228 BEAMS N0 ☑ Nm=02☑ M,-02☑ Figure 6.24:The forces and moments along the longitudinal edge of an open-section,thin-walled beam. the cross section are small compared with the beam's length,these in-plane forces and moments are approximately zero inside the wall: Nn=Nan=Mn=0. (6.88) We apply only axial strain and curvatures to the beam,but no twist.Therefore, the twist of the wall is zero: K5n=0. (6.89) We now recall the strain-force relationships given by Eq.(3.22)with the com- pliance matrix for an orthotropic material given by Eq.(3.37).To apply the re- lationships to the wall we replace x by g and y by n.The resulting generalized strain-force relationships are (orthotropic): C11 Q12 011B12 0 N 012 022 0 P21 P22 0 N 0 0 a66 0 0 B66 0 (6.90) KE P11 f21 d11 812 0 M Kn f12 f22 0 812 822 0 M KEn 0 0 B66 0 0 866 Substitution of Eqs.(6.88)and (6.89)into Eq.(6.90)yields 阁-别+日。 (6.91) kn=0=0×N+0×M+866Mm (6.92) From these equations we can obtain N,M,and Mn in terms of e and Ks. The resulting expressions are listed in the left column of Table 6.2(page 222) Step 3.The resultant force and moments in the beam are (Fig.6.25) N- Nedn (6.93) (S) zNgdn (6.94) (S) (5)
228 BEAMS Nη = 0 Nξη = 0 Mη = 0 Figure 6.24: The forces and moments along the longitudinal edge of an open-section, thin-walled beam. the cross section are small compared with the beam’s length, these in-plane forces and moments are approximately zero inside the wall: Nη = Nξ η = Mη = 0. (6.88) We apply only axial strain and curvatures to the beam, but no twist. Therefore, the twist of the wall is zero: κξ η = 0. (6.89) We now recall the strain–force relationships given by Eq. (3.22) with the compliance matrix for an orthotropic material given by Eq. (3.37). To apply the relationships to the wall we replace x by ξ and y by η. The resulting generalized strain–force relationships are (orthotropic): o ξ o η γ o ξ η κξ κη κξ η = α11 α12 0 β11 β12 0 α12 α22 0 β21 β22 0 0 0 α66 0 0 β66 β11 β21 0 δ11 δ12 0 β12 β22 0 δ12 δ22 0 0 0 β66 0 0 δ66 Nξ Nη Nξ η Mξ Mη Mξ η . (6.90) Substitution of Eqs. (6.88) and (6.89) into Eq. (6.90) yields o ξ κξ = α11 β11 β11 δ11 ! Nξ Mξ + 0 0 ! Mξ η (6.91) κξ η = 0 = 0 × Nξ + 0 × Mξ + δ66Mξ η. (6.92) From these equations we can obtain Nξ , Mξ , and Mξ η in terms of o ξ and κξ . The resulting expressions are listed in the left column of Table 6.2 (page 222). Step 3. The resultant force and moments in the beam are (Fig. 6.25) N = ) (S) Nξdη (6.93) My = ) (S) Mξ cos αkdη + ) (S) zNξdη (6.94)
6.3 THIN-WALLED,OPEN-SECTION ORTHOTROPIC OR SYMMETRICAL CROSS-SECTION BEAMS 229 M. Figure 6.25:The forces and moments acting on the beam and the forces and moments(per unit length)acting inside the wall. M:= -M sinakdn+yNdn (6.95) (S) (S) =2Mendn =0. (6.96 (S) where S is the length of the circumference.The torque resultant Tis zero because, for orthotropic beams,Mn is zero(Eq.6.92). Since Tis zero for orthotropic beams,P4,P4.and P34 are zero(see Eq.6.2) Consequently,there is no tension-bending-torsion coupling in orthotropic beams. Step 4.We axially elongate the beam such that 1/py and 1/p:remain zero.Thus, from Eqs.(6.86)and (6.87)we have e8=eK=0. (6.97) For a beam with orthotropic and unsymmetrical layup(arbitrary cross section) Eqs.(6.93)and(6.97),together with the expression for N in Table 6.2,left column, yield N- (6.98) (S9 EA The coordinates of the centroid are obtained by taking a moment about point O shown in Figure 6.22: M:sin akdn (6.99) (S) ( Ne=N+ M:cos akdn, (6.100) (S) (S)
6.3 THIN-WALLED, OPEN-SECTION ORTHOTROPIC OR SYMMETRICAL CROSS-SECTION BEAMS 229 Nξ Mξη N Mξ x My Mz Figure 6.25: The forces and moments acting on the beam and the forces and moments (per unit length) acting inside the wall. Mz = ) (S) −Mξ sin αkdη + ) (S) yNξdη (6.95) T = ) (S) 2Mξ ηdη = 0, (6.96) where S is the length of the circumference. The torque resultant T is zero because, for orthotropic beams, Mξ η is zero (Eq. 6.92). Since T is zero for orthotropic beams, P14, P24, and P34 are zero (see Eq. 6.2). Consequently, there is no tension–bending–torsion coupling in orthotropic beams. Step 4. We axially elongate the beam such that 1/ρy and 1/ρz remain zero. Thus, from Eqs. (6.86) and (6.87) we have o ξ = o x κξ = 0. (6.97) For a beam with orthotropic and unsymmetrical layup (arbitrary cross section) Eqs. (6.93) and (6.97), together with the expression for Nξ in Table 6.2, left column, yield N = ) (S) δ11 D dη % &' ( EA o x . (6.98) The coordinates of the centroid are obtained by taking a moment about point O shown in Figure 6.22: Ny c = ) (S) Nξ ydη − ) (S) Mξ sin αkdη (6.99) Nz c = ) (S) Nξ zdη + ) (S) Mξ cos αkdη, (6.100)
230 BEAMS where y and are the coordinates of a point on the wall's reference surface in the y,z coordinate system.By setting 1/py=1/p:=0 and by using Eqs.(6.86), (6.98),and the expression for N in Table 6.2 (page 222),left column,we obtain the coordinates of the centroid as follows: ∫(+psina)dn ∫（罗-骋cosa)dn (S9 2e= ( (6.101) 器d砌 岩d砌 (S9 (S) Next we determine the bending stiffnesses.First,we set e equal to zero in Eqs.(6.86)and(6.87).Then,Eqs.(6.86),(6.87),(6.94),and(6.95)together with the expressions in Table 6.2 give the following resultant moments of a beam with orthotropic and unsymmetrical layup(arbitrary cross section): 2p11 011 D dn y (S) El a cosa sina 1 dn 且g (6.102) [x+会--男 1 D cosa sin dn- Py Ely: 26 011 -y sina+ sin2a dn- (6.103) D D D (S) E且a The tensile stiffnesses,the coordinates of the centroid,and the bending stiff- nesses of beams with orthotropic and symmetrical layup(arbitrary cross section) and with arbitrary layup (symmetrical cross section)are obtained similarly.The main difference is in Step 2,where the appropriate stress-strain relationships must be used instead of Eq.(6.90).The results are given in Tables 6.3-6.5. Choice ofthe reference surface.The expressions of the replacement stiffnesses simplify when the properties are not evaluated at an arbitrary reference surface but at a"neutral"surface,where Bu(orthotropic layup-arbitrary cross section)or Bu(arbitrary layup-symmetrical cross section)is zero.The surface is"neutral" in the sense that a bending moment M does not cause axial strain es in this surface.(However,it is not a real neutral surface because,unlike in an isotropic beam,in this reference surface the strain perpendicular to the beam's axis en is not zero.)
230 BEAMS where y and z are the coordinates of a point on the wall’s reference surface in the x, y, z coordinate system. By setting 1/ρy = 1/ρz = 0 and by using Eqs. (6.86), (6.98), and the expression for Nξ in Table 6.2 (page 222), left column, we obtain the coordinates of the centroid as follows: yc = (S) yk δ11 D + β11 D sin α dη (S) δ11 D dη zc = (S) zk δ11 D − β11 D cos α dη (S) δ11 D dη . (6.101) Next we determine the bending stiffnesses. First, we set o x equal to zero in Eqs. (6.86) and (6.87). Then, Eqs. (6.86), (6.87), (6.94), and (6.95) together with the expressions in Table 6.2 give the following resultant moments of a beam with orthotropic and unsymmetrical layup (arbitrary cross section): My = ) (S) δ11 D z2 − 2β11 D zcos αk + α11 D cos2 α ! dη % &' ( EIyy 1 ρy + ) (S) δ11 (D) yz + β11 D (zsin α − y cos α) − α11 D cos α sin α ! dη % &' ( EIyz 1 ρz (6.102) Mz = ) (S) δ11 (D) yz + β11 D (zsin α − y cos α) − α11 D cos α sin α ! dη % &' ( EIyz 1 ρy + ) (S) δ11 D y2 + 2β11 D y sin α + α11 D sin2 α ! dη % &' ( EIzz 1 ρz . (6.103) The tensile stiffnesses, the coordinates of the centroid, and the bending stiffnesses of beams with orthotropic and symmetrical layup (arbitrary cross section) and with arbitrary layup (symmetrical cross section) are obtained similarly. The main difference is in Step 2, where the appropriate stress–strain relationships must be used instead of Eq. (6.90). The results are given in Tables 6.3–6.5. Choice of the reference surface. The expressions of the replacement stiffnesses simplify when the properties are not evaluated at an arbitrary reference surface but at a “neutral” surface, where β11 (orthotropic layup – arbitrary cross section) or β11(arbitrary layup – symmetrical cross section) is zero. The surface is “neutral” in the sense that a bending moment Mξ does not cause axial strain ξ in this surface. (However, it is not a real neutral surface because, unlike in an isotropic beam, in this reference surface the strain perpendicular to the beam’s axis η is not zero.)
6.3 THIN-WALLED,OPEN-SECTION ORTHOTROPIC OR SYMMETRICAL CROSS-SECTION BEAMS 231 Table 6.3.The tensile and bending stiffnesses and the coordinates of the centroid of open-and closed-section beams with curved walls. The cross section is arbitrary,and the layup of the wall is orthotropic and symmetrical;a and d are evaluated at the midsurface. arbitrary cross section orthotropic symmetrical layup Tensile stiffness EA=∫六d () Coordinates of the centroid k喝资 备 Bending stiffnesses iw=∫（品&+六cos'a)dn ( x=(品+六sim) (9 ：=∫（yz+六)d (9 The 11 component of the compliance matrix corresponding to the "neutral" surface is(Eq.3.48) 呢=B11+Q11, (6.104) where Bu and 8u are the components of the compliance matrices in the arbitrarily chosen reference surface and o is the location of the "neutral"surface(Fig.6.26). At the“neutral"surfaceβ陨=O,and the preceding equation yields 0=、Bu orthotropic layup (6.105) 811 arbitrary cross section. This equation applies to orthotropic beams with symmetrical as well as un- symmetrical cross sections. For beams with arbitrary layup(symmetrical cross section),Eq.(3.48)and the expression for Bu in Table 6.2(page 222)yield 品=（院+e8m）-B6+oi6i6=a1+p1. (6.106) 866
6.3 THIN-WALLED, OPEN-SECTION ORTHOTROPIC OR SYMMETRICAL CROSS-SECTION BEAMS 231 Table 6.3. The tensile and bending stiffnesses and the coordinates of the centroid of open- and closed-section beams with curved walls. The cross section is arbitrary, and the layup of the wall is orthotropic and symmetrical; a11 and d11 are evaluated at the midsurface. z y y z z y ζ η α zc yc arbitrary cross section orthotropic symmetrical layup Tensile stiffness EA = (S) 1 a11 dη Coordinates of the centroid yc = (S) y 1 a11 dη (S) 1 a11 dη zc = (S) z 1 a11 dη (S) 1 a11 dη Bending stiffnesses EI yy = (S) 1 a11 z2 + 1 d11 cos2α dη EI zz = (S) 1 a11 y2 + 1 d11 sin2 α dη EI yz = (S) 1 a11 yz + 1 d11 sin α cos α dη The 11 component of the compliance matrix corresponding to the “neutral” surface is (Eq. 3.48) β 11 = β11 + δ11, (6.104) where β11 and δ11 are the components of the compliance matrices in the arbitrarily chosen reference surface and is the location of the “neutral” surface (Fig. 6.26). At the “neutral” surface β 11 = 0, and the preceding equation yields = −β11 δ11 orthotropic layup arbitrary cross section. (6.105) This equation applies to orthotropic beams with symmetrical as well as unsymmetrical cross sections. For beams with arbitrary layup (symmetrical cross section), Eq. (3.48) and the expression for β11 in Table 6.2 (page 222) yield β 11 = β 11 + δ11 − (β16 + δ16) δ16 δ66 = β11 + δ11. (6.106)
232 BEAMS Table 6.4.The tensile and bending stiffnesses and the coordinates of the centroid of open-and closed-section beams with curved walls.The cross section is arbitrary,and the layup of the wall is orthotropic and unsymmetrical; 11,B1,and are evaluated in the wall's -n-coordinate system;D is defined in Table 6.2 and 611 and D are defined by Eq.(6.157). arbitrary cross section orthotropic unsymmetrical layup Tensile stiffness Open section:EA=∫兽dn Closed section:EA=∫是dh () () Coordinates of the centroid 告+告m西 告-告m ye= 2e= (9 Bending stiffnesses E面w=(告2-0 zcOSk+共cos2a）d 立=∫（供y2+骋ysina+-器sin2a）d () 面r=∫(yz+g((sina-ycosa)-咎cossin）d (S) Table 6.5.The tensile and bending stiffnesses and the coordinates of the centroid of open-section beams with symmetrical cross section.The layup of the wall is unsymmetrical and nonorthotropic;D,11,and B are defined in Table 6.2.The elements of the compliance matrix are evaluated in the wall's -n-coordinate system. symmetrical cross section nonorthotropic unsymmetrical layup Tensile stiffness EA=∫共d (S) Coordinates of the centroid %=0 之=要华mn 局翠山 Bending stiffness n=∫(2-cos+cos2a）动 (S)
232 BEAMS Table 6.4. The tensile and bending stiffnesses and the coordinates of the centroid of open- and closed-section beams with curved walls. The cross section is arbitrary, and the layup of the wall is orthotropic and unsymmetrical; δ11, β11, and α11 are evaluated in the wall’s ξ–η–ζ coordinate system; D is defined in Table 6.2 and δ11 and D are defined by Eq. (6.157). z y y z z y ζ η α zc yc arbitrary cross section orthotropic unsymmetrical layup Tensile stiffness Open section: ?EA= (S) δ11 D dη Closed section: ?EA= (S) δ11 D dη Coordinates of the centroid yc = (S) yk δ11 D + β11 D sin α dη (S) δ11 D dη zc = (S) zk δ11 D − β11 D cos α dη (S) δ11 D dη Bending stiffnesses EI yy = (S) δ11 D z2 − 2β11 D zcos αk + α11 D cos2 α dη EI zz = (S) δ11 D y2 + 2β11 D y sin α + α11 D sin2 α dη EI yz = (S) δ11 (D) yz + β11 D (zsin α − y cos α) − α11 D cos α sin α dη Table 6.5. The tensile and bending stiffnesses and the coordinates of the centroid of open-section beams with symmetrical cross section. The layup of the wall is unsymmetrical and nonorthotropic; D , δ11, and β11 are defined in Table 6.2. The elements of the compliance matrix are evaluated in the wall’s ξ–η–ζ coordinate system. z y y z z η α zc symmetrical cross section nonorthotropic unsymmetrical layup Tensile stiffness EA = (S) δ 11 D dη Coordinates of the centroid yc = 0 zc = (S) zk δ 11 D − β 11 D cos α dη (S) δ 11 D dη Bending stiffness EI yy = (S) δ 11 D z2 − 2β 11 D zcos αk + α11 D cos2 α dη