Copper rod in an aluminium sleeve with an initial gap: 16-Civ-B9 The Finite Element Method, May 2013, Q1

Problem 1: Copper rod in an aluminium sleeve 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 two-component axial assembly, loaded in compression through a rigid plate.

Quantity	Symbol	Value
Copper rod diameter	$d$	1.40 in
Aluminium sleeve bore	$d_i$	1.42 in
Aluminium sleeve wall thickness	$t$	0.20 in
Aluminium sleeve outside diameter	$D = d_i + 2t$	1.82 in
Sleeve length	$L_s$	10.000 in
Rod length (0.005 in longer)	$L_r$	10.005 in
Initial gap under the plate	$g$	0.005 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 stress carried by the copper rod and by the aluminium sleeve once the full 60,000 lb is on the plate.

[Figure not reproduced: Figure 1.1 — the assembly as printed on the paper. The rod stands 0.005 in proud of the sleeve, so the rigid plate touches the rod first. See the official exam paper.]

Figure 1.2 — the finite element idealisation: two one-dimensional bar elements in parallel between a fixed node and the single axial degree of freedom of the rigid plate, the sleeve element carrying an initial gap.

Approach. Idealise each component as a one-dimensional bar element of stiffness $k = AE/L$ running from the fixed support to the rigid plate, treat the 0.005 in overhang as an initial gap in the sleeve element, and solve the resulting single-degree-of-freedom stiffness equation for the plate travel before recovering the element forces.

Cross-sectional areas of the two elements. The sleeve outside diameter follows from its bore and wall thickness, $D = d_i + 2t = 1.42 + 2(0.20) = 1.82\ \text{in}$, so $$\begin{aligned}A_r&=\frac{\pi d^{2}}{4}=\frac{\pi (1.40)^{2}}{4}=1.5394\ \text{in}^{2} \\ A_s&=\frac{\pi (D^{2}-d_i^{2})}{4} =\frac{\pi\left(1.82^{2}-1.42^{2}\right)}{4}=1.0179\ \text{in}^{2}\end{aligned}$$
Element axial stiffnesses. A bar element carries only the axial term $k = AE/L$ of the stiffness matrix printed on the paper. Taking each component over its own length, $$k_r=\frac{A_rE_c}{L_r}=\frac{(1.5394)(17\times10^{6})}{10.005}=2.6156\times10^{6}\ \text{lb/in}$$ $$k_s=\frac{A_sE_a}{L_s}=\frac{(1.0179)(10\times10^{6})}{10.000}=1.0179\times10^{6}\ \text{lb/in}$$ The copper rod is roughly two and a half times the stiffer element, both because it is the larger area and because copper has the larger modulus.
Confirm that the gap actually closes. Until the plate has travelled 0.005 in the rod acts alone, so the load required just to bring the plate down onto the sleeve is $$P_{g}=k_r\,g=(2.6156\times10^{6})(0.005)=13{,}078\ \text{lb}$$ Since $P_g = 13{,}078\ \text{lb} \ll 60{,}000\ \text{lb}$, the sleeve is engaged and both elements share the remainder. Had the applied load been smaller than $P_g$ the sleeve would have carried nothing at all — this check is the whole content of the gap.
Assemble and solve the one active degree of freedom. Let $\Delta$ be the downward travel of the rigid plate, which is also the shortening of the rod. The sleeve, being 0.005 in shorter, shortens only $\Delta - g$. Vertical equilibrium of the plate gives $$k_r\Delta+k_s(\Delta-g)=P \;\Longrightarrow\; \Delta=\frac{P+k_s\,g}{k_r+k_s}$$ Substituting, $$\Delta=\frac{60{,}000+(1.0179\times10^{6})(0.005)}{(2.6156+1.0179)\times10^{6}} =\frac{65{,}089}{3.6335\times10^{6}}$$ $$\boxed{\Delta=0.017914\ \text{in}}$$
Recover the element forces by back-substitution. Each element force is its stiffness times its own shortening, $$F_r=k_r\Delta=(2.6156\times10^{6})(0.017914)=46{,}855\ \text{lb}$$ $$F_s=k_s(\Delta-g)=(1.0179\times10^{6})(0.012914)=13{,}145\ \text{lb}$$ and the two must add to the applied load: $46{,}855+13{,}145 = 60{,}000\ \text{lb}$, which they do. Both forces are compressive.
Convert to stresses. Dividing each element force by its own area, $$\begin{aligned}\sigma_r&=\frac{F_r}{A_r}=\frac{46{,}855}{1.5394} \\ \sigma_s&=\frac{F_s}{A_s}=\frac{13{,}145}{1.0179}\end{aligned}$$ $$\boxed{\sigma_r=30{,}438\ \text{psi}=30.44\ \text{ksi (compression)}}$$ $$\boxed{\sigma_s=12{,}914\ \text{psi}=12.91\ \text{ksi (compression)}}$$ The corresponding axial strains are $\varepsilon_r=\Delta/L_r=1.790\times10^{-3}$ and $\varepsilon_s=(\Delta-g)/L_s=1.291\times10^{-3}$; the difference between them is exactly the 0.005 in mismatch spread over the 10 in length.
Sanity check against the no-gap case. Had the two components been machined to the same length, the load would have split in proportion to $AE/L$ alone and the rod would have carried $\sigma_r = 28{,}062$ psi with the sleeve at 16,510 psi. The 0.005 in of pre-travel therefore shifts about 2.4 ksi of duty from the aluminium onto the copper, which is the practical message of the question: a manufacturing mismatch of five thousandths of an inch changes the stress split by roughly eight per cent.

Result	Symbol	Value
Travel of the rigid plate	$\Delta$	0.017914 in
Shortening of the sleeve	$\Delta-g$	0.012914 in
Force in the copper rod	$F_r$	46,855 lb (compression)
Force in the aluminium sleeve	$F_s$	13,145 lb (compression)
Stress in the copper rod	$\sigma_r$	30,438 psi = 30.44 ksi (compression)
Stress in the aluminium sleeve	$\sigma_s$	12,914 psi = 12.91 ksi (compression)
Strain, rod / sleeve	$\varepsilon_r,\ \varepsilon_s$	$1.790\times10^{-3}$ / $1.291\times10^{-3}$

Topic. One-dimensional bar elements in parallel with an initial gap — the simplest contact problem in the finite element method.

Key relations. Element stiffness $k=AE/L$; compatibility with an initial gap $\delta_s=\delta_r-g$; equilibrium at the shared node $k_r\Delta+k_s(\Delta-g)=P$; stress recovery $\sigma=F/A$.

Why this works. A bar element has one degree of freedom per node and a stiffness that is exactly $AE/L$, so a two-component assembly bearing on a rigid plate collapses to a single unknown — the plate travel. The gap makes the problem piecewise linear rather than nonlinear in any deep sense: below $P_g$ only one element is in the assembled stiffness, above it both are, and the transition is handled by carrying the gap as a prescribed initial displacement of the shorter element. Because the rigid plate enforces a single displacement it is the classic example of a multi-point constraint, and the same bookkeeping is what a commercial code does inside a gap or contact element.

Common pitfall. Two errors dominate. The first is to split the load in proportion to $AE/L$ from the outset, ignoring the gap entirely — that gives 28.1 ksi in the rod instead of 30.4 ksi. The second is to compute $A_s$ from the outside diameter alone, or to take the outside diameter as $1.42+0.2=1.62$ in instead of $1.42+2(0.2)=1.82$ in, because the wall thickness appears on both sides of the bore.

Source. Logan, A First Course in the Finite Element Method, 6th ed., Ch. 2–3 (bar element, assembly, stress recovery); Hibbeler, Structural Analysis, 10th ed., Ch. 16 for the equivalent classical statically-indeterminate treatment.

Question 1 of 4: Copper rod in an aluminium sleeve with an initial gap

Problem 1: Copper rod in an aluminium sleeve with an initial gap (equal value)