Question 8 of 8: Deriving the Equilibrium Equations and the Stiffness Matrix for a Gable Frame

Question 8: Deriving the Equilibrium Equations and the Stiffness Matrix for a Gable Frame (21 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.

Find. (a) the translation equilibrium equation at joint 3, (b) the two joint moment equilibrium equations, and (c) the terms of \([K]\) and \(\{P\}\). The equations are not to be solved.

Approach. Express the sway kinematics in terms of the single parameter \(\delta\), write the four end moments in the paper’s counter-clockwise-positive convention, then form one equation by virtual work on the sway mechanism and two by joint moment balance.

1. Sway kinematics of the gable. Neglecting axial strain, both rafters keep their length. Joint 1 is fixed and joint 3 has \(v_3 = 0,\ u_3 = \delta\). Writing the two inextensibility conditions,$$3u_2 + 4v_2 = 0,\qquad 3(\delta-u_2) - 4(0-v_2) = 0$$and solving the pair,$$u_2 = \boxed{\tfrac{\delta}{2}},\qquad v_2 = \boxed{-\tfrac{3\delta}{8}}$$so pushing the roller to the right lifts nothing — the apex moves right by half the roller movement and drops by three-eighths of it. Everything that follows hangs on these two numbers.
2. Chord rotations. The transverse component of the relative end displacement of each rafter, divided by its length, gives (counter-clockwise positive)$$\psi_{12} = -0.125\,\delta,\qquad \psi_{23} = +0.125\,\delta$$equal and opposite, as symmetry of the mechanism requires.
3. Fixed-end moments of the normal load. A load acting perpendicular to a member is a uniform transverse load on that member, so the standard formula applies with the member’s own length:$$\mathrm{FEM}_{12} = +\frac{wL^{2}}{12} = +\frac{9.6(5)^{2}}{12} = +20.0\ \text{kN}\cdot\text{m},\qquad \mathrm{FEM}_{21} = -20.0\ \text{kN}\cdot\text{m}$$in the counter-clockwise-positive convention the question specifies. No projection on to the horizontal is involved, because the load is normal to the member and not gravity-directed.
4. End moments with \(k = EI/L = EI/5\). With \(\theta_1 = 0\),$$M_{12}=2k(\theta_2+0.375\delta)+20.0,\qquad M_{21}=2k(2\theta_2+0.375\delta)-20.0$$$$M_{23}=2k(2\theta_2+\theta_3-0.375\delta),\qquad M_{32}=2k(2\theta_3+\theta_2-0.375\delta)$$
5. Part (b): moment equilibrium at joint 2. The apex is a rigid joint carrying no applied couple, so$$M_{21}+M_{23}=0 \;\Longrightarrow\; 2k(4\theta_2+\theta_3) = 20.0$$$$\boxed{8k\,\theta_2 + 2k\,\theta_3 = 20.0}$$The \(\delta\) terms cancel identically. That cancellation is a property to be verified, not assumed: it happens here because both rafters have the same length and the same \(EI\), so their equal and opposite chord rotations contribute equal and opposite \(3k\psi\) terms at the apex.
6. Part (b): moment equilibrium at joint 3. Joint 3 is a hinge on a roller, so the moment there is zero:$$M_{32}=0 \;\Longrightarrow\; \boxed{2k\,\theta_2 + 4k\,\theta_3 - 0.75k\,\delta = 0}$$
7. Part (a): the translation equation by virtual work. Impose the sway mechanism with \(\delta^{*} = 1\); then \(\psi^{*}_{12}=-0.125,\ \psi^{*}_{23}=+0.125\), and the mid-point of the loaded rafter moves \((0.25,\,-0.1875)\). The resultant of the normal load is \(9.6(5) = 48\) kN acting in the direction \((0.6,\,-0.8)\), so the external virtual work is$$W^{*} = 48\big[(0.6)(0.25)+(-0.8)(-0.1875)\big] = 48(0.3125) = 15.0\ \text{kN}\cdot\text{m}$$and the virtual-work statement \(\sum (M_{ij}+M_{ji})\psi^{*}_{ij} + W^{*} = 0\) becomes$$-0.125\,(M_{12}+M_{21}) + 0.125\,(M_{23}+M_{32}) + 15.0 = 0$$Substituting the end moments and collecting terms,$$\boxed{0.375k\,\delta - 0.75k\,\theta_3 = 15.0}$$The rotation \(\theta_2\) drops out of the translation equation, mirroring the way \(\delta\) dropped out of the joint-2 equation — the two cancellations are the same symmetry seen from opposite sides, and together they guarantee that \(K_{12} = K_{21} = 0\).
8. Part (c): assemble the matrix. With \(k = EI/5\), so that \(0.375k = 0.075EI\), \(0.75k = 0.15EI\), \(8k = 1.6EI\), \(4k = 0.8EI\) and \(2k = 0.4EI\), the three equations in the order (translation, joint 2, joint 3) are$$EI\begin{bmatrix} 0.075 & 0 & -0.150 \\ 0 & 1.600 & 0.400 \\ -0.150 & 0.400 & 0.800 \end{bmatrix}\begin{Bmatrix} \delta \\ \theta_2 \\ \theta_3 \end{Bmatrix} = \begin{Bmatrix} 15.0 \\ 20.0 \\ 0 \end{Bmatrix}$$or, entry by entry,$$\boxed{K_{11}=0.075EI,\; K_{12}=K_{21}=0,\; K_{13}=K_{31}=-0.150EI}$$$$\boxed{K_{22}=1.600EI,\; K_{23}=K_{32}=0.400EI,\; K_{33}=0.800EI}$$$$\boxed{\{P\} = \{15.0,\;20.0,\;0\}^{\mathsf{T}}\ \text{kN}\cdot\text{m}}$$As instructed, the equations are left unsolved.
9. Check the assembly without solving it. Three tests cost nothing. First, \([K]\) must be symmetric, and it is — an unsymmetric result would mean the translation row had been scaled differently from the moment rows. Second, every rotational term must be recognisable as \(4EI/L\), \(2EI/L\) or a sum of them: \(K_{22} = 8k = 4EI/L + 4EI/L\) is the two rafters meeting at the apex, and \(K_{23} = 2k = 2EI/L\) is the carry-over along member 2–3. Third, \(P_2 = 20.0\) is exactly the fixed-end moment \(wL^{2}/12\) released at the apex, and \(P_3 = 0\) because a roller cannot apply a couple.