Composite Bar (Core + Sleeve)

stressmass-cost

Bond a solid core inside a close-fitting sleeve of a different material, cap the ends with plates stiff enough to stay flat, and push the assembly with a centric axial load $P$. A steel reinforcing bar cast in a concrete column, a copper wire drawn through an aluminium conductor, a bolt clamping a gasketed flange, a shrink-fitted bushing carrying thrust — all are two members forced to act as one. The question is the same every time: how does the load split between them?

Equilibrium gives one equation — the two internal forces add up to the applied load, $P_1 + P_2 = P$ — but there are two unknowns. Like the propped cantilever, the bar is statically indeterminate, and the missing equation comes from geometry, not force balance.

The missing equation is equal elongation

The rigid end plates hold the core and the sleeve to the same elongation $\delta$. Both are the same length $L$, so both see the same strain $\varepsilon = \delta/L$. For a linear-elastic member $\delta = P_i L /(A_i E_i)$, and setting the core’s elongation equal to the sleeve’s is the compatibility condition. Solve the two equations together and the load splits in proportion to each member’s axial stiffness $A_i E_i$:

$P_1 = P\,\frac{A_1 E_1}{A_1 E_1 + A_2 E_2}, \qquad P_2 = P\,\frac{A_2 E_2}{A_1 E_1 + A_2 E_2}$

The build does not hand you these to trust. It certifies that the two relations (equilibrium and equal-elongation compatibility) form a system linear in the unknowns $\{P_1, P_2\}$, solves that $2\times2$ system exactly at build time, and checks the solution back through every relation. This is the same solveLinear capability the propped cantilever introduced — but with a crucial difference spelled out next. What the machine proves and what still rests on a book is on the verification page.

Here the material does not cancel out

In the propped cantilever the flexural rigidity $EI$ divided out of the compatibility equation, so its reactions were material-blind. Here it is the opposite: the stiffnesses $A_1 E_1$ and $A_2 E_2$ sit right in the load-share formula and in the system determinant $A_1 E_1 + A_2 E_2$, so the load split depends on what each member is made of. This is the whole reason to build a two-material member — and it makes the stress result almost paradoxical:

$\frac{\sigma_1}{\sigma_2} = \frac{E_1}{E_2}$

Because the two members share one strain, Hooke’s law $\sigma = E\varepsilon$ puts their stresses in the ratio of their moduli — independent of area. The stiffer material always carries the higher stress. Put a steel core (E $\approx 200$ GPa) in an aluminium sleeve (E $\approx 69$ GPa) and the steel runs about three times the stress of the aluminium, even though every fibre of both stretched by exactly the same amount. Swap the sleeve’s metal in the widget and watch the load migrate toward the stiffer member while $\delta$ and the margins all move — the load share, the stresses, and the deflection are three faces of one material choice:

• Young’s modulus $E$ sets the share and the stresses (above), and the elongation $\delta = PL/(A_1E_1 + A_2E_2)$ — the two members act as springs in parallel.
• Yield strength $\sigma_y$ sets each member’s own margin $\mathrm{SF}_i = \sigma_{yi}/\sigma_i$, and it is independent of stiffness: a stiff-but-weak member can be the first to yield.
• Density $\rho$ sets the mass $m = (\rho_1 A_1 + \rho_2 A_2)L$ and nothing else.

Sign convention and scope

Both members carry the same-sense centric axial load (all of $P_1, P_2, P > 0$), share the free length $L$, and stretch by the common elongation $\delta$; $\sigma_i = P_i/A_i$ is the uniform member stress and $\mathrm{SF}_i = \sigma_{yi}/\sigma_i$ its margin to first yield. The model assumes rigid end plates, a perfect bond (no slip), equal free lengths, and a centric load so neither member bends. It is linear-elastic: the moment either member reaches its yield stress the proportional load-share stops being the truth, and the widget says so.

Try it

Governing relations

Derivation

Steps marked modeling step are where physics enters by citation — every other line is machine-proven to follow from them, and the cited models are independently re-derived in the test pipeline where possible. See what is and isn't machine-verified.

How it fails

The widget guards first yield of each member separately — $\mathrm{SF}_1 = \sigma_{y1}/\sigma_1$ for the core and $\mathrm{SF}_2 = \sigma_{y2}/\sigma_2$ for the sleeve — and warns the moment either one exceeds its yield stress. Which member gets there first is the most interesting failure story this THING tells, and it is not obvious:

• The stiffer member often yields first — even if it is the “stronger” one. Two effects stack. The member with the greater axial stiffness $A_iE_i$ (area × modulus) carries the larger share of the load, and the stiffer material carries the higher stress at any given strain ($\sigma_i \propto E_i$, independent of area). A member that is both — high modulus and ample area — is doubly loaded. Whether it actually yields first depends on the race between its stress and its own yield strength. Put an A36 steel core in a 6061-T6 aluminium sleeve of larger area: the steel takes about two-thirds of the load and runs near $165$ MPa against a $\sim 248$ MPa yield, while the aluminium sits near $56$ MPa against a $\sim 276$ MPa yield. The steel yields first ($\mathrm{SF}\approx 1.5$ vs $\approx 4.9$) — the stiff member is also, here, the comparatively weak one. A stiff core is not automatically the safe core; check its margin, not your intuition.

• Past first yield, the elastic share is a lie. Every number on this page assumes both members are linear-elastic. Once one yields, it can no longer take more load as the assembly stretches further — the still-elastic member picks up the slack and the load redistributes toward it, exactly the redistribution a plastic analysis captures and these formulas cannot. The σ > σ_y warning marks where the linear load-share stops telling the truth.

• Loss of the bond — the assumption under everything. The whole solution rests on the two members straining together. If the interface debonds or slips (a failed shrink fit, a sheared key, a crushed adhesive layer), the equal-elongation compatibility is gone: each member is then on its own, and the one that was leaning on its partner can be suddenly overloaded. Bond shear is a separate limit state this axial model does not see.

• Buckling, if the load is compressive. In compression a slender member can buckle out of the assembly long before it yields — especially a thin sleeve around a short core, or a core that debonds and is no longer braced by the sleeve. That is a stability failure governed by geometry and $E$ (see the Euler column), not by the axial stress shown here.

• Thermal mismatch — a load case with no external load. Two bonded materials with different thermal expansion coefficients cannot expand freely: heat the assembly and the one that wants to grow more is put into compression by its partner, the other into tension, with zero applied $P$. A composite bar can therefore be pre-stressed — or crack — just by changing temperature. That coupling is its own THING; this page holds temperature fixed.

• Brittle members and fatigue. A brittle core (gray cast iron, a ceramic, concrete in tension) has no yield plateau — rate it against ultimate strength with a generous factor, not the yield-based margin here. And under cyclic load the interface and the stiffer member see the largest stress ranges; fatigue and interfacial fretting sit far below static yield and are outside this model.

Sources

• Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — ch. 2 (Axially Loaded Members): statically indeterminate bars, the flexibility/compatibility method, and members sharing load in proportion to axial stiffness A·E.
• Hibbeler, R. C., Mechanics of Materials, 10th ed., Pearson, 2017 — ch. 4 (Axial Load): statically indeterminate axially loaded members solved by the compatibility (equal displacement) method.