Transcript Title

SM1-07: Continuum Mechanics: State of stress
CONTINUUM MECHANICS
(STATE OF STRESS)
M.Chrzanowski: Strength of Materials
1/19
SM1-07: Continuum Mechanics: State of stress
Internal forces - stress
n
The sum of all internal forces acting on ΔA
Δw
{wI}
P1
A
A
I
n
Pn
{ZI}
Neighbourhood of
point A
M.Chrzanowski: Strength of Materials
ΔA
– area of point A neighbourhood
w
lim

A 0 A

p
A
Stress vector at A
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SM1-07: Continuum Mechanics: State of stress
w
lim

A0 A

p r , n
A
A
Stress matrix
Stress vector is a measure of internal forces intensity and
depends on the chosen point and cross section
Stress vectors:
n3
σ33
σ31
x3
n2
p1
rA
σ11
p3
σ32
σ23
σ13
n1
p1[σ11 , σ12 , σ13 ]
p2
σ22
p2[σ21 , σ22 , σ23 ]
p3[σ31 , σ32 , σ33 ]
σ11 , σ12 , σ13
Tσ
σ31 , σ32 , σ33
σ12 σ21
Stress matrix
x2
x1
Point A image
Components ij of matrix T are called stresses.
Stress measure is [N/m2] i.e. [Pa]
M.Chrzanowski: Strength of Materials
σ21 , σ22 , σ23
Tσ(σij)
i,j = 1,2,3
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SM1-07: Continuum Mechanics: State of stress
Stress matrix
n3
σ33
σ31
σ13
n1
x3
σ11
σ11 , σ12 , σ13
p3
σ32
n2
σ23
p2
σ22
p1
σ12 σ21
x2
x1
M.Chrzanowski: Strength of Materials
Tσ
σ21 , σ22 , σ23
σ31 , σ32 , σ33
Shear stresses
Normal stresses
Positive and negative stresses
Stress is defined as positive when the direction of stress
vector component and the direction of the outward
normal to the plane of cross-section are both in the
positive sense or both in the negative sense in relation to
the co-ordinate axes.
If this double conjunction of stress component and
normal vector does not occur – the stress component is
negative one.
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SM1-07: Continuum Mechanics: State of stress
n3
σ33
x3
σ31
{xi} x2
x1
σ13
n1
σ11
Stress transformation
p3
σ32
n2
σ23
p1
Tσ[σij]
p2
σ22
σ12 σ21
[ij]
T’σ[σ’ij]
{x’i}
3
3
 ij,   ik jl kl
k 1 l 1 summation
Here Einstein’s
convention has been applied
M.Chrzanowski: Strength of Materials
5/19
SM1-07: Continuum Mechanics: State of stress
Stresses on inclined plane
X1= 0
 ~11  A1  ~21  A2  ~31  A3  ~ 1  A  0
x3
ν(νi )
Ai  A i
A1 A   1
ΔAν
~21
~ 1
n2
x1
ΔA2
1
A
…, …
 ~11 1  ~21 2  ~31 3  ~ 1  0
~
p (~i )
 1  11 1   21 2   31 3
~
n1
 11
 1   j1j
ΔA1
~31
ΔA3
x2
If we assume:
then:
n3
M.Chrzanowski: Strength of Materials
i   jij
 ij   ji
lim
A0
Symmetry of
stress tensor
i   ijj
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SM1-07: Continuum Mechanics: State of stress
Stresses on inclined plane


p  T
i   ijj

p (i )  ?
 11  12  13   1 
 1,  2 ,  3 ,   21  22  23   2 
 31  32  33   3 
x3
j=1
j=2
i=1
i=2
i=3
j=3
x2
x1
i=1
 1 11 1  12 2  13 3
j=1
M.Chrzanowski: Strength of Materials
j=2
j=3
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SM1-07: Continuum Mechanics: State of stress
Stresses on inclined plane
x3
n3


σ33=2
σ31=1 σ32=-3
 2  4 / 3
n1

p
1
2 0
T  0  1  3


1  3 2 
i=2
i=3
i   ijj
j=1
2
i=1
j=3
3  i=1

3  i=2
3 i=3
 1  2 3  0 3  1 3  3 3
 2  0 3   1 3  3 3    4 3
 3  1 3   3 3   2 3  0

p (3 3 , 4
M.Chrzanowski: Strength of Materials
3

3
3 
1  1
2 0
0  1  3 1



,

,

,

1
2
3
σ23=-3


1  3 2  1
σ22= -1
x
j=2
σ13=1
σ 112=0
3/ 3
σ21=0
σ11=2
x1

p (i )  ?
n2
1
 
  1
1

3 ,0)
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SM1-07: Continuum Mechanics: State of stress

 n3
x3
σ33=2
σ31
=1  σ
432/=-3
3
2
n1
x1
1
2 0
T  0  1  3


1  3 2 




n
p   2

p (3 3 , 4
σ23=-3
σ13=1

 1 σ322/ = 3-1
p σ12=0
σ21=0
x2
σ11=2


p  


p  T
i   ijj
M.Chrzanowski: Strength of Materials

1
 
  1
1

3

3
3 
3 ,0)
On this plane none of the vector
components are perpendicular nor
parallel to the plane.

  T
i   ijj
We will look for such a plane to which
vector
will be perpendicular, thus
having no shear components.
 ijj  i  0
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SM1-07: Continuum Mechanics: State of stress
Principal stresses
 ijj
1 0 0 
 ij  0 1 0
 i  0


Kronecker’s delta 0 0 1
or
ij=
1
if i=j
0
if ij
We will use Kronecker’s delta to renumber normal vector components i
i  ijj  i1 1  i 2 2  i 3 3
 1  1 jj  11 1  12 2  13 3  1  1  0  0   1
ijj  jij  0
ij  ij j  0
Seeking are :
3 components of normal vector
and vector

p
size
M.Chrzanowski: Strength of Materials

 :
etc…
Three equations
 1, 2 , 3

  p
4 unknowns
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SM1-07: Continuum Mechanics: State of stress
Principal stresses

ij
 ij j  0
i=1
i=2
i=3
in the explicit form:
0
1
0
11  11  1  12  12  2  13  13  3  0
 21   21  1   22   22  2   23   23  3  0
 31   31  1   32   32  2   33   33  3  0
j=1
11    1 
 21 1 
 31 1 
j=2
12 2 
 22    2 
 32 2 
j=3
13 3  0
 23 3  0
 33    3  0
The above is set of 3 linear equations with respect to 3 unknowns i
with zero-valued constants . The necessary condition for non-zero solution
is vanishing of matrix main determinant composed of the coefficients of the
unknowns.
M.Chrzanowski: Strength of Materials
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SM1-07: Continuum Mechanics: State of stress
Principal stresses
 11  
 12
 13
 21
 22  
 23  0
 31
 32
 33  
where invariants I1 , I2 , I3 are following
determinants of σij matrix
 3  I1 2  I 2  I3  0
 11  12  13 

 22  23 
21


 31  32  33   31
I1  11   22   33
 13  11
I 2  11 22  12 21   22 33   23 32   3311   3113
I 3  11 ( 22 33   23 32 )  12 ( 23 31   21 33 )  13 ( 21 32   22 31 )
Solution of this algebraic equation of the 3rd order yields 3 roots being real
numbers due to symmetry of σij matrix
1, 2 , 3
These roots being eigenvalues of matrix σij are called
principal stresses
M.Chrzanowski: Strength of Materials
12/19
SM1-07: Continuum Mechanics: State of stress
Principal stresses
 11  12

 22
 21
0
 0
In the special case of plane stress state
 3  I1 2  I 2  I3  0
I1  11   22
I 2  11 22  12 21
 3  I1 2  I 2  0


I
and:
... 1, 2
 1, 2 
1
I1 
2
1
  11   22 
2 
M.Chrzanowski: Strength of Materials
2
2
1
0
0

0
I3  0
 I1  I 2   0
 0

 4 I 2   ...
 11   22 2  4 122 

13/19
SM1-07: Continuum Mechanics: State of stress
Principal stresses
Now, from the set of equations:
j


11  11  1  12  12  2  13  13  3  0
 21   21  1   22   22  2   23   23  3  0
 31   31  1   32   32  2   33   33  3  0
one can find out components of 3 eigenvectors, corresponding to each principal stress
 1


 1 ,  2 ,  3 
1
1
1
 2 



 1 ,  2 ,  3 
2 
2 
 3
2 
  ,  ,   
3
1
3
2
3
3
These vectors are normal to three perpendicular planes. The stress vectors on
these planes are also perpendicular to them and no shear components of stress
vector exist, whereas normal stresses are equal principal stresses.
 1

 1
M.Chrzanowski: Strength of Materials
 2 

 2
 3

 3
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SM1-07: Continuum Mechanics: State of stress
Principal stresses
It can be proved that  1 ,  2 ,  3 are extreme values of normal stresses
(stresses on a main diagonal of stress matrix) . Customary, these values are
ordered as follows
1   2   3
Surface of an ellipsoid with semi-axis (equatorial radii) equal to the values of
principal stresses represents all possible stress vectors in the chosen point and
under given loading.
2
3
M.Chrzanowski: Strength of Materials
1
15/19
SM1-07: Continuum Mechanics: State of stress
Principal stresses
With given stress matrix in the chosen point and given loading one can find 3
perpendicular planes such that stress vectors (principal stresses) have only normal
components (no shear components).
The coordinate system defined by the directions of principal stresses is called system of
principal axis.

n3
n1
 ( 3)  3
x3
σ33
p3
σ32
σ3111  12  13 

σ23
σ 
 23
13
21
22
p


1
 31  23  33 
σ11
σ12 σ21
n2
p2
σ22
x2
 (2)

 1 0 0 
0 

0
 (1)
2


 0 0  3 
1
2
x1
M.Chrzanowski: Strength of Materials
16/19
SM1-07: Continuum Mechanics: State of stress
Stresses on characteristic planes
i   ijj
 1 0 0 
0 
0
2


 0 0  3 
 1  31
2  0  0
 2 
0
2
 3 
0
2 0
3

 1,0,0
2
1 1


 1 2 ,0,1 2
3
3

2

2
2


p
M.Chrzanowski: Strength of Materials



p 1

 0,0,1
0 / 2  0 3
02  2 

2  3 / 2   3
2,0,23

2



2

2
  3
2  0  1 2 3 2  1
1
1
2
 2
 2
p   12 2   32 2
 2   2  p
2
2





3
2  p  2  12 2   32 2   1

2




  p   1
/4
1
1

2  13 2
 0,1,0
2 1
1
2
2   12 4   1 3   32 4 
 
1   3
1
 1   3 2
4
2
17/19
SM1-07: Continuum Mechanics: State of stress


 0,1 2 ,1 2
3



 1 0 0 
0 
0
2


 0 0  3 


 1 2 ,0,1 2
 1 2 ,1 2 ,0
3
2
1


p 0, 2
 
2 , 3
2  3
2
2  3
 
2
2

i   ijj
3

2
1

p 1

max σ
M.Chrzanowski: Strength of Materials
2 , 2
 

2,0
1   2
2
 2
  1
2
2
1


p 1
 
max 

2 , 2

2,0
1   3
2
  3
  1
2
18/19
SM1-07: Continuum Mechanics: State of stress

Mohr circles
– represent 3D state of stress in a given point –
on the plane of normal and shear stresses
1   3
2
2  3
2
1   2

2
0


p 0, 2
 
3
2 , 3
2  3
2

2
2  3
 
2
M.Chrzanowski: Strength of Materials
 2 1   2
2  3

1   3
2

p 1
2,22
 

2
2,0
1   2
2
1   2
 
2
1


p 1
 
max 
2, 0, 3
2

1   3
2
1   3
 
2
19/19
SM1-07: Continuum Mechanics: State of stress
stop
M.Chrzanowski: Strength of Materials
20/19