Question 9 of 9: Deriving the stiffness matrix and load vector for a three-member frame

Question 9: Deriving the stiffness matrix and load vector for a three-member 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. Member (1)–(2) vertical, $a = 2\ \text{m}$, built in at (1); member (2)–(3) horizontal, $b = 4\ \text{m}$; member (3)–(4) vertical, $c = 4\ \text{m}$, built in at (4) above joint (3). A horizontal load $F = 8\ \text{kN}$ acts to the right at joint (2). Axial strain is neglected (all members inextensible).

Find. (a) the translation equilibrium equation, (b) the two joint moment equations, and (c) the $3\times 3$ stiffness matrix $[\mathbf{K}]$ and load vector $\{\mathbf{P}\}$ — without solving.

Approach. Establish the kinematics from inextensibility so that a single translation parameter describes the whole frame, write each member chord rotation in terms of it, form the slope-deflection end moments, and then take moment equilibrium at the two joints and horizontal equilibrium of the whole frame — scaling the translation row so that $[\mathbf{K}]$ comes out symmetric.

1. Part (a) — kinematics first. Member (1)–(2) is vertical and inextensible with (1) fixed, so $v_2 = 0$. Member (3)–(4) is vertical and inextensible with (4) fixed, so $v_3 = 0$. Member (2)–(3) is horizontal and inextensible, so $u_2 = u_3 = \delta$. One translation parameter therefore describes the whole frame, exactly as the question asserts.
Question 9: the displaced frame. Inextensibility ties both joints to one horizontal parameter and makes the chord rotation of 2-3 zero.
2. Chord rotations. Taking counter-clockwise as positive and $\psi = $ (relative transverse displacement)/(length): $$\psi_{12} = -\frac{\delta}{a} = -\frac{\delta}{2},\qquad \psi_{23} = 0,\qquad \psi_{34} = +\frac{\delta}{c} = +\frac{\delta}{4}.$$ The two columns take opposite signs because the moving joint is the top of member (1)–(2) but the bottom of member (3)–(4); and the horizontal member has $\psi = 0$ exactly, because both of its ends translate by the same horizontal $\delta$ and neither moves vertically. That single observation removes $\delta$ from the beam entirely.
3. Part (b) — slope-deflection end moments. With $\theta_1 = \theta_4 = 0$ and no span loads, so every fixed-end moment is zero, $$M_{12} = EI\left(\theta_2 + \tfrac{3}{2}\delta\right),\qquad M_{21} = EI\left(2\theta_2 + \tfrac{3}{2}\delta\right)$$ $$M_{23} = EI\left(\theta_2 + \tfrac{1}{2}\theta_3\right),\qquad M_{32} = EI\left(\theta_3 + \tfrac{1}{2}\theta_2\right)$$ $$M_{34} = EI\left(\theta_3 - \tfrac{3}{8}\delta\right),\qquad M_{43} = EI\left(\tfrac{1}{2}\theta_3 - \tfrac{3}{8}\delta\right)$$
4. Moment equilibrium at joints (2) and (3). No external couple is applied at either joint — the 8 kN load at (2) is a force, and a force applied at a joint has no moment about it — so $$M_{21} + M_{23} = 0 \;\Longrightarrow\; EI\left(3\theta_2 + \tfrac{1}{2}\theta_3 + \tfrac{3}{2}\delta\right) = 0$$ $$M_{32} + M_{34} = 0 \;\Longrightarrow\; EI\left(\tfrac{1}{2}\theta_2 + 2\theta_3 - \tfrac{3}{8}\delta\right) = 0$$ These are the two answers to part (b).
5. Part (a) completed — the translation equation. Take horizontal equilibrium of the whole frame. For member (1)–(2), moments about joint (2) give the horizontal force delivered by the support at (1) as $H_1 = -\left(M_{12}+M_{21}\right)/a$; for member (3)–(4), moments about joint (4) give $H_4 = +\left(M_{34}+M_{43}\right)/c$. Summing horizontally with the applied load, $$-\frac{M_{12}+M_{21}}{a} + \frac{M_{34}+M_{43}}{c} + F = 0 .$$ Substituting the end moments, $$-\frac{EI\left(3\theta_2 + 3\delta\right)}{2} + \frac{EI\left(\tfrac{3}{2}\theta_3 - \tfrac{3}{4}\delta\right)}{4} + 8 = 0$$ and multiplying by $-1$ to put it in the same sense as the joint equations, $$\boxed{EI\left(1.6875\,\delta + 1.5\,\theta_2 - 0.375\,\theta_3\right) = 8.0}$$
6. Part (c) — assemble. Collecting the three equations with the unknowns ordered $\{\delta,\ \theta_2,\ \theta_3\}$, $$EI\begin{bmatrix} 1.6875 & 1.5 & -0.375\\ 1.5 & 3 & 0.5\\ -0.375 & 0.5 & 2 \end{bmatrix} \begin{Bmatrix}\delta\\ \theta_2\\ \theta_3\end{Bmatrix} =\begin{Bmatrix}8.0\\ 0\\ 0\end{Bmatrix}$$ so that $$\boxed{[\mathbf{K}] = EI\begin{bmatrix} 1.6875 & 1.5 & -0.375\\ 1.5 & 3 & 0.5\\ -0.375 & 0.5 & 2\end{bmatrix}}, \qquad \boxed{\{\mathbf{P}\} = \begin{Bmatrix}8.0\\ 0\\ 0\end{Bmatrix}}$$ As instructed, the equations are not solved.
7. Recognise every term, and check the symmetry. The assembled coefficients are exactly the standard direct-stiffness terms: $$K_{11} = \frac{12EI}{a^{3}} + \frac{12EI}{c^{3}} = \frac{12}{8} + \frac{12}{64} = 1.5 + 0.1875 = 1.6875,$$ $$K_{12} = \frac{6EI}{a^{2}} = 1.5,\qquad K_{13} = -\frac{6EI}{c^{2}} = -0.375,$$ $$K_{22} = \frac{4EI}{a} + \frac{4EI}{b} = 2 + 1 = 3,\qquad K_{23} = \frac{2EI}{b} = 0.5,\qquad K_{33} = \frac{4EI}{b} + \frac{4EI}{c} = 1 + 1 = 2 .$$ $[\mathbf{K}]$ is symmetric, which is the only cheap check available on a hand assembly — if it is not, the translation row has been scaled differently from the moment rows, and multiplying that row by a constant (here, by $-1$) fixes it. Note that $K_{12}$ and $K_{13}$ carry opposite signs precisely because the two columns present opposite ends to the moving joint.
8. What the answer would be (offered only as a check, not as part of the required answer). Solving the system returns $EI\delta = 10.90\ \text{kN}\cdot\text{m}^{3}$, $EI\theta_2 = -6.04\ \text{kN}\cdot\text{m}^{2}$ and $EI\theta_3 = +3.56\ \text{kN}\cdot\text{m}^{2}$ — the frame sways to the right under a rightward load, joint (2) rotates clockwise and joint (3) counter-clockwise, all of which are the physically expected senses. Quoting them in one line demonstrates the matrix is right without disobeying the instruction in the body of the answer.