Register to reply 
Hooke's law for a balloon? 
Share this thread: 
#1
Feb913, 09:51 AM

P: 373

Hi,
For a linear spring one has Hooke's law: [itex] F = k x [/itex] Is there an equivalent law for a spherical elastic balloon giving pressure inside as a function of radius? Thanks, John 


#2
Feb913, 10:52 AM

P: 505

Here's a good place to start
http://en.wikipedia.org/wiki/Bulk_modulus Bulk modulus would describe how the pressure of the balloon reacts to the additional pressure over atmosphere that the balloon exerts. However, I don't think it will tell you the actual pressure the balloon exerts. 


#3
Feb913, 08:00 PM

Sci Advisor
P: 2,210

I tried to see if I could derive such a law, and I got a result that is a little counterintuitive, so it's possible I made a mistake somewhere.
Here's the idea: Think of a little rectangle made of balloon material. It has a length and a width. If you stretch it to increase its length, it's just like stretching a spring. So it's going to require an amount energy [itex]\frac{1}{2} k L^2[/itex] to stretch it to length [itex]L[/itex], where [itex]k[/itex] is some spring constant that depends on the balloon and possibly how big the rectangle is. (That's actually not quite right; it really should be something like [itex]\frac{1}{2} k (LL_0)^2[/itex], where [itex]L_0[/itex] is the equilibrium length. I'm going to simplify things by taking the limit as [itex]L_0 \rightarrow 0[/itex]). Similiarly, stretching it to increase its width will require an amount of energy proportional to [itex]\frac{1}{2} k' W^2[/itex] (again ignoring the equilibrium width). So the total energy associated with a rectangle of length [itex]L[/itex] and width [itex]W[/itex] will be just [itex]\frac{1}{2} (k L^2 + k' W^2)[/itex]. If we assume that it's a square, so that [itex]L = W[/itex] and [itex]k = k'[/itex], then we get the total energy of a square with side [itex]L[/itex] will be [itex]k L^2[/itex]. That can be rewritten as [itex]k A[/itex], where [itex]A[/itex] is the area of the little square of balloon. Assuming that there is this same amount of energy stored in every little square of the balloon, we conclude that the same formula applies for the entire balloon: [itex]E \propto A \propto r^2[/itex] where I used [itex]A = 4 \pi r^2[/itex] for a spherical balloon. I'm using [itex]\propto[/itex] to mean "proportional to", which means I'm not keeping track of constants like [itex]\pi[/itex] and [itex]k[/itex], etc. So that's a pretty sensible result, it seems to me, the energy due to stretching for a balloon is proportional to the square of the radius. We can rewrite the above formula for [itex]E[/itex] to get it in terms of volume, using [itex]V \propto r^3[/itex] [itex]E \propto V^\frac{2}{3}[/itex] That's still sensible: The more you increase the volume of the balloon, the more energy is required. But here comes the counterintuitive part. The relationship between pressure (the force per unit area required to keep the balloon expanded) and energy is given by: [itex]P = \dfrac{dE}{dV}[/itex] Using our formula for [itex]E[/itex], we get: [itex]P \propto V^\frac{1}{3}= \dfrac{1}{V^\frac{1}{3}}[/itex] There's the counterintuitive part. As you expand the balloon more and more, the pressure needed to keep it expanded goes DOWN. I feel like that can't possibly be right. On the other hand, it is true from experience that when blowing up a balloon, it's hardest when the balloon is small, and gets easier when the balloon is bigger. But I always assumed that was a failure of Hooke's law, that the forces holding the balloon together got weaker as it stretched, instead of increasing linearly. Anyway, assuming that this analysis is right, the conclusion for the balloon version of Hooke's law would be (converting back from volume to radius): [itex]P = K/r[/itex] The pressure required to inflate a balloon to radius [itex]r[/itex] is inversely proportional to [itex]r[/itex]. 


#4
Feb1013, 12:30 AM

Mentor
P: 5,370

Hooke's law for a balloon?
A balloon is a rubber sheet that is undergoing large deformation biaxial stretching, and its "stressstrain behavior" is nonlinear. One of the key characteristics of rubber is that it is virtually incompressible, so the volume of the rubber sheet remains constant. If the balloon were a perfect sphere, the surface area of the rubber sheet would be [itex]4\pi r^2[/itex] and its thickness (assumed uniform) would be h, so its volume would be [itex]4\pi r^2h[/itex].
Initially, you would need a small (virtually insignificant) amount of initial pressure to snap the balloon into its initial spherical shape. If the initial radius was r_{0} and the initial thickness of the rubber was h_{0}, then the initial volume of the balloon rubber would be [itex]4\pi r_0^2h_0[/itex]. But, since rubber is incompressible, the initial and final volumes of the rubber would have to be the same, so that [tex]\frac{h}{h_0}=(\frac{r_0}{r})^2[/tex]So, as the balloon inflates, the thickness of the rubber decreases as the square of the radius. The amount that the balloon rubber stretches can be characterized by the biaxial stretch ratio. The rubber sheet surface stretches equally in all directions (for a sphere), and the distance along a great circle between any two points on the sphere increases in proportion to the ratio of the present radius to the initial radius. This ratio is called the stretch ratio λ:[tex]\lambda=\frac{r}{r_0}[/tex] So, in terms of the stretch ratio, the rubber thickness h is given by:[tex]h=\frac{h_0}{\lambda^2}[/tex] In general, rubber is a very nonlinear elastic material, and the tensile stress within the sheet σ (force per unit area) will be a nonlinear function of the stretch ratio λ: [tex]\sigma=\sigma(\lambda)[/tex] The next step in this development is to do an equilibrium force balance so that the stress can be expressed in terms of the pressure difference between the inside and outside of the balloon, the balloon sheet thickness h, and the balloon radius r. If you do an equilibrium force balance on the balloon, you will find that the biaxial tensile stress in the balloon rubber σ is related to the balloon radius r, the rubber thickness h, and the difference in pressure between inside and outside the balloon (p_{in}  p_{out)} by: [tex]\sigma(\lambda)=\frac{r(p_{in}p_{out})}{2h}[/tex] If we combine this equation with the equations I presented previously in the development, we get: [tex]\frac{\sigma(\lambda)}{{\lambda} ^3}=\frac{r_0(p_{in}p_{out})}{2h_0}[/tex] The functional relationship σ(λ) between the tensile stress σ and the stretch ratio λ is unique to the particular rubber comprising the balloon, and is independent of the geometry of the specific system under consideration ( as characterized by r_{0} and h_{0}). For this reason, σ(λ) is referred to as a material function for the particular rubber. Since the entire left hand side of the above equation is a function only of λ, it too is a material function for the rubber, now designated by [itex]\hat{\sigma}(\lambda)[/itex]: [tex]\hat{\sigma}(\lambda)=\frac{r_0(p_{in}p_{out})}{2h_0}[/tex] If the functional relationship [itex]\hat{\sigma}(\lambda)[/itex]between the stress parameter [itex]\hat{\sigma}[/itex] and the stretch ratio λ were known in advance, then we could use the above equation to predict, for any arbitrary balloon geometry (r_{0} and h_{0}), the relationship between the pressure difference [itex](p_{in}p_{out})[/itex] and the inflated balloon radius r. Alternately, we could, for a specific balloon, experimentally measure the right hand side of the equation as a function of the measured stretch ratio λ, and thereby determine the functional relationship [itex]\hat{\sigma}(\lambda)[/itex] experimentally. We could then use that relationship for all other balloons of different r_{0} and h_{0} involving the same rubber to predict its inflation behavior. We could also determine the required functionality by doing experiments on flat sheets of rubber. Chet 


#5
Feb1013, 07:11 AM

Sci Advisor
P: 2,210




#6
Feb1013, 08:21 AM

Mentor
P: 5,370

If you are just looking for a linear relationship for a balloon that applies in the limit of very small perturbations to an initially "uninflated" balloon, just take σ(λ) = E(λ1) (where E is an elastic material constant), and λ≈1 in the balloon equation I presented. However, what you end up with will not be very satisfying because the actual equation is very nonlinear, and the analysis will thus not extend to anything realistic. Chet 


#7
Feb1013, 09:14 AM

Sci Advisor
P: 2,210




#8
Feb1013, 11:57 AM

Mentor
P: 5,370

An important difference between Hooke's law for a metal and the corresponding deformation law for a rubber is that, even though the rubber behaves nonlinearly, it remains elastic up to much larger deformations.
In the case of an actual spring, because of the unique helical geometry of the spring, the strains in the metal are small, even though the overall change in length of the spring divided by its original axial length can be large. Each little increment in coiled metal twists a small amount as the overall spring is extended axially. But none of the strains in the metal are large enough to cause plastic deformation (unless you try to stretch the spring too far). 


Register to reply 
Related Discussions  
Does a balloon's temperature increase when a balloon is popped?  Introductory Physics Homework  4  
Balloon within a balloon question  Astronomy & Astrophysics  4  
Hooke's law  Classical Physics  2  
Hooke's law, again  General Physics  3  
Air Balloon ( 5.5 inch diameter) vs Helium gas balloon ( 11 inch diameter )?  Introductory Physics Homework  1 