Prob. 6.8
Assuming the material in a spherical rubber balloon can be
modeled as linearly elastic with modulus E and Poisson's
ratio ν = 0.5, show that the internal pressure p needed to
expand the balloon varies with the radial expansion ratio λ
= r/r0 as
pr0
1
1
= 2 − 3
4Eb0 λ r λ r
where b0 is the initial wall thickness. Plot this function
and determine its critical values.
The true stress as given by Eq. 6.1 is
pr
2b
Since the material is incompressible, the current wall thickness b is related to the original
thickness bo as
2
r
b
4πr 2 ⋅ b = 4πr02 ⋅ b0 ⇒ b = b0 0 = 20
r
λr
The stress is then
p r 2 p ro 3
σ=
λr =
λr
(1)
2 b0
2 b0
The strain is
2πr − 2πr0 r
εθ = εφ ≡ ε =
= − 1 ≡ λ r −1
(2)
2πr0
r0
If the material is linearly elastic, the strain and stress are related as
1
σ
σ
ε φ = σ φ − ν σ θ + σ r = 1 − 0.5 =
(3)
E
E
2E
Using (1) and (2) in (3):
1 p ro 3
pro
λ −1
λr −1 =
λr ⇒
= r3
2 E 2 b0
4Eb0
λr
pro
1
1
=
−
4Eb0 λ2r λ3r
Plot:
pstar:=1/lambda[r]^2 - 1/lambda[r]^3;
σθ = σφ ≡ σ =
F I
H K
b
pstar :=
plot(pstar,lambda[r]=1..5);
a
g
1
λr
2
−
1
λr
3
f
p
*
0.14
0.12
0.1
0.08
0.06
0.04
0.02
01
1
2
3
4
5
λr
Determine λr at maximum pressure:
'lambda[r,max]'=solve(diff(pstar,lambda[r])=0,lambda[r]);
λr, max =
3
2
The maximum normalized pressure is
Digits:=4;'pstar[max]'=evalf(subs(lambda[r]=3/2,pstar));
pstarmax = .1481
This maximum is commonly experienced as a yield-like phenomenon in blowing up a
balloon. However, its origin is geometrical and not a function of the material.
Prob. 6.9
Repeat the previous problem, but using the given
constitutive relation for rubber:
1
E 2
λx − 2 2
tσ x =
λ xλ y
3
The circumfrential extension ratio is:
2πr
λθ = λ φ =
= λr
2πr0
This is also both λx and λy in the given relation. From Eq. (1) of the previous solution we
can write
F
GH
σ=
I
JK
FG
H
p ro 3 E 2 1
λr =
λr − 4
λr
2 b0
3
FG
H
1 1
1
pro
=
− 7
4b0 E 6 λ r λ r
IJ
K
IJ
K
Plotting this along with the prevous result:
pstar1:= 1/lambda[r]^2 - 1/lambda[r]^3; pstar2:= (1/6)*(1/lambda[r] - 1/lambda[r]^7); plot({pstar1,pstar2},lambda[r]=1..5); 0.14
0.12
0.1
0.08
0.06
0.04
0.02
1
linear elastic
rubber elastic
2
3
4
5
λr
Extension at maximum pressure:
Digits:=4;'lambda[r,max]'=fsolve(diff(pstar2,lambda[r])=0,lambda[r]);
λr, max = -1.383
The maximum normalized pressure:
'pstar[max]'=evalf(subs(lambda[r]=1.383,pstar2));
pstarmax = .1033