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The assembly shown consists of a shell fully bonded to a core. The two materials have different...

Question:

The assembly shown consists of a shell fully bonded to a core. The two materials have different Young's modulus and coefficients of thermal expansion. The assembly is stress-free when just bonded at the initial temperature.

(a) Derive the axial stresses in the shell and core as functions of temperature change {eq}\Delta T {/eq}. (by considering Sin equilibrium, material law, and geometric constraint.)

(b) Now the assembly consists of an aluminum shell ({eq}E_{a}=10\times 10^{6}psi,\alpha _{o}=13\times 10^{-6}/^{\circ}F {/eq}) fully bonded to a steel core ( {eq}E_{s}=30\times 10^{6}psi,\alpha _{s}=6.5\times 10^{-6}/^{\circ}F {/eq}) and is unstressed in the beginning. If the stress in Aluminum is not to exceed 6ksi, what is the largest allowable temperature change?

(c) Determine the corresponding change in the length of the assembly.

Thermal Expansion Coefficient:

In reference to the mechanics of materials, the thermal expansion coefficient of any material is a constant which is fixed for the given material. The total change in length is a depending parameter on this coefficient.

Answer and Explanation:


Given Data

  • The modulus of elasticity of the aluminium is: {eq}{E_a} = 10 \times {10^6}\;{\rm{psi}} {/eq}.
  • The thermal coefficient of aluminium is: {eq}{\alpha _a} = 13 \times 1{0^{ - 6}}\;/^\circ {\rm{F}} {/eq}.
  • The modulus of elasticity of the steel is: {eq}{E_s} = 30 \times {10^6}\;{\rm{psi}} {/eq}.
  • The thermal coefficient of steel is: {eq}{\alpha _s} = 6.5 \times 1{0^{ - 6}}\;/^\circ {\rm{F}} {/eq}.
  • The maximum compressive stress in aluminium is: {eq}{\sigma _a} = - 6\;{\rm{ksi}} {/eq}.
  • The outer diameter of the shell is: {eq}{d_s} = 1.25\;{\rm{in}} {/eq}.
  • The outer diameter of the core is: {eq}{d_c} = 0.75\;{\rm{in}} {/eq}.
  • The length of the assembly is: {eq}l = 8\;{\rm{in}} {/eq}.


a)


The expression for the net axial force balance for the assembly is given as:

{eq}{\sigma _a}{A_a} + {\sigma _s}{A_s} = 0......(1) {/eq}

Here, {eq}{\sigma _s} {/eq} is the stress in steel, {eq}{A_a} {/eq} is the aluminium shell cross-section area and {eq}{A_s} {/eq} is the steel core cross-section area.


The expression for the total strain in aluminium shell is given as:

{eq}{\varepsilon _a} = \left( {{\alpha _a} \times \Delta T} \right) + \left( {\dfrac{{{\sigma _a}}}{{{E_a}}}} \right) {/eq}

Here, {eq}\Delta T {/eq} is the temperature change.


The expression for the total strain in steel core is given as:

{eq}{\varepsilon _s} = \left( {{\alpha _s} \times \Delta T} \right) + \left( {\dfrac{{{\sigma _s}}}{{{E_s}}}} \right) {/eq}


The expression for the assembly to remain in original shape is:

{eq}\begin{align*} {\varepsilon _a} &= {\varepsilon _s}\\ \left( {{\alpha _a} \times \Delta T} \right) + \left( {\dfrac{{{\sigma _a}}}{{{E_a}}}} \right) &= \left( {{\alpha _s} \times \Delta T} \right) + \left( {\dfrac{{{\sigma _s}}}{{{E_s}}}} \right)\\ \left( {\dfrac{{{\sigma _a}}}{{{E_a}}}} \right) - \left( {\dfrac{{{\sigma _s}}}{{{E_s}}}} \right) &= \left( {{\alpha _s} - {\alpha _a}} \right)\Delta T\\ \Delta T &= \dfrac{{\left\{ {\left( {\dfrac{{{\sigma _a}}}{{{E_a}}}} \right) - \left( {\dfrac{{{\sigma _s}}}{{{E_s}}}} \right)} \right\}}}{{\left( {{\alpha _s} - {\alpha _a}} \right)}}......(2) \end{align*} {/eq}


Thus, the relationship between the temperature change and the axial stress is {eq}\Delta T = \dfrac{{\left\{ {\left( {\dfrac{{{\sigma _a}}}{{{E_a}}}} \right) - \left( {\dfrac{{{\sigma _s}}}{{{E_s}}}} \right)} \right\}}}{{\left( {{\alpha _s} - {\alpha _a}} \right)}} {/eq}.


b)


Substitute all the values in the equation (1):

{eq}\begin{align*} 0 &= {\sigma _a} \times \dfrac{\pi }{4} \times \left\{ {{{\left( {{d_s}} \right)}^2} - {{\left( {{d_c}} \right)}^2}} \right\} + \left( {{\sigma _s} \times \dfrac{\pi }{4} \times {{\left( {{d_s}} \right)}^2}} \right)\\ 0 &= - 6\;{\rm{ksi}} \times \dfrac{\pi }{4} \times \left\{ {{{\left( {1.25\;{\rm{in}}} \right)}^2} - {{\left( {0.75\;{\rm{in}}} \right)}^2}} \right\} + \left( {{\sigma _s} \times \dfrac{\pi }{4} \times {{\left( {0.75\;{\rm{in}}} \right)}^2}} \right)\\ {\sigma _s} &= 10.667\;{\rm{ksi}} \end{align*} {/eq}


Substitute all the values in equation (2):

{eq}\begin{align*} \Delta T &= \dfrac{{\left\{ {\left( {\dfrac{{ - 6000\;{\rm{psi}}}}{{10 \times {{10}^6}\;{\rm{psi}}}}} \right) - \left( {\dfrac{{10667\;{\rm{psi}}}}{{30 \times {{10}^6}\;{\rm{psi}}}}} \right)} \right\}}}{{\left( {6.5 \times 1{0^{ - 6}}\;/^\circ {\rm{F}} - 13 \times 1{0^{ - 6}}\;/^\circ {\rm{F}}} \right)}}\\ &= 147.01\;^\circ {\rm{F}} \end{align*} {/eq}


Thus, the total temperature change is {eq}147.01\;^\circ {\rm{F}} {/eq}.


c)


The corresponding change in the length of the assembly can be calculated as:

{eq}\begin{align*} {\varepsilon _a} &= \left( {{\alpha _a} \times \Delta T} \right) + \left( {\dfrac{{{\sigma _a}}}{{{E_a}}}} \right)\\ \dfrac{{\Delta l}}{l} &= \left( {{\alpha _a} \times \Delta T} \right) + \left( {\dfrac{{{\sigma _a}}}{{{E_a}}}} \right)\\ \Delta l &= l\left\{ {\left( {{\alpha _a} \times \Delta T} \right) + \left( {\dfrac{{{\sigma _a}}}{{{E_a}}}} \right)} \right\}\\ &= 8\;{\rm{in}}\left\{ {\left( {13 \times {{10}^{ - 6}}/^\circ {\rm{F}} \times {\rm{147}}{\rm{.01}}^\circ {\rm{F}}} \right) + \left( {\dfrac{{ - 6000\;{\rm{psi}}}}{{10 \times {{10}^6}\;{\rm{psi}}}}} \right)} \right\}\\ &= 0.011\;{\rm{in}} \end{align*} {/eq}


Thus, the total change in length is {eq}0.011\;{\rm{in}} {/eq}.


Learn more about this topic:

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Thermal Expansion: Definition, Equation & Examples

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