ents, only the three normal ones and three of the six shear ones
are independent.
Because the stress tensor is symmetric, we usually write (8) as
T i =
j=
∑
1
3
σ ij n j = σ ij n j ,
(11)
or, in terms of the vectors rather than their components,
T = σ 4.
(12)
Stress has units of force per area. In the cgs system of units
based on the centimeter, gram, and second, force is given
in dynes (dyn), with 1 dyn = 1 g-cm/s
2
, so stress is given in
dyn/cm 2 , or bars, a unit equal to 10 6 dyn/cm 2 . The bar has the
convenient property that atmospheric pressure at sea level is
1.01 bars. In SI units based on the meter, kilogram, and second
(mks), force is given in Newtons (N), with 1 N = 1 kg-m/s 2 , so
stress is given in Pascals (Pa), a unit equal to 1 N/m 2 . The two
sets of units can be related by noting that 1 Pa = 10 5 dyn/
10 4 cm 2 = 10 dyn/cm 2 = 10 −5 bars, so 1 MPa equals 10 bars.
2.3.3 Stress as a tensor
We have been using the term “tensor” without defining it. Already, we saw that it came from a relation between the traction
and normal vectors, and is an entity with two subscripts that
has properties similar to those of vectors. Vectors are entities
that are independent of coordinate system, so that physical
laws written using them do not depend on the coordinate
system and can be analyzed using any convenient coordinate
system. We now show that tensors are similar entities.
Specifically, a vector is an entity that remains the same in two
coordinate systems (Section A.5.1), such that its components
in two different Cartesian coordinate systems are related by the
transformation matrix A. Hence, given two sets of axes (x 1 , x 2 ,
x 3 ) and (x′ 1 , x′ 2 , x′ 3 ), the components of a vector u are related by
u′ = Au.
(13)
The relation between the components of the stress tensor
in two Cartesian coordinate systems can be found using the
fact that it relates the traction and normal vectors in each
coordinate system. The components of the traction and normal
vectors in the two coordinate systems satisfy
T′ = AT, 4′ = A4.
(14)
The reverse transformation can be written using the inverse of
A which, because A is orthogonal, equals its transpose:
4 = A −1 4′ = A T 4′.
(15)
In the primed coordinate system, the traction is related to the
normal vector and the stress tensor by
x 2
x 1
f 1
f 2
12
σ
11
σ
22
σ
21
σ
dx 2
dx 1
22 +
dx 2
σ
∂ 22
σ
∂x 2
21 +
dx 2
σ
∂ 21
σ
∂x 2
12 +
dx 1
σ
∂ 12
σ
∂x 1
11 +
dx 1
σ
∂ 11
σ
∂x 1
Fig. 2.3-5 Clockwise and counterclockwise torques about the x 3 axis on a
rectangle due to the stress components and body forces. If the stress tensor
were not symmetric, σ 12 = σ 21 , a net torque would arise.
start to rotate if it is not already doing so. The net body force, if
any, is f i dx 1 dx 2 , where f i is the force at the center of the block.
Because a torque is the product of a force and a lever (or
moment) arm, the shear stresses σ 21 and σ 12 acting on the faces
along the x 1 and x 2 axes contribute no torque. The other stress
components cause torques equal to the product of the lever arm
and the traction, the stress component times the area of the
face. Thus the total counterclockwise torque is the sum of that
due to the shear tractions on the other two faces, with lever
arms dx 1 and dx 2 , the normal tractions on all four faces, with
lever arms dx 1 /2 and dx 2 /2, and the two body force components acting at the center of the block, with lever arms dx 1 /2
and dx 2 /2:
τ
σ
σ
σ
σ
3
1 2
12
1
1
1
2
2 1
21
2
2
1
2
=
+
⎛
⎝
⎜
⎞
⎠
⎟
−
+
⎛
⎝
⎜
⎞
⎠
⎟
∂
∂
∂
∂
x
dx dx dx
x
dx dx dx
−
+
⎛
⎝
⎜
⎞
⎠
⎟
+
σ
σ
σ
11
11
1
1
2
2
11
2
2
2
2
∂
∂x
dx dx
dx
dx
dx
+
+
⎛
⎝
⎜
⎞
⎠
⎟
−
σ
σ
σ
22
22
2
2
1
1
22
1
1
2
2
∂
∂x
dx dx
dx
dx
dx
+
−
.
f dx dx
dx
f dx dx
dx
2
1
2
1
1
1
2
2
2
2
(10)
Dividing by the area and letting dx 1 and dx 2 go to zero, we see
that for there to be no torque, σ 12 = σ 21 . The same argument for
the torque about the other two axes shows that σ 13 = σ 31 and
σ 23 = σ 32 . Thus, although the stress tensor has nine compon2.3 Stress and strain 41
are independent.
Because the stress tensor is symmetric, we usually write (8) as
T i =
j=
∑
1
3
σ ij n j = σ ij n j ,
(11)
or, in terms of the vectors rather than their components,
T = σ 4.
(12)
Stress has units of force per area. In the cgs system of units
based on the centimeter, gram, and second, force is given
in dynes (dyn), with 1 dyn = 1 g-cm/s
2
, so stress is given in
dyn/cm 2 , or bars, a unit equal to 10 6 dyn/cm 2 . The bar has the
convenient property that atmospheric pressure at sea level is
1.01 bars. In SI units based on the meter, kilogram, and second
(mks), force is given in Newtons (N), with 1 N = 1 kg-m/s 2 , so
stress is given in Pascals (Pa), a unit equal to 1 N/m 2 . The two
sets of units can be related by noting that 1 Pa = 10 5 dyn/
10 4 cm 2 = 10 dyn/cm 2 = 10 −5 bars, so 1 MPa equals 10 bars.
2.3.3 Stress as a tensor
We have been using the term “tensor” without defining it. Already, we saw that it came from a relation between the traction
and normal vectors, and is an entity with two subscripts that
has properties similar to those of vectors. Vectors are entities
that are independent of coordinate system, so that physical
laws written using them do not depend on the coordinate
system and can be analyzed using any convenient coordinate
system. We now show that tensors are similar entities.
Specifically, a vector is an entity that remains the same in two
coordinate systems (Section A.5.1), such that its components
in two different Cartesian coordinate systems are related by the
transformation matrix A. Hence, given two sets of axes (x 1 , x 2 ,
x 3 ) and (x′ 1 , x′ 2 , x′ 3 ), the components of a vector u are related by
u′ = Au.
(13)
The relation between the components of the stress tensor
in two Cartesian coordinate systems can be found using the
fact that it relates the traction and normal vectors in each
coordinate system. The components of the traction and normal
vectors in the two coordinate systems satisfy
T′ = AT, 4′ = A4.
(14)
The reverse transformation can be written using the inverse of
A which, because A is orthogonal, equals its transpose:
4 = A −1 4′ = A T 4′.
(15)
In the primed coordinate system, the traction is related to the
normal vector and the stress tensor by
x 2
x 1
f 1
f 2
12
σ
11
σ
22
σ
21
σ
dx 2
dx 1
22 +
dx 2
σ
∂ 22
σ
∂x 2
21 +
dx 2
σ
∂ 21
σ
∂x 2
12 +
dx 1
σ
∂ 12
σ
∂x 1
11 +
dx 1
σ
∂ 11
σ
∂x 1
Fig. 2.3-5 Clockwise and counterclockwise torques about the x 3 axis on a
rectangle due to the stress components and body forces. If the stress tensor
were not symmetric, σ 12 = σ 21 , a net torque would arise.
start to rotate if it is not already doing so. The net body force, if
any, is f i dx 1 dx 2 , where f i is the force at the center of the block.
Because a torque is the product of a force and a lever (or
moment) arm, the shear stresses σ 21 and σ 12 acting on the faces
along the x 1 and x 2 axes contribute no torque. The other stress
components cause torques equal to the product of the lever arm
and the traction, the stress component times the area of the
face. Thus the total counterclockwise torque is the sum of that
due to the shear tractions on the other two faces, with lever
arms dx 1 and dx 2 , the normal tractions on all four faces, with
lever arms dx 1 /2 and dx 2 /2, and the two body force components acting at the center of the block, with lever arms dx 1 /2
and dx 2 /2:
τ
σ
σ
σ
σ
3
1 2
12
1
1
1
2
2 1
21
2
2
1
2
=
+
⎛
⎝
⎜
⎞
⎠
⎟
−
+
⎛
⎝
⎜
⎞
⎠
⎟
∂
∂
∂
∂
x
dx dx dx
x
dx dx dx
−
+
⎛
⎝
⎜
⎞
⎠
⎟
+
σ
σ
σ
11
11
1
1
2
2
11
2
2
2
2
∂
∂x
dx dx
dx
dx
dx
+
+
⎛
⎝
⎜
⎞
⎠
⎟
−
σ
σ
σ
22
22
2
2
1
1
22
1
1
2
2
∂
∂x
dx dx
dx
dx
dx
+
−
.
f dx dx
dx
f dx dx
dx
2
1
2
1
1
1
2
2
2
2
(10)
Dividing by the area and letting dx 1 and dx 2 go to zero, we see
that for there to be no torque, σ 12 = σ 21 . The same argument for
the torque about the other two axes shows that σ 13 = σ 31 and
σ 23 = σ 32 . Thus, although the stress tensor has nine compon2.3 Stress and strain 41