tailieunhanh - Introduction to Continuum Mechanics 3 Episode 3

Tham khảo tài liệu 'introduction to continuum mechanics 3 episode 3', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 66 Curvilinear Coordinates Vv tor dr tof dr dr lpVf ì 1 pvr r w ỹ- sinớj in . 1 Itoo . _ J ỊdVự 1 ỞVự vr VflCQtfl r dỡ rsinớ dự r r iii div V Using the components of Vv obtained in ii we have . tor ldvfl 1 dv vr VQCOtỡ diw tr Vv dr r dỡ rsinớ d p r r _ 1 d r2 1 d v0SÌnứ 1 dr 2 dr rsinớ d0 rsinớ ty curlv iv curl V From the definition of the curl and Eq. we have 1 a v sin0 1 dvệ 1 dvr ld v l 1 d zvg 1 dvr dỡ rsinớ dtp erT rsinỡdíp r dr r dr r dỡ eự v Components of div T Using the definition of the divergence of a tensor Eq. with the vector a equal to the unit base vector er gives divT r div Trer -tr Ver Tr To evaluate the first term on the right-hand side we note that Trer T r T o T so that according to Eq. with vr 7jr vq T T p T p 1 . 1 sing a 7 div T er ỹ inớ 2D3-21 To evaluate the second term on the right-hand side of Eq. we first use Eq. with v er to obtain PartD Spherical Coordinates 67 0 0 0 0 0 0 VeA 0 7 0 r and VeATr T e Tẹẹ 7- r r r 0 0 7 Trf Tỹ r r r r so that tr VefTr Z From Eq. we obtain divT A l 2 r 1 5 7 sin6 1 afy Tẹẹ T r2 dr rsinớ rsinô dộ r In a similar manner we can obtain see Prob. 2D9 1 a 7 sing 1 cot0 divT ớ đivT ự 1 5 r3Tự r 1 d 7 flSÌnờ 1 dTýý T - 7 r TtyCPỈỠ J dr fsinớ aớ rsinớ 50 r 68 Problems PROBLEMS 2A1. Given 1 0 2 1 0 1 2 and a 2 3 0 3 3 evaluate a sib b SijSlp c SjkSkj d amam e smnaman 2A2. Determine which of these equations have an identical meaning with ũtị Qijdj a ap Qptna m b ap Qqpilq c am anQmn- 2A3. Given the following matrices k 1 2 3 0 0 3 1 0 Bij 0 5 1 Cy 1 0 2 2 0 2 1 2 4 3 Demonstrate the equivalence of the following subscripted equations and the corresponding matrix equations. a Djj Bịj ữ B r b 6 BW b B a c cy Byjứj c B a i s Bipfli s laflBiW e Dik BijCjk B C ĩ Dịk BịjCkj D B C 7 . 2A4. Given that Tịj Eỹ Ấựĩ show that a b 2A5. Given W E iE E 2 p TiịTiị 2EijEij Ekkfy 3i2 Ì 0 0 1 2 k 2 k 2 k 1 2 3 0 3 4 0