Introduction to Continuum Mechanics 3 Episode 5 docx

... respect to the basis at the reference configuration 3. 73. From Eqs. (3. 30.4a), obtain Eqs. (3. 30 .5) . 3. 74. Verify Eq. (3. 30.8b) and (3. 30.8d). 3. 75. Verify Eq.( 3. 30.9b) and (3. 30.9d). 3. 76. ... (3. 30.9d). 3. 76. Derive Eqs. (3. 30.10). 3. 77. Using Eqs. (3. 30.10) derive Eqs. (3. 30.12a) and (3. 30.12d). 3. 78. Verify Eqs. (3. 30. 13 a) and (3....

Ngày tải lên: 13/08/2014, 16:21

... from Eq. (3. 23. 4), we have That is That is similarly, Kinematics of a Continuum 133 thus, for this material element (d) For dX = dS^ and dX = dS 2 e 2 Example 3. 23. 3 Show that ... single-valued continuous solutions #1, #2 an ^ U 3 °f the six equation Eq. (3. 16.1) to Eq. (3. 16.6) are 118 Kinematics of a Continuum and Thus, the equation is sa...

Ngày tải lên: 13/08/2014, 16:21

... Eq. (5. 2.7) is given in Example 5. 2.1), it can be shown (see Example 5. 2.2 below) that Equations (5. 2.8) reduces the number of elastic coefficients from 36 to 21. Example 5. 2.1 (a)In ... vector on the left end face x\ ~ 0. 4.12. For any stress state T., we define the deviatoric stress S to be Stress Power 2 03 where Div denotes the divergen...

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... Incompressible Simple Fluid 52 0 8.22 Channel Flow 5 23 8. 23 Couette Flow 52 6 Problems 53 2 Appendix: Matrices 53 7 Answer to Problems 5 43 References 55 0 Index 55 2 12 Tensors Example 2B1.2 Let ... 33 1 Problems 33 5 Chapter 6 Newtonian Viscous Fluid 34 8 6.1 Fluids 34 8 6.2 Compressible and Incompressible Fluids 34 9 6 .3 Equations of Hydr...

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... Q 23 TII T^ 2 7 *33 ^ 13 0 23 033 T 31 T 32 T 33 031 032 033 or PartB Symmetric and Antisymmetric Tensors 35 This is the transformation law for the components of a vector. Thus, fy ... j2 11 =cos(e 1 ,ei)=cos30°=—, (2i2 =cos ( e i» e 2) =cosl2 0 0 = , (2i3 = cos(e 1 ,e 3 )=cos90 0 =G ^2i=cos(e2,ei)=cos60 0 =-,j322 == cos(e 2 ,ei)=cos30 0 =—,j2 23 ==c os(...

Ngày tải lên: 13/08/2014, 16:21

... the motion of a continuum can be described either by the pathlines equation Eq. (3. 1.1) or by its displacement vector field as given by Eq. (3. 5. 1). Example 3. 5. 1 The position ... Eq. (2D3.17) we have (v)Components ofdiv T Using the definition of the divergence of a tensor, Eq. (2C4 .3) , with the vector a equal to the unit base vector e r...

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... we have a unit normal vector n = (l/a)(x2*2 + X 3* $)- Therefore, the surface traction on the lateral surface Substituting from Eqs. (5. 13. 3) and (5. 13. 5) , we have Thus, in agreement ... Eq. (5 .3. 5) into Eq. (i) and since we have Or, denoting a + ft by 2^ , we have or, in direct notation where e = EM = first scalar invariant of E. In long for...

Ngày tải lên: 13/08/2014, 16:21

... Si and 52 planes automatically ensures the symmetry with respect to the 53 plane. (c) All planes that are perpendicular to the 3 plane have their normals parallel to the ... as shown in Fig. 5. 17, then i.e., In view of Eqs. (5. 19.2), Eq. (5. 19.4) can also be written as [see Prob. 5. 80] Equations (5. 19.2a) (5. 19.2b) and (5. 19.4) are...

Ngày tải lên: 13/08/2014, 16:21

... material. Fig. P5 .3 5. 52. Solve the previous problem if a\ - 3. 0 cm, a*i = 2 .5 cm, l\ - /2 = 75 cm, and M t = 700 N -m 5. 53 . For the circular shaft shown in Fig.P5 .3, determine ... (5 .33 .9) and (5 .33 .11), as those cor- responding to a change of rectangular Cartesian basis, then we come to the conclusion that the constitutive equat...

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... Eq. (6 .3. 5) , we have Newtonian Viscous Fluid 37 1 In the following sections, we restrict ourselves to the study of laminar flows only. It is therefore to be understood that ... 100,000. Newtonian Viscous Fluid 38 3 6.17 Dissipation Functions for Newtonian Fluids The rate of work done P by the stress vectors and the body forces on a material partic...

Ngày tải lên: 13/08/2014, 16:21