The torsional assembly consists of a solid 5 mm dia bronze [G=45GPa] segment (1) securely...

Question:

The torsional assembly consists of a solid {eq}5 mm {/eq} dia bronze {eq}[G=45GPa] {/eq} segment {eq}(1) {/eq} securely connected at flange {eq}B {/eq} to hollow ( {eq}75 mm {/eq} outer, {eq}50 mm {/eq} inner dia) stainless steel {eq}[G=86Gpa] {/eq} segments {eq}(2) {/eq} and {eq}(3) {/eq}. A concentrated torque {eq}T_C = 12 kN-m {/eq} is applied to the stainless pipe at {eq}C {/eq}. Determine the maximum shear stress in the assembly.

Shear stress in shaft

When a shaft is subjected to external torque then it gets twist, this gives rise to internal shear stress on different sections depending upon the intensity of torque. If this shear stress reaches a maximum permissible value then shaft may fail.

Answer and Explanation:


Given Data


  • The modulus of rigidity of bronze is: {eq}{G_{br}} = 45\;{\rm{GPa}} {/eq}
  • The modulus of elasticity of stainless steel is: {eq}{G_{st}} = 86\;{\rm{GPa}} {/eq}
  • The concentrated torque applied at C is: {eq}{T_C} = 12\;{\rm{kN}} \cdot {\rm{m}} {/eq}
  • The diameter of the bronze part is: {eq}{D_{br}} = 5\;{\rm{mm}} {/eq}
  • The outer diameter of the stainless steel part is: {eq}{D_o} = 75\;{\rm{mm}} {/eq}
  • The inner diameter of the stainless steel part is: {eq}{D_i} = 50\;{\rm{mm}} {/eq}
  • The length of section 1 is: {eq}{L_{AB}} = 400\;{\rm{mm}} {/eq}
  • The length of section2 is: {eq}{L_{BC}} = 750\;{\rm{mm}} {/eq}
  • The length of section 3 is: {eq}{L_{CD}} = 400\;{\rm{mm}} {/eq}


The polar moment of inertia of the bronze part is,

{eq}{J_{br}} = \dfrac{\pi }{{32}}{D_{br}}^4 {/eq}


Substitute the known values in the above expression.

{eq}\begin{align*} {J_{br}} &= \dfrac{\pi }{{32}}{\left( {75\;{\rm{mm}}} \right)^4}\\ &= 3106311.09\;{\rm{m}}{{\rm{m}}^4} \end{align*} {/eq}


The polar moment of inertia of the stainless steel part is,

{eq}{J_{st}} = \dfrac{\pi }{{32}}\left( {{D_o}^4 - {D_i}^4} \right) {/eq}


Substitute the known values in the above expression.

{eq}\begin{align*} {J_{st}} &= \dfrac{\pi }{{32}}\left[ {{{\left( {75\;{\rm{mm}}} \right)}^4} - {{\left( {50\;{\rm{mm}}} \right)}^4}} \right]\\ &= 2492718.78\;{\rm{m}}{{\rm{m}}^4} \end{align*} {/eq}


Let the reaction torque at A is {eq}{T_{br}} {/eq} and the reaction torque acting at D is {eq}{T_{st}} {/eq}.


The figure below shows the torque acting in different sections.

Torque in different sections.


The torque acting in section 1is,

{eq}{T_1} = {T_{br}} {/eq}


The twist in the section 1 is,

{eq}{\phi _1} = \dfrac{{{T_{br}}{L_{AB}}}}{{{G_{br}}{J_{br}}}} {/eq}


Substitute the known values in the above expression.

{eq}\begin{align*} {\phi _1} &= \dfrac{{{T_{br}}\left( {400\;{\rm{mm}}} \right)}}{{\left( {45\;{\rm{GPa}}} \right)\left( {\dfrac{{{{10}^3}\;{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {{\rm{m}}{{\rm{m}}^{\rm{2}}}}}} \right. } {{\rm{m}}{{\rm{m}}^{\rm{2}}}}}}}{{1\;{\rm{GPa}}}}} \right)\left( {3106311.09\;{\rm{m}}{{\rm{m}}^4}} \right)}}\\ {\phi _1} &= {T_{br}}\left( {2.86 \times {{10}^{ - 9}}\;{{{\rm{mm}}} {\left/ {\vphantom {{{\rm{mm}}} {\rm{N}}}} \right. } {\rm{N}}}} \right) \end{align*} {/eq}


The torque acting in section 2 is,

{eq}{T_2} = {T_{br}} {/eq}


The twist in the section 2 is,

{eq}{\phi _2} = \dfrac{{{T_{br}}{L_{BC}}}}{{{G_{st}}{J_{st}}}} {/eq}


Substitute the known values in the above expression.

{eq}\begin{align*} {\phi _2} &= \dfrac{{{T_{br}}\left( {750\;{\rm{mm}}} \right)}}{{\left( {86\;{\rm{GPa}}} \right)\left( {\dfrac{{{{10}^3}\;{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {{\rm{m}}{{\rm{m}}^{\rm{2}}}}}} \right. } {{\rm{m}}{{\rm{m}}^{\rm{2}}}}}}}{{1\;{\rm{GPa}}}}} \right)\left( {2492718.78\;{\rm{m}}{{\rm{m}}^4}} \right)}}\\ {\phi _2} &= {T_{br}}\left( {3.49 \times {{10}^{ - 9}}\;{{{\rm{mm}}} {\left/ {\vphantom {{{\rm{mm}}} {\rm{N}}}} \right. } {\rm{N}}}} \right) \end{align*} {/eq}


The torque acting in the section 3 is,

{eq}{T_3} = {T_{br}} - {T_C} {/eq}


The twist in the section 2 is,

{eq}{\phi _3} = \dfrac{{\left( {{T_C} - {T_{br}}} \right){L_{CD}}}}{{{G_{st}}{J_{st}}}} {/eq}


Substitute the known values in the above expression.

{eq}\begin{align*} {\phi _3} &= \dfrac{{\left( {{T_C} - {T_{br}}} \right)\left( {400\;{\rm{mm}}} \right)}}{{\left( {86\;{\rm{GPa}}} \right)\left( {\dfrac{{{{10}^3}\;{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {{\rm{m}}{{\rm{m}}^{\rm{2}}}}}} \right. } {{\rm{m}}{{\rm{m}}^{\rm{2}}}}}}}{{1\;{\rm{GPa}}}}} \right)\left( {2492718.78\;{\rm{m}}{{\rm{m}}^4}} \right)}}\\ {\phi _3} &= \left( {{T_C} - {T_{br}}} \right)\left( {1.86 \times {{10}^{ - 9}}\;{{{\rm{mm}}} {\left/ {\vphantom {{{\rm{mm}}} {\rm{N}}}} \right. } {\rm{N}}}} \right) \end{align*} {/eq}


Since the two ends of the torsion structure are rigidly fixed by the support at A and D. So, the sum of angle of twist of all three sections is equal to zero.


Therfore, the sum of angle of twist is,

{eq}{\phi _1} + {\phi _2} + {\phi _3} = 0 {/eq}


Substitute the known values in the above expression.

{eq}\begin{align*} {T_{br}}\left( {2.86 \times {{10}^{ - 9}}\;{{{\rm{mm}}} {\left/ {\vphantom {{{\rm{mm}}} {\rm{N}}}} \right. } {\rm{N}}}} \right) + {T_{br}}\left( {3.49 \times {{10}^{ - 9}}\;{{{\rm{mm}}} {\left/ {\vphantom {{{\rm{mm}}} {\rm{N}}}} \right. } {\rm{N}}}} \right) + \left( {{T_{br}} - {T_C}} \right)\left( {1.86 \times {{10}^{ - 9}}\;{{{\rm{mm}}} {\left/ {\vphantom {{{\rm{mm}}} {\rm{N}}}} \right. } {\rm{N}}}} \right) &= 0\\ {T_{br}}\left( {8.21 \times {{10}^{ - 9}}\;{{{\rm{mm}}} {\left/ {\vphantom {{{\rm{mm}}} {\rm{N}}}} \right. } {\rm{N}}}} \right) - {T_C}\left( {1.86 \times {{10}^{ - 9}}\;{{{\rm{mm}}} {\left/ {\vphantom {{{\rm{mm}}} {\rm{N}}}} \right. } {\rm{N}}}} \right) &= 0 \end{align*} {/eq}


Substitute the known values in the above expression.

{eq}\begin{align*} {T_{br}}\left( {8.21 \times {{10}^{ - 9}}\;{{{\rm{mm}}} {\left/ {\vphantom {{{\rm{mm}}} {\rm{N}}}} \right. } {\rm{N}}}} \right) - \left( {12\;{\rm{kN}} \cdot {\rm{m}}} \right)\left( {\dfrac{{{{10}^6}\;{\rm{N}} \cdot {\rm{mm}}}}{{1\;{\rm{kN}} \cdot {\rm{m}}}}} \right)\left( {1.86 \times {{10}^{ - 9}}\;{{{\rm{mm}}} {\left/ {\vphantom {{{\rm{mm}}} {\rm{N}}}} \right. } {\rm{N}}}} \right) &= 0\\ {T_{br}}\left( {8.21 \times {{10}^{ - 9}}} \right) &= \left( {22.32 \times {{10}^{ - 3}}} \right)\\ {T_{br}} &= \left( {2.7186 \times {{10}^6}\;{\rm{N}} \cdot {\rm{mm}}} \right)\left( {\dfrac{{{{10}^{ - 6}}\;{\rm{kN}} \cdot {\rm{m}}}}{{1\;{\rm{N}} \cdot {\rm{mm}}}}} \right)\\ {T_{br}} &= 2.7186\;{\rm{kN}} \cdot {\rm{m}} \end{align*} {/eq}


Thus, the torque in section 1 is,

{eq}{T_1} = 2.7186\;{\rm{kN}} \cdot {\rm{m}} {/eq}


The expression for shear stress in segment 1 is,

{eq}{\tau _1} = \dfrac{{{T_1}\left( {\dfrac{{{D_{br}}}}{2}} \right)}}{{{J_{br}}}} {/eq}


Substitute the known values in the above expression.

{eq}\begin{align*} {\tau _1} &= \dfrac{{\left( {2.7186\;{\rm{kN}} \cdot {\rm{m}}} \right)\left( {\dfrac{{{{10}^6}\;{\rm{N}} \cdot {\rm{mm}}}}{{1\;{\rm{kN}} \cdot {\rm{m}}}}} \right)\left( {\dfrac{{75\;{\rm{mm}}}}{2}} \right)}}{{3106311.09\;{\rm{m}}{{\rm{m}}^4}}}\\ &= 32.819\;{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {{\rm{m}}{{\rm{m}}^{\rm{2}}}}}} \right. } {{\rm{m}}{{\rm{m}}^{\rm{2}}}}} \end{align*} {/eq}


Thus, the torque in section 2 is,

{eq}{T_2} = 2.7186\;{\rm{kN}} \cdot {\rm{m}} {/eq}


The expression for shear stress in segment 2 is,

{eq}{\tau _2} = \dfrac{{{T_2}\left( {\dfrac{{{D_o}}}{2}} \right)}}{{{J_{st}}}} {/eq}


Substitute the known values in the above expression.

{eq}\begin{align*} {\tau _2} &= \dfrac{{\left( {2.7186\;{\rm{kN}} \cdot {\rm{m}}} \right)\left( {\dfrac{{{{10}^6}\;{\rm{N}} \cdot {\rm{mm}}}}{{1\;{\rm{kN}} \cdot {\rm{m}}}}} \right)\left( {\dfrac{{75\;{\rm{mm}}}}{2}} \right)}}{{2492718.78\;{\rm{m}}{{\rm{m}}^4}}}\\ &= 40.89\;{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {{\rm{m}}{{\rm{m}}^{\rm{2}}}}}} \right. } {{\rm{m}}{{\rm{m}}^{\rm{2}}}}} \end{align*} {/eq}


Thus, the torque in section 3 is,

{eq}{T_3} = {T_{br}} - {T_C} {/eq}


Substitute the known values in the above expression.

{eq}\begin{align*} {T_3} &= 2.7186\;{\rm{kN}} \cdot {\rm{m}} - 12\;{\rm{kN}} \cdot {\rm{m}}\\ &= {\rm{9}}{\rm{.28}}\;{\rm{kN}} \cdot {\rm{m}} \end{align*} {/eq}


The expression for shear stress in segment 3 is,

{eq}{\tau _3} = \dfrac{{{T_3}\left( {\dfrac{{{D_o}}}{2}} \right)}}{{{J_{st}}}} {/eq}


Substitute the known values in the above expression.

{eq}\begin{align*} {\tau _2} &= \dfrac{{\left( {{\rm{9}}{\rm{.28}}\;{\rm{kN}} \cdot {\rm{m}}} \right)\left( {\dfrac{{{{10}^6}\;{\rm{N}} \cdot {\rm{mm}}}}{{1\;{\rm{kN}} \cdot {\rm{m}}}}} \right)\left( {\dfrac{{75\;{\rm{mm}}}}{2}} \right)}}{{2492718.78\;{\rm{m}}{{\rm{m}}^4}}}\\ &= 139.6\;{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {{\rm{m}}{{\rm{m}}^{\rm{2}}}}}} \right. } {{\rm{m}}{{\rm{m}}^{\rm{2}}}}} \end{align*} {/eq}


Thus, the maximum shear stress in the shaft is {eq}139.6\;{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {{\rm{m}}{{\rm{m}}^{\rm{2}}}}}} \right. } {{\rm{m}}{{\rm{m}}^{\rm{2}}}}} {/eq} in section 3 .


Learn more about this topic:

Loading...
Torque: Concept, Equation & Example

from UExcel Physics: Study Guide & Test Prep

Chapter 7 / Lesson 4
19K

Related to this Question

Explore our homework questions and answers library