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 9.20P

(a)

To determine

The strain in the tube at time t=0 s .

(a)

Expert Solution
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Answer to Problem 9.20P

The strain in the tube at time t=0 s is 0.25×103 .

Explanation of Solution

Given:

Temperature of soda-lime glass is 395°C .

Tensile stress is 10MPa .

Elastic modulus of glass is 40GPa .

Concept used:

Write the expression for elastic strain rate in material at t=0 s .

  εe=σE …… (1)

Here, εe is the elastic strain rate in material at t=0 s , σ is the stress in material and E is the Elastic modulus of glass.

Calculation:

Substitute 10MPa for σ and 40GPa for E in equation (1).

  εe=10MPa40GPa( 10 3 MPa 1 GPa )=0.25×103

Conclusion:

Thus, the strain in the tube at time t=0 s is 0.25×103 .

(b)

To determine

Total strain in the tube after one year.

(b)

Expert Solution
Check Mark

Answer to Problem 9.20P

Total strain in the tube after one year is 1.05025 .

Explanation of Solution

Given:

Value of constant n is 1 .

Concept used:

Refer to Figure 9.3 “The temperature dependence of the viscosity of silica glass (SiO2) soda-lime glass, and boron oxide (B2O3) ” to obtain the viscosity of soda lime glass at 668 K as 1014 Pas .

Write the expression for plastic strain in soda-lime glass.

  dεpdt=σ3η …… (2)

Here, dεpdt is the plastic strain in soda-lime glass and η is the viscosity of material.

Write the expression for total plastic strain in material.

  εp=dεpdt(t) …… (3)

Here, εp is the total plastic strain in material and t is the time.

Write the expression for total strain.

  εt=εe+εp …… (4)

Here, εt is the total strain.

Calculation:

Substitute 1014 Pas for η and 10 MPa for σ in equation (2).

  dεpdt=10 MPa( 10 6 Pa 1MPa )3( 10 14  Pas)=3.33×108s1

Substitute 3.33×108s1 for dεpdt and 1 year for t in equation (3).

  εp=(3.33× 10 8s 1)(1 year)( 365 days 1 year)( 24h 1 day)( 3600 s 1 h)=1.05

Substitute 1.05 for εp and 0.25×103 for εe in equation (4).

  εt=0.25×103+1.05=1.05025

Conclusion:

Thus, the total strain in the tube after one year is 1.05025 .

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ISBN:9781111988609
Author:Charles Gilmore
Publisher:Cengage Learning