Question 9 of 9: Derive the stiffness matrix and load vector (24 marks)

Question 9: Derive the stiffness matrix and load vector (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 frame with fixed bases at joints 1 and 4. Joint 1 sits 3 m to the right of the vertical through joint 2 and 4 m below it; the top member 2–3 is horizontal and 5 m long; joint 4 sits 5 m to the right of joint 1 at base level. A 48 kN horizontal force acts at joint 2 and a 4.8 kN/m UDL acts downward on member 2–3.

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

Approach. Get the sway pattern from inextensibility first, then write the three slope-deflection members, then assemble two joint-moment equations and one virtual-work translation equation and scale them so that $[K]$ comes out symmetric.

1. Establish the sway pattern. Members 1–2 and 3–4 are parallel, both lying along the direction $(3,\,-4)/5$. With member 2–3 horizontal and inextensible, $u_{2}=u_{3}=\delta$; inextensibility of each leg then gives $$v_{2}=v_{3}=\tfrac{3}{4}\,\delta$$ so the whole top member translates rigidly along $(4,\,3)/5$, which is perpendicular to the legs. One translation unknown, as the question states.
2. Write the chord rotations. Using $\psi_{ij}=\left[(\mathbf{D}_{j}-\mathbf{D}_{i})\cdot\mathbf{n}\right]/L$, $$\psi_{12}=\psi_{34}=-\tfrac{1}{4}\,\delta,\qquad \psi_{23}=0$$ The top member has no chord rotation at all — the single most useful consequence of the legs being parallel.
3. Write the six end moments. With $2EI/L=0.4EI$ on every member and $\mathrm{FEM}_{23}=wL^{2}/12=4.8(5)^{2}/12=10$ kN·m, $$M_{12}=0.4EI\left(\theta_{2}+0.75\delta\right),\qquad M_{21}=0.4EI\left(2\theta_{2}+0.75\delta\right)$$ $$M_{23}=0.4EI\left(2\theta_{2}+\theta_{3}\right)+10,\qquad M_{32}=0.4EI\left(2\theta_{3}+\theta_{2}\right)-10$$ $$M_{34}=0.4EI\left(2\theta_{3}+0.75\delta\right),\qquad M_{43}=0.4EI\left(\theta_{3}+0.75\delta\right)$$
4. (b) Moment equilibrium at joints 2 and 3. Setting $\sum M_{ij}=0$ at each joint, $$0.3\,EI\delta+1.6\,EI\theta_{2}+0.4\,EI\theta_{3}=-10$$ $$0.3\,EI\delta+0.4\,EI\theta_{2}+1.6\,EI\theta_{3}=+10$$
5. (a) Translation equilibrium by virtual work. Give the sway mode a virtual value $\delta^{*}=1$, so that $\psi^{*}_{12}=\psi^{*}_{34}=-0.25$, $\psi^{*}_{23}=0$, and both joints move through $(1,\,0.75)$. Then $$\sum\left(M_{ij}+M_{ji}\right)\psi^{*}_{ij}+\sum \mathbf{F}\cdot\mathbf{d}^{*}=0$$ with external work $48(1)-24(0.75)=30$ kN, the 48 kN force moving with the sway and the 24 kN of UDL being carried upward by it.
6. Reduce the translation equation. Substituting the member equations, $$-0.25\left[0.4EI(3\theta_{2}+1.5\delta)+0.4EI(3\theta_{3}+1.5\delta)\right]+30=0$$ which, divided by 4 so that its coefficients match the joint equations, becomes $$0.3\,EI\delta+0.3\,EI\theta_{2}+0.3\,EI\theta_{3}=30$$
7. (c) Assemble. The three equations in the stated order $\{\delta,\theta_{2},\theta_{3}\}$ give $$[K]=EI\begin{bmatrix}0.30 & 0.30 & 0.30\\ 0.30 & 1.60 & 0.40\\ 0.30 & 0.40 & 1.60\end{bmatrix},\qquad \{P\}=\begin{Bmatrix}30\\ -10\\ 10\end{Bmatrix}$$

The equations are not to be solved, but they can and should be checked. Every term has a recognisable origin: $K_{22}=K_{33}=4EI/5+4EI/5=1.6EI$ is the sum of two $4EI/L$ terms, $K_{23}=2EI/5=0.4EI$ is the carry-over across the top member, $K_{12}=K_{13}=3(2EI/L)\psi=3(0.4EI)(0.25)=0.3EI$ couples each rotation to the sway, and $K_{11}=2\times 12EI\psi^{2}/L=2(0.4EI)(6)(0.25)^{2}=0.3EI$ is the transverse stiffness of the two legs working through the sway mode. The load vector is equally readable: $\mp 10$ kN·m are the fixed-end moments of the UDL, and $+30$ kN is the net external virtual work per unit sway. Above all, $[K]$ is symmetric, which it must be for any conservative structure; an unsymmetric matrix means the translation row has been scaled differently from the moment rows, and that scaling is the only judgement call in the whole derivation.