208 BEAMS mounted so that all the fibers are either parallel or perpendicular to the longi- tudinal x-axis or one of the symmetry axes of two adjacent unidirectional plies (treated as a single layer)must be parallel to the beam's longitudinal axis. It is shown subsequently (Section 6.3.3)that for an orthotropic beam P2 P3=P4=P24=P34 =0,and the force-strain relationship is N 0 0 0 应， 0 P2 P23 0 M. orthotropic. 0 P P33 1 (6.7) 0 0 0 0 P44 (P] From the preceding equation we see that there is no tension-bending-torsion coupling in an orthotropic beam. We designate the elements of the stiffness matrix by EA EL,GI and write 「EA 0 0 0 Elyy 0 0 Ely: El: 0 orthotropic. (6.8) 0 0 0 Principal direction.For isotropic beams,there is a coordinate system y' (Fig.6.6)in which the moment of inertia Iy is zero (Iy=0).The angle bet- ween the y'-axis of this coordinate system and the y-axis is tan 20 = 21z_ 2Elyz =-E4,-E (6.9) Lyy -I The relationships between the moments of inertia in the y-z and y'-z'coordinate systems are + (6.10) 2 = (6.11) 1y2=0 (6.12) The directions y'and z'are called principal directions. As in Eq.(6.9),for an orthotropic beam we write the angle between y'and y as tan 2o=- 2Elyz (6.13) Elyy-Elez 4 E.P.Popov,Engineering Mechanics of Solids.Prentice-Hall,Englewood Cliffs,New Jersey,1990. p.342
208 BEAMS mounted so that all the fibers are either parallel or perpendicular to the longitudinal x-axis or one of the symmetry axes of two adjacent unidirectional plies (treated as a single layer) must be parallel to the beam’s longitudinal axis. It is shown subsequently (Section 6.3.3) that for an orthotropic beam P12 = P13 = P14 = P24 = P34 = 0, and the force–strain relationship is N My Mz T = P11 000 0 P22 P23 0 0 P23 P33 0 000 P44 % &' ( [P] o x 1 ρy 1 ρz ϑ orthotropic. (6.7) From the preceding equation we see that there is no tension–bending–torsion coupling in an orthotropic beam. We designate the elements of the stiffness matrix by EA, EI , GI t and write [P] = EA 0 00 0 EI yy EI yz 0 0 EI yz EI zz 0 00 0 GI t orthotropic. (6.8) Principal direction. For isotropic beams, there is a coordinate system y –z (Fig. 6.6) in which the moment of inertia Iy z is zero (Iy z = 0). The angle between the y -axis of this coordinate system and the y-axis is4 tan 2ϕ = − 2Iyz Iyy − Izz = − 2EIyz EIyy − EIzz . (6.9) The relationships between the moments of inertia in the y–z and y –z coordinate systems are Iy y = Iyy + Izz 2 + , Iyy − Izz 2 2 + I2 yz (6.10) Iz z = Iyy + Izz 2 − , Iyy − Izz 2 2 + I2 yz (6.11) Iy z = 0. (6.12) The directions y and z are called principal directions. As in Eq. (6.9), for an orthotropic beam we write the angle between y and y as tan 2ϕ = − 2EI yz EI yy − EI zz . (6.13) 4 E. P. Popov, Engineering Mechanics of Solids. Prentice-Hall, Englewood Cliffs, New Jersey, 1990, p. 342.
6.1 GOVERNING EQUATIONS 209 Figure 6.6:The y,z'coordinate system in which ly is zero. By referring to Eqs.(6.10)-(6.12),we express the bending stiffnesses in the yz coordinate system in the forms Elyy+Ele (6.14) 2 +武 Elyy +Elz (色） +武 (6.15) Elyz=0. (6.16) 6.1.3 Compliance Matrix With respect to the x-y-z coordinate system attached to the centroid,the strain- force relationships are defined as W 0 0 Wi47 111 0 W22 W23 W24 应， (6.17) 0 W23 W33 W34 M. Wi4 W24 W34 W44 The W2i and W31 terms are zero because an axial force applied at the centroid does not cause bending,whereas Wi2 and Wi3 are zero because the compliance matrix is symmetrical.The compliance matrix [W]is the inverse of the stiffness matrix Wu 0 0 Wi4 P P2 3 0 W22 W23 W24 P2 P2 P3 P24 (6.18) 0 W23 W W34 P3 P3 P33 P34 Wi4 W24 W34 W44 P4 B4 P4 P44 We obtain the compliance matrix of an orthotropic beam by substituting the elements of the matrix [P]given in Eq.(6.8)into this expression.The
6.1 GOVERNING EQUATIONS 209 z y ϕ z ′ y ′ Figure 6.6: The y , z coordinate system in which Iyz is zero. By referring to Eqs. (6.10)–(6.12), we express the bending stiffnesses in the y –z coordinate system in the forms EI y y = EI yy + EI zz 2 + 7889 - EI yy − EI zz 2 .2 + EI 2 yz (6.14) EI z z = EI yy + EI zz 2 − 7889 - EI yy − EI zz 2 .2 + EI 2 yz (6.15) EI y z = 0. (6.16) 6.1.3 Compliance Matrix With respect to the x–y–z coordinate system attached to the centroid, the strain– force relationships are defined as o x 1 ρy 1 ρz ϑ = W11 0 0 W14 0 W22 W23 W24 0 W23 W33 W34 W14 W24 W34 W44 N My Mz T . (6.17) The W21 and W31 terms are zero because an axial force applied at the centroid does not cause bending, whereas W12 and W13 are zero because the compliance matrix is symmetrical. The compliance matrix [W] is the inverse of the stiffness matrix W11 0 0 W14 0 W22 W23 W24 0 W23 W33 W34 W14 W24 W34 W44 = P11 P12 P13 P14 P12 P22 P23 P24 P13 P23 P33 P34 P14 P24 P34 P44 −1 . (6.18) We obtain the compliance matrix of an orthotropic beam by substituting the elements of the matrix [P] given in Eq. (6.8) into this expression. The
210 BEAMS result is 0 0 0 0 且g 0 [W]= Elyy El -(Ely:) iwig-(i)月 Elyy orthotropic. 0 lwEx-(自P 0 0 0 1 (6.19) 6.1.4 Replacement Stiffnesses The parameters EA EI,GIt are referred to as replacement stiffnesses.By com- paring Eq.(6.5)with Eq.(6.6)and Egs.(6.4)with Egs.(6.7)and (6.8),we note the similarity in the force-strain relationships of isotropic and composite beams. Therefore,the strains(and consequently the displacements)of an orthotropic beam and of a composite beam with symmetrical cross section can be obtained by replacing EA EI,GI by EA,EI,GIt in the corresponding isotropic beam solution.For an orthotropic beam,the necessary substitutions are Isotropic beam Orthotropic beam EA→EA ELw,El,EL:→v,Eia,Ei GL→Gi For a composite beam with symmetrical cross section and loaded in the symmetry plane the following substitutions apply: Isotropic beam Composite beam symmetrical cross section symmetrical cross section EA.Elyy → EA.Elyy As the preceding discussion indicates,the displacements of orthotropic beams and of composite beams with symmetrical cross section are similar to the displacements of isotropic beams.However,the stresses are markedly different in isotropic and in composite beams.In an isotropic beam subjected to an axial load and pure bending,there is only axial stress.In a composite beam subjected to an axial load and pure bending,in addition to the axial stress,there are also transverse normal and shear stresses.Furthermore,in an isotropic beam subjected to pure torque,there is only shear stress,whereas in a composite beam there are also axial and transverse normal stresses. In this chapter the stresses are calculated by the laminate plate theory,which does not take into account the interlaminar stresses near free edges(page 166). 6.2 Rectangular,Solid Beams Subjected to Axial Load and Bending We consider rectangular laminated beams having solid cross sections with an axial force N acting at the centroid and a pure bending moment My acting in the
210 BEAMS result is [W] = 1 EA 0 00 0 EIzz EIyyEIzz − (EI yz)2 −EI yz EI yyEI zz − (EI yz)2 0 0 −EI yz EI yyEI zz − (EI yz)2 EI yy EI yyEI zz − (EI yz)2 0 00 0 1 GI t orthotropic. (6.19) 6.1.4 Replacement Stiffnesses The parameters EA, EI , GI t are referred to as replacement stiffnesses. By comparing Eq. (6.5) with Eq. (6.6) and Eqs. (6.4) with Eqs. (6.7) and (6.8), we note the similarity in the force–strain relationships of isotropic and composite beams. Therefore, the strains (and consequently the displacements) of an orthotropic beam and of a composite beam with symmetrical cross section can be obtained by replacing EA, EI, GIt by EA, EI , GI t in the corresponding isotropic beam solution. For an orthotropic beam, the necessary substitutions are Isotropic beam Orthotropic beam EA =⇒ EA EIyy, EIzz, EIyz =⇒ EI yy, EI zz, EI yz GIt =⇒ GI t For a composite beam with symmetrical cross section and loaded in the symmetry plane the following substitutions apply: Isotropic beam Composite beam symmetrical cross section symmetrical cross section EA, EIyy =⇒ EA, EI yy As the preceding discussion indicates, the displacements of orthotropic beams and of composite beams with symmetrical cross section are similar to the displacements of isotropic beams. However, the stresses are markedly different in isotropic and in composite beams. In an isotropic beam subjected to an axial load and pure bending, there is only axial stress. In a composite beam subjected to an axial load and pure bending, in addition to the axial stress, there are also transverse normal and shear stresses. Furthermore, in an isotropic beam subjected to pure torque, there is only shear stress, whereas in a composite beam there are also axial and transverse normal stresses. In this chapter the stresses are calculated by the laminate plate theory, which does not take into account the interlaminar stresses near free edges (page 166). 6.2 Rectangular, Solid Beams Subjected to Axial Load and Bending We consider rectangular laminated beams having solid cross sections with an axial force N acting at the centroid and a pure bending moment My acting in the
6.2 RECTANGULAR,SOLID BEAMS SUBJECTED TO AXIAL LOAD AND BENDING 211 Figure 6.7:Rectangular laminated beam. x-z plane (Fig.6.7).In this section,we develop expressions for calculating the displacements and the stresses. We treat the beam as a narrow plate and build the analysis on the results of laminate plate theory presented in Chapter 3.By convention,for plates along an edge parallel to the y-axis,the in-plane force per unit length is N,and the moment per unit length is M(Fig.6.8).For beams,the total force Nalong an edge parallel to y and the total moment in the x-plane My are specified.The total axial force in the beam Ncorresponds to bN,in the plate,and the total moment in the beam My corresponds to bM in the plate(where bis the width).Thus,we can apply the laminate plate theory expressions to beams by making the following substitutions: N Nx=b 应， Mr= b (6.20) 6.2.1 Displacements-Symmetrical Layup The layup of the beam is symmetrical.As noted in the preceding section,we analyze this beam by the laminate plate theory,according to which the midplane strain and curvature of the plate are (Eqs.3.31 and 3.32) e=anN plate. (6.21) Beam Plate Figure 6.8:Internal forces and curvatures in a beam and in the corresponding plate
6.2 RECTANGULAR, SOLID BEAMS SUBJECTED TO AXIAL LOAD AND BENDING 211 z y x b h N My Figure 6.7: Rectangular laminated beam. x–z plane (Fig. 6.7). In this section, we develop expressions for calculating the displacements and the stresses. We treat the beam as a narrow plate and build the analysis on the results of laminate plate theory presented in Chapter 3. By convention, for plates along an edge parallel to the y-axis, the in-plane force per unit length is Nx, and the moment per unit length is Mx (Fig. 6.8). For beams, the total force Nalong an edge parallel to y and the total moment in the x–z plane My are specified. The total axial force in the beam Ncorresponds to bNx in the plate, and the total moment in the beam My corresponds to bMx in the plate (where b is the width). Thus, we can apply the laminate plate theory expressions to beams by making the following substitutions: Nx = N b Mx = My b . (6.20) 6.2.1 Displacements – Symmetrical Layup The layup of the beam is symmetrical. As noted in the preceding section, we analyze this beam by the laminate plate theory, according to which the midplane strain and curvature of the plate are (Eqs. 3.31 and 3.32) o x = a11Nx κx = d11Mx plate. (6.21) x Nx y x N My b y Mx κx 1 ρy y ∂y w∂ o y o w w ψ Beam Plate x y x y z z Figure 6.8: Internal forces and curvatures in a beam and in the corresponding plate
212 BEAMS where au and du are the elements of the compliance matrices (Egs.3.29 and 3.30).We observe that the curvature of the plate in the x-z plane Kx corresponds to the curvature of the beam 1/p(Fig.6.8).If we replace Kx with 1/py,for a beam Egs.(6.20)and (6.21)yield e= ) composite b (6.22) beam. Wit 那2 By comparing this equation with Eq.(6.17),we see that the terms in paren- theses are the Wi and W2 elements of the compliance matrix. For a beam made of an isotropic material,the strain and curvature are(Eq.6.5) e=EA isotropic (6.23) beam. It follows from Eqs.(6.22)and (6.23)that the axial strain and the curvature of the axis (and consequently the displacements u and w)of a composite beam (symmetrical layup)can be calculated by replacing EAand EI byandin the relevant expressions for the corresponding isotropic beam. An isotropic beamsubjected to an axial force Nand a bending moment M,only bends in the x-z plane.On the other hand,under these loads the cross sections of a composite beam may also twist.To determine the amount of this twist we refer to the twisting of a plate.The out-of-plane curvature of a plate(symmetrical layup) is(Eq.3.32) Kxy d6Mr plate, (6.24) where Kxy is defined in Eq.(3.8)and is repeated below 282w080 Ky=- -=-2 ay (6.25) axay dx where wo is the deflection of the midplane.The expression aw/ay in the plate corresponds to in the beam(Fig.6.8).Thus,we have ay Kxy=-2 (6.26) 8x Equations (6.1)and(6.26)give the rate of twist of the beam as follows: 1 8=一2Kg (6.27) By combining Eqs.(6.20),(6.24),and (6.27),we obtain the rate of twist of a beam as follows: My composite 2b) (6.28) beam. 4 When only N and My act,the relevant elements of the compliance matrix are Wi1,W22,W23,Wi4,W24(Eq.6.17).The elements Wi1,W22,and W24 are given
212 BEAMS where a11 and d11 are the elements of the compliance matrices (Eqs. 3.29 and 3.30). We observe that the curvature of the plate in the x–z plane κx corresponds to the curvature of the beam 1/ρy (Fig. 6.8). If we replace κx with 1/ρy, for a beam Eqs. (6.20) and (6.21) yield o x = a11 b % &' ( W11 N 1 ρy = d11 b % &' ( W22 My composite beam. (6.22) By comparing this equation with Eq. (6.17), we see that the terms in parentheses are the W11 and W22 elements of the compliance matrix. For a beam made of an isotropic material, the strain and curvature are (Eq. 6.5) o x = 1 EAN 1 ρy = 1 EI My isotropic beam. (6.23) It follows from Eqs. (6.22) and (6.23) that the axial strain and the curvature of the axis (and consequently the displacements u and w) of a composite beam (symmetrical layup) can be calculated by replacing EAand EI by b a11 and b d11 in the relevant expressions for the corresponding isotropic beam. An isotropic beam subjected to an axial force Nand a bending moment My only bends in the x–z plane. On the other hand, under these loads the cross sections of a composite beam may also twist. To determine the amount of this twist we refer to the twisting of a plate. The out-of-plane curvature of a plate (symmetrical layup) is (Eq. 3.32) κxy = d16Mx plate, (6.24) where κxy is defined in Eq. (3.8) and is repeated below κxy = −2∂2wo ∂x∂y = −2 ∂ ∂wo ∂y ∂x , (6.25) where wo is the deflection of the midplane. The expression ∂wo/∂y in the plate corresponds to ψ in the beam (Fig. 6.8). Thus, we have κxy = −2 ∂ψ ∂x . (6.26) Equations (6.1) and (6.26) give the rate of twist of the beam as follows: ϑ = −1 2 κxy. (6.27) By combining Eqs. (6.20), (6.24), and (6.27), we obtain the rate of twist of a beam as follows: ϑ = −1 2 d16 b % &' ( W24 My composite beam. (6.28) When only Nand My act, the relevant elements of the compliance matrix are W11, W22, W23, W14, W24 (Eq. 6.17). The elements W11, W22, and W24 are given