Introduction to Continuum Mechanics 3 Episode 2 doc

... < 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 =-...

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 ... six strain components (^11 ^22 ^33 ^ 12^ 13^ 23 ) are arbitrarily prescribed in some region, in general, there may not exist three displacement functions (#i, 2 M 3)...

Ngày tải lên: 13/08/2014, 16:21

... (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. 30.13d). Kinematics of a Continuum 151 where ... 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. Ve...

Ngày tải lên: 13/08/2014, 16:21

... direction of ej + 2e 2 + 3e 3 and 63& apos;is in the direction of e 1 4-e 2 ~e 3 . 4.7. Consider the following stress distribution Linear Elastic Solid 2 23 where -r~- are infinitesimal. ... Sect. 3. 12, we obtained [see Eq. (3. 12. 4)], Since dx - F dX [ see Eq. (3. 7 .2) ], therefore Equation (i) is to be true for all dX, thus or, Using Eqs...

Ngày tải lên: 13/08/2014, 16:21

... transform the vector a = ej +26 2 +36 3? Solution. Using Eq. (2B3.1b) b i\ [ 23 -2] fll [2& quot; b 2 = -6 4 12 = 5 b 3 [ 4 -1 2J [3J [8 or b = 2e 1 +5e 2 +8e 3 2B4 Sum of Tensors Let ... 2B3.1 Given that a tensor T which transforms the base vectors as follows: Tej = 2e 1 -6e 2 +4e 3 T 02 = 3ej+4 62- 63 Te 3 = -26 J+ 62+ 2 63 How does this...

Ngày tải lên: 13/08/2014, 16:21

... that according to Eq. (2D3.18), with v r =T m v e =T^, 7^= To evaluate the second term on the right-hand side of Eq. (2D3 .20 ) we first use Eq. (2D3.17) with v=e r to obtain Geometrical ... elongation is 22 which is also zero. The decrease in angle between these elements is given by 2 ^2, which is equal to 2k, i.e., 2x 10~ 4 radians, (c)...

Ngày tải lên: 13/08/2014, 16:21

... "*" cos a 2 e 2& apos; x ' e « 2 ~ x i sma: 2 + X 2 cos 2 e / 2 = ±(cos 2 ei - sin 2 e 2) ( m< ) Fig. 5.4 (c) 25 2 The Elastic Solid Example 5.11.1 (a) Find ... 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),...

Ngày tải lên: 13/08/2014, 16:21

... 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 23 3 = 0 leads...

Ngày tải lên: 13/08/2014, 16:21

... kX 2 X 3 , u 2 = kX 1 X 3 , u 3 = kX 1 X 2t k = 10~ 4 5.16. Repeat Problem 5.14, except that the displacement components are: «i = -kX 3 X 2 , u 2 = kXiX-$, u 3 = ksinX 2 , ... 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...

Ngày tải lên: 13/08/2014, 16:21

... Navier-Stokes equations, Eqs. (6.7 .2) yield: Equations (6. 12. 1b) and (6. 12. 1c) state that p does not depend on* 2 and * 3 . If we differentiate Eq. (6. 12. la) with respect to ... vector on any plane at any point is normal to the plane. That is for any n, Newtonian Viscous Fluid 37 9 In terms of stress tensors, we have T- ' 62 = "...

Ngày tải lên: 13/08/2014, 16:21