Question 3 of 7: Rigid plane frame — nodal displacements, member forces and reactions
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2016 — 07-Mec-B10 Finite Element Analysis. Three hours, open book (any texts, references or notes; any non-communicating calculator). Seven equally weighted questions of 20 marks; candidates attempt any five, and every question is to be solved within the context of the finite element method. All seven are worked here, because the set is a study resource rather than a sitting.
Reference texts. Logan, A First Course in the Finite Element Method (6th ed.); Reddy, An Introduction to the Finite Element Method (4th ed.); Cook, Malkus, Plesha & Witt, Concepts and Applications of Finite Element Analysis (4th ed.); Zienkiewicz, Taylor & Zhu, The Finite Element Method: Its Basis and Fundamentals (7th ed.); Bathe, Finite Element Procedures (2nd ed.); Hutton, Fundamentals of Finite Element Analysis.
Question 3: Rigid plane frame — nodal displacements, member forces and reactions (20 marks)
Given. A two-element rigid frame fixed at nodes 1 and 3, with the geometry and section properties tabulated below; only node 2 is free.
Given data
Symbol
Value
Meaning
$E$
$30 \times 10^{6}$ psi
Young's modulus, both members
$A$
10 in²
cross-sectional area, both members
$I$
200 in⁴
second moment of area, both members
$w$
150 lb/ft = 12.5 lb/in
downward uniform load on element (2)
Node 1
(0, 540) in
fixed support, top of the inclined member
Node 2
(240, 0) in
free knee joint — three degrees of freedom
Node 3
(720, 0) in
fixed support at the right-hand end
Find. (a) the horizontal displacement, vertical displacement and rotation at node 2; (b) the local end forces in both elements and the six support reactions.
Two-element rigid plane frame: element (1) rises 45 ft over a 20 ft run, element (2) spans 40 ft and carries the 150 lb/ft downward load.
Approach. Treat each member as a two-noded plane-frame element (axial plus bending), transform both element stiffness matrices to global axes, assemble only the three free degrees of freedom at node 2, replace the distributed load by its work-equivalent nodal loads, solve, and recover member forces and reactions by back-substitution.
Establish member geometry. Working consistently in inches and pounds, the inclined member runs from (0, 540) to (240, 0):$$L^{(1)} = \sqrt{240^{2} + 540^{2}} = 590.93\ \text{in} \;(49.24\ \text{ft}),\qquad C = \frac{240}{590.93} = 0.40614,\quad S = \frac{-540}{590.93} = -0.91381$$ The horizontal member has $L^{(2)} = 480$ in, $C = 1$, $S = 0$.
Rotate to global axes. With $[T]$ built from $C$ and $S$, each global element matrix is $[k] = [T]^{T}[k^{\prime}][T]$. Element (2) is horizontal so $[T] = [I]$ and its local and global matrices coincide.
Assemble the free degrees of freedom. Nodes 1 and 3 are fixed, so only $\{u_{2}, v_{2}, \phi_{2}\}$ survive. Adding the node-2 blocks of both elements:$$[K_{\text{ff}}] = \begin{bmatrix}709\,031 & -188\,285 & -94\,207 \\-188\,285 & 424\,642 & 114\,380 \\-94\,207 & 114\,380 & 9.0614\times 10^{7}\end{bmatrix}\ \ \text{(lb/in, lb, lb\,in)}$$
Replace the distributed load by work-equivalent nodal loads. For a beam element carrying a downward intensity $w$ over span $L$, $\int N_{i} q \, dx$ gives half the total load and a fixed-end moment of $wL^{2}/12$ at each end:$$F_{y2} = -\frac{wL}{2} = -\frac{12.5 \times 480}{2} = -3\,000\ \text{lb},\qquad M_{2} = -\frac{wL^{2}}{12} = -\frac{12.5 \times 480^{2}}{12}= -240\,000\ \text{lb\,in}$$
Solve for the node-2 displacements. Inverting the $3\times 3$ system $[K_{\text{ff}}]\{d\} = \{F\}$:$$\boxed{\;u_{2} = -2.310\times 10^{-3}\ \text{in},\quad v_{2} = -7.378\times 10^{-3}\ \text{in},\quad\phi_{2} = -2.642\times 10^{-3}\ \text{rad}\;}$$ The knee settles about 0.0074 in, drifts 0.0023 in to the left and rotates clockwise — all consistent with a downward load on the right-hand span.
Recover the element end forces. For each element, $\{f^{\prime}\} = [k^{\prime}][T]\{d\}$, with the fixed-end forces added back for the loaded member. Element (1):$$\{f^{\prime(1)}\} = \{-2\,946.2\ \text{lb},\; -270.6\ \text{lb},\;-53\,118\ \text{lb\,in},\; +2\,946.2\ \text{lb},\; +270.6\ \text{lb},\;-106\,763\ \text{lb\,in}\}$$ and element (2):$$\{f^{\prime(2)}\} = \{-1\,443.8\ \text{lb},\; +2\,582.4\ \text{lb},\;+106\,763\ \text{lb\,in},\; +1\,443.8\ \text{lb},\; +3\,417.6\ \text{lb},\;-307\,195\ \text{lb\,in}\}$$ Element (1) therefore carries an axial tension of 2 946 lb, and the two member end moments at node 2 are equal and opposite, as they must be at an unloaded rigid joint.
Extract the reactions. Multiplying the full global stiffness by the displacement vector and subtracting the equivalent nodal loads gives$$\boxed{\;R_{1} = \{-1\,443.8\ \text{lb},\; +2\,582.4\ \text{lb},\;-53\,118\ \text{lb\,in}\},\qquad R_{3} = \{+1\,443.8\ \text{lb},\; +3\,417.6\ \text{lb},\;-307\,195\ \text{lb\,in}\}\;}$$
The reactions must be checked before they are reported, and here all three equilibrium equations close exactly. The vertical reactions sum to $2\,582.4 + 3\,417.6 = 6\,000$ lb, which is the full applied load $150 \times 40$ lb; the horizontal reactions are equal and opposite at $\pm 1\,443.8$ lb, as required since no horizontal load is applied; and taking moments about node 1 the support moments and the reaction couples cancel the moment of the 6 000 lb resultant acting 480 in to the right of node 1.
Physically the frame behaves as expected of a portal with one inclined leg. The 45 ft leg is stretched slightly because node 2 drops away from the fixed point above it, so the leg acts as a tension member as well as a bending member. The larger of the two fixed-end moments, 307 195 lb·in (25 600 lb·ft), occurs at node 3, where the loaded span meets a full fixity, and that section governs the design of the horizontal member.
