Robert Sayre
MANE 7100
HW1
1.
Write a brief essay highlighting key characteristics of the mechanical behavior of metals,
ceramics, polymer and reinforced composite materials. Discuss similarities and differences.
Metal mechanical properties typically include good compressive, shear, and tensile strength. Metals are
typically isotropic, and can be brittle or ductile depending on the metal. Metals have moderate thermal
resistance. The plastic deformation range is less than polymers but greater than ceramics.
Ceramic mechanical properties include an increased brittleness and less fracture resistance than metals,
composites, and polymers. Ceramics are much harder than metals, plastics, and composites. Ceramics
can retain mechanical properties at higher temperatures than metals and polymers, but are more
susceptible to thermal shock than metals.
Polymers mechanical properties change drastically with temperature. Polymers typically have less creep
resistance than metals and ceramics. The fracture strength of polymers is low compared to metals and
ceramics and the hardness is typically less than that of ceramics.
Composites consist of a matrix and reinforcement, the composition and mixture of the constituents
largely determine the mechanical properties. Composites can have excellent compressive and tensile
strength, and can be brittle or ductile depending. Composites can have excellent thermal resistance, on
par with ceramics. One large drawback to composites is they are typically anisotropic.
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 FC cell. How many atoms are
there in a single crystal shaped as a cube with side = 1 mm?
g
  8908
Density of nickel
3
m
23
ava  6.022 10 
Ni  58.6394
1
Avogadro’s number
mol
g
Atomic mass of nickel
mol
Calculate the mass of 1 nickel atom
atom 
Ni
ava
 23
atom  9.737529060112921  10
g
Calculate the mass of the unit cell
Since it is an FCC, there is 1/8 * 8 +1/2 *6 = 4 atoms in 1 cell
unit_cell  4 atom
 22
unit_cell  3.895011624045168  10
g
Calculate the volume using the unit cell mass and density
unit_cell
vol 
3
vol  43.724872295073716 nm

 17
vol  4.372487229507372  10
3
 mm
Calculate the side length from the volume
Since it is a cube the volume is equal to side cubed
1
side  ( vol)
9
3
side  3.523  10
m
side  3.523 nm
Calculate the radius
r 
9
side
r  1.246  10
2 2
m
r  1.246 nm
len  1 mm
Length of side of new cube
Calculate how many atoms fit in 1 length of a cube
num 
len
5
num  4.014  10
2 r
Calculate how many atoms total in the cube
6
3 num  1.204  10
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
where the Boltzmann constant k =1:38 10 – 23 J=moleK. Calculate and plot the value of Nv for the
temperature range 300 <= T <= 2042.
 23
k  1.38 10
boltzman constant
temp  300301 2042
Temperature range
  2.56 10 19 


kT


Nv ( T)  e
1.510
110
Fraction of vacant lattice sites
4
4
Nv( temp )
J
mol
510
5
0
0
500
110
3
temp
K
1.510
3
210
3
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.
16 1
dis  10 
2
m
Dislocation density
assume the dislocations are all edge type
Need to find the average length, in a square the dislocation density dictates the average
distance between dislocations per length of the side of a square. The number of
dislocations per side squared (since it is a square) = the area of the dislocation area.
side 
5.
8 1
side  1  10
m
dis
Review your basic knowledge of strength of materials and then consider a simply supported
beam of length L , elastic modulus E and moment of inertia 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.
Deflection = moment/ flexural modulus
2
M
d
E I
dx
2
y
Differentiating once yields:
3
V
d
E I
dx
3
Where V is the first derivative
y
Differentiating twice yields:
4
q
d
E I
dx
4
y
Where q is the loading
Integrating once yields:

E I
d
y
dx
Where θ is the slope
Integrating twice yields:
y
f ( x)
The integral forms of these equations in the bounds 0<x<l:
V
M

 q d x  C1


 V d x  C1 x  C2






y

3
2
  d x  C1 x  C2 x  C3 x  C4

M
2
d x  C1 x  C2 x  C3
E I
Summing the moments about the right hand side and summing the forces in the y direction.

Mz
R1 L 
0
P ( L  x)
2
2
i
R1
w ( L  x)
2
2 L
 Fy
0
R1  w ( L  x)  R2
i
R2
w ( L  x)
2
 w ( L  x)
2 L
loading function:
q
R1 ( x  0)
1
0
 P ( x  L)  R2 ( x  l)
1
Integrating:
V

 q dx

M

 Vdx



 M dx

y

  dx

Find the four constants of integration:
Since the reaction force and moments are included in the loading function C.1 and C.2 are zero
C1
0
C2
0
Since the deflection is zero at the supports, The BC's are as follows:
y( 0)
0
y( L)
0
Then
C4
C3
0
P
 ( L  x)  2 L  ( L  x)
4
24 L
2
2

The resulting deflection is then:
y
P
 2 ( l  x)   x  ( L  x)  2 L  ( L  x)   x  L ( x  x)
2
24 L E I
3
4
2
2
4

Max deflection occurs at the point where the slope curve is zero:

 R1
1
E I

2
P
2
x 
6
  x 


 C3

2

L
3
0
Plugging the value of x into the deflection eqn and simplifying yields:
3
y max_point 
P L
E I
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.
Deflection = moment/ flexural modulus
2
M
d
E I
dx
2
y
Differentiating once yields:
3
V
d
E I
dx
3
y
Where V is the first derivative
Differentiating twice yields:
4
q
d
E I
dx
4
y
Where q is the loading
Integrating once yields:

E I
d
y
dx
Where θ is the slope
Integrating twice yields:
y
f ( x)
The integral forms of these equations in the bounds 0<x<l:
V
M

 q d x  C1


 V d x  C1 x  C2






y

3
2
  d x  C1 x  C2 x  C3 x  C4

M
E I
2
d x  C1 x  C2 x  C3
Summing the moments about the right hand side and summing the forces in the y direction.

Mz
R1 L 
0
w ( L  a)
2
2
i
R1
w ( L  a)
2
2 L
 Fy
0
R1  w ( L  a)  R2
i
R2
w ( L  a)
2
 w ( L  a)
2 L
loading function:
q
R1 ( x  0)
1
0
 w ( x  a)  R2 ( x  l)
1
Integrating:
V

 q dx

M

 Vdx



 M dx

y

  dx

Find the four constants of integration:
Since the reaction force and moments are included in the loading function C.1 and C.2 are zero
C1
0
C2
0
Since the deflection is zero at the supports, The BC's are as follows:
y( 0)
0
y( L)
0
Then
C4
C3
0
w
24 L
 ( L  a)  2 L  ( L  a)
4
2
2

The resulting deflection is then:
y
w
24 L E I
 2 ( l  a)   x  ( L  a)  2 L  ( L  a)   x  L ( x  a)
2
3
4
2
2
4

Max deflection occurs at the point where the slope curve is zero:

 R1 2 w

3
 x   ( x  a)  C3
E I  2
6

1

0
Plugging the value of x into the deflection eqn and simplifying yields:
y max_uni 
5
3
w L

384 E I
c.- Consider a beam of square cross section a = b =0:1 m, moreover let L =1m and E =10 11Pa. Calculate
the values of the maximum beam deflection for cases (a) and (b) above if P =105N and Q =105N/m,
respectively.
a  0.1 m
Beam cross section
b  0.1m
L  1 m
Length
11
E  10  Pa
Modulus of elasticity
5 N
W  10 
Uniform loading
m
5
P  10  N
Point load
Moment of inertia
I 
a b
3
6 4
I  8.333  10
12
m
Part A: Point Load
3
y max_point 
P L
ymax_point  0.12m
E I
Part B: uniform load
Equivalent load
w  W L
5
w  1  10 N
y max_uni 
5
3

w L
384 E I
3
ymax_uni  1.562  10
m
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):
a.- Compute the values of the first three stress invariants of the stress tensor.
known
11  0
22  200
33  280
12  240
13  2.4
23  0
21  12
31  13
32  23
 11 12 13 


matrix   21 22 23 
   
 31 32 33 
Kroenecker delta
1 0 0
ij   0 1 0 


0 0 1
Three stress invariants
I1  11  22  33
I2 
 22 23 
 11 13 
 11 12 
 
 
 32 33 
 31 33 
 21 22 
I3  matrix
I1  80
5
I2  1.136  10
7
I3  1.613  10
b.- Solve the characteristic equation of the stress tensor and determine the values of the
principal stresses.
2.4 
  240


matrix    ij  240 200  
0


0
  280 
 2.4
Characteristic eqn
chareqn  matrix    ij
2
3
chareqn  113605.76   80.0   1.0   1.6126848e7
guess
  0
given
113605.76   80.0   1.0   1.6126848e7
2
z  find (  )
 280.03598988084296144 
z   159.96677937112552594 


 360.00276925196848739 
T
 280.03598988084296144 
ztrans   159.96677937112552594 


 360.00276925196848739 
Principal Stresses


1  360.003


3  280.036
1  max ztrans
3  min ztrans
2  I1  1  3
2  159.967
3
0
c.- Compute the hydrostatic stress and the components of the stress deviator tensor
Hydrostatic stress tensor:
Calculate the mean stress:
p 
I1
3
p  26.667
Then the hydrostatic stress tensor is:
0
0 
 26.667

hydro 
0
26.667
0 


0
26.667 
 0
hydro  p  ij
The stress deviator tensor is:
2.4 
 26.667 240


deviator  240 226.667
0


0
253.333 
 2.4
deviator  matrix  hydro
d.- Compute the values of the first three invariants of the stress deviator tensor.
J1  0
1
2
J2    I1  I2


3
J2  4  10
J3  deviator
J3  1.306  10
4
7
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.
Characteristic eqn
chareqn2  deviator   1 ij
3
chareqn2  115739.09333333333333  1  1.0  1  1.3059435140740740741e7
solve block
guess
 1  0
3
115739.09333333333333  1  1.0  1  1.3059435140740740741e7
Given
0
 
z1  Find  1
transpose the answer
 133.30011270445885928 
z1   386.66943591863515405 


 253.36932321417629477 
T
 133.30011270445885928 
ztrans_1   386.66943591863515405 


 253.36932321417629477 
Principal Deviator Stresses
s 1  max ztrans_1


s 1  386.669
s 3  min ztrans_1


s 3  253.369
s 2  J1  s 1  s 3
s 2  133.3
Since s kk  0 then the stress deviator tensor is in a state of pure shear, and the principle shear
stresses are the same as the stress deviator tensor principle stresses.