in which ny and n,are the cosines of the outward normal n,and dI represents element of frontier OD.If one assumes the absence of shear stresses applied over the lateral surface of the beam,then t n+tn=0 along the external frontier OD.Then for longitudinal equilibrium we have dNs= dx 引s+(+先s=0 where one recognizes the shear stress resultant: =n Then transforming the second integral into an integral over the external frontier OD of the domain D of the cross section°： 2张+no+=r=0 if one remarks that: ∫non4,+5.ar=jn-z动ar=jnar=p,Wm aD which is the transverse density of loading on the lateral surface of the beam, transverse equilibrium can be written as: +p,=0 dx 引。o.45+∫n-费+=0 5 We have neglected the body forces which appear in the local equation of equilibrium in the form of a function If these exist (inertia forces,centrifugal forces,or vibration inertia,for example),one obtains for the equilibrium:o in which Pds represents the longitudinal load density. 6 Note that the equalityds(dr is made possible due to the continuity of the expression n+-n.across the frontier lines between different phases (see Equation 15.6). 2003 by CRC Press LLCin which ny and nz are the cosines of the outward normal , and d G represents element of frontier ∂D. If one assumes the absence of shear stresses applied over the lateral surface of the beam, then txy ny + txz nz = 0 along the external frontier ∂D. Then for longitudinal equilibrium we have5 where one recognizes the shear stress resultant: Then transforming the second integral into an integral over the external frontier ∂D of the domain D of the cross section6 : if one remarks that: which is the transverse density of loading on the lateral surface of the beam, transverse equilibrium can be written as: 5 We have neglected the body forces which appear in the local equation of equilibrium in the form of a function fx. If these exist (inertia forces, centrifugal forces, or vibration inertia, for example), one obtains for the equilibrium: in which represents the longitudinal load density. 6 Note that the equality is made possible due to the continuity of the expression (syy ny + tyz nz ) across the frontier lines between different phases (see Equation 15.6). n dNx dx ------ + px = 0 px ÚD f = xdS dNx dx --------- = 0 d dx------ txydS ∂syy ∂y ---------- ∂t yz ∂z + --------- Ë ¯ Ê ˆ dS = 0 DÚ + DÚ Ty txydS. DÚ = ÚD ∂syy ∂y ------- ∂t yz ∂z ( ) + ------ dS Ú∂D syyny t + yznz = ( )dG ∂Ty ∂x -------- syyny + t yznz ( )d G ∂D Ú + = 0 syyny + t yznz ( )dG ∂D Ú y S( ) n dG y sdG ∂D Ú ◊ = ◊ ∂D Ú = = py ( ) N/m dTy dx -------- + py = 0 d dx------ –ysxxdS y ∂txy ∂y --------- ∂txz ∂z + --------- Ë ¯ Ê ˆ – dS = 0 DÚ + DÚ TX846_Frame_C15 Page 289 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC