Stresses in a bar due to rotation (Stress and Strain Part-2)

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

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.

Now let us see the expression for the elongation of element dx.

Stress = E* Strain

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