Stress Transformation
Shailendra P. Joshi
UH
ME3369
1
Relating components of stress tensor in one coordinate system to another coordinate
system
P
t=
n
2
2D state of stress
2
6
[ ]=4
3D state of stress
11
12
21
22
31
32
13
3
7
23 5
33
2D state of stress
(plane stress)
[ ]=
"
11
12
21
22
#
3
2D state of stress and stress transformation
a
x2
b
P
P
x1
b
a
Consider section on the left side of a-a
x01 x02
x2
P
x1
x01 x02
P cos ✓
n
P
A0 ✓
A0 / cos ✓
n = cos ✓e1 + sin ✓e2
P sin ✓
4
Writing the force P into two components
On a plane inclined at ✓ to the vertical, the traction vector along e1 is
5
Then, the stress components
acting on the plane a-a are
Along the unit normal n : tn · n ⌘ n
2
Normal stress
Tangent to the plane
Shear stress
6
Limiting cases
Maximum normal stress
On which plane is the shear stress the maximum?
Maximum
shear stress
7
Let’s consider the other section (b-b)
x2
P
x01
x02
P cos ✓
x1
x01 x02
A0 / sin ✓
90 ✓
⌧90
✓
P
P sin ✓
P sin ✓
P
=
=
sin2 ✓
A0 / sin ✓
A0
P cos ✓
P
=
=
sin ✓ cos ✓
A0 / sin ✓
A0
8
Bringing things together
a
b
x2
x1
b
a
P
cos2 ✓
A0
Note the directions of shear
stresses
In general, the normal
stresses on adjoining faces
are not equal.
However, shear stresses are
always equal on adjoining
faces
P
sin ✓ cos ✓
A0
(General 2D stress state)
P
sin ✓ cos ✓
A0
P
sin2 ✓
A0
9
2D stress transformation
✓ ⌧
11
x2
21
21
21
=
=
12
=
12
12
12
x1 (e1 )
i.e. relating
n
n
and ⌧ to
22
11 ,
22 , and
12
10
2D stress transformation
i.e. relating
n
and ⌧ to
11 ,
22 , and
12
x2
A
⌧
✓
dl cos ✓
11
21
21
21
=
n
=
12
=
dl
n = cos ✓e1 + sin ✓e2
12
12
B
✓
dl sin ✓
C
12
22
x1 (e1 )
11
Let us calculate the forces on the three planes and take their components along n
z
On plane AB :
On plane BC :
force
}|
{
11 (dl cos ✓)h cos ✓
| {z }
area
22 (dl sin ✓)h
z
sin ✓
On plane AC :
n
force
}|
{
12 (dl cos ✓)h sin ✓
| {z }
area
12 (dl sin ✓)h
cos ✓
dl h cos ✓
Equilibrium of forces along n gives
n
n
=
11
=
+
2
11
cos2 ✓ +
22
+
22
sin2 ✓ + 2
11
22
2
12
sin ✓ cos ✓
cos 2✓ +
12
sin 2✓ 12 (1)
Similarly, force equilibrium along the tangent to the inclined plane gives
⌧ =(
⌧=
22
12 ) cos ✓ sin ✓
11
22
2
+
12 (cos
sin 2✓ +
2
✓
12
sin2 ✓)
cos 2✓
(2)
13
General form of 2D stress transformation
n
=
11
⌧=
+
2
22
+
11
11
2
22
2
22
cos 2✓ +
sin 2✓ +
12
12
sin 2✓
cos 2✓
14
Principal stresses and principal planes
d n
= 0 ) tan 2✓p =
d✓
2
11
12
22
Solution has 2 roots: 180 degrees apart
One locates the plane on which the maximum normal stress acts;
The other locates the plane on which the minimum normal stress acts (90o from the first plane)
15
n
✓
Maximum and minimum principal stresses
= 0 ) tan 2✓p =
2
12
11
n
22
Max. normal stress
1
=
max
Min. normal stress
2
=
min
=
11
+
2
11 +
=
2
=
22
22
11
+
+
2
22
s✓
s✓
+
11
22
2
11
22
2
11
22
2
cos 2✓ +
12
◆2
+(
12 )
◆2
+(
12 )
sin 2✓
2
2
16
Maximum shear stress (in 2D)
⌧=
11
22
2
sin 2✓ +
12
d⌧
= 0 ) tan 2✓s =
d✓
cos 2✓
11
2
22
12
2 roots: 90 degrees apart (because, this equation is just the negative reciprocal of the one for
principal planes)
Hence, the max. and min. shear stresses form angle of 45 degrees with the planes of principal
stresses
⌧(max)or(min) = ±
s✓
11
22
2
◆2
+(
12 )
2
17
Summary (2D state of stress)
Normal and shear stress on a plane
n
=
11
+
2
22
+
11
22
2
cos 2✓ +
12
11
⌧=
sin 2✓
22
2
sin 2✓ +
12
s✓
22
cos 2✓
Principal stresses and planes
1
=
max
=
11
+
2
22
+
s✓
11
22
2
◆2
+(
12 )
2
2
=
min
=
11
+
2
22
11
2
◆2
+(
12 )
2
• Principal stresses are the maximum and the minimum normal stresses
• Principal planes are the two mutually perpendicular planes on which principal stresses act.
• Shear stress is always zero on the principal planes.
Maximum and minimum shear stresses
⌧(max)or(min) = ±
s✓
11
22
2
◆2
+(
12 )
2
• Max and min shear stresses differ only in sign
• Planes of max and min shear stresses are at 90 degrees to each other and at 45
degrees to the principal planes
• In general, normal stress is not zero on planes of max and min. shear stresses
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