Question 9 of 9: Deriving the stiffness equations for a frame with one sway freedom

Question 9: Deriving the stiffness equations for a frame with one sway freedom (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 three-member frame built in at joint 1 and at joint 4. Member 1–2 rises 4.8 m over a 3.6 m run, member 2–3 is horizontal and 6 m long, and member 3–4 rises 4.8 m over a further 3.6 m run, so all three members are 6 m long with direction cosines 0.6 and 0.8. A vertical load of 12 kN and a horizontal load of 6 kN act at joint 2. All members have the same $EI$ and axial strain is neglected. The unknowns are the translation $\delta$ of joint 2 in the direction shown, and the joint rotations $\theta_2$ and $\theta_3$.

Find. (a) the translational equilibrium equation, (b) the two joint moment equilibrium equations, and (c) the terms of $[K]$ and $\{P\}$. The equations are not to be solved.

Approach. Establish the sway pattern from member inextensibility, write the six slope-deflection end moments in terms of $\delta$, $\theta_2$ and $\theta_3$, then form one virtual-work equation for the translation and one moment equation at each joint.

1. Establish the sway pattern. Member 1–2 is inextensible and joint 1 is fixed, so joint 2 can move only perpendicular to that member, in the direction $\mathbf{n} = (0.8,\,-0.6)$ shown on the drawing: $\mathbf{u}_2 = \delta\,(0.8,\,-0.6)$. Member 3–4 is inextensible with joint 4 fixed, so $\mathbf{u}_3$ is perpendicular to member 3–4, and the horizontal member 2–3 forces $u_{3x}=u_{2x}$. Solving these together gives $$\mathbf{u}_3 = \delta\,(0.8,\,-0.6) = \mathbf{u}_2 ,$$ so joints 2 and 3 translate identically and a single translational freedom describes the whole frame.
2. Compute the chord rotations. Because $\mathbf{u}_3 = \mathbf{u}_2$ the horizontal member suffers no relative transverse movement at all. For the two inclined members the transverse relative movement is $\delta$ over a length of 6 m, with opposite senses: $$\psi_{12}=+\frac{\delta}{6},\qquad \psi_{23}=0,\qquad \psi_{34}=-\frac{\delta}{6} .$$
3. Write the six end moments. With $k=2EI/6=EI/3$ and $\theta_1=\theta_4=0$ (both ends built in), and no member carrying a span load, $$M_{12}=k\!\left(\theta_2-\tfrac{\delta}{2}\right),\qquad M_{21}=k\!\left(2\theta_2-\tfrac{\delta}{2}\right),$$ $$M_{23}=k\!\left(2\theta_2+\theta_3\right),\qquad M_{32}=k\!\left(2\theta_3+\theta_2\right),$$ $$M_{34}=k\!\left(2\theta_3+\tfrac{\delta}{2}\right),\qquad M_{43}=k\!\left(\theta_3+\tfrac{\delta}{2}\right).$$
4. (a) The translational equilibrium equation. Give the frame a virtual sway $\delta^{*}=1$ with both joint rotations held. Joints 2 and 3 each move $(0.8,\,-0.6)$, so the applied loads do virtual work $$W_{\text{ext}} = 6(0.8) + 12(0.6) = 12.0\ \text{kN} ,$$ the 12 kN load being downward and moving down. The virtual chord rotations are $\psi^{*}_{12}=\tfrac16$, $\psi^{*}_{23}=0$ and $\psi^{*}_{34}=-\tfrac16$, and the internal virtual work of the end moments is $\sum\left(M_{ij}+M_{ji}\right)\psi^{*}_{ij}$. Equating, $$-\frac{1}{6}\left(M_{12}+M_{21}\right)+\frac{1}{6}\left(M_{34}+M_{43}\right)=12.0 ,$$ which in terms of the unknowns is $$\boxed{\;EI\left[\frac{1}{9}\,\delta-\frac{1}{6}\,\theta_2+\frac{1}{6}\,\theta_3\right]=12.0\;}$$ (this is the “shear” equation that the sway degree of freedom demands, obtained here by virtual work so that the inclined geometry is handled automatically).
5. (b) Moment equilibrium at joint 2. No external couple acts there, so $$M_{21}+M_{23}=0 \quad\Longrightarrow\quad k\!\left(4\theta_2+\theta_3-\tfrac{\delta}{2}\right)=0 ,$$ $$\boxed{\;EI\left[-\frac{1}{6}\,\delta+\theta_2+\frac{1}{4}\,\theta_3\right]=0\;}$$ after dividing through by 4 to keep the matrix symmetric.
6. (b) Moment equilibrium at joint 3. Similarly $M_{32}+M_{34}=0$, so $k\!\left(\theta_2+4\theta_3+\tfrac{\delta}{2}\right)=0$ and $$\boxed{\;EI\left[\frac{1}{6}\,\delta+\frac{1}{4}\,\theta_2+\theta_3\right]=0 .\;}$$
7. (c) Assemble the matrix form. Collecting the three equations with the unknowns ordered $\{\delta,\ \theta_2,\ \theta_3\}$, $$EI\begin{bmatrix} \dfrac{1}{9} & -\dfrac{1}{6} & \dfrac{1}{6}\\[6pt] -\dfrac{1}{6} & \dfrac{4}{3} & \dfrac{1}{3}\\[6pt] \dfrac{1}{6} & \dfrac{1}{3} & \dfrac{4}{3} \end{bmatrix} \begin{Bmatrix}\delta\\ \theta_2\\ \theta_3\end{Bmatrix} =\begin{Bmatrix}12.0\\ 0\\ 0\end{Bmatrix} .$$ Note that the second and third equations have been scaled by $4/3$ relative to step 5 and 6 so that $[K]$ is the true stiffness matrix, i.e. the coefficient of each unknown is the generalised force conjugate to the corresponding generalised displacement. As required, the equations are not solved.
8. Check every term against its physical meaning. $[K]$ must be symmetric by Betti's reciprocal theorem, and it is. Each coefficient should be recognisable: $$K_{11}=2\times\frac{12EI}{L^{3}}=\frac{24EI}{216}=\frac{EI}{9},\qquad K_{12}=K_{21}=-\frac{6EI}{L^{2}}=-\frac{EI}{6},$$ $$K_{22}=K_{33}=\frac{4EI}{L}+\frac{4EI}{L}=\frac{4EI}{3},\qquad K_{23}=K_{32}=\frac{2EI}{L}=\frac{EI}{3},$$ with $L=6\text{ m}$ throughout — the sway stiffness of two members bent in double curvature, the standard sway-rotation coupling term, the sum of two rotational stiffnesses at each joint, and the carry-over stiffness of the member joining them. The load vector holds the component of the applied loads that works through the sway pattern, $12.0\ \text{kN}$, and zeros for the two joints, because no couples are applied and no member carries a span load.