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Symmetry of Stress Tensor
Consider moment equilibrium of differential element:
Taking moments about x
1
axis (i.e point C):
È
Â
M
1
=
0
:
2
s
23
d
x
3
x
1
)
2
Area of
È
(
x
2
˘

2
32
(
x
2
x
1
)
x
3
˘
=
0
Î˚
Î
2
˚
Moment
f
23
=
face
arm
32
Thus, in general
mn
=
nm
Stress tensor is symmetric. Six independent components of the stress tensor.
11
12
Ê=
21
ˆ
22
23
Á
Á
=
32
˜
˜
33
31
Ë =
13
¯
Note
a positive (tensile) component of stress acts on a face with a positive normal in a
positive direction. Thus a stress acting on a negative normal face, in a negative direction
is also positive. If the stresses do not vary over the infinitesimal element,
mn
acts on
opposite faces, in opposite directions but with equal magnitude.
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View Full Document But what happens if stress varies with position?
Stress Equations of Equilibrium
Consider equilibrium in x
1
direction. Let stresses vary across cube.
Also allow there to
be a body force (per unit volume) f
1
acting on the cube in the x1 direction:
x
3
s
11
d
x
2
∂s
31
d
x
3
s
+
∂
x
3
21
x
2
+
x
2
s
21
d
x
3
x
2
d
x
1
s
31
x
1
11
x
1
f
1
+
x
1
f
n
is a “body force”, e.g. due to the weight of the element (
r
g), centrifugal acceleration
(
r
r
w
2
), electromagnetic fields etc.
Taking equilibrium of forces in the x
1
direction gives:
Ê
ˆˆ
Á
11
x
1
˜
(
x
2
x
3
) 
11
(
x
2
x
3
) +
Á
Ê
21
+
21
x
2
˜
(
x
1
x
3
) 
21
(
x
1
x
3
)
Ë
11
+
x
1
¯Ë
x
2
¯
ˆ
+
Ê
Á
31
+
31
x
3
˜
(
x
1
x
2
)

31
(
x
1
x
2
)
+
f
1
(
x
1
x
2
x
3
)
=
0
Ë
x
3
¯
which simplifies to:
∂s
11
+
21
+
31
+
f
1
=
0
∂
x
1
x
2
x
3
similarly, equilibrium in the x
2
and x
3
directions yields:
23
12
+
22
+
32
+
f
2
=
0
and
13
+
+
33
+
f
3
=
0
x
1
x
2
x
3
x
1
x
2
x
3
Canceling out terms and dividing through by the common
dx
1
,
dx
2
,
dx
3
.
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This note was uploaded on 01/28/2012 for the course AERO 16.01 taught by Professor Markdrela during the Fall '05 term at MIT.
 Fall '05
 MarkDrela

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