Chapter 3, Problem 1P

A tensile member in a machine structure is subjected to a steady load of 4.50 kN. It has a length of 750 mm and is made from a steel tube having an outside diameter of 18 mm and an inside diameter of 12 mm. Compute the tensile stress in the tube and the axial deformation.

To determine

The value of tensile stress in the tube.

The axial deformation in the tube.

Answer to Problem 1P

The value of stress in a tube is, $\sigma =31.83MPa$ .

The axial deformation in the tube is, $\delta =0.12mm$ .

Explanation of Solution

Given:

Length of the tube, $L=750mm$

Inside diameter of the tube, $d_i=18mm$

Outside diameter of the tube, $d_o=12mm$

Steady load in the tensile member, $F=4.50KN$

Concept used:

Stress- It is the applied force per unit area.

  $\sigma =\frac{Force\left(F\right)}{Area\left(A\right)}$

Unit of the stress is N/m².

The axial deformation in the tube is, $\delta =\frac{\sigma \times L}{E}$

Where,

  $\delta$ is the axial deformation in the tube.

  $L$ is the length of the tube..

  $E$ is the young’s modulus,

  $\begin{array}{c}E=200GPa\\ =200\frac{N}{m{m}^2}\end{array}$ (For steel).

Calculation:

The cross sectional area of the tube is calculated as, $A=\frac{\pi }{4}\left(D^2-d^2\right)$ .

The cross sectional area of the tube is,

  $\begin{array}{c}A=\frac{\pi }{4}\left(D^2-d^2\right)\\ =\frac{\pi }{4}\left[{\left(18\right)}^2-{\left(12\right)}^2\right]m{m}^2\\ =\frac{3.142}{4}\times \left[324-144\right]m{m}^2\\ =141.39m{m}^2\end{array}$

Now the value of tensile stress is calculated as:

  $\begin{array}{c}Stress\left(\sigma \right)=\frac{Force\left(F\right)}{Area\left(A\right)}\\ =\frac{4.5\times {10}^{3}N}{141.39m{m}^2}\\ =31.83\frac{N}{m{m}^2}\end{array}$

As, $1\frac{N}{m{m}^2}$ is converted as $1MPa$ .

Thus, $31.83\frac{N}{m{m}^2}$ is converted as:

  $\begin{array}{c}\sigma =431.83\frac{N}{m{m}^2}\\ =31.83MPa\end{array}$

Hence, the value of stress in a tube is, $\sigma =31.83MPa$ .

Now calculate the axial deformation in the tube. $\delta =\frac{\sigma \times L}{E}$

The axial deformation in the tube is calculated as,

  $\begin{array}{l}\delta =\frac{\sigma \times L}{E}\\ =\frac{31.83\frac{N}{m{m}^2}\times 750mm}{200\times {10}^3\frac{N}{m{m}^2}}\\ =0.12mm\end{array}$

Hence, the axial deformation in the tube is, $\delta =0.12mm$ .