Mech 473 Assignment #1

Professor Rodney Herring

Question 1 Design of an Aerospace Component Because Mg is a low-density material ( r = 1.738 g/cm 3 ) with a low melting temperature (670 o C) and boiling point, it has been considered for use in an aerospace vehicle intended to enter the outer space environment. Is this a good material design?

In space, the pressure is low. Even at relatively low temperatures, solid magnesium can begin to change to a vapor, causing metal loss that could damage a space vehicle. In addition, solar radiation could cause the vehicle to heat, increasing the rate of magnesium loss.

A low density material with higher boiling point (lower vapor pressure) would be better. Al boils at 2494 o C and is a good alternative.

Question 2

Question 3

Calculate the theoretical density of Si. Assume a diamond cubic structure and the radius to be 1.176 Angstroms.

3

a o

 8

r

3

a o

 8  1 .

176  9 .

4 o 

a o

 5 .

427 o 

density



mass volume



8 ( 28 .

08 ) / 6 .

02



10

23

( 5 .

427



10

 8

cm )

3 

2 .

33 g/cm

3

Question 4

Calculate the theoretical density of GaAs, which is a compound semiconductor having the zinc blende structure, given that the atomic mass of Ga is 69.723 g/mole and As is 74.921 g/mole and that the radius of the Ga ion is 0.62 A and the radius of the As ion is 2.22 A. (2)

a o



2

r Ga



2

r As



2



0 .

62



2 .

22

  o

5 .

68 A

r 

mass volume

 

4

 

5 .

68 69 .

921

 

4



10

 8 3   

6 .

02 74 .

921

 

10

23  

5 .

245 g/cm

3

Question 5 The coefficient for linear thermal expansion for Al is 23.6x10exp(-6) / o C. If a 200 mm long, 100 mm diameter rod is heated from 20 o C to 100 o C, what are the dimensions (length and diameter) of the rod at 100 o C? Is there a stress on the rod? (2)

Question 6 If an application requires materials to support loads at 250 degrees C, determine whether gold, copper, iron and titanium will likely creep. The design temperature is 250 + 273 = 523 K. Both Au and Cu will likely creep since T/Tm is 0.39 and 0.38, respectively, and they are pure elements that have no creep strengthening. The melting temperatures Fe (2447 o C) and Ti (1668 o C ) are too high for creep at 250 C.

Question 7

D

: 0,1,1/2 – 100 = -1,1,1/2 = [ 2 21 ]

Question 8

Question 9

h



1 3 ( 2 h

 

k



) k



1 3 ( 2 k

 

h



) i



l

 

1 3 ( h

 

k



) l



Question 10

Question 11 What are the indices of the four directions of the form <111> that lie in the

 

plane of a cubic cell? (2)

Question 12 Calculate the length of the dislocation Burgers vector in BCC iron (low temperature), FCC iron, aluminum, and germanium.

BCC Fe :

b



repeat distance



1 2 3

a o



1 2 FCC Fe :

b



1 2 Al :

b



1 2 2

a o

2

a



1 2

o



1 2 2 ( 4 2 ( 3 .

050 ) .

589 )

 

2 2 .

86

o

A .

54

o

A Ge :

b



1 2 2

a o



1 2 2 ( 5.657)

 o

4 .

000 A 3 ( 2 .

866 )

 o

2 .

482 A

Question 13 Calculate the length of the dislocation Burgers vector for Silicon given that its (111) has width of 3.135 Angstroms.

a hkl



d hkl h

2 

k

2 

l

2 

d hkl

1 2  1 2  1 2 

d hkl

Si :

b

 repeat distance  1 2 Si :

b

 1 2 2

d hkl

1 2  1 2  1 2 2

d hkl

 1 2 2

h

2 

k

2  ( 3.135)

l

2 3 3  o 3.839

A

Question 14 A slight misalignment of an optoelectronic device during fabrication leaves a grain boundary passing through the optically active region having a width of 20 nm. If the two grains approach each other at an angle of 10 degrees, how many dislocations, N, could exist to short out the device?

sin

D



b



2 sin

b

/



2

D D



4 .

7 / 2 sin( 10 / 2 )

 o

27 .

0 A



2 .

7 nm N



20/2.7



7.4

dislocatio ns

( 110 )

Question 15 A single crystal of an BCC metal is oriented so that the [001] direction is parallel to an applied stress. If the critical resolved shear stress is 12,000 psi, calculate the magnitude of the applied stress required to cause slip to begin in the

 

Question 16 The strength of titanium is found to be 65,000 psi when the grain size is 17 x 10 -6 m and 82,000 psi when the grain size is 0.8 x 10 -6 m. Determine the constants in the Hall-Petch equation and the strength of the titanium when the grain size is reduced to 0.2 x 10 -6 m.

Question 17

Question 18

Question 19

f f

Question 20 – Part 1 The acrylic polymer from which the figure below was obtained is subjected to an alternating stress between 15 MPa and 0 MPa. The largest surface cracks initially detected by non-destructive testing are 0.001 mm in length. From the graph calculate the constants C and n for the steady state fatigue equation. If the critical fracture toughness of the polymer is 2 MPa sqrt(in), calculate the critical flaw size, a c and the number of cycles, N, required before failure occurs, which is given by the equation below. Let f = 1.0.

Question 20 – Part 2 The acrylic polymer from which the figure below was obtained is subjected to an alternating stress between 15 MPa and 0 MPa. The largest surface cracks initially detected by non-destructive testing are 0.001 mm in length. From the graph calculate the constants C and n for the steady state fatigue equation. If the critical fracture toughness of the polymer is 2 MPa sqrt(in), calculate the critical flaw size, a c and the number of cycles, N, required before failure occurs, which is given by the equation below. Let f = 1.0. From:

N

 2  (

a c

( 2 )  ( 2 

n

) /

n

)

Cf

2

n

  (

a i

 )

n

 ( 2 

n n

/ 2 ) / 2 

Question 21 In a fatigue test, the maximum stress applied in tension is 80,000 psi and the minimum stress in compression is 20,000 psi. Determine the stress range, stress amplitude, mean stress and stress ratio, R S max = 80,000 psi; S min = -20,000 psi Stress range, S r = 80,000 – (-20,000) = 100,000 psi Stress amplitude = S r /2 = 50,000 psi Mean stress = (80,000 + (-20,000))/2 = 30,000 psi Stress ratio = S max / S min = -20,000/80,000 = -0.25

Question 22 Calculate the fatigue limit of a steel that has an ultimate tensile strength of 160 psi when the stress ratio, R = 0 and a 50 per cent probability of enduring the stress for the life of the component. Use a safety factor of 2. Determine the diameter of the rod to be used to carry a fluctuating load of 20,000 lb-force.

The 50% probability of survival for the life of the component corresponds to the fatigue limit of the steel. Thus, Fatigue Limit = 0.5 UTS = 80 psi With a safety factor of 2, cycle stress = 80,000/2 = 40,000 psi For a cyclic load of 20,000 lb-force, the conservative diameter of the steel will be in push-pull mode and

40 , 000  20 , 000  2

D

4

D

 20 , 000

x

4 40 , 000   0 .

8 inch

Question 23 The activation energy for self-diffusion in copper is 49,300 cal/mol. A copper specimen creeps at 0.002 in/in –h when a stress of 15,000 psi is applied at 600 o C. If the creep rate is dependent on self-diffusion, determine the creep rate if the temperature is 800 o C.

Question 24