# What will the strain energy stored in the bar due to centrifugal effects?

I having trouble with this problem for a while now. Each time I land with same solution but it's not matching with the given answer. My answer: (6 ρ^2 Α L^5 ω^4)/(5Ε) Given answer: (2 ρ^2 Α L^5 ω^4)/(15Ε).

Let us calculate the strain energy for a portion $$AC$$ of the bar $$ACB$$.

For convenience, we set the origin of the reference frame at the axis of rotation $$C$$. Consider an infinitesimal mass element $$dx$$ of the bar at a distance $$x$$ from the origin. The centrifugal force acting on the mass element $$dx$$ located at $$x$$ is: $$dF = -\rho A \omega^2 x dx$$

Also, the infinitesimal (tensile) centrifugal stress due to element $$dx$$ at that location is: $$d\sigma = \frac{-dF}{A} = \rho \omega^2 x dx$$. (sign should be postive for tensile stress)

Now, integrate the expression for $$dF$$ from $$x$$ to $$L$$ to calculate the centrifugal force at the cross-section at $$x$$: \begin{align} \int_{x}^{L} dF &= \int_{x}^L \rho A \omega^2 x dx \end{align} Centrifugal force acting at the tip is zero, i.e,$$F(x=L) = 0$$. So the above equation evaluates to: $$F(x) = \rho A \omega^2\frac{(L^2 - x^2)}{2}$$ The centrifugal stress at the cross-section is : $$\sigma = \frac{F}{A} = \rho \omega^2\frac{(L^2 - x^2)}{2}$$

From strain-displacement relation: $$d\epsilon = \frac{u}{x}$$, where $$u$$ is the extension due to the centrifugal force for a segment of length $$x$$ from the origin. Now, $$u = x~d\epsilon$$. From Hooke's law, $$\sigma = E\epsilon~\implies d\sigma = Ed\epsilon$$, or $$d\epsilon = \frac{d\sigma}{E}$$; $$E$$ is Young's modulus. Substituting this in the the expression for $$u$$: $$u = x\frac{d\sigma}{E}$$

Let us now calculate the strain energy for the bar $$AC$$.

Infinitesimal strain energy stored in the element $$dx$$ is equal to the centrifugal force at a location $$x$$ multiplied by its corresponding extension $$u$$, i.e., $$dU = Fu = Fx\frac{d\sigma}{E}$$.

Substituting the expression $$\sigma$$ and $$F$$ in the above and integrating from $$x = 0$$ to $$L$$:

\begin{align} \int_0^U dU &= \int_0^L \rho A \omega^2\frac{(L^2 - x^2)}{2} x\frac{\rho \omega^2 x dx}{E} \\ U &= \frac{\rho^2 A \omega^4}{2E} \int_0^L (L^2x^2 - x^4) dx \\ \end{align}

On evaluating the above expression, $$U = \frac{\rho^2 A \omega^4 L^5}{15E}$$.

Strain energy for the complete bar $$ACB$$

Total strain energy stored in the bar $$ACB$$, $$U_{total} = 2U = \frac{2\rho^2 A \omega^4 L^5}{15E}$$.