Introduction to Fracture Mechanics phần 2 pptx

... are: . 1 1          u u u r    , 1 ,           u r r u r u rr rr 19 . 1 2 1              r u r u u r r r      The strain compatibility equation in polar coordinates is: .0 111 12 2 2 22 2 2 2                    rrrrrrrrr rrrr rr         Modes ... ): 0  zz  , rr E       ,       rrrr E . 2...

Ngày tải lên: 08/08/2014, 17:20

... Management Process Performance Monitoring Process Performance Monitoring Effect Effect Page 15 Methodology & tools z A modeling tool is more than a drawing tool  Aris is a framework for describing ... optimization. Page 12 The Process Repository foundation Strategic Management Measures & Standards Process Best Practices Process Dictionary Process Repository Operational Data Store(...

Ngày tải lên: 07/08/2014, 02:20

... elastic plate with an edge crack of length as shown below: , a 9 Introduction to Fracture Mechanics 1 Introduction to Fracture Mechanics From Suresh: Fatigue of Materials Using the results of ... The total system energy is then given by .4 ' 22 SSP aB E Ba WWU    6 Griffith noted that the critical condition for the onset of crack growth is: , 02 ' 2 ...

Ngày tải lên: 08/08/2014, 17:20

... satisfies the biharmonic equation:   . 11 ,0 2 2 22 2 222            rrrr The boundary conditions for this plane crack problem 22 The boundary conditions for this plane crack ... be bounded. 0  ij  2    2 2 r                 2coscos 12 11  BAr               rx rr r 1 25           2sin2sin1 11 ...

Ngày tải lên: 08/08/2014, 17:20

... applied to the first 11 aK I   33 crack of length 2a 2 . A stress σ 1 is applied to the first plate, and a stress σ 2 is applied to the second plate. If we choose σ 1 , σ 1 , σ 2 and σ 2 so ... which is very important in fracture mechanics as it allows results from laboratory scale tests to be applied to large scale fracture problems. )2( )1( II KK  Si 3 N...

Ngày tải lên: 08/08/2014, 17:20

... B Substitute in to the Mises yield condition: Plane strain:     2 2 2 2 2cos 121 sin 2 3 2 ys I r K           7 2 2 r      Plane stress: 22 2 2cossin 2 3 1 2 ys I r K           Plastic ... find that:        2 sin1 2 cos 2 1    r K I     K 6 (plane strain); σ 3 = 0 (plane stress). 2 cos 2 2 3    r K I   ...

Ngày tải lên: 08/08/2014, 17:20

... strain and anti-plane strain loading:   22 22 1 1       K EE KK   17 For the plane stress case (combination Mode I and Mode II): E KK 22     ... measurement? m 20 Example Problem Size Requirements K Ic Test ASTM standard E399 (1974) for K Ic testing (specimen dimensions large compared to plastic zone): 2 5 . 2 , ,        Ic K B b a  21 5 . 2...

Ngày tải lên: 08/08/2014, 17:20

... n1 2 nn 2C 2C 3 −− ++ Đại học Kinh tế Quốc dân 20 00 1n1 2n 22 3n33 nn n n n n C2 x C2 x C2 x Cx −− − + + ++ ha c 1n1 2n2 23 n3 n1n nn n n C2 2xC2 3xC2 nx C −− −− ++ ++ n x ợc n1 1 n1 2 ... x)dx+ ∫ = n3 n2 n1 21 2 (21 )21 n3 n2 n1 +++ − −− −+ + ++ = n1 441 21 2 n3n2 n1 n2 n3n + ⎛⎞⎛ −+ + −− ⎜⎟⎜ +++ +++ ⎝⎠⎝ 1 1 ⎞ ⎟ ⎠ = 2 n1 nn2 2 2 (n 1)(n + ++ ++ = 2) (n 3)...

Ngày tải lên: 13/12/2013, 17:15

... http://www.simpopdf.com Substitute in to the Mises yield condition: Plane strain:     2 2 2 2 2cos 121 sin 2 3 2 ys I r K           7 2 2 r      Plane stress: 22 2 2cossin 2 3 1 2 ys I r K           Simpo ... singular term (λ = 1 /2 and thus B 1 =A 1 /3) we find that:  29 we find that:                  2 5 2...

Ngày tải lên: 27/06/2014, 14:20