Question 9 of 9: Derivation of the Stiffness Matrix and Load Vector

Question 9: Derivation of the Stiffness Matrix and Load Vector (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.

Joints 1 and 4 are pin supports; a horizontal point load of 100 kN acts to the right at joint 3; axial strain is neglected. The overall span is \(3+9+3 = 15\ \text{m}\) and \(\delta\) is the translation of joint 2 perpendicular to member 1–2, positive down-and-to-the-right as drawn.

Find. The three equilibrium equations, presented as the stiffness matrix \([K]\) and load vector \(\{P\}\) for the unknown vector \(\{\theta_2,\theta_3,\delta\}\). The equations are not to be solved.

Approach. Derive the sway pattern from inextensibility first, express every chord rotation in terms of \(\delta\), write the four end moments in slope-deflection form using the modified stiffness at the two pins, and assemble the three equilibrium equations (two joint moments plus one virtual-work translation equation).

1. Establish the sway pattern. Member 1–2 has unit vector \((0.6,\,0.8)\) and joint 1 is a fixed point, so joint 2 must move perpendicular to it: \(\mathbf{d}_2 = \delta(0.8,\,-0.6)\). The horizontal member 2–3 is inextensible, so \(u_3=u_2=0.8\delta\). Member 3–4 has unit vector \((0.6,\,-0.8)\) with joint 4 fixed, so \(\mathbf{d}_3\) must be perpendicular to it, giving \(\mathbf{d}_3 = \delta(0.8,\,+0.6)\). The two knees therefore move outward by the same amount but one drops while the other rises — a purely anti-symmetric sway, which is exactly what a horizontal load at joint 3 should produce.
2. Compute the chord rotations per unit \(\delta\). Using \(\psi_{ij}=[(\mathbf{d}_j-\mathbf{d}_i)\cdot\mathbf{e}_\perp]/L\), $$\psi_{12}=-\frac{1}{5}=-0.2,\qquad \psi_{23}=\frac{0.6-(-0.6)}{9}=\frac{2}{15}=0.13\overline{3},\qquad \psi_{34}=-\frac{1}{5}=-0.2$$ each multiplied by \(\delta\).
3. Write the member stiffness constants and the fixed-end moment. Pins at joints 1 and 4 mean the modified stiffness applies to both legs, while the beam keeps the standard form: $$k_{12}=k_{34}=\frac{3(2EI)}{5}=1.2EI,\qquad k_{b}=\frac{2(2.7EI)}{9}=0.6EI$$ $$\mathrm{FEM}_{23}=+\frac{wL^{2}}{12}=+\frac{4(9)^{2}}{12}=+27.0\ \text{kN}\cdot\text{m},\qquad \mathrm{FEM}_{32}=-27.0\ \text{kN}\cdot\text{m}$$
4. Write the four end moments. $$M_{12}=0,\qquad M_{21}=1.2EI\left(\theta_2 + 0.2\delta\right)$$ $$M_{23}=0.6EI\left(2\theta_2+\theta_3-0.4\delta\right)+27.0$$ $$M_{32}=0.6EI\left(\theta_2+2\theta_3-0.4\delta\right)-27.0$$ $$M_{34}=1.2EI\left(\theta_3 + 0.2\delta\right),\qquad M_{43}=0$$
5. Part (b) — joint moment equilibrium. At joint 2, \(M_{21}+M_{23}=0\): $$2.4EI\,\theta_2 + 0.6EI\,\theta_3 + \left(0.24-0.24\right)EI\,\delta = -27.0$$ $$\boxed{2.4EI\,\theta_2 + 0.6EI\,\theta_3 + 0\cdot\delta = -27.0}$$ At joint 3, \(M_{32}+M_{34}=0\): $$\boxed{0.6EI\,\theta_2 + 2.4EI\,\theta_3 + 0\cdot\delta = +27.0}$$ The \(\delta\) terms cancel identically in both rows — a genuine property of this frame worth stating, not an accident of arithmetic.
6. Part (a) — the translation equation by virtual work. Impose the virtual sway \(\delta^{*}=1\) with \(\theta_2^{*}=\theta_3^{*}=0\) and equate internal to external virtual work: $$\sum \left(M_{ij}+M_{ji}\right)\psi^{*}_{ij} + \sum \mathbf{F}\cdot\mathbf{d}^{*} = 0$$ Substituting the end moments and the chord rotations, the \(\theta_2\) and \(\theta_3\) coefficients again cancel and $$0.16EI\,\delta = 100(0.8) + \left(-\tfrac{wL}{2}\right)(-0.6) + \left(-\tfrac{wL}{2}\right)(+0.6) = 80.0 + 10.8 - 10.8$$ $$\boxed{0 \cdot \theta_2 + 0\cdot\theta_3 + 0.16EI\,\delta = 80.0}$$ Note that the equivalent nodal loads from the uniformly distributed load contribute equally and oppositely and therefore drop out; only the 100 kN horizontal load feeds the sway.
7. Part (c) — assemble the matrix form. $$EI\begin{bmatrix}2.4 & 0.6 & 0\\[2pt] 0.6 & 2.4 & 0\\[2pt] 0 & 0 & 0.16\end{bmatrix}\begin{Bmatrix}\theta_2\\[2pt]\theta_3\\[2pt]\delta\end{Bmatrix}=\begin{Bmatrix}-27.0\\[2pt]+27.0\\[2pt]+80.0\end{Bmatrix}$$ As required, the equations are not solved. Two properties confirm the derivation: \([K]\) is symmetric, as any correctly scaled stiffness matrix must be, and the sway row is completely decoupled from the two rotation rows, so the translation could be found from a single division. (For information only, a direct-stiffness check of the same frame returns \(EI\theta_2=-15.0\), \(EI\theta_3=+15.0\) and \(EI\delta=500.0\), consistent with the matrix above.)