ICF13A

13th International Conference on Fracture June 16–21, 2013, Beijing, China -2- Here, A denotes the area of the cross section, is the shear coefficient, 0C and 1C are the surface elastic modulus and the Steigman-Ogden constant, respectively, S denotes the circumference of the cross section. The potential energy of the lateral and axial load is given by: 2 0 0 1 ( ) ( ) ( ) ( ) 2 L L f w U w f x w x dx N dx x       , (4) where         s f f x q x q x q x    . (5) Here, ( ) sq x stands for the equivalent lateral force due to the residual surface stress and ( ) f q x represents the distributed force arising from the substrate medium. To minimize the total potential energy, we apply the variational theory, ( ) ( ) 0 Total f U w U w       . (6) Finally, we get: 2 2 2 2 0 1 ( ) ( ) 0 ( ) (( * ) ) 0 w w GA f x N x x x w GA EI C I CS x x x                                 , (7) where 2 A I y dA  , 2 * S I y dS  , 2 * y S S n dS   . Take partial derivative of the second equation with respect tox,and substitute the first equation into the second equation, we have: 4 2 4 2 ( )* ( )* ( )* ( )* ( ) ( ) 0 p w p w EI EI w EI w EI H N C N C H C C w x GA GA x GA x                             . (8) The general solution of Eq. (8) can be written as: () mx w x Ae  . (9) Substituting Eq. (9) into Eq. (8), we can solvem. The boundary conditions of a simply supported NW are given as follows: (0) 0 w  , '' (0) 0 w  , (10a) ( ) 0 w L  , '' ( ) 0 w L  . (10b) Substitute Eqs.(10a,b) into Eq. (9), we can get the characteristic equation to solve the critical buckling force. The corresponding buckling mode is given by Eq. (9). 3. Results and Discussions

RkJQdWJsaXNoZXIy MjM0NDE=