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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).

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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 $ACB$, $U_{total} = 2U = \frac{2\rho^2 A \omega^4 L^5}{15E}$.

Hope this answers your question!

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  • $\begingroup$ Centrifugal force should be zero at the center isn't it? $\endgroup$ – huministic Apr 29 at 10:30
  • $\begingroup$ No, the centrifugal force is maximum at the center. This is due to the entire mass pulling outwards at the center. Centrifugal force is zero at the tip. $\endgroup$ – G R Krishna Chand Avatar Apr 29 at 10:32
  • $\begingroup$ Ok, so had it been it had asked for centripetal effect then what will be the integration limits for dF, 0 to x? If so in that case result will be different. $\endgroup$ – huministic Apr 29 at 10:47
  • $\begingroup$ The integration limit would be from x to L to find the centrifugal force at x. This is because the mass from the location x to the tip is pulling it outwards and thus generating centrifugal force. Hope it helps you. $\endgroup$ – G R Krishna Chand Avatar Apr 29 at 10:50

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