Page 112 - The Mechatronics Handbook

0066_Frame_C08 Page 3 Wednesday, January 9, 2002 3:48 PM

In this section we will review the fundamental equations describing the motion of deformable bodies.
These equations also result from the balance of linear and angular momentum applied to an infinitesi-
mally small portion of the material volume dV. Each element dV is subjected not only to external body
force f, but also to internal forces originating from the rest of the body. These internal forces are described
by a second order tensor T, called stress tensor. The balance of linear momentum can then be stated in
integral form for an arbitrary portion of the body occupying volume V as

d
⋅
d
----- ∫ rv v = ∫ Tn A + ∫ f v (8.8)
d
d
dt V ∂V V
where ρ is the mass density, v is the velocity of the element dV, and f is the force per unit volume acting
upon dV. The above balance law states that the rate of change of linear momentum is equal to the sum of
the internal force flux (stress) acting on the boundary of V and the external body force, distributed inside
V. Applying the transport theorem to (8.8) along with the mass conservation law reduces the above to
∫ rv ˙ v = ∫ ∇ T v + ∫ f v (8.9)
⋅
d
d
d
V V V
Since (8.9) is valid for an arbitrary volume, it follows that the integrands are also equal. Thus the local
(differential) form of linear momentum balance is
⋅
v ˙
rv ˙ = ∇ T + f or with index notation ρ = T + f i (8.10)
ij,j
i
Using analogous procedure, the balance of angular momentum can be shown to reduce to a simple
symmetry condition of the stress tensor
T ij = T ji (8.11)

which is valid for materials without external body couples. It should be mentioned that in certain
anisotropic materials, the polarization or magnetization vectors can develop body couples, for example
when E × P ≠ 0. In these cases the stress tensor is nonsymmetric and its vector invariant is equal to the
body couple. Equations (8.10) are usually used in one of the three most common coordinate systems.
For example, using rectangular coordinates we have
---------- + ---------- + ---------- + f x = ra x , T xy = T yx
∂T xx
∂T xz
∂T xy
∂x ∂y ∂z
---------- + ---------- + ---------- + f y = ra y , T yz = T zy (8.12)
∂T yz
∂T yy
∂T yx
∂x ∂y ∂z
---------- + ---------- + ---------- + f z = ra z , T xz = T zx
∂T zx
∂T zy
∂T zz
∂x ∂y ∂z
and in cylindrical coordinates
---------- + T rr – T qq ------------ + ---------- + f r = ra r , T rq =
-------------------- +
∂T rz
∂T rr
1∂T rq
∂r r r ∂q ∂z T qr
∂T rq
1∂T qq
---------- + 2 ------------- + ---------- + f q = ra q , T qz = T zq (8.13)
--T rq +
∂T qz
∂r r r ∂q ∂z
---------- + 1 ------------ + ---------- + f z = ra z , T rz = T zr
--T rz +
1∂T qz
∂T rz
∂T zz
∂r r r ∂q ∂z
where (x, y, z) and (r, θ, z) are the three coordinates, f ’s are the corresponding body force densities, and
a’s are the accelerations. In addition to Eqs. (8.12) or (8.13), a relation between the stress and the
displacement is needed in order to determine the deformation. Since the rigid body translations and
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