A shaft has a diameter of 20 mm and is made of an aluminum alloy with yield stress Sigma_YP=300...

Question:

A shaft has a diameter of 20 mm and is made of an aluminum alloy with yield stress {eq}{{\sigma }_{\text{yp}}}=300\text{ MPa} {/eq}. The shaft is subjected to an axial load {eq}\text{P}=50.0\text{ kN} {/eq}. Use the maximum energy of distortion failure criterion of failure to determine

(a) torque T that can be applied to the shaft to initiate yielding

(b) torque T that can be applied to the shaft, if the shaft is designed with a factor of safety 1.75 for both P and T against initiation of yielding.

Torque

When an effect occurs on the body by force act on the body to rotate about its axis of rotation that is called torque. Its measurable unit is Newton meter. It has magnitude with the indication of direction, so it is a vector quantity.

Answer and Explanation:


Given Data:

  • The diameter of shaft is: {eq}{d_s} = 20\;{\rm{mm}} = 0.02\;{\rm{m}} {/eq}
  • The yield stress is: , {eq}{\sigma _{yp}} = 300\;{\rm{MPa}} {/eq}
  • The shaft subjected to axially load is: {eq}P = 50\;{\rm{kN}} {/eq}


(a)

The expression for area of shaft is

{eq}{A_s} = \dfrac{\pi }{4}{\left( {{d_s}} \right)^2} {/eq}


Substitute the value in above expression

{eq}\begin{align*} {A_s} &= \dfrac{\pi }{4}{\left( {0.02\;{\rm{m}}} \right)^2}\\ &= 0.0003141\;{{\rm{m}}^2} \end{align*} {/eq}


The expression for normal stress is

{eq}{\sigma _s} = \dfrac{P}{{{A_s}}} {/eq}


Substitute the value in above expression

{eq}\begin{align*} {\sigma _s} &= \dfrac{{50\;{\rm{kN}}}}{{0.0003141\;{{\rm{m}}^2}}}\\ &= 159184.97\;{\rm{kPa}}\left( {\dfrac{{1\;{\rm{MPa}}}}{{{{10}^3}\;{\rm{kPa}}}}} \right)\\ &= 159.187\;{\rm{MPa}} \end{align*} {/eq}


According to maximum energy off distortion, octahedral shear stress expression is

{eq}{\tau _s} = \dfrac{{\sqrt {{{\left( {{\sigma _{yp}}} \right)}^2} - {{\left( {{\sigma _s}} \right)}^2}} }}{{\sqrt 3 }} {/eq}


Substitute the value in above expression

{eq}\begin{align*} {\tau _s} &= \dfrac{{\sqrt {{{\left( {300\;{\rm{MPa}}} \right)}^2} - {{\left( {159.187\;{\rm{MPa}}} \right)}^2}} }}{{\sqrt 3 }}\\ &= 146.80\;{\rm{MPa}} \cdots\cdots\rm{(I)} \end{align*} {/eq}


The expression for polar moment of inertia is

{eq}{J_s} = \dfrac{\pi }{{32}}{\left( {{d_s}} \right)^4} {/eq}


Substitute the value in above expression

{eq}\begin{align*} {J_s} &= \dfrac{\pi }{{32}}{\left( {0.02\;{\rm{m}}} \right)^4}\\ &= 1.571 \times {10^{ - 8}}\;{{\rm{m}}^{\rm{4}}} \end{align*} {/eq}


The expression for shear stress when torque {T_s} applied to shaft is

{eq}{\tau _s} = \dfrac{{{T_s}\left( {\dfrac{{{d_s}}}{2}} \right)}}{{{J_s}}} {/eq}


Substitute the value in above expression

{eq}\begin{align*} {\tau _s} &= \dfrac{{{T_s}\left( {\dfrac{{0.02\;{\rm{m}}}}{2}} \right)}}{{1.571 \times {{10}^{ - 8}}\;{{\rm{m}}^{\rm{4}}}}}\\ &= 6.365 \times {10^5}{T_s}\;{{\rm{m}}^{{\rm{ - 3}}}} \cdots\cdots\rm{(II)} \end{align*} {/eq}


Substitute the value from expression (I) in expression (II)

{eq}\begin{align*} 146.80\;{\rm{MPa}} &= 6.365 \times {10^5}{T_s}\;{{\rm{m}}^{{\rm{ - 3}}}}\\ {T_s} &= 230.636\;{\rm{kN}} \cdot {\rm{m}} \end{align*} {/eq}

Thus the torque applied on shaft is {eq}230.636\;{\rm{kN}} \cdot {\rm{m}} {/eq}


(b)

When factor of safety is: {eq}{f_s} = 1.75 {/eq}

The expression for normal stress is

{eq}{\sigma _s} = {f_s} \times \dfrac{P}{{{A_s}}} {/eq}


Substitute the value in above expression

{eq}\begin{align*} {\sigma _s} &= 1.75 \times \dfrac{{50\;{\rm{kN}}}}{{0.0003141\;{{\rm{m}}^2}}}\\ &= 278573.69\;{\rm{kPa}}\left( {\dfrac{{1\;{\rm{MPa}}}}{{{{10}^3}\;{\rm{kPa}}}}} \right)\\ &= 278.573\;{\rm{MPa}} \end{align*} {/eq}


According to maximum energy off distortion, octahedral shear stress expression is

{eq}{\tau _s} = \dfrac{{\sqrt {{{\left( {{\sigma _{yp}}} \right)}^2} - {{\left( {{\sigma _s}} \right)}^2}} }}{{\sqrt 3 }} {/eq}


Substitute the value in above expression

{eq}\begin{align*} {\tau _s} &= \dfrac{{\sqrt {{{\left( {300\;{\rm{MPa}}} \right)}^2} - {{\left( {278.573\;{\rm{MPa}}} \right)}^2}} }}{{\sqrt 3 }}\\ &= 64.283\;{\rm{MPa}} \cdots\cdots\rm{(III)} \end{align*} {/eq}


The expression for shear stress when torque {T_s} applied to shaft is

{eq}{\tau _s} = \dfrac{{{f_s}{T_s}\left( {\dfrac{{{d_s}}}{2}} \right)}}{{{J_s}}} {/eq}


Substitute the value in above expression

{eq}\begin{align*} {\tau _s} &= \dfrac{{1.75 \times {T_s}\left( {\dfrac{{0.02\;{\rm{m}}}}{2}} \right)}}{{1.571 \times {{10}^{ - 8}}\;{{\rm{m}}^{\rm{4}}}}}\\ &= 11.138 \times {10^5}{T_s}\;{{\rm{m}}^{{\rm{ - 3}}}} \end{align*} {/eq}

Substitute the value {eq}{\tau _s} {/eq} in expression (II)

{eq}\begin{align*} 11.138 \times {10^5}{T_s}\;{{\rm{m}}^{{\rm{ - 3}}}} &= 64.283\;{\rm{MPa}}\\ {{\rm{T}}_s} &= 57.71\;{\rm{kN}} \cdot {\rm{m}} \end{align*} {/eq}

Thus the torque applied on shaft when factor of safety consider is {eq}57.71\;{\rm{kN}} \cdot {\rm{m}} {/eq}


Learn more about this topic:

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Torque in Physics: Equation, Examples & Problems

from Physics: Middle School

Chapter 3 / Lesson 13
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