218 BEAMS symmetrical.An axial force N and a bending moment M,act at the centroid of the beam.The origin of the x-y-z coordinate system is attached to this centroid. In addition we use the i-m-i and 2-n2-2 coordinate systems attached to the midpoints of the flange and the web,respectively.The flange and the web are designated by the subscripts 1 and 2. In the following we derive the replacement stiffnesses.The displacements of a T-beam are obtained by substituting these replacement stiffnesses into the ex- pressions for the displacements of the corresponding isotropic beam.Because the cross section is symmetrical with respect to the z-axis and the loads act in the x-z plane,the beam bends only about the y-axis.Consequently,only EAand Elyy are of interest. The calculation proceeds in four steps.In Step 1 we deform the axis of the beam and calculate the strains in,and the curvatures of,each wall segment;in Step 2 we calculate the forces and moments in each wall segment;in Step 3 we calculate the resultant forces and moments acting on the beam;in Step 4 we determine the replacement stiffnesses. We treat both the flange and the web as thin plates and analyze them by the laminate plate theory. Tensile stiffness EA and centroid. Step 1.The axis of the beam(passing through the centroid)is elongated,and the strain of the axis is denoted by e.The beam does not bend,and the axial strains are the same across the cross section, 0=e=, (6.49) where e and e are the axial strains in the midplanes of the flange and the web, respectively Step 2.The axial strains result in distributed axial forces(per unit length)N, N2 in the midplanes of the flange and the web(Fig.6.14,left).These forces are (see Eq.3.31) 1 M1= @ N2= (6.50) Figure 6.14:The distributed forces and the force resultants
218 BEAMS symmetrical. An axial force N and a bending moment My act at the centroid of the beam. The origin of the x–y–z coordinate system is attached to this centroid. In addition we use the ξ1–η1–ζ1 and ξ2–η2–ζ2 coordinate systems attached to the midpoints of the flange and the web, respectively. The flange and the web are designated by the subscripts 1 and 2. In the following we derive the replacement stiffnesses. The displacements of a T-beam are obtained by substituting these replacement stiffnesses into the expressions for the displacements of the corresponding isotropic beam. Because the cross section is symmetrical with respect to the z-axis and the loads act in the x–z plane, the beam bends only about the y-axis. Consequently, only EA and EI yy are of interest. The calculation proceeds in four steps. In Step 1 we deform the axis of the beam and calculate the strains in, and the curvatures of, each wall segment; in Step 2 we calculate the forces and moments in each wall segment; in Step 3 we calculate the resultant forces and moments acting on the beam; in Step 4 we determine the replacement stiffnesses. We treat both the flange and the web as thin plates and analyze them by the laminate plate theory. Tensile stiffness EA and centroid. Step 1. The axis of the beam (passing through the centroid) is elongated, and the strain of the axis is denoted by o x . The beam does not bend, and the axial strains are the same across the cross section, o ξ1 = o ξ2 = o x , (6.49) where o ξ1 and o ξ2 are the axial strains in the midplanes of the flange and the web, respectively. Step 2. The axial strains result in distributed axial forces (per unit length) Nξ1, Nξ2 in the midplanes of the flange and the web (Fig. 6.14, left). These forces are (see Eq. 3.31) Nξ1 = 1 (a11)1 o ξ1 Nξ2 = 1 (a11)2 o ξ2, (6.50) N Nξ1 Nξ2 b N2 ξ2 b N1 ξ1 2 z z1 zc Figure 6.14: The distributed forces and the force resultants.
6.3 THIN-WALLED,OPEN-SECTION ORTHOTROPIC OR SYMMETRICAL CROSS-SECTION BEAMS 219 where (an)1 and(a)2 are evaluated in the g-n-coordinate system at the mid- planes of the flange and the web. Step 3.The total forces acting in the flange and in the web are bi Ni and b2 N2, respectively,and the total force acting on the beam is(Fig.6.14,right) N=b Ne1 +b2 Ng2. (6.51) Step 4.Equations (6.49)-(6.51)give b2 (6.52) (a11)1(a11)2 EA The term indicated by the bracket is the replacement tensile stiffness EA The coordinate of the centroid ze is calculated by a moment balance about the bottom edge of the web (Fig.6.14,right)as follows: zeN =z1bI Ng1 +7b Ne2. (6.53) The distances and 32 are shown in Figure 6.14.By combining this equation with Egs.(6.49)-(6.52),we obtain the position of the centroid: 品+五品正 b2 Zc = (6.54) +品 b2 The coordinates of the centers of the flange and the web with respect to the centroid are(Fig.6.13) 21=z1-2e22=22-2c. (6.55) Depending on where the centroid is located,z2 is either positive or negative. In Figure 6.13,z2 is negative. Bending stiffness ELyy.The beam is bent about the y-axis passing through the centroid(Fig.6.15).The radius of curvature of the beam is denoted by py. Step 1.The strain varies linearly with z as follows: 1 ex=Z. (6.56) Figure 6.15:The radius of curvature py,the strain distribution,and the deformed shapes of the midplanes of the flange and the web
6.3 THIN-WALLED, OPEN-SECTION ORTHOTROPIC OR SYMMETRICAL CROSS-SECTION BEAMS 219 where (a11)1 and (a11)2 are evaluated in the ξ–η–ζ coordinate system at the midplanes of the flange and the web. Step 3. The total forces acting in the flange and in the web are b1Nξ1 and b2Nξ2, respectively, and the total force acting on the beam is (Fig. 6.14, right) N = b1Nξ1 + b2Nξ2. (6.51) Step 4. Equations (6.49)–(6.51) give N = b1 (a11)1 + b2 (a11)2 % &' ( EA o x . (6.52) The term indicated by the bracket is the replacement tensile stiffness EA. The coordinate of the centroid zc is calculated by a moment balance about the bottom edge of the web (Fig. 6.14, right) as follows: zcN = z1b1Nξ1 + z2b2Nξ2. (6.53) The distances z1 and z2 are shown in Figure 6.14. By combining this equation with Eqs. (6.49)–(6.52), we obtain the position of the centroid: zc = z1 b1 (a11)1 + z2 b2 (a11)2 b1 (a11)1 + b2 (a11)2 . (6.54) The coordinates of the centers of the flange and the web with respect to the centroid are (Fig. 6.13) z1 = z1 − zc z2 = z2 − zc. (6.55) Depending on where the centroid is located, z2 is either positive or negative. In Figure 6.13, z2 is negative. Bending stiffness EIyy. The beam is bent about the y-axis passing through the centroid (Fig. 6.15). The radius of curvature of the beam is denoted by ρy. Step 1. The strain varies linearly with z as follows: x = 1 ρy z. (6.56) ρy 1 ρy z 1 1 z1 ρy z y z x Figure 6.15: The radius of curvature ρy, the strain distribution, and the deformed shapes of the midplanes of the flange and the web.
220 BEAMS At the midplane of the flange z=z and the strain is 1 =2a1 (6.57) Py The curvature of the midplane of the flange about the y-axis(by definition)is 1 K51= (6.58) Py In the midplane of the web,the strain e is g=2 (6.59) Py The midplane of the web remains flat,and its curvature is zero (Fig.6.15)as denoted by K2=0. (6.60) Step 2.The axial force N and the bending moment M(per unit length)acting on the flange are (Eqs.3.31 and 3.32) 1 1 N1=ai号 Ma=(du) 1 (6.61) The preceding force and moment are illustrated in Figure 6.16(left).The terms (a)1 and(du)are evaluated in the coordinate system at the midplane of the flange. The layup of the web is symmetrical,and K2=0.Consequently,M2 is zero. Thus,the only force acting in the web is N2(Fig.6.16,right),and this force is 1 N2=an方最. (6.62) where (a)2 is evaluated in the -n-coordinate system at the midplane of the web. Step 3.The resultant bending moment about the y-axis is (Fig.6.16) My =bi Neiz1+bi Me Nezdz, (6.63) (b) where b,b2,and z are shown in Figure 6.13. Na ⊙ 9 Figure 6.16:The distributed forces and moments(per unit length)in the flange and in the web of a T-beam bent about the y-axis
220 BEAMS At the midplane of the flange z = z1 and the strain is o ξ1 = 1 ρy z1. (6.57) The curvature of the midplane of the flange about the y-axis (by definition) is κξ1 = 1 ρy . (6.58) In the midplane of the web, the strain o ξ2 is o ξ2 = 1 ρy z. (6.59) The midplane of the web remains flat, and its curvature is zero (Fig. 6.15) as denoted by κξ2 = 0. (6.60) Step 2. The axial force Nξ1 and the bending moment Mξ1 (per unit length) acting on the flange are (Eqs. 3.31 and 3.32) Nξ1 = 1 (a11)1 o ξ1 Mξ1 = 1 (d11)1 κξ1. (6.61) The preceding force and moment are illustrated in Figure 6.16 (left). The terms (a11)1 and (d11)1 are evaluated in the ξ–η–ζ coordinate system at the midplane of the flange. The layup of the web is symmetrical, and κξ2 = 0. Consequently, Mξ2 is zero. Thus, the only force acting in the web is Nξ2 (Fig. 6.16, right), and this force is Nξ2 = 1 (a11)2 o ξ2, (6.62) where (a11)2 is evaluated in the ξ–η–ζ coordinate system at the midplane of the web. Step 3. The resultant bending moment about the y-axis is (Fig. 6.16) My = b1Nξ1z1 + b1Mξ1 + ) (b2) Nξ2zdz, (6.63) where b1, b2, and z are shown in Figure 6.13. Nξ1 Mξ1 Nξ2 z y y Figure 6.16: The distributed forces and moments (per unit length) in the flange and in the web of a T-beam bent about the y-axis
6.3 THIN-WALLED,OPEN-SECTION ORTHOTROPIC OR SYMMETRICAL CROSS-SECTION BEAMS 221 Step 4.Equations (6.57)-(6.63)yield My= b1 12 +动） (6.64) (a11h (d1)1 (a11)2 Ely The term indicated by the bracket is the replacement bending stiffnessE 6.3.2 Displacements of L-Beams We consider an L-beam.The walls of the flanges are orthotropic,and one of the axes of orthotropy is aligned with the axis of the beam.The layups of the flanges need not be symmetrical with respect to the flanges'midplanes.The beam is subjected to an axial force N and bending moments M,and M.acting at the centroid (Fig.6.17). We use an x-y-z coordinate system with its origin attached to the centroid of the beam and the -m-and 2-2-2 coordinate systems attached to the midpoints of the arbitrarily chosen reference planes in the horizontal and vertical flanges,respectively.The horizontal and vertical flanges are designated by the subscripts 1 and 2. In this section the replacement stiffnesses are determined.The displacements are obtained by substituting these replacement stiffnesses into the expressions for the displacements of the corresponding isotropic beam.Because the cross section of the L-beam is unsymmetrical,all three bending stiffnesses EI.Elyy,and Ely as well as the tensile stiffness EAare needed to determine the displacements. We treat the flanges as thin plates and employ the laminate plate theory equa- tions.The calculation proceeds along the four steps used in the analysis of T-beams (page218). Tensile stiffness EA and centroid.The tensile stiffness is obtained by consid- ering the elongation of the beam while the axis of the beam remains straight Step 1.The axis of the beam(passing through the centroid)is elongated,and the strain of the axis is denoted by e.In the absence of bending,the axial strains are M ① h ⊙ ② Figure 6.17:L-beam
6.3 THIN-WALLED, OPEN-SECTION ORTHOTROPIC OR SYMMETRICAL CROSS-SECTION BEAMS 221 Step 4. Equations (6.57)–(6.63) yield My = 1 ρy z2 1 b1 (a11)1 + b1 (d11)1 + 1 (a11)2 b3 2 12 + z2 2b2 ! % &' ( EIyy . (6.64) The term indicated by the bracket is the replacement bending stiffness EI yy. 6.3.2 Displacements of L-Beams We consider an L-beam. The walls of the flanges are orthotropic, and one of the axes of orthotropy is aligned with the axis of the beam. The layups of the flanges need not be symmetrical with respect to the flanges’ midplanes. The beam is subjected to an axial force N and bending moments My and Mz acting at the centroid (Fig. 6.17). We use an x–y–z coordinate system with its origin attached to the centroid of the beam and the ξ1–η1–ζ1 and ξ2–η2–ζ2 coordinate systems attached to the midpoints of the arbitrarily chosen reference planes in the horizontal and vertical flanges, respectively. The horizontal and vertical flanges are designated by the subscripts 1 and 2. In this section the replacement stiffnesses are determined. The displacements are obtained by substituting these replacement stiffnesses into the expressions for the displacements of the corresponding isotropic beam. Because the cross section of the L-beam is unsymmetrical, all three bending stiffnesses EI zz, EI yy, and EI yz as well as the tensile stiffness EA are needed to determine the displacements. We treat the flanges as thin plates and employ the laminate plate theory equations. The calculation proceeds along the four steps used in the analysis of T-beams (page 218). Tensile stiffness EA and centroid. The tensile stiffness is obtained by considering the elongation of the beam while the axis of the beam remains straight. Step 1. The axis of the beam (passing through the centroid) is elongated, and the strain of the axis is denoted by o x . In the absence of bending, the axial strains are y z h2 b2 h1 b1 My ζ2 ζ1 η 1 1 2 η2 2 z 1z Mz y1 y2 2 z z1 y1 y2 x N O zc yc Figure 6.17: L-beam.
222 BEAMS Ma Na ⊙ Centroid Figure 6.18:The distributed forces and moments acting in an L-beam with unsymmetrical layup when the axis of the beam is elongated. the same across the cross section, 0=e2=e, (6.65) where e and are the axial strains of the reference planes of the two flanges, respectively.The locations of the reference planes may be chosen arbitrarily. Step 2.The axial forces N,N2 and the bending moments M,M2 (per unit length)in the flanges (Fig.6.18)are expressed in terms of the strains eThe derivation of these expressions is discussed subsequently on pages 227-228.Here we quote the results,which are N1= ( (D) N2= (2 (6.66) (D)2 M1=- (D)1 M2= (B112o (D)2 2, (6.67) where D is defined in Table 6.2,and 8u and Bu are evaluated in the -coor- dinate system. Table 6.2.The axial force and moments inside the wall (N=N=M=0).The elements of the compliance matrix11,616, 666,B1,and s are evaluated in the wall's -n-coordinate system. Symmetrical Arbitrary cross section cross section Orthotropic Orthotropic and Arbitrary unsymmetrical layup symmetrical layup layup N=告-碧 六 碧-碧4 M=-咎+岩烤 聘g+骋 Man =0 0 -尝N-尝M where au 可1=(-） 811= D=a11d11-3 D=a41511-
222 BEAMS Nξ1 Mξ1 Nξ2 z1 Mξ2 Centroid Figure 6.18: The distributed forces and moments acting in an L-beam with unsymmetrical layup when the axis of the beam is elongated. the same across the cross section, o ξ1 = o ξ2 = o x , (6.65) where o ξ1 and o ξ2 are the axial strains of the reference planes of the two flanges, respectively. The locations of the reference planes may be chosen arbitrarily. Step 2. The axial forces Nξ1, Nξ2 and the bending moments Mξ1, Mξ2 (per unit length) in the flanges (Fig. 6.18) are expressed in terms of the strains o ξ1, o ξ2. The derivation of these expressions is discussed subsequently on pages 227–228. Here we quote the results, which are Nξ1 = (δ11)1 (D)1 o ξ1 Nξ2 = (δ11)2 (D)2 o ξ2 (6.66) Mξ1 = −(β11)1 (D)1 o ξ1 Mξ2 = −(β11)2 (D)2 o ξ2, (6.67) where D is defined in Table 6.2, and δ11 and β11 are evaluated in the ξ–η–ζ coordinate system. Table 6.2. The axial force and moments inside the wall (N η = Nξη = Mη = 0). The elements of the compliance matrix δ11, δ16, δ66, β11, and β16 are evaluated in the wall’s ξ–η–ζ coordinate system. Symmetrical Arbitrary cross section cross section Orthotropic Orthotropic and Arbitrary unsymmetrical layup symmetrical layup layup Nξ = δ11 D o ξ − β11 D κξ 1 a11 o ξ δ 11 D o ξ − β 11 D κξ Mξ = −β11 D o ξ + α11 D κξ 1 d11 κξ −β 11 D o ξ + α11 D κξ Mξ η = 0 0 −β16 δ66 Nξ − δ16 δ66 Mξ where α11 = α11 − β2 16 δ66 β11 = β11 − β16δ16 δ66 δ11 = δ11 − δ2 16 δ66 D = α11δ11 − β2 11 D = α11δ11 − β2 11