Notes on the Linear Analysis of Thin-walled Beams

Transcription

1 Notes on the Linear Analysis of Thin-walled Beams Walter D. Pilkey and Levent Kitiş Department of Mechanical Engineering University of Virginia Charlottesville, Virginia c August 1996

2 Contents Chapter I. SAINT-VENANT TORSION OF THIN-WALLED BEAMS Fundamental Equations of Saint-Venant Torsion Saint-Venant s Warping Function Thin-walled Open Sections Thin-walled Closed Sections 11 Chapter II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION Geometry of Deformation 15.. Properties of the Warping Function Stress-Strain Relations 3.4. Equations of Equilibrium Stress Resultants Shear Center Calculation of the Angle of Twist Stress Analysis 39 Chapter III. THIN-WALLED ELASTIC BEAMS OF CLOSED CROSS SECTION Geometry of Deformation Equations of Equilibrium A Multicell Analysis Eample Cross Sections with Open and Closed Parts 66 i

3 ii CONTENTS

4 CHAPTER I SAINT-VENANT TORSION OF THIN-WALLED BEAMS 1.1. Fundamental Equations of Saint-Venant Torsion In this section, the fundamental equations of pure torsion are derived, starting from Prandtl s assumptions about the stresses, for a prismatic beam of arbitrarily shaped cross section, made of isotropic, homogeneous material for which Hooke s law is valid. The beam is subjected to end torques T as shown in Figure 1.1. The coordinate ais is chosen to lie along the beam ais, and the y, z coordinates in the plane of the cross section. The coordinate origin is the centroid C of one of the end sections. T T y C z FIGURE 1.1 Prismatic beam in torsion When a beam is in this state of uniform torque, it is found that only the shear stresses y and z are nonzero y z yz In the absence of body forces, the equations of equilibrium y z y

5 I. SAINT-VENANT TORSION OF THIN-WALLED BEAMS These equations show that the stresses are independent of, which means that the shear stress distribution is the same over all cross sections. By Hooke s law of linear elasticity, only the shear strains y and z are nonzero y z yz The nonzero shear stresses are related to the stresses by y y G z z G where G is the shear modulus. Only the following two of the si compatibility equations are not trivially satisfied for this state In view of Hooke s law, these compatibility conditions can be written in terms of the shear stresses ; @z z Since neither stress depends on, the parenthesized quantity in the preceding equation is independent of, y, and z. ;C (1.1) where C is a constant. This equation and the first of the equations of equilibrium form a set of two first-order partial differential equations to be solved, with the applicable boundary conditions, for the stresses. Prandtl s stress function (y z) is @y ; z Stresses calculated from the stress function satisfy the equations of equilibrium, and Eq. @ ;C Because no forces are applied to the surface of the beam, the equality of the shear stress at the surface and the shear stress component perpendicular the boundary line of the cross section implies that this component is zero at all points of the cross-sectional boundary. Let the curvilinear coordinate s trace the boundary as shown in Figure 1.. The s ais is tangential to the boundary in the direction of increasing s. The positive direction of the normal n to the boundary is chosen to

6 1.1. FUNDAMENTAL EQUATIONS OF SAINT-VENANT TORSION 3 y C y n s z s z FIGURE 1. Boundary condition for shear stress make n, s, and a right-handed orthogonal system of aes. The boundary condition for the shear stress is n y cos z sin where is the angle from the positive y ais to the positive n ais. Since dy ds ; sin the boundary condition can be rewritten as dz y ds ; dy z which, in terms of the stress function, @y dz ds cos ds This shows that the value of the stress function on the boundary remains constant. When the boundary of the cross section is a single closed curve, the stress function assumes a single constant value on it, and this value may be set equal to zero. When the boundary contains several closed curves, however, an arbitrary value can be assigned to the stress function only on one of these curves. On the remaining boundary curves, the stress function assumes different values. The stress resultants over the cross section are the two transverse shear forces V y, V z and the torque T, which are calculated from the shear stress distribution

7 4 I. SAINT-VENANT TORSION OF THIN-WALLED BEAMS over the cross-sectional area A V y V z T y z da (y z ; z y )da @z da The first two integrals are zero by Green s theorem, which transforms them to line integrals over the boundary curves, where is constant. The third integral is evaluated as follows, by another application of Green s theorem, assuming that the boundary value of has been set equal to zero @z da da Let u, v, and w be the displacements of a point of the cross section in the, y, and z directions, respectively. The linear strain-displacement @y because all normal strains are zero, and the assumption of zero shear strain yz @z The functional dependence of the displacements on the coordinates, y, z determined by Eq. (1.3) is u u(y z) v v( z) w w( y) According to Eq. (1.4), the partial derivative of v with respect to z has no z dependence, and the partial derivative of w with respect to y has no y dependence. This implies that v is linear function of z and w is a linear function of y. Because the shear stresses y and z are independent of, so are the strains y and z. The @ z imply that the dependence of v and w on is linear. The longitudinal displacement u(y z) is called the warping displacement. The warping displacement in Saint-Venant torsion has the same value for all cross sections. For this theory to be applicable, the beam must be unrestrained in the longitudinal direction. A cantilever beam, for instance, has a fied end, which is not free to undergo the same warping displacement as the other sections. If eternal torque is applied to the free end of such a beam, normal warping stresses are developed, and Saint-Venant s solution is not applicable. Because the in-plane shear strain yz and all normal strains are zero, the components of the displacement in the plane of the cross section are those of a plane rigid body moving in the yz plane. It will be assumed that the ais of twist is a

8 1.1. FUNDAMENTAL EQUATIONS OF SAINT-VENANT TORSION 5 e y P Q Q e z FIGURE 1.3 Displacement of a point of the cross section line parallel to the beam ais and passes through the point P whose coordinates in the centroidal Cyz system are y P and z P. It will also be assumed that the section at is restrained against rotation. The in-plane displacement of a point Q of the cross section is as shown in Figure 1.3. The point Q moves to Q by a rotation about P r Q P ; r QP ve y we z where r QP denotes the position vector of Q measured from P, and e y, e z are unit vectors in the y, z directions. For small angles of twist, the displacement v is calculated as follows v r Q P e y ; r QP e y r cos( ) ; r cos r(cos cos ; sin sin ; cos ) ;r sin ;(z ; z P ) where r is the length of r QP. A similar calculation gives w, and the displacements in the yz plane are determined to be v( z) ; ()(z ; z P ) w( y) ()(y ; y P ) (1.5) As mentioned earlier, the dependence of v and w on is linear. Therefore for some constant K. () K

9 6 I. SAINT-VENANT TORSION OF THIN-WALLED BEAMS The constant C appearing in Eq. (1.1) can be evaluated in terms of the angle of twist C y @ If Eq. (1.) is solved for C G 1and the solution is, the solution corresponding to a rate of twist of is G and the torque is given by T da 4G The torsional constant J of the beam is defined as J 4 da da The torsional constant, which is obtained by solving Eq. (1.) with right hand side equal to unity and boundary conditions that depend only on the cross-sectional shape and dimensions, is a geometrical property of the cross section. The relationship between the applied torque and the angle of twist is, therefore, T GJ (1.6) 1.. Saint-Venant s Warping Function An alternative to the stress-function approach of the preceding section is Saint- Venant s classical solution, which starts from hypotheses about the displacement field. Saint-Venant made the assumption that the cross sections rotate about the ais of twist, and even though the cross section warps out of its original plane, the projection of the deformed cross section on the yz plane retains its original shape and dimensions. The same conclusion, epressed by Eq. (1.5), was reached by the stress-function approach. If the ais of twist is taken to be the beam ais, Eq. (1.5) becomes v( z) ; ()z w( y) ()y If the rate of change of the angle of twist is assumed constant, and the end of the beam at is assumed to be restrained against rotation, then Saint- Venant s displacement hypothesis about the v and w displacement components can be epressed by v( z) ; z w( y) y (1.7) The aial displacement u is assumed to be the same for all cross sections, so that it is a function of y, z only. It is also assumed that u is directly proportional to the rate of twist u(y z)!(y z) (1.8) where!(y z) is an unknown function called the warping function.

10 1.. SAINT-VENANT S WARPING FUNCTION 7 The strain-displacement relations determine the strains from the assumed @y @z y The stresses are then given by Hooke s law y z yz ; z y The ratio of the change in volume to the original volume, called the cubical dilatation, is zero e y z and all surface forces are zero, so that the displacement formulation of the equations of elasticity reduces to where r is the Laplacian r u r v r The partial differential equations for the displacement components v and w are trivially satisfied. The equation for the warping displacement u @! The boundary condition for the shear stresses on the cylindrical surface of the beam, shown in Figure 1., is y cos z sin

11 8 I. SAINT-VENANT TORSION OF THIN-WALLED BEAMS which, in terms of the warping @y ; z y sin (1.1) The torque at any section is T (y z ; z y y y ; z da The integral in the preceding equation is identified as the torsional constant J ; The area integral can be transformed into a line integral over the boundary by applying Green s theorem I I da ;!(zdz ydy) @y where r denotes the position vector from the centroid to points on the boundary of the cross section. If the cross section is multiply connected, then the boundary integral is the sum of the line integrals along individual parts of the boundary. The torsional constant is given by J ; I!r dr (1.11) y ' C r dr z FIGURE 1.4 Solid elliptic cross section Closed-form solutions for the warping function! are known only for simple and regular geometric shapes, such as the solid elliptic cross section shown in Figure 1.4. The equation of the elliptical boundary is y a z b 1 The warping function is known from the theory of elasticity! ; a ; b a b yz

12 1.3. THIN-WALLED OPEN SECTIONS 9 The line integral in Eq. (1.11) can be evaluated with the parametric representation of the ellipse in terms of the angle ' I I!r dr ; a ; b yz(ydy a b zdz) ; a ; b (ba a b cos ' sin ')(;a b )d' ab(b ; a ) 4(a b ) The area moments of inertia are and the torsional constant is 1 4 ab3 1 4 ba3 J ; ab(b ; a ) 4(a b ) a3 b 3 a b 1.3. Thin-walled Open Sections A thin-walled section is called open if the centerline of its walls is not a closed curve. Equivalently, a thin-walled section whose boundary is a single piecewise continuous closed curve is an open section. The simplest open thin-walled section is the narrow rectangular strip shown in Figure 1.5, for which the wall thickness t is less than one-tenth of the length h. An approimate value for the torsional t z y C h z FIGURE 1.5 Narrow rectangular cross section constant J for this section will be obtained by assuming that y is negligibly small and the shear stress z varies linearly across the wall thickness y z ma y t

13 1 I. SAINT-VENANT TORSION OF THIN-WALLED BEAMS The equation to be solved is Eq. (1.1), which, with these assumptions, becomes G ma t This determines the differential equation for the stress function ; d dy G z y When this equation is integrated and the stress function is set equal to zero on the longer edges of the rectangle, the result is (y) G t ; 4 y and the torsional constant is found to be t J 4 da ; 4 y da t A ; I z t3 h ; h t3 1 t3 h 3 where A is the area and is the area moment of inertia about the z ais. In this solution, it is not possible to set the value of the stress function to zero on the shorter edges of the rectangle. Consequently, the stress distribution Ty z J is not valid near the shorter edges, where the boundary conditions require that the shear stress be zero. In addition, the torque due to z is one-half the actual torque T. This is partially because the neglected shear stresses y are concentrated near the shorter edges and have longer moment arms than the stresses z. s s 1 y C n s s 3 z FIGURE 1.6 Horseshoe section The approimate results obtained for a narrow rectangular strip can be applied to more complicated thin-walled open sections, such as the horseshoe section shown in Figure 1.6. Saint-Venant s approimation for the torsional constant

14 is 1.4. THIN-WALLED CLOSED SECTIONS 11 J 1 3 t 3 (s)ds (1.1) where s is the coordinate that traces the median line of the section and t(s) is the wall thickness. The shear stress distribution is Tn z (1.13) J where n is the normal coordinate measured from the median line. The maimum shear stress occurs at the maimum wall thickness t ma Tt ma ma (1.14) J 1.4. Thin-walled Closed Sections A thin-walled section is called closed if the centerline of its walls is a closed curve. Equivalently, a thin-walled section whose boundary is formed by two piecewise continuous closed curves is a closed section. The boundary of a multicell closed section is made up of more than two closed curves. y C s s n z FIGURE 1.7 Closed thin-walled section A closed thin-walled cross section is shown in Figure 1.7. The tangential and normal coordinates, s and n, are chosen so that the aes n, s, form a right-handed triad. The coordinate s traces the median line starting from an arbitrarily selected origin, and the y, z coordinates of any point on the median line are functions of s.

15 1 I. SAINT-VENANT TORSION OF THIN-WALLED BEAMS The normal coordinate n of any point of the median line is zero. The angle (s) is measured from the positive y ais to the positive n ais. It will be assumed that the shear stress is tangent to the median line and does not vary across the wall thickness. The shear flow q due to the shear stress s, defined by, q t(s) s (s) will be assumed constant. Then, at any s, the derivative of the stress function with respect to @ dn ; z cos y sin ; s ; q t(s) The stress function is then determined by setting its value equal to zero on the outer boundary of the section (n q s) 1 ; n t(s) The derivatives of the aial y G (z ; z z G ; (y ; y P ) where P is a point on the ais of twist. Thus, on the median line, the derivative of u with respect to s G (z ; z P )dy ds ; (y ; y P )dz ds Let r P (s) denote the position vector of the point at s measured from the point P r P (y ; y P )e y (z ; z P )e z where e y, e z are unit vectors in the positive y, z directions, respectively. The unit tangent vector e s at the point with coordinates y, z is e z The unit normal vector e n is, therefore, given by e n e e e z The projection of the position vector r P onto the unit normal vector is r P e n (y ; P ; (z ; z P The derivative of the warping displacement with respect to s G ; r P e n (1.15)

16 1.4. THIN-WALLED CLOSED SECTIONS 13 The line integral of this derivative over the closed path formed by the median line of the cross section is zero 1 G I s ds ; I r P e n ds (1.16) y P r P e s r P e n ds s e n z FIGURE 1.8 Definition of sectorial area In the preceding equation, the integral I r P e n ds is interpreted as twice the area enclosed by the median line. Figure 1.8 shows that the differential quantity under the integral is twice the area of the triangle with base length ds. As the position vector sweeps through the entire median line, the integral gives the twice the area enclosed by the median line. In terms of the constant shear flow q, Eq. (1.16) is rewritten as I q ds G ; t(s)

17 14 I. SAINT-VENANT TORSION OF THIN-WALLED BEAMS The torque resultant is calculated as the moment due to the shear stress about the point P T e q This equation gives the shear stress and the torsional constant r P s tdse s q r P e n ds G q H ds T t(s) s (s) t(s) J T G H ds t(s) r P (e s e )ds For constant wall thickness, the torsional constant becomes J t S where S denotes the length of the median line. (1.17) (1.18)

18 CHAPTER II THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION.1. Geometry of Deformation Figure.1 shows a prismatic thin-walled beam and its cross section. The beam ais, which is defined as the line of centroids of the cross sections, is chosen to lie along the ais. Points on a particular cross section are specified by defining their y and z coordinates. The coordinate s traces the median line of the cross section. Each value of s corresponds to a well-defined point of the median line, so that the coordinates y and z of a point on the median line are functions of s. y C s z FIGURE.1 A thin-walled beam and its cross section It will be assumed that the shape of the median line and its dimensions remain unchanged in the yz plane when the beam undergoes a deformation under static loads. This means that the transverse displacements, which are defined as the displacement components in the plane of the undeformed cross section, of a point on the median line are those of a point belonging to a plane rigid curve constrained to move in its own plane. Let A and B be arbitrarily chosen points of such a plane rigid body in its initial position. After the body undergoes a displacement, the points A and B occupy new positions in space. Let A, B be the projections of these new positions onto the yz plane, as shown in Figure.. Let r BA be the position vector of point B measured from point A. The vector r B is given by A r B A r BA cos e r BA sin (.1) where e is the unit vector in the direction of the positive ais, and is the angle measured from the vector r AB to the vector r A B, ; 6 6, with the vector

19 16 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION r AB translated such that the points A and A become coincident. The rotation is counterclockwise, if the cross product r BA r B, evaluated according to the A right-hand rule, has the same direction as e. Since r BA r B A r BA (e r BA )sin jr BA j sin e the rotation is counterclockwise for sin >. This establishes that the sign of is positive when the sense of rotation from AB to A B is counterclockwise. For small rotations, Eq. (.1) becomes r B A r BA e r BA (.) Let v A and w A be the displacement components of point A along the y and z aes, and let v B and w B be the corresponding displacement components of point B. Let the transverse displacement vector be denoted by u A for point A and by u B for point B u A r A A From the vector polygon A ABB in Figure. r B A u B r B B ; r BA r B B ; r A A u B ; u A (v B ; v A )e y (w B ; w A )e z and Eq. (.) can be written in terms of displacement as u B u A e r BA (.3) The scalar components of this equation are v B v A (z B ; z A ) (.4) w B w A ; (y B ; y A ) y A r BA e r BA B B r B A A z FIGURE. Geometry of plane rigid body motion To apply the foregoing considerations to a thin-walled beam, let A be an arbitrarily chosen reference point, which need not be a material point of the median

20 .1. GEOMETRY OF DEFORMATION 17 line, but which is assumed to be displaced as if it were rigidly attached to the median line. Let e s be a unit vector tangent to the median line in the direction of inreasing s as shown in Figure.3. The unit normal vector e n is defined so as to make the triad e n, e s, and e a right-handed set of orthogonal vectors e e n e s e n e s e Let (s) denote the tangential component of the displacement of the point of the median line at the coordinate s. This component is given by Eq. (.3) (s) u A e s e s (e r A (s)) u A e s (e s e ) r A (s) where r A (s) is the position vector of the point at s measured from A. In terms of the angle between the s and y aes, this equation becomes (s) v A cos w A sin r A e n Let r n A denote the projection of the vector r A onto the unit normal r n A r A e n The tangential component of the displacement of the point at s is given by ( s) v A ()cos(s) w A () sin (s) ()r n A (s) (.5) y A n r A (s) s s z FIGURE.3 Tangential and normal components of displacement It will now be assumed that the shear strain s is negligible in a thin-walled beam with open cross section. This means that longitudinal fibers of the beam material remain orthogonal to the fibers along the median line. This assumption can be @

21 18 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION where u is the displacement of the point at s along the ais. This @ ;v A ()cos(s) ; w A () sin (s) ; ()rn A (s) in which prime denotes differentiation with respect to. Integration with respect to s gives u( s) ;v s A () cos (s)ds ; w s A () sin (s)ds ; s () r n A (s)ds where ;v A ()y(s) ; w A ()z(s) ; ()! A (s) u () s! A (s) r n A (s)ds s r A (s) e n ds is the sectorial area and u () is the longitudinal displacement of the point of the median line at s. The longitudinal strain is calculated by differentiating the longitudinal displacement with ( s) u () ; v A ()y(s) ; w A ()z(s) ; ()! A (s) (.6) The first three terms of this equation are consistent with the Navier-Bernoulli hypothesis that plane sections remain plane. The contribution of the warping of the section is epressed by the last term. For this reason the sectorial area! A is called the warping function. The warping function depends on the sectorial origin, which is the origin chosen for the coordinate s, and on the reference point A, termed the pole of the warping function... Properties of the Warping Function The warping function! A, with pole at point A and origin at s s, is defined as the integral s! A (s) s r A (s) e n (s)ds where r A (s) is the position vector of the point at s of the median line measured from the pole A, as shown in Figure.4. The direction of e n (s) is determined from the convention that the aes (n s ) are right-handed, with the s ais in the direction of increasing s. The magnitude of the differential quantity r A (s)e n (s)ds is twice the area of the triangle with base ds and height r A (s) e n (s). The sign of this quantity is positive if the projection of the position vector onto the unit normal vector e n (s) is positive. This sign is more conveniently determined on the basis of the sense of rotation of the position vector as it sweeps through the area in the direction of increasing s. If this rotation is clockwise, the contribution to the integral is negative, and if it is counterclockwise, the contribution is positive. This is verified for any point of the median line by writing the projection of the position vector onto the unit normal in the form r A e n r A (e s e )(r A e s ) e Thus, when the rotation of the vector r A is counterclockwise as its tip moves in the direction of e s, the cross product r A e s is in the same direction as e, making r A e n positive.

22 .. PROPERTIES OF THE WARPING FUNCTION 19 y A r A e s r A e n ds s e n z FIGURE.4 Definition of sectorial area Let A and B be two arbitrarily selected poles for the warping function. Suppose that the origin for! A is chosen to be at s s and the origin for! B at s s 1, as shown in Figure.5. The relationship between! A and! B is found by the computation s! A (s) s1 s r B e n ds s r B e n ds s s s s ;! B (s )! B (s) s (r B r BA ) e n ds s 1 r B e n ds s s r BA e n ds s r BA (sin e y ; cos e z )ds ;! B (s )! B (s) r BA (dze y ; dye z ) s ;! B (s )! B (s) (y B ; y A )(z(s) ; z ) ; (z B ; z A )(y(s) ; y )

23 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION y B A r BA s 1 r B r A s s z FIGURE.5 Poles and origins of warping functions In this calculation, is the angle between the s and y aes as shown in Figure.3, and y, z are the coordinates of the origin chosen for! A y y(s ) z z(s ) The equation for finding the warping function! A with origin s from the the warping function! B with origin s 1 is, therefore,! A (s)! B (s) ;! B (s )(z A ; z B )(y(s) ; y ) ; (y A ; y B )(z(s) ; z ) (.7) When A and B are coincident points, but s and s 1 are two distinct origins, the transformation equation becomes! A (s)! B (s) ;! B (s ) (.8) showing that the effect of changing the origin of a warping function without changing its pole is to add a constant to it. If, on the other hand, the origins s, s 1 are the same and the poles A, B are different, the transformation equation is! A (s)! B (s) (z A ; z B )(y(s) ; y ) ; (y A ; y B )(z(s) ; z ) (.9) since, in this case,! B (s )! B (s 1 )

24 .. PROPERTIES OF THE WARPING FUNCTION 1 If!(s) is a warping function for a particular pole and origin, the area integral Q!!(s)dA is called the first sectorial moment. The area integrals!! y(s)!(s)da z(s)!(s)da are known as the sectorial products of area. These definitions are analogous to the definitions of the first, second, and product moments of area Q y Q z z zda yda z da y da yzda A pole for which the the sectorial products of area are both zero is called a principal pole. Let A and B be two poles for the warping function with origins at s and s 1, respectively. By multiplying both sides of Eq. (.7) by y and integrating both sides of the result over the cross sectional area, one obtains!a!b ;! B (s )Q z (z A ; z B )( ; y Q z ) ; (y A ; y B )(z ; z Q z ) Since the origin of the y, z aes is the centroid C of the cross section, the first moments Q y and Q z are both zero, so that!a!b (z A ; z B ) ; (y A ; y B )z (.1) A similar calculation gives!a!b (z A ; z B )z ; (y A ; y B ) (.11) The conditions for A to be a principal pole!a!a are solved for the coordinates of the pole!b ;!B z y A y B ; Iyz (.1)!B z ;!B z A z B (.13) ; Iyz These epressions do not depend on the origin chosen for the pole B, because if this origin is shifted, the resulting warping function! B differs from! B by a

25 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION constant value, say K, and!b y(! B K)dA!B KQ z!b Hence, the sectorial products of area for the warping function! B remain the same when the sectorial origin is changed, and the coordinates determined by Eqs. (.1) and (.13) are independent of this origin. It is important to realize that the coordinates given by Eqs. (.1) and (.13) are also independent of the pole B. If D is any arbitary pole, its sectorial products of area are related to those of the pole B by!b!d (z B ; z D ) ; (y B ; y D )z Then!B!D (z B ; z D )z ; (y B ; y D )!B ;!B z!d ;!D z ; (y B ; y D )( ; I yz ) which, when substituted into Eq. (.1), gives!d ;!D z y A y D ; Iyz The right side of this equation is the epression for the y coordinate of A with the pole D. Similarly, the z coordinate of A remains the same regardless of the pole used to find it. Thus, the principal pole depends only on the cross-sectional shape and dimensions; it is a cross-sectional property. If, for a given pole A, there is a sectorial origin s such that Q!A! A (s)da the point s is termed a principal origin. To determine s, let B be a pole coincident with A but with a known origin s 1. According to Eq. (.8)! A (s)! B (s) ;! B (s ) so that the condition for s to be a principal origin is Q!A Q!B ;! B (s )A (.14) This equation determines the principal origin s in terms of the arbitrarily selected origin s 1. The eistence of at least one s is ensured, for most cross sectional shapes, by the mean value theorem for integrals. There may be multiple solutions for s, in which case any one solution can be selected as the principal origin. If another pole D, which is coincident with A and B but has its origin at s, is used instead of B in determining the principal origin s, then Q!D Q!B ;! B (s )A and the condition for s to be a principal origin, written in terms of D, becomes Q!A Q!D ;! D (s )A Q!B ;! B (s )A ; (! B (s ) ;! B (s ))A Q!B ;! B (s )A This shows that the same principal origin is obtained regardless of where the reference origin s 1 is placed. Hence, the principal origin is a cross-sectional property.

26 .. PROPERTIES OF THE WARPING FUNCTION 3 Let A be the principal pole for the warping function! A, whose origin has been selected arbitrarily. When this origin is changed to a principal origin, the sectorial products of area for the warping function remain zero, because, as mentioned above, these products are independent of the sectorial origin as long as the centroid is used as the origin of the coordinates y and z. Hence, for a given cross section, it is possible to find a pole A and an origin s such that Q!A,!A, and!a are zero. A warping function satisfying these conditions is termed a principal warping function. Principal warping functions will henceforth be written without a subscript. b s t w h s 1 n y O C n 3 O n 1 z h s 3 s 4 t f n 4 FIGURE.6 Symmetric channel section For the symmetric channel section shown in Figure.6, the warping function with pole and origin both at the point of intersection O of the y ais and the median line is given by! O (s 1 )! O (s ); h s! O (s 3 )! O (s h 4 ) s 4 where the signs are determined from the sense of rotation of the vector from O to points on the median line. For instance,! O (s ) is negative, because the position vector rotates clockwise as it traces the median line of the upper flange.

27 4 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION Let A be the principal pole of the cross section shown in Figure.6. The coordinates of A will be found by using O both as a reference pole and as the sectorial origin. Since z and z, the coordinates of A are given, according to Eqs. (.1) and (.13), by y A y O!O z A ;! O The sectorial product of area!o is zero by symmetry, and the sectorial product of area!o is!o z(s)! O (s)da b The area moment of inertia for this section is (; h b )(;h s )t f ds h ( )(hs )t 4 f ds b h t f 4 4 6bh t f h 3 t w 1 The coordinates of the principal pole A are found to be 3b t f y A y O 6bt f ht w z A The principal origin s for the principal pole A of the symmetric channel section is determined from the condition Q!A ;! A (s )A where! A has the arbitrarily selected origin O and is given by! A (s 1 )(y A ; y O )s 1! A (s )(y A ; y O ) h ; h s! A (s 3 );(y A ; y O )s 3! A (s 4 );(y A ; y O ) h h s 4 The first moment of sectorial area with pole A and origin at O happens to be zero by symmetry Q!A! A (s)da so that the point O is the principal origin for the principal pole A, and! A is the principal warping function. The warping constant I! is defined as the sectorial moment of inertia of the principal warping function I!! (s)da

28 .. PROPERTIES OF THE WARPING FUNCTION 5 Because the section shown in Figure.6 is symmetric with respect to the y ais, the warping constant is calculated as follows h I!! A (s 1 ) t w ds 1 b b 3 h t! A (s ) f (3bt f ht w ) t f ds 1(6bt f ht w ) b 1 s t y C h d e s 1 O s 3 b z FIGURE.7 Unsymmetric channel section For the unsymmetric channel section shown in Figure.7 with the dimensions b 1 b b b h b and constant thickness t, the centroid C is at a horizontal distance d and a vertical distance e from the intersection O of the lower flange and the web d b e 4b 5 The area moments of inertia and the area product of inertia are 5tb3 15 7tb3 4 z ;tb 3 If the point O is used both as the pole and the sectorial origin, the warping function is! O (s 1 )! O (s );hs! O (s 3 )

29 6 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION To find the principal pole, the values of the sectorial products of area are needed!o!o y! O da z! O da b b (d ; s )! O (s )tds ;(h ; e)! O (s )tds The principal pole A has the coordinates!o ;!O z y A y O ; Iyz!O z ;!O z A z O ; Iyz b b tbs (s ; b)ds tb4 18b 19 18b 85 1tb s The warping function with principal pole A and origin O is h Q!A! A (s 1 )(y A ; y O )s 1 17b 38 s 1! A (s )(y A ; y O )h ; (h ; e z A )s 17b 19 ; 94b! A (s 3 )(e ; z A )s 3 b The first sectorial area moment is! A (s 1 )tds 1 b1 57 s 3! A (s )tds The condition for s to be a principal origin is b 5 ds 6tb s! A (s 3 )tds 3 5tb3 3 6 or Q!A ;! A (s )A 5tb3 3 ; 5tb! A (s )! A (s ) b 3 According to Eq. (.8), the shift of the origin to s gives the principal warping function!(s)! A (s) ;! A (s )! A (s) ; b 3 which, when written out for the web and the flanges, yields!(s 1 )! A (s 1 ) ; b 17b 3 s ; b !(s )! A (s ) ; b 3 ; 94b s ; 3b 57 57!(s 3 )! A (s 3 ) ; b b 3 s ; b The principal warping function!(s) shown sketched to scale in Figure.8. The function is zero at three distinct points of the median line of the section. Any one of these points can be regarded as the principal sectorial origin s. If a cross section has a symmetry ais, say the y ais, this is a principal ais, so that z. In calculating the coordinates of the principal pole A, using Eqs. (.1) and (.13), the choice of the reference pole B is arbitrary. When B is chosen to

30 .. PROPERTIES OF THE WARPING FUNCTION 7 3b 57 3b 57 ; ; 6b 57 7b 19 ; ; b 3 ; ; b 3 FIGURE.8 Warping function for the unsymmetric channel section be any point on the symmetry ais y, it is readily verified that!b coordinate z A is then. The!B z ;!B z A z B ; Iyz This shows that the principal pole lies on the symmetry ais. In addition, the point of intersection of the y ais with the median line is a principal sectorial origin. For a doubly symmetric cross section, the point of intersection of the two aes of symmetry is the principal pole, the principal origin, and the centroid.

31 8 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION A formula for the warping constant in terms of the warping function! B, whose pole and origin are arbitrary, can be derived starting from the transformation equation Eq. (.7)! A! B ;! B (s )(z A ; z B )(y ; y ) ; (y A ; y B )(z ; z ) (.15) Since the origin s for! A is a principal origin, the area integral of Eq. (.15) is Q!A Q!B ;! B (s )A ; y (z A ; z B )A z (y A ; y B )A (.16) The first moments of area Q y and Q z have been set equal to zero in the calculation of the right side of this equation because y and z are centroidal aes. The result of Eq. (.16) allows Eq. (.15) to be rewritten as! A! B ; Q! B A (z A ; z B )y ; (y A ; y B )z (.17) The conditions for A to be a principal pole are then!a!b (z A ; z B ) ; (y A ; y B )z (.18) The warping constant is given by!a!b (z A ; z B )z ; (y A ; y B ) (.19) I! I!B (z A ; z B )!B ; (y A ; y B )!B ; Q! B A (y A ; y B ) (z A ; z B ) ; (y A ; y B )(z A ; z B )z which is simplified by using Eqs. (.18) and (.19) to I! I!B ; Q! B A ; (y A ; y B ) (y A ; y B )(z A ; z B )z ; (z A ; z B ) (.) The channel section with double flanges shown in Figure.9 has uniform thickness t, and it is symmetric with respect to the y ais. The point O, which is the point of intersection of the median line and the y ais, is chosen as a convenient pole and origin for the warping function! O (s 1 ) 6 s 1 6 h! O (s ); h 1 s 6 s 6 b! O (s 3 ); h s 6 s b The area moment of inertia is calculated using the centerline dimensions th3 1 tbh 1 tbh and the sectorial product of area!o is found, taking advantage of symmetry, b!o ; h b 1! O (s )tds 1 1 tb 4 (h 1 h ) ; h! O (s 1 )tds 1

32 .. PROPERTIES OF THE WARPING FUNCTION 9 b s 3 s s 1 y A C h 1 h O z FIGURE.9 Channel section with double flanges The principal pole A is on the symmetry ais y y A y O!O y O 3b (h 1 h ) h 6b(h 1 h ) The warping constant is given by Eq. (.) I! I!O ; (y A ; y O ) tb3 (h 1 h )(h3 3b(h 1 h )) 1(h 3 6b(h 1 h )) Since point O is the principal origin, the principal warping function can be written as!(s 1 )(y A ; y O )s 1!(s )(y A ; y O ) h 1 ; h 1 s!(s 3 )(y A ; y O ) h ; h s 3 from which the value found for I! from Eq. (.) can be verified by evaluating h I!!(s 1 ) tds 1 b!(s ) tds b!(s 3 ) tds 3

33 3 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION.3. Stress-Strain Relations The only significant stresses will be assumed to be the normal stress and the shear stress s. Although the strain s was taken to be negligible in Section.1, the corresponding stress s is not assumed to be zero, and it will be noticed that this contradicts Hooke s law. The distribution of normal stress across the wall thickness will be assumed to be uniform. The shear stress due to unrestrained, or Saint-Venant, torsion will be assumed to be linear, with a zero value at the median line. All other shear stresses will be assumed to be constant across the wall thickness. According to the kinematic assumption that the median line of the cross section is inetensible, the strain s is zero s 1 E ( s ; ) where E is the modulus of elasticity and is Poisson s ratio. The longitudinal strain is 1 E ( ; s 1 ; ) The normal stress is written, using the kinematical epression for given in Eq. (.6), as ; E u () ; v A ()y(s) ; w ()z(s) ; A ()! A (s) (.1) in which the material constant E is E E 1 ; ds)d tds d d)tds qd FIGURE.1 Forces on a wall element The shear stress is determined from the component of the force equilibrium equation for the wall element shown in Figure.1. If it is assumed that there is

34 .4. EQUATIONS OF EQUILIBRIUM 31 no longitudinal applied load on @ where t is the thickness of the wall, and the shear flow q is defined by q( s) s ( s)t(s) The shear flow is found by integrating the equilibrium equation with respect to s q( s) q () ; t(s)ds where q () is the shear ; flow at s. By substituting E u () ; v()y(s) ; w()z(s) ; A A ()! A (s) (.3) into Eq. (.) and writing da t(s)ds for the element of cross-sectional area, the shear flow can be written ; as q( s) q () E v A ()Q (s) w z A ()Q (s) y ()Q (s)! A ; u ()A(s) (.4) where s A(s) s Q y (s) s Q z (s) s Q!A (s) t(s)ds z(s)da y(s)da! A (s)da s da.4. Equations of Equilibrium The equations of equilibrium will be written for a differential element of length d of the beam. It will be assumed that the coordinate aes y, z are centroidal, and that only the principal the warping function! is being used. Let p be applied force per unit length of the beam in the longitudinal direction. The normal stress resultant on the differential element d da ; da and the equilibrium of forces in the direction da p d The first term is evaluated by integrating the epression on the right side of Eq. (.3) over the cross sectional area. The result is which simplifies to ; E u ()A ; v A ()Q z ; w A ()Q y! ()Q p Eu ()A p

35 3 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION Let p y denote the applied force in the y direction per unit length of the beam. The direct shear force in the y direction balances the applied force in cos dsd p yd where is the angle between the y and s aes, and q is the direct shear flow. Since dy y @ ds da where it has been assumed that the edges are free of shear stresses. The last term is evaluated by differentiating the epression on the right side of Eq. (.3) with respect to and integrating the result over the cross sectional area da E ; u ()Q z ; viv A () ; wiv A ()z iv ()! E ;v iv A () ; wiv A ()z Hence, the equation of equilibrium in the y direction becomes E v iv A () EI yz w iv A () p y (.5) Let p z denote the applied force in the z direction per unit length of the beam. The direct shear force in the z direction balances the applied force in sin dsd p d z where is the angle between the y and s aes, and q is the direct shear flow. Since dz z t da where it has been assumed that the edges are free of shear stresses. The last term is evaluated using Eq. da E ; u ()Q y ; viv A ()z ; wiv A () iv ()! E ; ;v iv A ()z ; wiv A () The equation of equilibrium in the z direction is Ez v iv A () EI y w iv A () p (.6) z The total torque at the section is the sum of two parts T T t T! The torque T t is due to the shear stresses resulting from pure, or unrestrained, torsion. It is related to the angle of rotation by T t GJ ()

36 .5. STRESS RESULTANTS 33 The torque T! is called the warping torque. It is due to the shear flow q. For a beam element of length d, equilibrium of the torques about the pole A gives e ddse s GJ d m()d where r A is the vector from the pole A to the point at s, and m() is the applied torsional moment per unit length. Since (r A e s ) e r A (e s e );r A e n the first term in the torque equation can be rewritten as e ddse dr A e n dd! An integration by @s ds so that the torsional equilibrium equation becomes da m() which is brought to its final form using Eq. (.3) t(s)ds ; GJ m() (.7).5. Stress Resultants The stress resultants for the normal stress are the aial force N, the bending moments M y and M z, and the bimoment M!, which are defined by N M y M z ; M! The normal stress is given by Eq. (.1) da z da y da! da ; E u () ; v A ()y(s) ; w ()z(s) ; A ()!(s) The stress resultants are evaluated, recalling that the origin of the coordinates y, z is the centroid, and that! is the principal warping function B@ N M y M z M! 1 CA E 1 1 B@ A ;z ; z CA B@ u v ACA ;I! w A

37 34 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION Hence Eu N () () A Ev A () z M y () M z () ; I yz Ew A () ; M y () z M z () ; I yz E () ;M! () I! The normal stress is found in terms of the stress resultants by using these epressions in Eq. (.1) N A ; z M y M z M y z M z y ; Iyz z M!! ; Iyz (.8) I! In Eq. (.4), let the point s be placed at the free edge so that the shear flow q () is zero, and suppose that there is no longitudinal eternal load p on the beam. Then Eq. (.4) shows that u () is zero, and the shear flow is given by ; q( s) E v A ()Q z (s) w A ()Q y (s) ()Q! (s) (.9) Let the shear stress resultants V y, V z, and T! be defined by V y V z T! The definition of Q z (s) is q( s)ds cos (s) q( s)ds sin (s) q( s)d! s Q z (s) Integration by parts gives Q z dz zq z ; A Similarly, and Q z dy yq z ; A Q z d!!q z ; A y(s)da zdq z ; ydq z ; q( s)dy q( s)dz zyda ;z y da ;!yda ;! (.3)

38 The corresponding integrals for Q y (s) are.6. SHEAR CENTER 35 Q y dy ;z Q y dz ; Q y d! The definition of Q! (s) is Integration by parts gives Q! dz zq! ; A Similarly, and s Q! (s) zdq! ;!(s)da Q! dy ;! Q! d!!q! ; A z!da ;!! da ;I! The stress resultants are evaluated using Eq. (.3) and Eq. (.9) V y ; E ; v A () I w yz V z ; E z v A () I w y A () T! ; EI! () A () Substitution of these results into Eq. (.9) gives the shear flow q( s) ; Q z (s) ; z Q y (s) ; I yz V y ; Q y (s) ; z Q z (s) V ; Iyz z ; Q! (s) T I! (.31)! The total shear stress is found by adding Saint-Venant s torsional stress to the contribution from q( s) T t n q s (.3) J t where n is the coordinate measured from the median line in the normal direction..6. Shear Center A beam is said to be in pure fleure if the angle of twist () is identically zero. The shear center S is defined as the point, in the plane of the cross section, through which the line of action of the transverse shear forces must pass for the beam to be in pure fleure. Suppose that a beam is in pure fleure under the action of transverse shear forces in the y direction only, as shown in Figure.11. The line of action of V y passes through the shear center S. From Figure.11, the moment

39 36 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION y V y S q e s r A e n A z FIGURE.11 Shear center calculation of the shear stresses about point A, which is the principal pole of the warping function!, is calculated by integrating dm A e (r A qdse s )r A (e s e )qds r A e n qds qd! over the cross-sectional area M A q( s)d! T! ; EI! () This moment is equal to the moment of V y about A M A e (r SA V y e y )(z A ; z S )V y which shows that the z coordinates of the shear center and the principal pole are identical. A similar computation with the beam cross section subjected only to V z shows that the y coordinates of A and S are also identical. The shear center and the principal pole are, therefore, the same point..7. Calculation of the Angle of Twist The warping stresses in a thin-walled beam depend on the bimoment M! and the warping torsion T! M! () ; EI! () T! () ; EI! () The torque equilibrium equation.7, solved with the applicable boundary conditions, determines the angle of twist as as function of. With the definition c GJ EI!

40 .7. CALCULATION OF THE ANGLE OF TWIST 37 Eq. (.7) is rewritten in the form d 4 d 4 ; c d d m() EI! (.33) The most common boundary conditions on the angle of twist are those for fied, simple, free or beam supports. At a fied support, no twisting or warping occurs. These kinematical conditions are epressed by where the second condition is obtained by setting equal to zero the warping component, which is proportional to, of the longitudinal displacement u( s). A simple support does not allow twisting and is free of normal stress where the second condition epresses that the bimoment is zero M!! da At a free support there are two statical conditions, one epressing that there is no normal stress, and the other that the total torque is zero. The second of these conditions is T t T! GJ ; EI! Thus, for a free support, the boundary conditions are The general solution of Eq. (.33) is EI! (c ; ) c ; () C 1 C C 3 cosh c C 4 sinh c ; 1 cgj [c( ; ) ; sinh c( ; )]m()d (.34) where C k, 1 6 k 6 4, are the constants of integration, and one end of the beam is assumed to be at. The bimoment is obtained from Eq. (.34) by two differentiations M! () ;JG(C 3 cosh c C 4 sinh c) ; 1 c The warping torque is the derivative of M! () T! () ;cgj (C 3 sinh c C 4 cosh c) ; The pure torsion torque is T t () GJ (C cc 3 sinh c cc 4 cosh c) ; and the total torque is given by T () GJC ; m() sinh c( ; )d (.35) m() cosh c( ; )d (.36) m()[1 ; cosh c( ; )]d (.37) m()d (.38) A concentrated torque is applied at the unsupported end of the cantilever beam shown in Figure.1. For this loading condition, the distributed torque

41 38 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION T L FIGURE.1 Cantilever beam with end torque m() is zero, because there is no eternal torque for the cross sections that lie between the two end sections. The eternal torque is set equal to the total torque at L T (L) GJC T The other boundary condition at L is that the cross section is free of normal stress M! (L) ;GJ (C 3 cosh cl C 4 sinh cl) At the fied end the boundary conditions are () C 1 C 3 () C cc 4 The angle of twist is determined from these boundary conditions () T cgj ; c ; sinh c ; tanh cl(1 ; cosh c) a T L FIGURE.13 Cantilever beam with torque at a If the torque is applied at a point a<las shown in Figure.13, the distributed torque is no longer zero. The concentrated torque of magnitude T can be epressed as a distributed torque in terms of the Dirac delta function m() T ( ; a) The angle of twist is calculated from Eq. (.34) as () C 1 C C 3 cosh c C 4 sinh c ; T [c( ; a) ; sinh c( ; a)]u ( ; a) cgj

42 .8. STRESS ANALYSIS 39 where U denotes the unit step function. The boundary conditions are () C 1 C 3 () C cc 4 T (L) GJC ; T M! (L) ;JG(C 3 cosh cl C 4 sinh cl) ; T c The angle of twist for 6 6 a is L T () c ; sinh c (1; cosh c); sinh c(l ; a) cgj cosh cl and for a 6 6 L sinh c(l ; a) R () L () ; T [c( ; a) ; sinh c( ; a)] cgj ; tanh cl.8. Stress Analysis As a first eample, stresses in a cantilever beam of length L, with its fied end at and its free end at L, will be analyzed. The cross section of the beam, shown in Figure.14, is symmetric with respect to the y ais and is of constant thickness t. The load is a single vertical force of magnitude P applied at the free end of the beam. The point of application of P on the cross section is the lower end of the left flange. The centroid is located by the dimension a The area moments of inertia are t 1 (b3 1 b3 ) a h(b h) (h b 1 b ) tb 1 a tb (h ; a) t 3 (a3 (h ; a) 3 ) z The warping function whose origin and pole are both chosen to be point O can be written from Figure.14 as! O (s 4 );hs 4! O (s 5 )hs 5 The warping function! O is zero on the branches s 1, s, and s 3. To calculate the y coordinate of the shear center using Eq. (.1), its is necessary to evaluate the sectorial product of area!o z(s)! O (s)da b1 The shear center location is then given by Eq. (.1) as y S y O!O y O hb3 1 b 3 1 b3 hs 4 tds 4 thb3 1 1

43 4 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION h y b 1 C S b a P z s 1 s 4 s 3 O s 5 s FIGURE.14 I-beam cross section or y S ;(h ; a) hb3 1 b 3 1 ; h (b ; b)hb b 1 1 (b ; b 1 ) b3 (b 1 b h)(b 3 1 b3 ) which shows that S is to the left of the centroid for b >b 1. The warping constant I! is found from Eq. (.), which for this cross section becomes I! I!O ; (y S ; y O ) b1 h s tds ; h b 6 1 t 4 4 (b 3 1 b3 ) 1 (b3 1 b3 )th 1 b 3 1 b3 b 3 1 b3 The principal warping function, the origin of which can be taken at O, is found by transforming! O according to Eq. (.9)!(s)! O (s) ; (y S ; y O )z(s)