6.3 THIN-WALLED,OPEN-SECTION ORTHOTROPIC OR SYMMETRICAL CROSS-SECTION BEAMS 223 Step 3.The resultant axial force is (Fig.6.18) N=b Ne +b2Ng2, (6.68) where bi and b2 are shown in Figure 6.17. Step 4.Equations (6.65)-(6.68)give N-22 台(Dk (6.69) EA The term indicated by the bracket is the tensile stiffness EA.The terms (D)g and (811)k are evaluated in the coordinate systems attached to the reference planes of the horizontal (k 1)and vertical (k 2)flanges,respectively. The coordinates of the centroid ze,ye are calculated by moment equilibria about point O(Fig.6.17).With the symbols defined in Figures 6.17 and 6.18, moment equilibria give zeN=zib Ner +b Ng2 +b Me (6.70) yeN=y bi Ng1 +y2b2 Ng2 +b2 Mg2. (6.71) By combining Egs.(6.65)-(6.71),we obtain the coordinates of the centroid (Fig.6.17): b(*- h 22(③2 (D2 Ze= (6.72) (③法 (D)k +(器-盛〉 (D1 (D 儿= (6.73) 君物 The coordinates of the centers of the flanges with respect to the centroid are (Fig.6.17) y1=1-火 21=z1-2c (6.74) y2=2-火2=z2- (6.75) The location of the centroid determines whether y,y2,z1,and z2 are positive or negative.In Figure 6.17,y2 and zi are positive and yi and z2 are negative. Bending stiffnesses EL,EL.To determine the bending stiffnesses Elyy and El the beam is bent about the y-axis,which passes through the centroid.The radius of curvature is py (Fig.6.19,left).This bending results in moments My and M
6.3 THIN-WALLED, OPEN-SECTION ORTHOTROPIC OR SYMMETRICAL CROSS-SECTION BEAMS 223 Step 3. The resultant axial force is (Fig. 6.18) N = b1Nξ1 + b2Nξ2, (6.68) where b1 and b2 are shown in Figure 6.17. Step 4. Equations (6.65)–(6.68) give N = * 2 k=1 bk(δ11)k (D)k % &' ( EA o x . (6.69) The term indicated by the bracket is the tensile stiffness EA. The terms (D)k and (δ11)k are evaluated in the coordinate systems attached to the reference planes of the horizontal (k = 1) and vertical (k = 2) flanges, respectively. The coordinates of the centroid zc, yc are calculated by moment equilibria about point O (Fig. 6.17). With the symbols defined in Figures 6.17 and 6.18, moment equilibria give zcN = z1b1Nξ1 + z2b2Nξ2 + b1Mξ1 (6.70) ycN = y1b1Nξ1 + y2b2Nξ2 + b2Mξ2. (6.71) By combining Eqs. (6.65)–(6.71), we obtain the coordinates of the centroid (Fig. 6.17): zc = b1 z1 (δ11)1 (D)1 − (β11)1 (D)1 + z2b2(δ11)2 (D)2 2 2 k=1 bk(δ11)k (D)k (6.72) yc = y1b1(δ11)1 (D)1 + b2 y2 (δ11)2 (D)2 − (β11)2 (D)2 2 2 k=1 bk(δ11)k (D)k . (6.73) The coordinates of the centers of the flanges with respect to the centroid are (Fig. 6.17) y1 = y1 − yc z1 = z1 − zc (6.74) y2 = y2 − yc z2 = z2 − zc. (6.75) The location of the centroid determines whether y1, y2, z1, and z2 are positive or negative. In Figure 6.17, y2 and z1 are positive and y1 and z2 are negative. Bending stiffnesses EIyy, EIyz. To determine the bending stiffnesses EI yy and EI yz the beam is bent about the y-axis, which passes through the centroid. The radius of curvature is ρy (Fig. 6.19, left). This bending results in moments My and Mz.
224 BEAMS P: Figure 6.19:The radii of curvatures py and p:and illustration of the axial strain distributions. The bending stiffness E is determined below. Step 1.The strains and curvatures of the two flanges are identical to those of the T-beam (see Egs.6.57-6.60).Hence,we write 8= 2 K1= (6.76) Py Py 1 =。z K2=0. (6.77) Py Step 2.The axial forces N,N2 and the bending moments M,M2 in the flanges (Fig.6.20)are expressed in terms of the strainseeand curvatures,K2.The derivation of these expressions is discussed subsequently on pages 227-228.Here we quote the results,which are N1= 0一 (Bu) (D)1 -KE1 M1=-+ cuK (6.78) (D) (D)1 (D)1 N2= (③112 (B112 最- KE2 M2=-+ -(D)2 KE2- (6.79) (D)2 Ma 44 Na ⊙ ⊙ Ne Me C Figure 6.20:Forces and moments(per unit length)acting in the L-beam with unsymmetrical layup bent about the y-axis
224 BEAMS ρy ρz z y z y ρy z 1 1 ρy 1 ρz 1 ρz y 1 2 x z y x Figure 6.19: The radii of curvatures ρy and ρz and illustration of the axial strain distributions. The bending stiffness EI yy is determined below. Step 1. The strains and curvatures of the two flanges are identical to those of the T-beam (see Eqs. 6.57–6.60). Hence, we write o ξ1 = 1 ρy z1 κξ1 = 1 ρy (6.76) o ξ2 = 1 ρy z κξ2 = 0. (6.77) Step 2. The axial forces Nξ1, Nξ2 and the bending moments Mξ1, Mξ2 in the flanges (Fig. 6.20) are expressed in terms of the strains o ξ1, o ξ2 and curvatures κξ1, κξ2. 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 − (β11)1 (D)1 κξ1 Mξ1 = −(β11)1 (D)1 o ξ1 + (α11)1 (D)1 κξ1 (6.78) Nξ2 = (δ11)2 (D)2 o ξ2 − (β11)2 (D)2 κξ2 Mξ2 = −(β11)2 (D)2 o ξ2 + (α11)2 (D)2 κξ2. (6.79) Nξ1 Mξ1 Nξ2 Mξ2 y y z Figure 6.20: Forces and moments (per unit length) acting in the L-beam with unsymmetrical layup bent about the y-axis.
6.3 THIN-WALLED,OPEN-SECTION ORTHOTROPIC OR SYMMETRICAL CROSS-SECTION BEAMS 225 Step 3.The resulting bending moment M,about the y-axis is My =bi Neiz +bi Me+ Nezdz. (6.80) (b2 Step 4.Equations(6.76)-(6.80)give M,= 1 、2 (611h P11h (c111 (811)2 b (D)1 D)1 21十 (D) (D)2 12 Elyy (6.81) The term indicated by the bracket is the bending stiffness Elyy. The bending stiffness Elvz is determined below. Steps 1 and 2.In the calculation of the bending moment Ma,the first two steps are identical to those given for the calculation of My. Step 3.When the beam is bent about the y-axis,the resultant bending moment M.about the z-axis is(Figs.6.20 and 6.17) M:bi Naly+ M:dz (6.82) (b) () Step 4.Equations (6.76)-(6.79)and(6.82)give 应2= 4- (B111 (6.83) (D)1(D)1 (D)2 (D)2 且 The term indicated by the bracket is the bending stiffness E Bending stiffnesses EL ELy.To determine the bending stiffnesses EL:and El,the beam is bent about the z-axis with a radius of curvature p:(Fig.6.19, right).Expressions for these bending moments can be derived in the same way as Egs.(6.81)and (6.83).The results are 应= (611)2 2(a12 (a11)2 (⑧11)h + b (D (D2 (D)2 (D1 Ela (6.84) M,= (11,_(B1力 -Z1 b1y1+ (112_(B112 (D)1 (6.85) (D)1 (D)2 (D)2 El The terms indicated by the brackets are the bending stiffnesses EI and E (=Ei)
6.3 THIN-WALLED, OPEN-SECTION ORTHOTROPIC OR SYMMETRICAL CROSS-SECTION BEAMS 225 Step 3. The resulting bending moment My about the y-axis is My = b1Nξ1z1 + b1Mξ1 + ) (b2) Nξ2zdz. (6.80) Step 4. Equations (6.76)–(6.80) give My = 1 ρy b1 (δ11)1 (D)1 z2 1 − 2 (β11)1 (D)1 z1 + (α11)1 (D)1 + (δ11)2 (D)2 b3 2 12 + z2 2b2 ! % &' ( EIyy . (6.81) The term indicated by the bracket is the bending stiffness EI yy. The bending stiffness EI yz is determined below. Steps 1 and 2. In the calculation of the bending moment Mz, the first two steps are identical to those given for the calculation of My. Step 3. When the beam is bent about the y-axis, the resultant bending moment Mz about the z-axis is (Figs. 6.20 and 6.17) Mz = b1Nξ1 y1 + ) (b2) Nξ2 y2dz + ) (b2) Mξ2dz. (6.82) Step 4. Equations (6.76)–(6.79) and (6.82) give Mz = 1 ρy (δ11)1 (D)1 z1 − (β11)1 (D)1 b1 y1 + (δ11)2 (D)2 y2 − (β11)2 (D)2 z2b2 ! % &' ( EIyz . (6.83) The term indicated by the bracket is the bending stiffness EI yz. Bending stiffnesses EIzz, EIzy. To determine the bending stiffnesses EI zz and EI zy, the beam is bent about the z-axis with a radius of curvature ρz (Fig. 6.19, right). Expressions for these bending moments can be derived in the same way as Eqs. (6.81) and (6.83). The results are Mz = 1 ρz b2 (δ11)2 (D)2 y2 2 − 2 (β11)2 (D)2 y2 + (α11)2 (D)2 + (δ11)1 (D)1 b3 1 12 + y2 1b1 ! % &' ( EIzz (6.84) My = 1 ρz (δ11)1 (D)1 z1 − (β11)1 (D)1 b1 y1 + (δ11)2 (D)2 y2 − (β11)2 (D)2 z2b2 ! % &' ( EIzy . (6.85) The terms indicated by the brackets are the bending stiffnesses EI zz and EI zy (= EI yz).
226 BEAMS Figure 6.21:Illustration ofthin-walled,open-section beams with symmetrical and unsymmetrical cross sections. 6.3.3 Displacements of Arbitrary Cross-Section Beams In this section we treat thin-walled,open-section beams.The beams are sub- jected to an axial force N and to bending moments My,M:acting at the centroid (Fig.6.21). Three types of beams are considered: 1.The layup of the wall is orthotropic but unsymmetrical;the beam's cross section is arbitrary.5 2.The layup of the wall is orthotropic and symmetrical;the beam's cross section is arbitrary. 3.The layup of the wall is arbitrary;the cross section is symmetrical with respect to the z-axis,and the loads N and M.are applied in the x-z symmetry plane (Fig.6.21,right). There are no tension-bending-torsion couplings in any of the preceding three types of beams.In the first two types of beams these couplings are not present because the beam is orthotropic(page 207);in the third type of beam couplings are not present because the cross section is symmetrical and the loads act in the symmetry plane. In the following we derive the replacement stiffnesses.The displacements are then obtained by substituting these replacement stiffnesses into the expressions for the displacements of the corresponding isotropic beam. We perform the analysis for beams of the first type,that is,the layup is or- thotropic but is not necessarily symmetrical and the cross section is arbitrary.We then generalize the results to the other two types of beams in the list above. We employ three coordinate systems(Fig.6.22).For the beam we use the x-y-z coordinate system with the origin at the centroid and the r-y-z coordinate system with the origin at an arbitrarily chosen point.We also define a-coordinate system with the origin at the reference plane of the wall.At each point in the wall is parallel to the x coordinate,n is tangential to the circumference of the wall, and is perpendicular to the circumference. The calculation proceeds in four steps.In Step 1 we deform the axis of the beam(axial strain e and curvatures 1/py and 1/p.)and calculate the strains in, 5 J.C.Massa and E.J.Barbero,A Strength of Materials Formulation for Thin Walled Composite Beams with Torsion.Journal of Composite Materials,Vol.32,1560-1594,1998
226 BEAMS N My x y M z z N My x y z Figure 6.21: Illustration of thin-walled, open-section beams with symmetrical and unsymmetrical cross sections. 6.3.3 Displacements of Arbitrary Cross-Section Beams In this section we treat thin-walled, open-section beams. The beams are subjected to an axial force Nand to bending moments My, Mz acting at the centroid (Fig. 6.21). Three types of beams are considered: 1. The layup of the wall is orthotropic but unsymmetrical; the beam’s cross section is arbitrary.5 2. The layup of the wall is orthotropic and symmetrical; the beam’s cross section is arbitrary. 3. The layup of the wall is arbitrary; the cross section is symmetrical with respect to the z-axis, and the loads N and Mz are applied in the x–z symmetry plane (Fig. 6.21, right). There are no tension–bending–torsion couplings in any of the preceding three types of beams. In the first two types of beams these couplings are not present because the beam is orthotropic (page 207); in the third type of beam couplings are not present because the cross section is symmetrical and the loads act in the symmetry plane. In the following we derive the replacement stiffnesses. The displacements are then obtained by substituting these replacement stiffnesses into the expressions for the displacements of the corresponding isotropic beam. We perform the analysis for beams of the first type, that is, the layup is orthotropic but is not necessarily symmetrical and the cross section is arbitrary. We then generalize the results to the other two types of beams in the list above. We employ three coordinate systems (Fig. 6.22). For the beam we use the x–y–z coordinate system with the origin at the centroid and the x–y–z coordinate system with the origin at an arbitrarily chosen point. We also define a ξ–η–ζ coordinate system with the origin at the reference plane of the wall. At each point in the wall ξ is parallel to the x coordinate, η is tangential to the circumference of the wall, and ζ is perpendicular to the circumference. The calculation proceeds in four steps. In Step 1 we deform the axis of the beam (axial strain o x and curvatures 1/ρy and 1/ρz) and calculate the strains in, 5 J. C. Massa and E. J. Barbero, A Strength of Materials Formulation for Thin Walled Composite Beams with Torsion. Journal of Composite Materials, Vol. 32, 1560–1594, 1998
6.3 THIN-WALLED,OPEN-SECTION ORTHOTROPIC OR SYMMETRICAL CROSS-SECTION BEAMS 227 Figure 6.22:The coordinate systems. and the curvatures of,the wall;in Step 2 we calculate the forces and moments in the wall;in Step 3 we calculate the resultant forces and moments acting on the beam;in Step 4 we determine the stiffnesses. Step 1.The Bernoulli-Navier hypothesis states that the axial strain varies linearly with the curvatures of the beam.Thus,the axial strain e at a point on the arbitrarily chosen reference surface of the wall is related to the axial strain e and curvatures 1/py and 1/p.of the beam by 11 g=e+z。+y二， (6.86) Py Pz where z and y are the coordinates of the point on the wall's reference surface. From geometry,it can be shown that the wall's curvature Ka in the-plane is, (Fig.6.23) 1 1 K5=一Cosa- sin a, (6.87) Py where a is the angle between the n-and y-coordinate axes(Fig.6.23). Step 2.In this step we express the axial force N and bending moments M and Men in terms of e and K.To derive the necessary expressions we observe that along the free longitudinal edges of the beam the in-plane forces and moments (per unit length)are zero:N=Nn=M=0(Fig.6.24).Since the dimensions of P. Figure 6.23:Curvatures of the beam(left)and the curvature of the wall(middle)and its vector representation(right)
6.3 THIN-WALLED, OPEN-SECTION ORTHOTROPIC OR SYMMETRICAL CROSS-SECTION BEAMS 227 x y z z O ξ ζ η α x y zc yc Figure 6.22: The coordinate systems. and the curvatures of, the wall; in Step 2 we calculate the forces and moments in the wall; in Step 3 we calculate the resultant forces and moments acting on the beam; in Step 4 we determine the stiffnesses. Step 1. The Bernoulli–Navier hypothesis states that the axial strain varies linearly with the curvatures of the beam. Thus, the axial strain o ξ at a point on the arbitrarily chosen reference surface of the wall is related to the axial strain o x and curvatures 1/ρy and 1/ρz of the beam by o ξ = o x + z 1 ρy + y 1 ρz , (6.86) where z and y are the coordinates of the point on the wall’s reference surface. From geometry, it can be shown that the wall’s curvature κξ in the ξ–ζ plane is, (Fig. 6.23) κξ = 1 ρy cos α − 1 ρz sin α, (6.87) where α is the angle between the η- and y-coordinate axes (Fig. 6.23). Step 2. In this step we express the axial force Nξ and bending moments Mξ and Mξ η in terms of o ξ and κξ . To derive the necessary expressions we observe that along the free longitudinal edges of the beam the in-plane forces and moments (per unit length) are zero: Nη = Nξ η = Mη = 0 (Fig. 6.24). Since the dimensions of y z ρy 1 ρz 1 α η ξ ζ κξ κξ 1 = η Figure 6.23: Curvatures of the beam (left) and the curvature of the wall (middle) and its vector representation (right)