Chapter IV: Stress 9Stress on the sloping plane øø ng suÊt trªn mÆt ph¼ng nghiªng It can be demonstrated, that stress state of on point is completely determined if all term of stress ten
Trang 1Chapter 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
Stress Components on a Plane
P
∆A
plane
body
: normal of plane
: tangent to plane
: simplest static equivalent force on
n
t
r
r
ur
Normal stress component
at point P on given plane:
Shear stress component
at point P on given plane:
0
A
F d F
σ
∆ →
∆
∆
0
A
F d F
τ
∆ →
∆
∆
Trang 2Chapter IV: Stress 3
Stress General notion of Stress (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
P5
P n
Engineering versus True Stress
l
0
dl
Assuming that the axial stress is ‘uniform’ over the cross-secction
we can define:
Engineering Stress:
True Stress:
0 0
F A
F A
σ = We use only true stressesin plasticity!
Trang 3Chapter IV: Stress 5
τ x y = τ yx ; τ x z = τ zx ; τ zy = τ yz
σ
x x y x z
z
•
Co-ordination sysytem of Decac
State of Stress at a Point
01.061
01.10.2001
The state of stress at point P:
Because of moment equilibrium:
xy yx yz zy zx xz
Note: All components shown
are positive!
Trang 4Chapter 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
xx xy
Uniaxial Stress State:
Only σxx ≠ ⇒ 0
0 0
0 0 0
0 0 0
xx ij
σ σ
Stress Vector
Let’s define the “true (Cauchy or Euler) stress tensor” in “vector
notation”:
{ }
x y z xy
zx yx zy xz
σ σ σ τ τ τ τ τ τ
[ ]
Trang 5Chapter IV: Stress 9
Stress on the sloping plane (øø 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) = az
A B
C
Call ∆F is an area of sloping plane ABC, area of rest planes of tetrahedron (khối tứ diện)
coresponding to their position will be ∆Fx, ∆Fy and ∆Fz (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
σnand tangential stress τ.
The terms of total stress S following the directions of co-ordinations in succession
(lần lượt) are sx, syvµ sz.
The tetrahedron OABC lies on the equilibrium state, if it satisfies (thỏa mãn) the following
conditions:
σn = σxax2 +σyay2 +σzaz2 + τxya ax y + τyza ay z + τzxa az x
τ2 =S2 −σ2n
Trang 6Chapter 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 σ1; σ2; σ3
Tσ σij
σ σ σ
11 22 33
0
T ij
x x x y x z
yy yz zz
σ
This is a linear homogeneous (thuần nhất) equation system This system has the roots
(nghiệm số) if:
x
y
z
−
−
−
=
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
2 2 2
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
- Invariant I1: first grade
I1 = σx +σy +σz = const
I2 = − σ σx y +σ σy z +σ σz x + τ2xy + τ2yz + τ2zx = const
I3 = σ σ σx y z + 2 τ τ τxy yz zx − σ τx 2yz − σ τy zx2 − σ τz 2xy = const
1
2
Stress Tensor has also 3 independent stress invariants:
- Invariant I2: second grade
- Invariant I3: third grade
Therefore, we get an equation:
Solve this equation, we get 3 principal normal stresses σ1;σ2 ;σ3
Trang 7Chapter IV: Stress 13
Stress Invariants
Three independent stress invariants are:
1
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
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
=
The principal stresses can be also found as the root of the equation:
Main tangential Stresses (principal tangential stresses)
Determination: on which planes the tangential stresses reach maximum ???
1
2
1 2 2 2 2 2 3 2 3 2
1 1 2
2 2 2
3 3
2 2
a 1 a a
2
2
2
3 2 1
1 2 1 2 2 2 2 2 3 2 1 2 2 2
1 1 2
2 2 2
2 2
2 2
= a + a + − a − a − a + a + − a − a
To determine the extrema, we derivative to a1,a2,a3and assign them to 0
Trang 8Chapter IV: Stress 15
τ σ σ
τ σ σ
τ σ σ
12
23
31
2
1 2
1
2 0
1 2
1 2 2
1
2 0
1 2
= ± − = ± = ± =
= ± − = = ± = ±
= ± − = ± = = ±
Value of principal tangential stresses:
Main tangential Stresses (principal tangential stresses)
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
Trang 9Chapter IV: Stress 17
Transformation of Stress Components
σ
τ (cw)
τ (ccw) σmean =(σxx+σyy)/2
σmax
σmin
( σ τyy,yx)
( σ τxx, xy)
2θ
τmax
τxy
σyy
σxx
τyx
x y
σmax
σmin
θ
Three Dimensional Mohr’s Circle
τ locus of all possible
stress components
1 2 3
ij
σ
σ
Principal Stress State:
Trang 10Chapter 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
Local Equilibrium Equations
0, 0, 0.
x
y
z
F F F
=
=
=
∑
∑
∑
Assumption:
No body forces and body moments
Trang 11Chapter IV: Stress 21
Equilibrium
in x-Axis
Direction (1)
0
yx xx
zx
dz dx dy dy dz dy dx dz dx z
τ σ
τ
∂
∂
∂
∂
∑
Equilibrium
in x-Axis
Direction (2)
F = σ
yx area
stress
dx dy dz x
σ
τ
∂
∂
yx
area stress
zx
dy dx dz y
τ
τ
∂
∂
+
1442443
zx
xx
dz dx dy z
τ
σ
∂
∂
⋅dy dz− τzx ⋅dy dx−τyx ⋅dz dx=0
Trang 12Chapter IV: Stress 23
Equilibrium
in x-Axis
Direction (3)
xx
x
x
σ
∂
= ⋅
∂
dx dy dz y
τ
∂
+ ⋅
∂
zx
dx dy dz z
τ
∂
+ ⋅
∂
0
yx
τ
Complete Equilibrium Equations
in Three Dimensions
yx
x
y
yz
z
F
F
F
τ
τ
∂
∂
∑
∑
∑
Assumption:
No body forces and body moments
Trang 13Chapter IV: Stress 25
Complete Equilibrium Equations
in Two Dimensions
yx xx
x
xy yy y
F
F
τ σ
∂
∂
∑
∑
Assumption:
No body forces and body moments
2 2
yy xx
σ
∂
=
or:
Hydrostatic Stress
By definition hydrostatic stress is computed as:
3
xx yy zz h
h
The hydrostatic
stress state is:
h h
h
σ
σ
Trang 14Chapter 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
Deviatoric Stress Tensor
The true stress tensor can be splitted additively as:
h
ij ij ij
where, the deviatoric stress tensor is simply:
h
It is the deviatoric stress state which deforms material plastically!
Trang 15Chapter IV: Stress 29
Deviatoric Stress Invariants
Three independent stress invariants are:
1
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
=
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
′ ′ ′
=
=
The principal stresses can be also found as the root of the equation:
3
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
ij
σ