How Does Inflation Affect the Thickness of a Plastic Ball?

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hatchelhoff
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Homework Statement



A plastic ball is inflated enough to produce tangential stresses. σX = σY = 2000Kpa
The radial thickness of the material is 1.2mm brfore inflation. Find the thickness after inflation if the tensile modulus of elasticity is 3.4Gpa and the shear modulus is 1.4Gpa.


Homework Equations


σ = Stress, ε= Strain, E = Modulus of Elasticity, G = Shear Modulus. v = Poisons Ratio

σ = F/A, ε = Elongation/original Length, σ = Modulus of Elasticity x ε
Elongation = (Force x Length) / (AE), v = -Lateral Strain/axial Strain
E = 2G(1 + v), ε1 = σ1/E - vσ2/E, σ1 = (E(ε1 + vε2))/(1-vxv)


The Attempt at a Solution


E = 2G(1 + v)
v = (E-2G)/2G
v = 0.2143

ε1 = σ1/E - vσ2/E
ε1 = 0.00046

ε = Elongation/original Length
Elongation = ε x original Length
= 0.00046 x 1.2
 

Discussion

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Everything you have matches what I have so far.

For your final answer, remember that since the ball is expanding, the cross-sectional area of the ball is also increasing. Based on Hooke's Law, normal stress, and normal strain equations, you can derive a mathematical relationship between thickness t and the cross-sectional area, A.

σ=εE, σ=P/A, and ε=(L-t_i)/t_i (t_i = initial thickness)

Therefore: [(L/t_1)-1]*E=P/A

Since there is an inverse relationship between t and A, and you know that A is increasing, we know that is decreasing. This means that for this problem, your final t value must be smaller than your initial t value.

Thus, t_final = t_i - L
 
hatchelhoff: (1) The unit symbol for kilopascal is spelled kPa, not Kpa. Uppercase K means kelvin, and lowercase p means pico. The unit symbol for gigapascal is spelled GPa, not Gpa. Always use correct capitalization of units[/color] and prefixes[/color].

(2) Always leave a space between a numeric value and its following unit symbol. E.g., 1.2 mm, not 1.2mm. See the international standard for writing units[/color] (ISO 31-0[/color]).

(3) Try to avoid using the letter "x" for the multiplication operator, because it cannot necessarily be distinguished from the variable x. Usually use an asterisk (*), instead.

(4) Two quantities multiplied together must be separated by an asterisk, middle dot (·), space, or parentheses. E.g., v*sigma2, not vsigma2.

(5) For exponentiation, use the caret (^) symbol. E.g., v^2, instead of v*v.

(6) Generally always maintain at least four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits, unless the first significant digit of the final answer is 1, in which case round the final answer to four significant digits.

(7) I did not understand post 2 yet, because L is undefined.

(8) hatchelhoff, I currently get a different answer from your current thickness answer in post 1. In post 1, it appears you are currently pretending there is strain in only one in-plane direction, which is not the case. I think you might be able to use generalized Hooke's law to derive the transverse (out-of-plane) strain. Give it a try.

(9) Also, I currently do not know why you are computing a transverse strain in post 1 with no Poisson effect. It does not seem to make sense yet.

(10) We are not allowed to give you the relevant equations for your homework, nor tell you how to solve your homework problem. We can only check your math. Keep trying.
 
Last edited:
I forgot to define L as elongation, or the change in thickness.
 
Thanks gOest. I agree with your logic that the Thickness must be less as the pressure increases. But according to the book I am using the thickness after inflation is 1.9 mm
 
Thanks very much for your comments nvn
 
hatchelhoff: If the answer in the book is 1.9 mm, then the book answer is wrong. Hint 1: Use generalized Hooke's law to compute eps1 and eps2. Hint 2: Using generalized Hooke's law again, add the three equations of generalized Hooke's law together, to derive eps3.

By the way, even though the general rule in item 6 of post 3 is to maintain at least four significant digits, strains are an exception. Because strains are so small, I usually maintain at least six significant digits throughout all strain calculations, then round the final answer to four or five significant digits.
 
Last edited:
What do you mean by eps
 
I have now calculated [itex]\epsilon[/itex]1 and [itex]\epsilon[/itex]2 and [itex]\epsilon[/itex]3 using the following equations
[itex]\epsilon[/itex]1 = ([itex]\sigma[/itex]1/E) -(v.[itex]\sigma[/itex]2/E)

[itex]\epsilon[/itex]2 = ([itex]\sigma[/itex]2/E) -(v.[itex]\sigma[/itex]1/E)

[itex]\epsilon[/itex]3 = v.([itex]\sigma[/itex]1/E) - v.([itex]\sigma[/itex]2/E)
There fore
[itex]\epsilon[/itex]1 = 4.6*10^-4
[itex]\epsilon[/itex]2 = 4.6*10^-4
[itex]\epsilon[/itex]3 = -7.14*10^-4

I am not sure what the next step is.
 
hatchelhoff: Are you missing a negative sign in your third equation? Try again. Ignore hint 2 in post 7; although hint 2 is not wrong, it does not make sense here, and is not needed. Please see item 6 in post 3 (and see the second paragraph of post 7). Afterwards, the next step is to compute the final thickness of the plastic.
 
Thanks nvn. I was missing a Minus sign in the equation for ε3
the new value is
ε3 = -2.52116*10^-4.

Im not sure how to deal with these strains which I have calculated.
Can I just simply add them together to find a total strain, and feed this total into a an equation
to find the thickness.
 
Thanks nvn. Why did you choose ε1 as opposed to ε2 or ε3?
 
Sorry, I ment to say eps 3
 
Thanks nvn
Here is my final solution
ε = L*ti/ti
L = ti + ti*ε
L = 1.2 mm + 1.2 mm * -2.52116 *10^-4
L= 1.1997 mm
 
Thats great. Thanks very much nvn
 

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