Equilibrium equations and the stiffness matrix of a sway frame: 16-Civ-B1 Advanced Structural Analysis, May 2015, Q9

Question 9: Equilibrium equations and the stiffness matrix of a sway frame (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. Column 1–2 of height $10\ \text{m}$ built in at joint 1; beam 2–3 of span $10\ \text{m}$ carrying $w=12\ \text{kN/m}$; column 4–3 of height $5\ \text{m}$ built in at joint 4; a horizontal load of $60\ \text{kN}$ applied at joint 2; all members of the same $EI$ and axially rigid.

Find. The translation equilibrium equation, the two joint moment equations, and the resulting $[K]$ and $\{P\}$. The equations are not to be solved.

Question 9: the sway frame; the translation d at joint 3 is positive to the right.

Approach. Establish that the axially rigid members leave exactly one sway freedom, write the six slope-deflection expressions, then take moment equilibrium at joints 2 and 3 and a virtual-work equation for the sway, scaling the sway equation so that the assembled matrix is symmetric.

Kinematics: one sway freedom. Both columns are axially rigid and built in at their bases, so $v_2=v_3=0$; the beam is axially rigid, so $u_2=u_3=\delta$. Joints 2 and 3 translate together by $\delta$ and the beam has no chord rotation: $$\psi_{12}=-\frac{\delta}{10},\qquad \psi_{43}=-\frac{\delta}{5},\qquad \psi_{23}=0$$ using $\psi_{ij}=(\mathbf{D}_j-\mathbf{D}_i)\cdot\mathbf{e}_2/L$ with $\mathbf{e}_2$ the member axis turned 90 degrees counter-clockwise (for a column numbered upward, $\mathbf{e}_2$ points to the left, which is why both chord rotations are negative for a rightward sway).
Fixed-end moments. Only the beam is loaded: $$M^{F}_{23}=+\frac{wL^{2}}{12}=+\frac{12(10)^{2}}{12}=+100\ \text{kN}\cdot\text{m},\qquad M^{F}_{32}=-100\ \text{kN}\cdot\text{m}$$
Slope-deflection expressions. $$M_{12}=\frac{2EI}{10}\left(\theta_2+\frac{3\delta}{10}\right) =0.2EI\,\theta_2+0.06EI\,\delta$$ $$M_{21}=\frac{2EI}{10}\left(2\theta_2+\frac{3\delta}{10}\right) =0.4EI\,\theta_2+0.06EI\,\delta$$ $$M_{43}=\frac{2EI}{5}\left(\theta_3+\frac{3\delta}{5}\right) =0.4EI\,\theta_3+0.24EI\,\delta$$ $$M_{34}=\frac{2EI}{5}\left(2\theta_3+\frac{3\delta}{5}\right) =0.8EI\,\theta_3+0.24EI\,\delta$$ $$M_{23}=0.4EI\,\theta_2+0.2EI\,\theta_3+100,\qquad M_{32}=0.4EI\,\theta_3+0.2EI\,\theta_2-100$$
(a) Translation equation. Give the frame a virtual sway $\delta^{*}=1$ to the right. The virtual chord rotations are $\psi^{*}_{12}=-1/10$ and $\psi^{*}_{43}=-1/5$, the beam does not rotate, and the only external force that moves is the 60 kN load: $$\left(M_{12}+M_{21}\right)\left(-\tfrac{1}{10}\right) +\left(M_{43}+M_{34}\right)\left(-\tfrac{1}{5}\right)+60=0$$ Substituting the expressions above and multiplying through by $-1$, $$\boxed{0.108\,EI\,\delta+0.06\,EI\,\theta_2+0.24\,EI\,\theta_3=60}$$ This is the storey-shear equation: the two column shears, $(M_{12}+M_{21})/10$ and $(M_{43}+M_{34})/5$, must together carry the 60 kN.
(b) Moment equilibrium at joint 2. The column and the beam meet rigidly and no couple is applied there, so $M_{21}+M_{23}=0$: $$0.4EI\,\theta_2+0.06EI\,\delta+0.4EI\,\theta_2+0.2EI\,\theta_3+100=0$$ $$\boxed{0.06\,EI\,\delta+0.8\,EI\,\theta_2+0.2\,EI\,\theta_3=-100}$$
Moment equilibrium at joint 3. Similarly $M_{32}+M_{34}=0$: $$0.4EI\,\theta_3+0.2EI\,\theta_2-100+0.8EI\,\theta_3+0.24EI\,\delta=0$$ $$\boxed{0.24\,EI\,\delta+0.2\,EI\,\theta_2+1.2\,EI\,\theta_3=+100}$$
(c) Matrix form. Collecting the three boxed equations, $$EI\begin{bmatrix} 0.108 & 0.06 & 0.24\\ 0.06 & 0.80 & 0.20\\ 0.24 & 0.20 & 1.20 \end{bmatrix} \begin{Bmatrix}\delta\\ \theta_2\\ \theta_3\end{Bmatrix} =\begin{Bmatrix}60\\ -100\\ 100\end{Bmatrix}$$ so the individual terms are $$K_{11}=\frac{12EI}{10^{3}}+\frac{12EI}{5^{3}}=0.108EI,\qquad K_{12}=K_{21}=\frac{6EI}{10^{2}}=0.06EI,\qquad K_{13}=K_{31}=\frac{6EI}{5^{2}}=0.24EI$$ $$K_{22}=\frac{4EI}{10}+\frac{4EI}{10}=0.80EI,\qquad K_{23}=K_{32}=\frac{2EI}{10}=0.20EI,\qquad K_{33}=\frac{4EI}{5}+\frac{4EI}{10}=1.20EI$$ $$P_1=60\ \text{kN},\qquad P_2=-\frac{wL^{2}}{12}=-100,\qquad P_3=+\frac{wL^{2}}{12}=+100\ \text{kN}\cdot\text{m}$$ The equations are not solved, as instructed.
Verify the matrix without solving it. Every term can be recognised as a standard stiffness coefficient: $12EI/h^{3}$ and $6EI/h^{2}$ for the sway of a member built in at one end, and $4EI/L$ and $2EI/L$ for the rotations. $[K]$ is symmetric, as any correctly scaled stiffness matrix must be — if the translation row had been left with the sign it carried before multiplying by $-1$, the matrix would have been unsymmetric, which is the standard signal that the sway equation has been scaled differently from the moment equations. The diagonal is positive and dominant, and $\{P\}$ contains the applied force work-conjugate to $\delta$ together with the equivalent joint moments, which are the fixed-end moments reversed.

Question 9 — the requested terms
Term	Expression	Value
$K_{11}$	$12EI/10^{3}+12EI/5^{3}$	$0.108\,EI$
$K_{12}=K_{21}$	$6EI/10^{2}$	$0.06\,EI$
$K_{13}=K_{31}$	$6EI/5^{2}$	$0.24\,EI$
$K_{22}$	$4EI/10+4EI/10$	$0.80\,EI$
$K_{23}=K_{32}$	$2EI/10$	$0.20\,EI$
$K_{33}$	$4EI/5+4EI/10$	$1.20\,EI$
$P_1$	applied horizontal load	$60\ \text{kN}$
$P_2$	$-wL^{2}/12$	$-100\ \text{kN}\cdot\text{m}$
$P_3$	$+wL^{2}/12$	$+100\ \text{kN}\cdot\text{m}$

Topic. Assembling the reduced stiffness equations of a one-storey sway frame with unequal column heights.

Key relations. Slope-deflection $M_{ij}=\frac{2EI}{L}(2\theta_i+\theta_j-3\psi_{ij})+M^{F}_{ij}$; virtual-work translation equation $\sum(M_{ij}+M_{ji})\psi^{*}_{ij}+W_{\text{ext}}=0$; the standard coefficients $12EI/h^{3}$, $6EI/h^{2}$, $4EI/L$, $2EI/L$.

Why this works. Neglecting axial strain is what reduces six joint freedoms to three: the two columns fix the joint levels and the beam locks the two horizontal movements together. Each slope-deflection expression is linear in the three remaining unknowns, so the three equilibrium statements — one force equation for the storey and one moment equation per rigid joint — are exactly the reduced stiffness equations. Because the columns have different heights, the sway couples much more strongly to joint 3 ($6EI/25$) than to joint 2 ($6EI/100$): the short column is four times as stiff in sway and attracts most of the storey shear.

Common pitfall. Writing the sway equation from free bodies and using the total applied load rather than the load that actually moves through the virtual displacement; the distributed load on the beam does no work in a horizontal sway and must not appear in $P_1$. The other classic error is an unsymmetric $[K]$, which always means the translation row has been scaled inconsistently — check symmetry before reporting the matrix.

Source. Kassimali, Structural Analysis, ch. 16 and 18 (slope deflection with sidesway; the matrix stiffness formulation); Hibbeler, ch. 11 and 16.

Question 9 of 9: Equilibrium equations and the stiffness matrix of a sway frame

Question 9: Equilibrium equations and the stiffness matrix of a sway frame (24 marks)