5$ and $X=3\psi_{12}$, and remembering $\theta_1=\theta_4=0$ and zero fixed-end moments, $$M_{12}=k(\theta_2-X),\quad M_{21}=k(2\theta_2-X),\quad M_{23}=k(2\theta_2+\theta_3)$$ $$M_{32}=k(2\theta_3+\theta_2),\quad M_{43}=k(\theta_3-X),\quad M_{34}=k(2\theta_3-X)$$
4. Joint equilibrium. $M_{21}+M_{23}=0$ and $M_{32}+M_{34}=0$ give $$4\theta_2+\theta_3=X,\qquad 4\theta_3+\theta_2=X$$ Subtracting shows $\theta_2=\theta_3=\theta$, and then $5\theta=X$, so $\theta=X/5$.
5. Sway equation by virtual work. Give the frame a unit sway ($\Delta^{*}=1$, so $\mathbf{d}^{*}_2=\mathbf{d}^{*}_3=(1,\,2.4)$ and $\psi^{*}_{12}=\psi^{*}_{43}=0.4$). The applied loads do virtual work $$\sum \mathbf{F}\cdot\mathbf{d}^{*}=(7.2)(1)+(-20)(2.4)+(7.2)(1)=-33.6$$ and the end moments do internal work $\bigl[(M_{12}+M_{21})+(M_{43}+M_{34})\bigr](0.4)=0.8k(3\theta-2X)$, so $$0.8k(3\theta-2X)-33.6=0$$
6. Solve. Substituting $\theta=X/5$ makes $3\theta-2X=-1.4X$, hence $$-1.12\,kX=33.6\;\Longrightarrow\;\boxed{kX=-30\ \text{kN}\cdot\text{m}}$$ The rotations and sway themselves are not required for the diagrams, because every end moment is a multiple of $kX$.
7. End moments. Back-substituting $\theta=0.2X$, $$M_{12}=-0.8kX=+24.0,\quad M_{21}=-0.6kX=+18.0,\quad M_{23}=+0.6kX=-18.0$$ $$M_{32}=-18.0,\quad M_{34}=+18.0,\quad M_{43}=+24.0\ \text{kN}\cdot\text{m}$$ $$\boxed{\text{base moments }24.0\ \text{kN}\cdot\text{m at (1) and (4)};\quad \text{joint moments }18.0\ \text{kN}\cdot\text{m at (2) and (3)}}$$
8. Member shears. No member carries a span load, so each shear is constant and equal to the sum of its end moments divided by its length: $$V_{12}=V_{43}=\frac{24.0+18.0}{6.5}=6.4615\ \text{kN},\qquad V_{23}=\frac{18.0+18.0}{6.5}=5.5385\ \text{kN}$$
9. Reactions and global checks. Resolving the member end forces at the two fixed bases, $$\text{At (1): } H=12.7385\ \text{kN (leftward)},\ V=12.3077\ \text{kN},\ M=24.0\ \text{kN}\cdot\text{m}$$ $$\text{At (4): } H=1.6615\ \text{kN (leftward)},\ V=7.6923\ \text{kN},\ M=24.0\ \text{kN}\cdot\text{m}$$ $$\textstyle\sum F_x:\;-12.7385-1.6615+7.2+7.2=0,\qquad \sum F_y:\;12.3077+7.6923-20=0\quad\checkmark$$

Both fixed bases finish with the same moment, 24.0 kN·m, and both interior joints with 18.0 kN·m. That is a consequence of the frame's point symmetry about the mid-point of member (2)–(3): the geometry maps onto itself under a half-turn, so the two inclined members must respond identically.