a f b g 1 = + = = 1 0 5 . (3) r - - PDF document

• 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 pr Eb

r r 2 3

4 1 1 = − λ λ 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 σ σ σ

θ φ

= ≡ = pr b 2 Since the material is incompressible, the current wall thickness b is related to the original thickness bo as 4 4

2 2 2 2

π π λ r b r b b b r r b

r

⋅ = ⋅ ⇒ = F

H I K =

The stress is then σ λ λ = = p r b p r b

r

• r

2 2

2 3

(1) The strain is ε ε ε π π π λ

θ φ

= ≡ = − = − ≡ − 2 2 2 1 1 r r r r r

r

(2) If the material is linearly elastic, the strain and stress are related as ε σ ν σ σ σ σ

φ φ θ

= − + = − = 1 1 0 5 2 E E E

r

b g a f

. (3) Using (1) and (2) in (3): λ λ λ λ

r

• r
• r

r

E p r b pr Eb − = ⇒ = − 1 1 2 2 4 1

3 3

pr Eb

• r

r

4 1 1

2 3

= − λ λ Plot: pstar:=1/lambda[r]^2 - 1/lambda[r]^3; := pstar − 1 λr

2

1 λr

3

plot(pstar,lambda[r]=1..5);

λr 5 4 3 2 1 0.14 0.12 0.1 0.08 0.06 0.04 0.02 1 p

*

Determine λr at maximum pressure: 'lambda[r,max]'=solve(diff(pstar,lambda[r])=0,lambda[r]); = λ ,

r 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:

t x x x y

E σ λ λ λ = −

F H G I K J

3 1

2 2 2

The circumfrential extension ratio is: λ λ π π λ

θ φ

= = = 2 2 r r

r

This is also both λx and λy in the given relation. From Eq. (1) of the previous solution we can write σ λ λ λ = = −

F H G I K J

p r b E

• r

r r

2 3 1

3 2 4

pr b E

• r

r

4 1 6 1 1

7

= −

F H G I K J

λ λ

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);

linear elastic λr rubber elastic 0.14 0.12 0.1 0.08 0.06 0.04 0.02 5 4 3 2 1 1

Extension at maximum pressure: Digits:=4;'lambda[r,max]'=fsolve(diff(pstar2,lambda[r])=0,lambda[r]); = λ ,

r max

• 1.383

The maximum normalized pressure: 'pstar[max]'=evalf(subs(lambda[r]=1.383,pstar2)); = pstarmax .1033