Deriving the stiffness matrix and load vector: 16-Civ-B1 Advanced Structural Analysis, May 2017, Q9

Question 9: Deriving the stiffness matrix and load vector (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 6 m built in at 1; beam 2–3 of 12 m carrying $w=12$ kN/m; column 3–4 of 3 m pinned at 4; all members the same $EI$ and axially rigid. Unknowns $\delta$ (translation of joint 3, positive right), $\theta_{2}$ and $\theta_{3}$ (counter-clockwise positive).

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

Both columns are vertical and both joints sit on the same horizontal beam, so the beam chord rotation is exactly zero and the two columns share one sway.

Approach. Express the chord rotations in terms of $\delta$, write the three member moment equations, then assemble the two joint-moment equations and one storey-shear equation and read off the coefficients.

Establish the kinematics. Axial rigidity of the columns fixes $v_{2}=v_{3}=0$, and axial rigidity of the beam gives $u_{2}=u_{3}=\delta$. Hence$$\psi_{12}=-\frac{\delta}{6},\qquad \psi_{23}=0,\qquad \psi_{34}=-\frac{\delta}{3}.$$The beam chord rotation is exactly zero because both of its ends translate horizontally by the same amount and neither moves vertically — a point worth stating explicitly, since it removes $\delta$ from the beam equations altogether.
Write the member equations. The far end of column 3–4 is a pin, so that member takes the modified form:$$M_{12}=\frac{EI}{3}\!\left(\theta_{2}+\frac{\delta}{2}\right),\qquad M_{21}=\frac{EI}{3}\!\left(2\theta_{2}+\frac{\delta}{2}\right),$$$$M_{23}=\frac{EI}{6}\!\left(2\theta_{2}+\theta_{3}\right)+\frac{wL^{2}}{12},\qquad M_{32}=\frac{EI}{6}\!\left(2\theta_{3}+\theta_{2}\right)-\frac{wL^{2}}{12},$$$$M_{34}=EI\!\left(\theta_{3}+\frac{\delta}{3}\right),\qquad M_{43}=0,$$with $wL^{2}/12=12(12)^{2}/12=144$ kN·m.
Part (b): moment equilibrium at joints 2 and 3. Summing member end moments at each joint against the applied joint couple (zero here) gives $M_{21}+M_{23}=0$ and $M_{32}+M_{34}=0$, that is$$EI\left[\tfrac{1}{6}\delta+\theta_{2}+\tfrac{1}{6}\theta_{3}\right]=-144,$$$$EI\left[\tfrac{1}{3}\delta+\tfrac{1}{6}\theta_{2}+\tfrac{4}{3}\theta_{3}\right]=+144 .$$The fixed-end moments appear on the right-hand side with opposite signs, which is the algebraic statement that the uniform load tries to rotate the two joints in opposite senses.
Part (a): the translation equation. There is no applied horizontal load, so horizontal equilibrium of the whole frame requires the two column shears to cancel. Since the equivalent nodal loads of the uniform load are vertical, they do no work in the sway pattern, and virtual work on that pattern gives$$\frac{M_{12}+M_{21}}{6}+\frac{M_{34}+M_{43}}{3}=0 .$$Substituting the member equations,$$EI\left[\tfrac{1}{6}\delta+\tfrac{1}{6}\theta_{2}+\tfrac{1}{3}\theta_{3}\right]=0 .$$This is the equation asked for in part (a).
Part (c): assemble and check symmetry. Collecting the three equations in the order $(\delta,\theta_{2},\theta_{3})$,$$EI\begin{bmatrix}\tfrac{1}{6}&\tfrac{1}{6}&\tfrac{1}{3}\\[2pt]\tfrac{1}{6}&1&\tfrac{1}{6}\\[2pt]\tfrac{1}{3}&\tfrac{1}{6}&\tfrac{4}{3}\end{bmatrix}\begin{Bmatrix}\delta\\ \theta_{2}\\ \theta_{3}\end{Bmatrix}=\begin{Bmatrix}0\\ -144\\ +144\end{Bmatrix}.$$Equivalently, multiplying through by 6,$$\boxed{\;\frac{EI}{6}\begin{bmatrix}1&1&2\\ 1&6&1\\ 2&1&8\end{bmatrix}\begin{Bmatrix}\delta\\ \theta_{2}\\ \theta_{3}\end{Bmatrix}=\begin{Bmatrix}0\\ -144\\ +144\end{Bmatrix}\;}$$The matrix is symmetric, which is the only cheap check available on a hand assembly: if it is not, the translation row has been scaled differently from the moment rows. Term by term the coefficients are $K_{11}=12/h_{1}^{3}+3/h_{2}^{3}$, $K_{12}=6/h_{1}^{2}$, $K_{13}=3/h_{2}^{2}$, $K_{22}=4/h_{1}+4/L$, $K_{23}=2/L$ and $K_{33}=4/L+3/h_{2}$, each recognisable as one of the standard $12EI/L^{3}$, $6EI/L^{2}$, $4EI/L$, $2EI/L$ or $3EI/L$ stiffnesses.

The question stops here by instruction, but it is worth noting what the matrix says. The off-diagonal terms $K_{12}$ and $K_{13}$ are non-zero, so sway and rotation are coupled: unlike a symmetric portal under symmetric load, this frame cannot be solved rotation-first. The larger $K_{13}$ reflects the short 3 m column, which is by far the stiffer of the two against sway.

Question 9 — terms of $[K]$ and $\{P\}$ (common factor $EI$)
Term	Expression	Value
$K_{11}$	$12/h_{1}^{3}+3/h_{2}^{3}$	$1/6$
$K_{12}=K_{21}$	$6/h_{1}^{2}$	$1/6$
$K_{13}=K_{31}$	$3/h_{2}^{2}$	$1/3$
$K_{22}$	$4/h_{1}+4/L$	$1$
$K_{23}=K_{32}$	$2/L$	$1/6$
$K_{33}$	$4/L+3/h_{2}$	$4/3$
$P_{1}$ (translation)	no horizontal load	$0$
$P_{2}$ (joint 2)	$-wL^{2}/12$	$-144$ kN·m
$P_{3}$ (joint 3)	$+wL^{2}/12$	$+144$ kN·m

Topic. Assembling the slope-deflection equations of a sway frame into a symmetric stiffness matrix and load vector.

Key relations. Chord rotations from axial rigidity; slope-deflection with the modified $3EI/L$ form at a pinned far end; joint equations $\sum M_{ij}=$ applied joint couple; translation equation from virtual work on the sway pattern; symmetry of $[K]$ as the assembly check.

Why this works. The matrix form is nothing more than the three equilibrium equations written with the unknowns collected. Its symmetry is guaranteed by Betti's theorem provided each equation is the true work-conjugate of its unknown — the moment equations pair with rotations, and the translation equation must be the force equation obtained from virtual work on a unit sway, not an arbitrarily scaled multiple of it. Recognising each entry as a standard element stiffness is the fastest way to spot a slip.

Common pitfall. Scaling the translation row differently from the moment rows, which destroys symmetry and is the single most common error in this question. A close second is assuming the beam has a chord rotation because the frame sways: here both beam ends move horizontally by the same $\delta$ and neither moves vertically, so $\psi_{23}=0$ exactly, and the load vector for the sway equation is zero because the uniform load's equivalent nodal forces are purely vertical.

Source. Hibbeler, Structural Analysis, 10th ed., §11.4–11.5 and Ch. 16 (the stiffness method for frames); Kassimali, 6th ed., Ch. 16 and §18.1; McGuire, Gallagher & Ziemian, Matrix Structural Analysis, 2nd ed., §4.4.

Question 9 of 9: Deriving the stiffness matrix and load vector

Question 9: Deriving the stiffness matrix and load vector (24 marks)