Question 9 of 9: Deriving the stiffness equations of a frame with an inclined member

Question 9: Deriving the stiffness equations of a frame with an inclined member (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. A frame with joint 1 built in at the origin, an inclined member 1–2 of relative stiffness $2.5EI$ rising 6 m over an 8 m horizontal run, a horizontal beam 2–3 of relative stiffness $EI$ and length 4 m carrying 12 kN/m, and a vertical column 3–4 of relative stiffness $1.5EI$ and height 6 m built in at joint 4. Axial strain is neglected. The unknowns are $\delta$ (translation of joint 3, positive to the right), $\theta_2$ and $\theta_3$ (counter-clockwise positive).

Find. The three equilibrium equations and the terms of $[K]$ and $\{P\}$ in $[K]\{\delta,\ \theta_2,\ \theta_3\}^{T} = \{P\}$. The equations are not to be solved.

Approach. Impose inextensibility to express every joint displacement in terms of the single translation $\delta$, convert those into member chord rotations, write the six slope-deflection moments, and form two joint-moment equations plus one virtual-work translation equation.

1. Part (a) — establish the sway pattern from inextensibility. Member 3–4 is vertical and inextensible, so joint 3 cannot move vertically. Member 2–3 is horizontal and inextensible, so joints 2 and 3 share the same horizontal movement, $u_2 = u_3 = \delta$. Member 1–2 is inextensible with joint 1 fixed, so the displacement of joint 2 must be perpendicular to that member, whose direction cosines are $(0.8,\ 0.6)$: $$0.8\,u_2 + 0.6\,v_2 = 0 \qquad\Longrightarrow\qquad \boxed{v_2 = -\tfrac{4}{3}\,\delta}$$ Joint 2 therefore drops as it moves right — a single parameter drives the whole sway.
2. Part (a) — convert the sway into chord rotations. Taking $\psi = (\text{transverse displacement of the far end relative to the near end})/L$ with the transverse direction obtained by rotating the member axis through $+90^{\circ}$, $$\psi_{12} = \frac{-\sqrt{1 + (4/3)^{2}}\;\delta}{10} = -\frac{\delta}{6}, \qquad \psi_{23} = \frac{0 - (-\tfrac{4}{3}\delta)}{4} = +\frac{\delta}{3}, \qquad \psi_{34} = \frac{0 - \delta}{6} = -\frac{\delta}{6}$$ The two end members happen to share the same chord rotation, which is a consequence of the 8–6–10 geometry rather than a general rule.
3. Part (b) — write the six slope-deflection moments. With $2EI_m/L$ equal to $0.5EI$ for every member (a numerical coincidence of the given proportions) and the only fixed-end moments coming from the 12 kN/m on member 2–3, $\text{FEM}_{23} = +wL^{2}/12 = +16$ kN·m and $\text{FEM}_{32} = -16$ kN·m: $$M_{12} = 0.5EI\left(\theta_2 + \tfrac{1}{2}\delta\right), \qquad M_{21} = 0.5EI\left(2\theta_2 + \tfrac{1}{2}\delta\right)$$ $$M_{23} = 0.5EI\left(2\theta_2 + \theta_3 - \delta\right) + 16, \qquad M_{32} = 0.5EI\left(2\theta_3 + \theta_2 - \delta\right) - 16$$ $$M_{34} = 0.5EI\left(2\theta_3 + \tfrac{1}{2}\delta\right), \qquad M_{43} = 0.5EI\left(\theta_3 + \tfrac{1}{2}\delta\right)$$
4. Part (b) — joint moment equilibrium. Joints 2 and 3 each carry two members and no applied couple: $$M_{21} + M_{23} = 0 \qquad\Longrightarrow\qquad EI\left(2\theta_2 + 0.5\theta_3 - 0.25\delta\right) = -16$$ $$M_{32} + M_{34} = 0 \qquad\Longrightarrow\qquad EI\left(0.5\theta_2 + 2\theta_3 - 0.25\delta\right) = +16$$ These are the required equations for part (b).
5. Part (a) — the translation equation by virtual work. Give the frame the virtual sway $\delta^{*} = 1$ with the joints not rotating. The internal virtual work of the end moments and the external virtual work of the loads must balance: $$\sum_{\text{members}}\left(M_{ij} + M_{ji}\right)\psi^{*}_{ij} + \sum F\,d^{*} = 0$$ Under $\delta^{*} = 1$ joint 2 moves $(1,\ -\tfrac{4}{3})$ and joint 3 moves $(1,\ 0)$. Using the equivalent nodal loads of the UDL, $wL/2 = 24$ kN at each end of member 2–3, only the load at joint 2 does work: $$\sum F\,d^{*} = 24\left(\tfrac{4}{3}\right) + 24(0) = 32\ \text{kN}\cdot\text{m}$$
6. Part (a) — expand the translation equation. Substituting the member-moment sums $M_{12}+M_{21} = 1.5EI\theta_2 + 0.5EI\delta$, $M_{23}+M_{32} = 1.5EI\theta_2 + 1.5EI\theta_3 - EI\delta$ (the two fixed-end moments cancel) and $M_{34}+M_{43} = 1.5EI\theta_3 + 0.5EI\delta$ into the virtual-work statement with $\psi^{*} = -\tfrac{1}{6},\ +\tfrac{1}{3},\ -\tfrac{1}{6}$ gives, after collecting terms and multiplying through by $-1$, $$\boxed{EI\left(0.5\,\delta - 0.25\,\theta_2 - 0.25\,\theta_3\right) = 32}$$ The sign flip is deliberate: it is what makes the assembled matrix symmetric.
7. Part (c) — assemble the matrix form. Collecting the three equations in the order $\{\delta,\ \theta_2,\ \theta_3\}$, $$EI\begin{bmatrix} 0.50 & -0.25 & -0.25 \\ -0.25 & 2.00 & 0.50 \\ -0.25 & 0.50 & 2.00 \end{bmatrix} \begin{Bmatrix} \delta \\ \theta_2 \\ \theta_3 \end{Bmatrix} = \begin{Bmatrix} 32 \\ -16 \\ 16 \end{Bmatrix}$$ with $\delta$ in metres, $\theta$ in radians, $EI$ in kN·m$^{2}$, the first load term in kN and the other two in kN·m. The equations are not solved, as instructed.
8. Verify the assembly without solving it. Three properties can be checked by inspection. $[K]$ is symmetric, as any stiffness matrix derived from a single strain energy must be. Every diagonal term is positive, as a stable structure requires. And the load vector is consistent: the two moment entries are exactly $\mp$ the fixed-end moments, while the translation entry is the virtual work of the equivalent nodal loads. As a numerical audit only (the paper forbids solving), the system returns $EI\delta = 71.11$, $EI\theta_2 = -3.56$ and $EI\theta_3 = 17.78$ kN·m$^{2}$, from which $M_{12} = 16.0$ and $M_{43} = 26.67$ kN·m — values that satisfy both joint equations identically.