I have a Structures question about determining strain energy that I am unable to solve. I have attempted it and I have the answer, I would appreciate if if someone can just tell me where I am going wrong.

Q: A solid conical bar of circular cross-section is suspended vertically. The length of the bar is $L$, the diameter at the base is $D$ and the weight per unit volume is $\gamma$ (equivalent to density x gravity). Determine the strain energy of the bar due to its own weight.

$$\text{Answer} = U = \dfrac{\pi D^2 \gamma^2 L^3}{360E}$$

My work is as follows:

$$\begin{align} U &= \dfrac{\gamma((1/3)\pi r^2 L)^2)L}{2E\pi r^2} \\ U &= \dfrac{\gamma(\pi^2r^4(1/9)L^2)L}{2E\pi r^2} \\ U &= \dfrac{\gamma\pi r^2L^3}{9\cdot2E} \\ U &= \dfrac{\gamma\pi(d^2/4)L^3}{18E} \\ U &= \dfrac{\gamma\pi d^2L^3}{18\cdot4E} \\ U &= \dfrac{\pi d^2 \gamma L^3}{72E} \\ \end{align}$$

I am incorrect by a factor of 1/5. Where in my working/logic did I go wrong?

I know:

$$U = \dfrac{(\rho gAL)^2L}{2EA}$$

I assume:

$$AL = \text{Volume (of a cone)} = (1/3)\pi r^2 h$$

I am confused as to whether the A in the denominator is the base of the cone or the volume divided by the height.

I can derive the correct numerator, however I am not sure how my lecturer arrived at 360E for the denominator.

  • $\begingroup$ This is an integration problem. The bar tapers to a cross-sectional area of zero at the top, and the volumetric strain energy changes throughout the length. Find the volumetric, differential strain energy in a small horizontal sliver of the bar with thickness dz (this will depend on the cross-section area at that point and the weight of the remaining bar underneath) and integrate that differential strain energy over the length of the bar. $\endgroup$ – Chemomechanics Mar 8 '18 at 19:09
  • $\begingroup$ Hi, Thank you so much for your reply! Here is my new work. Unfortunately I am still not getting the correct answer. $\endgroup$ – gunter Mar 8 '18 at 20:28
  • $\begingroup$ U = integral from 0 to L: (((y * V)^2) / (2 * E * A)) * dx $\endgroup$ – gunter Mar 8 '18 at 20:30
  • $\begingroup$ V = (1/3) * pi * (r^2) $\endgroup$ – gunter Mar 8 '18 at 20:30
  • $\begingroup$ A = (pi * (r^2)) $\endgroup$ – gunter Mar 8 '18 at 20:30

We assume that the wide end of the cone is up, of course—otherwise the stress would go to infinity as the full weight of the cone is applied to a cross-sectional area that tapers to zero. (I incorrectly specified the opposite configuration in my comment above.)

Measure the coordinate $z$ from the bottom (i.e., from the tip of the cone). Then, the diameter at any location $z$ is $$D(z)=\frac{Dz}{L}.$$ Therefore, the corresponding area is $$A(z)=\frac{\pi D(z)^2}{4}.$$ The volume hanging underneath location $z$ is $$V(z)=\frac{\pi D(z)^2z}{12},$$ with corresponding weight $W=\gamma V(z)$ applied across the cross section. Therefore, the stress on any horizontal infinitesimal slice of thickness $dz$ at location $z$ is $$\sigma(z)=\frac{W}{A(z)},$$ and the volumetric strain energy within that slice is thus $$dU=\left(\frac{\sigma(z)^2}{2E}\right)A(z)dz,$$ where I've applied the linear elasticity result that the differential strain energy per unit volume can always be expressed as $du=\sigma\,d\epsilon=\frac{\sigma}{E}d\sigma$ by Hooke's Law, or, integrated, $u=\frac{\sigma^2}{2E}$. Now, integrate $dU$ from $z=1$ to $L$.

  • $\begingroup$ Thank you so much for your assistance, you really helped me. $\endgroup$ – gunter Mar 9 '18 at 6:20

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