328 BEAMS WITH SHEAR DEFORMATION oo=b br2=b Figure 7.8:Sandwich beam and the equivalent I-beam. The presence of the core has little effect (i)on the bending stiffnesses in the x-y plane,(ii)on the tensile stiffness,(iii)on the warping stiffness,and (iv)on every element but the S element of the shear compliance matrix.We utilize this observation and determine these properties by neglecting the core. When the core is neglected,the sandwich beam behaves like an l-beam with a thin web (Fig.7.8).We now use the expressions for an I-beam in Tables A.2,A.5, and A.8 and substitute b for both b and bp,the superscripts t and b for subscripts fl and f2,and set the terms referring to the web to zero.The resulting bending, tensile,and warping stiffnesses are 凤=分+1公 EA= 1 b+- 1 b31 亚+点五 ed, (7.69) where and are evaluated in the coordinate systems whose origins are at the "neutral"reference planes (which is at 2,Eq.6.105)in the top and bottom flanges,respectively.When the core thickness is so low as to prevent distortion of the cross section,the warping stiffness is lower than the value given in Eq.(7.69). A conservative estimate of the warping stiffness is its lower limit,which is El=0. The locations of the centroid ze and the shear center e are (Fig.7.9) 1 =a(d) e=d- 、+=之· (7.70) "Neutral"plane d C Neutral”plane Figure 7.9:Location of the shear center and the centroid of a sandwich beam
328 BEAMS WITH SHEAR DEFORMATION x b b f2= x b Vy Tω N Mz Mz Vy Tω N b b f1= Figure 7.8: Sandwich beam and the equivalent I-beam. The presence of the core has little effect (i) on the bending stiffnesses in the x–y plane, (ii) on the tensile stiffness, (iii) on the warping stiffness, and (iv) on every element but the szz element of the shear compliance matrix. We utilize this observation and determine these properties by neglecting the core. When the core is neglected, the sandwich beam behaves like an I-beam with a thin web (Fig. 7.8). We now use the expressions for an I-beam in Tables A.2, A.5, and A.8 and substitute b for both bf1 and bf2, the superscripts t and b for subscripts f1 and f2, and set the terms referring to the web to zero. The resulting bending, tensile, and warping stiffnesses are EI zz = 1 αt 11 b3 12 + 1 αb 11 b3 12 ?EA= 1 αt 11 b + 1 αb 11 b EI ω = b3 1 αt 11 12 ed, (7.69) where αt 11 and αb 11 are evaluated in the coordinate systems whose origins are at the “neutral” reference planes (which is at , Eq. 6.105) in the top and bottom flanges, respectively. When the core thickness is so low as to prevent distortion of the cross section, the warping stiffness is lower than the value given in Eq. (7.69). A conservative estimate of the warping stiffness is its lower limit, which is EI ω = 0. The locations of the centroid zc and the shear center e are (Fig. 7.9) zc = 1 EA 1 αt 11 bd e = d 1 αt 11 1 αt 11 + 1 αb 11 = zc. (7.70) zc d “ ” Neutral plane C e SC “ ” Neutral plane Figure 7.9: Location of the shear center and the centroid of a sandwich beam.
7.3 TRANSVERSELY LOADED BEAMS 329 The shear compliances are 1.2 $w= ( (7.71) b (1+) (1+)月 =品(e2'+w） 1.2 (7.72) 如 1.2/ a) (7.73) 1+81+ s2=0Sw=0, (7.74) where is given by Eq.(6.196)and sc=(d-e)/d.The elements of the shear stiffness matrix are determined by substituting the elements of the shear compli- ance matrix into Eq.(7.38). Next we present upper and lower bounds of the torsional stiffness of a sandwich beam.The upper bound is the torsional stiffness of a solid(not sandwich)beam with the same cross section as the sandwich beam(Eq.6.169) 4b GI upper bound. (7.75) 866 The lower bound of the torsional stiffness corresponds to the torsional stiffness of a sandwich beam in which the shear stiffness of the core in the y-z plane is neglected.In this case we need to consider only the two facesheets.The torsional stiffnesses of the facesheets are(Eq.6.169)4b/and b/where superscriptst and brefer to the top and bottom facesheets.The torsional stiffness of the sandwich beam with two facesheets is GI= 4b4b +。 lower bound. (7.76) 7.3 Transversely Loaded Beams In this section we present the deflections of orthotropic beams,with cross sections symmetrical with respect to the x-z plane.A transverse load pa(per unit length) acts in the plane of symmetry(Fig.6.1).The relevant equilibrium equations are (Eq.7.39) av dx +P:=0 (7.77) da-=0. (7.78) dx where V.is the shear force,and M,is the bending moment in the x-z plane. For beams with symmetrical cross sections,Ely=0,Sy:=0,andS=0,and
7.3 TRANSVERSELY LOADED BEAMS 329 The shear compliances are syy = 1.2 b αν 66t (1 + δsc) 2 + αν 66b 1 + 1 δsc 2 (7.71) sωω = 1.2 d2b αν 66t + αν 66b (7.72) syω = 1.2 db - − αν 66t 1 + δsc + αν 66b 1 + 1 δsc . (7.73) syz = 0 szω = 0, (7.74) where αν 66 is given by Eq. (6.196) and δsc = (d − e) /d. The elements of the shear stiffness matrix are determined by substituting the elements of the shear compliance matrix into Eq. (7.38). Next we present upper and lower bounds of the torsional stiffness of a sandwich beam. The upper bound is the torsional stiffness of a solid (not sandwich) beam with the same cross section as the sandwich beam (Eq. 6.169) GI t = 4b δ66 upper bound. (7.75) The lower bound of the torsional stiffness corresponds to the torsional stiffness of a sandwich beam in which the shear stiffness of the core in the y–z plane is neglected. In this case we need to consider only the two facesheets. The torsional stiffnesses of the facesheets are (Eq. 6.169) 4b/δt 66 and 4b/δb 66, where superscripts t and b refer to the top and bottom facesheets. The torsional stiffness of the sandwich beam with two facesheets is GI t = 4b δt 66 + 4b δb 66 lower bound. (7.76) 7.3 Transversely Loaded Beams In this section we present the deflections of orthotropic beams, with cross sections symmetrical with respect to the x–z plane. A transverse load pz (per unit length) acts in the plane of symmetry (Fig. 6.1). The relevant equilibrium equations are (Eq. 7.39) dV z dx + pz = 0 (7.77) dMy dx − V z = 0, (7.78) where V z is the shear force, and My is the bending moment in the x–z plane. For beams with symmetrical cross sections, EI yz = 0, Syz = 0, and Szω = 0, and
330 BEAMS WITH SHEAR DEFORMATION Egs.(7.32)and (7.36)give 成=-a,器 V:=StY (7.79) For simplicity,we introduce the following notations: 应，=M = Elyy=Ei S=S (7.80) X:三X Ya=y Pa=p. With these notations we have M=-E立4以 =y. (7.81) dx Equations (7.78)and (7.81)yield Eidx y=-§d (7.82) By combining Eqs.(7.77),(7.78),(7.79)and by using Eq.(7.2),we obtain x +p=0 (7.83) +（-x)- (7.84) Equations (7.83)and(7.84),together with the boundary conditions listed in Table 7.1(page 321),provide the deflection w and the rotation of the cross sec- tion X. Known bending moment distribution.When the bending moment distribution along the beam is known,the deflection may be obtained by the following proce- dure.The deflection is due to both bending and shear deformation w wB+ws. (7.85) We recall that the deflection is related to the rotation of the cross section and to the shear strain by (Eq.7.2) dw =x+y. (7.86 We combine the two preceding equations and write dw dwB dws +dx=x+y. (7.87) dx dx Thus,we have dwB dws dx=x dx=y. (7.88) Substitution of the first of these expressions into Eq.(7.79)gives 2wB M (7.89)
330 BEAMS WITH SHEAR DEFORMATION Eqs. (7.32) and (7.36) give My = −EI yy dχz dx V z = Szzγz. (7.79) For simplicity, we introduce the following notations: My ≡ M V z ≡ V EI yy ≡ EI Szz ≡ S χz ≡ χ γz ≡ γ pz = p. (7.80) With these notations we have M = −EI dχ dx V = Sγ. (7.81) Equations (7.78) and (7.81) yield γ = − EI S d2χ dx2 . (7.82) By combining Eqs. (7.77), (7.78), (7.79) and by using Eq. (7.2), we obtain −EI d3χ dx3 + p = 0 (7.83) EI d2χ dx2 + S dw dx − χ = 0. (7.84) Equations (7.83) and (7.84), together with the boundary conditions listed in Table 7.1 (page 321), provide the deflection w and the rotation of the cross section χ. Known bending moment distribution. When the bending moment distribution along the beam is known, the deflection may be obtained by the following procedure. The deflection is due to both bending and shear deformation w = wB + wS. (7.85) We recall that the deflection is related to the rotation of the cross section and to the shear strain by (Eq. 7.2) dw dx = χ + γ. (7.86) We combine the two preceding equations and write dw dx = dwB dx + dwS dx = χ + γ. (7.87) Thus, we have dwB dx = χ dwS dx = γ. (7.88) Substitution of the first of these expressions into Eq. (7.79) gives d2wB dx2 = − M EI . (7.89)