3.5 Balance Laws
115
Angular Equation of Motion in 2D
In 1D, rotation is not an issue, since motion can be only linear, along a single axis.
In 2D, we consider again the differential element shown in Fig. 3.18. If we choose
the center of the element to be point o, then the moments of the normal stresses and
body forces vanish, leaving only the contributions of the shear stresses. In addition,
for a rectangular element of mass dm = ρ dxdydz, the mass moment of inertia is
I o = (dx 2 + dy 2 )dm/12. Equations (3.154) and (3.157) give
(σ xy dydz)(dx/2) +
σ xy +
∂σ xy
∂x
dx
dydz
(dx/2)
−(σ yx dxdz)(dy/2) −
σ yx +
∂σ yx
∂y
dy
dxdz
(dy/2)
= [(dx
2
+ dy
2 )ρ dxdydz/12]α z .
Expanding this relation, dividing by dxdydz, and letting dx, dy → 0 yield
σ xy = σ yx .
(3.158)
In other words, the 2D Cauchy stress tensor is symmetric.
Angular Equation of Motion in 3D
You probably can guess from the above result that the Cauchy stress tensor also is
symmetric in the general 3D case, possibly making the details of a tensor analysis
more painful than they are worth. Nevertheless, they are included here for the
doubters, as well as those who love tensors.
For a three-dimensional solid body subjected to surface tractions and body forces
(Fig. 3.19), Eq. (3.156) becomes
A
r × T dA +
V
r × b dV =
d
dt
V
(r × ρv) dV .
(3.159)
To transform the surface integral into a volume integral, we first need to put the
integrand into a suitable form. Substituting Eqs. (3.111) and (3.112) gives
r × T = r × (n · σ ) = r × (n · e i T i ) = (n · e i ) r × T i
= n · (e i r × T i ),
in which (n·e i ) could be moved because it is a scalar. With this result, the divergence
theorem (2.70) yields
A
r × T dA =
V
∇ · (e i r × T i ) dV .
115
Angular Equation of Motion in 2D
In 1D, rotation is not an issue, since motion can be only linear, along a single axis.
In 2D, we consider again the differential element shown in Fig. 3.18. If we choose
the center of the element to be point o, then the moments of the normal stresses and
body forces vanish, leaving only the contributions of the shear stresses. In addition,
for a rectangular element of mass dm = ρ dxdydz, the mass moment of inertia is
I o = (dx 2 + dy 2 )dm/12. Equations (3.154) and (3.157) give
(σ xy dydz)(dx/2) +
σ xy +
∂σ xy
∂x
dx
dydz
(dx/2)
−(σ yx dxdz)(dy/2) −
σ yx +
∂σ yx
∂y
dy
dxdz
(dy/2)
= [(dx
2
+ dy
2 )ρ dxdydz/12]α z .
Expanding this relation, dividing by dxdydz, and letting dx, dy → 0 yield
σ xy = σ yx .
(3.158)
In other words, the 2D Cauchy stress tensor is symmetric.
Angular Equation of Motion in 3D
You probably can guess from the above result that the Cauchy stress tensor also is
symmetric in the general 3D case, possibly making the details of a tensor analysis
more painful than they are worth. Nevertheless, they are included here for the
doubters, as well as those who love tensors.
For a three-dimensional solid body subjected to surface tractions and body forces
(Fig. 3.19), Eq. (3.156) becomes
A
r × T dA +
V
r × b dV =
d
dt
V
(r × ρv) dV .
(3.159)
To transform the surface integral into a volume integral, we first need to put the
integrand into a suitable form. Substituting Eqs. (3.111) and (3.112) gives
r × T = r × (n · σ ) = r × (n · e i T i ) = (n · e i ) r × T i
= n · (e i r × T i ),
in which (n·e i ) could be moved because it is a scalar. With this result, the divergence
theorem (2.70) yields
A
r × T dA =
V
∇ · (e i r × T i ) dV .