Chapter 19, Problem 19P

An aluminum cylinder (E_a = 70 MPa, v_a = 0.33) with an outer diameter of 150 mm and inner diameter of 100 mm is to be press-fitted over a stainless-steel cylinder (E_s = 190 MPa, v_s = 0.30) with an outer diameter of 100.20 mm and inner diameter of 50 mm. Determine (a) the interface pressure p and (b) the maximum tangential stresses in the cylinders.

To determine

The pressure at the interface.

Answer to Problem 19P

The pressure at the interface is $0.029\text{MPa}$.

Explanation of Solution

Write the expression for the radial interface.

    $dr=\left({\left(r_0\right)}_s-{\left(r_i\right)}_a\right)$                                                                              (I)

Here, the outer radius of steel cylinder is ${\left(r_0\right)}_s$, the inner radius of the aluminum cylinder is ${\left(r_i\right)}_a$,and the radial interface is $dr$.

Write the expression for the interface pressure.

    $P=\frac{dr}{\left(\left(\frac{{\left(r_i\right)}_s}{E_s}\right)\left(\frac{{\left(r_0\right)}_s^2+{\left(r_i\right)}_s^2}{{\left(r_0\right)}_s^2-{\left(r_i\right)}_s^2}\right)\right)+\left(\left(\frac{{\left(r_0\right)}_a}{E_a}\right)\left(\frac{{\left(r_0\right)}_a^2+{\left(r_i\right)}_a^2}{{\left(r_0\right)}_a^2-{\left(r_i\right)}_a^2}\right)\right)}$                  (II)

Here, the inner radius of steel cylinder is ${\left(r_i\right)}_s$, the inner radius of aluminum cylinder is ${\left(r_i\right)}_a$, the modulus of elasticity of steel is $E_s$ and the modulus of elasticity of aluminum cylinder is $E_a$.

Conclusion:

Substitute $50.1\text{mm}$ for ${\left(r_0\right)}_s$ and $50\text{mm}$ for ${\left(r_i\right)}_a$ in Equation (I).

    $\begin{array}{c}dr=50.1\text{mm}-50\text{mm}\\ =\text{0}\text{.1}\text{mm}\end{array}$

Substitute $50.1\text{mm}$ for ${\left(r_0\right)}_s$ and $50\text{mm}$ for ${\left(r_i\right)}_a$, $0.1\text{mm}$ for $dr$, $70\text{MPa}$ for $E_a$, $190\text{MPa}$ for $E_s$, $75\text{mm}$ for ${\left(r_0\right)}_a$ and $25\text{mm}$ for ${\left(r_i\right)}_s$ in Equation (II).

    $\begin{array}{c}P=\left[\frac{0.1\text{mm}}{\begin{array}{l}\left(\left(\frac{25\text{mm}}{190\text{MPa}}\right)\left(\frac{{\left(50.1\text{mm}\right)}^2+{\left(25\text{mm}\right)}^2}{{\left(50.1\text{mm}\right)}^2-{\left(25\text{mm}\right)}^2}\right)\right)\\ +\left(\left(\frac{75\text{mm}}{70\text{MPa}}\right)\left(\frac{{\left(75\text{mm}\right)}^2+{\left(50\text{mm}\right)}^2}{{\left(75\text{mm}\right)}^2-{\left(50\text{mm}\right)}^2}\right)\right)\end{array}}\right]\\ =\frac{0.1\text{mm}}{\left(\left(\frac{25\text{mm}}{190\text{MPa}}\right)\left(1.663{\text{mm}}^2\right)\right)+\left(\left(\frac{75\text{mm}}{70\text{MPa}}\right)\left(2.6{\text{mm}}^2\right)\right)}\\ =0.029\text{MPa}\end{array}$

Thus the pressure at the interface is $0.029\text{MPa}$.

To determine

The maximum tangential stress in aluminum cylinder.

The maximum tangential stress in steel cylinder.

Answer to Problem 19P

The maximum tangential stress in aluminum cylinder is $0.0754\text{MPa}$.

The maximum tangential stress in steel cylinder is $0.0482\text{MPa}$.

Explanation of Solution

Write the expression for maximum stress in aluminum cylinder.

    ${\sigma }_a=P\left(\frac{{\left(r_0\right)}_a^2+{\left(r_i\right)}_a^2}{{\left(r_0\right)}_a^2-{\left(r_i\right)}_a^2}\right)$                                 (III)

Here, the pressure at interface is $P$, and the maximum stress in aluminum cylinder is ${\sigma }_a$.

Write the expression for maximum stress in steel cylinder.

    ${\sigma }_s=P\left(\frac{{\left(r_0\right)}_s^2+{\left(r_i\right)}_s^2}{{\left(r_0\right)}_s^2-{\left(r_i\right)}_s^2}\right)$                                   (IV)

Here, the maximum stress in steel cylinder is ${\sigma }_s$.

Conclusion:

Substitute $75\text{mm}$ for ${\left(r_0\right)}_a$, $50\text{mm}$ for ${\left(r_i\right)}_a$, and $0.029\text{MPa}$ for $P$ in Equation (III).

    $\begin{array}{c}{\sigma }_a=\left(0.029\text{MPa}\right)\times \left(\frac{{\left(75\text{mm}\right)}^2+{\left(50\text{mm}\right)}^2}{{\left(75\text{mm}\right)}^2-{\left(50\text{mm}\right)}^2}\right)\\ =\left(0.029\text{MPa}\right)\times 2.6\\ =0.0754\text{MPa}\end{array}$

Thus, the maximum tangential stress in aluminum cylinder is $0.0754\text{MPa}$.

Substitute $25\text{mm}$ for ${\left(r_i\right)}_s$, $50.1\text{mm}$ for ${\left(r_0\right)}_a$, and $0.029\text{MPa}$ for $P$ in Equation (IV).

    $\begin{array}{c}{\sigma }_a=\left(0.029\text{MPa}\right)\times \left(\frac{{\left(50.1\text{mm}\right)}^2+{\left(25\text{mm}\right)}^2}{{\left(50.1\text{mm}\right)}^2-{\left(25\text{mm}\right)}^2}\right)\\ =\left(0.029\text{MPa}\right)\times 1.663\\ =0.0482\text{MPa}\end{array}$

Thus, the maximum tangential stress in steel cylinder is $0.0482\text{MPa}$.