Linear Momentum Principle Equation of Motion Momentum Principle
22 The momentum principle states that the time rate of change of the total momentum of a given set of particles equals the vector sum of all external forceps acting on the particles of the set, provided Newton's Third Law applies. The continuum form of this principle is a basic postulate of continuum mechanics.
Then we substitute ti = Tijnj and apply the divergence theorm to obtain dTi i v \ dxj
or for an arbitrary volume dTij dvi __ , dv
which is Cauchy's (first) equation of motion, or the linear momentum principle, or more simply equilibrium equation.
23 When expanded in 3D, this equation yields:
dTii 
dTi2 
dTi3  
+ 
+  
dxi 
dx2 
dx3  
dT2i 
dT22 
dT23  
dxi 
+ 
dx2 
+ 
dx3  
dT3i 
+ 
dT32 
+ 
dT33  
dxi 
dx2 
24 We note that these equations could also have been derived from the free body diagram shown in Fig. 6.2 with the assumption of equilibrium (via Newton's second law) considering an infinitesimal element of dimensions dxi x dx2 x dx3. Writing the summation of forces, will yield where p is the density, bi is the body force (including inertia). Example 61: Equilibrium Equation In the absence of body forces, does the following stress distribution x\ + v (xi— xX) 2vxix2 where v is a constant, satisfy equilibrium? Solution: j dxj dT dx2 dT2 dxdT32 dx2 dx3 dT23 dx3 dT33 dx3 Therefore, equilibrium is satisfied. 8c yy yy 8t yx 8 cx Figure 6.2: Equilibrium of Stresses, Cartesian Coordinates 6.3.2 Moment of Momentum Principle 25 The moment of momentum principle states that the time rate of change of the total moment of momentum of a given set of particles equals the vector sum of the moments of all external forces acting on the particles of the set. 26 Thus, in the absence of distributed couples (this theory of Cosserat will not be covered in this course) we postulate the same principle for a continuum as 6.3.2.1 Symmetry of the Stress Tensor 27 We observe that the preceding equation does not furnish any new differential equation of motion. If we substitute tn = Tn and the symmetry of the tensor is assumed, then the linear momentum principle (Eq. 6.24) is satisfied. 28 Alternatively, we may start by using Eq. 1.18 (ci = £ijkajbk) to express the cross product in indicial form and substitute above: /(8rmnXmtn)dS + I (8rmnXmbnp^)dV J( I (8rmnXmpvn)dV S J V dtJ V we then substitute tn = Tjnnj, and apply Gauss theorem to obtain dXmT jn j + xmpbn BXn dV 8rmn~T¡ (xmvn) pdV V dt but since df^x ^^ / dt = vm, this becomes m but £rmnvmvn = 0 since vmvn is symmetric in the indeces mn while ermn is antisymmetric, and the last term on the right cancels with the first term on the left, and finally with 5mjTjn = Tmn we are left with ^rmnTmndV 0 y or for an arbitrary volume V, at each point, and this yields

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