In summary,the function (y,2)is the solution of the problem: 7Φ=0 in domain D of the section ∂Φ an zny-ynz on the external boundaryoD with the internal continuity: Φ，=Φ along the internal 4-(2m,+m,=G a-(zn,+n,） boundaries y and the condition of uniqueness: ∫，Eoas=0 16.1.4 Energy Interpretation The strain energy of an elementary segment of a beam with thickness da is written as: dm=2(+aaw={.c+aa西 then,taking into account the displacement field in Equation 16.1: 器=”j+院+} which can be rewritten as: 器=器c9-架++- +c恩*爱川r 3 In effect,one has,for example: -架-器引叫佛 2003 by CRC Press LLCIn summary, the function F(y,z) is the solution of the problem: with the internal continuity: and the condition of uniqueness: 16.1.4 Energy Interpretation The strain energy of an elementary segment of a beam with thickness dx is written as: then, taking into account the displacement field in Equation 16.1: which can be rewritten as2 : 2 In effect, one has, for example: —2 F = 0 in domain D of the section ∂ F ∂n-------- zny ynz = – on the external boundary∂D Ó Ô Ì Ô Ï Fi = Fj Gi ∂Fi ∂n-------- zny + ynz – ( ) Ë ¯ Ê ˆ Gj ∂Fj ∂n-------- zny + ynz – ( ) Ë ¯ Ê ˆ = ˛ Ô ˝ Ô ¸ along the internal boundaries
ij EiF dS DÚ = 0 dW 1 2 -- 2 txyexy + txzexz ( )dV Ú 1 2 -- Gi g xy 2 g xz 2 ( ) + dS DÚ Ó ˛ Ì ˝ Ï ¸ = = dx dW dx -------- 1 2 -- dqx dx -------- Ë ¯ Ê ˆ 2 Gi ∂F ∂y ------- – z Ë ¯ Ê ˆ 2 ∂F ∂z ------- + y Ë ¯ Ê ˆ 2 + Ó ˛ Ì ˝ Ï ¸ dS DÚ = ∂F ∂y ------ Ë ¯ Ê ˆ 2 z ∂F ∂y – ------ ∂F ∂y ------ ∂F ∂y ------ – z Ë ¯ Ê ˆ ∂ ∂y ----- F ∂F ∂y ------ – z Ë ¯ Ê ˆ Ó ˛ Ì ˝ Ï ¸ F∂ 2 F ∂y 2 = = – -------- dW dx -------- 1 2 -- dqx dx -------- Ë ¯ Ê ˆ 2 Gi y ∂F ∂z ------- z ∂F ∂y ------- y 2 z2 – + + Ó ˛ Ì ˝ Ï ¸ S GiF—2 F dSº DÚ d – DÚÓ Ì Ï = º GiF ∂F ∂y ------- – z Ë ¯ Ê ˆ ny ∂F ∂z ------- + y Ë ¯ Ê ˆ + nz Ó ˛ Ì ˝ Ï ¸ dG ∂D Ú + ˛ ˝ ¸ TX846_Frame_C16 Page 311 Monday, November 18, 2002 12:32 PM © 2003 by CRC Press LLC