Difficult Toy Ball Analysis

Question

Note, as I have my degree in chemistry, I denote "=>" as the next step in the equation. Also, sorry for any obvious mistakes that I made - as I mentioned, my degree is not in engineering. Thanks!

I am considering a design for a PET spherical toy ball. The ball will have a radius of 0.07 m that will have pressurized air in it at a pressure of 5.81 atm (588698.3 Pa). Also, the ball will include a 35 nm coating of Reduced Graphene Oxide (RGO) as a durable coating on the outside of the ball.

To get the thickness of the walls of the ball, I know that the ultimate tensile strength of PET is 55 MPa. Because I want a 15% safety factor for the ball, I do the following calculation: $$(1-15\%)\cdot55 = 46.75\text{ MPa} = 46750000\text{ Pa}$$

I then use the reduced ultimate tensile strength to find the thickness of the wall using the formula for a thin-walled sphere where $p$ is pressure, $t$ is thickness, $r$ is radius, and $\sigma$ is stress:

$$\begin{gather} \sigma = \dfrac{pr}{2t} \\ 46750000 = \dfrac{588698.3 \cdot 0.07}{2t} \\ \therefore t = 4.41\times 10^{-4}\text{ m} \end{gather}$$

I can calculate the burst pressure of the sphere $P$ for pressure (gauge) inside sphere, $FS$ for factor of safety, $\sigma$ for allowable stress and additionally $R_i$ for inner radius, $R_o$ for outer radius:

$$P = \dfrac{(R_o^2-R_i^2)\sigma}{(R_i^2)FS}$$

For calculation of bursting pressure, we take $\sigma$ as ultimate stress for a given material and put $FS=1$.

Note here that I chose the express the ultimate tensile strength of PET as $5.5 \times 10^7$:

$$P = \dfrac{((0.07 + 4.41 \times 10^{-4})^2 - 0.07^2)\cdot 5.5 \times 10^7}{0.07^2 \times 1} = 695183.67\text{ Pa} = 6.86\text{ atm}$$

I now want to calculate the volume of PET needed to create a sphere of the desired dimensions:

$$\begin{align} V &= \dfrac{4}{3}\pi \left(\left(\dfrac{d_o}{2}\right)^3 - \left(\dfrac{d_i}{2}\right)^3\right) \\ &= \dfrac{4}{3}\pi \left(\left(\dfrac{0.140 + 4.41 \times 10^{-4}}{2}\right)^3 - \left(\dfrac{0.140}{2}\right)^3\right) \\ &= 1.36 \times 10^{-5}\text{ m}^3 \end{align}$$

I now want to calculate the volume of Reduced Graphene Oxide (RGO) to apply to the surface of the sphere:

$$\begin{align} V &= \dfrac{4}{3}\pi \left(\left(\dfrac{d_o}{2}\right)^3 - \left(\dfrac{d_i}{2}\right)^3\right) \\ &= \dfrac{4}{3}\pi \left(\left(\dfrac{0.140 + 4.41 \times 10^{-4} + 35 \times 10^{-9}}{2}\right)^3 - \left(\dfrac{0.140 + 4.41 \times 10^{-4}}{2}\right)^3\right) \\ &= 1.084366 \times 10^{-9}\text{ m}^3 \end{align}$$

Can you see any mistakes in my calculations?

Please do not ask why I am using PET as a material or other design decisions like that. There are good reasons for them.

Answer 1

I don't understand what you're trying to achieve, whether a 15% factor of safety is correct, whether having a factor of safety only on the material strength (instead of a combination of strength-minimizing and load-maximizing factors) or any such thing is adequate.

I also didn't bother checking your calculations (as in, punching numbers into a calculator). But just looking at your reasoning, it seems fine enough.

The only mistake I spotted was in the volume calculations, where you calculate the volume using diameters, but then add the wall thicknesses only once. If you're using diameters, you should add the thicknesses twice, once for each side of the ball.

Or just simplify your life and use radii instead (as you did for the rest of the question):

\begin{align} V &= \dfrac{4}{3}\pi \left(r_o^3 - r_i^3\right) \\ &= \dfrac{4}{3}\pi \left(\left(0.07 + 4.41 \times 10^{-4}\right)^3 - 0.07^3\right) \\ &= 2.73\times 10^{-5}\text{ m}^3\text{ of PET} \\ V &= \dfrac{4}{3}\pi \left(\left(0.07 + 4.41 \times 10^{-4} + 35\times 10^{−9} \right)^3 - \left(0.07 + 4.41 \times 10^{-4}\right)^3\right) \\ &= 2.18\times 10^{-9}\text{ m}^3\text{ of RGO} \end{align}

Note that I did not check that your calculations for the values needed for the above (such as 4.41x10^-4 m needed for the wall thickness) were correct.