##### I am assuming you have read Stress and Strain Part-1 and have a knowledge of basic definitions and terms, If not please click the link above and have a look at it.

#### In exams, we may directly have to use the formula for stress in a bar or any cross-section due to the rotation. To have a better understanding let us understand the derivation.

#### So that we can never forget how we came to that expression.

**Stress in a bar due to rotation**

#### Let us assume a bar of length *l* revolving about Y-axis at an angular speed of ω rad/s as can be seen in the figure

*l*

#### In the figure above consider a small strip dx at a distance x from the center. As the bar is rotating along axis Y-Y’ there will be a tensile force generated in the element AB or dx which will be given by the expression

##### Force on element AB (dx) = Centrifugal force Fc on the part BC.

First let us take some assumptions:

Mass of part BC = M

Area of Cross-section = Ac

Density of material = ρ

as shown in the figure distance of centre of Mass BC from Y-Y’ axis = r

With these assumptions let’s proceed

#### Centrifugal force Fc on the part BC, Fc = **ω²rM ** ————–**( 1 )**

We all know that Mass = ρ x Volume

Now we will put the values of equation (2) , (3) in equation (1) , we get

This expression is very important because in exams if there is a question of rotating beams or bars, you can directly use this expression to calculate stress at some general point.

**Let us see some special cases, which might be asked in the exams.**

We have already seen in figure that x is the distance from rotational axis.