6.2 RECTANGULAR,SOLID BEAMS SUBJECTED TO AXIAL LOAD AND BENDING 213 "Neutral"”plane Midplane Figure 6.9:Unsymmetrical solid rectangular beam. by Egs.(6.22)and (6.28).Elements Wi4 and W23 are zero because the layup is symmetrical. When the beam is orthotropic,di6 is zero(Eq.3.37),and the rate of twist is zero(=0)when the beam is subjected to an My bending moment. 6.2.2 Displacements-Unsymmetrical Layup The layup of the laminated beam is unsymmetrical.The centroid is located at the a-a plane at a distance o from an arbitrary reference plane.This reference plane may be chosen as the midplane (Fig.6.9).The location of this a-a plane is such that an axial load does not cause a change in the curvature 1/py,and a bending moment does not cause axial strain e.Thus,the a-a plane is analogous to the neutral plane(which,for beams made of isotropic materials,passes through the center of gravity). We now consider a laminated plate(unsymmetrical layup)on which an axial force Nr and a moment M(per unit length)act in the a-a plane.For this plate, the following relationships hold in the a-a plane(Eq.3.22): ee=aN.+唱Mkx=唱N.+8品，M,plate, (6.29) where B,and are evaluated at the a-a"neutral"plane.According to the aforementioned definition of the "neutral"plane,e depends only on M,and Kx depends only on M.Therefore,Be must be equal to zero.Thus,we write (see Eq.3.48) 唱=B11+Qd11=0, (6.30) where Bu and 8u are evaluated at the arbitrarily chosen reference plane that here we have taken to be the midplane.The consequence of B=0 is twofold.The first consequence is that the distance e is 0、 (6.31) 811 The second consequence is that Eq.(6.29)reduces to e=号NKr=8唱M plate. (6.32)
6.2 RECTANGULAR, SOLID BEAMS SUBJECTED TO AXIAL LOAD AND BENDING 213 z y My x a a a a Midplane N b h “Neutral” plane Figure 6.9: Unsymmetrical solid rectangular beam. by Eqs. (6.22) and (6.28). Elements W14 and W23 are zero because the layup is symmetrical. When the beam is orthotropic, d16 is zero (Eq. 3.37), and the rate of twist is zero (ϑ = 0) when the beam is subjected to an My bending moment. 6.2.2 Displacements – Unsymmetrical Layup The layup of the laminated beam is unsymmetrical. The centroid is located at the a–a plane at a distance from an arbitrary reference plane. This reference plane may be chosen as the midplane (Fig. 6.9). The location of this a–a plane is such that an axial load does not cause a change in the curvature 1/ρy, and a bending moment does not cause axial strain o x . Thus, the a–a plane is analogous to the neutral plane (which, for beams made of isotropic materials, passes through the center of gravity). We now consider a laminated plate (unsymmetrical layup) on which an axial force Nx and a moment Mx (per unit length) act in the a–a plane. For this plate, the following relationships hold in the a–a plane (Eq. 3.22): o x = α 11Nx + β 11Mx κx = β 11Nx + δ 11Mx plate, (6.29) where α 11, β 11, and δ 11 are evaluated at the a–a “neutral” plane. According to the aforementioned definition of the “neutral” plane, o x depends only on Nx, and κx depends only on Mx. Therefore, β 11 must be equal to zero. Thus, we write (see Eq. 3.48) β 11 = β11 + δ11 = 0, (6.30) where β11 and δ11 are evaluated at the arbitrarily chosen reference plane that here we have taken to be the midplane. The consequence of β 11 = 0 is twofold. The first consequence is that the distance is = −β11 δ11 . (6.31) The second consequence is that Eq. (6.29) reduces to o x = α 11Nx κx = δ 11Mx plate. (6.32)
214 BEAMS We note again that the curvature of the plate K,corresponds to the curvature of the beam 1/py.Then,for a beam,Eqs.(6.20)and(6.32)yield composite (6.33) beam. Wi1 肠2 By comparing this equation with Eq.(6.17),we see that the terms in paren- theses are the Wi and W22 elements of the compliance matrix. It follows from Egs.(6.33)and (6.23)that the axial strain and curvature of the axis (and consequently the displacements u and w)of a composite beam （unsymmetrical layup)be calulated by replacing EAand EI by÷and条 in the relevant expressions for the corresponding isotropic beam. When an isotropic beam is subjected to an axial force N and to a bending moment My,it will only bend in the x-z plane.The cross section of a composite beam subjected to N and M,may also twist.To determine the rate of twist of a composite beam,we again refer to a laminated plate.The out-of-plane curvature of a laminated plate (unsymmetrical layup)is(see Eq.3.22) Kxy B6N:+846Mr plate (6.34) where B,and 8%are evaluated at the a-a"neutral"plane.From Eq.(3.48)we have 6=B16+Q616 86=816. (6.35) From Eqs.(6.20),(6.27),and(6.34)we obtain the rate of twist of the beam: -()+( 2b M composite (6.36) beam. Wia By comparing this equation with Eq.(6.17),we see that the terms in the paren- theses are the Wi4 and W24 elements of the compliance matrix. When only N and My act,the relevant elements of the compliance matrix are Wi1,W22,W3,Wi4,W24 (Eq.6.17).Elements W,W,Wi4,and W24 are given by Eqs.(6.33)and(6.36),and W23 is zero. When the beam is orthotropic and one of the orthotropy axes is aligned with the beam's axis,816 and B16 are zero(Eq.3.37).Therefore,(see Eqs.6.35 and 6.36), the rate of twist is zero (=0)when an orthotropic beam is subjected to an axial force N and a bending moment My. 6.2.3 Stresses and Strains The stresses and strains are given in this section for a rectangular thin beam whose thickness h is small compared with its width b(Fig.6.9).We consider only regions away from the edges and employ the laminate plate theory expressions
214 BEAMS We note again that the curvature of the plate κx corresponds to the curvature of the beam 1/ρy. Then, for a beam, Eqs. (6.20) and (6.32) yield o x = α 11 b % &' ( W11 N 1 ρy = δ 11 b % &' ( W22 My composite beam. (6.33) 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. It follows from Eqs. (6.33) and (6.23) that the axial strain and curvature of the axis (and consequently the displacements u and w) of a composite beam (unsymmetrical layup) can be calculated by replacing EAand EI by b α 11 and b δ 11 in the relevant expressions for the corresponding isotropic beam. When an isotropic beam is subjected to an axial force N and to a bending moment My, it will only bend in the x–z plane. The cross section of a composite beam subjected to N and My may also twist. To determine the rate of twist of a composite beam, we again refer to a laminated plate. The out-of-plane curvature of a laminated plate (unsymmetrical layup) is (see Eq. 3.22) κxy = β 16Nx + δ 16Mx plate, (6.34) where β 16, and δ 16 are evaluated at the a–a “neutral” plane. From Eq. (3.48) we have β 16 = β16 + δ16 δ 16 = δ16. (6.35) From Eqs. (6.20), (6.27), and (6.34) we obtain the rate of twist of the beam: ϑ = −1 2 β 16 b % &' ( W14 N + −1 2 δ 16 b % &' ( W24 My composite beam. (6.36) By comparing this equation with Eq. (6.17), we see that the terms in the parentheses are the W14 and W24 elements of the compliance matrix. When only Nand My act, the relevant elements of the compliance matrix are W11, W22, W23, W14, W24 (Eq. 6.17). Elements W11, W22, W14, and W24 are given by Eqs. (6.33) and (6.36), and W23 is zero. When the beam is orthotropic and one of the orthotropy axes is aligned with the beam’s axis, δ16 and β16 are zero (Eq. 3.37). Therefore, (see Eqs. 6.35 and 6.36), the rate of twist is zero (ϑ = 0) when an orthotropic beam is subjected to an axial force N and a bending moment My. 6.2.3 Stresses and Strains The stresses and strains are given in this section for a rectangular thin beam whose thickness h is small compared with its width b (Fig. 6.9). We consider only regions away from the edges and employ the laminate plate theory expressions
6.2 RECTANGULAR,SOLID BEAMS SUBJECTED TO AXIAL LOAD AND BENDING 215 Symmetrical layup.At the neutral plane (which for a symmetrical laminate coincides with the midplane),the strains and curvatures are (see Eqs.3.31,3.32, and6.20) 11 Kx d 12 d2 } (6.37) 16 In a ply at a distance z from the midplane,the strains are (see Eq.3.7) (6.38) Kxy The ply stresses are given by (Eq.3.11) 012 Q16 Ex (6.39) Unsymmetrical layup.The strains and curvatures of the axis passing through the centroid (which is in the "neutral"plane)are (see Egs.3.22 and 6.20) 07 Kx 0 (6.40) Ky 開 唱 Kxy 6J where ae and Be are evaluated at the "neutral"plane (Fig.6.9).The location of the "neutral"plane is given by Eq.(6.31). In a ply at a distance z from the "neutral"plane,the strains and stresses are given by Eqs.(6.38)and (6.39). 6.1 Example.An L 0.2-m-long and b =0.02-m-wide beam,with the cross section shown in Fig.6.10,is made of graphite epoxy.The material properties are given in Table 3.6(page 81).The layup is [455/02/+45].The beam,simply supported at each end,is loaded uniformly (p=100 N/m).Calculate the maximum bending moment,the maximum deflection,and the ply stresses and strains. p=1000N/m 个个个个个个个个 h=2 mm L =200 mm b=20mm 长 Figure 6.10:The beam in Example 6.1
6.2 RECTANGULAR, SOLID BEAMS SUBJECTED TO AXIAL LOAD AND BENDING 215 Symmetrical layup. At the neutral plane (which for a symmetrical laminate coincides with the midplane), the strains and curvatures are (see Eqs. 3.31, 3.32, and 6.20) o x o y γ o xy = a11 a12 a16 = N b > κx κy κxy = d11 d12 d16 = My b > . (6.37) In a ply at a distance z from the midplane, the strains are (see Eq. 3.7) x y γxy = o x o y γ o xy + z κx κy κxy . (6.38) The ply stresses are given by (Eq. 3.11) σx σy τxy = Q11 Q12 Q16 Q21 Q22 Q26 Q61 Q62 Q66 x y γxy . (6.39) Unsymmetrical layup. The strains and curvatures of the axis passing through the centroid (which is in the “neutral” plane) are (see Eqs. 3.22 and 6.20) o x o y γ o xy κx κy κxy = α 11 0 α 12 β 21 α 16 β 61 0 δ 11 β 12 δ 12 β 16 δ 16 N b My b , (6.40) where α and β are evaluated at the “neutral” plane (Fig. 6.9). The location of the “neutral” plane is given by Eq. (6.31). In a ply at a distance z from the “neutral” plane, the strains and stresses are given by Eqs. (6.38) and (6.39). 6.1 Example. An L = 0.2-m-long and b = 0.02-m-wide beam, with the cross section shown in Fig. 6.10, is made of graphite epoxy. The material properties are given in Table 3.6 (page 81). The layup is [±45f 2/012/±45f 2]. The beam, simply supported at each end, is loaded uniformly ( p = 1 000 N/m). Calculate the maximum bending moment, the maximum deflection, and the ply stresses and strains. L = 200 mm b = 20 mm h = 2 mm x p = 1 000 N/m z y Figure 6.10: The beam in Example 6.1
216 BEAMS Solution.The maximum bending moment at the middle of the beam is(Table 7.3, page 332) ,= PL2 =5N.m. 8 (6.41) The maximum deflection of the corresponding isotropic beam is (Table 7.3) 0二 5 pL 384E1 (6.42) The maximum deflection of the composite beam is obtained by replacing EI by(see page 212) 0 5 pLA 3网品 (6.43) With the value of du=33.10x 10-3(Table 3.8,page 85),the maximum deflection is 0=0.0345m=34.5mm. (6.44) The axial force is zero(N=0).Thus,with the compliance matrices given in Table 3.8,the midplane strains and curvatures are (see Egs.6.37 and 6.41) (6.45) 33.10 8.28 1 -6.40 (6.46) 0 where Kx and Ky are illustrated in Figure 6.11.In a ply,at a distance z from the midplane,the strains are given by Eq.(6.38)as follows: (6.47) Figure 6.11:Illustration of the curvatures of the beam in Example 6.1
216 BEAMS Solution. The maximum bending moment at the middle of the beam is (Table 7.3, page 332) My = pL2 8 = 5 N · m. (6.41) The maximum deflection of the corresponding isotropic beam is (Table 7.3) w = 5 384 pL4 EI . (6.42) The maximum deflection of the composite beam is obtained by replacing EI by b d11 (see page 212) w = 5 384 pL4 b d11 . (6.43) With the value of d11 = 33.10 × 10−3 1 N · m (Table 3.8, page 85), the maximum deflection is w = 0.0345 m = 34.5 mm. (6.44) The axial force is zero (N = 0). Thus, with the compliance matrices given in Table 3.8, the midplane strains and curvatures are (see Eqs. 6.37 and 6.41) o x o y γ o xy = a11 a12 a16 N b = 0 0 0 (6.45) κx κy κxy = d11 d12 d16 My b = 33.10 −25.59 0 10−3 5 0.02 = 8.28 −6.40 0 1 m, (6.46) where κx and κy are illustrated in Figure 6.11. In a ply, at a distance z from the midplane, the strains are given by Eq. (6.38) as follows: x y γxy = z κx κy κxy = z 8.27 −6.40 0 . (6.47) κx 1 κy 1 L b Figure 6.11: Illustration of the curvatures of the beam in Example 6.1.
6.3 THIN-WALLED,OPEN-SECTION ORTHOTROPIC OR SYMMETRICAL CROSS-SECTION BEAMS 217 ±45 012 0 0 0.5ex 0.5 5000 20 (%) ±452 Figure 6.12:The nonzero stresses and strains in the beam in Example 6.1.The unit of o is 10N/m2. The stresses in the kth ply are (Eq.6.39) 1 12 16 Oy Q21 Q26 (6.48) k 61 亘 26]k Yxy The elements of the [O]matrix of the 0-degree ply are given by Eq.(3.65)and of the woven fabric by Eq.(3.66).The stresses are calculated with these values of the [O]matrices and with the strains given in Eq.(6.47).The results are shown in Figure 6.12.We note again that these stresses are valid only in regions away from the edges. 6.3 Thin-Walled,Open-Section Orthotropic or Symmetrical Cross-Section Beams Subjected to Axial Load and Bending In this section we present the deflections of,and stresses and strains in,thin- walled,open-section beams.The method is first illustrated via orthotropic T-beams with symmetrical cross section and via orthotropic L-beams.The analysis is then extended to orthotropic beams with arbitrary cross section and to nonorthotropic beams with symmetrical cross section. 6.3.1 Displacements of T-Beams We consider a T-beam whose cross section is symmetrical with respect to the z-axis(Fig.6.13).The layup of both the flange and the web is orthotropic and b2 Figure 6.13:Illustration of the T-beam
6.3 THIN-WALLED, OPEN-SECTION ORTHOTROPIC OR SYMMETRICAL CROSS-SECTION BEAMS 217 z 0 −1 1 0.5 εx z 0 −1 1 500 σx z 0 −1 1 20 σy (%) 2 mm z 0 −1 1 0.5 εy (%) 012 45f +_ 2 45f +_ 2 Figure 6.12: The nonzero stresses and strains in the beam in Example 6.1. The unit of σ is 106 N/m2. The stresses in the kth ply are (Eq. 6.39) σx σy τxy k = Q11 Q12 Q16 Q21 Q22 Q26 Q61 Q62 Q66 k x y γxy k . (6.48) The elements of the [Q] matrix of the 0-degree ply are given by Eq. (3.65) and of the woven fabric by Eq. (3.66). The stresses are calculated with these values of the [Q] matrices and with the strains given in Eq. (6.47). The results are shown in Figure 6.12. We note again that these stresses are valid only in regions away from the edges. 6.3 Thin-Walled, Open-Section Orthotropic or Symmetrical Cross-Section Beams Subjected to Axial Load and Bending In this section we present the deflections of, and stresses and strains in, thinwalled, open-section beams. The method is first illustrated via orthotropic T-beams with symmetrical cross section and via orthotropic L-beams. The analysis is then extended to orthotropic beams with arbitrary cross section and to nonorthotropic beams with symmetrical cross section. 6.3.1 Displacements of T-Beams We consider a T-beam whose cross section is symmetrical with respect to the z-axis (Fig. 6.13). The layup of both the flange and the web is orthotropic and y z h2 b2 h1 b1 My ζ2 ξ2 ζ1 ξ1 η1 1 2 η2 2 z 1z 2 z 1z x N zc Figure 6.13: Illustration of the T-beam