Holpzaphel_-_Nonlinear-Solid-Mechanics-a-Contin
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3.4 Alternative Stress Tensors |
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2.The Cauchy stress components at a certain point are given with respe-ct to a x1., :1;rcoordinate system with the associated set {e0 }, a = 1, 2, ·of orthonormal basis vectors as
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1 ] kN/m2 . |
[u] = . |
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Assume a .rotation of {e0 } into the new set {e0 } of orthonormal bas.is vectors according to
Compute the components of the corotated Cauchy stress tensor u 0 •
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4 Balance Principles |
SURROUNDINGS
Boundary or wall of the system
Control
surface
Fixed amount of mass; |
Fixed amount of volume; |
.Energy can cross the boundary |
Mass and energy can cross |
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the control boundary |
Figure 4.1 |
Closed and open systems~ |
to be fixed. If even energy does not interact between the system and its surroundings, then we say that the boundary is insulated. Such a system is called mechanically and thermally isolated, which is an idealization for a physical system. There always exist electromagnetic and other types of forces which permeate the space. Note that no physical system .is truly isolated.
An open syste.m (or control volume) consists of a fixed amount of volume of
a properly selected reg.ion flc which .is independent of time t (see .Figure 4.1). The enclosing boundary of a control volume, over which both mass and energy can cross (enter or leave), is called a control surface, which we denote by anc.
Conservation of mass. In non-relativistic physics mass cannot be produced or
destroyed. It is assumed that during a motion there are neither mass sources (reservoirs that supply mass) nor mass sinks (reservoirs that abs-orb mass), so that the mass m of a body is a ·conserved quant-ity. Hence, if a particle has a certain mass in the reference
4.1 Conservation of M·ass |
135 |
By assuming that Vis an arbitrc11J' volume of region n0 , we condude that the integrand in (4.9) must vanish everywhere. Hence,
Po(X) = p(x(X, t_L t).J(X, t) , |
(4.l 0) |
holds for .all X E n0 • It represents the continuity mass equation (continuity stands for constancy of mass) in the material (or Lagrangian) description which is the appropriate descri pt.ion in solid mechanics.
Since the reference mass dens.ity p0 is independent of time we find simply from
(4.10) that
Dpo(X) |
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. (X) _ O |
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(4. l l) |
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{Jo |
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which is the rate form of (4.10) in the material description.
EXAMPLE 4.1 Show how the spatial mass density p = p(x~ t) changes with time. In particular, derive the rate form of continuity mass equation in the spatial (or Eulerian) description, which .is (expressed in terms of the velocity components)
rJ(x, t) + p(x, t)divv{x, t) = 0 |
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j1 + p~;~: = |
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(4.12) |
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or in the two equivalent forms |
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Dp~~,t) + gradp(x, t) · v(x, t) + p(x, t)divv(x, t) = 0 |
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(4.13) |
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8p(x, t) |
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(4.14) |
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for all x E H and for all times t. |
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Solution. Since 110 = 0 we obtain from (4.10) that |
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D ( |
pJ |
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(4.15) |
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Dt |
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(for simplicity written without arguments of the scalar quantities). In order to express eq. (4.15) .in terms of the spatial velocity components we find using the product rule
~_nd J = Jdivv, i.e. eq. (2.175){;, that
pJ = i)J + p} = J(p + pdivv) = 0 , |
(4.16) |
where the material time derivative of the spatial density function p is, having regard to (225), g.iven by the explicit expression
. Dp |
Dp |
(4.17) |
p= Dt |
= Dt +gradp·v |
4.1 Conservation of Mass |
137 |
EXERCISES
1. A velocity field of a plane motion has components of the form
V3 = Q ,
where a and /3 are positive constants. Assume that the spatial mass density p is independent of the curren_t position x so that gradp = o.
(a)Express p so that the continuity mass equation :is satisfied.
(b)Find a condition for which the given motion is isochoric.
2.Tuo motions of a continuum body are given in the form
x = (1 + a:(t)t)X ,
with the scalar function a:(t) and the set {ea}, a = 1, 2_, 3, of orthogonal unit vectors.
Find expressions for the spatial mass density pin terms of p0 so that the continuity mass equation is satisfied.
3.Consider a spatial scalar field <I> = <I>(x, t) and a spatial vector field u = u(x, t). Use the rate form of the continuity -mass -equation and obtain the identities
p - = |
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p Du = a~u) + div(pv ® u) , (4.21) |
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D<I> |
D(p<I>) |
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Dt |
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Dt |
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where v denotes the spatial velocity field.
4.An irrotational motion of an incompressible continuum is given by the s.patial velocity field v = -grad<I>. Show that for this case the scalar field <I> is harmonic.
5.By means of continuity mass equation (4..12) and identity (2.181.) show that the vorticity vector 2w is .related to the spatial mass density p and to the spatial velocity vector v by
D(2w) |
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+ (gradv)2w , |
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p -. |
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which is known as the Beltrami vorticity ·equation.
.138 |
4 Balance Principles |
4.2 Reynolds' Transport Theorem
Suppose we have a spatial scalar field <I> = <I>(.x, t) describing some physical quantity (for example, mass, internal energy, entropy, heat or entropy sources) of a particle in space per unit volume at time t:.
Assume <I> to be smooth, so that it is continuously differentiable. Hence, the present status of a continuum body in some three-dimensional region n with volume v at given time t may be characterized by the scalar-valued function
l(t) = / <l>(x, t)dv . |
(422) |
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The aim is now to compute the material time derivative of the volume integral I(t). Since the region of integration n depends on time t, integration and time differentiation do not commute. Therefore, as a first step I(t) ·must be transformed to the reference configuration. By .changing variables using the motion x = x(X, -t) and the relation dv = J(X_, t)dll we find the time rate of change of I(t) to be
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(4.23) |
<I>(x, t)d'IJ = Dt. |
<I>(x(X, t), t)J(X, t)dl' . |
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l(t) = |
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Since the region of integration is now time-independent, integration and differentiation commute. Hence, as a second step, from (4.23h we obtain, using the product rule of differentiation,
gt / <I>(x, t)dv = |
/ [<i>(x(X, t), t)J(X, t) + <I>(x(X, t), t).i(X, t)] dV , (4.24) |
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where 4~ denotes the material time derivative of the spatial scalar fie.Id <I> .according to relation (2..25). In a last step we undo the change of variables and convert the volume integral back to the current configuration. By means of eq. (2.l 75)fi, dv = J(X, t)d\/ and motion x = x(X, t), we find finally that
D ;·<I>(x, t)clv = |
;·(<i>(x(X, t), t) + <I>(x(X, t), t) ·~~~·t~]J(X,t)dF |
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}(X, t) J |
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<I>(x, t) + cl>(x, t) J(X, t) dv |
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(<i> (x, t) + cl>(x, t}divv(x, t)]clv |
(4.25) |
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