One deduces,in integrating over the domain occupied by the cross section of the beam: eEds=∫noas-Jn Oy+:2)dS Taking into account the form of the displacements in Equation 15.3,one can write B=装=-影6+nE+别n which leads,with the notation in Equation 15.1,to the relation: N.=张+∫，vo+a.a西 (15.8) SyeEds-Jyods+Sv.ds Taking into account the form of the displacements in Equation 15.3 one can write e品=∫E-u-m. This leads,with the notation in Equation 15.1,to the relation: M=(喂-∫no+a n2eds=jnws (15.9) Taking into account the form of the displacements in Equation 15.3,one can write ∫n2a=j(等+装}=-ncds +∫cs+cs+n which gives with the notation in Equation 15.1,the relation: 五=(c偎-+∫，c映s (15.10) 2003 by CRC Press LLCOne deduces, in integrating over the domain occupied by the cross section of the beam: Taking into account the form of the displacements in Equation 15.3, one can write which leads, with the notation in Equation 15.1, to the relation: (15.8) Taking into account the form of the displacements in Equation 15.3 one can write This leads, with the notation in Equation 15.1, to the relation: (15.9) Taking into account the form of the displacements in Equation 15.3, one can write which gives with the notation in Equation 15.1, the relation: (15.10) exx EidS sxxdS ni syy + szz ( )dS DÚ – DÚ = DÚ exx Ei dS ∂ux ∂x --------Ei dS dqz dx –-------- yEi dS du dx ------ Ei dS ∂ ∂x ------ Eihx dS DÚ + DÚ + DÚ = DÚ = DÚ Nx · Ò ES du dx ------ ni syy + szz ( )dS DÚ = + –yexxEi dS y– sxx dS ni y syy + szz ( )dS DÚ + DÚ = DÚ –yexxEi dS dqz dx -------- Ei y 2 dS du dx ------ Ei y dS ∂ ∂x ------ Ei yhx dS DÚ – DÚ – DÚ = DÚ Mz EIz · Òdqz dx -------- ni y syy + szz ( )dS DÚ = – 2exyGidS txy dS DÚ = DÚ 2exyGidS ∂ux ∂y -------- ∂uy ∂x + -------- Ë ¯ Ê ˆ GidS –qz GidSº DÚ = DÚ = DÚ º Gi ∂hx ∂y --------dS DÚ dv dx------ GidS ∂ ∂x ------ hyGidS DÚ + DÚ + + Ty · Ò GS dv dx------ – qz Ë ¯ Ê ˆ Gi ∂hx ∂y --------dS DÚ = + TX846_Frame_C15 Page 291 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC