07-Str-B3 · May 2013

Question 1 of 4: Copper rod in an aluminium sleeve — a two-bar model with an initial gap

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Examinations — May 2013 — 07-Str-B3 Applications of the Finite Element Method. Three hours, closed book; one of the two approved calculators (any Casio or Sharp model) and one 8.5 in by 11 in aid sheet written on both sides are permitted. The paper prints four problems, all of equal value, and instructs the candidate to answer only three (3) problems out of the four (4) proposed, the first three appearing in the answer book being the ones marked. Candidates are urged to submit a clear statement of any assumption made where a question admits more than one reading. All four problems are worked below, because this set is intended as a study resource rather than as a single exam sitting.

Reference texts: Logan, D.L., A First Course in the Finite Element Method (6th ed., Cengage) — bar and beam elements, the constant-strain triangle, and the isoparametric quadrilateral, in the same notation this paper uses; Cook, R.D., Malkus, D.S., Plesha, M.E. & Witt, R.J., Concepts and Applications of Finite Element Analysis (4th ed., Wiley) — quadrature, spurious zero-energy (hourglass) modes and element quality; Chandrupatla, T.R. & Belegundu, A.D., Introduction to Finite Elements in Engineering (4th ed., Pearson) — the CST gradient matrix in the beta/alpha form printed on page 4 of this paper; Bathe, K.-J., Finite Element Procedures (2nd ed., Prentice Hall) — isoparametric formulation and numerical integration; Zienkiewicz, O.C. & Taylor, R.L., The Finite Element Method: Its Basis and Fundamentals (7th ed., Butterworth-Heinemann) — general theory; Przemieniecki, J.S., Theory of Matrix Structural Analysis (Dover) — the plane-frame element stiffness matrix printed on page 3; Hibbeler, R.C., Mechanics of Materials (10th ed., Pearson) — statically indeterminate axially loaded members with an initial clearance; Ghali, A., Neville, A.M. & Brown, T.G., Structural Analysis: A Unified Classical and Matrix Approach (7th ed., CRC) — symmetry conditions in closed frames.

Check — the two lengths in Figure 1. Problem 1 dimensions the assembly once, as 10 in, and separately states that the copper rod is 0.005 in longer than the aluminium sleeve. The dimension line runs from the rigid support to the face of the rigid bearing plate, i.e. to the end of the rod, so the rod is taken as 10.000 in long and the sleeve as 9.995 in. Reading it the other way (sleeve 10.000 in, rod 10.005 in) changes every stress below by less than 0.1 per cent, which is far inside the precision of the data; the choice is therefore not load-bearing. Both members are treated as prismatic two-node bar elements sharing one axial degree of freedom at the loaded end.

Problem 1: Copper rod in an aluminium sleeve — a two-bar model with an initial gap (equal value)

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 copper rod and a concentric aluminium sleeve stand between a rigid support and a rigid bearing plate, the rod protruding 0.005 in beyond the sleeve, and the plate is pushed on by a 60,000 lb compressive load.

Given data
Quantity	Symbol	Value
Copper rod diameter	$d_c$	1.4 in
Sleeve inside diameter	$d_i$	1.42 in
Sleeve wall thickness	$t$	0.2 in
Initial gap (rod protrusion)	$\delta_0$	0.005 in
Rod length / sleeve length	$L_c$ / $L_a$	10.000 in / 9.995 in
Applied load	$P$	60,000 lb (compression)
Modulus, copper	$E_c$	$17\times10^{6}$ psi
Modulus, aluminium	$E_a$	$10\times10^{6}$ psi

Find. The axial normal stress carried by the copper rod and by the aluminium sleeve under the full 60,000 lb load, allowing for the fact that the sleeve carries nothing until the 0.005 in gap has closed.

[Figure not reproduced: Figure 1 (redrawn) — copper rod inside the aluminium sleeve, loaded through a rigid bearing plate. The sleeve is 0.005 in short of the plate before loading. See the official exam paper.]

Approach. Model each member as a two-node bar element, compute the load needed to close the 0.005 in gap with the rod acting alone, then share the remaining load between rod and sleeve in proportion to their axial stiffnesses, which is exactly what a two-element parallel assembly of the bar stiffness matrix does.

Two-bar (parallel-spring) idealisation. Node 1 is fixed at the rigid support; node 2 is the rigid bearing plate. The aluminium branch is inactive until the 0.005 in gap closes.

Form the element areas. The rod is solid and the sleeve is an annulus whose outside diameter is the inside diameter plus twice the wall thickness, $d_o = d_i + 2t = 1.42 + 2(0.2) = 1.82\ \text{in}$, so $$A_c=\frac{\pi d_c^{2}}{4}=\frac{\pi (1.4)^{2}}{4}=1.5394\ \text{in}^{2},\ \ A_a=\frac{\pi}{4}\left(d_o^{2}-d_i^{2}\right)=\frac{\pi}{4}\left(1.82^{2}-1.42^{2}\right)=1.0179\ \text{in}^{2}.$$ The 1.42 in bore against a 1.4 in rod leaves a 0.01 in radial clearance, which is a running fit and carries no load; only the axial behaviour matters.
Form the two bar-element stiffnesses. A two-node bar element of area $A$, modulus $E$ and length $L$ has the element stiffness matrix $$[k]=\frac{AE}{L}\begin{bmatrix}1 & -1\\ -1 & 1\end{bmatrix},$$ so its scalar axial stiffness is $k = AE/L$. Substituting, $$k_c=\frac{A_cE_c}{L_c}=\frac{(1.5394)(17\times10^{6})}{10.000}=2.6169\times10^{6}\ \text{lb/in},$$ $$k_a=\frac{A_aE_a}{L_a}=\frac{(1.0179)(10\times10^{6})}{9.995}=1.0184\times10^{6}\ \text{lb/in}.$$ The copper rod is about 2.57 times the stiffer member, mostly because it is 51 per cent larger in area and 70 per cent stiffer in material.
Check whether the gap closes at all. While the plate travels the first 0.005 in the sleeve is not touched, so the rod alone resists. The load that just brings the plate onto the sleeve is $$P_{\text{gap}}=k_c\,\delta_0=(2.6169\times10^{6})(0.005)=\boxed{13{,}085\ \text{lb}}.$$ Since $13{,}085\ \text{lb}$ is well below the applied 60,000 lb, contact does occur and the problem is genuinely two-phase. Had the applied load been smaller than this value, the sleeve would have remained unstressed and the whole load would have gone into the rod.
Assemble the two-element system for the second phase. Once contact is made, both elements connect node 1 (fixed) to node 2 (the rigid plate), so their stiffnesses add directly on the single free degree of freedom $u_2$: $$K_{22}=k_c+k_a=2.6169\times10^{6}+1.0184\times10^{6}=3.6353\times10^{6}\ \text{lb/in}.$$ The remaining load to be shared after the gap closes is $\Delta P = P - P_{\text{gap}} = 60{,}000 - 13{,}085 = 46{,}915\ \text{lb}$.
Share the remaining load in proportion to stiffness. Solving $K_{22}\,\Delta u = \Delta P$ and recovering each element force as $k\,\Delta u$ gives the classical stiffness-ratio split, $$\Delta P_c=\Delta P\frac{k_c}{k_c+k_a}=46{,}915\,(0.7198)=33{,}772\ \text{lb},\ \ \Delta P_a=46{,}915\,(0.2802)=13{,}143\ \text{lb}.$$ The rod therefore takes about 72 per cent of everything added after contact, and the sleeve about 28 per cent.
Recover the total element forces. The rod carries the gap-closing load as well as its share of the remainder, while the sleeve carries only its share: $$P_c=P_{\text{gap}}+\Delta P_c=13{,}085+33{,}772=\boxed{46{,}857\ \text{lb}},\ \ P_a=\Delta P_a=\boxed{13{,}143\ \text{lb}}.$$ Adding the two recovers the applied load, $46{,}857 + 13{,}143 = 60{,}000\ \text{lb}$, which is the equilibrium check that must always be made after a stiffness solution.
Convert element forces to stresses. A constant-strain bar carries uniform stress $\sigma = P/A$, so $$\sigma_c=\frac{P_c}{A_c}=\frac{46{,}857}{1.5394}=\boxed{30{,}439\ \text{psi (compression)}}$$ $$\sigma_a=\frac{P_a}{A_a}=\frac{13{,}143}{1.0179}=\boxed{12{,}912\ \text{psi (compression)}}$$ that is 30.4 ksi in the copper and 12.9 ksi in the aluminium, both compressive.
Confirm the answer from the displacements instead of the forces. The plate travels $u_2 = \delta_0 + \Delta P/(k_c+k_a) = 0.005 + 46{,}915/(3.6353\times10^{6}) = 0.01791\ \text{in}$, so the rod shortens 0.01791 in and the sleeve 0.01291 in. The implied stresses are $$\sigma_c=E_c\frac{u_2}{L_c}=17\times10^{6}\frac{0.01791}{10.000}=30{,}439\ \text{psi},\ \ \sigma_a=E_a\frac{u_2-\delta_0}{L_a}=10\times10^{6}\frac{0.01291}{9.995}=12{,}912\ \text{psi},$$ identical to Step 7. Agreement between the force route and the displacement route is the standard internal check on a bar-element solution.

Problem 1 — final results
Quantity	Value
Copper area $A_c$ / aluminium area $A_a$	1.5394 in² / 1.0179 in²
Bar stiffnesses $k_c$ / $k_a$	$2.617\times10^{6}$ lb/in / $1.018\times10^{6}$ lb/in
Load to close the 0.005 in gap	13,085 lb
Force in the copper rod $P_c$	46,857 lb (compression)
Force in the aluminium sleeve $P_a$	13,143 lb (compression)
Stress in the copper rod $\sigma_c$	30,439 psi = 30.4 ksi (compression)
Stress in the aluminium sleeve $\sigma_a$	12,912 psi = 12.9 ksi (compression)
Shortening of the assembly $u_2$	0.01791 in

Topic. Statically indeterminate axial members with an initial clearance, solved as a two-element bar assembly — the smallest genuine finite element problem there is, and the standard vehicle for introducing gap or contact non-linearity.

Key relations. Bar element stiffness $[k]=\dfrac{AE}{L}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$; parallel assembly $K = \sum k_i$ on a shared degree of freedom; gap-closing load $P_{\text{gap}} = k\,\delta_0$; stiffness-ratio load share $P_i = P\,k_i/\sum k_j$; element stress recovery $\sigma = E\,(u_2-u_1)/L$.

Why this works. Two members that span the same two nodes are in parallel, so they share a displacement and add their stiffnesses; the load each one takes is therefore proportional to its own stiffness. The initial 0.005 in protrusion makes the assembly piecewise linear rather than linear: the structure has one stiffness, $k_c$, while the gap is open and a different, larger one, $k_c+k_a$, once it has shut. Because both phases are individually linear, superposing them — the whole gap-closing load into the rod, then the remainder shared — is exact and no iteration is needed. This is precisely how a commercial code treats a gap element, except that the code has to discover the switching load iteratively rather than being handed it.

Common pitfall. Splitting the entire 60,000 lb by the stiffness ratio and ignoring the gap, which overstates the sleeve stress by about 27 per cent and understates the rod stress; and computing the sleeve area from the 1.42 in bore and the 1.4 in rod rather than from the 1.82 in outside diameter that the 0.2 in wall thickness implies.

Source. Logan, A First Course in the Finite Element Method (6th ed.), Ch. 2 — the bar element and direct assembly — and Ch. 3.10 on inclined and constrained supports; Hibbeler, Mechanics of Materials (10th ed.), Ch. 4.4–4.5, statically indeterminate axially loaded members and the method of superposition with an initial gap.

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