Essentials Of Materials Science And Engineering
Essentials Of Materials Science And Engineering
4th Edition
ISBN: 9781337385497
Author: WRIGHT, Wendelin J.
Publisher: Cengage,
Question
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Chapter 7, Problem 7.50P
Interpretation Introduction

Interpretation:

The values of temperature should be determined for specimens made up of tin, molybdenum, iron, gold, zinc, and chromium.

Concept introduction:

At constant stress and elevated temperature, when a progressive deformation occurs in any material or component, it is known as creep deformation.

It is also known as time-dependent deformation because this deformation occurs depending on time for how long the materials are exposed to stress.

Expert Solution & Answer
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Answer to Problem 7.50P

The values of temperature are (71.04)C0, 885.4C0, 451.4C0, 261.96C0, 4.2C0, and 599C0 at which creep deformation occurs in the specimens made up of tin, molybdenum, iron, gold, zinc, and chromium respectively.

Explanation of Solution

With respect to absolute melting temperature, we have below-mentioned formula which can be taken into consideration for determining creep deformation.

T=0.4Tm......(1)

Here,

Tm = absolute temperature at which metal melts

With the help of Appendix-A, 'Selected Physical Properties of Metals', one can get the values of melting points of the given specimens of materials as follow.

Tin, ( T m)Sn=231.9C0Molybdenum, ( T m)Mo=2623C0Iron, ( T m)Fe=1538C0Gold, ( T m)Au=1064.4C0Zinc, ( T m)Zn=420C0Chromium, ( T m)Cr=1907C0

Now, by using equation (1), determine the temperature at which the given materials will experience creep deformations.

  1. For Tin,
  2. T=0.4TmT=0.4( T m)SnT=0.4[231.9C0+273]KT=201.96KT=(201.93273)C0T=71.04C0
  3. For Molybdenum,
  4. T=0.4TmT=0.4( T m)MoT=0.4[2623C0+273]KT=1158.4KT=(1158.4273)C0T=885.4C0

  5. For Iron,
  6. T=0.4TmT=0.4( T m)FeT=0.4[1538C0+273]KT=724.4KT=(724.4273)C0T=451.4C0

  7. For Gold,
  8. T=0.4TmT=0.4( T m)AuT=0.4[1064.4C0+273]KT=534.96KT=(534.96273)C0T=261.96C0

  9. For Zinc,
  10. T=0.4TmT=0.4( T m)ZnT=0.4[420C0+273]KT=277.2KT=(277.2273)C0T=4.2C0

  11. For Chromium,
  12. T=0.4TmT=0.4( T m)CrT=0.4[1907C0+273]KT=872KT=(872273)C0T=599C0

Conclusion

Thus, different temperatures corresponding to given specimen of materials of tin, molybdenum, iron, gold, zinc and chromium at which they experiences creep deformation are (71.04)C0, 885.4C0, 451.4C0, 261.96C0, 4.2C0, and 599C0.

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