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Stresses and Stress TensorForces and stressesS T R E S S E SS T R A I N STransformation of the stress tensor
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2
Stresses and Stress TensorForces and stresses
Two basic types of forces act on a body to produce stresses:
- Forces of the first type are called surface forces and act on the surface of the body. Surface forces are generally exerted when one body comes in contact with another.
- Forces of the second type are called body forces since they act on each particle of the body. Body forces are commonly produced by centrifugal, gravitational, or other force fields.
A solid body of arbitrary size, shape, and material acted upon by an equilibrium system of surface and /or body forces will respond so as to develop a system of internal forces also in equilibrium at any point within the body. If the body were cut by an arbitrary imaginary plane, these internal forces would in general be distributed continuouslyover the cut surface and would vary across the surface in both direction and intensity. Furthermore, the internal force distribution would also be a function of the orientation of the plane chosen for investigation.
• Stress is the term used to define the intensity and direction of the internal forces acting at given point on a particular plane.
• Stress has a magnitude and a sense, and is also associated with an area (or plane)over which it acts. Such mathematical objects are known as TENSORS and are of higher order than vectors. However, the components of the stress on the same plane (area) are vectors. In other words the Tensor is made of Vectors.
•The intensity of force (vector) normal to the plane (section) is called the normal stress σ.
•The intensity of force acting parallel to the plane (section) is called shear stress τ.
3
∆Fz
∆Fy
∆Fx
∆Fzz
∆Fzx
∆Fzy
The meaning of stress tensor
0lim
kl
ij ijij A
kl k l
F FA x x
σ∆ →
∆ ∆= ≅
∆ ∆ ∆
4
0 0
0 0
lim lim
lim lim
n n
A A
t t
A A
F normal force FA surface area A
F shear force FA surface area A
σ
τ
∆ → ∆ →
∆ → ∆ →
∆ ∆= =
∆ ∆∆ ∆
= =∆ ∆
Consider first a surface (plane) whose outer normal is the positive direction of axis z.
A complete description of the magnitudes and directions of stresses on all possible planes through a point constitute the stress state at a point. As illustrated in the Figure below, it is possible to resolve force ∆F into two components: one ∆Fn normal to the surface and the other ∆Ft tangential to the surface. The resultant force ∆F associated with a particular plane (surface) can be further resolved into three Cartesian components along the x, y, z axes i.e., ∆Fx, ∆Fy, ∆Fz. Subsequently, the Cartesian stress components on this plane can be defined as:
0
0
0
lim
lim
lim
zzz A
xzx A
yzy A
FA
FAFA
σ
τ
τ
∆ →
∆ →
∆ →
∆=
∆∆
=∆∆
=∆
All the components are associated with the plane having the outer normal parallel to the z axis.
If the same procedure is followed using surfaces whose outer normals are in the positive x and y directions, two more sets ofCartesian components σxx, τxy, τxz and σyy, τyx, τyz respectively can be obtained.
6
xx xy xz
ij yx yy yz
zx zy zz
ε ε εε ε ε ε
ε ε ε
⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦
S T R A I N S
!
2ij
ij
ij ij
Note
for i j
for i j
γε
ε ε
= ≠
= =
7
( )
( )
( )( ) ( ) ( )
( ' )1
1
1
2 1 2 1 2 1
xx xx yy zz
yy yy zz xx
zz zz xx yy
xy xy yz yz zx zx
Strains in terms of stresses Hooke s law
E
E
E
E E E
ε σ ν σ σ
ε σ ν σ σ
ε σ ν σ σ
ν ν νγ τ γ τ γ τ
⎡ ⎤= − +⎣ ⎦
⎡ ⎤= − +⎣ ⎦
⎡ ⎤= − +⎣ ⎦
+ + += = =
( )( ) ( ) ( )
( )( ) ( ) ( )
( )( ) ( ) ( )
( ) ( ) ( )
( ' )
11 1 2
11 1 2
11 1 2
2 1 2 1 2 1
xx xx yy zz
yy yy zz xx
zz zz xx yy
xy xy yz yz zx zx
Stresses in terms of strains Hooke s lawE
E
E
E E E
σ ν ε ν ε εν ν
σ ν ε ν ε εν ν
σ ν ε ν ε εν ν
τ γ τ γ τ γν ν ν
⎡ ⎤= + + +⎣ ⎦+ −
⎡ ⎤= + + +⎣ ⎦+ −
⎡ ⎤= + + +⎣ ⎦+ −
= = =+ + +
8
• Uni-axial stress state, σ11=σ33=σ12=σ23=σ31=0 and σ22≠0
22 22 2211 22 33
12 23 31
11
22 22
33
; ; ;
0; 0; 0;
0 0 0 0 00 0 ; 0 00 0 0 0 0
ij ij
E E Eνσ σ νσ
ε ε ε
ε ε ε
εε ε σ σ
ε
= − = = −
= = =
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
σ22
Thin plate under uniaxialtension or bending
9
• Plane stress state in principal axes, σ33=σ12=σ23=σ31=0 and σ11≠0, σ22≠0
( )11 2211 22 22 1111 22 33
12 23 31
11 11
22 22
33
; ;
0; 0; 0;
0 00 ; 0 ;
0 0 0 0 0ij ij
E E Eν σ σσ νσ σ νσ
ε ε ε
ε ε ε
ε σε ε σ σ
ε
+− −= = = −
= = =
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
• Plane stress state in principal axes, σ33=σ12=σ23=σ31=0 and σ11≠0, σ22≠0
ω
σ22
σ11
Thin rotating disk
10
• General plane strain state, σ23=σ31=0, ε33=0 and σ11≠0, σ22≠0, σ33≠0, σ12≠0
( ) ( ) ( )
( ) ( )
( ) ( ) ( ) ( )
11 22 33 22 11 33 33 11 2211 22 33
12 23 31
33 11 2233 11 22
2 211 22 22 11
11 22 33
12 23 31
; ; 0
0; 0; 0;
0
1 1 1 1; ; 0
0; 0; 0;
E E E
thenE
E E
σ ν σ σ σ ν σ σ σ ν σ σε ε ε
ε ε ε
σ ν σ σσ ν σ σ
σ ν νσ ν σ ν νσ νε ε ε
ε ε ε
− + − + − += = = =
= = =
− += ⇒ = +
− − + − − += = =
= = =
( )
11 11
22 22
33
11 11
22 22
11 22
0 0 0 00 0 ; 0 00 0 0 0 0
0 0 0 00 0 ; 0 00 0 0 0 0
ij ij
ij ij
ε σε ε σ σ
σ
ε σε ε σ σ
ν σ σ
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
⎡ ⎤⎡ ⎤⎢ ⎥⎢ ⎥= = ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥ +⎣ ⎦ ⎣ ⎦
or
11
( )
( )
( )( ) ( ) ( )
( ' )1
1
1
2 1 2 1 2 1
xx xx yy zz
yy yy zz xx
zz zz xx yy
xy xy yz yz zx zx
Strains in terms of stresses Hooke s law
E
E
E
E E E
ε σ ν σ σ
ε σ ν σ σ
ε σ ν σ σ
ν ν νγ τ γ τ γ τ
⎡ ⎤= − +⎣ ⎦
⎡ ⎤= − +⎣ ⎦
⎡ ⎤= − +⎣ ⎦
+ + += = =
( )( ) ( ) ( )
( )( ) ( ) ( )
( )( ) ( ) ( )
( ) ( ) ( )
( ' )
11 1 2
11 1 2
11 1 2
2 1 2 1 2 1
xx xx yy zz
yy yy zz xx
zz zz xx yy
xy xy yz yz zx zx
Stresses in terms of strains Hooke s lawE
E
E
E E E
σ ν ε ν ε εν ν
σ ν ε ν ε εν ν
σ ν ε ν ε εν ν
τ γ τ γ τ γν ν ν
⎡ ⎤= + + +⎣ ⎦+ −
⎡ ⎤= + + +⎣ ⎦+ −
⎡ ⎤= + + +⎣ ⎦+ −
= = =+ + +
12
Transformation of the stress tensor
xxσ
yyσ
xyτ
ϕ
x
y
,xxσ
,yyσ
,xyτ
ϕ
x’
y’
The stress components σxx, σyy, and τxy can be calculated on planes x and y, i.e. in x-y co-ordinates.
However we need to know stress components σ’xx, σ’yy, and τ’
xy on planes x’ and y’ being at angle ϕ with respect to the original system of co-ordinates x and y.
The stress components σxx, σyy, and τxy can be calculated on planes x and y, i.e. in x-y co-ordinates.
It is possible to determine stress components σ’xx, σ’yy, and τ’
xy by transformation of the original stress components σxx, σyy, and τxy.
13
xxxσ
yyσ
xyτ
y
ϕ,
xxσ
,yyσ
,xyτ
x’
y’
,
,
,
cos 2 sin 22 2
cos 2 sin 22 2
sin 2 cos 22
xx yy xx yyxy
xx yy xx yyxy
xx yyx
xx
yy
xy y
σ σ σ σϕ τ ϕ
σ σ σ σϕ τ ϕ
σ σ
σ
σ
ϕ τ ϕτ
+ −= + +
+ −= − −
−= − +
Transformation of the stress tensor – the mathematics
14
yyσ
xyτ
x
y
,xxσ
,yyσ
,xyτ
x’
y’
xxσ
A(σxx, τxy)
B(σyy, τyx)
A’(σ’xx, τ’xy)
B’(σ’yy, τ’yx)
ϕ
The Mohr circle
A(σxx, τxy)
A’(σ’xx, τ’xy)
B(σyy, τyx)
B’(σ’yy, τ’yx)
τxy
τ’xy
τyx
σyy
σ’xx
σ’yy
σxx
τ’yx
2ϕ
α
(σxx+ σyy)/2
O
R
0σ
τ
15
The Mohr circle - interpretation
Position of the Mohr circle center:
O [ (σxx+ σyy)/2, 0]
Radius of the Mohr circle:
( )2
2
2xx yy
xyRσ σ
τ−⎛ ⎞
= +⎜ ⎟⎝ ⎠
A(σxx, τxy)
A’(σ’xx, τ’xy)
B(σyy, τyx)
B’(σ’yy, τ’yx)
τxy
τ’xy
τyx
σyy
σ’xx
σ’yy
σxx
τ’yx
2ϕ
α
(σxx+ σyy)/2
O
R
0 σ
τ
16
The Mohr circle - interpretation
sin( ) / 2
xy
xx yy
τα α
σ σ= ⇒
−
,
,
,
cos(2 )2
cos(2 )2
sin(2 )
xx yy
xx
xx
yy
xy
yy
R
R
R
σ σϕ α
σ σϕ ασ
ϕ α
σ
τ
+= + −
+= − −
= −
A(σxx, τxy)
A’(σ’xx, τ’xy)
B(σyy, τyx)
B’(σ’yy, τ’yx)
τxy
τ’xy
τyx
σyy
σ’xx
σ’yy
σxx
τ’yx
2ϕ
α
(σxx+ σyy)/2
O
R
0σ
τ
17
The Mohr circle – Principal stresses
A(σxx, τxy)
A’(σ’xx, τ’xy)
B(σyy, τyx)
B’(σ’yy, τ’yx)
τxy
τ’xy
τyx
σyy
σ’xx
σ’yy
σxx
τ’yx
2ϕ
α
(σxx+ σyy)/2
O
R
0σ1σ2
σ
τ
( )
( )
22
1
22
1
2 2 2
2 2 2
sin ;( ) / 2
xx yy xx yy xx yyxy
xx yy xx yy xx yyxy
xy
xx yy
R
R
angle of principal axes
σ σ σ σ σ σσ τ
σ σ σ σ σ σσ τ
τα α
σ σ
+ + −⎛ ⎞= + = + +⎜ ⎟
⎝ ⎠
+ + −⎛ ⎞= − = − +⎜ ⎟
⎝ ⎠
= ⇒−
18
6-8 mm
σ22
1
2
3
Smooth laboratory specimens used for thedetermination of the σ − ε
Stress and strain state in specimens used for the determination of material properties
σ22
σ22
ε22
ε11
ε11= ε33 = −νε 22
ε33
ε
ε
εε
ij=
11
22
33
0 00 00 0
σ σij
=0 0 00 00 0 0
22
Measurement of basic mechanical properties of engineering materials
19
Engineering strain
Eng
inee
ring
stre
ss
σuts
σys
Typical engineering stress-strain behavior to fracture point F. The ultimate tensile strength σuts is indicated at point M. The inserts show the geometry of the deformed specimen at various points along the curve.
20
Ductile material
Brittle material
Typical stress-strain behavior of engineering materials
σuts
σys
σf
σf
σys
σ0.2
21
σu
Stre
ss, σ
Strain, εεf
Stre
ss, σ
σu
Strain, εεf
Stre
ss, σ
σu
εf
Mild steel
Elastic brittle material
High strength steel
Strain, ε
σY
M O D E L S
Stre
ss, σ
Strain, ε
Stre
ss, σ
Strain, ε
Stre
ss, σ
Ideal elastic-plastic material
Ideal elastic material
Strain hardening material
Strain, ε
E E E
σY
Eε σ= YE forε σ σ σ= <
1'
'
n
E Kσ σε ⎛ ⎞= + ⎜ ⎟
⎝ ⎠
Stre
ss, σ
Strain, ε
σY
Ideal rigid-plastic material
0 Yforε σ σ= < A
F F
FA
σ =
Specimen
Macroscopic behavior of engineering material under uni-axial tension
22
MONOTONIC STRESS-STRAIN BEHAVIOUR
Basic Definitions. A monotonic tension test of a smooth specimen
is usually used to determine the engineering stress-strain behaviour
of a material where
o
PS engineering stressA
= =
o
o o
l l le engineering strainl l− ∆
= = =
where:
P = applied load
Lo = original length
do = original diameter
Ao = original area
l = instantaneous length
d = instantaneous diameter
A = instantaneous area
Due to changes in cross-sectional area during deformation the true
stress in tension is larger than the engineering stress,
Ptrue stressA
σ = =
True or natural strain, based on an instantaneous gage length l, is
defined as
0
lnl
ol
dl ltrue strainl l
ε ⌠⎮⌡
= = =
23
F - applied load; l0 - original length
d0 - original diameter; A0 - original area
l - instantaneous length
d - instantaneous diameter
A0 - instantaneous area
( ) ( )1 ln 1S e and eσ ε= + = +
0
0
0
0 0
0
lnl
l
FS engineering stressAl l le engineering strain
l lFtrue stressA
dl ltrue strainl l
σ
ε
= =
− ∆= = =
= =
= = =⌠⎮⌡
d ll0d0
A
A0
F
F
24
.
True stress and strain can be related to engineering stress and
strain.
ln(1 )eε = +
(1 )S eσ = +
The equation above is only valid up to necking. At necking the
strain is no longer uniform throughout the gage length.
The total true strain ε in a tension test can be separated into elastic
and plastic components:
Linear elastic strain: that portion of the strain, which is recovered
upon unloading, εe.
Plastic strain (nonlinear): that portion which cannot be recovered on
unloading, εp.
Stated in equation form,
e pε ε ε= +
For most metals a log-log plot of true stress versus true plastic
strain is modelled as a straight line. Consequently, this curve can
be expressed using a power function.
1/ n
p Kσε ⎛ ⎞= ⎜ ⎟
⎝ ⎠
where: K is the strength coefficient and n is the strain hardening
exponent.
25
At fracture two important quantities can be defined. These are true
fracture strength and true fracture ductility. True fracture strength,
σf, is the true stress at final fracture.
ff
f
PA
σ =
where: Af is the area at fracture and Pf is the load at fracture.
True fracture ductility, εf, is the true strain at final fracture. This
value can be defined in terms of the initial cross-sectional area and
the area at fracture.
1ln ln1
of
f
o f
o
AA RA
A ARA reduction in area
A
ε = =−
−= =
The strength coefficient, K, can be defined in terms of the true
stress at fracture, σf, and the true strain at fracture, εf.
nf
fKεσ
=
We can also define plastic strain in terms of these quantities.1/ n
p ff
σε εσ
⎛ ⎞= ⎜ ⎟
⎝ ⎠
The total strain can be expressed as:
e pε ε ε= +
26
Static Strength Theories
The material properties such as the yield limit, Sys (σys), and the ultimate strength, Sut (σut), are obtained in simple uni-axial tensile tests.
However the analyst or designing engineer must asses the critical loading conditions for machine and structural components which are most often subjected to multiaxial stress states. Therefore a theory is needed (or failure criterion) allowing to translate the mutiaxial stress state into an equivalent uniaxial stresses state making possible the use of uniaxial material properties.
Maximum Normal Stress Theory
Failure is predicted to occur in the multiaxial stress state of stress when the maximum principal normal stress becomes equal to or exceeds the maximum normal stress at the time of failure in a simple uni-axial stress test using a specimen of the same material, i.e.,
failure is predicted by the maximum normal stress theory to occur if:
1 2 3
1 2 3
; ; ;
; ; ;
f t f t f
f c f c f
t
c
or
σ σ σσ σ σ
σ σσ σσσ
− −
− − −
−≥ ≥ ≥
≤ ≤ ≤
Where: σf-t is the uni-axial tensile strength of the material
σf-c is the uni-axial compressive strength of the material
The maximum normal stress theory provides good results for brittle materials.
27
Graphical representation of the maximum normal stress failure criterion
General bi-axial plane stress state, σ33=σ12=σ31=0 and σ11≠0, σ22≠0, σ23≠0 reduces two two non-zero principal stresses: σ1≠0, σ2≠0 and σ3=0
σ1
σ2
σf-t
σf-t
σf-c
σf-c 0
( )
( )
2211 22 11 22
1 12
2211 22 11 22
2 12
3
2 2
2 20
σ σ σ σσ σ
σ σ σ σσ σ
σ
+ −⎛ ⎞= + +⎜ ⎟⎝ ⎠
+ −⎛ ⎞= − +⎜ ⎟⎝ ⎠
=
principal stresses:
28
Maximum Shearing Stress Theory (Tresca)
Failure is predicted to occur in the multiaxial stress state of stress when the maximum shearing stress becomes equal to or exceeds themaximum shearing stress magnitude at the time of failure in a simple uni-axial stress test using a specimen of the same material, i.e.,
failure is predicted by the maximum normal stress theory to occur if:
1 2 3; ; ;f f fτ ττ τ τ τ≥ ≥ ≥
Where: τf - is the shear strength of the material obtained in uni-axial tension test.
The maximum normal stress theory provides good results for ductile materials.
;2ys
f
στ =
The Tresca theory in principal stresses:
General bi-axial plane stress state, σ33=σ12=σ31=0 and σ11≠0, σ22≠0, σ23≠0 reduces two two non-zero principal stresses: σ1≠0, σ2≠0 and σ3=0
1 21 1 2
22 2
13 1
; ;2 2
0; ;
2 20
; ;2 2
ysf ys
ysf ys
ysf ys
σ στ σ σ
στ σ
στ σ
στ
τ σ
σ
τ σσ σ
−≥ ≥ − ≥
−≥ ≥ ≥
−≥ ≥ ≥
29
Graphical representation of the Maximum Shearing Stress Theory (Tresca)
σ1
σ2
0
σ1 =- σysσ2 = 0
safe region
σ1 = - σysσ2 = - σys
σ1 = 0σ2 = - σys
σ1 = σysσ2 = 0
σ1 = σys/2σ2 = - σys/2
σ1 = σysσ2 = σys
σ1 = 0σ2 = σys
σ1 = - σys/2σ2 = σys/2
30
Distortion Energy Theory (Huber-von Mises-Hencky)
Failure is predicted to occur in the multiaxial stress state of stress when the distortion energy per unit volume becomes equal to or exceeds the distortion energy per unit volume at the time of failure in a simple uni-axial stress test using a specimen of the same material, i.e.,
failure is predicted by the Distortion Energy theory to occur if:
yeq sσ σ≥Where: - for general 3-D stress state:
( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 211 22 22 33 33 11 12 31 23
1 62eqσ σ σ σ σ σ σ σ σ σ⎡ ⎤= − + − + − + + +⎣ ⎦
- for 3-D stress state in principal stresses:
( ) ( ) ( )
( ) ( ) ( )
2 2 21 2 2 3 3 1
2 2 21 2 3 1 2 2 3 3 1
12eqσ σ σ σ σ σ σ
σ σ σ σ σ σ σ σ σ
= − + − + −
= + + − − −
- for general plane stress state:
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
2 2 2 2 2 211 22 22 11 12
2 2 211 11 22 22 12
1 0 0 6 0 02
3
eqσ σ σ σ σ σ
σ σ σ σ σ
⎡ ⎤= − + − + − + + + +⎣ ⎦
= − + +
- for plane stress state in principal stresses:
( ) ( ) ( )
( ) ( )
2 2 21 2 2 1
2 21 2 1 2
1 0 02eqσ σ σ σ σ
σ σ σ σ
= − + − + −
= + −
31
- for plane strain state in principal stresses:
( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
2 2 21 2 3 1 2 2 3 3 1
2 2 21 2 1 2 1 2 2 1 2 1 1 2
2 2 2 21 2 1 21 1 2 2
eqσ σ σ σ σ σ σ σ σ σ
σ σ ν σ σ σ σ σ ν σ σ σ ν σ σ
σ σ ν ν σ σ ν ν
= + + − − −
= + + + − − + − +⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎡ ⎤= + − + − + −⎣ ⎦
σ1
σ2
0
σ1 =- σysσ2 = 0
safe region
σ1 = - σysσ2 = - σys
σ1 = 0σ2 = - σys
σ1 = σysσ2 = 0
σ1 = σys/√3σ2 = - σys/√3
σ1 = σysσ2 = σys
σ1 = 0σ2 = σys
σ1 = - σys/√3σ2 = σys/√3
Graphical representation of the Maximum Distortion Energy Theory (H-M-H)
32
Comparison of the Tresca and H-M-H failure criterion of for plane stress states
σ1
σ2
σ1 =- σysσ2 = 0
σ1 = - σysσ2 = - σys
σ1 = 0σ2 = - σys
σ1 = σysσ2 = 0
σ1 = σys/√3σ2 = - σys/√3
σ1 = σysσ2 = σys
σ1 = 0σ2 = σys
σ1 = - σys/√3σ2 = σys/√3
σ1 = σys/2σ2 = - σys/2
0
safe region
σ1 = - σys/2σ2 = σys/2
33
Example:
A thick-wall cylindrical pressure vessel made of carbon steel plate (ASME SA-285M, Grade C, σut = 65 ksi, σys = 30 ksi) is pressurized internally. Find the maximum internal pressure, pi, at the initiation of plastic yielding in the cylinder’s wall. The internal and external diameter is 8 in and 12 in respectively. Use the Tresca and H-M-H theory. Assume open ended cylinder with:
a) unrestrained ends,
b) with restrained ends
a) Internally pressurized thick wall cylinder with unrestrained ends,
ab
p
σ22 = σθθ
σ11= -pσθθ
Di = 2a = 8 in.
Do = 2b =12 in.
34
2 2 2 2
2 2 2 2 2 21 ; 1
0
rr
zz
p a b p a bb a r b a rθθσ σ
σ
⎡ ⎤ ⎡ ⎤⋅ ⋅= − = +⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦=
Radial and hoop stresses in an open ended internally pressurizedthick wall cylinder (with un-restrained ends)
Maximum stresses at the inner surface of the cylinder:
2 2
( ) ( )2 2 2
2 2 2 2
( ) 2 2 2 2 2
2 2 2 2
1 ( ) 2 2 2 2
2
3 ( )
1 ; 0
1
6 4 2.66 4
0
rr at r a zz at r a
at r a
at r a
rr at r a
p a b pb a r
p a b b apb a r b a
then
b ap p pb a
p
θθ
θθ
σ σ
σ
σ σ
σσ σ
= =
=
=
=
⎡ ⎤⋅= − = − =⎢ ⎥− ⎣ ⎦
⎡ ⎤ ⎛ ⎞⋅ += + = ⎜ ⎟⎢ ⎥− −⎣ ⎦ ⎝ ⎠
⎛ ⎞ ⎛ ⎞+ += = = =⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠== = −
Plastic yielding will commence according to the Tresca criterion when:
( )1 3
2
2.6 30 3.6 30
8.777 8777.
ys p p ksi p ksi
lbp ksi pin
σ σσ− = ⇒ − − = ⇒ =
= ⇒ =
35
Plastic yielding will commence according to the H-M-H criterion when:
yeq sσ σ≥
( ) ( ) ( )
( ) ( ) ( ) ( )( )
2 2 21 2 3 1 2 2 3 3 1
2 2 2
2
2.6 0 2.6 0
3
0 3.218
3.218 0
9.322 9322
eq
eq ys
p p p p p
plbp ksi pin
ksi
σ σ σ σ σ σ σ σ σ σ
σσ
= + + − − −
= − + + − − − − =
=
=
= ⇒ =
36
b) Internally pressurized thick wall cylinder with restrained ends,
Thick cylinder under internal pressure and torsion with fixed ends (no axial expansion)
σ33
σθθ
σ22 = σθθ
σ11= -p
p
( )
2 2 2 2
2 2 2 2 2 21 ; 1 ;
; 0
rr
zz rr zz
p a b p a bb a r b a r
because
θθ
θθ
σ σ
σ ν σ σ ε
⎡ ⎤ ⎡ ⎤⋅ ⋅= − = +⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦
= + =
Radial and hoop and axial stresses in an open ended internally pressurized thick wall cylinder with restrained ends
37
Maximum stresses at the inner surface of the cylinder:
( )
2 2
( ) 2 2 2
2 2 2 2
( ) 2 2 2 2 2
2
( ) 2 2
2 2 2 2
1 ( ) 2 2 2 2
2
2 ( ) 2 2
1
1
2
6 4 2.66 4
2 0.
rr at r a
at r a
zz at r a rr
at r a
zz at r a
p a b pb a r
p a b b apb a r b a
ap thenb a
b ap p pb a
apb a
θθ
θθ
θθ
σ
σ
σ ν σ σ ν
σ σ
σ σ ν
=
=
=
=
=
⎡ ⎤⋅= − = −⎢ ⎥− ⎣ ⎦
⎡ ⎤ ⎛ ⎞⋅ += + = ⎜ ⎟⎢ ⎥− −⎣ ⎦ ⎝ ⎠
= + =−
⎛ ⎞ ⎛ ⎞+ += = = =⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠
= = =−
2
2 2
3 ( )
2 43 0.486 4
rr at r a
p p
pσ σ =
⋅=
−= = −
Plastic yielding will commence according to the Tresca criterion when:
( )1 3
2
2.6 30 3.6 30
8.777 8777.
ys p p ksi p ksi
lbp ksi pin
σ σσ− = ⇒ − − = ⇒ =
= ⇒ =
38
Plastic yielding will commence according to the H-M-H criterion when:
yeq sσ σ≥
( ) ( ) ( )
( ) ( ) ( )( )( ) ( )( ) ( )( )
2 2 21 2 3 1 2 2 3 3 1
2 2 2
2
2.6 0.483.134
2.6 2.6 0.
3
48 0.48
3.134
9.572 9572
0
ys
eq
eq
p p pp
p p p p p p
plbp ksi
k
pin
si
σ σ σ σ σ σ σ σ σ σ
σ σ
= + + − − −
− + + −= =
− − − −
=
=
= ⇒ =
39
The effect of the mutiaxiality of the stress state on the tendency to plastic yielding
The change of the maximum stress necessary for plastic deformation of a ductile material under various stress states: a) uni-axial tension, b) tension with transverse compression, c) bi-axial tension, and d) hydrostatic compression (after N. Dowling)