HW 1
Andrew Biehl
10/2/2012
1. Write a brief essay highlighting key characteristics of the mechanical behavior of metals, ceramics,
polymer and reinforced composite materials. Discuss similarities and differences
Most materials can be classified as a metal, a ceramic, or a polymer. Materials can further be combined
to create a composite material. Each of these sets of materials share specific material properties which
can be attributed to their unique atomic structures. Metals have relatively weak atomic bonds, which
results in a high capacity for plastic deformation (ductility) and good thermal and electrical conductivity.
Ceramics have stronger ionic bonds, which result in low ductility values, high hardness values and high
coefficients of thermal expansion. Ceramics are also brittle materials which tend to fail by fracturing.
Polymers consist of long molecule chains which are bonded via weak van der Waals forces. Polymer
chains can either be linear (thermoplastic) or cross-linked (thermoset). Thermoplastics are easily
molded when the material is heated up and the chains easily slide past one another. Thermosets cannot
be molded like thermoplastics, due to the cross-linking of the polymer chains. Reinforced composite
materials provide the opportunity generate a material which has the combined properties of its
constituents. While their properties widely vary, reinforced composites are often known for their high
strength and stiffness, and low weight.
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?
𝑔
Atomic mass: 𝑀𝑁𝑖 = 58.69 𝑚𝑜𝑙
Density: 𝜌𝑁𝑖 = 8.912
𝑔
𝑐𝑚3
Avogadro’s Number: 6.02 × 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙
Determine number of atoms/cell for FCC crystal:
4
1 𝑎𝑡𝑜𝑚
8 𝑙𝑎𝑡𝑡𝑖𝑐𝑒 𝑝𝑡
1 𝑎𝑡𝑜𝑚
× 8 + 2 𝑓𝑎𝑐𝑒 × 6 = 4𝑎𝑡𝑜𝑚𝑠
𝑎𝑡𝑜𝑚𝑠
1
58.69 𝑔/𝑚𝑜𝑙
×
×
= 4.3758 𝑐𝑚3 /𝑐𝑒𝑙𝑙
𝑐𝑒𝑙𝑙
8.912𝑔/𝑐𝑚^3 6.02 × 1023 𝑎𝑡/𝑚𝑜𝑙
Lattice parameter of Nickel: 𝑎3 = 4.3758𝑐𝑚3
3
𝑎 = √4.3758𝑐𝑚3 = 3.52385 × 10−8 𝑐𝑚
Radius of Nickel atom: 𝑟 = 2
𝑎
√2
=
3.52385×10−8 𝑐𝑚
2√2
= 2.49174 × 10−8 𝑐𝑚
1𝑐𝑚3
Number of atoms in a 1mm crystal: 9.14129𝑎𝑡/𝑐𝑚3 × 1000𝑚𝑚3 = 9.14129 × 1019 𝑎𝑡𝑜𝑚𝑠
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:
𝑵𝒗 = 𝐞𝐱𝐩 (
−𝟐. 𝟓𝟔 × 𝟏𝟎−𝟏𝟗
)
𝒌𝑻
Where the Boltzmann constant 𝒌 = 𝟏. 𝟑𝟖 × 𝟏𝟎−𝟐𝟑 𝑱/𝒎𝒐𝒍 𝑲. Calculate and plot the value of 𝑵𝒗 for
the temperature range 𝟑𝟎𝟎 ≤ 𝑻 ≤ 𝟐𝟎𝟒𝟐.
T (K)
300
500
700
900
1100
1300
1500
1700
1900
2042
Nv
1.39661E-27
7.70984E-17
3.09561E-12
1.11779E-09
4.74165E-08
6.34906E-07
4.25613E-06
1.82346E-05
5.75107E-05
0.0001134
0.00012
0.0001
Nv
0.00008
0.00006
0.00004
0.00002
0
0
500
1000
1500
2000
2500
T (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.
Average space between dislocation lines: 𝐿 =
1
√𝜌
=
1
√1016
= 1 × 10−8
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.
The standard moment – curvature equation can be integrated twice to get an equation for beam
deflection:
𝐸𝐼
𝑑2 𝜈
=𝑀
𝑑𝑥 2
The following equation for the moment in the beam from 0 ≤ 𝑥 ≤ 𝐿/2 can be determined from the
principle of equilibrium:
𝑀=
𝑃𝑥
2
𝑑2 𝜈
Therefore:
𝐸𝐼 𝑑𝑥 2 =
Integrage:
𝑑𝜈
𝑑𝑥
BC1:
𝑑𝑣 𝐿
( )
𝑑𝑥 2
= 0 therefore:
Integrate again:
𝑣=
BC2: 𝜐(0) = 0 therefore:
𝑃𝑥
2
𝑃𝑥 2
4𝐸𝐼
+ 𝑐1
0=
𝑃𝐿2
16𝐸𝐼
+ 𝑐1
𝑃𝑥 3
12𝐸𝐼
−
𝑣=
=
𝑃𝐿2
𝑥
16𝐸𝐼
𝑃
𝑥3
12𝐸𝐼
−
+ 𝑐2
𝑃𝐿2
𝑥
16𝐸𝐼
The maximum deflection occurs when the deflection angle,
𝑣𝑚𝑎𝑥 =
𝑑𝑣
,
𝑑𝑥
is zero at L/2. Therefore:
𝑃𝐿3
𝑃𝐿3
𝑃𝐿3
−
=−
96𝐸𝐼 32𝐸𝐼
48𝐸𝐼
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.
Following the same procedure as was followed above:
𝑀=
𝑑2 𝜐
𝐸𝐼 𝑑𝑥 2 =
Therefore:
𝑑𝜐
𝑑𝑥
Integrate:
BC1:
𝑑𝑣 𝐿
( )
𝑑𝑥 2
𝑞𝐿
𝑞
𝑥 − 𝑥2
2
2
= 0 therefore:
=
𝑞𝐿
𝑥
2
𝑞𝐿 2
𝑥
4𝐸𝐼
𝑞𝐿3
−
𝑞
− 2 𝑥2
𝑞
𝑥3
6𝐸𝐼
𝑞𝐿3
+ 𝑐1
0 = 16𝐸𝐼 − 48𝐸𝐼 + 𝑐1
𝑐1 = −
Integrate again:
𝑞𝐿
𝑞𝐿3
24𝐸𝐼
𝑞𝐿3
𝑞
𝜐 = 12𝐸𝐼 𝑥 3 − 24𝐸𝐼 𝑥 4 − 24𝐸𝐼 𝑥 + 𝑐2
𝑞𝐿
𝑞𝐿3
𝑞
𝜐 = 12𝐸𝐼 𝑥 3 − 24𝐸𝐼 𝑥 4 − 24𝐸𝐼 𝑥
BC2: 𝜐(0) = 0 therefore:
𝑑𝑣
The maximum deflection occurs when the deflection angle, 𝑑𝑥, is zero at L/2. Therefore:
𝑣𝑚𝑎𝑥 =
𝑞𝐿4
𝑞𝐿4
𝑞𝐿4
5𝑞𝐿4
−
−
=−
96𝐸𝐼 384𝐸𝐼 48𝐸𝐼
384𝐸𝐼
c. Consider a beam of square cross section a = b = 0.1m, moreover let L = 1m and E = 10 11 Pa.
Calculate the values of the maximum beam deflection for cases (a) and (b) above if P = 105
and Q = 105 N/m respectively.
First the moment of inertia is calculated: 𝐼𝑥 =
𝑏ℎ 3
12
=
(0.1𝑚)(0.1𝑚)3
12
= 8.333 × 10−3 𝑚4
(105 𝑁)(1𝑚)3
For point load:
𝜐𝑚𝑎𝑥 =
For distributed load:
𝜐𝑚𝑎𝑥 = 384(1011 𝑃𝑎)(8.333×10−3 𝑚4 ) = −.00156𝑚
48(1011 𝑃𝑎)(8.333×10−3 𝑚4 )
= −.0025𝑚
5×(105 𝑁)(1𝑚)3
6. The state of stress at a point in a structural component is specified by the provided components of
the stress tensor.
a. Compute the values of the first three stress invariants of the stress tensor.
As calculated in the attached excel spreadsheet:
𝐼1 = −80, 𝐼2 = −113606, 𝐼3 = 16126848
b. Solve the characteristic equation of the stress tensor and determine the values of the
principal stresses.
The characteristic equation, 𝜆3 − 𝜆2 𝐼1 + 𝜆𝐼2 − 𝐼3 = 0 , was solved via graphing:
(𝜆 − 360.0)(𝜆 + 159.9)(𝜆 + 280.0) = 0
Therefore 𝜎1 = 360.0𝑀𝑃𝑎, 𝜎2 = −159.9𝑀𝑃𝑎, 𝜎3 = −280.0𝑀𝑃𝑎
c. Compute the hydrostatic stress and the components of the stress deviator tensor.
The stress tensor can be expressed as the sum of the hydrostatic stress and the stress deviator tensor.
First the mean stress is calculated:
1
1
𝜎0 = 𝐼1 = (−80) = −26.667𝑀𝑃𝑎
3
3
The stress deviation tensor is calculated using the equation:
′
𝜏𝑖𝑗
= 𝜏𝑖𝑗 − 𝜎0 𝛿𝑖𝑗
In matrix form, the stress deviation tensor is:
26.66667
-240
-240 226.6666667
-2.4
0
-2.4
0
-253.333
d. Compute the values of the first three invariants of the stress deviator tensor.
𝐽1 = 0
𝐽2 = 3𝜎02 − 𝐼2 = 115739.1
𝐽3 = 𝐼3 + 𝐽2 𝜎0 − 𝜎03 = 13059435
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.
The characteristic equation, 𝜆3 − 𝜆2 𝐽1 − 𝜆𝐽2 − 𝐽3 = 0 , was solved via graphing:
(𝜆 − 386.7)(𝜆 + 133.3)(𝜆 − 253.4) = 0
Therefore 𝜎1 = 386.7𝑀𝑃𝑎, 𝜎2 = −133.3𝑀𝑃𝑎, 𝜎3 = −253.4𝑀𝑃𝑎
Principal shearing stresses:
𝜏13 =
𝜎1 − 𝜎3 386.7 − (−253.4)
=
= 320𝑀𝑃𝑎
2
2
𝜏12 =
𝜎1 − 𝜎2 386.7 − (−133.3)
=
= 260𝑀𝑃𝑎
2
2
𝜏23 =
𝜎2 − 𝜎3 −133.3 − (−253.4)
=
= 60𝑀𝑃𝑎
2
2