Chapter 3, Problem 3.8.14P

The composite shaft shown in the figure is manufactured by shrink-Fitting a steel sleeve over a brass core so that the two parts act as a single solid bar in torsion. The outer diameters of the two parts are d_Y= 40 mm for the brass core and d₂= 50 mm for the steel sleeve. The shear moduli of elasticity are G_b= 36 GPa for the brass and G_s= 80 GPa for the steel.

(a)

Assuming that the allowable shear stresses

in the brass and steel are r_b= 48 MPa and

t_s= 80 MPa, respectively, determine the maximum permissible torque T_max that may be applied to the shaft.

(b)

If the applied torque T = 2500 kN · m, find

the required diameter d₂so that allowable shear

stress t₃is reached in the steel.

  

  

To determine

The maximum permissible torque applied to the shaft.

Answer to Problem 3.8.14P

The maximum permissible torque applied to the shaft is $1520.21\text{N}\cdot \text{m}$ .

Explanation of Solution

Given information:

The outer diameter of steel sleeve is $50\text{mm}$ , the outside diameter of the brass core is $40\text{mm}$ . is $2.25\text{in}$ , the shear modulus of the steel is $80\text{GPa}$ , the shear modulus for the brass is $36\text{GPa}$ , allowable shear stress in brass is $48\text{MPa}$ and allowable shear stress in steel is $80\text{MPa}$ .

Write the expression for total torque of the system.

  $T=T_b+T_s$ ...... (I)

Here, the total torque in the system is $T$ , torque in the brass core is $T_b$ and torque in the steel sleeve is $T_s$ .

Write the expression for polar moment of inertia of the brass core.

  $I_{pb}=\frac{\pi }{32}\left(d_1^4\right)$...... (I)

Here, the polar moment of inertia of brass core is $I_{Pb}$ , the diameter of the brass core is $d_1$ .

Write the expression for polar moment of inertia of steel sleeve.

  $I_{ps}=\frac{\pi }{32}\left(d_2^4-d_1^4\right)$...... (II)

Here, the polar moment of inertia of steel sleeve is $I_{Ps}$ , the diameter of the steel sleeve is $d_2$ .

Write the expression for torque on the brass core

  $T_b=T\left(\frac{G_b{I}_{pb}}{G_b{I}_{pb}+G_s{I}_{ps}}\right)$...... (III)

Here, the torque in the bar is $T_b$ , shear modulus of steel sleeve is $G_s$ , the shear modulus of brass core is $G_b$ and total twisted torque is $T$ .

Write the expression for torque in the steel sleeve.

  $T_s=T\left(\frac{G_s{I}_{ps}}{G_b{I}_{pb}+G_s{I}_{ps}}\right)$ ...... (IV)

Here, the torque in the bar is $T_s$ .

Write the expression for maximum shear stress in brass core.

  ${\tau }_b=\frac{T_b\left(\frac{d_1}{2}\right)}{I_{pb}}$...... (V)

Here, the maximum shear stress in the brass core is ${\tau }_b$ .

Write the expression for maximum shear stress in the steel sleeve.

  ${\tau }_s=\frac{T_s\left(\frac{d_2}{2}\right)}{I_{ps}}$...... (VI)

Here, the maximum shear stress in the bar is ${\tau }_s$ .

Calculation:

Substitute $40\text{mm}$ for $d_1$ in Equation (I).

  $\begin{array}{c}{I}_{pb}=\frac{\pi }{32}{\left(40\text{mm}\right)}^4\\ =\frac{\pi }{32}{\left(40\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^4\\ =\frac{\pi }{32}{\left(0.04\text{m}\right)}^4\\ =2.51\times {10}^{-7}{\text{m}}^{\text{4}}\end{array}$

Substitute $50\text{mm}$ for $d_2$ and $40\text{mm}$ for $d_1$in Equation (II).

  $\begin{array}{c}{I}_{ps}=\frac{\pi }{32}\left({\left(50\text{mm}\right)}^4-{\left(40\text{mm}\right)}^4\right)\\ =\frac{\pi }{32}\left({\left(50\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^4-{\left(40\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^4\right)\\ =\frac{\pi }{32}\left(6.25\times {10}^{-6}{\text{m}}^4-2.56\times {10}^{-6}{\text{m}}^4\right)\\ =3.62\times {10}^{-7}{\text{m}}^4\end{array}$

Substitute $3.62\times {10}^{-7}{\text{m}}^4$ for $I_{ps}$ , $2.51\times {10}^{-7}{\text{m}}^{\text{4}}$ for $I_{pb}$ , $80\text{GPa}$ for $G_s$ and $36\text{GPa}$ for $G_b$ in Equation (III).

  $\begin{array}{c}{T}_b=T\left(\frac{36\text{GPa}\left(2.51\times {10}^{-7}{\text{m}}^{\text{4}}\right)}{36\text{GPa}\left(2.51\times {10}^{-7}{\text{m}}^{\text{4}}\right)+80\text{GPa}\left(3.62\times {10}^{-7}{\text{m}}^4\right)}\right)\\ =T\left(\frac{36\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\left(2.51\times {10}^{-7}{\text{m}}^{\text{4}}\right)}{36\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\left(2.51\times {10}^{-7}{\text{m}}^{\text{4}}\right)+80\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\left(3.62\times {10}^{-7}{\text{m}}^4\right)}\right)\\ =T\left(\frac{9036\text{Pa}\cdot {\text{m}}^{\text{4}}}{9036\text{Pa}\cdot {\text{m}}^{\text{4}}+28960\text{Pa}\cdot {\text{m}}^{\text{4}}}\right)\\ =T\left(0.237\right)\end{array}$

Substitute $3.62\times {10}^{-7}{\text{m}}^4$ for $I_{ps}$ , $2.51\times {10}^{-7}{\text{m}}^{\text{4}}$ for $I_{pb}$ , $80\text{GPa}$ for $G_s$ and $36\text{GPa}$ for $G_b$ in Equation (IV).

  $\begin{array}{c}{T}_s=T\left(\frac{80\text{GPa}\left(3.62\times {10}^{-7}{\text{m}}^4\right)}{36\text{GPa}\left(2.51\times {10}^{-7}{\text{m}}^{\text{4}}\right)+80\text{GPa}\left(3.62\times {10}^{-7}{\text{m}}^4\right)}\right)\\ =T\left(\frac{80\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\left(3.62\times {10}^{-7}{\text{m}}^4\right)}{36\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\left(2.51\times {10}^{-7}{\text{m}}^{\text{4}}\right)+80\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\left(3.62\times {10}^{-7}{\text{m}}^4\right)}\right)\\ =T\left(\frac{28960\text{Pa}\cdot {\text{m}}^{\text{4}}}{9036\text{Pa}\cdot {\text{m}}^{\text{4}}+28960\text{Pa}\cdot {\text{m}}^{\text{4}}}\right)\\ =T\left(0.762\right)\end{array}$

Substitute $3.62\times {10}^{-7}{\text{m}}^4$ for $I_{ps}$ , $T\left(0.237\right)$ for $T_b$ , $48\text{MPa}$ for ${\tau }_b$ and $40\text{mm}$ for $d_1$ in Equation (V).

  $\begin{array}{l}48\text{MPa}=\frac{\left(T\left(0.237\right)\right)\left(\frac{\left(40\text{mm}\right)}{2}\right)}{2.51\times {10}^{-7}{\text{m}}^{\text{4}}}\\ 48\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)=\frac{\left(T\left(0.237\right)\right)\left(40\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}{2\left(2.51\times {10}^{-7}{\text{m}}^{\text{4}}\right)}\\ 48\times {10}^6\text{Pa}=\left(T\left(0.237\right)\right)\left(79681.2\right)\\ \frac{48\times {10}^6\text{Pa}}{\left(79681.2\right)\left(0.237\right)}=T\end{array}$

  $T=2541.77\text{N}\cdot \text{m}$

Substitute $3.62\times {10}^{-7}{\text{m}}^4$ for $I_{pb}$ , $T\left(0.762\right)$ for $T_s$ , $80\text{MPa}$ for ${\tau }_s$ and $50\text{mm}$ for $d_2$ in Equation (VI).

  $\begin{array}{l}80\text{MPa}=\frac{\left(T\left(0.762\right)\right)\left(\frac{\left(50\text{mm}\right)}{2}\right)}{3.62\times {10}^{-7}{\text{m}}^4}\\ 80\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)=\frac{\left(T\left(0.762\right)\right)\left(50\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}{2\left(3.62\times {10}^{-7}{\text{m}}^4\right)}\\ 80\times {10}^6\text{Pa}=\left(T\left(0.762\right)\right)\left(69060.77\right)\\ \frac{80\times {10}^6\text{Pa}}{\left(69060.77\right)\left(0.762\right)}=T\end{array}$

  $T=1520.21\text{N}\cdot \text{m}$

Conclusion:

The maximum permissible torque applied to the shaft is $1520.21\text{N}\cdot \text{m}$ .

To determine

The diameter of the steel sleeve.

Answer to Problem 3.8.14P

The diameter of the steel sleeve is $0.0569\text{m}$ .

Explanation of Solution

Given information:

The applied torque is $2500\text{kN}\cdot \text{m}$ .

Substitute, $40\text{mm}$ for $d_1$ in Equation (I)

  $\begin{array}{c}{I}_{ps}=\frac{\pi }{32}\left(d_2^4-{\left(40\text{mm}\right)}^4\right)\\ =\frac{\pi }{32}\left(d_2^4-{\left(40\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^4\right)\\ =\left(0.098\right)\left(d_2^4-2.56\times {10}^{-6}{\text{m}}^4\right)\end{array}$

Substitute $T\left(\frac{G_s{I}_{ps}}{G_b{I}_{pb}+G_s{I}_{ps}}\right)$ for $T_s$ in Equation (VI).

  $\begin{array}{c}{\tau }_s=\frac{\left(T\left(\frac{G_s{I}_{ps}}{G_b{I}_{pb}+G_s{I}_{ps}}\right)\right)\left(\frac{d_2}{2}\right)}{I_{ps}}\\ =\left(\left(\frac{G_{s}T{I}_{ps}}{G_b{I}_{pb}+G_s{I}_{ps}}\right)\right)\left(\frac{d_2}{2}\right)\left(\frac{1}{I_{ps}}\right)\\ =\frac{G_{s}T{d}_2}{2\left(G_b{I}_{pb}+G_s{I}_{ps}\right)}\end{array}$ ...... (VII)

Calculation:

Substitute $2500\text{kN}\cdot \text{m}$ for $T$ , $80\text{GPa}$ for $G_s$ , $36\text{GPa}$ for $G_b$ and $2.51\times {10}^{-7}{\text{m}}^{\text{4}}$ for $I_{pb}$ , $\left(0.098\right)\left(d_2^4-2.56\times {10}^{-6}{\text{m}}^4\right)$ for $I_{ps}$ and $80\text{MPa}$ for ${\tau }_s$ in Equation (VII).

$80\text{MPa}=\frac{\left(80\text{GPa}\right)\left(2500\text{kN}\cdot \text{m}\right)d_2}{2\left(\begin{array}{l}\left(36\text{GPa}\right)\left(2.51\times {10}^{-7}{\text{m}}^{\text{4}}\right)\\ +\left(36\text{GPa}\right)\left(\left(0.098\right)\left(d_2^4-2.56\times {10}^{-6}{\text{m}}^4\right)\right)\end{array}\right)}$

$80\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)=\frac{\left(80\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\right)\left(2500\text{N}\cdot \text{m}\right)d_2}{2\left(\begin{array}{l}\left(36\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\right)\left(2.51\times {10}^{-7}{\text{m}}^{\text{4}}\right)\\ +\left(80\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\right)\left(\left(0.098\right)\left(d_2^4-2.56\times {10}^{-6}{\text{m}}^4\right)\right)\end{array}\right)}$

$80\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)=\frac{\left(2\times {10}^{14}\text{N}\cdot \text{m}\right)d_2}{2\left(9036\text{Pa}\cdot {\text{m}}^{\text{4}}+\left(\begin{array}{l}80\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)\\ \times \left(\left(0.098\right)\left(d_2^4-2.56\times {10}^{-6}{\text{m}}^4\right)\right)\end{array}\right)\right)}$

$6283.2d_2^4-d_2-8.8467\times {10}^{-3}=0$

  $d_2=0.0569\text{m}$

Conclusion:

The diameter of the steel sleeve is $0.0569\text{m}$ .