Deriving the stiffness equations: 07-Str-A4, May 2014, Q9

Question 9: Deriving the stiffness equations (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, encastré at ① and at ④, with a UDL on the horizontal member.

Given data
Item	Value
Member ①–② (vertical)	3 m, $1.5EI$
Member ②–③ (horizontal)	4 m, $EI$
Member ③–④ (inclined)	run 4 m, drop 3 m, length 5 m, $1.25EI$
Supports	encastré at ① and ④
Load	3 kN/m downward on ②–③
Unknowns	$\delta$ (positive left), $\Theta_2$, $\Theta_3$ (counter-clockwise positive)
Axial strain	neglected (all members inextensible)

Find. The three equilibrium equations and the terms of $[K]$ and $\{P\}$. The equations are not to be solved.

Question 9: frame with one sway degree of freedom and two joint rotations.

Approach. Establish the sway mode from inextensibility first — that is what makes a single $\delta$ describe the translation of both joints — then write the six slope-deflection expressions, assemble two joint-moment equations directly and the translation equation by virtual work, and scale the result so that $[K]$ comes out symmetric.

Derive the sway mode. Member ①–② is vertical with a fixed base, so $v_2=0$; member ②–③ is horizontal and inextensible, so $u_3=u_2$; member ③–④ is inextensible with ④ fixed, so $(\mathbf{d}_4-\mathbf{d}_3)\cdot\mathbf{e}_{34}=0$, which with $\mathbf{e}_{34}=(4,-3)/5$ gives $v_3=\tfrac{4}{3}u_3$.
Express the displacements in $\delta$. Taking $\delta$ positive to the left, $u_2=u_3=-\delta$, $v_2=0$ and $v_3=-\tfrac{4}{3}\delta$: joint ③ moves left and down, sliding along the perpendicular to member ③–④ in the ratio 3 : 4.
Compute the three chord rotations. Using $\psi_{ij}=[(\mathbf{d}_j-\mathbf{d}_i)\cdot\mathbf{e}_2]/L$ gives $\psi_{12}=+\delta/3$, $\psi_{23}=-\delta/3$ and $\psi_{34}=+\delta/3$ — equal magnitudes, but the sign on the beam is opposite.
Write the six slope-deflection expressions. The stiffness coefficients are $2E(1.5I)/3=EI$, $2EI/4=0.5EI$ and $2E(1.25I)/5=0.5EI$, and the only fixed-end moments are $\mathrm{FEM}_{23}=+3(4)^{2}/12=+4$ and $\mathrm{FEM}_{32}=-4$ kN·m.
Assemble the two joint equations (part b). $M_{21}+M_{23}=0$ and $M_{32}+M_{34}=0$, collecting terms in $\delta$, $\Theta_2$ and $\Theta_3$.
Assemble the translation equation (part a) by virtual work. Give the frame a unit virtual sway $\delta^{*}=1$; then $\sum(M_{ij}+M_{ji})\psi^{*}_{ij}+W^{*}_{\text{ext}}=0$, where the external term is the work of the 12 kN resultant moving through the average virtual drop of the beam, $\tfrac{1}{2}(0+\tfrac{4}{3})=\tfrac{2}{3}$ m, i.e. $W^{*}=+8$ kN·m.
Check symmetry. If the three equations are written in this order and this scaling, $[K]$ must come out symmetric; that is the only self-check available on a question that forbids solving.

(a) Translation equation at joint ②. With $\psi^{*}_{12}=\psi^{*}_{34}=\tfrac13$ and $\psi^{*}_{23}=-\tfrac13$,

$$EI\left(\tfrac{4}{3}\delta-\tfrac{1}{2}\Theta_2\right)=8\text{ kN}\cdot\text{m}$$

(b) Moment equilibrium at joints ② and ③. From $M_{21}+M_{23}=0$ and $M_{32}+M_{34}=0$,

$$EI\left(-\tfrac{1}{2}\delta+3\Theta_2+\tfrac{1}{2}\Theta_3\right)=-4\text{ kN}\cdot\text{m}$$ $$EI\left(\tfrac{1}{2}\Theta_2+2\Theta_3\right)=+4\text{ kN}\cdot\text{m}$$

Note that $\delta$ has dropped out of the joint-③ equation entirely: the $+\delta/3$ of member ③–④ and the $-\delta/3$ of the beam cancel exactly, which is both a genuine feature of this geometry and the reason $[K]$ has a zero in its corner.

(c) Matrix form.

$$EI\begin{bmatrix}\tfrac{4}{3} & -\tfrac{1}{2} & 0\\[2pt]-\tfrac{1}{2} & 3 & \tfrac{1}{2}\\[2pt]0 & \tfrac{1}{2} & 2\end{bmatrix}\begin{Bmatrix}\delta\\ \Theta_2\\ \Theta_3\end{Bmatrix}=\begin{Bmatrix}8\\ -4\\ 4\end{Bmatrix}$$

so that

$$\boxed{[K]=EI\begin{bmatrix}1.3333 & -0.5 & 0\\ -0.5 & 3.0 & 0.5\\ 0 & 0.5 & 2.0\end{bmatrix},\qquad \{P\}=\begin{Bmatrix}8\\ -4\\ 4\end{Bmatrix}\text{ kN}\cdot\text{m}}$$

The matrix is symmetric, as Betti’s law demands, and its diagonal is positive — the two properties to check before handing the question in. As instructed, the equations are left unsolved.

Check: the support at ④ is drawn as an encastré wall set perpendicular to the inclined member, so $\Theta_4=0$ and $\mathbf{d}_4=\mathbf{0}$ have been assumed. If that support were instead a roller on the inclined plane, member ③–④ would contribute $3EI/L$ modified stiffness and $K_{33}$ would fall from $2.0EI$ to $1.75EI$; the sway kinematics would be unchanged.

Term	Value	Origin
$K_{11}$	$1.3333EI$	translation equation, $\delta$ term
$K_{12}=K_{21}$	$-0.5EI$	sway–rotation coupling at ②
$K_{13}=K_{31}$	$0$	$\psi_{23}$ and $\psi_{34}$ cancel at joint ③
$K_{22}$	$3.0EI$	$EI+0.5EI$ summed at joint ②
$K_{23}=K_{32}$	$0.5EI$	carry-over along the beam
$K_{33}$	$2.0EI$	$0.5EI+0.5EI$ summed at joint ③
$\{P\}$	$\{8,\,-4,\,+4\}$ kN·m	virtual work of the UDL; fixed-end moments $\mp 4$

Topic. Assembling the displacement-method equations for a frame with one sway degree of freedom.

Key relations. Kinematics: inextensibility gives one constraint per member, $(\mathbf{d}_j-\mathbf{d}_i)\cdot\mathbf{e}_1=0$. Statics: $\sum_{\text{members}}M_{ij}=M^{\text{ext}}_i$ at each joint, and $\sum(M_{ij}+M_{ji})\psi^{*}_{ij}+W^{*}_{\text{ext}}=0$ for each independent sway mode. Diagonal terms scale as $4EI/L$, off-diagonal rotation terms as $2EI/L$, sway terms as $6EI/L^{2}$ and $12EI/L^{3}$.

Why this works. The examiner asks for the equations, not the answer, because setting them up correctly is the whole skill. Everything hinges on the sway mode: once inextensibility is used to show that a single $\delta$ drives $u_2=u_3=-\delta$ and $v_3=-\tfrac{4}{3}\delta$, the three chord rotations follow mechanically and the rest is bookkeeping. Building the translation equation by virtual work rather than by free bodies is worth the habit: it is shorter, it automatically produces the row that makes $[K]$ symmetric, and it handles the inclined member without a separate resolution of forces.

Common pitfall. Assuming the inclined member sways by the same amount as the beam, which misses the vertical component $v_3$ entirely, and quoting an unsymmetric $[K]$ — a sure sign that the translation equation has been scaled differently from the joint equations.

Source. McGuire, Gallagher and Ziemian, Matrix Structural Analysis, 2nd ed., Ch. 4–5; Kassimali, Structural Analysis, 6th ed., §16.1–16.3; Ghali, Neville and Brown, 7th ed., Ch. 3.

Question 9 of 9: Deriving the stiffness equations

Question 9: Deriving the stiffness equations (24 marks)