Deriving the stiffness matrix and load vector of a parallel-leg frame: 16-Civ-B1 Advanced Structural Analysis, December 2018, Q9

Question 9: Deriving the stiffness matrix and load vector of a parallel-leg frame (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 built-in bases at ① and ④. Joint ② sits 6 m above ground and 2.5 m to the left of base ①; joint ③ is 6.5 m to the right of ②; base ④ is 6.5 m to the right of base ①. Each leg therefore runs 2.5 m horizontally and 6 m vertically — length 6.5 m — and the two legs are parallel. A horizontal point load of 32.5 kN acts to the right at joint ②, and a uniformly distributed load of 2 kN/m acts downward on the 6.5 m beam ②–③. The arrow labelled δ at joint ③ is drawn perpendicular to the legs.

Given data
Quantity	Symbol	Value
Leg length (2.5 m run, 6.0 m rise)	a	6.5 m
Beam length	b	6.5 m
Distributed load on the beam	w	2.0 kN/m
Horizontal point load at joint ②	F	32.5 kN
Flexural rigidity	EI	same for all members

Find. The translation equilibrium equation at joint ③, the two joint-moment equations, and the assembled [K] and {P}. The equations are not to be solved.

Question 9 — the two legs are parallel (each 2.5 m across and 6.0 m up), which is what makes δ normal to them and the beam chord rotation exactly zero.

Approach. Establish the sway pattern from inextensibility first, then write the three slope-deflection expressions, form the two joint equations directly and the translation equation by virtual work, and finally scale the translation row so that [K] comes out symmetric.

Establish the sway pattern. Each leg is inextensible with a built-in far end, so the joint it carries can only move perpendicular to it. The legs are parallel, so joints ② and ③ move in the same direction, along the unit vector $$\mathbf{e}_{\delta} = \dfrac{1}{6.5}(6.0,\;2.5) = (0.9231,\;0.3846)$$ and, because both must move perpendicular to parallel legs, they move by the same amount δ. This is the key structural fact of the question, and it is the reason the δ arrow is drawn at that angle rather than horizontally.
Chord rotations follow immediately. Joints ② and ③ translate by the same vector, so the beam between them does not rotate at all: $$\psi_{23} = 0 \ \text{exactly}, \qquad \psi_{12} = -\dfrac{\delta}{a}, \qquad \psi_{43} = -\dfrac{\delta}{a}$$ Both legs feel $\delta/a$ regardless of how steep they are — only the load term feels the inclination.
(b) Slope-deflection expressions. With $k_{l} = 2EI/a$, $k_{b} = 2EI/b$, $\theta_{1} = \theta_{4} = 0$ and the beam fixed-end moments $\pm wb^{2}/12$, $$M_{12} = k_{l}\!\left(\theta_{2} + \dfrac{3\delta}{a}\right),\quad M_{21} = k_{l}\!\left(2\theta_{2} + \dfrac{3\delta}{a}\right)$$ $$M_{23} = k_{b}(2\theta_{2} + \theta_{3}) + \dfrac{wb^{2}}{12},\quad M_{32} = k_{b}(2\theta_{3} + \theta_{2}) - \dfrac{wb^{2}}{12}$$ $$M_{43} = k_{l}\!\left(\theta_{3} + \dfrac{3\delta}{a}\right),\quad M_{34} = k_{l}\!\left(2\theta_{3} + \dfrac{3\delta}{a}\right)$$
(b) Joint ②. $M_{21} + M_{23} = 0$ gives $$\dfrac{6EI}{a^{2}}\,\delta + \left(\dfrac{4EI}{a} + \dfrac{4EI}{b}\right)\theta_{2} + \dfrac{2EI}{b}\,\theta_{3} = -\dfrac{wb^{2}}{12}$$
(b) Joint ③. $M_{32} + M_{34} = 0$ gives the mirror equation $$\dfrac{6EI}{a^{2}}\,\delta + \dfrac{2EI}{b}\,\theta_{2} + \left(\dfrac{4EI}{a} + \dfrac{4EI}{b}\right)\theta_{3} = +\dfrac{wb^{2}}{12}$$
(a) Translation equation by virtual work. Give the frame a virtual $\delta^{*} = 1$ along $\mathbf{e}_{\delta}$, for which $\psi^{*}_{12} = \psi^{*}_{43} = -1/a$ and $\psi^{*}_{23} = 0$. Then $\sum(M_{ij}+M_{ji})\psi^{*}_{ij} + \sum F\!\cdot\!\mathbf{d}^{*} = 0$ gives, after multiplying through by −1 so that the row is work-conjugate with the two moment rows, $$\dfrac{24EI}{a^{3}}\,\delta + \dfrac{6EI}{a^{2}}\,\theta_{2} + \dfrac{6EI}{a^{2}}\,\theta_{3} = F\cos\alpha - wb\sin\alpha$$
Evaluate the load terms. The point load does work only through the horizontal component of the joint movement, and the beam’s two equivalent nodal forces $wb/2$ work through the vertical component: $$F\cos\alpha = 32.5 \times \dfrac{6.0}{6.5} = 30.0\ \text{kN}, \qquad wb\sin\alpha = 2 \times 6.5 \times \dfrac{2.5}{6.5} = 5.0\ \text{kN}$$ $$P_{1} = 30.0 - 5.0 = \boxed{25.0\ \text{kN}}, \qquad P_{2} = -\dfrac{wb^{2}}{12} = -7.042, \qquad P_{3} = +7.042\ \text{kN}\cdot\text{m}$$
(c) Assemble. With $a = b = 6.5$ m the general form $$[K] = EI\begin{bmatrix} 24/a^{3} & 6/a^{2} & 6/a^{2} \\ 6/a^{2} & 4/a + 4/b & 2/b \\ 6/a^{2} & 2/b & 4/a + 4/b \end{bmatrix}$$ evaluates to $$[K] = EI\begin{bmatrix} 0.087392 & 0.142012 & 0.142012 \\ 0.142012 & 1.230769 & 0.307692 \\ 0.142012 & 0.307692 & 1.230769 \end{bmatrix}, \qquad \{P\} = \begin{Bmatrix} 25.000 \\ -7.042 \\ 7.042 \end{Bmatrix}$$ The matrix is symmetric, which is the only cheap check available on a hand assembly — and it is symmetric only because the translation row was rescaled in step 6. As instructed, the system is left unsolved.

Question 9 — assembled system (EI factored out of [K]; {P} in kN and kN·m)
Term	General form	Value
K₁₁	24EI/a³	0.087392 EI
K₁₂ = K₁₃	6EI/a²	0.142012 EI
K₂₂ = K₃₃	4EI/a + 4EI/b	1.230769 EI
K₂₃	2EI/b	0.307692 EI
P₁	F cosα − wb sinα	25.000 kN
P₂	−wb²/12	−7.042 kN·m
P₃	+wb²/12	+7.042 kN·m
Sway direction	(6.0, 2.5)/6.5	normal to both legs
ψ₂₃	—	exactly 0

Topic. Assembling [K] and {P} for a sway frame with inclined, parallel legs.

Key relations. Slope-deflection $M_{ij} = \frac{2EI}{L}(2\theta_{i}+\theta_{j}-3\psi_{ij}) + \text{FEM}_{ij}$; translation equation $\sum(M_{ij}+M_{ji})\psi^{*} + \sum F\!\cdot\!\mathbf{d}^{*} = 0$; parallel-leg frame stiffness $K_{11} = 24EI/a^{3}$, $K_{12} = K_{13} = 6EI/a^{2}$, $K_{22} = K_{33} = 4EI/a + 4EI/b$, $K_{23} = 2EI/b$.

Why this works. The whole question turns on one kinematic observation. Inextensible legs with built-in bases force each upper joint to move on a circle about its base, which to first order is a straight line perpendicular to the leg; parallel legs make those two directions identical, so both joints travel together and the beam between them suffers no chord rotation whatever the leg inclination. The consequence is a stiffness matrix that depends only on the member lengths, with the inclination confined entirely to the load vector. It also explains why the δ arrow is drawn at an angle: a horizontal sway assumption is kinematically impossible for this frame, and assuming one produces a set of equations that cannot be satisfied. Because the question is examined on the derivation rather than the answer, the symmetry of [K] is the marker’s quickest way to see whether the translation row was formed correctly.

Common pitfall. Leaving the translation row in the raw sign and scale that virtual work delivers. The virtual-work statement gives an equation that is correct but not work-conjugate with the moment rows, so [K] comes out unsymmetric; multiplying that row by −1 fixes it here, and on other proportions a division by the leg length is also needed. Derive the row and then check $K_{12}$ against the δ coefficient of the joint equation rather than assuming a factor. The second pitfall is putting the full 32.5 kN into $P_{1}$: only the component of the joint movement along the load does work, so 30.0 kN is the correct contribution, less the 5.0 kN the beam load takes back.

Source. Hibbeler, Structural Analysis, 10th ed., §11.5 and Ch. 16 (matrix stiffness for frames); Kassimali, Structural Analysis, 6th ed., §13.6 and Ch. 18; Ghali, Neville and Brown, 7th ed., Ch. 5.

Question 9 of 9: Deriving the stiffness matrix and load vector of a parallel-leg frame

Question 9: Deriving the stiffness matrix and load vector of a parallel-leg frame (22 marks)