Spring assemblage by the principle of minimum potential energy: 22-Mec-B10 Finite Element Analysis, December 2014, Q1

Question 1: Spring assemblage by the principle of minimum potential energy (20 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. Four linear springs join four nodes on a single horizontal degree-of-freedom axis; node 1 is built into the rigid wall, and nodes 2, 3 and 4 ride on rollers so each carries one axial displacement. Node 2 is displaced 2 mm in the positive x-direction by an external agency, and a 1.5 kN load acts at node 4 in the negative x-direction.

Given data (read from the figure)
Quantity	Symbol	Value
Element (1), nodes 1–2	k₁	250 kN/m
Element (2), nodes 2–3	k₂	500 kN/m
Element (3), nodes 3–4	k₃	600 kN/m
Element (4), nodes 1–4	k₄	1000 kN/m
Prescribed displacement at node 2	u₂	2 mm = 0.002 m
Applied load at node 4	F_4x	−1.5 kN
Support at node 1	u₁	0 (fixed wall)

Find. The total potential energy of the assemblage, the equilibrium equations that stationarity of that energy produces, the assembled matrix equation, the unknown displacements with the associated reactions, and the internal forces carried by element (2).

Question 1 — spring assemblage: four springs, four nodes, node 1 fixed to the wall, node 2 given a prescribed 2 mm displacement, and a 1.5 kN load applied at node 4.

Approach. Write the strain energy stored in each spring in terms of the four nodal displacements, subtract the work of the applied loads, set the first variation with respect to each free displacement to zero, assemble the resulting equations, impose the two known displacements and solve.

Set up the element connectivity. Each spring is a two-node bar carrying one axial degree of freedom per node, so the element elongation is the difference of its two end displacements: $\delta^{(1)}=u_2-u_1$, $\delta^{(2)}=u_3-u_2$, $\delta^{(3)}=u_4-u_3$ and $\delta^{(4)}=u_4-u_1$. Element (4) spans from the wall directly to node 4, which is what makes the assemblage a closed loop rather than a chain.
(a) Write the total potential energy. The total potential energy is the sum of the spring strain energies minus the work done by the applied nodal forces, $\Pi_p=U-W$:$$\Pi_p=\tfrac{1}{2}k_1(u_2-u_1)^2+\tfrac{1}{2}k_2(u_3-u_2)^2+\tfrac{1}{2}k_3(u_4-u_3)^2+\tfrac{1}{2}k_4(u_4-u_1)^2-\big(F_{1x}u_1+F_{2x}u_2+F_{3x}u_3+F_{4x}u_4\big)$$Substituting the stiffnesses in kN/m and the single applied load $F_{4x}=-1.5$ kN (with $F_{3x}=0$, while $F_{1x}$ and $F_{2x}$ are the as-yet unknown reactions at the two restrained nodes):$$\Pi_p=125(u_2-u_1)^2+250(u_3-u_2)^2+300(u_4-u_3)^2+500(u_4-u_1)^2-F_{1x}u_1-F_{2x}u_2+1.5\,u_4$$
(b) Apply the principle of minimum potential energy. Equilibrium requires $\Pi_p$ to be stationary with respect to every nodal displacement, $\partial\Pi_p/\partial u_i=0$ for $i=1\ldots 4$. Differentiating term by term gives four equations:$$\frac{\partial\Pi_p}{\partial u_1}=-k_1(u_2-u_1)-k_4(u_4-u_1)-F_{1x}=0$$$$\frac{\partial\Pi_p}{\partial u_2}=k_1(u_2-u_1)-k_2(u_3-u_2)-F_{2x}=0$$$$\frac{\partial\Pi_p}{\partial u_3}=k_2(u_3-u_2)-k_3(u_4-u_3)-F_{3x}=0$$$$\frac{\partial\Pi_p}{\partial u_4}=k_3(u_4-u_3)+k_4(u_4-u_1)-F_{4x}=0$$Each equation is simply the axial force balance at that node, which is the expected result: minimising the energy reproduces nodal equilibrium.
(c) Collect the equations in matrix form. Grouping the coefficients of $u_1\ldots u_4$ gives $[K]\{u\}=\{F\}$ with the symmetric, singular (before restraint) global stiffness matrix$$\begin{bmatrix}k_1+k_4&-k_1&0&-k_4\\-k_1&k_1+k_2&-k_2&0\\0&-k_2&k_2+k_3&-k_3\\-k_4&0&-k_3&k_3+k_4\end{bmatrix}\begin{Bmatrix}u_1\\u_2\\u_3\\u_4\end{Bmatrix}=\begin{Bmatrix}F_{1x}\\F_{2x}\\F_{3x}\\F_{4x}\end{Bmatrix}$$and, in kN/m,$$\boxed{\begin{bmatrix}1250&-250&0&-1000\\-250&750&-500&0\\0&-500&1100&-600\\-1000&0&-600&1600\end{bmatrix}\begin{Bmatrix}u_1\\u_2\\u_3\\u_4\end{Bmatrix}=\begin{Bmatrix}F_{1x}\\F_{2x}\\0\\-1.5\end{Bmatrix}}$$The $-k_4$ entries in the corners are the signature of element (4) tying node 1 directly to node 4.
(d) Impose the boundary conditions and solve. Two displacements are known, $u_1=0$ and $u_2=0.002$ m, so rows 3 and 4 (the rows belonging to the free degrees of freedom) are solved first with the known displacements carried to the right-hand side:$$1100\,u_3-600\,u_4=500(0.002)=1.0\ \text{kN}$$$$-600\,u_3+1600\,u_4=-1.5\ \text{kN}$$Eliminating $u_3$ gives $1272.73\,u_4=-0.9545$, hence$$\boxed{u_3=5.00\times10^{-4}\ \text{m}=0.50\ \text{mm},\qquad u_4=-7.50\times10^{-4}\ \text{m}=-0.75\ \text{mm}}$$Node 3 follows node 2 to the right, whereas node 4 is pulled to the left by the applied load, which the stiff 1000 kN/m spring resists.
Recover the reactions. Rows 1 and 2 of the assembled system, now that all four displacements are known, return the forces at the restrained nodes:$$F_{1x}=1250(0)-250(0.002)+0-1000(-0.00075)=-0.50+0.75$$$$F_{2x}=-250(0)+750(0.002)-500(0.0005)+0=1.50-0.25$$so that$$\boxed{F_{1x}=+0.25\ \text{kN},\qquad F_{2x}=+1.25\ \text{kN}}$$Checking global equilibrium, $0.25+1.25+0-1.5=0$, confirms the solution. $F_{2x}$ is the force the external agency must apply at node 2 to hold the prescribed 2 mm; $F_{1x}$ is the wall reaction.
(e) Extract the element (2) forces. For any spring the element equation is $\{f\}=k\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\{u\}$, so with $u_2=0.002$ m and $u_3=0.0005$ m$$\begin{Bmatrix}f_{2x}^{(2)}\\f_{3x}^{(2)}\end{Bmatrix}=500\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\begin{Bmatrix}0.002\\0.0005\end{Bmatrix}$$$$\boxed{f_{2x}^{(2)}=+0.75\ \text{kN},\qquad f_{3x}^{(2)}=-0.75\ \text{kN}}$$The two forces are equal and opposite, as element equilibrium demands. Because the elongation $u_3-u_2=-1.5$ mm is negative the spring is shortened, so element (2) carries 0.75 kN in compression.

Final results — Question 1
Quantity	Value
Displacement of node 3, u₃	+0.50 mm
Displacement of node 4, u₄	−0.75 mm
Reaction at node 1 (wall), F_1x	+0.25 kN
Force required at node 2 to hold 2 mm, F_2x	+1.25 kN
Element (2) nodal forces	+0.75 kN at node 2, −0.75 kN at node 3
Element (2) internal force	0.75 kN compression

Topic. Minimum potential energy as the variational basis of the direct stiffness method.

Key relations. $\Pi_p=U-W=\sum_e\tfrac12 k_e\delta_e^2-\sum_i F_iu_i$; stationarity $\partial\Pi_p/\partial u_i=0$ yields $[K]\{u\}=\{F\}$; element law $\{f\}=k[1,-1;-1,1]\{u\}$.

Why this works. The strain energy of a linear spring is a quadratic form in the nodal displacements, so its first derivative is linear and its second derivative is the stiffness matrix itself. Minimising the energy therefore produces exactly the nodal force balances that the direct stiffness method writes down by inspection — the same matrix, obtained from an energy statement rather than from equilibrium. The matrix is singular until at least one displacement is prescribed, because a rigid-body translation of the whole assemblage stores no energy; here the wall at node 1 removes that mode. A prescribed non-zero displacement is handled by moving its known column to the load side rather than by deleting the equation, since the corresponding row still carries useful information (the reaction).

Common pitfall. Deleting the row and the column of a node whose displacement is prescribed but non-zero. The column must be multiplied by the known displacement and carried to the right-hand side, otherwise the 2 mm imposed at node 2 never enters the load vector and both u₃ and u₄ come out wrong. A second trap is reading the 1.5 kN arrow as acting at node 3 rather than at node 4 — the leader line runs to the node-4 rigid bar.

Source. Logan, A First Course in the Finite Element Method (6th ed.), §2.1–2.6 (spring assemblages, minimum potential energy) and §2.7 (prescribed non-zero displacements).

Question 1 of 7: Spring assemblage by the principle of minimum potential energy

Question 1: Spring assemblage by the principle of minimum potential energy (20 marks)