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 ... dsp where m and n are unit vectors having an angle of ft between them, then Eq. (3. 23. 4) gives That is Similarly Kinematics of a Continuum 1 23 (b)At the deformed...

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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. 3...

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... normal to the undeformed area. From Eqs. (4. 10.8a) (4. 10.8b) and (4. 10.9). we have We also have (see Eqs. (4. 10 .3) and (4. 10 .4) Comparing Eqs. (i) and (ii), we have Equation (4. 10.10) ... - 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...

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... a Material Volume 43 3 7 .4 Reynolds Transport Theorem 43 5 7.5 Principle of Conservation of Mass 43 7 7.6 Principle of Linear Momentum 44 0 7.7 Moving Frames 44 7 7.8 Control Volume ... 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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... 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 ... (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(e2,e 3 )=cos90 0 =0 (Q 3 i=cos(e 3 ,ei)=cos90°=0, £> 3 2=c...

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... 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 gives To evaluate ... 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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... 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 ... corresponding to this displacement field can be obtained from Eqs. (5.6 .3) , with e = 3A : 30 4 The Elastic Solid We now show that the condition given in Eq. (...

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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) ... function f. In a change of frame, (see Example 5 .32 .4, Eq. (5 .32 . 13) ), Thus, That is 34 2 The Elastic Solid 5.57. Compare the twisting torque which can be transm...

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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...

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