Review
When we use matrix notation to show the
stresses on an element
The rows represent the axis which the face
is parallel to
Chapter Four – States of Stress
Part Three
“When making your choice in life,
do not neglect to live.”
Samuel Johnson
Wednesday, October 16,
2002
Review
Meeting Twenty One - States of Stress III
X-face
Y-face
Z-face
2
Review
The columns represent the axis which the
stress is directed along or parallel to
X-axis
Y-axis
If we look at a element in the shape of a
cube we can draw the stresses on that
element
Z-axis
σ x τ xy τ xz 


τ
σ
τ
yx
y
yz


τ zx τ zy σ z 


Wednesday, October 16,
2002
σ x τ xy τ xz 


τ
σ
τ
yx
y
yz


τ zx τ zy σ z 


Meeting Twenty One - States of Stress III
σ x τxy τ xz 


τ yx σ y τ yz 
τ zx τ zy σ z 


3
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III
4
1
Review
General State of Stress
If all the stresses act in one plane, we
have plane stress
For the xy plane we would have
σ x τ xy
τ
 yx σ y
 0 0
Wednesday, October 16,
2002
If we return to our original definition of an
element that has stresses on all faces of
the cube we have
0
0 
0 
Meeting Twenty One - States of Stress III
σ x τxy τ xz 


τ yx σ y τ yz 
τ zx τ zy σ z 


5
General State of Stress
σ x τxy τ xz 


τ yx σ y τ yz 
τ zx τ zy σ z 


Meeting Twenty One - States of Stress III
Meeting Twenty One - States of Stress III
6
General State of Stress
In our previous discussions we rotated the
orientation of the axis so that we were
able to locate the maximum stresses on
each face
Wednesday, October 16,
2002
Wednesday, October 16,
2002
In three dimensions, we have to rotate our
axis system is a more complex fashion
than the case with plane stress
σ x τxy τ xz 


τ yx σ y τ yz 
τ zx τ zy σ z 


7
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III
8
2
General State of Stress
General State of Stress
If we rotate our axis system to some new
prime position, the stresses on the new
σ
τ
τ 


faces will also be transformed
τ σ τ 
x
xy
xz
yx
y
yz
τ zx τ zy σ z 
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III
9
General State of Stress
Wednesday, October 16,
2002
τ ' xz  σ 1 0

τ ' yz  =  0 σ 2

σ 'z   0 0
Meeting Twenty One - States of Stress III
x
xy
xz
yx
y
yz
τ zx τ zy σ z 
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III
10
General State of Stress
In this case we will have the situation
where
σ 'x τ 'xy

τ ' yx σ 'y
τ ' zx τ 'zy

There exists some transformation to a new
coordinate system (x’, y’, z’) such that all
σ
τ
τ 
the shear stresses will be equal to 0 τ σ τ 
The axis (x’, y’, and z’) are call the
principal axis
σ 'x τ 'xy

τ ' yx σ 'y
τ ' zx τ 'zy

0
0

σ 3 
11
Wednesday, October 16,
2002
τ ' xz  σ 1 0

τ ' yz  =  0 σ 2

σ 'z   0 0
Meeting Twenty One - States of Stress III
0
0

σ 3 
12
3
General State of Stress
General State of Stress
The axial stresses σ1, σ2 and σ3 are the
principal stresses
σ 'x τ 'xy

τ ' yx σ 'y
τ ' zx τ 'zy

Wednesday, October 16,
2002
τ ' xz  σ 1 0

τ ' yz  =  0 σ 2

σ 'z   0 0
Meeting Twenty One - States of Stress III
In order to solve for the principal stresses
we must solve for the roots of a 3rd order
equation
13
General State of Stress
Wednesday, October 16,
2002
τ ' xz  σ 1 0
0
 

τ ' yz  = 0 σ 2 0


σ 'z   0 0 σ 3 
Meeting Twenty One - States of Stress III
14
There are multiple possible orientations
that can give the principal stresses on
alternate faces
σ 'x τ 'xy

τ ' yx σ 'y
τ ' zx τ 'zy

σ 3 − I1σ 2 + I2σ − I3 = 0
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III
General State of Stress
The coefficients, I’s, in the expression are
known as stress invariants and do not
depend on the orientation of the
transformed coordinate system
σ 'x τ 'xy

τ ' yx σ 'y
τ ' zx τ 'zy

σ 'x τ 'xy τ ' xz  σ 1 0
0

 
τ
'
σ
'
τ
'
=
0
σ
0
y
yz 
2
 yx


τ ' zx τ 'zy σ 'z   0 0 σ 3 


σ 3 − I1σ 2 + I2σ − I3 = 0
0
0

σ 3 
τ ' xz  σ 1 0
0
 

τ ' yz  = 0 σ 2 0


σ 'z   0 0 σ 3 
σ 3 − I1σ 2 + I2σ − I3 = 0
15
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III
16
4
General State of Stress
Problem 5-4.10
The three stress invariants are given by
σ 'x τ 'xy

τ ' yx σ 'y
τ ' zx τ 'zy

τ ' xz  σ 1 0 0 
 

τ ' yz  = 0 σ 2 0


σ 'z   0 0 σ 3 
Given the following state of stress (all units
are ksi)
σ x := 300
σ y := 200
σ z := − 200
τ xy := 150
τ yz := 100
τ zx := − 100
σ 3 − I1σ 2 + I2σ − I3 = 0
I1 = σ x + σ y + σ z
I 2 = σ xσ y + σ yσ z + σ zσ x − τxy2 − τ 2yz − τ 2zx
I 3 = σ xσ y σ z − σ xτ yz − σ yτ zx − σ zτ xy + 2τ xyτ yz τ zx
2
Wednesday, October 16,
2002
2
2
Meeting Twenty One - States of Stress III
17
Problem 5-4.10
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III
Problem 5-4.10
We can use our expressions to calculate
the stress invariants
Substituting we have
σ x := 300
σ y := 200
σ z := − 200
σ x := 300
σ y := 200
σ z := − 200
τ xy := 150
τ yz := 100
τ zx := − 100
τ xy := 150
τ yz := 100
τ zx := − 100
I1 : = σ x + σ y + σ z
I1 : = σ x + σ y + σ z
I 2 : = σ x ⋅σ y + σ y ⋅σ z + σ z ⋅σ x − τxy
I 2 : = σ x ⋅σ y + σ y ⋅σ z + σ z ⋅σ x − τxy
( ) 2 − ( τyz ) 2 − ( τzx ) 2
2
2
2
I 3 : = σ x ⋅σ y ⋅ σ z − σ x ⋅( τyz ) − σ y ⋅( τzx ) − σ z ⋅( τxy ) + 2 ⋅τxy ⋅τ yz ⋅τ zx
( ) 2 − ( τyz ) 2 − ( τzx ) 2
2
2
2
I 3 : = σ x ⋅σ y ⋅ σ z − σ x ⋅( τyz ) − σ y ⋅( τzx ) − σ z ⋅( τxy ) + 2 ⋅τxy ⋅τ yz ⋅τ zx
I1 = 300
Wednesday, October 16,
2002
18
Meeting Twenty One - States of Stress III
19
Wednesday, October 16,
2002
I 2 = −8.25 × 10
4
I3 = −1.55 × 10
Meeting Twenty One - States of Stress III
7
20
5
Problem 5-4.10
Problem 5-4.10
We then solve for the values of σ that
solve the expression
I1 = 300
I 2 = −8.25 × 10
4
I3 = −1.55 × 10
So the principal stresses on the element
are
σ 3 − I1 σ 2 + I 2σ − I 3 = 0
 −I3

 I2
Vector : = 
 −I1

 1
σ 1 = −256.678
Wednesday, October 16,
2002







σ 3 = 409.051
Meeting Twenty One - States of Stress III
Wednesday, October 16,
2002







σ : = polyroots (Vector )
σ 2 = 147.627
σ 3 = 409.051
Meeting Twenty One - States of Stress III
22
Triaxial stress is a state of stress that can
be defined as
σ 3 = 409.051
σ2 − σ 3
σ 3 − σ1 
 σ 1 − σ2
τmax := max
,
,

2
2
2


Wednesday, October 16,
2002
409.05 ksi
Triaxial Stress
The maximum shear stress on the element
can then be found by
σ 2 = 147.627
n
σ 1 = −256.678
21
Problem 5-4.10
σ 1 = −256.678
-256.68 ksi
147.63 ksi
 −I3

 I2
Vector : = 
 −I1

 1
σ : = polyroots (Vector )
σ 2 = 147.627
n
n
7
Meeting Twenty One - States of Stress III
σ x
0

 0
τmax = 332.864
23
Wednesday, October 16,
2002
0
σy
0
0
0 
σ z 
Meeting Twenty One - States of Stress III
24
6
Triaxial Stress
Triaxial Stress
These stresses are by definition also the
principal stresses
σ x
0

 0
Wednesday, October 16,
2002
0
σy
0
0
0 
σ z 
Meeting Twenty One - States of Stress III
 σ x − σ y σ y − σ z σ z −σ x
Maximum 
,
,

2
2
2

25
Triaxial Stress
Meeting Twenty One - States of Stress III
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III




26
Triaxial Stress
Triaxial stress becomes important when we
consider pressure vessels with circular
cross sections
Wednesday, October 16,
2002
So we can define the maximum shear
stress on an element in triaxial stress by
27
Triaxial stress becomes important when we
consider pressure vessels with circular
cross sections
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III
28
7
Spherical Pressure Vessels
Spherical Pressure Vessels
We can start by assuming we have a
spherical pressure vessel with a wall
thickness, t, and a radius, R
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III
We must make the assumption that the
thickness, t, is much smaller than the
radius of the vessel, R
29
Spherical Pressure Vessels
Meeting Twenty One - States of Stress III
Meeting Twenty One - States of Stress III
30
Spherical Pressure Vessels
The confined gas within the pressure
vessel is at some pressure pi while the
external pressure on the vessel is po
Wednesday, October 16,
2002
Wednesday, October 16,
2002
If we cut through the center of the vessel,
we can look at the forces developed on the
vessel
Remember we are cutting a sphere in half
31
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III
32
8
Spherical Pressure Vessels
Spherical Pressure Vessels
The force to the right exerted by the
external pressure, po, is equal to the
product of the pressure times the area
projected
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III
Think of the projected area as what you
would see is you were looking along an
axis straight at the half of the ball.
33
Spherical Pressure Vessels
Wednesday, October 16,
2002
By the same idea, the force exerted by the
internal pressure will be equal to
(
Fpi = pi π ( R − t )
Fp0 = p 0 (π R 2 )
Meeting Twenty One - States of Stress III
34
Spherical Pressure Vessels
You would see a circular area with a radius
of R
So the force exerted by the external
pressure would be equal to
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III
35
2
)
Fp0 = p 0 (π R 2 )
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III
36
9
Spherical Pressure Vessels
Spherical Pressure Vessels
But since R is >> t, we can make a good
approximation to the force as
Fpi = p i (π R 2 )
So the net force from the pressure
difference would be
Fp0 − Fp i = p 0 (π R 2 ) − p i (π R 2 )
Fp0 = p 0 (π R 2 )
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III
37
Spherical Pressure Vessels
F − F = p 0 (π R 2 ) − p i (π R 2 )
Meeting Twenty One - States of Stress III
Meeting Twenty One - States of Stress III
38
Spherical Pressure Vessels
In order for the system to be in
equilibrium, the difference between the
outward force Fpi and the inward force Fpo
must be made up by the walls of the
vessel
0
Wednesday,pOctober
16,p i
2002
Wednesday, October 16,
2002
39
This force is developed as the stress in the
wall times the area
The area over which the stress acts is
A = π R2 −π ( R − t )
2
F − F = p 0 (π R 2 ) − p i (π R 2 )
0
Wednesday,pOctober
16,p i
2002
Meeting Twenty One - States of Stress III
40
10
Spherical Pressure Vessels
Spherical Pressure Vessels
Expanding the second term we have
Since t is very small, we can ignore the t2
term and we have
A = π R 2 − π ( R 2 − 2 Rt − t 2 )
A = π R 2 − π ( R 2 − 2Rt )
A = 2π Rt
F − F = p 0 (π R 2 ) − p i (π R 2 )
0
Wednesday,pOctober
16,p i
2002
Meeting Twenty One - States of Stress III
41
Spherical Pressure Vessels
Meeting Twenty One - States of Stress III
42
So the sum of the forces is equal to
0 = p0 (π R 2 )− pi (π R 2 ) + 2σπ Rt
Fconfining = 2σπ Rt
F − F = p 0 (π R 2 ) − p i (π R 2 )
Meeting Twenty One - States of Stress III
Spherical Pressure Vessels
So the force required to generate the
confining stress σ is equal to
0
Wednesday,pOctober
16,p i
2002
F − F = p 0 (π R 2 ) − p i (π R 2 )
0
Wednesday,pOctober
16,p i
2002
43
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III
44
11
Spherical Pressure Vessels
Spherical Pressure Vessels
Isolating the stress we have
− p0 (π R ) + pi (π R
2
2π Rt
Wednesday, October 16,
2002
2
And reducing
) =σ
R ( pi − p0 )
2t
Meeting Twenty One - States of Stress III
45
Spherical Pressure Vessels
2t
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III
Meeting Twenty One - States of Stress III
46
Spherical Pressure Vessels
The stress σ is the same at every point on
the vessel
R ( pi − p0 )
Wednesday, October 16,
2002
=σ
The state of stress is different on the inner
and outer wall of the vessel
R ( pi − p 0 )
=σ
2t
47
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III
=σ
48
12
Spherical Pressure Vessels
Spherical Pressure Vessels
If we look at the outer wall, we have the
pressures acting as a axial stress along
one axis and the stress σ acting along the
other two axis
σ
0

 0
Wednesday, October 16,
2002
σ
0

 0
0 
0 
− po 
0
σ
0
Meeting Twenty One - States of Stress III
And on the inner wall we have
49
Wednesday, October 16,
2002
0
σ
0
0 
0 
− p i 
Meeting Twenty One - States of Stress III
50
Homework
Read Section on Cylindrical Vessel and
Allowable Stress
Problem 5-5.1
Problem 5-5.2
Problem 5-5.4
Wednesday, October 16,
2002
Meeting Twenty One - States of Stress III
51
13