Question

1. A metallic rod, characterised by a gauge length of 150 mm and a diameter of 20 mm, undergoes uniaxial tensile testing to failure. The loading direction is parallel to the main axis of the rod, and the Poisson's ratio of the material is 0.33. Yielding is observed when an applied force of 55 kN is reached, for an engineering strain along the loading direction of 0.2%. Calculate: (a) The Young's modulus of the material. (4 marks) (b) The length and the diameter of the rod at yielding. (4 marks) (c) The elastic energy per unit volume and the total stored elastic energy in the specimen. (4 marks) (d) Explain why dislocations move on specific crystal slip systems and how this is defined? (3 marks) (e) Define Griffith's criteria for measuring a materials toughness. (3 marks) (f) What thermally activated process gives rise to the glass transition temperature in polymers? (2 marks)

          1.
A metallic rod, characterised by a gauge length of 150 mm and a diameter
of 20 mm, undergoes uniaxial tensile testing to failure. The loading
direction is parallel to the main axis of the rod, and the Poisson's ratio of
the material is 0.33. Yielding is observed when an applied force of 55 kN
is reached, for an engineering strain along the loading direction of 0.2%.
Calculate:
(a) The Young's modulus of the material.
(4 marks)
(b) The length and the diameter of the rod at yielding.
(4 marks)
(c) The elastic energy per unit volume and the total stored elastic energy
in the specimen.
(4 marks)
(d) Explain why dislocations move on specific crystal slip systems and
how this is defined?
(3 marks)
(e) Define Griffith's criteria for measuring a materials toughness.
(3 marks)
(f) What thermally activated process gives rise to the glass transition
temperature in polymers?
(2 marks)

Show more…

1.
A metallic rod, characterised by a gauge length of 150 mm and a diameter
of 20 mm, undergoes uniaxial tensile testing to failure. The loading
direction is parallel to the main axis of the rod, and the Poisson's ratio of
the material is 0.33. Yielding is observed when an applied force of 55 kN
is reached, for an engineering strain along the loading direction of 0.2%.
Calculate:
(a) The Young's modulus of the material.
(4 marks)
(b) The length and the diameter of the rod at yielding.
(4 marks)
(c) The elastic energy per unit volume and the total stored elastic energy
in the specimen.
(4 marks)
(d) Explain why dislocations move on specific crystal slip systems and
how this is defined?
(3 marks)
(e) Define Griffith's criteria for measuring a materials toughness.
(3 marks)
(f) What thermally activated process gives rise to the glass transition
temperature in polymers?
(2 marks)

Added by Nathaniel D.

Materials Science and Engineering: An Introduction

William D. Callister, Jr. David G.… 7th Edition

Chapter 6

Instant Answer

Solved by Expert Supreeta N

06/06/2023

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### (a) Calculate the Young's modulus of the material ** Show more…

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Transcript

00:01 Hello students in this question given as a length of the rod is 150 mm diameter is 20 mm.

00:10 Poisson's ratio is given as 0 .33 applied force is equal to 55 kilo newton strain is given as 0 .2 percent is that will be equal to the 0 .002.

00:31 Now in the first part of the question, we have to calculate the young modulus for this.

00:37 We will first of all we will calculate the stress which is equal to the applied force upon area cross -sectional area.

00:44 The force is given as 55 into 10 to the power 3 newton upon area is pi by 4 into t square.

00:54 T is 20 into 10 to the power minus 3 meter whole square.

01:00 So after calculation, we will get sigma is equal to that when the stress is equal to 1 .75 into 10 to the power 8 newton per meter square.

01:14 So young modulus will be equal to the stress upon strain stress is 1 .75 into the power 8 newton per meter square upon a strain is which is given in the question 0 .002.

01:26 So this will be equal to the young modulus is equal to 8 .75 into 10 to the power 10 newton per meter square or we can say that 87 .5 giga pascal in the second part of the question.

01:55 We will calculate the length and diameter.

01:57 So a stress strain is equal to change in length.

02:01 That means l minus original length.

02:07 So original length is given as 150 mm upon 150 and the strain value is given as 0 .002.

02:18 So the length will be equal to the 150 .3 mm.

02:29 Similarly, we have to calculate the diameter for diameter...

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