264
25 Elastic Displacements and Waves
k =i
f ki =
f ki −→
((V )
df .
(318)
(V ) once again stands for the surface of the volume V , df is the force acting on
the vectorial surface element dA. This defines the stress tensor σ according to
df = σ · dA , i. e.
⎛
⎝
d f x
d f y
d f z
⎞
⎠ =
⎛
⎝
σ xx σ xy σ xz
σ yx σ yy σ yz
σ zx σ zy σ zz
⎞
⎠
⎛
⎝
d A x
d A y
d A z
⎞
⎠ .
(319)
In place of the stress τ of the rod, the far more complicated quantity σ arises when
we deal with three-dimensional problems. According to definition (319), σ is a force
per unit area. Written down in detail, Eq. (319) looks like
d f x = σ xx d A x + σ xy d A y + σ xz d A z ,
d f y = σ yx d A x + σ yy d A y + σ yz d A z ,
d f z = σ zx d A x + σ zy d A y + σ zz d A z .
⎫
⎬
⎭
(320)
We have indicated all quantities with x, y, z in order to enable easier recognition. We
use the equivalent indicating with numbers, especially when dealing with simplifications of sums over individual components according to 1 ↔ x, 2 ↔ y, 3 ↔ z, thus
d f 1 in place of d f x , σ 23 in place of σ yz etc.
The individual components of σ have the following meaning: If one dissects the
continuum along the area dA, then one has to use the force df so that the intersection
planes do not move. It is accepted for the sign that a traction is positive and a pressure
is negative. If one chooses for example that dA is orthogonal to the x-axis, then σ xx ,
σ xy and σ xz are the Cartesian components of df divided by the amount of the area
|dA|.
An important property of the stress tensor σ still has to be discussed. In order
to do this we consider a small, cubic volume element V positioned parallel to the
axes. We assume at the areas of the cubes surface σ xy > 0 as well as σ yx > 0. Then
σ xy causes a clockwise rotation of the cube along the z-axis and σ yx a rotation in
the opposite direction, in other words, for σ xy − σ yx = 0 the cube begins to rotate.
However, we received σ = σ (x, t) as the limit of the interaction forces on an
arbitrarily chosen mass point at x, in other words σ = σ(x, t) describes the motion
of the mass point if we let V go to zero. A point cannot however rotate. The
following important law follows
2 :
2 For the case of elastic coupled mass points which we consider exclusively here, this property for
the stress tensor σ of the equivalent continuum is necessary. There are however other models where
the symmetry of the stress tensor is lost. This is the case for example, if one calculates using small
rigid solids instead of mass points (for example, as a model of approximation for molecules). For
the resulting continua new limits have to be reformulated.
25 Elastic Displacements and Waves
k =i
f ki =
f ki −→
((V )
df .
(318)
(V ) once again stands for the surface of the volume V , df is the force acting on
the vectorial surface element dA. This defines the stress tensor σ according to
df = σ · dA , i. e.
⎛
⎝
d f x
d f y
d f z
⎞
⎠ =
⎛
⎝
σ xx σ xy σ xz
σ yx σ yy σ yz
σ zx σ zy σ zz
⎞
⎠
⎛
⎝
d A x
d A y
d A z
⎞
⎠ .
(319)
In place of the stress τ of the rod, the far more complicated quantity σ arises when
we deal with three-dimensional problems. According to definition (319), σ is a force
per unit area. Written down in detail, Eq. (319) looks like
d f x = σ xx d A x + σ xy d A y + σ xz d A z ,
d f y = σ yx d A x + σ yy d A y + σ yz d A z ,
d f z = σ zx d A x + σ zy d A y + σ zz d A z .
⎫
⎬
⎭
(320)
We have indicated all quantities with x, y, z in order to enable easier recognition. We
use the equivalent indicating with numbers, especially when dealing with simplifications of sums over individual components according to 1 ↔ x, 2 ↔ y, 3 ↔ z, thus
d f 1 in place of d f x , σ 23 in place of σ yz etc.
The individual components of σ have the following meaning: If one dissects the
continuum along the area dA, then one has to use the force df so that the intersection
planes do not move. It is accepted for the sign that a traction is positive and a pressure
is negative. If one chooses for example that dA is orthogonal to the x-axis, then σ xx ,
σ xy and σ xz are the Cartesian components of df divided by the amount of the area
|dA|.
An important property of the stress tensor σ still has to be discussed. In order
to do this we consider a small, cubic volume element V positioned parallel to the
axes. We assume at the areas of the cubes surface σ xy > 0 as well as σ yx > 0. Then
σ xy causes a clockwise rotation of the cube along the z-axis and σ yx a rotation in
the opposite direction, in other words, for σ xy − σ yx = 0 the cube begins to rotate.
However, we received σ = σ (x, t) as the limit of the interaction forces on an
arbitrarily chosen mass point at x, in other words σ = σ(x, t) describes the motion
of the mass point if we let V go to zero. A point cannot however rotate. The
following important law follows
2 :
2 For the case of elastic coupled mass points which we consider exclusively here, this property for
the stress tensor σ of the equivalent continuum is necessary. There are however other models where
the symmetry of the stress tensor is lost. This is the case for example, if one calculates using small
rigid solids instead of mass points (for example, as a model of approximation for molecules). For
the resulting continua new limits have to be reformulated.