Question 9 of 9: Deriving the stiffness matrix and load vector of a frame with parallel legs (24 marks)

Question 9: Deriving the stiffness matrix and load vector of a frame with parallel legs (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 three-member frame built in at joints 1 and 4, with a 3 kN/m downward load on the horizontal member.

Find. (a) the translation equilibrium equation at joint 2, (b) the moment equilibrium equations at joints 2 and 3, and (c) the terms of [K] and {P}. The equations are not to be solved.

Approach. Establish the sway pattern from inextensibility first, write the six slope-deflection expressions, then take three equilibrium equations — two joint moments and one virtual-work translation equation — and scale the last so that [K] comes out symmetric.

1. Fix the sway pattern. Member 1–2 runs along the unit vector $(0.6,\,0.8)$ and is inextensible with joint 1 built in, so joint 2 can only move perpendicular to it: $\mathbf{D}_{2}=\delta\,(0.8,\,-0.6)$, which is the direction drawn on the figure. Member 2–3 is horizontal and inextensible, so $u_{3}=u_{2}=0.8\delta$. Member 3–4 is parallel to 1–2, so joint 3 must also move perpendicular to $(0.6,\,0.8)$, and matching the horizontal component gives $$\mathbf{D}_{3}=\delta\,(0.8,\,-0.6)=\mathbf{D}_{2}.$$ The two joints translate by the same vector.
2. Read off the chord rotations. With $\psi_{ij}=\bigl(\mathbf{D}_{j}-\mathbf{D}_{i}\bigr)\cdot\mathbf{e}_{2}/L$ and $\mathbf{e}_{2}=(-0.8,\,0.6)$ for both legs, $$\psi_{12}=-\frac{\delta}{4},\qquad \psi_{23}=0,\qquad \psi_{34}=+\frac{\delta}{4}.$$ The horizontal member has exactly zero chord rotation, which is the single most useful consequence of the legs being parallel: it drops out of the sway equation entirely.
3. Fixed-end moments. Only member 2–3 is loaded, and with counter-clockwise-positive end moments a downward uniform load gives $$M^{F}_{23}=+\frac{wL^{2}}{12}=+\frac{3(4)^{2}}{12}=+4.00\ \text{kN}\cdot\text{m},\qquad M^{F}_{32}=-4.00\ \text{kN}\cdot\text{m}.$$
4. Write the six slope-deflection expressions. Every member is 4.00 m long, so $2EI/L=0.5EI$ throughout, and $\theta_{1}=\theta_{4}=0$: $$M_{12}=0.5EI\bigl(\theta_{2}+0.75\delta\bigr),\qquad M_{21}=0.5EI\bigl(2\theta_{2}+0.75\delta\bigr),$$ $$M_{23}=0.5EI\bigl(2\theta_{2}+\theta_{3}\bigr)+4,\qquad M_{32}=0.5EI\bigl(2\theta_{3}+\theta_{2}\bigr)-4,$$ $$M_{34}=0.5EI\bigl(2\theta_{3}-0.75\delta\bigr),\qquad M_{43}=0.5EI\bigl(\theta_{3}-0.75\delta\bigr).$$
5. (b) Moment equilibrium at joints 2 and 3. Setting the sum of the member end moments at each joint to zero, $$M_{21}+M_{23}=0\;\Rightarrow\;0.375\,EI\,\delta+2\,EI\,\theta_{2}+0.5\,EI\,\theta_{3}=-4.00,$$ $$M_{32}+M_{34}=0\;\Rightarrow\;-0.375\,EI\,\delta+0.5\,EI\,\theta_{2}+2\,EI\,\theta_{3}=+4.00.$$ These are the two equations part (b) asks for.
6. (a) Translation equilibrium at joint 2, by virtual work. Give the frame the unit sway pattern $\delta^{*}=1$ with all joint rotations held at zero. The chord rotations are then $\psi^{*}_{12}=-\tfrac14$, $\psi^{*}_{23}=0$, $\psi^{*}_{34}=+\tfrac14$, and the whole horizontal member translates rigidly through $(0.8,\,-0.6)$, so the 12.0 kN of applied load does $W^{*}_{\text{ext}}=12.0(0.6)=7.20$ kN·m of virtual work. The equation is $$\sum_{\text{members}}\bigl(M_{ij}+M_{ji}\bigr)\psi^{*}_{ij}+W^{*}_{\text{ext}}=0.$$
7. Expand the translation equation. With $M_{12}+M_{21}=1.5EI\theta_{2}+0.75EI\delta$ and $M_{34}+M_{43}=1.5EI\theta_{3}-0.75EI\delta$, $$-\tfrac14\bigl(1.5EI\theta_{2}+0.75EI\delta\bigr)+\tfrac14\bigl(1.5EI\theta_{3}-0.75EI\delta\bigr)+7.20=0,$$ and multiplying through by $-1$ so that the $\delta$ coefficient is positive, $$\boxed{0.375\,EI\,\delta+0.375\,EI\,\theta_{2}-0.375\,EI\,\theta_{3}=7.20}$$ That sign choice is not cosmetic: it is what makes the assembled matrix symmetric, and symmetry is the only cheap check available on a hand assembly.
8. (c) Assemble [K] and {P}. Collecting the three equations in the order $\{\delta,\ \theta_{2},\ \theta_{3}\}$, $$EI\begin{bmatrix} 0.375 & 0.375 & -0.375 \\ 0.375 & 2.000 & 0.500 \\ -0.375 & 0.500 & 2.000 \end{bmatrix}\begin{Bmatrix} \delta \\ \theta_{2} \\ \theta_{3} \end{Bmatrix}=\begin{Bmatrix} 7.20 \\ -4.00 \\ 4.00 \end{Bmatrix}$$ The matrix is symmetric, $K_{12}=K_{21}=0.375EI$ and $K_{13}=K_{31}=-0.375EI$, and every term traces to a $12EI/L^{3}$, $6EI/L^{2}$, $4EI/L$ or $2EI/L$ contribution: $K_{11}=12EI/L^{3}\cdot 2\cdot(\text{leg})$, $K_{22}=K_{33}=4EI/L+4EI/L$, $K_{23}=2EI/L$. The equations are not solved, as instructed.