# Determine the strain energy of a bar

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.

• 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. Commented Mar 8, 2018 at 19:09
• Hi, Thank you so much for your reply! Here is my new work. Unfortunately I am still not getting the correct answer. Commented Mar 8, 2018 at 20:28
• U = integral from 0 to L: (((y * V)^2) / (2 * E * A)) * dx Commented Mar 8, 2018 at 20:30
• V = (1/3) * pi * (r^2) Commented Mar 8, 2018 at 20:30
• A = (pi * (r^2)) Commented Mar 8, 2018 at 20:30

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 $L$.