n Figure 15.3 Interface between Two Phases 15.1.2.3 Stresses The stress vector=E(),where E represents the stress tensor,remains con- tinuous across an element of the interface with normal as: Txyny+Txznz Txyny+Txznz ① ① ① ① Oyny+Tyznz Oyny+Tyznz (15.6 ① ① ① ① Tyzny+Ozzn:Tyzny+Ozzn: 15.1.3 Equilibrium Relations Starting from the local equilibrium,in the absence of body forces,we have ∂oy=0 xj By integrating over the cross section: +(紧+是}s=0 where the normal stress resultant Nx appears as: N-Soud5. Then,transforming the second integral to an integral over the frontier aD of D': 袋+小+r=0 Note that equalitydsis made possible due to the conti- nuity of the expression (tn+tn)across the interfaces between the different phases (see Equation 15.6). 2003 by CRC Press LLC
15.1.2.3 Stresses The stress vector , where S represents the stress tensor, remains continuous across an element of the interface with normal as: (15.6) 15.1.3 Equilibrium Relations Starting from the local equilibrium, in the absence of body forces, we have By integrating over the cross section: where the normal stress resultant Nx appears as: Then, transforming the second integral to an integral over the frontier ∂D of D4 : Figure 15.3 Interface between Two Phases 4 Note that equality is made possible due to the continuity of the expression (txy ny + txz nz) across the interfaces between the different phases (see Equation 15.6). s = S( ) n n txyny + txznz = txyny + txznz ii jj syyny + t yznz = syyny + t yznz ii jj t yzny + szznz = t yzny + szznz i i jj ∂sij ∂xj --------- = 0 d dx------ sxxdS ∂txy ∂y --------- ∂txz ∂z + --------- Ë ¯ Ê ˆ dS = 0 DÚ + DÚ Nx sxx dS . DÚ = ÚD ∂t xy ∂y ------ ∂t xz ∂z ( ) + ------ dS Ú∂D t xyny t + xznz = ( )dG dNx dx --------- txyny + txznz ( )dG ∂D Ú + = 0 TX846_Frame_C15 Page 288 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC
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 LLC
in 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
where appears the moment resultant: M.=∫p-yos Then transforming the second integral': 0+∫n%+dr+ns=0 where one notes that: ∫n-pcg%,+.,)r=jn-定.2（动ar=jnar=hmNm which can be called a density moment on the beam.Then one obtains the equilibrium relation: M+T,+4:=0 dx The case where a density moment could exist in statics is practically nil,we therefore assume that u=0.In summary,one obtains for the equations of equilibrium: s= dx +p,=0 (15.7) dx dM: dx +T,=0 15.1.4 Constitutive Relations Taking into account the isotropic nature of the different phases,the constitutive relation can be written in tensor form for Phase i as: tr( E= (I=unity tensor) E Analogous note for the continuity of the expression (across the lines of internal interfaces (Equation 15.6). 2003 by CRC Press LLC
where appears the moment resultant: Then transforming the second integral7 : where one notes that: which can be called a density moment on the beam. Then one obtains the equilibrium relation: The case where a density moment could exist in statics is practically nil, we therefore assume that mz = 0. In summary, one obtains for the equations of equilibrium: (15.7) 15.1.4 Constitutive Relations Taking into account the isotropic nature of the different phases, the constitutive relation can be written in tensor form for Phase i as: 7 Analogous note for the continuity of the expression (txy ny + txz nz) across the lines of internal interfaces (Equation 15.6). Mz –ysxxdS. DÚ = dMz dx ---------- y txyny + txznz – ( )dG txydS DÚ + ∂D Ú + = 0 y txyny + txznz – ( )dG ∂D Ú –yx S( ) n dGº –y( ) s ◊ x dG ∂D Ú ◊ = ∂D Ú = = mz( ) mN/m dMz dx ---------- + + Ty mz = 0 dNx dx --------- = 0 dTy dx -------- + py = 0 dMz dx ---------- + Ty = 0 e 1 + ni Ei -------------S ni Ei = – ----tr( ) S I (I = unity tensor) TX846_Frame_C15 Page 290 Monday, November 18, 2002 12:30 PM © 2003 by CRC Press LLC