Materials Science And Engineering Properties
Materials Science And Engineering Properties
1st Edition
ISBN: 9781111988609
Author: Charles Gilmore
Publisher: Cengage Learning
Question
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Chapter 9, Problem 9.22P

(a)

To determine

The maximum possible elastic modulus in the polymer that will result in the maximum allowable clamping force.

(a)

Expert Solution
Check Mark

Answer to Problem 9.22P

The maximum possible elastic modulus in the polymer that will result in the maximum allowable clamping force is 2GPa .

Explanation of Solution

Given:

The length of the polymer is 0.05m .

The change in length of the polymer is 0.0001m .

The maximum clamping force is 1×104N .

Formula Used:

Write the expression for the strain in the polymer as:

  ε=Δll …… (I)

Here, ε is the elastic strain in the material, Δl is the change in length of the polymer and l is the length of the polymer.

Write the expression for the stress acting on the polymer as:

  σ=FA …… (II)

Here, F is the clamping force and A is the cross sectional area.

Write the expression for the elastic modulus in the polymer as:

  E=σε …… (III)

Here, σ is the stress acting on the polymer.

Calculation:

Substitute 0.05m for l and 0.0001m for Δl in equation (I)

  ε=0.0001m0.05m=0.02

Substitute 1×104N for F and (0.05m)2 for A in equation (II)

  σ=1×104N(0.05m)2=4×106Pa=4MPa

Substitute 4×106Pa for σ and 0.02 for ε in equation (III)

  E=4×106Pa0.002=2×109Pa×109GPa1Pa=2GPa

Conclusion:

Thus, the maximum possible elastic modulus in the polymer that will result in the maximum allowable clamping force is 2GPa .

(b)

To determine

The minimum allowable viscosity of the polymer.

(b)

Expert Solution
Check Mark

Answer to Problem 9.22P

The minimum allowable viscosity of the polymer is 30×1015Pas .

Explanation of Solution

Given:

The length of the polymer is 0.05m .

The change in length of the polymer is 0.0001m .

The maximum clamping force is 1×104N .

Formula Used:

Write the expression for the relation of viscosity and decay of stress.

  σσ0=exp(E3ηt) …… (IV)

Here, σ is the stress in the polymer after one year, σ0 is the initial stress in the polymer, η is the viscosity of the polymer and t is the expected life of the polymer.

Calculation:

Substitute 4×106Pa for σ0 , 2×106Pa for σ , 2×109Pa for E and 365days for t in equation (IV).

  2×106Pa4×106Pa=exp(2×109Pa3η×(365days)×(24h1day)×(60min1h)×(60s1min))0.5=exp(2.1×1016η)ln(0.5)=ln[exp(2.1×1016η)]η=30×1015Pas

Conclusion:

Thus, the minimum allowable viscosity of the polymer is 30×1015Pas .

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