Chapter 3, Problem 3.4.21P

A nonprismatic bar ABC with a solid circular cross section is loaded by distributed torques (sec figure). The intensity of the torques, that is, the torque per unit distance, is denoted t(x) and varies linearly from zero at A to a maximum value T₀/L at B. Segment BC has linearly distributed torque of intensity r(x) = T₀/3L of opposite sign to that applied along AB. Also, the polar moment of inertia of AB is twice that of BC and the shear modulus of elasticity of the material is G.

1. Find the reaction torque R_A.

Find internal torsional moments T(x) in segments AB and BC.

Find the rotation t₀

Find the maximum shear stress t_maxand its location along the bar, Draw the torsional moment diagram (TMD:T(x),0 < x < L).

  

To determine

The reaction torque $R_A$.

Answer to Problem 3.4.21P

The reaction torque $R_A$is = $\frac{-T_o}{6}$.

Explanation of Solution

Given information:

The intensity of torque at point (A) is $0$, the intensity of torque at point (B) is $\frac{T_o}{L}$, intensity of torque in segment $BC$is $\frac{T_o}{3L}$, the polar moment of inertia is twice in section $AB$as compared to section $BC$

Write the expression for equilibrium torque equation.

  $R_A+\frac{1}{2}\left(\frac{T_o}{L}\right)\left(\frac{L}{2}\right)-\frac{1}{2}\left(\frac{T_o}{3L}\right)\left(\frac{L}{2}\right)=0$..... (I).

Here, the reaction torque is at point (A) is $R_A$.and the maximum torque per distance is $\left(\frac{T_o}{L}\right)$.

Write the expression for the reaction torque is at point (A) from Equation (I)

  $\begin{array}{l}{R}_A+\left(\frac{T_o}{4}\right)-\left(\frac{T_o}{12}\right)=0\\ R_A+\frac{3T_o-T_o}{12}\\ R_A=\frac{-T_o}{6}\end{array}$

Conclusion:

The reaction torque $R_A$is = $\frac{-T_o}{6}$.

To determine

The internal torsional moment in the segment $AB$.

The internal torsional moment in the segment $BC$

Answer to Problem 3.4.21P

The internal torsional moment in the segment $AB$is $\frac{T_o}{6}-\frac{x^2{T}_o}{L^2}$.

The internal torsional moment in the segment $BC$is $\left[\frac{-{\left(L-x\right)}^2{T}_o}{3L^2}\right]$.

Explanation of Solution

Write the expression for internal torsional moment in section $AB$.

  $\begin{array}{c}{T}_{AB}\left(x\right)=\frac{T_o}{6}-\left(\frac{x}{\frac{L}{2}}\right)\left(\frac{T_o}{L}\right)\frac{x}{2}\\ =\frac{T_o}{6}-\frac{x^2{T}_o}{L^2}\end{array}$

Write the expression for internal torsional moment in section $BC$.

  $\begin{array}{c}{T}_{BC}\left(x\right)=\frac{-\left(L-x\right)}{\left(\frac{L}{2}\right)}\left(\frac{T_o}{3L}\right)\left(\frac{1}{2}\left(L-x\right)\right)\\ =-\left[{\left(\frac{x-L}{L}\right)}^2\left(\frac{T_o}{3}\right)\right]\\ =\left[\frac{-{\left(L-x\right)}^2{T}_o}{3L^2}\right]\end{array}$

Conclusion:

The internal torsional moment in the segment $AB$is $\frac{T_o}{6}-\frac{x^2{T}_o}{L^2}$.

The internal torsional moment in the segment $BC$is $\left[\frac{-{\left(L-x\right)}^2{T}_o}{3L^2}\right]$.

To determine

The twist rotation at point (C).

Answer to Problem 3.4.21P

The twist rotation at point (C) is $\frac{T_{o}L}{144G{I}_p}$.

Explanation of Solution

Write the expression angle of twist at point (C).

  ${\varphi }_C={\displaystyle \underset{0}{\overset{L}{\int }}\frac{T\left(x\right)dx}{G{I}_P\left(x\right)}}$...... (II)

Here, the angle of twist at point (C) is ${\varphi }_C$.

Write the expression for polar moment of inertia of the bar.

  ${\varphi }_C={\varphi }_{AB}+{\varphi }_{BC}$..... (III)

Write the expression for polar moment of inertia in the section $AB$.

  ${\varphi }_{AB}={\displaystyle \underset{0}{\overset{\frac{L}{2}}{\int }}\frac{\frac{T_o}{6}-\frac{x^2{T}_o}{L^2}}{2G{I}_p}}$..... (IV)

Here, polar moment of inertia is in the section $AB$is ${\varphi }_{AB}$.

Integrate Equation (IV).

  $\begin{array}{c}{\varphi }_{AB}=\frac{1}{2G{I}_p}{\left[\frac{T_o}{6}x-\frac{x^3{T}_o}{3L^2}\right]}_0^{\frac{L}{2}}\\ =\frac{1}{2G{I}_p}\left[\frac{T_{o}L}{12}-\frac{T_{o}L}{24}\right]\\ =\frac{1}{2G{I}_p}\left[\frac{2T_{o}L-T_{o}L}{24}\right]\\ =\frac{T_{o}L}{48G{I}_p}\end{array}$..... (V)

Write the expression for angle of twist in $BC$.

  ${\varphi }_{BC}={\displaystyle \underset{\frac{L}{2}}{\overset{L}{\int }}\frac{T_{BC}\left(x\right)}{G{\left(I_p\right)}_{BC}}}dx$...... (VI)

Here, angle of twist in section $BC$is ${\varphi }_{BC}$.

Substitute $\left[\frac{-{\left(L-x\right)}^2{T}_o}{3L^2}\right]$for $T_{BC}\left(x\right)$in Equation (VI).

  $\begin{array}{c}{\varphi }_{BC}={\displaystyle \underset{\frac{L}{2}}{\overset{L}{\int }}\frac{\left[\frac{-{\left(L-x\right)}^2{T}_o}{3L^2}\right]}{G\left(I_p\right)}}dx\\ =\frac{T_o}{G\left(I_p\right)}{\displaystyle \underset{\frac{L}{2}}{\overset{L}{\int }}\frac{{\left(L-x\right)}^2}{3L^2}dx}\\ =\frac{T_o}{G\left(I_p\right)}{\left[\frac{{\left(L-x\right)}^3}{9L^2}\right]}_{L/2}^L\\ =\frac{-T_{o}L}{72G\left(I_p\right)}\end{array}$

Substitute $\frac{T_{o}L}{48G{I}_p}$for ${\varphi }_{AB}$and $\frac{-T_{o}L}{72G\left(I_p\right)}$for ${\varphi }_{BC}$in Equation (III)

  $\begin{array}{c}{\varphi }_C=\frac{T_{o}L}{48G{I}_p}+\frac{-T_{o}L}{72G\left(I_p\right)}\\ =\frac{T_{o}L}{144G{I}_p}\end{array}$

Conclusion:

The twist rotation at point (C) is $\frac{T_{o}L}{144G{I}_p}$.

To determine

The maximum stress and the location of maximum stress in the bar.

Answer to Problem 3.4.21P

The maximum stress is $\frac{2.67T_o}{\pi d_{AB}^3}$.

The maximum stress in the bar is at point (A).

Explanation of Solution

Write the expression for the polar moment of inertia of section $AB$.

  $2I_p=\frac{\pi }{32}{d}_{AB}^4$...... (VII)

Here, the polar moment of inertia is $I_p$and the diameter of the section $AB$is $d_{AB}$.

Write the expression for the polar moment of inertia of section $BC$.

  $I_p=\frac{\pi }{32}{d}_{BC}^4$..... (VIII)

Here, the polar moment of inertia is $I_p$and the diameter of the section $BC$is $d_{BC}$.

Write the expression for the shear stress at point (A) from torsional equation.

  ${\tau }_A=\frac{T{r}_A}{I_p}$...... (IX)

Here, the shear stress is ${\tau }_A$and the radius of the shaft is $r$.

Write the expression for the shear stress at point (A) from torsional equation.

  ${\tau }_B=\frac{T{r}_B}{I_p}$...... (X)

Here, the shear stress is ${\tau }_B$and the radius of the shaft is $r_B$.

Substitute $\frac{T_o}{6}$for $T$, $\frac{d_{AB}}{2}$for $r_A$and $\frac{\pi }{32}{d}_{AB}^4$for $I_p$in Equation (IX).

  $\begin{array}{c}{\tau }_A=\frac{\left(\frac{T_o}{6}\right)\left(\frac{d_{AB}}{2}\right)r}{\left(\frac{\pi }{32}{d}_{AB}^4\right)}\\ =\frac{8T_o}{3\pi d_{AB}^3\left(\frac{\pi }{32}{d}_{AB}^4\right)}\\ =\frac{2.67T_o}{\pi d_{AB}^3}\end{array}$

Substitute $\frac{T_o}{12}$for $T$, $\frac{d_{BC}}{2}$for $r_B$and $\frac{\pi }{32}{d}_{BC}^4$for $I_p$in Equation (X).

  $\begin{array}{c}{\tau }_B=\frac{\left(\frac{T_o}{12}\right)\left(\frac{d_{BC}}{2}\right)}{\left(\frac{\pi }{32}{d}_{BC}^4\right)}\\ =\frac{4T_o}{3\pi d_{BC}^3}\end{array}$...... (XI)

Substitute, $\left(\frac{1}{2}\right)\frac{\pi }{32}{d}_{AB}^4$for $I_p$in Equation (VIII)

  $\begin{array}{l}\left(\frac{1}{2}\right)\frac{\pi }{32}{d}_{AB}^4=\frac{\pi }{32}{d}_{BC}^4\\ \left(\frac{1}{2}\right)d_{AB}^4=d_{BC}^4\\ d_{BC}^{}={\left(\frac{1}{2}\right)}^{\frac{1}{4}}{d}_{AB}\end{array}$

Substitute ${\left(\frac{1}{2}\right)}^{\frac{1}{4}}{d}_{AB}$for $d_{BC}^{}$in Equation

  $\begin{array}{c}{\tau }_B=\frac{4T_o}{3\pi {\left({\left(\frac{1}{2}\right)}^{\frac{1}{4}}{d}_{AB}\right)}^3}\\ =\frac{4T_o}{3\pi \left({\left(\frac{1}{2}\right)}^{\frac{3}{4}}{d}_{{}_{AB}}^3\right)}\\ =\frac{2.242T_o}{\pi d_{AB}^3}\end{array}$

Conclusion:

The maximum stress is= $\frac{2.67T_o}{\pi d_{AB}^3}$.

The maximum stress in the bar is at point (A).

To determine

The torsional moment diagrams.

Explanation of Solution

Write the torsional moment at point (A)

  $R_A=-\frac{T_o}{6}$

Write the torsional moment at point (B).

  $R_B=\frac{T_o}{12}$

The following diagram shows the torsional moment diagram:

  

       Figure-(1)