A composite bar of length L has a central core of copper loosely inserted in a sleeve of steel; the ends of the steel and copper are attached to each other by rigid plates. If the bar is subjected to a temperature rise ΔT, determine the stress in the steel and in the copper and the extension of the

Question

A composite bar of length L has a central core of copper loosely inserted in a sleeve of steel; the ends of the steel and copper are attached to each other by rigid plates. If the bar is subjected to a temperature rise ΔT, determine the stress in the steel and in the copper and the extension of the composite bar. The copper core has a Young’s modulus [latex]E_{c}[/latex], a cross-sectional area [latex]A_{c}[/latex], and a coefficient of linear expansion [latex]\alpha_{c}[/latex]; the corresponding values for the steel are [latex]E_{s}, A_{s}[/latex], and [latex]\alpha_{s} .[/latex]

Accepted Answer

Assume that [latex]\alpha_{c}>\alpha_{s}[/latex]. If the copper core and steel sleeve are allowed to expand freely, their final lengths would be different, since they have different values of the coefficient of linear expansion. However, since they are rigidly attached at their ends, one restrains the other and an axial stress is induced in each. Suppose that this stress is [latex]\sigma_{x}[/latex]. Then, in Eqs. (1.58). [latex]\sigma_{x}=\sigma_{c}[/latex] or [latex]\sigma_{s}[/latex] and [latex]\sigma_{y}=\sigma_{z}=0[/latex]; the total strain in the copper and steel is then, respectively,

[latex]\left.\begin{array}{rl} \varepsilon_{x} & =\frac{1}{E}\left[\sigma_{x}-v\left(\sigma_{y}+\sigma_{z} ight) ight]+\alpha \Delta T \ \varepsilon_{y} & =\frac{1}{E}\left[\sigma_{y}-v\left(\sigma_{x}+\sigma_{z} ight) ight]+\alpha \Delta T \ \varepsilon_{z} & =\frac{1}{E}\left[\sigma_{z}-v\left(\sigma_{x}+\sigma_{y} ight) ight]+\alpha \Delta T \end{array} ight\}[/latex] (1.58)

[latex]\varepsilon_{c}=\frac{\sigma_{c}}{E_{c}}+\alpha_{c} \Delta T[/latex] (i)

[latex]\varepsilon_{s}=\frac{\sigma_{s}}{E_{s}}+\alpha_{s} \Delta T[/latex] (ii)

The total strain in the copper and steel is the same, since their ends are rigidly attached to each other. Therefore, from compatibility of displacement,

[latex]\frac{\sigma_{c}}{E_{c}}+\alpha_{c} \Delta T=\frac{\sigma_{s}}{E_{s}}+\alpha_{s} \Delta T[/latex] (iii)

No external axial load is applied to the bar, so that

[latex]\sigma_{c} A_{c}+\sigma_{s} A_{s}=0[/latex]

that is,

[latex]\sigma_{s}=-\frac{A_{c}}{A_{s}} \sigma_{c}[/latex] (iv)

Substituting for [latex]\sigma_{ S }[/latex] in Eq. (iii) gives

[latex]\sigma_{ c }\left(\frac{1}{E_{ c }}+\frac{A_{ c }}{A_{ s } E_{ s }} ight)=\Delta T\left(\alpha_{ s }-\alpha_{ c } ight)[/latex]

from which

[latex]\sigma_{c}=\frac{\Delta T\left(\alpha_{s}-\alpha_{c} ight) A_{s} E_{s} E_{c}}{A_{s} E_{s}+A_{c} E_{c}}[/latex] (v)

Also [latex]\alpha_{ c }>\sigma_{s}[/latex], so that [latex]\sigma_{c}[/latex] is negative and therefore compressive. Now substituting for [latex]\sigma_{c}[/latex] in Eq. (iv),

[latex]\sigma_{s}=-\frac{\Delta T\left(\alpha_{s}-\alpha_{c} ight) A_{c} E_{s} E_{c}}{A_{s} E_{s}+A_{c} E_{c}}[/latex] (vi)

which is positive and therefore tensile, as would be expected by a physical appreciation of the situation. Finally, the extension of the compound bar, [latex]\delta[/latex], is found by substituting for [latex]\sigma_{c}[/latex] in Eq. (i) or for [latex]\sigma_{s}[/latex] in Eq. (ii). Then,

[latex]\delta=\Delta T L\left(\frac{\alpha_{c} A_{c} E_{c}+\alpha_{s} A_{s} E_{s}}{A_{s} E_{s}+A_{c} E_{c}} ight)[/latex] (vii)

Question 1.6: A composite bar of length L has a central core of copper loo...

Related Answered Questions