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/cm3) with a low melting
temperature (670 oC) 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 oC 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.
3ao  8r
o
3ao  8 1.176  9.4 
o
ao  5.427 
mass
8(28.08) / 6.02 1023
3
density 


2
.
33
g/cm
volume
(5.427 108 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)
ao  2rGa  2rAs  20.62  22.22  5.68 A
o
r
4  69.921  4  74.921  5.245 g/cm 3
mass

8 3
volume
5.68  10
6.02  1023



Question 5
The coefficient for linear thermal expansion for Al is 23.6x10exp(-6) /oC. If a 200
mm long, 100 mm diameter rod is heated from 20 oC to 100 oC, what are the
dimensions (length and diameter) of the rod at 100 oC? 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 oC) and Ti (1668 oC) are too high for creep at
250 C.
Question 7
D: 0,1,1/2 – 100 = -1,1,1/2
= [ 221]
Question 8
Question 9
1
h  (2 h  k )
3
1
k  (2 k  h )
3
1
i   (h  k )
3
l  l
Question 10
Question 11
What are the indices of the four directions of the form <111> that lie in the
1 01 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.
o
1
1
BCC Fe : b  repeat distance 
3 ao 
3 (2.866)  2.482 A
2
2
o
1
1
FCC Fe : b 
2 ao 
2 (3.589)  2.54 A
2
2
o
1
1
Al : b 
2 ao 
2 (4.050)  2.86 A
2
2
o
1
1
Ge : b 
2 ao 
2 (5.657)  4.000 A
2
2
Question 13
Calculate the length of the dislocation Burgers vector for Silicon given
that its (111) has width of 3.135 Angstroms.
ahkl  d hkl h 2  k 2  l 2  d hkl 12  12  12  d hkl 3
1
Si : b  repeat distance 
2 d hkl h 2  k 2  l 2
2
o
1
1
2
2
2
Si : b 
2 d hkl 1  1  1 
2 (3.135) 3  3.839 A
2
2
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    b / 2 D
D  b 2 sin   
o
D  4.7 / 2 sin( 10 / 2)  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
) (10
1) planes.
1 1 1 in the
direction
, (110, ) (011
and
slip
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, ac 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, ac and the number of cycles, N, required before failure occurs,
which is given by the equation below. Let f = 1.0.
From:

2 (ac )( 2n ) / 2  (ai )( 2n) / 2
N
(2  n)Cf n  n n / 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
Smax = 80,000 psi;
Smin = -20,000 psi
Stress range, Sr = 80,000 – (-20,000) = 100,000 psi
Stress amplitude = Sr/2 = 50,000 psi
Mean stress = (80,000 + (-20,000))/2 = 30,000 psi
Stress ratio = Smax/ Smin = -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
20,000
 2
D
4
20,000 x 4
D
 0.8 inch
40,000
40,000 
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 oC. If the creep rate is dependent on self-diffusion, determine the creep
rate if the temperature is 800 oC.
Question 24