Question 9 of 9: Derivation of the Stiffness Matrix and Load Vector for an L-Frame

Question 9: Derivation of the Stiffness Matrix and Load Vector for an L-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. An L-frame with a built-in base, a roller at the far end of the beam, both members of length 8 m and the same $EI$, carrying a uniformly distributed load $w$ on the beam. Axial strain is neglected.

Find. The three equilibrium equations — one translation equation at joint 3 and two joint-moment equations — expressed as $[K]\{\delta,\theta_2,\theta_3\}^{\mathsf T}=\{P\}$, with every term of $[K]$ and $\{P\}$ given. The system is not to be solved.

Approach. Establish the kinematics first (which displacements the inextensibility assumption ties together), write the four member-end moments in slope-deflection form, then obtain the translation equation by virtual work and the two moment equations by joint equilibrium. Scale the translation equation so that $[K]$ comes out symmetric.

1. Kinematics — establish what $\delta$ actually moves. The column 1–2 is vertical and inextensible with joint 1 fixed, so $v_2=0$. The beam 2–3 is horizontal and inextensible, so $u_2=u_3=\delta$: the single translation unknown sways the column and slides the beam together. The roller holds $v_3=0$. Hence $$\psi_{12}=\frac{(\mathbf{D}_2-\mathbf{D}_1)\cdot\mathbf{e}_2}{8} =-\frac{\delta}{8},\qquad \psi_{23}=\frac{v_3-v_2}{8}=0.$$ The beam has no chord rotation, which is what keeps $\delta$ out of the joint-3 equation entirely.
2. Member-end moments. With $k=2EI/8=EI/4$ for both members, $\theta_1=0$, and the beam fixed-end moments $\mathrm{FEM}_{23}=+w(8)^{2}/12=+\tfrac{16w}{3}$, $\mathrm{FEM}_{32}=-\tfrac{16w}{3}$: $$M_{12}=\frac{EI}{4}\left(\theta_2+\frac{3\delta}{8}\right),\qquad M_{21}=\frac{EI}{4}\left(2\theta_2+\frac{3\delta}{8}\right),$$ $$M_{23}=\frac{EI}{4}\left(2\theta_2+\theta_3\right)+\frac{16w}{3},\qquad M_{32}=\frac{EI}{4}\left(2\theta_3+\theta_2\right)-\frac{16w}{3}.$$
3. (a) Translation equation at joint 3, by virtual work. Give the structure a virtual sway $\delta^{*}=1$; the virtual chord rotations are $\psi^{*}_{12}=-1/8$ and $\psi^{*}_{23}=0$. The uniformly distributed load is vertical and does no work on a horizontal virtual displacement, and no horizontal load is applied, so $$\sum\left(M_{ij}+M_{ji}\right)\psi^{*}_{ij}+W_{\text{ext}}=0 \;\Longrightarrow\;-\frac{M_{12}+M_{21}}{8}=0.$$ Physically this is the storey-shear statement that the column carries no shear, which had to be true: the roller at joint 3 offers no horizontal restraint, so the only horizontal reaction available is at joint 1, and there is nothing for it to balance. Multiplying by $-1$ and substituting, $$\boxed{\frac{3EI}{128}\,\delta+\frac{3EI}{32}\,\theta_2=0.}$$
4. (b) Moment equilibrium at joint 2. Only the column and the beam meet there, and no external couple is applied, so $M_{21}+M_{23}=0$: $$\boxed{\frac{3EI}{32}\,\delta+EI\,\theta_2+\frac{EI}{4}\,\theta_3 =-\frac{16w}{3}.}$$
5. (b) Moment equilibrium at joint 3. Joint 3 carries only the beam and a roller, which supplies no moment restraint, so $M_{32}=0$: $$\boxed{\frac{EI}{4}\,\theta_2+\frac{EI}{2}\,\theta_3 =+\frac{16w}{3}.}$$
6. (c) Assemble in matrix form. Collecting the three equations in the prescribed order $\{\delta,\ \theta_2,\ \theta_3\}$, $$EI\begin{bmatrix} \dfrac{3}{128} & \dfrac{3}{32} & 0\\[6pt] \dfrac{3}{32} & 1 & \dfrac{1}{4}\\[6pt] 0 & \dfrac{1}{4} & \dfrac{1}{2}\end{bmatrix} \begin{Bmatrix}\delta\\ \theta_2\\ \theta_3\end{Bmatrix} =\begin{Bmatrix}0\\[4pt] -\dfrac{16w}{3}\\[4pt] +\dfrac{16w}{3}\end{Bmatrix}.$$ Two structural checks confirm the assembly. First, $[K]$ is symmetric, as Maxwell–Betti requires — this is the sole reason the translation equation had to be scaled by $-1/8$ rather than written as a bare column-shear balance. Second, $K_{13}=K_{31}=0$: joint 3's rotation is uncoupled from the sway because $\psi_{23}=0$, so the beam's chord contributes nothing to either equation. The load vector contains only the beam's fixed-end moments, entering with opposite signs at the two ends, and $P_1=0$ because no horizontal load acts.