Silva
Kevin Silva
MANE 7100 – HOMEWORK ASSIGNMENT 1
Problem 1
Question:
Write a brief essay highlighting key characteristics of the mechanical behavior of metals, ceramics,
polymer and reinforced composite materials. Discuss the similarities and differences.
Solution:
Ductility is a key characteristic property that affects the mechanical behavior of metals. Ductility is a
solid materials ability to deform under stress. This characteristic is important in describing how metals
react when they are under a load. Metals can typically deform a significant amount before failure or
rupture. There are two types of deformations. The first is an elastic deformation (material obtains original
shape/dimensions after a load is removed) which can be described by Hooke’s Law where stress is
linearly proportional to strain. The second type is plastic deformation where the material is loaded to
certain point where when unloaded it will not return to its original shape/dimensions.
Ceramic materials typically have either ionic (bonding with electrostatic attraction) or covalent bond
(bonding with sharing pairs of electrons). A material with these type of bonds fracture before any plastic
deformation occurs. This key characteristic property of ceramics is called brittleness. In addition, ceramic
materials tend to be porous, the pores act as stress concentrators which further reduces tensile strength.
Polymers are a large molecule composed of repeating structural units. In general, the tensile strength of a
polymer increases with polymer chain length and cross linking of polymer chains. When polymers are
subjected to tensile stresses in the elastic region, the chain molecules are elongated in the direction of the
applied stress. When a polymer is subjected to a plastic deformation, the chains of molecules slip past
each other and align in the loading direction. In general, polymers are stronger in compression than in
other loading conditions.
Reinforced composites are engineered or naturally occurring materials made from two or more materials
with reinforcing fibers. Physical properties of reinforced composite materials are dependent on the
direction of the applied force. This is due to how the material was manufactured in which the fiber
reinforcement is aligned.
A similarity between metals, ceramics, polymers and reinforced composites is the temperature affects the
mechanical properties of the material. Metals and reinforced composites are more effective under tensile
loading than in compression. However, physical properties of metals are independent of the direction of
the force whereas reinforced composites mechanical properties are dependent due to the direction of the
reinforcing material. Ceramic and polymer materials are more effective under compression than in
tension. Polymers have favorable strength to weight ratios and manufacturing benefits compared to metals
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make plastics preferable in many applications. All the materials discussed behave different ways when
they are subjected to a tensile or compressive load.
Problem 2
Question:
Nickel has the face centered cubic (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 = 1 mm?
Solution:
The density and standard atomic weight of nickel are 7.81 g/cm3 and 58.6934 amu, respectively (Source:
http://en.wikipedia.org/wiki/Nickel).
𝜌𝑁𝑖 = 8.908
𝑔
𝑐𝑚3
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐴𝑡𝑜𝑚𝑖𝑐 𝑊𝑒𝑖𝑔ℎ𝑡 = 58.6934 𝑎𝑚𝑢 = 58.6934
𝑔
𝑚𝑜𝑙
To determine the mass of one Ni atom
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑁𝑖 𝐴𝑡𝑜𝑚 =
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐴𝑡𝑜𝑚𝑖𝑐 𝑊𝑒𝑖𝑔ℎ𝑡
𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜′𝑠 𝑁𝑢𝑚𝑏𝑒𝑟
Where:
𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜 ′ 𝑠 𝑁𝑢𝑚𝑏𝑒𝑟 = 6.023 𝑥 1023
𝑎𝑡𝑜𝑚𝑠
𝑚𝑜𝑙
𝑔
𝑔
𝑚𝑜𝑙
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑁𝑖 𝐴𝑡𝑜𝑚 =
= 9.745 𝑥10−23
𝑎𝑡𝑜𝑚𝑠
𝑎𝑡𝑜𝑚
6.023 𝑥 1023
𝑚𝑜𝑙
58.6934
As stated in the problem statement, Ni has a FCC crystal structure. Figure 1 shows the FCC crystal
structure.
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Figure 1 – Face Centered Cubic Structure
(Source: http://www.chemprofessor.com/solids_files/image019.jpg)
Each face of the unit cell has 1/2 an atom and each corner of the unit cell has a 1/8 of an atom, therefore:
(6
𝑓𝑎𝑐𝑒
1 𝑎𝑡𝑜𝑚
𝑎𝑡𝑜𝑚
)∗(
) = (3
)
𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
2 𝑓𝑎𝑐𝑒
𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
And
(8
𝑐𝑜𝑟𝑛𝑒𝑟
1 𝑎𝑡𝑜𝑚
𝑎𝑡𝑜𝑚
)∗(
) = (1
)
𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
8 𝑐𝑜𝑟𝑛𝑒𝑟
𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
(3
𝑎𝑡𝑜𝑚
𝑎𝑡𝑜𝑚
𝑎𝑡𝑜𝑚
) + (1
) = (4
) 𝑖𝑛 𝑎 𝐹𝐶𝐶 𝑠𝑡𝑟𝑢𝑐𝑡𝑢𝑟𝑒
𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
The total mass of a crystal structure is determined by multiplying the mass of a Ni atom by number of
atoms are in a unit cell.
(9.745 𝑥10−23
𝑔
𝑎𝑡𝑜𝑚
𝑔
) ∗ (4
) = 3.898 𝑥10−22
𝑁𝑖 𝑎𝑡𝑜𝑚
𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
The total volume of a unit cell is determined by dividing the mass of the cell by the density of Ni.
𝑉𝑐𝑒𝑙𝑙
𝑔
−22
3
𝑚𝑐𝑒𝑙𝑙 3.898 𝑥10
𝑢𝑛𝑖𝑡
𝑐𝑒𝑙𝑙 = 4.376𝑥10−23 𝑐𝑚
=
=
𝑔
𝜌𝑁𝑖
𝑢𝑛𝑖𝑡 𝑐𝑒𝑙𝑙
8.908
𝑐𝑚3
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To determine the length of each leg of the cube (a, as shown in Figure 1) we solve the equation.
𝑎3 = 4.376𝑥10−23
3
𝑎 = √4.376𝑥10−23 = 3.524 𝑥 10−8 𝑐𝑚
From Figure 1,
4𝑟 = √𝑎2 + 𝑎2
or
𝑟=
√𝑎2 + 𝑎2 √(3.524 𝑥 10−8 𝑐𝑚)2 + (3.524 𝑥 10−8 𝑐𝑚)2
=
= 1.246𝑥10−8 𝑐𝑚
4
4
Radius of a nickel atom = 1.246𝑥10−8 𝑐𝑚
a is the lattice parameter
𝑎 = 3.524 𝑥 10−8 𝑐𝑚
To determine how many atoms there are in a single crystal shaped as a cube with side = 1 mm
The volume of a cube 1mm = .1 cm
𝑉𝑐𝑢𝑏𝑒 = (. 1𝑐𝑚)3 = .001 𝑐𝑚3
4 𝑎𝑡𝑜𝑚𝑠
𝑋 𝑎𝑡𝑜𝑚𝑠
=
−23
3
4.376𝑥10 𝑐𝑚
. 001 𝑐𝑚3
(𝑋 𝑎𝑡𝑜𝑚𝑠) ∗ (4.376𝑥10−23 𝑐𝑚3 ) = (4 𝑎𝑡𝑜𝑚𝑠) ∗ (.001 𝑐𝑚3 )
𝑋 𝑎𝑡𝑜𝑚𝑠 =
(4 𝑎𝑡𝑜𝑚𝑠) ∗ (.001 𝑐𝑚3 )
= 8.014𝑥1019 𝑎𝑡𝑜𝑚𝑠
(4.376𝑥10−23 𝑐𝑚3 )
9.141x1019 atoms in a 1mm cube
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Problem 3
Question:
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 𝑥 10−19
)
𝑘𝑇
𝐽
Where the Boltzmann constant 𝑘 = 1.38𝑥10−23 𝑚𝑜𝑙𝑒𝐾. Calculate and plot the value of 𝑁𝑣 for the
temperature range 300 ≤ 𝑇 ≤ 2042.
Solution:
Fraction of Latice Vacant Sites (Nv)
Nv in a Pt
0.00012
0.0001
0.00008
0.00006
Nv
0.00004
0.00002
0
250
450
650
850
1050
1250
1450
1650
1850
2050
Temperature (Kelvin)
Figure 2 – Fraction of Vacant Lattice Sites
Temperature
300
350
400
450
500
Nv
1.39661E-27
9.58282E-24
7.22448E-21
1.24944E-18
7.70984E-17
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550
600
650
700
750
800
850
900
950
1000
1050
1100
1150
1200
1250
1300
1350
1400
1450
1500
1550
1600
1650
1700
1750
1800
1850
1900
1950
2000
2042
2.24833E-15
3.73713E-14
4.03106E-13
3.09561E-12
1.81147E-11
8.49969E-11
3.325E-10
1.11779E-09
3.30748E-09
8.78057E-09
2.12405E-08
4.74165E-08
9.87114E-08
1.93316E-07
3.58772E-07
6.34906E-07
1.07706E-06
1.75944E-06
2.77849E-06
4.25613E-06
6.34268E-06
9.21938E-06
1.31005E-05
1.82346E-05
2.49057E-05
3.34333E-05
4.4172E-05
5.75107E-05
7.38709E-05
9.37047E-05
0.0001134
Problem 4
Question:
The density of dislocations in heavily worked nickel is of the order of 106 dislocation lines per square
meter. Assume that the dislocations are randomly arranged and estimate the average space between
neighboring dislocation lines.
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Solution:
1
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝐵𝑒𝑡𝑤𝑒𝑒𝑛 𝐷𝑖𝑠𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛𝑠 = √
𝜌
Where 𝜌 is the density of the dislocation lines per square meter.
𝜌 = 106 𝑑𝑖𝑠𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛 𝑙𝑖𝑛𝑒𝑠 𝑝𝑒𝑟 𝑠𝑞𝑢𝑎𝑟𝑒 𝑚𝑒𝑡𝑒𝑟
1
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝐵𝑒𝑡𝑤𝑒𝑒𝑛 𝐷𝑖𝑠𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛𝑠 = √ 6
= 1𝑥10−3 𝑚
10 𝑑𝑖𝑠𝑙𝑜𝑐𝑎𝑡𝑖𝑜𝑛𝑠
𝑚2
1𝑥10−3 𝑚 is the average space
between dislocation lines
Problem 5
Question:
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.
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 = 0.1 m, moreover let L = 1 m and
E = 1011 Pa. Calculate the values of the maximum beam deflection for cases (a) and (b) above if
P = 105 N and Q = 105 N=m, respectively.
Solution:
Case 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.
Figure 3 shows a supported beam on either end with a point load in the center.
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Figure 3 – Beam with a Point Load in the Center (Assume W=P)
The governing differential equation for the elastic curve is
𝑑2 𝑦 𝑀(𝑥)
=
𝑑𝑥 2
𝐸𝐼
𝑒𝑞(1)
Where M is the moment, E is the modulus of elasticity and I is moment of inertia. In this case the moment
is
𝑀=−
𝑃𝑥
2
𝑒𝑞(2)
Plug eq (2) into eq (1)
𝑑2 𝑦
𝑃𝑥 1
=−
2
𝑑𝑥
2 𝐸𝐼
𝑒𝑞(3)
To solve the second order differential equation you need two boundary conditions (B.C.)
B.C. 1
𝑦(0) = 0
B.C. 2
𝑦′(𝐿/2) = 0
Integrate eq (3)
𝑑𝑦
𝑃𝑥 2
=−
+ 𝐶1
𝑑𝑥
4𝐸𝐼
𝑒𝑞 (4)
Where C1 is a constant of integration, plug in B.C. 2
0=−
𝐶1 =
𝐿 2
𝑃 (2)
4𝐸𝐼
+ 𝐶1
𝑃𝐿2
16𝐸𝐼
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Plug C1 into eq (4)
𝑑𝑦
𝑃𝑥 2
𝑃𝐿2
=−
+
𝑑𝑥
4𝐸𝐼 16𝐸𝐼
𝑒𝑞 (5)
Integrate eq (5)
𝑃𝑥 3 𝑃𝐿2 𝑥
𝑦(𝑥) = −
+
+ 𝐶2
12𝐸𝐼 16𝐸𝐼
𝑒𝑞 (6)
Where C2 is a constant of integration, plug in B.C. 1
𝑃(0)3 𝑃𝐿2 (0)
0=−
+
+ 𝐶2
12𝐸𝐼
16𝐸𝐼
𝐶2 = 0
Plug C2 into eq (6)
𝑦(𝑥) = −
𝑃𝑥 3 𝑃𝐿2 𝑥
+
12𝐸𝐼 16𝐸𝐼
𝑒𝑞 (7)
𝑃𝑥 3 𝑃𝐿2 𝑥
𝑦(𝑥) = −
+
12𝐸𝐼 16𝐸𝐼
Maximum deflection occurs at center of beam or when x=L/2
𝐿 3
𝐿
3
3
3
𝑃( )
𝑃𝐿2 ( )
2
2 = − 𝑃𝐿 + 𝑃𝐿 = 𝑃𝐿
𝑦(𝑥) = −
+
12𝐸𝐼
16𝐸𝐼
96𝐸𝐼 32𝐸𝐼 48𝐸𝐼
Case 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.
Figure 4 shows a supported beam on either end with a distributed load.
Figure 4 - Beam with a Distributed Load (assume distributed load Q)
The governing differential equation for the elastic curve is
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𝑑2 𝑦 𝑀(𝑥)
=
𝑑𝑥 2
𝐸𝐼
𝑒𝑞(8)
Where M is the moment, E is the modulus of elasticity and I is moment of inertia. In this case the moment
is
𝑄𝑥 2 𝑄𝐿𝑥
𝑀=
−
2
2
𝑒𝑞(9)
Plug eq (2) into eq (1)
𝑑2 𝑦
𝑄𝑥 2 𝑄𝐿𝑥 1
=(
−
)
𝑑𝑥 2
2
2 𝐸𝐼
𝑒𝑞(10)
To solve the second order differential equation you need two boundary conditions (B.C.)
B.C. 1
𝑦(0) = 0
B.C. 2
𝑦′(𝐿/2) = 0
Integrate eq (10)
𝑑𝑦
𝑄𝑥 3 𝑄𝐿𝑥 2 1
=(
−
) + 𝐶1
𝑑𝑥
6
4
𝐸𝐼
𝑒𝑞 (11)
Where C1 is a constant of integration, plug in B.C. 2
𝐿 3
𝐿 2
𝑄 (2)
𝑄𝐿 (2)
1
0=(
−
) + 𝐶1
6
4
𝐸𝐼
0=(
𝐶1 =
𝑄𝐿3 𝑄𝐿3 1
−
) + 𝐶1
48
16 𝐸𝐼
𝑄𝐿3
24𝐸𝐼
Plug C1 into eq (11)
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𝑑𝑦
𝑄𝑥 3 𝑄𝐿𝑥 2 1
𝑄𝐿3
=(
−
) +
𝑑𝑥
6
4
𝐸𝐼 24𝐸𝐼
𝑒𝑞 (12)
Integrate eq (12)
𝑦(𝑥) = (
𝑄𝑥 4 𝑄𝐿𝑥 3 1 𝑄𝐿3 𝑥
−
+ 𝐶2
) +
24
12 𝐸𝐼 24𝐸𝐼
𝑒𝑞 (12)
Where C2 is a constant of integration, plug in B.C. 1
𝑄(0)4 𝑄𝐿(0)3 1 𝑄𝐿3 (0)
0=(
−
+ 𝐶2
) +
24
12
𝐸𝐼
24𝐸𝐼
𝐶2 = 0
Plug C2 into eq (12)
𝑦(𝑥) =
𝑄𝑥 4 𝑄𝐿𝑥 3 𝑄𝐿3 𝑥
−
+
24𝐸𝐼 12𝐸𝐼 24𝐸𝐼
𝑒𝑞 (13)
𝑦(𝑥) =
𝑄𝑥 4 𝑄𝐿𝑥 3 𝑄𝐿3 𝑥
−
+
24𝐸𝐼 12𝐸𝐼 24𝐸𝐼
Maximum deflection occurs at center of beam or when x=L/2
𝑦(𝑥) =
𝐿 4
𝑄 (2)
24𝐸𝐼
−
𝐿 3
𝑄𝐿 (2)
12𝐸𝐼
+
𝐿
𝑄𝐿3 (2)
24𝐸𝐼
=
𝑄𝐿4
𝑄𝐿4
𝑄𝐿4
5𝑄𝐿4
−
+
=
384𝐸𝐼 96𝐸𝐼 48𝐸𝐼
384𝐸𝐼
Case c. Consider a beam of square cross section a = b = 0.1 m, moreover let L = 1 m and
E = 1011 Pa. Calculate the values of the maximum beam deflection for cases (a) and (b) above if
P = 105 N and Q = 105 N/m, respectively.
Moment of inertia for a beam with a square cross section is given by
𝐼𝑥 =
𝑎𝑏 3
12
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For case a
𝑃𝐿3
12𝑃𝐿3
12(105 𝑁)(1𝑚)3
𝑦(𝑥) =
=
=
= .0025𝑚
48𝐸𝐼 48𝐸𝑎𝑏 3 48(1011 𝑃𝑎)(0.1𝑚)(0.1𝑚)3
For case a, max deflection y(x) = .0025 m
For case b
𝑦(𝑥) =
5𝑄𝐿4
125𝑄𝐿4
12 ∗ 5(105 𝑁)(1𝑚)4
=
=
= .0016𝑚
384𝐸𝐼 384𝐸𝑎𝑏 3 384(1011 𝑃𝑎)(0.1𝑚)(0.1𝑚)3
For case b, max deflection y(x) = .0016 m
Problem 6
Question:
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 tensor.
e. Solve the characteristic equation of the stress deviator tensor and determine the values of the
principle stress deviations and give the values of the principle shearing stress.
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Solution:
Case a. Compute the values of the first three stress invariants of the stress tensor.
𝐼1 = 𝜎11 + 𝜎22 + 𝜎33
𝐼1 = 0 + 200 − 280 = −80
I1 = -80
𝜎22
𝐼2 = [𝜎
23
𝜎32
𝜎11
𝜎33 ] + [𝜎12
𝜎21
𝜎11
𝜎22 ] + [𝜎13
𝜎31
𝜎33 ]
𝐼2 = (𝜎22 𝜎33 − 𝜎32 𝜎23 ) + (𝜎11 𝜎22 − 𝜎21 𝜎12 ) + (𝜎11 𝜎33 − 𝜎31 𝜎13 )
𝐼2 = ((200)(−280) − (0)(0)) + ((0)(200) − (−240)(−240)) + ((0)(−280) − (−2.4)(−2.4))
𝐼2 = 113606
I2 = -113606
𝜎11
𝐼3 = [𝜎21
𝜎31
𝜎12
𝜎22
𝜎32
𝜎13
𝜎23 ]
𝜎33
𝐼3 = 𝜎11 (𝜎22 𝜎33 − 𝜎23 𝜎32 ) − 𝜎12 (𝜎21 𝜎33 − 𝜎23 𝜎31 ) + 𝜎13 (𝜎21 𝜎32 − 𝜎22 𝜎31 )
𝐼3 = 0((200)(−280) − (0)(0)) − −240((−240)(−280) − (0)(−2.4))
+ −2.4((−240)(0) − (200)(−2.4))
𝐼3 = 0 − (−16128000) + (−1152) = 16126848
I3 = 16126848
Case b. Solve the characteristic equation of the stress tensor and determine the values of the principal
stresses.
Characteristic equation of the stress tensor
𝜎 3 − 𝜎 2 𝐼1 + 𝜎 1 𝐼2 − 𝐼3 = 0
𝑒𝑞(14)
Plug invariants into eq (14)
𝜎 3 − 𝜎 2 (−80) + 𝜎(−113606) − (16126848) = 0
𝑒𝑞 (15)
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Solved eq (15) for the three roots using a TI-89 calculator
Principle Stresses are: σ1 = 360 MPa, σ2 = -160 MPa, σ3 = -280 MPa
Case c. Compute the hydrostatic stress and the components of the stress deviator tensor.
The hydrostatic stress is the mean stress
𝜎1 + 𝜎2 + 𝜎3
360 𝑀𝑃𝑎 + (−160 𝑀𝑃𝑎) + (−280 𝑀𝑃𝑎)
80 𝑀𝑃𝑎
𝜎𝑚𝑒𝑎𝑛 = 𝜎𝑜 =
=
=
3
3
3
= −26.67 𝑀𝑃𝑎
Hydrostatic Stress = Mean Stress = -26.67 MPa
Stress deviator tensor
𝜎𝑖𝑗′ = 𝜎𝑖𝑗 − 𝜎𝑜 𝛿𝑖𝑗
𝜎𝑜
𝜎𝑜 𝛿𝑖𝑗 = [𝜎𝑜
𝜎𝑜
𝜎′ =
𝜎11 − 𝜎𝑜
𝜎21
𝜎31
𝜎𝑜
𝜎𝑜
𝜎𝑜
𝜎𝑜 1 0
𝜎𝑜 ] [0 1
𝜎𝑜 0 0
𝜎12
𝜎22 − 𝜎𝑜
𝜎32
𝜎𝑜
0
0] = [ 0
0
1
0
𝜎𝑜
0
0
0]
𝜎𝑜
𝜎13
𝜎23
𝜎33 − 𝜎𝑜
0 − (−26.67)
−240
−2.4
−240
200 − (−26.67)
0
𝜎′ =
−2.4
0
−280 − (−26.67)
26.67 −240
−2.4
𝜎′ = −240 226.67
0
−2.4
0
−253.33
Stress Deviator Tensor
𝜎′11 = 26.67
𝜎′22 = 226.67
𝜎′33 = −253.33
𝜎′12 = 𝜎′21 = −240
𝜎′13 = 𝜎′31 = −2.4
𝜎′23 = 𝜎′32 = 0
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Case d. Compute the values of the first three invariants of the stress tensor.
The invariants of the stress deviation tensor are
𝐽1 = 0
𝐽2 = 3𝜎𝑜2 − 𝐼2 = 3(−26.67)2 − (−113606) = 115740
𝐽2 = 115739
𝐽3 = 𝐼3 + 𝐽2 𝜎𝑜 − 𝜎𝑜3 = 16126848 + (115739)(−26.67) − (−26.67)3 = 13059058
𝐽3 = 13059058
Case e. Solve the characteristic equation of the stress deviator tensor and determine the values of the
principle stress deviations and give the values of the principle shearing stress.
Characteristic equation of the stress deviator tensor
𝜎 3 − 𝜎 2 𝐽1 − 𝜎 1 𝐽2 − 𝐽3 = 0
𝑒𝑞 (16)
Plug stress deciator invariants into eq (16)
𝜎 3 − 𝜎 2 (0) − 𝜎 1 (115739) − 13059058 = 0
𝜎 3 − 𝜎 1 (115739) − 13059058 = 0
𝑒𝑞 (17)
𝑒𝑞 (17)
Solved eq (17) for the three roots using a TI-89 calculator
Principle Stress Deviations are: σ’1 = 387 MPa, σ’2 = -133 MPa, σ’3 = -253 MPa
The principle shearing stresses are
𝜎1 − 𝜎3 (387𝑀𝑃𝑎) − (−253𝑀𝑃𝑎)
=
= 320𝑀𝑃𝑎
2
2
𝜎1 − 𝜎2 (387𝑀𝑃𝑎) − (−133𝑀𝑃𝑎)
𝜏2 =
=
= 260𝑀𝑃𝑎
2
2
𝜎2 − 𝜎3 (−133𝑀𝑃𝑎) − (−253𝑀𝑃𝑎)
𝜏3 =
=
= 60𝑀𝑃𝑎
2
2
𝜏1 =
Principle Shearing Stress are: 𝜏1 = 320𝑀𝑃𝑎, 𝜏2 = 260𝑀𝑃𝑎, 𝜏3 = 60𝑀𝑃𝑎
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