Introduction to Continuum Mechanics 3 Episode 9 pptx

... (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 ... Spectra 4 73 8.4 Generalized Linear Maxwell Fluid with a Continuous Relaxation Spectrum 474 Part A Permutation Symbol 7 d 11 = d 22 = d 33 = 1 d 12 =d 13...

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... < 2B 13 - lb ) T 2l T 22 T 23 ~ Ql2 Qn 0 32 T 21 T 22 T 23 Qzi Ql2 Q 23 TII T^ 2 7 *33 ^ 13 0 23 033 T 31 T 32 T 33 031 0 32 033 or PartB Symmetric and Antisymmetric Tensors 35 This is the ... have //jr 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 =...

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... 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...

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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 ... 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 deform...

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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....

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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 ... normal stress acting. 4 .6. For the following state of stress findTn' and T 1 ^' where e x ' is in the direction of ej + 2e 2 + 3e 3 and 63 & apos;is i...

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... Solid 2 27 Example 5 .3. 1 Find the components of stress at a point if the strain matrix is and the material is steel with A = 119.2 GPa ( 17 .3 xl() b psi) and p = 79 .2 GPa (11.5xl0 6 psi). Solution. ... 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)...

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... 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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... (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) ... because both Eqs. (5 .33 .9) and (5 .33 .11) have the same function f. We note that the special case where a is a constant, is called a Hookean Solid. 5 ,34 Consti...

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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 ... upper limit can be as high as 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...

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... which the gate will start to fall. 63. The liquids in the U-tube shown in Fig.P6 .3 is in equilibrium. Find h^ as a function of Pi'P2»P3» hi an d ^3- The liquids are immiscible. 6.4. ... Problems 4 23 630 . Referring to Problem 6.29, consider a pipe having an elliptic cross section given by y 2 /a 2 + z 2 / b 2 = 1. Assuming that find A and B...

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... [xxdivT], we have OJtn Example 7.2 .3 Referring to Example 7.2.2, show that the total power (rate of work done) by the stress vector on S is given by, 436 Integral Formulation of ... Eq. (7.7.8) to the surface and body force terms. 7.8 Control Volume Fixed with respect to a Moving Frame If a control volume is chosen to be fixed with respect t...

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... Section 3. 13, Eq. (3. 13. 6a), that 1 T where D = :r[Vv + (Vv) ] is the rate of deformation tensor. Thus, £ Next, from Eq. (8.10.4), Non-Newtonian Fluids 489 Example 8.9 .3 Consider ... are similar to those of the deformation tensors using a fixed reference time. [See Chapter 3, Section 3. 18 to 3. 29]. Indeed by polar decomposition theorem (Se...

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... example into Eq. (8.19.4), we obtain 544 Answers 2B31. Eigenvector of T isn, Eigenvector of T is r^ (c) No, the first invariants are not equal Non-Newtonian Fluids 5 13 Since ... be shown to be objective stress rates. [See Prob. 23] These are called the Oldroyd upper convected derivatives. and note that one can derive Non-Newtonian Fluids 531 Thus, fr...

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... material, 32 2 Creep function, 466 volume change, 146 Current configuration as reference con- Finite deformation tensor, 121,128, 134 , figuration, 476 136 , 138 ,141 ,151 ,1 53, 155 -156 ,206 ,31 8 -32 1 in ... 508 Dissipation functions, 38 3 First Piola Kirchhoff stress tensor, 202 Divergence theorem, 430 Flow Dual vector, 36 ,94 channel flow, 37 2,5 23 Dummy...

Ngày tải lên: 13/08/2014, 16:21