Introduction to Continuum Mechanics 3 Episode 7 potx

... 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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... 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 to a frame ... [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 Inte...

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... 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 ... a Material Volume 433 7. 4 Reynolds Transport Theorem 435 7. 5 Principle of Conservation of Mass 4 37 7. 6 Principle of Linear Momentum 440 7. 7 Moving Fram...

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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 ... then ±62 are eigenvectors with corresponding eigenvalue T22 and if 7i3 =7* 23= 0, then ± 63 are eigenvectors with corresponding eigenvalue T 33. Example 2B 17 ....

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... 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 ... 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...

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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...

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... following Solution. In 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. (4.12 .7) and (4.12.10b), ... 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 Elas...

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... dependence). The Elastic Solid 279 From the compatibility equations, Eqs. (3. 16.8), (3. 16.9) and (3. 16 .7) , we have Thus, G(XI £1) = a K\ + J3x2 + y. In the absence of body ... 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...

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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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