Derivation of angular momentum balance law using the cauchy stress tensor measure. (HW#4, MAE 295. UCI)

Transcription

1 Derivation of angular momentum balance law using the cauchy stress tensor measure. (HW#4, MAE 295. UCI) by Nasser Abbasi February 28, 2006 Problem Derive the angular momentum balance (AMB) equation for a soli boy using the Cauchy stress tensor measure. 1 olution Consier a boy in the eforme con guration uner the action of surface traction forces represente by a vector t (force per unit area), an assume there exist boy forces acting on the boy an represente by the vector b (force per unit mass). Let be the eforme volume. Let be the surrouning surface enclosing the boy. Let p be the angular momentum of a i erential mass element m with an associate i erential volume element. The angular momentum of the element m is measure with reference to rotation aroun the origin O of the reference frame. Let v be the current velocity vector of m: We start by formulating the AMB law. The formulation is base on the rule that any balance equation of a physical quantity can generally be state as follows: 1 References: 1. Professor Atluri N class notes. MAE 295, soli mechanics. UCI. 2. Brown university EN175 Avance Mechanics of olis web site class notes 1

2 The time rate of change of the physical quantity associate with a mass m of volume an surface actions of the surrouning on the surface + actions of the surrouning on the mass particles that make up the boy itself. In this problem, the physical quantity whose balance law we wish to obtain is the angular momentum of the eforme boy. The rate at which the angular momentum changes is the left han sie of the balance equation. The right han sie are the forces which ten to change the angular momentum, an hence these will be the torques generate by forces acting on the surface in aition to torques generate by forces acting on the boy itself. The forces acting on the surface are the traction forces. The forces acting on the boy itself are calle boy forces such as gravitational forces an electromagnetic forces. The following iagram illustrates the eforme boy uner the actions of the external forces t an b. The angular momentum balance law can now be state as follows: Rate of change of angular momentum sum of all the torques prouce by the surface traction forces + sum of all the torques prouce by the boy forces. In mathematical terms this can be written as t r t a + m r b m Where is the angular momentum of the whole eforme boy. Now, assuming the boy is mae of homogeneous material, its ensity is the same for each m; hence we write m 2

3 Where is the i erential volume associate with the i erential mass m: The balance equation becomes t r t a + r b (0) We start the solution by eriving an expression for the left han sie of the equation (0). Recall that the angular momentum of the boy aroun O is e ne as the moment of the linear momentum of the boy aroun O. Hence, assuming that p is the vector that represents the angular momentum of a i erential element m with a position vector r, an that L is the vector that represents the linear moment of the same element we write p r L (1) Now, by Newton s secon law, the linear momentum L of a mass m is L m v (2) ubstitute (2) into (1) we obtain an expression for the p p r L r (m v) Hence the rate of change of the angular momentum of an element m is t p [r (m v)] t Now assuming the mass of the boy is conserve, hence m 0 an the above can be simpli e2 t to t p m (r v) t The above expression is the rate of change of the angular momentum of each m. To obtain the rate of change of the angular momentum of the whole boy which is ; we sum over m t 2 t m v m t v + v t m m t v since t m 0 t X t p i i X p i t i X m i (r i v i ) (3) t i 3

4 An since we are assuming the boy is mae of homogeneous material so its ensity is the same for each m then m i i an (3) becomes t t X i (r i v i ) i In the limit as is mae smaller an smaller, the above sum becomes an integral an we write R (r v) t t Which is the left han sie of the AMB law. ubstitute the above into equation (0) an rearrange the integrals, we obtain 0 (r t (r v) t r (r v) t 1 r t a + r b A r t a b r b r t a (5) To procee further, an to be able to convert the integral over the surface to an integral over a volume using Gauss ivergence theorem, we convert all the expressions from vector notation to component notation as follows: r v r j v k where is the permutation elta operator. imilarly r t r j t k an r b r j b k o now the AMB law as given in equation (5) can be written in component form as 4

5 rv r j v k t rb r j b k 1 A rt r j t k a t (r j v k ) r j b k r j t k a (6) But t (r r j j v k ) v k t r j v k t v k + r j t + r j a k where a k is the acceleration vector a of the i erential element m written in component notation. Now r j is the same as v t j which is the j component of the velocity vector v of the element m. Hence the above equation becomes t (r j v k ) 0 v k v j z} { r j t 1 + r j a k C A (v k v j + r j a k ) 0 v k v j + r j a k But the rst term above in the RH is zero, since v k v j represents the cross prouct of a parallel vector. Hence the equation above simpli es to t r j v k r j a k ubstitute the above result for the rst term in the LH of equation (6) an take r j as a common factor, we obtain ( r j a k r j b k ) r j (a k b k ) r j t k a r j t k a (7) Now express the traction vector t in terms of the Cauchy stress an a unit normal to the area a 5

6 t a n a :~ t a n :~ a o t n :~ or in component form t k mk n m substitute the above into equation (7) we obtain r j (a k b k ) r j t k mk n m a (8) Now apply the ivergence theorem on the RH of the above equation. ( r j mk ) n m ( r j mk ) (9) ubstitute (9) into the RH of (8) r j (a k b k ( r j mk (r j mk ) r j r j r j + m + mk jm + mk jm Now move the rst term in the RH of the above equation to the left sie an factor out the common r j term, we obtain 6

7 r j (a k b k ) r j r j (a k b k ) r j r j a k b k mk jm mk jm mk jm (10) Now recall that the strong form of the linear momentum balance equation: a k b k + Hence the LH of equation 10 is zero. Hence (10) becomes 8 9 Linear momentum Balance > mk r j a k b >: m >; 0 mk jm mk jm Hence the strong form of the above integral is that each component in the RH is itself zero. In other wors: Hence the above can be written as jm mk 0 jk 0 imn jk 0 (11) But Hence (11) becomes jm kn mk nj imn 7

8 ( jm kn mk nj ) jk 0 jm kn jk mk nj jk 0 mn nm 0 Hence the Cauchy stress tensor is symmetrical, i.e. ~ ~ T Hence the AMB law implies that the Cauchy stress tensor is symmetrical. 8