1. Write a brief essay highlighting key characteristics of the mechanical behavior of metals, ceramics, polymer
and reinforced composite materials. Discuss Similarities and differences.
Metals appear mostly in three structures (FCC, HCP, BCC). With metals you can see creep and fracture. Whereas
ceramics are ionic compounds and exist in many different structures and they show signs of brittle behavior.
Polymers are constituted by assemblies of large chain molecules with a distribution of molecular weights.
Polymers consist of plastics, elastomers, and biopolymers. Composite materials consist of dispersions of multiple
phases in intimate contact. (From notes sections 3.3 and 3.4 in Chapter 1)
2. Nickel has the FCC crystal structure. Find the density and atomic mass of nickel and determine the radius of
atoms in a single crystal and also the lattice parameter of the FCC cell. How many atoms are there in a single
crystal shaped as a cube with side = 1mm?
𝜌𝑁𝑖 = 8.908
𝑔
𝑐𝑚3
Standard Atomic Weight is 58.6934 g/mole
If 1 mole=6.02214179x1023atoms then a single atoms weighs
58.6934
6.02214179𝑥1023
= 9.74627𝑥10−23 𝑔
Since there is 4 atoms in FCC the mass of a unit cell is 9.74627x10-23g*4=3.89851x10-22g
The unit cell volume can then be calculated by 3.89851x10-22g/8.908g/cm3=0.043
Lattice parameter is 0.043(1/3)m3=0.35m
Using the FCC structure a diagonal would be of the length √2𝑎
3. The fraction of vacant lattice sites Nv, in an otherwise perfect crystal of Pt is given as a function of the
absolute temperature T (in Kelvin) by the expression
𝟐.𝟓𝟔×𝟏𝟎−𝟏𝟗
)
𝒌𝑻
𝑵𝒗 = 𝒆(−
where the Boltzmannconstant k=1.38x10-23 J/moleK. Calculate and plot the value of Nv for the temperature
range 300≤T≤2042.
Nv
1.39661E-27
7.22448E-21
7.70984E-17
3.73713E-14
3.09561E-12
8.49969E-11
1.11779E-09
8.78057E-09
4.74165E-08
1.93316E-07
6.34906E-07
1.75944E-06
4.25613E-06
9.21938E-06
1.82346E-05
3.34333E-05
5.75107E-05
9.37047E-05
0.0001134
Fraction of Vacant Lattice Sites (Nv)
Temperature (K)
300
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
2000
2042
0.00012
0.0001
0.00008
0.00006
0.00004
0.00002
0
0
500
1000
1500
2000
2500
Temperature (K)
4. The density of dislocations in heavily worked nickel is of the order of 1016 dislocation lines per square meter.
Assume that the dislocations are randomly arranged and estimate the average space between neighboring
dislocation lines.
1
1 2
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑎𝑐𝑒 𝐵𝑒𝑡𝑤𝑒𝑒𝑛 𝐷𝑖𝑠𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝐿𝑖𝑛𝑒𝑠 = ( 16 ) = 1 × 10−8 [𝑚]
10
5. Review your basic knowledge of strength of materials and then consider a simply supported beam of length L,
elastic modulus E and moment of intertia I:
a. Obtain an expression for the maximum deflection of the beam due to the application of a point load P
downwards, applied in the middle of the upper surface of the beam.
b. Obtain an expression for the maximum deflection of the beam due to the application of a distributed
load per unit length Q, downwards, applied on the upper surface of the beam.
c. Consider a beam of square cross section a=b=0.1m, moreover let L=1m and E=10^11Pa. Calculate the
values of the maximum beam deflection for cases (a) and (b) above if P=10^5 N and Q=10^5 N/m,
respectively.
𝑃×𝐿3
a. 𝜎 = 𝐸×𝐼×48 From (http://www.engineeringtoolbox.com/beam-stress-deflection-d_1312.html)
5×𝑄×𝐿4
b. 𝜎 = 𝐸×𝐼×348 From (http://www.engineeringtoolbox.com/beam-stress-deflection-d_1312.html)
c. For a point Load:
𝜎=
For a distributed load: 𝜎 =
𝑃×𝐿3
𝐸×𝐼×48
=
105 ×13
1011 ×
5×105 ×14
1011 ×
0.14
×348
12
0.14
×48
12
= 0.0025𝑚
= 0.0017𝑚
6. The state of stress at a point in a structural component is specified by the following components of the stress
tensor (all in MPa):
𝝈𝟏𝟏 = 𝟎
𝝈𝟐𝟐 = 𝟐𝟎𝟎
𝝈𝟑𝟑 = −𝟐𝟖𝟎
𝝈𝟏𝟐 = 𝝈𝟐𝟏 = −𝟐𝟒𝟎
𝝈𝟏𝟑 = 𝝈𝟑𝟏 = −𝟐. 𝟒
𝝈𝟐𝟑 = 𝝈𝟑𝟐 = 𝟎
a. Compute the values of the first three stress invariants of the stress tensor.
b. Solve the characteristic equation of the stress tensor and determine the values of the principal
stresses.
c. Compute the hydrostatic stress and the components of the stress deviator tensor.
d. Compute the values of the first three invariants of the stress deviator tensor.
e. Solve the characteristic equation of the stress deviator tensor and determine the values of the
principal stress deviations and give the values of the principal shearing stresses.
𝝈𝟏𝟏 + 𝝈𝟐𝟐 + 𝝈𝟑𝟑 = 𝑰𝟏 = 𝟎 + 𝟐𝟎𝟎 − 𝟐𝟖𝟎 = −𝟖𝟎
𝝈𝟏𝟏 × 𝝈𝟐𝟐 + 𝝈𝟐𝟐 × 𝝈𝟑𝟑 + 𝝈𝟑𝟑 × 𝝈𝟏𝟏 − 𝝈𝟏𝟐 𝟐 − 𝝈𝟐𝟑 𝟐 − 𝝈𝟏𝟑 𝟐 = 𝑰𝟐 = −𝟏𝟏𝟑𝟔𝟎𝟔
𝝈𝟏𝟏 × 𝝈𝟐𝟐 × 𝝈𝟑𝟑 + 𝟐 × 𝝈𝟏𝟐 × 𝝈𝟐𝟑 × 𝝈𝟑𝟏 − 𝝈𝟏𝟏 × 𝝈𝟐𝟑 𝟐 − 𝝈𝟐𝟐 × 𝝈𝟏𝟑 𝟐 − 𝝈𝟑𝟑 × 𝝈𝟏𝟐 𝟐 = 𝑰𝟑 = −𝟏𝟔𝟏𝟐𝟔𝟖𝟎𝟎