1.7 Stress Tensor - MIT OpenCourseWare

1 - Marine Hydrodynamics, Spring 2005. Lecture 3. Marine Hydrodynamics Lecture 3. Stress Tensor Stress Tensor ij The Stress (force per unit area) at a point in a uid needs nine components to be completely speci ed, since each component of the Stress must be de ned not only by the direction in which it acts but also the orientation of the surface upon which it is acting. The rst index i speci es the direction in which the Stress component acts, and the second index j identi es the orientation of the surface upon which it is acting. Therefore, the ith component of the force acting on a surface whose outward normal points in the j th direction is ij . X2. 22. 12. 32. 21. 23. 11. 13 31. 33 X1. X3. Figure 1: Shear stresses on an in nitesimal cube whose surfaces are parallel to the coordinate system. 1. X2. 2. P n . A1. 1. A3. X1. Q. 3. R. area A0. A2. X3. Figure 2: In nitesimal body with surface PQR that is not perpendicular to any of the Cartesian axis.

2 Consider an in nitesimal body at rest with a surface PQR that is not perpendicular to any of the Cartesian axis. The unit normal vector to the surface PQR is n = n1 x 1 +n2 x 2 +n3 x 3 . The area of the surface = A0 , and the area of each surface perpendicular to Xi is Ai = A0 ni , for i = 1, 2, 3.. Newton's law: Fi = (volume force)i for i = 1, 2, 3. on all 4 faces Note: If is the typical dimension of the body : surface forces 2. : volume forces 3. An example of surface forces is the shear force and an example of volumetric forces is the gravity force. At equilibrium, the surface forces and volumetric forces are in balance. As the body gets smaller, the mass of the body goes to zero, which makes the volumetric equal to zero and leaving the sum of the surface forces equal zero. So, as . forces 0, all4f aces Fi = 0 for i = 1, 2, 3 and i A0 = i1 A1 + i2 A2 + i3 A3 = ij Aj . But the area surface to Xi is Ai = A0 ni.

3 Therefore i A0 = ij Aj = ij (A0 nj ), where ij Aj is of each the notation (represents the sum of all components). Thus i = ij nj for i = 1, 2, 3, where i is the component of Stress in the ith direction on a surface with a normal n . We call i the Stress vector and we call ij the Stress matrix or Tensor . 2. Example: Pascal's Law for Hydrostatics In a static uid, the Stress vector cannot be di erent for di erent directions of the surface normal since there is no preferred direction in the uid. Therefore, at any point in the uid, the Stress vector must have the same direction as the normal vector n and the same magnitude for all directions of n . no summation . Pascal's Law for hydrostatics: ij = (pi ) ( ij ).. p1 0 0. = 0 p2 0 . 0 0 p3. where pi is the pressure acting perpendicular to the ith surface. If p0 is the pressure acting perpendicular to the surface PQR, then i = ni p0 , but: i = ij nj = (pi ) ij nj = (pi )(ni ).

4 Therefore po = pi , i = 1, 2, 3 and n is arbitrary. 3. Symmetry of the Stress Tensor To prove the symmetry of the Stress Tensor we follow the steps: j ij ji ji o i ij Figure 3: Material element under tangential Stress .. 1. The of surface forces = body forces + mass acceleration. Assume no symmetry. Balance of the forces in the ith direction gives: ( )( ij )T OP ( )( ij )BOT T OM = O( 2 ), since surface forces are 2 , where the O( 2 ) terms include the body forces per unit depth. Then, as 0, ( ij )T OP = ( ij )BOT T OM .. 2. The of surface torque = body moment + angular acceleration. Assume no sym . metry. Balance of moments about o gives: ( ji ) ( ij ) = O( 3 ), since the body moment is proportional to 3 . As 0 , ij = ji . 4. Mass and Momentum Conservation Consider a material volume m and recall that a material volume is a xed mass of mate . rial. A material volume always encloses the same uid particles despite a change in size, position, volume or surface area over time.

5 Mass Conservation The mass inside the material volume is: M ( m ) = d . m(t). Sm(t). m ( t ). Figure 4: Material volume m (t) with surface Sm (t). Therefore the time rate of increase of mass inside the material volume is: d d M ( m ) = d = 0, dt dt m (t). which is the integral form of mass conservation for the material volume m . 5. Momentum Conservation The uid velocity inside the material volume in the ith direction is denoted as ui . Linear momentum of the material volume in the ith direction is ui d . m(t). Newton's law of motion: The time rate of change of momentum of the uid in the material control volume must equal the sum of all the forces acting on the uid in that volume. Thus: d (momentum)i =(body force)i + (surface force)i dt d ui d = Fi d + ij nj dS. dt . m (t) m (t) Sm (t) i Divergence Theorems For vectors: v d = . v .n dS.. vj S vj nj xj ij For tensors: d = ij nj dS. xj S.

6 Using the divergence theorems we obtain . d ij ui d = Fi + d . dt xj m(t) m(t). which is the integral form of momentum conservation for the material volume m . 6. Kinematic Transport Theorems Consider a ow through some moving control volume (t) during a small time interval t. Let f ( x, t) be any (Eulerian) uid property per unit volume of uid ( mass, momentum, etc.). Consider the integral I(t): I(t) = f ( x, t) d . (t). According to the de nition of the derivative, we can write d I(t + t) I(t). I(t) = lim dt t 0. t .. 1 . = lim f ( x, t + t)d f ( x, t)d . t 0 t .. (t+ t) (t). S(t+ t). ( t + t ). ( t ) S(t). Figure 5: Control volume and its bounding surface S at instants t and t + t. 7. Next, we consider the steps 1. Taylor series expansion of f ( x, t + t) about ( x, t). f f ( x, t + t) = f ( x, t) + t ( x, t) + O(( t)2 ). t 2. d = d + d . (t+ t) (t) . where, d = [Un ( x, t) t] dS and Un ( x, t) is the normal velocity of S(t).

7 S(t). S(t+ t). S(t). v U n ( x, t ) t + O( t ) 2 dS. Figure 6: Element of the surface S at instants t and t + t. Putting everything together: .. d 1 f I(t) = lim d f + t d + t dSUn f d f + O( t)2 (1). dt t 0 . t t .. (t) (t) S(t) (t). 8. From Equation (1) we obtain the Kinematic Transport Theorem (KTT), which is equivalent to Leibnitz rule in 3D. d f ( x, t). f ( x, t)d = d + f ( x, t)Un ( x, t)dS. dt t (t) (t) S(t). For the special case that the control volume is a material volume it is (t) = m (t) and Un = v n , where v is the uid particle velocity. The Kinematic Transport Theorem (KTT), then takes the form d f ( x, t). f ( x, t)d = d + f ( x, t)( v n )dS. dt t . m (t) m (t) Sm (t) f (vi ni ). (Einstein Notation). Using the divergence theorem, . d = . n dS.. S i ni xi i we obtain the 1st Kinematic Transport Theorem (KTT).. d f ( x, t). f ( x, t) d = + (f v ) d , dt t . m (t) m (t).

8 (f vi ). xi where f is some uid property per unit volume. 9. Continuity Equation for Incompressible Flow Di erential form of conservation of mass for all uids Let the uid property per unit volume that appears in the 1st KTT be mass per unit volume ( f = ): . d . 0 = d = + ( v ) d . dt t conservation m (t) 1st KTT m (t). of mass But since m is arbitrary the integrand must be 0 everywhere. Therefore: . + ( v ) = 0. t . + [ v + v ] = 0. t . D . Dt Leading to the di erential form of D . Conservation of Mass: + v = 0. Dt 10. Continuity equation Conservation of mass for incompressible ow In general it is = (p, T, ..), but we consider the special case of an incompressible ow, D . Dt = 0 (Lecture 2). Note: For a ow to be incompressible, the density of the entire ow need not be constant ( ( x, t) = const). As an example consider a ow of more than one incom . pressible uids, like water and oil, as illustrated in the picture below.

9 Constant . fluid particle 2. fluid particle oil water 1. Figure 7: Interface of two uids (oil-water). Since for incompressible ows D . Dt = 0, substituting into the di erential form of the conservation of mass we obtain the vi Continuity Equation: v = 0. xi . rate of volume dilatation 11. Euler's Equation (Di erential Form of Conservation of Momentum). 2nd Kinematic Transport Theorem 1st KTT + di erential form of conservation of mass for all uids. If G = uid property per unit mass, then G = uid property per unit volume . d . Gd = ( G) + ( G v ) d . dt t m (t) 1st KT T m (t).. G . = G + v + + v G . t t d .. m (t). =0 from mass conservation = DG. Dt The 2nd Kinematic Transport Theorem (KTT) follows: d DG. Gd = d . dt Dt m m Note: The 2nd KTT is obtained from the 1st KTT (mathematical identity) and the only assumption used is that mass is conserved. 12. Euler's Equation We consider G as the ith momentum per unit mass (vi ).

10 Then, . ij d Dvi Fi + d = vi d = d . xj dt Dt m (t) conservation m (t) 2nd KTT m (t). of momentum But m (t) is an arbitrary material volume, therefore the integral identity gives Euler's equation: . Dvi vi ij . + v vi . = Fi +. Dt t xj v vj xi j And in vector Tensor form: . D v v + v v = F + . Dt t 13.