Lecture 2
Theory of Stress
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Print version Lecture on Theory of Elasticity and Plasticity of
Dr. D. Dinev, Department of Structural Mechanics, UACEG
2.1
Contents
1 Stress at a point 1
2 Stress vector 2
3 Stress tensor 2
4 Traction on an arbitrary plane 4
5 Stress transformation 6
6 Principal stresses and principal planes 7
7 Equilibrium equations 8 2.2
1 Stress at a point
Stress at a point
Mechanics of material approach (MoM)
Tensile test
2.3
Stress at a point
Mechanics of material approach (MoM)
Stress denition- Average force per unit area
Stress =
F
A
where
F- average tensile force
A- cross-section area
2.4
1
Stress at a point
Theory of elasticity approach (ToE)
Consider a body in equilibrium subject to a set of external forces
Cut the body by a plane trough point O dividing it into two parts
The forces acting on one part alone are in equilibrium with internal forces
Lets analyze the internal force F acting on area A
2.5
2 Stress vector
Stress vector
Denition
The stress vector (traction vector) can be dened as
t = lim
A0
F
A
2.6
Stress vector
Stress vector components
The traction vector t can be decomposed into components of an arbitrary coordinate system
t
x
= lim
A0
F
x
A
, t
y
= lim
A0
F
y
A
, t
z
= lim
A0
F
z
A
2.7
3 Stress tensor
Stress tensor
2
Denition
The stress state at a point can be specied by nine components of the stress tensor
2.8
Stress tensor
Denition
Consider a stress cube with dimensions of dx, dy and dz
Stress tensor
=
xx
xy
xz
yx
yy
yz
zx
zy
zz
2.9
Stress tensor
Stress components
ii
- normal stresses
i j
- shear stresses
2.10
Stress tensor
Indexes
Index notation
1-st index- indicates the direction of a normal to the plane
2-nd index- relates to the direction of the stress
2.11
Stress tensor
Who? Where?
i j
i- Where is the crime scene?
j- Who is the killer?
2.12
3
Stress tensor
Indexes
xy
x-plane
y-direction
2.13
Stress tensor
Stress tensor Stress vector
The relation between a stress tensor and a stress vector is
=
xx
xy
xz
yx
yy
yz
zx
zy
zz
t
x
t
y
t
z
2.14
Stress tensor
Symmetry of the stress tensor
Shear stresses on orthogonal planes are equal
Symmetry of the stress tensor
Applying M
z
= 0 we obtain that
xy
=
yx
Similarly,
xz
=
zx
yz
=
zy
2.15
Stress tensor
Vector representation
Vector representation of the stress tensor components
=
xx
yy
zz
xy
yz
zx
2.16
4 Traction on an arbitrary plane
Traction on an arbitrary plane
4
Stress equilibrium
Consider a tetrahedron with applied loads (surface traction) in equilibriumwith the internal
stresses
2.17
Traction on an arbitrary plane
Stress equilibrium
Traction components and direction cosines
t =
t
x
t
y
t
z
n =
n
x
n
y
n
z
cos(n, x)
cos(n, y)
cos(n, z)
2.18
Traction on an arbitrary plane
Stress equilibrium
The sides of planes are
A
x
= An
x
A
y
= An
y
A
z
= An
z
2.19
Traction on an arbitrary plane
Stress equilibrium
Force equilibrium of the tetrahedron gives
t
x
t
y
t
z
xx
yx
zx
xy
yy
zy
xz
yz
zz
n
x
n
y
n
z
2.20
Traction on an arbitrary plane
Stress equilibrium
Matrix form
t =
T
n
These equations represent stress boundary conditions
Cauchy stress formula
2.21
5
Traction on an arbitrary plane
Cauchy stress formula
Augustin-Louis Cauchy (17891857)
2.22
Traction on an arbitrary plane
Stress equilibrium
The magnitude of the component of the stress vector normal to the plane is
t
n
= t n =t
i
n
i
The component of t perpendicular of n is
t
s
=
|t|
2
t
2
n
2.23
Traction on an arbitrary plane
Example
The stress state at point P is given by
=
200 400 300
400 0 0
300 0 100
Determine the stress vector t and its normal and shear components at the point P on plane
(x) = x +2y +2z
Hint
The unit normal to the plane is given by
n =
||
2.24
5 Stress transformation
Stress transformation
Tensor transformation
The stress tensor given in one coordinate system can be transformed in any other coordi-
nate system by
= NN
T
where the transformation matrix is
N =
n
xx
n
xy
n
xz
n
yx
n
yy
n
yz
n
zx
n
zy
n
zz
and n
i j
= cos(x
i
, x
j
) = e
i
e
j
The prime coordinates belong to the new coordinate system
2.25
6
Stress transformation
Tensor transformation
The stress state at a point in (x, y, z)- coordinates is given by
=
2 2 0
2
2 0
0 0
Determine the stress tensor
for the rotated coordinates (x
, y
, z
) using the transforma-
tion tensor
N =
0
1
2
1
2
1
2
1
2
1
2
2
1
2
1
2
2.26
6 Principal stresses and principal planes
Principal stresses and principal planes
Principal stresses
The engineering design require to know the maximum stress values for a given problem
because the failures occur when the magnitude of stresses exceed the allowable stress
values called strengths of the material.
The normal component of the traction vector is largest when t is parallel to the normal n.
Lets denote the values of these components with and we can write
t = n or n = n
( I)n = 0
2.27
Principal stresses and principal planes
Principal stresses
The nontrivial solution is
det( I) = 0
or
3
I
1
2
+I
2
I
3
= 0
The above equation is called characteristic equation
2.28
Principal stresses and principal planes
Principal stresses
The invariants of the stress tensor are
I
1
=
xx
+
yy
+
zz
= tr
I
2
=
xx
xy
yx
yy
xx
xz
zx
zz
yy
yz
zy
zz
=
1
2
(tr)
2
tr(
2
)
I
3
=
xx
xy
xz
yx
yy
yz
zx
zy
zz
= det
2.29
7
Principal stresses and principal planes
Principal planes
The principal planes can be obtained by
(
i
I)n
i
= 0
2.30
Principal stresses and principal planes
Principal directions
The comparison of the general and the principal stress states shows that for the princi-
pal coordinate system all shear stresses vanish and the stress state includes only normal
stresses
2.31
Principal stresses and principal planes
Example
The stress state at a point is given by
=
3 1 1
1 0 2
1 2 0
Determine the principal values and the principal directions
2.32
7 Equilibrium equations
Equilibrium equations
Problem statement
Consider a closed subdomain with volume V and surface S within a body in equilibrium.
This region also has a general distribution of surface tractions t and body forces f
The static equilibrium implies that the forces acting on this region are balanced and thus
the resultant force must vanish
2.33
8
Equilibrium equations
Static equilibrium
Static equilibrium
S
tdS+
V
fdV = 0
Using the Cauchy stress formula and applying the divergence theorem we obtain
+f = 0
2.34
Equilibrium equations
Static equilibrium
In index notation
ji, j
+ f
i
= 0
In scalar notation
xx
x
+
yx
y
+
zx
z
+ f
x
= 0
xy
x
+
yy
y
+
zy
z
+ f
y
= 0
xz
x
+
yz
y
+
zz
z
+ f
z
= 0
2.35
Equilibrium equations
Example
Determine the body forces which equilibrate the following stress eld
xx
= 2x
2
3y
2
5z
yy
= 2y
2
+7
zz
= 4x +y +3z 5
xy
= z +4xy 6
xz
= 3x +2y +1
yz
= 0
2.36
Equilibrium equations
The End
Any questions, opinions, discussions?
2.37
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