16 COMPOSITE BEAMS IN TORSION As in the previous chapter,we consider here the composite beams made of isotropic phases.Extension to transversely isotropic phases is straightforward.The study of orthotropic phases with one principal direction parallel to the axis of the beam,the other two principal directions in the plane of a cross section,does not present fundamental difficulties. 16.1 UNIFORM TORSION We will keep the conventions and notations of the previous chapter.On Figure 16.1, 0 is the elastic center,x,y,and z are the principal axes.The beam is slender and uniformly twisted,this means that every cross section is subjected to a pure and constant torsion moment,along the x axis,denoted as M. Then,under the application of this moment,each line in the beam,initially parallel to the x axis,becomes a helicoid curve,including (in the absence of symmetry in the cross section)the line which,initially,was coinciding with the elastic x axis itself.The only line which remains rectilinear is cutting the plane of all sections at a point which will be called torsion center and denoted as C, with coordinates yc and zc in the principal axes (see Figure 16.1). 16.1.1 Torsional Degree of Freedom By definition,this is the rotation of each section about thexaxis,denoted as 0 The torsional moment M,being constant,the angle 6 evolves along the x axis in such a manner that,for any pair of cross sections spaced with a distance d, one can observe a same increment of rotation de;then: de: constant dx T Here it is not necessary to define the rotation by means of an integral of displacements. as in the previous chapter relating to flexure.In effect,we will see in the following that the displacement field associated with this pure rotation of the sections leads to the exact solution of the problem in the elastic domain (at least for the case of uniform warping). 2003 by CRC Press LLC
16 COMPOSITE BEAMS IN TORSION As in the previous chapter, we consider here the composite beams made of isotropic phases. Extension to transversely isotropic phases is straightforward. The study of orthotropic phases with one principal direction parallel to the axis of the beam, the other two principal directions in the plane of a cross section, does not present fundamental difficulties. 16.1 UNIFORM TORSION We will keep the conventions and notations of the previous chapter. On Figure 16.1, 0 is the elastic center, x,y, and z are the principal axes. The beam is slender and uniformly twisted, this means that every cross section is subjected to a pure and constant torsion moment, along the x axis, denoted as Mx. Then, under the application of this moment, each line in the beam, initially parallel to the x axis, becomes a helicoid curve, including (in the absence of symmetry in the cross section) the line which, initially, was coinciding with the elastic x axis itself. The only line which remains rectilinear is cutting the plane of all sections at a point which will be called torsion center and denoted as C, with coordinates yC and zC in the principal axes (see Figure 16.1). 16.1.1 Torsional Degree of Freedom By definition, this is the rotation of each section about the x axis, denoted as qx. 1 The torsional moment Mx being constant, the angle qx evolves along the x axis in such a manner that, for any pair of cross sections spaced with a distance dx, one can observe a same increment of rotation dqx; then: 1 Here it is not necessary to define the rotation qx by means of an integral of displacements, as in the previous chapter relating to flexure. In effect, we will see in the following that the displacement field associated with this pure rotation of the sections leads to the exact solution of the problem in the elastic domain (at least for the case of uniform warping). dqx dx -------- = constant TX846_Frame_C16 Page 307 Monday, November 18, 2002 12:32 PM © 2003 by CRC Press LLC
Figure 16.1 Elastic Center (O),Torsion Center(C),and Principal Axes From this it comes that the angle of rotation of the sections varies linearly along the longitudinal axis x.As a consequence,we assume a priori the compo- nents of the displacement field uu u to be written as: de dx ×p0y,z) (16.1) 4,=-(z-z)6x u2=(y-yc)8 where the function denoted as o(y,z)is characteristic of the cross section shape and of the materials that constitute the section.This is called the warping function for torsion. 16.1.2 Constitutive Relation With the displacement field in Equation 16.1 the only nonzero strains are written as: d架-(z-z - a=能+0-) The only nonzero stresses are then the shear stresses to and t The torsional moment can be deduced by integration over the domain of the straight section as: =j-s-c空-刘小 (佛+++ 2003 by CRC Press LLC
From this it comes that the angle of rotation of the sections varies linearly along the longitudinal axis x. As a consequence, we assume a priori the components of the displacement field ux, uy, uz, to be written as: (16.1) where the function denoted as j(y,z) is characteristic of the cross section shape and of the materials that constitute the section. This is called the warping function for torsion. 16.1.2 Constitutive Relation With the displacement field in Equation 16.1 the only nonzero strains are written as: The only nonzero stresses are then the shear stresses txy and txz. The torsional moment can be deduced by integration over the domain of the straight section as: Figure 16.1 Elastic Center (O), Torsion Center (C), and Principal Axes ux dqx dx = -------- ¥ j( ) y, z uy z z – c = –( )qx uz y y – c = ( )qx g xy dqx dx -------- ∂j ∂y ------ z z – c – ( ) Ë ¯ Ê ˆ = g xz dqx dx -------- ∂j ∂z ------ y y – c + ( ) Ë ¯ Ê ˆ = Mx ytxz – ztxy ( )dS DÚ dqx dx -------- Gi y ∂j ∂z ------ – yc Ë ¯ Ê ˆº Ó Ì Ï DÚ = = º z ∂j ∂ y ------ + zc Ë ¯ Ê ˆ – y 2 z2 + + ˛ ˝ ¸ dS TX846_Frame_C16 Page 308 Monday, November 18, 2002 12:32 PM © 2003 by CRC Press LLC
Substituting to the function (y,2)the function (y,z)such that: Φ0y,=p(0y,2z)+yz-zy (16.2) it becomes: M-6-++s In this expression,it is possible to define an equivalent stiffness in torsion with the form: ∂Φ ∂ +y2+2 (16.3) One obtains then for the constitutive relation: 08 M.=(G) 16.1.3 Determination of the Function (yz) 16.1.3.1 Local Equilibrium Local equilibrium is written as: Otx0 then with the displacement field in Equation 16.1: 72p=0 and with form 16.2 of the functionΦ： 7Φ=0 16.1.3.2 Conditions at the External Boundary The lateral surface being free of stresses,one can write along the external boundary aD: 含7=0. With the displacement field in Equation 16.1: 器-e-l+能+-w=0 2003 by CRC Press LLC
Substituting to the function j(y,z) the function F(y,z) such that: (16.2) it becomes: In this expression, it is possible to define an equivalent stiffness in torsion with the form: (16.3) One obtains then for the constitutive relation: 16.1.3 Determination of the Function F(y,z) 16.1.3.1 Local Equilibrium Local equilibrium is written as: then with the displacement field in Equation 16.1: —2 j =0 and with form 16.2 of the function F: —2 F =0 16.1.3.2 Conditions at the External Boundary The lateral surface being free of stresses, one can write along the external boundary ∂D: . With the displacement field in Equation 16.1: F( ) y, z = j ( ) y, z + yzc – zyc Mx dqx dx -------- Gi y ∂F ∂z ------- z ∂F ∂ y ------- y 2 z2 – + + Ë ¯ Ê ˆ DÚ = ¥ dS · Ò GJ Gi y ∂F ∂z ------- z ∂F ∂ y ------- y 2 z2 – + + Ë ¯ Ê ˆ dS DÚ = Mx · Ò GJ ∂qx ∂ x = ------- ∂txy ∂y --------- ∂txz ∂z + --------- = 0 t . n = 0 ∂f ∂y ------ z z – c – ( ) Ó ˛ Ì ˝ Ï ¸ ny ∂f ∂z ------ y y – c + ( ) Ó ˛ Ì ˝ Ï ¸ + nz = 0 TX846_Frame_C16 Page 309 Monday, November 18, 2002 12:32 PM © 2003 by CRC Press LLC
then again: Φ.，aΦ 苏，+正n:=m-m: 16.1.3.3 Conditions at the Internal Boundaries The continuity conditions of Section 15.1.2 are verified for uy and u.At an interfacial line ly between two phases i and j,the continuity of uy leads to Φ，=中 The continuity relations in Equation 15.6 lead to the continuity of (Tny+tn) when crossing the lines l,that produces the continuity of c(-,+c(+n 16.1.3.4 Uniqueness of the Function If one superimposes torsion and bending,by using the degrees of freedom for flexure defined in the previous chapter,the displacement component u becomes: dφ+n✉ ux=4-y9+z8,+ The longitudinal displacement u(x)has to respond to its definition (Section 15.1.1),meaning u= E ux ds s-y+y This requires that: Eps=0 JD Then,taking into account the form in Equation 16.2 of and the properties of the elastic center: 2003 by CRC Press LLC
then again: 16.1.3.3 Conditions at the Internal Boundaries The continuity conditions of Section 15.1.2 are verified for uy and uz. At an interfacial line lij between two phases i and j, the continuity of ux leads to Fi = Fj The continuity relations in Equation 15.6 lead to the continuity of (txyny + txz nz) when crossing the lines lij, that produces the continuity of 16.1.3.4 Uniqueness of the Function F If one superimposes torsion and bending, by using the degrees of freedom for flexure defined in the previous chapter, the displacement component ux becomes: The longitudinal displacement u(x) has to respond to its definition (Section 15.1.1), meaning This requires that: Then, taking into account the form in Equation 16.2 of F and the properties of the elastic center: ∂F ∂y -------ny ∂F ∂z + -------nz = zny – ynz Gi ∂Fi ∂y -------- – z Ë ¯ Ê ˆ ny Gi ∂Fi ∂z -------- + y Ë ¯ Ê ˆ + nz ux u y – qz zqy dqx dx = + + --------j h + x u 1 · Ò ES ----------- Eiux dS DÚ = u 1 · Ò ES ----------- u Ei S qz Eiy S qy Eiz dS DÚ d + DÚ d – DÚ Ó Ì = Ï º º dqx dx -------- Eij dS Eihx dS DÚ + DÚ + ˛ ˝ ¸ Eij dS DÚ = 0 EiF dS DÚ = 0 TX846_Frame_C16 Page 310 Monday, November 18, 2002 12:32 PM © 2003 by CRC Press LLC
In summary,the function (y,2)is the solution of the problem: 7Φ=0 in domain D of the section ∂Φ an zny-ynz on the external boundaryoD with the internal continuity: Φ，=Φ along the internal 4-(2m,+m,=G a-(zn,+n,） boundaries y and the condition of uniqueness: ∫，Eoas=0 16.1.4 Energy Interpretation The strain energy of an elementary segment of a beam with thickness da is written as: dm=2(+aaw={.c+aa西 then,taking into account the displacement field in Equation 16.1: 器=”j+院+} which can be rewritten as: 器=器c9-架++- +c恩*爱川r 3 In effect,one has,for example: -架-器引叫佛 2003 by CRC Press LLC
In summary, the function F(y,z) is the solution of the problem: with the internal continuity: and the condition of uniqueness: 16.1.4 Energy Interpretation The strain energy of an elementary segment of a beam with thickness dx is written as: then, taking into account the displacement field in Equation 16.1: which can be rewritten as2 : 2 In effect, one has, for example: —2 F = 0 in domain D of the section ∂ F ∂n-------- zny ynz = – on the external boundary∂D Ó Ô Ì Ô Ï Fi = Fj Gi ∂Fi ∂n-------- zny + ynz – ( ) Ë ¯ Ê ˆ Gj ∂Fj ∂n-------- zny + ynz – ( ) Ë ¯ Ê ˆ = ˛ Ô ˝ Ô ¸ along the internal boundaries ij EiF dS DÚ = 0 dW 1 2 -- 2 txyexy + txzexz ( )dV Ú 1 2 -- Gi g xy 2 g xz 2 ( ) + dS DÚ Ó ˛ Ì ˝ Ï ¸ = = dx dW dx -------- 1 2 -- dqx dx -------- Ë ¯ Ê ˆ 2 Gi ∂F ∂y ------- – z Ë ¯ Ê ˆ 2 ∂F ∂z ------- + y Ë ¯ Ê ˆ 2 + Ó ˛ Ì ˝ Ï ¸ dS DÚ = ∂F ∂y ------ Ë ¯ Ê ˆ 2 z ∂F ∂y – ------ ∂F ∂y ------ ∂F ∂y ------ – z Ë ¯ Ê ˆ ∂ ∂y ----- F ∂F ∂y ------ – z Ë ¯ Ê ˆ Ó ˛ Ì ˝ Ï ¸ F∂ 2 F ∂y 2 = = – -------- dW dx -------- 1 2 -- dqx dx -------- Ë ¯ Ê ˆ 2 Gi y ∂F ∂z ------- z ∂F ∂y ------- y 2 z2 – + + Ó ˛ Ì ˝ Ï ¸ S GiF—2 F dSº DÚ d – DÚÓ Ì Ï = º GiF ∂F ∂y ------- – z Ë ¯ Ê ˆ ny ∂F ∂z ------- + y Ë ¯ Ê ˆ + nz Ó ˛ Ì ˝ Ï ¸ dG ∂D Ú + ˛ ˝ ¸ TX846_Frame_C16 Page 311 Monday, November 18, 2002 12:32 PM © 2003 by CRC Press LLC
where we note the presence of the stiffness in torsion (G)defined by Equation 16.3.Thus, 16.2 LOCATION OF THE TORSION CENTER Consider the cantilever beam that is clamped at its left end as shown schematically in Figure 16.2,and more particularly the segment limited by the cross sections denoted by Do and D.In the section D,0 is the elastic center and C is the torsion center the position of which we wish to determine. With this objective,we will apply on the cross section D the two following successive loadings: Loading No.1:One applies on the torsion center C of the cross section D a force F situated in the plane of the section. Loading No.2:One applies on the same cross section D a torsional moment denoted as M(see Figure 16.2). When one applies these two loads successively,the final state is independent of the order of the application.As a consequence for the external forces acting on the isolated segment (Do D,the work corresponding to loading No.1 on the displacements created by loading No.2 is equal to the work corresponding to loading No.2 on the displacements created by loading No.1.This can be written in the following form: W(loading Ix displacement 2)W(loading x displacement 1 Now we evaluate these works: a)W (loading 1 x displacement 2) On D creates the bending moments M and thus a normal stress distribution given in the principal axes by Equation 15.17 as: My M Do Do case n°1 case°2 Figure 16.2 Cantilever Beam with Two Successive Loadings 2003 by CRC Press LLC
where we note the presence of the stiffness in torsion ·GJÒ defined by Equation 16.3. Thus, 16.2 LOCATION OF THE TORSION CENTER Consider the cantilever beam that is clamped at its left end as shown schematically in Figure 16.2, and more particularly the segment limited by the cross sections denoted by D0 and D1. In the section D1,0 is the elastic center and C is the torsion center the position of which we wish to determine. With this objective, we will apply on the cross section D1 the two following successive loadings: Loading No. 1: One applies on the torsion center C of the cross section D1 a force situated in the plane of the section. Loading No. 2: One applies on the same cross section D1 a torsional moment denoted as Mx (see Figure 16.2). When one applies these two loads successively, the final state is independent of the order of the application. As a consequence for the external forces acting on the isolated segment (D0 D1), the work corresponding to loading No. 1 on the displacements created by loading No. 2 is equal to the work corresponding to loading No. 2 on the displacements created by loading No. 1. This can be written in the following form: W (loading 1¥ displacement 2) = W (loading 2¥ displacement 1) Now we evaluate these works: a) W (loading 1 ¥ displacement 2) On D0: creates the bending moments Mz and My, thus a normal stress distribution given in the principal axes by Equation 15.17 as: Figure 16.2 Cantilever Beam with Two Successive Loadings dW dx -------- 1 2 -- · Ò GJ dqx dx -------- Ë ¯ Ê ˆ 2 or 1 2 -- Mx 2 · Ò GJ = = ------------ F F sxx ( )1 Ei Mz EIz · Ò ------------ y Ei My EIy · Ò = – ¥ + ------------ ¥ z TX846_Frame_C16 Page 312 Monday, November 18, 2002 12:32 PM © 2003 by CRC Press LLC
Then,taking into account the displacement field in Equation 16.1,the work done on Do is On D:The torsion center C does not move in the plane of the cross section during torsion.The work done by the force F in the displacement field of torsion is nil. b)W (loading 2 X displacement 1) Forceas applied to the torson center does not lead to the rotation of the cross sections around the longitudinal axis x.From this the torsional moment M does not work on the bending displacement field due to F. The equality of the two works is then written as: {高×y+备×@-%+as=0 then: 告w-:t+gus +高J+Ak=0 This relation has to be verified when the force applied at C varies in magnitude and direction in the plane of the section.One can deduce from there that the relation is valid no matter what the values of M,and M are.Both the above integrals are then nil.One extracts from this property the coordinates of the torsion center: 1 立=而Bw电ds In summary,the uniform torsion of a cylindrical composite beam made of perfectly bonded isotropic phases can be characterized by a homogenized 2003 by CRC Press LLC
Then, taking into account the displacement field in Equation 16.1, the work done on D0 is On D1: The torsion center C does not move in the plane of the cross section during torsion. The work done by the force in the displacement field of torsion is nil. b) W (loading 2 ¥ displacement 1) Force as applied to the torsion center C does not lead to the rotation of the cross sections around the longitudinal axis x. From this the torsional moment Mx does not work on the bending displacement field due to . The equality of the two works is then written as: then: This relation has to be verified when the force applied at C varies in magnitude and direction in the plane of the section. One can deduce from there that the relation is valid no matter what the values of Mz and My are. Both the above integrals are then nil. One extracts from this property the coordinates of the torsion center: In summary, the uniform torsion of a cylindrical composite beam made of perfectly bonded isotropic phases can be characterized by a homogenized sxx ( )1 ux ( )2 ¥ dS DÚ Ei Mz EIz · Ò – ------------ ¥ y Ei My EIy · Ò + ------------ ¥ z Ó ˛ Ì ˝ Ï ¸ dqx dx --------j dS DÚ = dqx dx -------- Ei Mz EIz · Ò – ------------ ¥ y Ei My EIy · Ò + ------------ ¥ z Ó ˛ Ì ˝ Ï ¸ F – yzc + zyc ( )dS DÚ = F F F dqx dx -------- Ei Mz EIz · Ò – ------------ ¥ y Ei My EIy · Ò + ------------ ¥ z Ó ˛ Ì ˝ Ï ¸ F – yzc + zyc ( )dS DÚ = 0 Mz EIz · Ò ------------ Eiy F Eiy 2 – zc + Eiyzyc ( )dSº DÚ º My EIy · Ò ------------ Eiz F Eiz2 + yc – Eiyzzc ( )ds DÚ + = 0 yc 1 EIy · Ò ------------ Eiz F dS DÚ = – zc 1 EIz · Ò ------------ Ei y F dS DÚ = TX846_Frame_C16 Page 313 Monday, November 18, 2002 12:32 PM © 2003 by CRC Press LLC
formulation-equivalent to that of a classical homogeneous beam-in the following manner: degree of freedom:about x axis:0 ·elastic center0:it is such that∫EydS=∫E,zdS=0 .principal axes:tbey are sucb tbat Eyzds equivalent stiffnesses: 〈E）=∑Bli；〈E)=∑PB n=小c(-9++s torsion center:coordinates in principal axes: 1 ye=-(E 「E,zΦdS Ze= 高505 dM, equilibrium relation:- =0 (M.constant) dx de ·constitutive relation:M.=(Gj)） dx 2-2 d6xa 。sbear stresses t=Gra示 (16.4 T:=( .function (y,z):it is the solution to the problem: 子Φ，子Φ =0 in domain D of tbe section. dΦ dn zny-yn:on the external boundary dD. witb internal continuity: ①，=① along internal dΦ -zm,+ym=G aΦ zny +yn: boundaries ti and the uniqueness condition:E ds=0 。strain energy density dw1M R=动 2003 by CRC Press LLC
formulation—equivalent to that of a classical homogeneous beam —in the following manner: • degree of freedom: about x axis: • elastic center 0: it is such that • principal axes: they are such that • equivalent stiffnesses: • torsion center: coordinates in principal axes: • equilibrium relation: • constitutive relation: • shear stresses (16.4) • function F(y, z): it is the solution to the problem: with internal continuity: and the uniqueness condition: • strain energy density qx Ei y dS DÚ Ei z dS DÚ = = 0 Ei yz dS DÚ = 0 EIz · Ò EiIzi ; EIy · Ò i  EiIyi i = =  · Ò GJ Gi y ∂F ∂z ------- z∂F ∂ y ------- y 2 z2 – + + Ë ¯ Ê ˆ dS DÚ = yc 1 EIy · Ò ------------ Eiz F dS DÚ = – zc 1 EIz · Ò ------------- Ei y F dS DÚ = dMx dx ---------- = 0 ( ) Mx = constant Mx · Ò GJ dqx dx = -------- txy Gi dqx dx -------- ∂F ∂y ------- – z Ë ¯ Ê ˆ = txz Gi dqx dx -------- ∂F ∂z ------- + y Ë ¯ Ê ˆ = ∂ 2 F ∂y 2 --------- ∂ 2 F ∂z2 + --------- = 0 in domain D of the section. ∂F ∂n ------- = zny – ynz on the external boundary ∂D. Ó Ô Ô Ì Ô Ô Ï Fi = Fj Gi ∂Fi ∂n-------- – zny + ynz Ë ¯ Ê ˆ Gj ∂Fj ∂n-------- – zny + ynz Ë ¯ Ê ˆ = ˛ Ô ˝ Ô ¸ along internal boundaries ij EiF dS DÚ = 0 dW dx -------- 1 2 -- Mx 2 · Ò GJ = ------------ TX846_Frame_C16 Page 314 Monday, November 18, 2002 12:32 PM © 2003 by CRC Press LLC
Remarks: A finite element computer program for the classical homogeneous beams is usable2 with the condition that the equivalent rigidity in torsion (G/) can be available.This requires the numerical calculation of the function .The latter is the solution of a Laplace type problem,which can be noted in relations 16.4.An equivalent functional is possible to define, which leads to the calculation of d by the finite element method,by means of discretization of the cross section. Flexion-torsion coupling:When,due to the loads applied on the beam, there exists simultaneously bending and torsion of the beam,the approach of the previous chapter is always valid.Keeping the definitions in Sections 15.1.1 and 15.2 for the degrees of freedom u,v,e,0 one arrives at the following displacement field: 4=u-y8:+z8,+pd uy v-z0x+ny uz=W+y6x+门z The torsion being uniform,the equilibrium relations in 15.19 become more restrictive and can be reduced to: d =0; =0; dT: dx =0 dx (16.5) dM. dx =0; dx +T=0: dMy-T:=0 dx Except if the considered application requires the calculation of the shear stresses ina cross section. 4 One has to solve an analogous problem for the homogeneous beams,when one desires to calculate the torsional Saint-Venant stiffness: =∫n”++s 2003 by CRC Press LLC
Remarks: A finite element computer program for the classical homogeneous beams is usable3 with the condition that the equivalent rigidity in torsion ·GJ Ò can be available. This requires the numerical calculation of the function F. 4 The latter is the solution of a Laplace type problem, which can be noted in relations 16.4. An equivalent functional is possible to define, which leads to the calculation of F by the finite element method, by means of discretization of the cross section. Flexion-torsion coupling: When, due to the loads applied on the beam, there exists simultaneously bending and torsion of the beam, the approach of the previous chapter is always valid. Keeping the definitions in Sections 15.1.1 and 15.2 for the degrees of freedom u, v, qx, qy, one arrives at the following displacement field: The torsion being uniform, the equilibrium relations in 15.19 become more restrictive and can be reduced to: (16.5) 3 Except if the considered application requires the calculation of the shear stresses in a cross section. 4 One has to solve an analogous problem for the homogeneous beams, when one desires to calculate the torsional Saint-Venant stiffness: J y . ∂F ∂z ------- z∂F ∂y ------- y2 z2 – + + Ë ¯ Ê ˆ dS DÚ = ux u y – qz zqy j dqx dx = + + -------- + hx uy = v z – qx + hy uz = w y + + qx hz Ó Ô Ô Ì Ô Ô Ï dNx dx --------- 0; dTy dx -------- 0; dTz dx === -------- 0 dMx dx ---------- 0; dMz dx ---------- + Ty 0; dMy dx === ---------- – Tz 0 TX846_Frame_C16 Page 315 Monday, November 18, 2002 12:32 PM © 2003 by CRC Press LLC
Taking into account six degrees of freedom also leads to six constitutive relations.One finds': 水=尝 =偿+8+) (16.6 M=n -ZcTy+ycTz 4跟 M=(a喂 5 In each of the relations marked with ()there appears a supplementary coupling term connected to the existence of a third coefficient denoted as k The complete form is then: 55+,=偿--狼》 乌：+=(G偎+8+碧） This secondary coupling has been neglected in the indicated form. 2003 by CRC Press LLC
Taking into account six degrees of freedom also leads to six constitutive relations. One finds5 : (16.6) 5 In each of the relations marked with (*), there appears a supplementary coupling term connected to the existence of a third coefficient denoted as kyz. The complete form is then: This secondary coupling has been neglected in the indicated form. kyTy + kyzTz · Ò GS dv dx------ qz zc dqx dx – – -------- Ë ¯ Ê ˆ = kyzTy + kzTz · Ò GS dw dx ------- qy yc dqx dx + + -------- Ë ¯ Ê ˆ = Nx · Ò ES du dx = ------ Ty · Ò GS ky ------------- dv dx------ – qz zc dqx dx – -------- Ë ¯ Ê ˆ = (*) Tz · Ò GS kz ------------- dw dx ------- qy yc dqx dx + + -------- Ë ¯ Ê ˆ = (*) Mx · Ò GJ dqx dx = -------- – zcTy + ycTz My EIy · Òdqy dx = -------- Mz EIz · Òdqz dx = -------- TX846_Frame_C16 Page 316 Monday, November 18, 2002 12:32 PM © 2003 by CRC Press LLC