Chapter 9, Problem 9.89RP

To determine

The principal stresses at any point located on the surface of the shaft.

Answer to Problem 9.89RP

The principal stresses at any point located on the surface of the shaft are $\underset{\_}{{\sigma }_1=10.7\text{MPa and}{\sigma }_2=-35.8\text{MPa}}$.

Explanation of Solution

Given information:

Rotation of the propeller is $\omega =15\text{rad}/\mathrm{s}$.

The engine develops the power of 900 kW.

Thrust on the shaft is $F=1.23\text{MN}$.

The outer diameter of the shaft is 250 mm.

Calculation:

Find the power transmission as shown below.

$T_0=\frac{P}{\omega }$

Here, P is the power transmission and $\omega$ is the rotation.

Substitute 900 kW for P and $15\text{rad}/\mathrm{s}$ for $\omega$.

$\begin{array}{c}{T}_0=\frac{900\text{kW}\times \frac{{10}^3\text{N}\cdot \text{m}/\text{s}}{1\text{kW}}}{15\text{rad}/\mathrm{s}}\\ =60\times {10}^3\text{N}\cdot \text{m}\end{array}$

Sketch the internal torque and forces for the propeller shaft as shown in Figure 1.

Find the section properties as shown below.

Find the cross sectional area as shown below.

Here, d is the diameter of the shaft.

Substitute 250 mm for d.

$\begin{array}{c}A=\frac{\pi {\left(250\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^2}{4}\\ =0.04909{\text{m}}^2\end{array}$

Find the polar moment of inertia as shown below.

$J=\frac{\pi d^4}{32}$

Substitute 250 mm for d.

$\begin{array}{c}J=\frac{\pi {\left(250\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}^4}{32}\\ =3.835\times {10}^{-4}{\text{m}}^4\end{array}$

Find the normal stress as shown below.

$\sigma =\frac{N}{A}$

Here, N is the normal force and A is the cross sectional area.

Substitute $-1.23\text{MN}$ for N and $0.04909{\text{m}}^2$ for A.

$\begin{array}{c}\sigma =-\frac{1.23\text{MN}\times \frac{{10}^6\text{N}}{1\text{MN}}}{0.04909{\text{m}}^2}\\ =-25.056\times {10}^6\text{N}/{\text{m}}^2\end{array}$

Convert the unit from $\text{N}/{\text{m}}^2$ to MPa.

$\begin{array}{c}\sigma =-25.056\times {10}^6\text{N}/{\text{m}}^2\times \frac{1\text{MPa}}{{10}^6\text{N}/{\text{m}}^2}\\ =-25.06\text{MPa}\end{array}$

Find the shear stress as shown below.

$\tau =\frac{Tc}{J}$

Here, T is the torque, J is the polar moment of inertia, and c is the centroidal distance between the neutral axis and the extreme fibre.

Substitute $60\times {10}^3\text{N}\cdot \text{m}$ for T, 0.125 m for c, and $3.835\times {10}^{-4}{\text{m}}^4$ for J.

$\begin{array}{c}\tau =\frac{60\times {10}^3\times 0.125}{3.835\times {10}^{-4}}\\ =19.557\times 10{}^6\text{N}/{\text{m}}^2\end{array}$

Convert the unit from $\text{N}/{\text{m}}^2$ to MPa.

$\begin{array}{c}\tau =19.557\times {10}^6\text{N}/{\text{m}}^2\times \frac{1\text{MPa}}{{10}^6\text{N}/{\text{m}}^2}\\ =19.56\text{MPa}\end{array}$

Sketch the state of stress as shown in Figure 2.

Refer to Figure 2.

$\begin{array}{l}{\sigma }_x=-25.06\text{MPa}\\ {\sigma }_y=0\\ {\tau }_{xy}=19.56\text{MPa}\end{array}$

Find the in-plane principal stresses as shown below.

${\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}$

Here, ${\sigma }_x$ is the normal stress along x direction, ${\sigma }_y$ is the normal stress along y direction, and ${\tau }_{xy}$ is the shear stress.

Substitute $-25.06\text{MPa}$ for ${\sigma }_x$, 0 for ${\sigma }_y$, and 19.56 MPa for ${\tau }_{xy}$.

$\begin{array}{c}{\sigma }_{1,2}=\frac{-25.06+0}{2}\pm \sqrt{{\left(\frac{-25.06+0}{2}\right)}^2+{19.56}^2}\\ =-12.53\pm \sqrt{539.5945}\\ =-12.53\pm 23.23\end{array}$

The principal stresses are,

$\begin{array}{c}{\sigma }_1=-12.53+23.23\\ =10.7\text{MPa}\\ {\sigma }_2=-12.53-23.23\\ =-35.76\text{MPa}\approx -35.8\text{MPa}\end{array}$

Therefore, the principal stresses at any point located on the surface of the shaft are $\underset{\_}{{\sigma }_1=10.7\text{MPa and}{\sigma }_2=-35.8\text{MPa}}$.