Introduction to Continuum Mechanics 3 Episode 8 pps
... respect to 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 ... — 0 is automatically satisfied together with On the other hand, since Q& = 1, we have This requirement leads to That is, Similarly, the equation Ci 233 = 0 leads...
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... (2A1.2) 3 etc. x Contents 6.17 Dissipation Functions for Newtonian Fluids 38 3 6. 18 Energy Equation for a Newtonian Fluid 38 4 6.19 Vorticity Vector 38 7 6.20 Irrotational Flow 39 0 6.21 ... Block 32 5 5 .37 Bending of a Incompressible Rectangular Bar. 32 7 5 . 38 Torsion and Tension of an Incompressible Solid Cylinder 33 1 Problems 33 5 Chapter 6 Newtonian...
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... Eq. (2C4 .3) , with the vector a equal to the unit base vector e r gives To evaluate the first term on the right-hand side, we note that so that according to Eq. (2D3. 18) , with ... lines of every particle in a continuum can be described by a vector equation of the Fig. 3. 1 form where x = x^i +*2 e 2 +JC 3 e 3 * s tne position vector at ti...
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Introduction to Continuum Mechanics 3 Episode 2 doc
... 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 ... Eq. (2B 13. 1a) reads Tn T'n T^\ \Q n Q 2l 0 31 1 |"r n T n T 13 \ \Q n Q n Qi3\ < 2B 13 - lb ) T 2l T 22 T 23 ~ Ql2 Qn 032 T 21 T 22...
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Introduction to Continuum Mechanics 3 Episode 4 docx
... 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 1 18 Kinematics of a Continuum and Thus, the equation is s...
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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. 3...
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Introduction to Continuum Mechanics 3 Episode 6 docx
... cylindrical wall is subjected to a uniform pressure p 0 . 4 38 . Verify that Eq. (4 .8. 2b) and (4 .8. 2c) are satisfied by the stress field given in Example 4.9.2. 439 . In Example 4.9.2, ... 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 dive...
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Introduction to Continuum Mechanics 3 Episode 7 potx
... 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 ... the lateral surface vanishes. The unit vector on the plane *3 = a is 63, so that the surface traction for the stress tensor of Eq. (5. 13. 1) is given by Similarly, there...
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Introduction to Continuum Mechanics 3 Episode 9 pptx
... (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 equation given by Eq. (5 .33 .6) ... Eqs. (5 . 38 . 18) and (5 . 38 .19), we can obtain Equation (5 . 38 .20) is known as "Rivlin's Universal relation". This equation gives, for small twistin...
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Introduction to Continuum Mechanics 3 Episode 10 pptx
... 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 particle of a continuum ... 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 b...
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