Conservation of Momentum | Physics for EmSAT Achieve PDF Download

Conservation of momentum
There are two kinds of forces that act in a body. They are:
1. Contact forces
2. Body forces
Contact forces arise due to contact between two bodies. Body forces are action at a distance forces like gravitational force. Similarly, while contact forces act per unit area of the boundary of the body, body forces act per unit mass of the body. Both these forces result in the generation of stresses.
Before deriving these conservation laws of momentum rigorously, we obtain the same using using an approximate analysis, as we did for the transformation of curves, areas and volumes. Moreover, we obtain these for a 2D body. Consider an 2D infinitesimal element in the current configuration with dimensions ∆x along the x direction and ∆y along the y direction, as shown in figure 5.1. We assume that the center of the infinitesimal element, O is located at (x + 0.5∆x, y + 0.5∆y).
We assume that the Cauchy stress, σ varies over the infinitesimal element in the current configuration, σ =  (x, y). That is we have assumed Eulerian description for the stress. Since, we are considering only 2D state of stress, the relevant Cartesian components of the stress are σ_xx, σ_yy, σ_xy and σ_yx. Expanding the Cartesian components of the stresses using Taylor’s series up to first order we obtain:



Also seen in figure 5.1 are the Cartesian components of the Cauchy stress acting on various faces of the infinitesimal element, obtained using the equations (5.18) through (5.21).
Now, we are interested in the equilibrium of the infinitesimal element in the deformed configuration under the action of these stresses and the body


Figure 5.1: 2D Infinitesimal element showing the variation of Cartesian components of the stresses on the various faces.

force, b with Cartesian components b_x and b_y along the x and y direction respectively. Also, we shall assume that the infinitesimal element is accelerating with acceleration, a with Cartesian components ax and ay along the x and y direction respectively. Thus, force equilibrium along the x direction requires:



where ρ is the density and the element is assumed to have a unit thickness. Note that here ρ(∆x)(∆y)(1) gives the infinitesimal mass of the element and the stresses are multiplied by the respective areas over which they act to get the forces; since only the forces should satisfy the equilibrium equations. Simplifying, (5.22) we obtain:



Similarly, writing the force equilibrium equation along the y direction:



The above equation simplifies to:

Finally, we appeal to the moment equilibrium. Here we assume that there are no body couples or contact couples acting in the infinitesimal element. Since, we can take moment equilibrium about any point, for convenience, we do it about the point O, marked in figure 5.1, to obtain:



Simplifying the above equation we obtain:


which in the limit ∆x and ∆y tending to zero, the limit that we are interested in, reduces to requiring

σxy = σyx.                                                                              (5.28)

Thus, a plane Cauchy stress field should satisfy the equations (5.23), (5.25) and (5.28) for it to be admissible. Generalizing it for 3D, the Cauchy stress field should satisfy:



and

σ_xy = σ_yx, σ_xz = σ_zx, σ_yz = σ_zy.                                               (5.32)

The equations (5.29) through (5.31) are written succinctly as

ρa = div(σ) + ρb,                                                                     (5.33)

The equations (5.29) through (5.31) are written succinctly as ρa = div(σ) + ρb, (5.33) and equations (5.32) tells us that the Cauchy stress should be symmetric. Similarly, requiring a sector of cylindrical shell to be in equilibrium, under the action of spatially varying stresses, it can be shown that



and

σ_rθ = σ_θr, σ_rz = σ_zr, σ_θz = σ_zθ.                                                               (5.37)

has to hold.


Conservation of linear momentum 
Let At be an arbitrary sub-region in the configuration B_t of a body B. The total linear momentum, Γ(A_t), of the material particles occupying A_t is defined by



Next, we find the forces acting on the body. Since, the part A_t has been isolated from its surroundings, traction t_(n)(x), introduced in the last chapter, acts on the boundary of the At . In addition to traction, which arises between parts of the body that are in contact, there exist forces like gravity which act on the material particles not through contact and are called body force. The body forces are denoted by b and are defined per unit mass. Hence, the resultant force, z acting on A_t is



Now, using Newton’s second law of motion, which states that the rate of change of linear momentum of the body must equal the resultant force that acts on the body in both magnitude and direction we obtain



Using (3.30), definition of acceleration, a and (5.16) the above equation can be written as



Next, using Cauchy’s stress theorem (4.3) and the divergence theorem (2.263) the above equation further can be reduced to

  

Since, the above equation has to hold for Bt and any subset, At , of it

ρa = div(σ) + ρb.                                                                          (5.43)

Many a times, we are interested in cases for which a = o, then (5.43) becomes

div(σ) + ρb = o,                                                                             (5.44)

which is referred to as Cauchy’s equation of equilibrium. Further, there arises scenarios where the body forces can be neglected and in those cases div(σ) = o. Thus, divergence free stress fields are called self equilibrated stress fields.

Equation (5.43) is in spatial form, that is it assumes that     are known. But on most occasions, especially in solid mechanics, we know only the material form of these fields that is  and in those occasions it is difficult to obtain the spatial divergence of these fields. To make it easier at these instances, we seek a statement of balance of linear momentum in material form. For this, equation (5.42a) has to be modified. Towards this, we obtain



for which we have successively used equation (3.75), (5.14), (3.72) and definition of Piola-Kirchhoff stress (4.42) and the divergence theorem. Since, (5.45) has to hold for any arbitrary subpart of the reference configuration, we obtain the material differential form of the balance of linear momentum as

ρ_ra = Div(P) + ρ_rb.                                                                                (5.46)

It is worthwhile to emphasize that irrespective of the choice of the independent variable, the equilibrium of the forces is established only in the current configuration. In other words, whether we use material or spatial description for the stress, the forces and moments have to be equilibrated in the deformed configuration and not the reference configuration.


Conservation of angular momentum 
Let At be an arbitrary sub-region in the configuration B_t of a body B. The total angular momentum, Ω(A_t), of the material particles occupying At is defined by



where xo is the point about which the moment is taken and p is the intrinsic angular momentum per unit volume in the current configuration. Next, we find the moment due to the forces acting on the body. As discussed before there are two kind of forces, surface or contact forces and body forces act on the body which contribute to the moment. Apart from moment arising due to applied forces, there could exist couples distributed per unit surface area, m and body couples c, defined per unit mass. Thus, m =  and c =  Hence, the resultant moment, ω acting on At is



However, in non-polar bodies that are the subject matter of the study here, there exist no body couples or couples distributed per unit surface area or intrinsic angular momentum, i.e., c = m = p = o. The balance of angular momentum states that the rate of change of the angular momentum must equal the applied momentum in both direction and magnitude. Hence,

                                                                                          (5.49)

Further simplification of this balance principle involves some tensor algebra. First, we begin by calculating the left hand side of the above equation:



Since, v ∧ v = o and using equations (5.13) and (5.9), (5.50) reduces to



Next, we simplify the right hand side of the equation (5.49). Towards that, we first show that



where τ is the axial vector of (σ − σ^t ). We begin by establishing a few vector identities which enable us to establish (5.52).

Identity - 1: If u and v are arbitrary vectors then u ∧ v is the axial vector of the skew-symmetric tensor v ⊗ u − u ⊗ v.
Proof: It is straightforward to verify that v ⊗ u − u ⊗ v is skewsymmetric. Then we observe that

(v ⊗ u − u ⊗ v)(u ∧ v) = {u � (u ∧ v)}v − {v � (u ∧ v)}u = o,                         (5.53)

hence, the axial vector of v⊗u−u⊗v is a scalar multiple of u∧v. Accordingly,

(v ⊗ u − u ⊗ v)a = (a � u)v − (a � v)u = α(u ∧ v) ∧ a,                                   (5.54)

for any vector a. Now, we have to show that α = 1. Setting a = u in the above equation and then forming a scalar product on each side with v we have

(u � u)(v � v) − (u � v)² = α{(u ∧ v) ∧ u} � v                                                     (5.55)

Using (2.36) we find that

α{(u ∧ v) ∧ u} � v = α[(u � u)(v � v) − (u � v)² ].                                               (5.56)

Comparing equations (5.55) and (5.56) we find that α = 1. Identity - 2: For any vector a, b, c in the vector space

a ∧ (b ∧ c) = (b ⊗ c − c ⊗ b)a.                                                                             (5.57)

This identity follows immediately from the above identity.

Identity - 3: Let At be a regular region with boundary ∂A_t , let n be the outward unit normal to ∂A_t and let u be a vector field and σ a tensor field, each continuous in A_t and continuously differentiable in the interior of A_t . Then



Towards proving this identity, note that



where a is an arbitrary constant vector. To obtain the above we have made use of the identities



and the definition of the gradient. Further, since a is arbitrary in (5.59) we have the identity



Taking the transpose of the above identity, we obtain the required vector identity (5.58).
Now, we are in a position to prove (5.52). For this we replace b by (x−xo) and c by σn in (5.57) and then integrating each side over the surface ∂At we obtain



a being an arbitrary vector. From (5.58) it follows that



Substituting the above equations in (5.63), we obtain

Again using (5.57) and the definition of axial vector to a skew-symmetric tensor (2.98) the above equations can be reduced to a

   

and hence the required identity (5.52), since a is an arbitrary vector. Substituting (5.51) and (5.52) in (5.49) yields


Since, the body also satisfies the balance of linear momentum (5.43), the first term on the RHS is zero. Hence, the balance of angular momentum requires




Here we are interested only in non-polar bodies and hence, p = c = m = o. Therefore, (5.69) simplifies to requiring



Since, this has to hold for any arbitrary subparts of the body, B_t , τ = o. Consequently,

σ = σ^t ,                                                              (5.71)

hat is the Cauchy stress tensor has to be symmetric in non-polar bodies.