Introduction to Continuum Mechanics 3 Episode 3 ppsx

... 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 Viscous Fluid 34 8 6.1 ... Description 83 3 .3 Material Derivative 85 3. 4 Acceleration of a Particle in a Continuum 87 3. 5 Displacement Field 92 3. 6 Kinematic Equations For Rigid Body...

Ngày tải lên: 13/08/2014, 16:21

... 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 time t for the particle ... 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...

Ngày tải lên: 13/08/2014, 16:21

... 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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... 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...

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... 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. 30...

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... 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 divergence with respect to ... - 2X2 + 3* 3 = 4. 4.4. The stress distribution in a certain body is given by Find the stress vector acting on a plane which passes through the point (1/2, V3~/2 ,3)...

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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 ... 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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... 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 plane formed ... — 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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... (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) ... Cauchy-Green tensor B corresponding to this deformation field can be calculated using Eqs. (3. 30.12): [Note / 3 = a c = 1] The inverse of B can be obtained to be T...

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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 particle...

Ngày tải lên: 13/08/2014, 16:21