GIÁO TRÌNH KHAI THÁC PHẦN mềm TRONG GIA CÔNG KHUÔN mẫu chapter IV stress

1
Chapter IV: Stress
1
Content
• Stress Components
• State of Stress (2D & 3D)
• Transformation of Stress Components
• Principal Stresses
• Stress Invariants
• Local Equilibrium Equations
• Hydrostatic Stress Components
• Deviatoric Stress Components
01.058
01.10.2001
Chapter IV: Stress
2
Stress Components on a Plane
P
∆
A
plane
body
: normal of plane
: tangent to plane
: simplest static equivalent force on
n
t
F A
∆ ∆
r
r

ur
Normal stress component
at point P on given plane:
Shear stress component
at point P on given plane:
0
lim
n n
A
F d F
A dA
σ
∆ →
∆
= =
∆
uur uur
0
lim
t t
A
F d F
A dA
τ
∆ →
∆
= =
∆
uur uur
2

Chapter IV: Stress
3
Stress
StressStress
Stress
General notion of Stress
General notion of Stress General notion of Stress
General notion of Stress (Kh¸i niÖm chung)
(Kh¸i niÖm chung)(Kh¸i niÖm chung)
(Kh¸i niÖm chung)
Internal forces and stresses in the body
Definition of Stress at the point:
Normal and tangential Stresses
Stresses in 3 directions:
On the one plane there
are 3 terms of stresses:
- 1 normal stress
- 2 tangential stresses
P
1
P
2
P
3
P
4
P
5
P
n

Chapter IV: Stress
4
Engineering versus True Stress
l
0
F
F
dl
A
0
A
Assuming that the axial stress is ‘uniform’ over the cross-secction
we can define:
Engineering Stress:
True Stress:
0
0
F
A
σ
=
F
A
σ
=
We use only true stresses
in plasticity!
3
Chapter IV: Stress
5

Stresses on the co-ordination planes (ø
øø
øng suÊt trªn c¸c mÆt to¹ ®é)
ng suÊt trªn c¸c mÆt to¹ ®é)ng suÊt trªn c¸c mÆt to¹ ®é)
ng suÊt trªn c¸c mÆt to¹ ®é)
σ σ σ
σ σ σ
σ σ σ
xx xy xz
yx yy yz
zx zy zz










τ
τ
τ
τ
τ
τ
xy yx xz zx zy yz
=
=

=
; ;
σ τ τ
σ τ
σ
x x y x z
y yz
z
•
• •










Co-ordination sysytem of Decac
Chapter IV: Stress
6
State of Stress at a Point
01.061
01.10.2001
xx xy xz
ij yx yy yz
zx zy zz
σ τ τ

σ τ σ τ
τ τ σ
 
 
=
 
 
 
The state of stress at point P:
Because of moment equilibrium:
= , ,
xy yx yz zy zx xz
τ τ τ τ τ τ
= =
Note: All components shown
are positive!
4
Chapter IV: Stress
7
Special Stress States
Plane Stress State ( Biaxial Stress State):
0
0
0
zz
zx
zy
σ
τ
τ


=

= ⇒


=

0
0
0 0 0
xx xy
ij yx yy
σ τ
σ τ σ
 
 
=
 
 
 
Uniaxial Stress State:
Only 0
xx
σ
≠ ⇒
0 0
0 0 0
0 0 0
xx

ij
σ
σ
 
 
=
 
 
 
Chapter IV: Stress
8
Stress Vector
Let’s define the “true (Cauchy or Euler) stress tensor” in “vector
notation”:
{ }
x
y
z
xy
i yz
zx
yx
zy
xz
T T
σ
σ
σ
τ
τ

τ
τ
τ
τ
 
 
 
 
 
 
 
= =
 
 
 
 
 
 
 
 
[ ]
x xy xz
ij yx y yz
zx zy z
σ τ τ
σ σ τ σ τ
τ τ σ
 
 
= =

 
 
 
5
Chapter IV: Stress
9
Stress on the sloping plane (ø
øø
øng suÊt trªn mÆt ph¼ng nghiªng)
ng suÊt trªn mÆt ph¼ng nghiªng)ng suÊt trªn mÆt ph¼ng nghiªng)
ng suÊt trªn mÆt ph¼ng nghiªng)
It can be demonstrated, that stress state of
on point is completely determined if all term
of stress tensor drawing through this point
are known.
It means, from the term of known stresses,
we can calculate the normal and tangential
stresses of any sloping plane drawing
throung the point.
Call N is a normal line of sloping plane. The position of N can be
determined by cosin directions:
cosα = cos(N,x) = a
x
cosβ = cos(N,y) = a
y
cosγ = cos(N,z) = a
z
A
B
C

Chapter IV: Stress
10
Call ∆F is an area of sloping plane ABC, area of rest planes of tetrahedron (khối tứ diện)
coresponding to their position will be ∆F
x
, ∆F
y
and ∆F
z
. (OAB is projection of ABC on
plane Oxy)
Supposing (giả thiết) that the total stress on sloping plane is S, in which normal stress is
σ
n
and tangential stress τ.
The terms of total stress S following the directions of co-ordinations in succession
(lần lượt) are s
x
, s
y
vµ s
z
.
The tetrahedron OABC lies on the equilibrium state, if it satisfies (thỏa mãn) the following
conditions:
X S F F F F
x x x xy y xz z
∑
= − − − =∆ ∆ ∆ ∆σ τ τ 0
Y S F F F F

y yx x y y yz z
∑
= − − − =∆ ∆ ∆ ∆τ σ τ 0
Z S F F F F
z zx x zy y z z
∑
= − − − =∆ ∆ ∆ ∆τ τ σ 0
S a a a
S a a a
S a a a
x x x xy y x z z
y yx x y y yz z
x zx x zy y z z
= + +
= + +
= + +





σ τ τ
τ σ τ
τ τ σ
σ σ σ σ τ τ τ
n x x y y z z xy x y yz y z zx z x
a a a a a a a a a= + + + + +
2 2 2
2 2 2
τ σ

2 2 2
= −S
n
6
Chapter IV: Stress
11
Main normal Stresses (principal stresses)
Through on point in the stress state, it can be always found 3 planes perpendiculary
(vuông góc) to each other, on which there are only normal stresses, the tangential
stresses are 0.
3 normal stresses are called main normal stresses. σ σ σ
1 2 3
; ;
T
ijσ
σ
σ
σ
σ
= = •
• •











11
22
33
0 0
0
T
ij
xx xy xz
yy yz
zz
σ
σ
σ σ σ
σ σ
σ
= = •
• •










This is a linear homogeneous (thuần nhất) equation system. This system has the roots
(nghiệm số) if:

(
)
( )
( )
σ σ τ τ
τ σ σ τ
τ τ σ σ
x
y
z
−
−
−
=
xy xz
yx yz
zx zy
0
(
)
(
)
( )
σ σ σ σ σ σ σ σ σ σ σ σ τ τ τ
σ σ σ τ τ τ σ τ σ τ σ τ
3 2 2 2 2
2 2 2
2 0
− + + − − − − + + + −
− + − − − =

x y z x y y z z x xy yz zx
x y z xy yz zx x yz y zx z xy
Stress Invariants
Chapter IV: Stress
12
- Invariant I
1
: first grade
I const
x y z1
=
+
+
=
σ
σ
σ
(
)
I const
x y y z z x xy yz zx2
2 2 2
= − + + + + + =σ σ σ σ σ σ τ τ τ
I const
x y z xy yz zx x yz y zx z xy3
2 2 2
2= + − − − =σ σ σ τ τ τ σ τ σ τ σ τ
σ σ σ
3
1

2
2 3
0− − − =I I I
Stress Tensor has also 3 independent stress invariants:
- Invariant I
2
: second grade
- Invariant I
3
: third grade
Therefore, we get an equation:
Solve this equation, we get 3 principal normal stresses
σ σ σ
1 2 3
; ;
7
Chapter IV: Stress
13
Stress Invariants
Three independent stress invariants are:
1
2 2 2
2
2 2 2
3
,
,
2
xx yy zz
xy yz zx xx yy yy zz zz xx

xx yy zz xy yz zx xx yz yy zx zz xy
I
I
I
σ σ σ
τ τ τ σ σ σ σ σ σ
σ σ σ τ τ τ σ τ σ τ σ τ
= + +
= + + − − −
= + − − −
In terms of principal stresses:
( )
1 1 2 3
2 1 2 2 3 3 1
3 1 2 3
,
,
I
I
I
σ σ σ
σ σ σ σ σ σ
σ σ σ
= + +
= − + +
=
The principal stresses can be also found as the root of the equation:
3 2
1 2 3
0

I I I
σ σ σ
− − − =
Chapter IV: Stress
14
Main tangential Stresses (principal tangential stresses)
Determination: on which planes the tangential stresses reach maximum ???
τ σ σ σ σ σ σ
2
1
2
1
2
2
2
2
2
3
2
3
2
1 1
2
2 2
2
3 3
2 2
= + + − + +a a a a a a( )
a a a
1

2
2
2
3
2
1+ + =
(
)
(
)
[
]
τ σ σ σ σ σ σ
2
1
2
1
2
2
2
2
2
3
2
1
2
2
2
1 1
2

2 2
2
3 1
2
2
2
2
1 1= + + − − − + + − −a a a a a a a a
To determine the extrema, we derivative to a
1
,a
2
,a
3
and assign them to 0.
8
Chapter IV: Stress
15
τ
σ
σ
τ
σ σ
τ
σ
σ
12
1 2
1
2

3
23
2
3
1
2
3
31
3
1
1
1
2
3
2
1
2
1
2
0
2
0
1
2
1
2
2
1
2
0

1
2
=
±
−
=
±
=
±
=






= ±
−
= = ± = ±






=
±
−
=
±

=
=
±

















;
;
;
;
;
;
;
;
;
a

a
a
a
a
a
a
a
a
Value of principal tangential stresses:
Main tangential Stresses (principal tangential stresses)
Chapter IV: Stress
16
Some Remarks on Stress
Components
Theorem 3.1:
The limits of force divided by area for diminishing area exist.
Theorem 3.2:
The complete internal force state at a point can be represented
fully by the stress components on three mutually orthogonal
planes passing through point P.
9
Chapter IV: Stress
17
Transformation of Stress Components
σ
τ
(cw)
τ
(ccw)
σ σ σ

mean
=( + )/2
xx yy
σ
max
σ
min
( , )
σ τ
yy yx
( , )
σ τ
xx xy
2
θ
τ
max
τ
xy
σ
yy
σ
xx
τ
yx
x
y
σ
max
σ

min
θ
Chapter IV: Stress
18
Three Dimensional Mohr’s Circle
σ
3
σ
2
σ
1
σ
τ
locus of all possible
stress components
1
2
3
0 0
0 0
0 0
ij
σ
σ σ
σ
 
 
=
 
 

 
Principal Stress State:
10
Chapter IV: Stress
19
Example (1)
Example 3.1:
Consider the stress state at a point given by its components:
a) Determine the principal stresses and their orientation.
b) Determine the largest shear stress.
100 MPa, 50 MPa, 0,
30 MPa, 0
xx yy zz
xy xz yz
σ σ σ
τ τ τ
= = =
= = =
Chapter IV: Stress
20
Local Equilibrium Equations
0,
0,
0.
x
y
z
F
F
F

=
=
=
∑
∑
∑
Assumption:
No body forces and
body moments
11
Chapter IV: Stress
21
Equilibrium
in x-Axis
Direction (1)
{
{
0
yx
xx
x xx yx
area area
stress
stress
zx
zx xx zx yx
F dx dy dz dy dx dz
x y
dz dx dy dy dz dy dx dz dx
z

τ
σ
σ τ
τ
τ σ τ τ
∂
 
∂
 
= + ⋅ + + ⋅ +
 
 
∂ ∂
 
 
∂
 
+ + ⋅ − ⋅ − ⋅ − ⋅ =
 
∂
 
∑
1442443
1442443
Chapter IV: Stress
22
Equilibrium
in x-Axis
Direction (2)
x xx

F
σ
=
∑
{
xx
yx
area
stress
dx dy dz
x
σ
τ
∂
 
+ ⋅ +
 
∂
 
144424443
{
yx
area
stress
zx
dy dx dz
y
τ
τ
∂

 
+ ⋅ +
 
∂
 
+
1442443
zx
xx
dz dx dy
z
τ
σ
∂
 
+ ⋅ −
 
∂
 
zx
dy dz
τ
⋅ −
yx
dy dx
τ
⋅ −
0
dz dx
⋅ =

12
Chapter IV: Stress
23
Equilibrium
in x-Axis
Direction (3)
xx
x
F dx dy dz
x
σ
∂
 
= ⋅
 
∂
 
∑
yx
dx dy dz
y
τ
∂
 
+ ⋅
 
∂
 
zx
dx dy dz

z
τ
∂
 
+ ⋅
 
∂
 
0
=
0
yx
xx zx
x y z
τ
σ τ
∂
∂ ∂
+ + =
∂ ∂ ∂
Chapter IV: Stress
24
Complete Equilibrium Equations
in Three Dimensions
0 : 0
0 : 0
0 : 0
yx
xx zx
x

xy yy zy
y
yz
xz zz
z
F
x y z
F
x y z
F
x y z
τ
σ τ
τ σ τ
τ
τ σ
∂
∂ ∂
= + + =
∂ ∂ ∂
∂ ∂ ∂
= + + =
∂ ∂ ∂
∂
∂ ∂
= + + =
∂ ∂ ∂
∑
∑
∑

Assumption:
No body forces and
body moments
13
Chapter IV: Stress
25
Complete Equilibrium Equations
in Two Dimensions
0 : 0
0 : 0
yx
xx
x
xy yy
y
F
x y
F
x y
τ
σ
τ σ
∂
∂
= + =
∂ ∂
∂ ∂
= + =
∂ ∂
∑

∑
Assumption:
No body forces and
body moments
2
2
2 2
yy
xx
x y
σ
σ
∂
∂
=
∂ ∂
or:
Chapter IV: Stress
26
Hydrostatic Stress
By definition hydrostatic stress is computed as:
3
xx yy zz
h
σ σ σ
σ
+ +
=
Note that:
1 1 2 3

3 3
h
I
σ σ σ
σ
+ +
= =
The hydrostatic
stress state is:
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
h
h
ij h h
h
σ
σ σ σ
σ
   
   
= =
   
   
   
14
Chapter IV: Stress
27
Remarks on Hydrostatic Stress
It is observed that a material exposed to a hydrostatic

stress state does not deform plastically.
However, the formability of metals increase with increasing
compressive hydrostatic stress.
Chapter IV: Stress
28
Deviatoric Stress Tensor
The true stress tensor can be splitted additively as:
h
ij ij ij
σ σ σ
′
= +
where, the deviatoric stress tensor is simply:
(
)
( )
( )
-
-
-
xx h xy xz
h
ij ij ij yx yy h yz
zx zy zz h
σ σ τ τ
σ σ σ τ σ σ τ
τ τ σ σ
 
 
′

= − =
 
 
 
 
It is the deviatoric stress state which deforms material plastically!
15
Chapter IV: Stress
29
Deviatoric Stress Invariants
Three independent stress invariants are:
1
2 2 2
2
2 2 2
3
,
,
2
0
xx yy zz
xy yz zx xx yy yy zz zz xx
xx yy zz xy yz zx xx yz yy zx zz xy
J
J
J
σ σ σ
τ τ τ σ σ σ σ σ σ
σ σ σ τ τ τ σ τ σ τ σ τ
′ ′ ′

= + +
′ ′ ′ ′ ′ ′
= + + − − −
′ ′ ′ ′ ′ ′
= + − − −
=
In terms of principal stresses:
( )
1 1 2 3
2 1 2 2 3 3 1
3 1 2 3
,
0,J
J
J
σ σ σ
σ σ σ σ σ σ
σ σ σ
′ ′ ′
= + +
′ ′ ′ ′ ′ ′
= − + +
′ ′ ′
=
=
The principal stresses can be also found as the root of the equation:
3
2 3
0
J J

σ σ
′ ′
− − =
Chapter IV: Stress
30
Example (2)
Example 3.2:
Consider the given uniaxial stress state as it is present in a
uniaxial specimen:
Find the hydrostatic and deviatoric stress tensors.
240 MPa 0 0
0 0 0
0 0 0
ij
σ
 
 
=
 
 
 