Notes on the Linear Analysis of Thin-walled Beams Walter D. Pilkey and Levent Kiti¸s Department of Mechanical Engineering University of Virginia Charlottesville, Virginia c August 1996 Contents Chapter I. SAINT-VENANT TORSION OF THIN-WALLED BEAMS 1.1. Fundamental Equations of Saint-Venant Torsion 1.2. Saint-Venant’s Warping Function 1.3. Thin-walled Open Sections 1.4. Thin-walled Closed Sections 1 1 6 9 11 Chapter II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION 2.1. Geometry of Deformation 2.2. Properties of the Warping Function 2.3. Stress-Strain Relations 2.4. Equations of Equilibrium 2.5. Stress Resultants 2.6. Shear Center 2.7. Calculation of the Angle of Twist 2.8. Stress Analysis 15 15 18 30 31 33 35 36 39 Chapter III. 3.1. 3.2. 3.3. 3.4. THIN-WALLED ELASTIC BEAMS OF CLOSED CROSS SECTION 55 Geometry of Deformation 55 Equations of Equilibrium 59 A Multicell Analysis Example 62 Cross Sections with Open and Closed Parts 66 i ii CONTENTS 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 x coordinate axis is chosen to lie along the beam axis, 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 x T y C z F IGURE 1.1 Prismatic beam in torsion When a beam is in this state of uniform torque, it is found that only the shear stresses xy and xz are nonzero x = y = z = yz = 0 In the absence of body forces, the equations of equilibrium become @ xy @ xz @y + @z = 0 @ xy @x = 0 @ xz = 0 @x 2 I. SAINT-VENANT TORSION OF THIN-WALLED BEAMS These equations show that the stresses are independent of x, 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 xy and xz are nonzero x = y = z = yz = 0 The nonzero shear stresses are related to the stresses by xy = Gxy xz = Gxz where G is the shear modulus. Only the following two of the six compatibility equations are not trivially satisfied for this state of strain @ @y @ @z @xz @xy ; @y + @z = 0 ; @@zxy + @@yxz = 0 In view of Hooke’s law, these compatibility conditions can be written in terms of the shear stresses as @ @y @ @z ; @@yxz + @@zxy = 0 ; @@yxz + @@zxy = 0 Since neither stress depends on x, the parenthesized quantity in the preceding equation is independent of x, y, and z . Consequently @ xy @ xz @z ; @y = ;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 defined by @ = @z xy @ = ; xz @y Stresses calculated from the stress function satisfy the equations of equilibrium, and Eq. (1.1) becomes @ 2 + @ 2 = ;C @y2 @z 2 (1.2) 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 1.2. The s axis is tangential to the boundary in the direction of increasing s. The positive direction of the normal n to the boundary is chosen to 1.1. FUNDAMENTAL EQUATIONS OF SAINT-VENANT TORSION 3 C y xy n s xz s z F IGURE 1.2 Boundary condition for shear stress make n, s, and x a right-handed orthogonal system of axes. The boundary condition for the shear stress is xn = xy cos + xz sin = 0 where is the angle from the positive y axis to the positive n axis. Since dy = ; sin ds dz = cos ds the boundary condition can be rewritten as dz ; xy ds dy = 0 xz ds which, in terms of the stress function, becomes @ dz + @ dy = @ = 0 @z ds @y ds @s 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 Vy , Vz and the torque T , which are calculated from the shear stress distribution 4 I. SAINT-VENANT TORSION OF THIN-WALLED BEAMS over the cross-sectional area A Vy = Z Z xy dA = Z @ dA = 0 Z@z@ ; @y dA Z= 0 @ @ T = (y xz ; z xy )dA = ; y @y + z @z dA Vz = Z xz 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 T= Z Z y) ; @ (z ) dA = 2 dA 2 ; @ (@y @z Let u, v, and w be the displacements of a point of the cross section in the x, y, and z directions, respectively. The linear strain-displacement relations give @u = @v = @w = 0 @x @y @z (1.3) @v + @w = 0 @z @y (1.4) because all normal strains are zero, and the assumption of zero shear strain yz means that The functional dependence of the displacements on the coordinates x, y, mined by Eq. (1.3) is u = u(y z ) v = v(x z ) z deter- w = w(x 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 xy and xz are independent of x, so are the strains xy and xz . The strain-displacement relations @v xy = @u + @y @x @w xz = @x + @u @z imply that the dependence of v and w on x 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 fixed end, which is not free to undergo the same warping displacement as the other sections. If external torque is applied to the free end of such a beam, normal warping stresses x 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 axis of twist is a 1.1. FUNDAMENTAL EQUATIONS OF SAINT-VENANT TORSION ey 5 P Q x Q 0 ez F IGURE 1.3 Displacement of a point of the cross section line parallel to the beam axis and passes through the point P whose coordinates in the centroidal Cyz system are yP and zP . It will also be assumed that the section at x = 0 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 Q0 by a rotation about P rQ0 P ; rQP = vey + wez where rQP denotes the position vector of Q measured from P , and ey , ez are unit vectors in the y, z directions. For small angles of twist x , the displacement v is calculated as follows v = rQ P ey ; rQP ey = r cos( + x ) ; r cos x = r(cos cos x ; sin sin x ; cos x ) = ;r sin x = ;(z ; zP )x 0 where r is the length of rQP . A similar calculation gives w, and the displacements in the yz plane are determined to be v(x z ) = ;x (x)(z ; zP ) w(x y) = x (x)(y ; yP ) As mentioned earlier, the dependence of v and w on x is linear. Therefore x (x) = Kx for some constant K . (1.5) 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 = ; @zxy + @yxz @ @u @w @ @u @v = ;G @z @y + @x + G @y @z + @x = 2Gx If Eq. (1.2) is solved for C = 2Gx = 1 and the solution is , the solution corresponding to a rate of twist of x is = 2Gx 0 0 0 0 and the torque is given by Z Z T = 2 dA = 4Gx dA 0 The torsional constant J of the beam is defined as Z J = 4 dA The torsional constant, which is obtained by solving Eq. (1.2) 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 = GJx 0 (1.6) 1.2. Saint-Venant’s Warping Function An alternative to the stress-function approach of the preceding section is SaintVenant’s classical solution, which starts from hypotheses about the displacement field. Saint-Venant made the assumption that the cross sections rotate about the axis 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, expressed by Eq. (1.5), was reached by the stress-function approach. If the axis of twist is taken to be the beam axis, Eq. (1.5) becomes v(x z ) = ;x (x)z w(x y) = x (x)y If the rate of change x of the angle of twist is assumed constant, and the end of the beam at x = 0 is assumed to be restrained against rotation, then SaintVenant’s displacement hypothesis about the v and w displacement components 0 can be expressed by v(x z ) = ;x xz w(x y) = x xy (1.7) The axial 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 0 rate of twist 0 u(y z ) = x !(y z ) 0 where !(y z ) is an unknown function called the warping function. (1.8) 1.2. SAINT-VENANT’S WARPING FUNCTION 7 The strain-displacement relations determine the strains from the assumed displacement field x = @u @x = 0 y = @v @y = 0 z = @w @z = 0 @w = ; x + x = 0 + yz = @v x x @z @y @u @v @! xy = @y + @x = x @y ; z @u @w @! xz = @z + @x = x @z + y 0 0 0 0 The stresses are then given by Hooke’s law x = 0 y = 0 z = 0 yz =0 xy = Gx 0 xz = Gx 0 @! ; z @y @! + y @z The ratio of the change in volume to the original volume, called the cubical dilatation, is zero e = x + y + z = 0 and all surface forces are zero, so that the displacement formulation of the equations of elasticity reduces to where r2 is the Laplacian r2u = r2v = r2w = 0 2 2 2 r2 = @x@ 2 + @y@ 2 + @z@ 2 The partial differential equations for the displacement components v and trivially satisfied. The equation for the warping displacement u gives 2 2 r2! = @@y!2 + @@z!2 = 0 w are (1.9) The boundary condition for the shear stresses on the cylindrical surface of the beam, shown in Figure 1.2, is xy cos + xz sin = 0 8 I. SAINT-VENANT TORSION OF THIN-WALLED BEAMS which, in terms of the warping function, becomes @! ; z cos + @! + y sin = 0 @y @z (1.10) Z @! @! T = (y xz ; z xy )dA = Gx @z + y y ; @y ; z z dA Z The torque at any section is 0 The integral in the preceding equation is identified as the torsional constant J J = Iy + Iz + Z @! dA y @! ; z @z @y The area integral can be transformed into a line integral over the boundary by applying Green’s theorem Z @(y!) @(z!) I I ; dA = ; ! ( zdz + ydy ) = ; ! r dr @z @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 I J = Iy + Iz ; !r dr y ' (1.11) C r dr z F IGURE 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 y2 + z 2 = 1 a2 b2 The warping function is known from the theory of elasticity 2 b2 ! = ; aa2 ; + b2 yz 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 2 b2 I !r dr = ; aa2 ; + b2 Z yz (ydy + zdz ) 2 b2 2 2 2 2 2 = ; aa2 ; + b2 0 (ba cos ' sin ')(;a + b )d' 2 22 = ab4((ba2;+ab2)) The area moments of inertia are Iy = 41 ab3 Iz = 14 ba3 and the torsional constant is 2 2 2 3 b3 J = Iy + Iz ; ab4((ba2;+ab2)) = aa 2 + b2 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 approximate value for the torsional t xz C y h z F IGURE 1.5 Narrow rectangular cross section constant J for this section will be obtained by assuming that xy is negligibly small and the shear stress xz varies linearly across the wall thickness xy =0 xz y = 2 max t 10 I. SAINT-VENANT TORSION OF THIN-WALLED BEAMS The equation to be solved is Eq. (1.1), which, with these assumptions, becomes 2Gx = C = @@yxz = 2 max t 0 This determines the differential equation for the stress function ; ddy = xz = 2Gxy 0 When this equation is integrated and the stress function is set equal to zero on the longer edges of the rectangle, the result is 2 (y) = Gx t4 ; y2 0 Z Z t2 2 3 3 3 2 dA = t A ; 2I = t h ; 2 t h = t h ; y z 4 2 2 12 3 and the torsional constant is found to be J = 4 dA = 2 where A is the area and Iz is the area moment of inertia about the z axis. 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 xz = 2Ty 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 xz is one-half the actual torque T . This is partially because the neglected shear stresses xy are concentrated near the shorter edges and have longer moment arms than the stresses xz . s1 s2 C y n s s3 z F IGURE 1.6 Horseshoe section The approximate 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 approximation for the torsional constant 1.4. THIN-WALLED CLOSED SECTIONS is J = 13 Z 11 t3 (s)ds (1.12) 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 xz = 2Tn J (1.13) = T tJmax (1.14) where n is the normal coordinate measured from the median line. The maximum shear stress occurs at the maximum wall thickness tmax max 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. C y xs s n z F IGURE 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 axes n, s, x 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. 12 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 axis to the positive n axis. 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 xs , defined by, q = t(s) xs (s) will be assumed constant. Then, at any s, the derivative of the stress function with respect to n is @ = @ dy + @ dz = ; cos + sin xz xy @n @y dn @z dn q = ; xs = ; t(s) The stress function is then determined by setting its value equal to zero on the outer boundary of the section (n s) = 2q 1 ; t2(sn) The derivatives of the axial displacement are @u = ; @v = @y xy @x @u = ; @w = @z xz @x xy G + x (z ; zP ) xz G ; x (y ; yP ) 0 0 where P is a point on the axis of twist. Thus, on the median line, the derivative of u with respect to s is @u = xs + (z ; z ) dy ; (y ; y ) dz P ds x P ds @s G x Let rP (s) denote the position vector of the point at s measured from the point P rP = (y ; yP )ey + (z ; zP )ez where ey , ez are unit vectors in the positive y, z directions, respectively. The unit tangent vector es at the point with coordinates y, z is @y e + @z e es = @s y @s z 0 0 The unit normal vector en is, therefore, given by en = es @y e ey ; ex = @z @s @s z The projection of the position vector rP onto the unit normal vector is rP en = (y @y ; yP ) @z @s ; (z ; zP ) @s The derivative of the warping displacement with respect to s becomes @u = xs ; r e @s G x P n 0 (1.15) 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 1I G xs ds ; x 0 I rP ends = 0 (1.16) y P rP es rP en s ds en z F IGURE 1.8 Definition of sectorial area In the preceding equation, the integral = I rP ends 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 q I ds ; = 0 G t(s) x 0 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 = ex =q Z Z rP xs tdses = q rP en ds = q = This equation gives the shear stress xs(s) = and the torsional constant Z rP Gx 2 H ds t(s) (es ex)ds 0 T t(s) T = 2 J = G x H ds t(s) 0 For constant wall thickness, the torsional constant becomes J = S t 2 where S denotes the length of the median line. (1.17) (1.18) CHAPTER II THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION 2.1. Geometry of Deformation Figure 2.1 shows a prismatic thin-walled beam and its cross section. The beam axis, which is defined as the line of centroids of the cross sections, is chosen to lie along the x axis. 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. x C y s z F IGURE 2.1 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 A0 , B 0 be the projections of these new positions onto the yz plane, as shown in Figure 2.2. Let rBA be the position vector of point B measured from point A. The vector rB0 A0 is given by rB0 A0 = rBA cos x + ex rBA sin x (2.1) where ex is the unit vector in the direction of the positive x axis, and x is the angle measured from the vector rAB to the vector rA0 B0 , 6 x 6 , with the vector ; 16 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION rAB translated such that the points A and A 0 become coincident. The rotation is counterclockwise, if the cross product rBA rB0 A0 , evaluated according to the right-hand rule, has the same direction as ex. Since rBA rB A = rBA (ex rBA) sin x = jrBAj2 sin x ex 0 0 the rotation is counterclockwise for sin x > 0. This establishes that the sign of x is positive when the sense of rotation from AB to A 0 B 0 is counterclockwise. For small rotations, Eq. (2.1) becomes rB0 A0 = rBA + x ex rBA (2.2) Let vA and wA be the displacement components of point A along the y and z axes, and let vB and wB be the corresponding displacement components of point B . Let the transverse displacement vector be denoted by uA for point A and by uB for point B uA = rA0 A uB = rB B 0 From the vector polygon A ABB in Figure 2.2 0 rB0 A0 0 ; rBA = rB B ; rA A = uB ; uA = (vB ; vA)ey + (wB ; wA)ez 0 0 and Eq. (2.2) can be written in terms of displacement as uB = uA + x ex rBA (2.3) The scalar components of this equation are vB = vA + (zB ; zA )x wB = wA ; (yB ; yA )x (2.4) y A rBA ex B rBA B 0 rB0 A0 A 0 x z F IGURE 2.2 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 2.1. GEOMETRY OF DEFORMATION 17 line, but which is assumed to be displaced as if it were rigidly attached to the median line. Let es be a unit vector tangent to the median line in the direction of inreasing s as shown in Figure 2.3. The unit normal vector en is defined so as to make the triad en, es , and ex a right-handed set of orthogonal vectors ex = en es en = es ex 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. (2.3) (s) = uA es + x es (ex rA(s)) = uA es + x (es ex ) rA (s) where rA (s) is the position vector of the point at s measured from A. In terms of the angle between the s and y axes, this equation becomes (s) = vA cos + wA sin + x rA en n Let rA denote the projection of the vector rA onto the unit normal rAn = rA en The tangential component of the displacement of the point at s is given by (x s) = vA (x) cos (s) + wA(x) sin (s) + x (x)rAn (s) (2.5) y A rA (s) n s s z F IGURE 2.3 Tangential and normal components of displacement It will now be assumed that the shear strain xs 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 written as @ xs = @u @s + @x = 0 18 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION where u is the displacement of the point at s along the x axis. This leads to @u = ; @ = ;v (x) cos (s) ; w (x) sin (s) ; (x)rn (s) A A x A @s @x in which prime denotes differentiation with respect to x. Integration with respect to s gives 0 0 Zs 0 Zs u(x s) = ;vA (x) cos (s)ds ; wA (x) sin (s)ds ; x (x) 0 0 = ;vA (x)y(s) ; wA (x)z (s) ; x (x)!A (s) + u0 (x) 0 0 0 0 0 Zs 0 0 where !A (s) = Zs 0 rAn (s)ds = Zs 0 rAn (s)ds rA (s) ends is the sectorial area and u0 (x) is the longitudinal displacement of the point of the median line at s = 0. The longitudinal strain is calculated by differentiating the longitudinal displacement with respect to x @u = (x s) = u (x) ; v (x)y(s) ; w (x)z (s) ; (x)! (s) 0 A A x A @x x 0 00 00 00 (2.6) 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 expressed 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. 2.2. Properties of the Warping Function The warping function !A , with pole at point A and origin at s = s0 , is defined as the integral !A (s) = Zs r (s) en(s)ds s0 A where rA(s) is the position vector of the point at s of the median line measured from the pole A, as shown in Figure 2.4. The direction of en (s) is determined from the convention that the axes (n s x) are right-handed, with the s axis in the direction of increasing s. The magnitude of the differential quantity rA (s) en (s)ds is twice the area of the triangle with base ds and height rA(s) en (s). The sign of this quantity is positive if the projection of the position vector onto the unit normal vector en(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 (es ex) = (rA es) ex Thus, when the rotation of the vector rA is counterclockwise as its tip moves in the direction of es, the cross product rA es is in the same direction as ex , making rA en positive. rA en = rA 2.2. PROPERTIES OF THE WARPING FUNCTION 19 y A rA es rA en s ds en z F IGURE 2.4 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 = s0 and the origin for !B at s = s1 , as shown in Figure 2.5. The relationship between !A and !B is found by the computation !A (s) = = Zs rB ends = Zs (rB + rBA ) en ds s 0 Zs Zs rB ends + rB ends + r e ds s0 s1 Z s0 BA n s !B (s0 ) + !B (s) + rBA (sin ey cos ez )ds Zs0s !B (s0 ) + !B (s) + rBA (dz ey dyez ) s Zs0s1 =; ; =; ; 0 = ;!B (s0 ) + !B (s) + (yB ; yA )(z (s) ; z0 ) ; (zB ; zA )(y(s) ; y0 ) 20 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION y B A rBA s1 rB rA s s0 z F IGURE 2.5 Poles and origins of warping functions In this calculation, is the angle between the s and y axes as shown in Figure 2.3, and y0 , z0 are the coordinates of the origin chosen for !A y0 = y(s0 ) z 0 = z (s 0 ) The equation for finding the warping function !A with origin s0 from the the warping function !B with origin s1 is, therefore, !A (s) = !B (s) ; !B (s0 ) + (zA ; zB )(y(s) ; y0 ) ; (yA ; yB )(z (s) ; z0 ) (2.7) When A and B are coincident points, but s0 and s1 are two distinct origins, the transformation equation becomes !A (s) = !B (s) ; !B (s0 ) (2.8) 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 s0 , s1 are the same and the poles A, B are different, the transformation equation is !A (s) = !B (s) + (zA ; zB )(y(s) ; y0 ) ; (yA ; yB )(z (s) ; z0 ) since, in this case, !B (s0 ) = !B (s1 ) = 0 (2.9) 2.2. PROPERTIES OF THE WARPING FUNCTION 21 If !(s) is a warping function for a particular pole and origin, the area integral Z Q! = !(s)dA is called the first sectorial moment. The area integrals Z Iy! = y(s)!(s)dA Z Iz! = 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 Qy = Qz = Iy = Iz = Iyz = Z zdA Z Z Z Z ydA z 2dA y2 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 s0 and s1 , respectively. By multiplying both sides of Eq. (2.7) by y and integrating both sides of the result over the cross sectional area, one obtains Iy!A = Iy!B ; !B (s0 )Qz + (zA ; zB )(Iz ; y0 Qz ) ; (yA ; yB )(Iyz ; z0 Qz ) Since the origin of the y, z axes is the centroid C of the cross section, the first moments Qy and Qz are both zero, so that Iy!A = Iy!B + (zA ; zB )Iz ; (yA ; yB )Iyz (2.10) A similar calculation gives Iz!A = Iz!B + (zA ; zB )Iyz ; (yA ; yB )Iy The conditions for A to be a principal pole Iy!A = Iz!A = 0 (2.11) are solved for the coordinates of the pole I I ; I B Iyz yA = yB + z!BI zI ; y! 2 y z Iyz I I ; Iy!B Iy zA = zB + z!BI yz I ; I2 y z yz (2.12) (2.13) These expressions 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 22 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION constant value, say K , and Iy!B = Z y(!B + K )dA = Iy!B + KQz = Iy!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. (2.12) and (2.13) are independent of this origin. It is important to realize that the coordinates given by Eqs. (2.12) and (2.13) 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 Iy!B = Iy!D + (zB ; zD )Iz ; (yB ; yD )Iyz Iz!B = Iz!D + (zB ; zD )Iyz ; (yB ; yD )Iy Then 2) Iz!B Iz ; Iy!B Iyz = Iz!D Iz ; Iy!D Iyz ; (yB ; yD )(Iy Iz ; Iyz which, when substituted into Eq. (2.12), gives I I ; I D Iyz yA = yD + z!DI zI ; y! 2 y z Iyz The right side of this equation is the expression 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 s0 such that Q!A = Z !A (s)dA = 0 the point s0 is termed a principal origin. To determine s0 , let B be a pole coincident with A but with a known origin s1 . According to Eq. (2.8) !A (s) = !B (s) ; !B (s0 ) so that the condition for s0 to be a principal origin is Q!A = Q!B ; !B (s0 )A = 0 (2.14) This equation determines the principal origin s0 in terms of the arbitrarily selected origin s1 . The existence of at least one s0 is ensured, for most cross sectional shapes, by the mean value theorem for integrals. There may be multiple solutions for s0 , 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 s2 , is used instead of B in determining the principal origin s0 , then Q!D = Q!B ; !B (s2 )A and the condition for s0 to be a principal origin, written in terms of D, becomes Q!A = Q!D ; !D (s0 )A = Q!B ; !B (s2 )A ; (!B (s0 ) ; !B (s2 ))A = Q!B ; !B (s0 )A = 0 This shows that the same principal origin is obtained regardless of where the reference origin s1 is placed. Hence, the principal origin is a cross-sectional property. 2.2. PROPERTIES OF THE WARPING FUNCTION 23 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 s0 such that Q!A , Iy!A , and Iz!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 s2 tw h s1 2 y O C O n3 h z n2 n1 s3 2 s4 tf n4 F IGURE 2.6 Symmetric channel section For the symmetric channel section shown in Figure 2.6, the warping function with pole and origin both at the point of intersection O of the y axis and the median line is given by !O (s1 ) = 0 !O (s2 ) = ; h2 s2 !O (s3 ) = 0 ! (s ) = h s O 4 2 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 (s2 ) is negative, because the position vector rotates clockwise as it traces the median line of the upper flange. 24 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION Let A be the principal pole of the cross section shown in Figure 2.6. The coordinates of A will be found by using O both as a reference pole and as the sectorial origin. Since z0 = 0 and Iyz = 0, the coordinates of A are given, according to Eqs. (2.12) and (2.13), by I O yA = yO + z! Iy I O zA = ; y! Iz The sectorial product of area Iy!O is zero by symmetry, and the sectorial product of area Iz!O is Z Iz!O = z (s)!O (s)dA = Zb h h Zb b2h2 t (; 2 )(; 2 s2 )tf ds2 + ( h2 )( h2 s4 )tf ds4 = 4 f 0 0 The area moment of inertia Iy for this section is Iy = 6bh2 tf + h3tw 12 The coordinates of the principal pole A are found to be 3b2t yA = yO + 6bt + fht f w zA = 0 The principal origin s0 for the principal pole A of the symmetric channel sec- tion is determined from the condition Q!A ; !A (s0 )A = 0 where !A has the arbitrarily selected origin O and is given by !A (s1 ) = (yA ; yO )s1 !A (s2 ) = (yA ; yO ) h2 ; h2 s2 !A (s3 ) = ;(yA ; yO )s3 ! (s ) = ;(y ; y ) h + h s A 4 A O 2 2 4 The first moment of sectorial area with pole A and origin at O happens to be zero by symmetry Q!A = Z !A (s)dA = 0 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 Z I! = !2 (s)dA 2.2. PROPERTIES OF THE WARPING FUNCTION 25 Because the section shown in Figure 2.6 is symmetric with respect to the y axis, the warping constant is calculated as follows I! = 2 Z h=2 0 )2 t !A (s1 w ds1 + 2 Zb 0 b3 h2tf (3btf + 2htw ) 12(6btf + htw ) !A (s2 )2 tf ds2 = b1 s2 t h C y e d s1 O s3 b2 z F IGURE 2.7 Unsymmetric channel section For the unsymmetric channel section shown in Figure 2.7 with the dimensions b1 = b b2 = 2b h = 2b 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 2 5 The area moments of inertia and the area product of inertia are Iy = 5215tb 3 Iz = 7tb4 3 Iyz = ;tb3 If the point O is used both as the pole and the sectorial origin, the warping function is !O (s1 ) = 0 !O (s2 ) = ;hs !O (s3 ) = 0 26 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION To find the principal pole, the values of the sectorial products of area are needed Z Zb Zb 4 tbs2 (2s2 ; b)ds2 = tb6 Z Z0b Z0 b 12tb2s 4 2 ds = 6tb Iz!O = z!O dA = ;(h ; e)!O (s2 )tds2 = 2 5 5 Iy!O = y!O dA = (d ; s2 )!O (s2 )tds2 = 0 0 The principal pole A has the coordinates I I ; I O Iyz 18b yA = yO + z!OI zI ; y! = 19 2 y z Iyz I I ; Iy!O Iy 128b zA = zO + z!OI yz = 285 2 y Iz ; Iyz The warping function with principal pole A and origin O is ! (s ) = (y ; y )s = 17b s A 1 A O 1 38 1 !A (s2 ) = (yA ; yO )h ; (h ; e + zA )s2 = 1719b ! (s ) = (e ; z )s = 20b s A 3 57 A 3 Zh Zb The first sectorial area moment is 2 ; 9457b s2 3 Zb !A(s1 )tds1 + !A (s2 )tds2 + !A (s3 )tds3 = 5tb3 0 0 0 The condition for s0 to be a principal origin is 3 Q!A ; !A (s0 )A = 5tb3 ; 5tb!A(s0 ) = 0 Q!A = 1 or 3 2 2 !A(s0 ) = b3 According to Eq. (2.8), the shift of the origin to s0 gives the principal warping function 2 !(s) = !A (s) ; !A (s0 ) = !A (s) ; b3 which, when written out for the web and the flanges, yields 2 b s ; b2 !(s1 ) = !A(s1 ) ; b3 = 17 38 1 3 2 b s ; 32b2 !(s2 ) = !A(s2 ) ; b3 = ; 94 57 2 57 2 2 !(s ) = ! (s ) ; b = 20b s ; b 3 A 3 3 57 3 3 The principal warping function !(s) shown sketched to scale in Figure 2.8. 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 s0 . If a cross section has a symmetry axis, say the y axis, this is a principal axis, so that Iyz = 0. In calculating the coordinates of the principal pole A, using Eqs. (2.12) and (2.13), the choice of the reference pole B is arbitrary. When B is chosen to 2.2. PROPERTIES OF THE WARPING FUNCTION 27 32b2 57 + 32b2 57 + ; ; 6257b 2 7b2 19 ; b3 2 ; + ; ; b3 2 F IGURE 2.8 Warping function for the unsymmetric channel section be any point on the symmetry axis y, it is readily verified that Iy!B coordinate zA is then = 0. The I I ; Iy!B Iy =0 zA = zB + z!BI yz 2 y Iz ; Iyz This shows that the principal pole lies on the symmetry axis. In addition, the point of intersection of the y axis with the median line is a principal sectorial origin. For a doubly symmetric cross section, the point of intersection of the two axes of symmetry is the principal pole, the principal origin, and the centroid. 28 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. (2.7) !A = !B ; !B (s0 ) + (zA ; zB )(y ; y0 ) ; (yA ; yB )(z ; z0 ) (2.15) Since the origin s0 for !A is a principal origin, the area integral of Eq. (2.15) is 0 = Q!A = Q!B ; !B (s0 )A ; y0 (zA ; zB )A + z0 (yA ; yB )A (2.16) The first moments of area Qy and Qz have been set equal to zero in the calculation of the right side of this equation because y and z are centroidal axes. The result of Eq. (2.16) allows Eq. (2.15) to be rewritten as Q !A = !B ; A!B + (zA ; zB )y ; (yA ; yB )z The conditions for A to be a principal pole are then Iy!A = Iy!B + (zA ; zB )Iz ; (yA ; yB )Iyz = 0 Iz!A = Iz!B + (zA ; zB )Iyz ; (yA ; yB )Iy = 0 (2.17) (2.18) (2.19) The warping constant is given by Q2 I! = I!B + 2(zA ; zB )Iy!B ; 2(yA ; yB )Iz!B ; A!B + (yA ; yB )2 Iy + (zA ; zB )2Iz ; 2(yA ; yB )(zA ; zB )Iyz which is simplified by using Eqs. (2.18) and (2.19) to Q2 I! = I!B ; A!B ; (yA ; yB )2Iy + 2(yA ; yB )(zA ; zB )Iyz ; (zA ; zB )2Iz (2.20) The channel section with double flanges shown in Figure 2.9 has uniform thickness t, and it is symmetric with respect to the y axis. The point O, which is the point of intersection of the median line and the y axis, is chosen as a convenient pole and origin for the warping function !O (s1 ) = 0 h !O (s2 ) = ; 21 s2 ! (s ) = ; h2 s 2 O 3 3 0 6 s1 6 h22 0 6 s2 6 b 0 6 s3 6 b The area moment of inertia Iy is calculated using the centerline dimensions 3 tbh2 tbh2 2 1 2 Iy = th 12 + 2 + 2 and the sectorial product of area Iz!O is found, taking advantage of symmetry, Zb h Zb h Iz!O = 2 ; 21 !O (s1 )tds1 + 2 ; 22 !O (s1 )tds1 0 0 2 tb 2 2 = (h + h ) 4 1 2 2.2. PROPERTIES OF THE WARPING FUNCTION 29 b s3 s1 y C A h1 h2 s2 O z F IGURE 2.9 Channel section with double flanges The principal pole A is on the symmetry axis y I O 3b2(h21 + h22) = y + yA = yO + z! O h2 + 6b(h2 + h2 ) Iy 2 1 2 The warping constant is given by Eq. (2.20) I! = I!O ; (yA ; yO )2 Iy = tb3 (h21 + h22 )(2h32 + 3b(h21 + h22 )) 12(h32 + 6b(h21 + h22)) Since point O is the principal origin, the principal warping function can be written as !(s1 ) = (yA ; yO )s1 !(s ) = (y ; y ) h1 ; h1 s 2 A O 2 2 2 h h !(s3 ) = (yA ; yO ) 22 ; 22 s3 from which the value found for I! from Eq. (2.20) can be verified by evaluating I! = 2 Z h =2 2 0 !(s1 )2 tds 1+2 Zb 0 !(s2 )2 tds 2+2 Zb 0 !(s3 )2 tds3 30 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION 2.3. Stress-Strain Relations The only significant stresses will be assumed to be the normal stress x and the shear stress xs . Although the strain xs was taken to be negligible in Section 2.1, the corresponding stress xs 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 inextensible, the strain s is zero s = E1 (s ; x) = 0 where E is the modulus of elasticity and is Poisson’s ratio. strain is The longitudinal x = E1 (x ; s) = 1 ;E x 2 The normal stress x is written, using the kinematical expression for x given in Eq. (2.6), as ; x = E u0(x) ; vA (x)y(s) ; wA (x)z (s) ; x (x)!A (s) is in which the material constant E E = 1 ;E 2 0 00 00 00 (2.21) @q ds)dx (q + @s x tds ds dx x (x + @ @x dx)tds qdx F IGURE 2.10 Forces on a wall element The shear stress is determined from the x component of the force equilibrium equation for the wall element shown in Figure 2.10. If it is assumed that there is 2.4. EQUATIONS OF EQUILIBRIUM 31 no longitudinal applied load on the surface @q + t(s) @x = 0 @s @x where t is the thickness of the wall, and the shear flow q is defined by q(x s) = xs (x s)t(s) to s The shear flow is found by integrating the equilibrium equation with respect q(x s) = q0(x) ; Z s @ x @x t(s)ds 0 (2.22) where q0 (x) is the shear flow at s = 0. By substituting the expression @x ; (2.23) = E u ( x ) ; v ( x ) y ( s ) ; w ( x ) z ( s ) ; ( x ) ! ( s ) 0 A A x A @x into Eq. (2.22) and writing dA = t(s)ds for the element of cross-sectional area, the 00 ; 000 000 000 shear flow can be written as q(x s) = q0(x) + E vA (x)Qz (s) + wA (x)Qy (s) + x (x)Q!A (s) ; u0 (x)A(s) 000 000 000 00 (2.24) where A(s) = Qy (s) = Qz (s) = Q!A (s) = Zs Z0s Z0s Zs 0 0 t(s)ds = Zs z (s)dA 0 dA y(s)dA !A (s)dA 2.4. Equations of Equilibrium The equations of equilibrium will be written for a differential element of length dx of the beam. It will be assumed that the coordinate axes y, z are centroidal, and that only the principal the warping function ! is being used. Let px be applied force per unit length of the beam in the longitudinal direction. The normal stress resultant on the differential element is Z; Z Z x dxdA ; dA = dx @x dA x + @ x @x @x and the equilibrium of forces in the x direction gives Z @ dx @xx dA + px dx = 0 The first term is evaluated by integrating the expression on the right side of Eq. (2.23) over the cross sectional area. The result is ; E u0 (x)A ; vA (x)Qz ; wA (x)Qy + x (x)Q! + px = 0 00 which simplifies to 000 000 0 (x)A + px = 0 Eu 00 000 32 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION Let py 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 this direction Z @q @x cos dsdx + py dx = 0 where is the angle between the y and s axes, and q is the direct shear flow. Since dy = ds cos , Z @q @q @x cos ds = @x y Z @ @q Z @ @ Z @2 ; y @s @x ds = y @x t @xx ds = y @x2x dA edges where it has been assumed that the edges are free of shear stresses. The last term is evaluated by differentiating the expression on the right side of Eq. (2.23) with respect to x and integrating the result over the cross sectional area Z @2 ; y @x2x dA = E u0 (x)Qz ; vAiv (x)Iz ; wAiv (x)Iyz + xiv (x)Iy! ; = E ;viv (x)I ; wiv (x)I 000 A z A yz Hence, the equation of equilibrium in the y direction becomes z vAiv (x) + EI yz wAiv (x) = py EI (2.25) Let pz 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 this direction Z @q @x sin dsdx + pz dx = 0 where is the angle between the y and s axes, and q is the direct shear flow. Since dz = ds sin , Z @q @q z ; Z z @ @q ds = Z z @ ;t @x ds = Z z @ 2 x dA sin ds = @x @x edges @s @x @x @x @x2 where it has been assumed that the edges are free of shear stresses. The last term is evaluated using Eq. (2.23) Z @2 ; z @x2x dA = E u0 (x)Qy ; vAiv (x)Iyz ; wAiv (x)Iy + xiv (x)Iz! ; = E ;viv (x)I ; wiv (x)I 000 A yz A The equation of equilibrium in the z direction is y yz vAiv (x) + EI y wAiv (x) = pz EI (2.26) The total torque at the section is the sum of two parts T = Tt + T! The torque Tt is due to the shear stresses resulting from pure, or unrestrained, torsion. It is related to the angle x of rotation by Tt = GJx (x) 0 2.5. 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 dx, equilibrium of the torques about the pole A gives ex Z rA @q dxdse + GJ dx + m(x)dx = 0 @x s x 00 where rA is the vector from the pole A to the point at s, and torsional moment per unit length. Since m(x) is the applied (rA es ) ex = rA (es ex ) = ;rA en the first term in the torque equation can be rewritten as ex Z Z @q Z @q @q rA @x dxdses = ; @x dxrA en ds = ; @x dxd! An integration by parts gives Z @q @q @x d! = ! @x edges ; Z @ @q Z @2 ! @s @x ds = ! @x2x t(s)ds so that the torsional equilibrium equation becomes ; Z @2 ! @x2x dA + GJx + m(x) = 0 00 which is brought to its final form using Eq. (2.23) ! xiv ; GJx = m(x) EI 00 (2.27) 2.5. Stress Resultants The stress resultants for the normal stress x are the axial force N , the bending moments My and Mz , and the bimoment M! , which are defined by N= My = Z Z xdA zx dA Z Mz = ; yx dA M! = Z !x dA The normal stress is given by Eq. (2.21) ; x = E u0(x) ; vA (x)y(s) ; wA (x)z (s) ; x (x)!(s) 0 00 00 00 The stress resultants are evaluated, recalling that the origin of the coordinates y, z is the centroid, and that ! is the principal warping function 0 N 1 0A 0 0 0 1 0 u 1 BBMy CC = E BB 0 ;Iyz ;Iy 0 CC BB vA0 CC @ Mz A @ 0 Iz Iyz 0 A @wAA 0 00 M! 0 0 0 ;I! 00 x 00 34 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION Hence 0 (x) = N (x) Eu A I A (x) = yz My (x) + Iy2Mz (x) Ev Iy Iz ; Iyz A (x) = ; Iz My (x) + Iyz2Mz (x) Ew Iy Iz ; Iyz x (x) = ; M! (x) E I! 0 00 00 00 The normal stress is found in terms of the stress resultants by using these expressions in Eq. (2.21) Iyz My + Iy Mz Iz My + Iyz Mz M! ! ; x = N (2.28) 2 y + I I ; I2 z + I A Iy Iz ; Iyz y z yz ! In Eq. (2.24), let the point s = 0 be placed at the free edge so that the shear flow q0(x) is zero, and suppose that there is no longitudinal external load px on the beam. Then Eq. (2.4) shows that u0 (x) is zero, and the shear flow is given by ; q(x s) = E vA (x)Qz (s) + wA (x)Qy (s) + x (x)Q! (s) (2.29) Let the shear stress resultants V y , Vz , and T! be defined by 00 000 Vy = Vz = T! = Z Z Z 000 q(x s)ds cos (s) = q(x s)ds sin (s) = Qz (s) = Z Similarly, and Z Z Z q(x s)dy q(x s)dz q(x s)d! The definition of Qz (s) is Integration by parts gives 000 Z Qz dz = zQz ; A Zs 0 y(s)dA Z zdQz = ; zydA = ;Iyz Z Z Qz dy = yQz ; ydQz = ; y2 dA = ;Iz A Z Z ; !ydA = ;Iy! = 0 Qz d! = !Qz A (2.30) 2.6. SHEAR CENTER 35 The corresponding integrals for Qy (s) are Z Z Z Qy dy = ;Iyz Qy dz = ;Iy Qy d! = 0 The definition of Q! (s) is Zs Q! (s) = Integration by parts gives Z 0 !(s)dA Z Z ; zdQ! = ; z!dA = ;Iz! = 0 Q! dz = zQ! A Similarly, and Z Z Q! dy = ;Iy! = 0 Z Q! d! = !Q! ; !2 dA = ;I! A The stress resultants are evaluated using Eq. (2.30) and Eq. (2.29) ; Vy = ;E Iz vA (x) + Iyz wA (x) ; Vz = ;E Iyz vA (x) + Iy wA (x) ! x (x) T! = ;EI 000 000 000 000 000 Substitution of these results into Eq. (2.29) gives the shear flow I Q (s) ; I Qy (s) Iz Qy (s) ; Iyz Qz (s) Q! (s) q(x s) = ; y zI I ; yz V ; V ; y z 2 2 Iy Iz ; Iyz I! T! y z Iyz (2.31) The total shear stress is found by adding Saint-Venant’s torsional stress to the contribution from q(x s) xs = 2Tt n + q J t (2.32) where n is the coordinate measured from the median line in the normal direction. 2.6. Shear Center A beam is said to be in pure flexure if the angle of twist x (x) 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 flexure. Suppose that a beam is in pure flexure under the action of transverse shear forces in the y direction only, as shown in Figure 2.11. The line of action of Vy passes through the shear center S . From Figure 2.11, the moment 36 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION y Vy S q rA es en A z F IGURE 2.11 Shear center calculation of the shear stresses about point A, which is the principal pole of the warping function !, is calculated by integrating dMA = ex (rA qdses) = rA (es ex )qds = rA enqds = qd! Z over the cross-sectional area MA = ! x (x) = 0 q(x s)d! = T! = ;EI 000 This moment is equal to the moment of V y about A MA = ex (rSA Vy ey ) = (zA ; zS )Vy = 0 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 Vz shows that the y coordinates of A and S are also identical. The shear center and the principal pole are, therefore, the same point. 2.7. Calculation of the Angle of Twist The warping stresses in a thin-walled beam depend on the bimoment M! and the warping torsion T! ! x (x) M! (x) = ;EI ! x (x) T! (x) = ;EI 00 000 The torque equilibrium equation 2.27, solved with the applicable boundary conditions, determines the angle of twist as as function of x. With the definition GJ c2 = EI ! 2.7. CALCULATION OF THE ANGLE OF TWIST 37 Eq. (2.27) is rewritten in the form d4x 2 d2x m(x) ! dx4 ; c dx2 = EI (2.33) The most common boundary conditions on the angle of twist are those for fixed, simple, free or beam supports. At a fixed support, no twisting or warping occurs. These kinematical conditions are expressed by x = 0 x = 0 0 where the second condition is obtained by setting equal to zero the warping component, which is proportional to x0 , of the longitudinal displacement u(x s). A simple support does not allow twisting and is free of normal stress x = 0 Z x = 0 00 where the second condition expresses that the bimoment is zero M! = !x dA = 0 At a free support there are two statical conditions, one expressing that there is no normal stress, and the other that the total torque is zero. The second of these conditions is ! x = EI ! (c2 x ; x ) = 0 Tt + T! = GJx ; EI 0 000 0 000 Thus, for a free support, the boundary conditions are x = 0 00 c2x ; x = 0 0 000 Zx 1 x (x) = C1 + C2x + C3 cosh cx + C4 sinh cx ; cGJ c(x ; ) ; sinh c(x ; )]m( )d The general solution of Eq. (2.33) is 0 (2.34) where Ck , 1 6 k 6 4, are the constants of integration, and one end of the beam is assumed to be at x = 0. The bimoment is obtained from Eq. (2.34) by two differentiations M! (x) = ;JG(C3 cosh cx + C4 sinh cx) ; 1c The warping torque is the derivative of M! (x) T! (x) = ;cGJ (C3 sinh cx + C4 cosh cx) ; Zx 0 Zx 0 The pure torsion torque is Tt (x) = GJ (C2 + cC3 sinh cx + cC4 cosh cx) ; and the total torque is given by T (x) = GJC2 ; Zx 0 m( ) sinh c(x ; )d m( ) cosh c(x ; )d Zx 0 (2.35) (2.36) m( )1 ; cosh c(x ; )]d m( )d (2.37) (2.38) A concentrated torque is applied at the unsupported end of the cantilever beam shown in Figure 2.12. For this loading condition, the distributed torque 38 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION T0 x L F IGURE 2.12 Cantilever beam with end torque m(x) is zero, because there is no external torque for the cross sections that lie between the two end sections. The external torque is set equal to the total torque at x=L T (L) = GJC2 = T0 The other boundary condition at x = L is that the cross section is free of normal stress M! (L) = ;GJ (C3 cosh cL + C4 sinh cL) = 0 At the fixed end the boundary conditions are x (0) = C1 + C3 = 0 x (0) = C2 + cC4 = 0 0 The angle of twist is determined from these boundary conditions T0 ; x (x) = cGJ cx ; sinh cx ; tanh cL(1 ; cosh cx) a T0 x L F IGURE 2.13 Cantilever beam with torque at x = a If the torque is applied at a point x = a < L as shown in Figure 2.13, the distributed torque is no longer zero. The concentrated torque of magnitude T0 can be expressed as a distributed torque in terms of the Dirac delta function m(x) = T0 (x ; a) The angle of twist is calculated from Eq. (2.34) as T0 x (x) = C1 + C2x + C3 cosh cx + C4 sinh cx ; cGJ c(x ; a) ; sinh c(x ; a)]U (x ; a) 2.8. STRESS ANALYSIS 39 where U denotes the unit step function. The boundary conditions are x (0) = C1 + C3 = 0 x (0) = C2 + cC4 = 0 T (L) = GJC2 ; T0 = 0 0 T M! (L) = ;JG(C3 cosh cL + C4 sinh cL) ; c0 sinh c(L ; a) = 0 The angle of twist for 0 6 x 6 a is ; T0 sinh c ( L ; a ) L x (x) = cGJ cx ; sinh cx + (1 ; cosh cx) cosh cL ; tanh cL and for a 6 x 6 L T0 c(x ; a) ; sinh c(x ; a)] xR (x) = xL (x) ; cGJ 2.8. Stress Analysis As a first example, stresses in a cantilever beam of length L, with its fixed end at x = 0 and its free end at x = L, will be analyzed. The cross section of the beam, shown in Figure 2.14, is symmetric with respect to the y axis 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 h(2b + h) a = 2(h + 2b + b ) 1 2 The area moments of inertia are Iy = 12t (b31 + b32) Iz = tb1 a2 + tb2(h ; a)2 + 3t (a3 + (h ; a)3) Iyz = 0 The warping function whose origin and pole are both chosen to be point O can be written from Figure 2.14 as !O (s4 ) = ;hs4 !O (s5 ) = hs5 The warping function !O is zero on the branches s1 , s2 , and s3 . To calculate the y coordinate of the shear center using Eq. (2.12), its is necessary to evaluate the sectorial product of area Iz!O = Z z (s)!O (s)dA = 2 Z b =2 1 0 thb3 hs24 tds4 = 121 The shear center location is then given by Eq. (2.12) as I O hb31 = y + yS = yO + z! O b3 + b3 Iy 1 2 40 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION h b1 C y S b2 a z P s1 s4 O s3 s5 s2 F IGURE 2.14 I-beam cross section or 2 b2 ; b2) + 2hb b (b2 ; b2 ) hb3 2 1 1 2 2 1 yS = ;(h ; a) + b3 +1b3 = ; h (2( b + b + h)(b3 + b3 ) 1 2 1 2 1 2 which shows that S is to the left of the centroid for b 2 > b1. The warping constant I! is found from Eq. (2.20), which for this cross section becomes I! = I!O ; (yS ; yO y = 2 )2 I Z b =2 1 0 2 b3b3 h2 b6 1 2 h2s24 tds4 ; (b3 + b13)2 12t (b31 + b32) = th 12 b3 + b3 1 2 1 2 The principal warping function, the origin of which can be taken at O, is found by transforming !O according to Eq. (2.9) !(s) = !O (s) ; (yS ; yO )z (s) 2.8. STRESS ANALYSIS In terms of the branch coordinates sk , principal warping function is 41 1 6 k 6 5, defined in Figure 2.14, the !(s1 ) = ;(yS ; yO )z (s1 ) = (yS ; yO )s1 !(s2 ) = ;(yS ; yO )z (s2 ) = ;(yS ; yO )s2 !(s3 ) = 0 !(s4 ) = ;hs4 ; (yS ; yO )z (s4 ) = ;(h ; yS + yO )s4 !(s5 ) = hs5 ; (yS ; yO )z (s5 ) = (h ; yS + yO )s5 The applied force P at the free end of the beam does not pass through the shear center. The force-couple equivalent of P at the shear center S is the force P and the torsional moment T0 of P about S Phb3 T0 = (a + jyS j)P = b3 + b23 1 2 The angle of twist of the beam is, therefore, determined as for the beam of Figure 2.12 T0 ; x (x) = cGJ cx ; sinh cx ; tanh cL(1 ; cosh cx) For the torsional constant J , Saint-Venant’s approximation can be used 3 J = t3 (h + b1 + b2) The constant c depends on material constants and cross-sectional dimensions GJ = Gt3(h + b1 + b2) (1 + b32 ) c2 = EI b3 3Eh ! 1 The internal forces at the clamped end are My = ;P L Vz = P T = T0 The torsional shear stress, which is proportional to x , is zero at the clamped end. The shear stress distribution over the cross section at the fixed end x = 0 is 0 given by Eq. (2.32) as xs(s) = ; P QtIy(s) ; T! QtI! (s) y ! (2.39) In Eq. (2.39), the first moment area Qy (s) is calculated using section cuts such as those indicated in Figure 2.15. For instance, with the section cut on the right flange Qy (z ) = Zz ; b2=2 ; b2 ztdz = 2t z 2 ; 42 The corresponding shear stress distribution is P ; b22 ; z 2 ( z ) = 1 2Iy 4 42 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION b1 C y S b2 z F IGURE 2.15 Transverse shear flow directions and section cuts Similarly, the shear stress in the left flange is given by P ; b21 ; z 2 ( z ) = 1 2Iy 4 and the shear stress in the web is zero. The second contribution to the shear stress in Eq. (2.39) is the warping shear stress 2(s) = ; T! QtI! (s) ! where T! is equal to the entire torque T0 , since the pure torsion torque Tt is zero at the clamped end. The first sectorial area moment Q! (s) of the principal warping function is calculated for the part of the cross section cut off at s, remembering that the integration starts at a free edge. The section cuts indicated in Figure 2.15 may be used for this calculation. For instance, with the section cut on the right flange Q! (s2 ) = Zs Similarly Q! (s1 ) = Zs 1 Q! (s3 ) = 0 b2=2 Zs 2 ; (yS ; yO )s2 tds2 = t(yS ;2 yO ) ( b42 ; s22 ) b =2 2 2 (yS ; yO )s1 tds1 = ; t(yS 2; yO ) ( b42 ; s21 ) 2 2 (yS ; yO ; h)s4 tds4 = t(h ; y2S + yO ) ( b41 ; s24 ) Zb1s=52 2 Q! (s5 ) = (h + yO ; yS )s5 tds5 = ; t(h ; y2S + yO ) ( b41 ; s25 ) b1=2 Q! (s4 ) = 4 2.8. STRESS ANALYSIS 43 The shear stresses are expressed by 6T0 ( b22 ; s2 ) ( s ) = 2 1 thb32 4 1 6T0 b22 2 2 (s2 ) = ; thb3 ( 4 ; s2 ) 2 ( s ) = 0 2 3 6T0 ( b21 ; s2 ) ( s ) = ; 2 4 thb31 4 4 6T0 b21 2 2 (s5 ) = thb3 ( 4 ; s5 ) 1 The sign of 2 (s1 ) is positive, which means that the shear flow is in the direction of increasing s1 , hence upward. Similarly, the sign of 2(s2 ) is negative, so the shear flow is in the direction of decreasing s2 , hence upward. Thus, warping shear stresses on the right flange are directed upward, but the signs of 2 (s4 ) and 2 (s5 ) show that the warping shear stresses on the left flange are directed downward. The normal stress at the fixed end due to bending is Mz 1(z ) = Iy = ; PLz Iy y The normal warping stress is M !(s) 2(s) = !I ! where M! is the bimoment at the fixed end ! x (0) = ; T0 tanh cL M! = ;EI c 00 To calculate the stresses numerically, the following dimensions will be assumed b h = b2 = 2b L = 20b t = 10 and the Poisson’s ratio will be taken to be = 0:25. The modulus of elasticity E shear modulus G are then E = 1 ;E 2 = 1615E G = 2(1E+ ) = 25E b1 = b The transverse and warping shear stress distributions at the clamped end of the beam are sketched in Figure 2.16. The force-couple equivalent of the transverse shear stress 1 at the shear center S is a single force of magnitude Vz = P . The warping shear stress 2 is statically equivalent to a couple. The total shear stress on the right flange is zero, so that, at the fixed end of the beam, all shear stresses are carried by the left flange. The bending and warping normal stresses are shown in Figure 2.17. The maximum normal warping stress exceeds the maximum bending stress. The bending stresses are statically equivalent to the bending moment My = PL. The warping ; 44 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION stresses are statically equivalent to zero force and zero couple. When considered separately for the two flanges, these stresses are equivalent to two equal and opposite bending moments. The maximum stresses are shown in Table 2.1. The reference stress 0 is defined as 0 = bP2 1 1 + + 2 2 + ; F IGURE 2.16 Transverse and warping shear stresses at the fixed end As a second example, the stress distribution in the simply supported beam shown in Figure 2.18 will be determined. The load is a vertical force of magnitude P at midspan. The cross section, whose centroid C is also the shear center S , is shown in Figure 2.19. The wall thickness t is the same for the flanges and the web. 2.8. STRESS ANALYSIS 45 1 + 1 ; + ; 2 ; + 2 + ; F IGURE 2.17 Bending and warping normal stresses at the fixed end The area moments of inertia and the area product of inertia for this cross section are Iy = 2tb3 3 3 tbh2 Iz = th 12 + 2 Iyz = thb2 2 The torsional constant, calculated from Saint-Venant’s approximation, is 3 3 3 J = ht3 + 2bt3 = t (h 3+ 2b) 46 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION MAXIMUM STRESS 1 =0 (Transverse) 2 =0 (Warping) 1=0 (Bending) 2=0 (Warping) RIGHT FLANGE LEFT FLANGE 6.7 6.7 266.7 91.3 1.7 13.3 133.3 365.0 TABLE 2.1 Maximum stresses at the clamped end L x L 2 P F IGURE 2.18 Simply supported beam with midspan load The warping function with pole and origin both at point O is !O (s1 ) = 0 !O (s2 ) = 0 !O (s3 ) = hs3 The principal warping function is given by Eq. (2.17) as Q !(s) = !O (s) ; A!O Since Q!O = Zb 0 ; (yS ; yO )z(s) hs3 tds3 = thb2 2 and A = t(h + 2b) the principal warping function is 2 !(s1 ) = hs21 ; 2(hhb+ 2b) 2 !(s2 ) = ; 2(hhb+ 2b) 2 hs !(s3 ) = 23 ; 2(hhb+ 2b) 2.8. STRESS ANALYSIS 47 b h y C b z s1 P s3 C s2 O F IGURE 2.19 Cross section of the simply supported beam The warping constant is found from Eq. (2.20) 2 3 Q2 I! = I!O ; A!O ; (yS ; yO )2Iy = th12(b h(b++22b)h) The applied load at x = L=2 is equivalent to a torsional couple T0 and a transverse force P at the shear center T0 = Ph 2 The applied torque per unit length can be written in terms of the Dirac delta function m(x) = T0 (x ; L2 ) The angle of twist is calculated from Eq. (2.34) as x (x) = C1 + C2x + C3 cosh cx + C4 sinh cx ; (x) where (x) = 0 for 0 6 x 6 L=2 and T0 (x) = cGJ c(x ; L2 ) ; sinh c(x ; L2 ) for L=2 6 x 6 L. 48 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION 0.6 0.5 0.4 x 0.3 0.2 0.1 0 0 0.2 0.4 0.6 0.8 x=L F IGURE 2.20 Angle of twist for the simply supported beam The boundary conditions at the two simple supports x (0) = x (L) = 0 (0) = (L) = 0 00 00 are solved for the integration constants C1 = 0 0 C2 = 2TGJ C3 = 0 0 sinh cL=2 C4 = ; TcGJ sinh cL For the left half of the beam T0 sinh cL= 2 x (x) = 2cGJ cx ; 2 sinh cL sinh cx T M! (x) = c sinh0 cL sinh cL 2 sinh cx T T sinh cL=2 Tt (x) = 20 ; 0 sinh cL cosh cx cL=2 T! (x) = T0 sinh sinh cL cosh cx The qualitative behavior of these functions over the entire span of the beam can be seen in Figures 2.20, 2.21, and 2.22. In Figure 2.22, the torques Tt and T! are shown 1 2.8. STRESS ANALYSIS 49 0.4 0.3 M! 0.2 0.1 0 0 0.2 0.4 x=L 0.6 0.8 F IGURE 2.21 Bimoment for the simply supported beam as fractions of the applied torque T0 . The total torque is the sum of Tt and T! T T (x) = 20 T (x) = ; T20 if x < L 2 L if x > 2 The stresses at x = L=2 at the section just to the left of the applied torque will be calculated. The transverse shear stress at this section is 1 (s) = ; Iz Qty((Is)I ;;IyzI 2Q)z (s) Vz y z yz From Figure 2.19, on the right flange, the first moments of area are Zs t(b2 ; s21 ) ; s1 tds1 = 2 Zb s h ht ( b ; s1) Qz (s1 ) = ; tds = 1 2 2 b Qy (s1 ) = 1 1 and 1(s1 ) = 3P (s1 ; b)(bh + hs1 + 6bs1) 4tb3(2h + 3b) 1 50 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION T! 0.4 Tt 0.2 0 -0.2 -0.4 0 0.2 0.4 0.6 0.8 x=L F IGURE 2.22 Pure torsion and warping torques for the simply supported beam The value of 1 (s1 ) for s1 < b is negative, which means that the the stress is in the negative s1 , or the positive z , direction. For the web, the first moments of area are Qy (s2 ) = ; tb2 2 Q (s ) = ; t(s2 ; hs2 ; bh) 2 z 2 The shear stress in the web is 1 (s2 ) = 2 3P (6s22 ; 6hs2 + h2 4tbh(2h + 3b) Similarly, on the left flange, 3P (b ; s3 )(bh + 6bs3 + hs3 ) 4tb3(2h + 3b) The warping shear stress at x = L=2 is (s) = ; Q! (s) T 1(s3 ) = 2 tI! ! 1 2.8. STRESS ANALYSIS 51 where the warping torque T! is T0 =2, because the pure torsion torque Tt is zero at midspan. The expressions for the warping shear stresses are T0 h(b ; s1 )(bh + hs1 + 2bs1 ) 8I! (h + 2b) 2 T0 hb (2s2 ; h) 2 (s2 ) = 8I (h + 2b) ! T0 h(b ; s3 )(bh + hs3 + 2bs3 ) 3 (s3 ) = 8I! (h + 2b) The normal stress distribution at x = L=2 due to bending is I M IM 1 = ; I Iyz; Iy 2 y + I I z ; yI 2 z y z yz y z yz where My = PL=4. The normal stress due to warping is M! 2 = I! ! 2 (s1 ) = where T M! = 2c0 tanh cL 2 The shear stress distribution at x = L=2 is sketched in Figure 2.23. The force resultant of the transverse shear stress 1 over the two flanges is equal to the total shear force P=2. The transverse shear stresses over the web are statically equivalent to a zero force-couple. The warping shear stress 2 is equivalent a torsional moment. The transverse shear stress adds to the warping shear stress over the left flange, but subtracts from it over the right flange. The normal stress distribution at x = L=2 is sketched in Figure 2.24. The normal stress 1 due to bending is statically equivalent to a bending moment about the y axis. The warping stress 2 is statically equivalent to a zero force-couple. The bending and warping stresses are additive over the left flange. 52 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION + 1 + + ; + ; ; + 2 + F IGURE 2.23 Shear stresses for the simply supported beam 2.8. STRESS ANALYSIS 53 ; ; ; 1 + + + + 2 ; ; ; + F IGURE 2.24 Normal stresses for the simply supported beam 54 II. THIN-WALLED ELASTIC BEAMS OF OPEN CROSS SECTION CHAPTER III THIN-WALLED ELASTIC BEAMS OF CLOSED CROSS SECTION 3.1. Geometry of Deformation A closed thin-walled cross section is shown in Figure 3.1. The tangential and normal coordinates, s and n, are chosen so that the axes n, s, x 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. The normal coordinate n of any point of the median line is zero. The angle (s) is measured from the positive y axis to the positive n axis. C y xs s n z F IGURE 3.1 Closed thin-walled section As in Chapter II, 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 56 III. THIN-WALLED ELASTIC BEAMS OF CLOSED CROSS SECTION 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 S be the shear center of the cross section shown in Figure 3.2 and let (s) denote the tangential component of the point of the median line at the coordinate s. As shown in Chapter II, (s) can be written as (x s) = vS (x) cos (s) + wS (x) sin (s) + x (x)h(s) (3.1) where vS , wS are the displacements of the shear center in the y, z directions, and h is the projection, onto the unit normal vector en , of the position vector r(s) of the point at s h = r en y S r(s) n s s z F IGURE 3.2 Tangential and normal components of displacement It will now be assumed that the shear strain xs of the median line is equal to its value found in Saint-Venant torsion. This assumption can be written as @ xs qt xs = @u @s + @x = G = tG where qt is the constant shear flow of Saint-Venant torsion T qt = t and denotes twice the area enclosed by the median line I = hds 3.1. GEOMETRY OF DEFORMATION 57 The derivative of u with respect to s is @u = Tt ; v cos ; w sin ; h S x @s tG S from which the displacement of the point at s along the x axis is obtained by inte0 gration 0 0 Zs T Zs t u = u0 + tG ds ; x hds ; vS y ; wS z 0 0 0 0 0 From Eq. (1.18) Tt x G = H ds t(s) 0 with which the axial displacement becomes Z s ds Z s u = u0 + x H ds t ; x hds ; vS y ; wS z 0 0 t 0 0 0 0 The warping function for a closed section is defined by !(s) = Zs 0 hds ; H ds t Z s ds 0 (3.2) t The first term of the preceding equation will be recognized as the sectorial area, or the warping function for an open section. The warping displacement can now be written in the same form as it was in Chapter II for open cross sections u(x s) = u0(x) ; vS (x)y(s) ; wS (x)z (s) ; x (x)!(s) 0 0 0 (3.3) It is easily verified that the presence of the second integral in Eq. (3.2) does not change Eq. (2.7) for changing the pole of the warping function from B to A !A (s) = !B (s) ; !B (s0 ) + (zA ; zB )(y(s) ; y0 ) ; (yA ; yB )(z (s) ; z0 ) (3.4) The equations for finding the principal pole, or the shear center, also remain the same as those for open sections I I ; I B Iyz yS = yB + z!BI zI ; y! 2 y z Iyz I I ; Iy!B Iy zS = zB + z!BI yz 2 y Iz ; Iyz (3.5) (3.6) The principal warping function is given in terms of a warping function !B , whose pole and origin are arbitrarily chosen, by Eq. (2.17) Q !(s) = !B (s) ; A!B + (zS ; zB )y(s) ; (yS ; yB )z (s) (3.7) Similarly, the warping constant is given by Q2 I! = I!B ; A!B ; (yS ; yB )2 Iy + 2(yS ; yB )(zS ; zB )Iyz ; (zS ; zB )2 Iz (3.8) 58 III. THIN-WALLED ELASTIC BEAMS OF CLOSED CROSS SECTION a C y b z s1 O s2 s3 s4 F IGURE 3.3 Thin-walled box section As an example, consider the thin-walled rectangular cross section of uniform thickness t shown in Figure 3.3. If point O is used both as pole and origin, the warping function is !O (s1 ) = ; a ab + b s1 2 !O (s2 ) = a a+ b (s2 ; b) 2 !O (s3 ) = a b+ b s3 !O (s4 ) = a ab + b (b ; s4 ) 3.2. EQUATIONS OF EQUILIBRIUM The area moments of inertia are Iy = tb (b 6+ 3a) 59 2 Iz = ta (a6+ 3b) 2 The shear center is at the centroid of the rectangle. The principal warping function is found by an application of Eq. (3.7) b(b ; a) (2s ; a) !(s1 ) = 4( a + b) 1 ; b) (2s ; b) !(s2 ) = a4((aa + b) 2 b(b ; a) (2s ; a) !(s3 ) = 4( a + b) 3 ; b) (2s ; b) !(s4 ) = a4((aa + b) 4 The principal warping function is zero for a square cross section, which according to the theory being described here, is free of warping. The warping constant for the rectangular box section is 2 b2(b ; a)2 I! = ta24( a + b) In the theory developed by Benscoter for closed thin-walled sections, the rate of angle of twist x0 in Eq. (3.3) is replaced by an arbitrary function # of x, so that the fundamental kinematical assumption for the warping displacement becomes u(x s) = u0(x) ; vS (x)y(s) ; wS (x)z (s) ; #(x)!(s) The normal strain x is written as x = u0 ; vS y ; wS z ; # ! The shear strain xs is @ = h ; # @! = h ; # h ; xs = @u + H @s @x x @s x t dst 0 0 0 00 00 0 0 0 (3.9) (3.10) (3.11) 3.2. Equations of Equilibrium The normal stress x is obtained from Hooke’s law as x = E (u0 ; vS y ; wS z ; # !) x = E 0 00 00 0 The shear stress is the sum of the bending and the torsional contributions xs = b + t The torsional contribution is t = Gxs = Gx h 0 where the abbreviation ; G#(h ; k) k = H ds t t has been introduced. As for open cross sections, the bending shear stress has no corresponding shear strain. 60 III. THIN-WALLED ELASTIC BEAMS OF CLOSED CROSS SECTION The stress resultants for the normal stress x are the axial force N , the bending moments My and Mz , and the bimoment M! , which are defined by Z N= Z My = xdA zdA Z Mz = ; ydA Z M! = !dA The stress resultants are evaluated, recalling that the origin of the coordinates y, z is the centroid, and that ! is the principal warping function 0 N 1 0A 0 0 0 1 0 u 1 BB My CC = E BB 0 ;Iyz ;Iy 0 CC BB vS0 CC @ Mz A @ 0 Iz Iyz 0 A @wS A 0 00 0 M! Hence 0 0 ;I! 00 #x 0 0(x) = N (x) Eu A I yz S (x) = My (x) + Iy2Mz (x) Ev Iy Iz ; Iyz S (x) = ; Iz My (x) + Iyz2Mz (x) Ew Iy Iz ; Iyz x(x) = ; M! (x) E# I! 0 00 00 0 The normal stress is found in terms of the stress resultants by using these expressions in Eq. (2.21) Iyz My + Iy Mz Iz My + Iyz Mz M! ! x = N ; 2 y + I I ; I2 z + I A Iy Iz ; Iyz y z yz ! The total torque T is T= Z h xs dA = Gx 0 Z h2 dA ; G# Z h2 dA ; Z hkdA (3.12) The last part of this expression can be evaluated as Z hkdA = Z hdA 2 H = H ds = J t ds Z t t A cross-sectional property, sometimes called the polar constant, is defined by Ih = so that the torque T is h2 dA T = GIh x ; G#(Ih ; J ) 0 3.2. EQUATIONS OF EQUILIBRIUM 61 The torque equilibrium equation for a length dx of the beam gives dT + m(x) = GI ; G# (I ; J ) + m(x) = 0 h x h dx where m(x) is the applied torque about the shear center S per unit length of the 00 0 beam. As in Chapter II, the equilibrium equation in the longitudinal direction, in the absence of applied axial load px , gives x @q t @ @x + @s = 0 The equilibrium of forces in the longitudinal direction remains the same 0 (x)A + px = 0 Eu 00 which gives u000 (x) = 0. Hence @q = ;t @x = tE (v y + w z + # !) S S @s @x 000 ; The shear flow is 000 00 q(x s) = q0 (x) + E vS (x)Qz (s) + wS (x)Qy (s) + #x (x)Q! (s) As in Chapter II, the shear stress resultants Vy , Vz , and T! be defined by 000 000 00 (3.13) Z Vy = q(x s)dy Z Vz = q(x s)dz Z T! = q(x s)d! The warping torque is calculated by integrating both sides of Eq. (3.13) with respect to ! x Q! (s)d! = ;E# x I! T! = E# 00 Z 00 because, as shown Chapter II by an integration by parts, Since it follows that Z Q! (s)d! = ;I! q(h ; k)ds = G(Ih ; J )(x ; #) 0 ;EI ! # 000 = G(Ih ; J )(x ; # ) 00 0 The equation for the angle of twist is found by eliminating # from GIh x ; G# (Ih ; J ) + m(x) = 0 !# = 0 G(Ih ; J )(x ; # ) + EI 00 0 00 0 From the first of these equations 000 I # = I ;h J x + G(I m; J ) h h 0 00 62 III. THIN-WALLED ELASTIC BEAMS OF CLOSED CROSS SECTION The differential equation for the angle of twist is ! Ih iv ! EI EI ; GJ = m ( x ) ; x x Ih ; J G(Ih ; J ) m (x) 00 00 (3.14) When attempting to use Eq. (3.14), it is possible to encounter cross sections for which Ih and J are equal. For the rectangular box section of Figure 3.3, the polar constant Ih is and the torsional constant J is Ih = tab(a2 + b) 2 2 J = 2ata+ bb For a square cross section, with a = b, Ih = J = tb3 and Eq. (3.14) cannot be used. In general, when the polar constant is the identical to the torsional constant, the cross section is free of warping, and the warping function is everywhere zero. The differential equation for the angle of twist then reduces to GJx + m(x) = 0 00 This is the governing equation for Saint-Venant torsion with a variable distributed moment m(x). 3.3. A Multicell Analysis Example If the area enclosed by the outer wall of a cross section is subdivided into any number of other closed thin-walled sections, the beam is a multicell structure, an example of which is shown in Figure 3.4. For such cross sections, the condition obtained in Chapter I by taking the line integral of the derivative of the longitudinal displacement u with respect to s around a closed contour is used 1 I q ds ; = 0 G i t(s) x i 0 (3.15) where the integral is taken around the contour of the ith cell, and i is twice the area enclosed by the contour of the ith cell. In pure torsion the shear flow in each cell has a constant value. On a shared wall, such as the one of length b in Figure 3.4, the shear flows are additive q12 = q2 ; q1 where q1, q2 are the individual shear flows in the two cells and q12 is the shear flow in the shared part of the wall. The condition in Eq. (3.15) is applied to the two cells, assuming that the thickness t is uniform throughout the cross section, d + e) + q2b = 0 x 1 ; 2q1(Gt Gt q b 2 q ( a + 1 2 x 2 + Gt ; Gt b) = 0 0 0 3.3. A MULTICELL ANALYSIS EXAMPLE 63 e t y d C S cz a b cy z F IGURE 3.4 Two-cell thin-walled cross section The directions assumed for the shear flows q1, q2 are indicated in Figure 3.5. The total torque in the section is T = q1 1 + q2 2 The torsional constant can be calculated from T = q1 1 + q2 2 J = G G 0 x as 0 x The warping function with respect to any arbitrarily chosen pole O is written !O (s) = Zs 0 Z s ds 1 hds ; G q t 0 where the first integral is the sectorial area. In the second integral, q denotes the shear flow corresponding to x0 = 1. For the example two-cell section, with the s coordinates defined in Figure 3.5, the warping function !O , whose pole and origin 64 III. THIN-WALLED ELASTIC BEAMS OF CLOSED CROSS SECTION s6 s7 q1 C y S s5 s1 s2 s8 z O q2 s4 s3 F IGURE 3.5 Definitions for the two-cell thin-walled section are both chosen to be point O, is written as q 2s1 !O(1) (s1 ) = ; Gt ( !O(2) (s2 ) = !O(1) (a) + as2 ; q2 ;Gtq1 )s2 2s3 !O(3) (s3 ) = !O(2) (b) + bs3 ; qGt 2 s4 !O(4) (s4 ) = !O(3) (a) ; qGt 1 s5 !O(5) (s5 ) = !O(1) (a) ; as5 ; qGt 1s6 !O(6) (s6 ) = !O(5) (d ; b) + (d ; b)s6 ; qGt 1s7 !O(7) (s7 ) = !O(6) (e) + (a + e)s7 ; qGt q1 s8 !O(8) (s8 ) = !O(7) (d) + bs8 ; Gt 3.3. A MULTICELL ANALYSIS EXAMPLE 65 Let the dimensions of the cross section be d = 100 mm e = 40 mm b = 20 mm a = 30 mm t = 0:25 mm Then, based on centerline dimensions, the location of the centroid C is defined by cy = 28:61 mm cz = 41:11 mm The area moments of inertia and the area product of inertia, based on centerline dimensions, are found to be Iy = 118222 mm4 Iz = 47993 mm4 Iyz = ;24111 mm4 The sectorial areas 1 = 2ed = 8000 mm2 2 = 2ab = 1200 mm2 are needed in the calculation of the torsional constant J . The shear flows are q1 = 51569Gx 0 q2 = 31069Gx 0 where the numerical values in the numerators are in mm2 . With the shear flows determined, the torsional constant can be calculated J= q1 1 + q2 2 = 65101 mm4 Gx 0 and the warping function !O becomes 1240s !O (s1 ) = ; 69 1 10(289s2 ; 3720) !O (s2 ) = 69 20(7 s3 + 1030) !O (s3 ) = 69 1240(20 ; s4) ! (s ) = 69 10(3720 + 413s5 ) !O (s5 ) = ; 69 20(173 s 6 ; 18380) !O (s6 ) = 69 10(277 s 7 ; 22920) !O (s7 ) = 69 40(1195 ; 17s8) !O (s8 ) = 69 O 4 where the dimension of the s coordinates is in mm, and the dimension of !O is in mm2 . 66 III. THIN-WALLED ELASTIC BEAMS OF CLOSED CROSS SECTION Next the section properties dependent on !O are calculated Q!O = I!O = Iy!O = Iz!O = Z Z Z Z !O dA = 129076 mm4 !O2 dA = 482:161 (106) mm6 y!O dA = ;575148 mm5 z!O dA = 5703540 mm5 The shear center coordinates are I I ; I O Iyz yS = yO + z!OI zI ; y! = 9:64 mm 2 y z Iyz I I ; Iy!O Iy zS = yO + z!OI yz = 7:46 mm I ; I2 y z yz The principal warping function, whose pole is the shear center S , can be obtained from !O by the transformation Q !(s) = !O (s) ; A!O + (zS ; zO )y(s) ; (yS ; yO )z (s) where A is the cross-sectional area A = t(2e + 2d + b + 2a) = 90 mm2 The warping constant I! can be determined either by integrating the square of the principal warping function over the cross-sectional area, or by the transformation formula Q2 I! = I!O ; A!O ; (yS ; yO )Iy ; (zS ; zO )Iz + 2(yS ; yO )(zS ; zO )Iyz = 13:8514 (106) mm6 3.4. Cross Sections with Open and Closed Parts Some cross sections contain both closed cells and open branches. An example is shown in Figure 3.7. In analyzing such cross sections, the warping functions for the open branches are found as described in Chapter II. The warping functions for the closed cells are found as described in the two-cell example of the preceding section. The contribution of the open branches to the torsional constant J is usually negligibly small, so that J can be calculated for the closed cells of the cross section alone. For the cross section shown in Figure 3.7, the origin of the user coordinate system is placed at point O, with the y axis horizontal and pointing left, the z axis vertically downward. The node coordinates are shown in Table 3.1 below. The wall thicknesses are in Table 3.2, each line entry of which lists two nodes and the thickness of the wall segment between them. The properties of the cross section are listed in Table 3.3. A qualitative idea of the distributions of normal stress and strain due to warping is provided by the principal warping function. Because this section has straight walls, the warping 3.4. CROSS SECTIONS WITH OPEN AND CLOSED PARTS 6 67 5 S 1 2 7 4 3 F IGURE 3.6 Principal warping function for the two-cell thin-walled section 10 9 6 O C 8 7 3 2 1 S 4 5 F IGURE 3.7 Thin-walled section with three cells function is piecewise linear. Table 3.4 lists numerical values of the warping function, and Figure 3.8 shows how it varies along the median line. As a final example, the cross section shown in Figure 3.9, which resembles certain thin-walled sections found in automobile frames, will be analyzed. The user coordinate system for this cross section has its origin at point O, with the y axis horizontal and directed toward the left and the z axis vertically downward. The position coordinates of the nodes of the section are listed in Table 3.5. The thickness is uniform for the entire cross section. The principal warping function for the section is sketched in Figure 3.10. The results derived above by the approximate linear theory of beams with thin-walled cross section are compared with the results calculated by the the computer program BEAMSTRESS in Table 3.7. When the wall thickness is very small, 68 III. THIN-WALLED ELASTIC BEAMS OF CLOSED CROSS SECTION y coordinate (mm) z coordinate (mm) Node ;250 ;150 ;50 ;50 ;100 1 2 3 4 5 6 7 8 9 10 50 50 100 150 250 0 0 0 100 100 0 100 100 0 0 TABLE 3.1 Nodal coordinates of the cross section shown in Figure 3.7 First Node 1 2 3 6 9 8 7 4 5 8 7 4 Second Node 2 3 6 9 10 9 6 3 2 7 4 5 Thickness (mm) 10 10 10 10 10 5 5 5 5 12 12 12 TABLE 3.2 Wall thicknesses of the cross section shown in Figure 3.7 the linear theory agrees well with the results obtained by this program, which provides a finite-element calculation based on the elasticity formulation. As the wall thickness increases, the linear theory results become less accurate, and it is possible to have very large errors in the section properties, especially in the warping constant. As the thickness is changed, the warping function of the linear theory does not change, because the median line of the section determines this function. The elasticity formulation considers the warping function as a function of y and z , and when the boundary of the section is changed, the warping function also changes. The large errors in Iyz are a consequence of the assumption in the linear theory that the elements of cross-sectional area are entirely concentrated at the median line. 3.4. CROSS SECTIONS WITH OPEN AND CLOSED PARTS Cross-Sectional Area (mm2 ) Centroid yC (User Coordinates) (mm) Centroid zC (User Coordinates) (mm) Shear Center yS (Centroidal Coordinates) (mm) Shear Center zS (Centroidal Coordinates) (mm) Shear Center yS (User Coordinates) (mm) Shear Center zS (User Coordinates) (mm) Area Moment of Inertia Iy (mm4) Area Moment of Inertia Iz (mm4 ) Area Product of Inertia Iyz (mm4 ) Polar Constant Ih (mm4 ) Torsional Constant J (mm4) Warping Constant I! (mm6 ) 9518:0 0 36:34 0 10:90 0 47:24 18:49 106 132:37 106 0 34:63 106 29:17 106 10:41 109 TABLE 3.3 Properties of the cross section in Figure 3.7 Node m 2 3 4 5 3 6 7 6 9 8 1 9 Node n 3 4 5 2 6 7 4 9 8 7 2 10 !m (mm2 ) !n (mm2 ) 1483:29 1102:45 1102:45 ;249:421 ;249:421 261:184 261:184 1483:29 1102:45 ;1102:45 ;1102:45 249:421 249:421 ;249:421 ;1102:45 ;1483:29 ;1483:29 ;261:184 ;261:184 249:421 ;3241:12 1483:29 ;1483:29 3241:12 TABLE 3.4 Values of the principal warping function for the section in Figure 3.7 69 70 III. THIN-WALLED ELASTIC BEAMS OF CLOSED CROSS SECTION 9 10 6 3 1 2 S 8 4 7 5 F IGURE 3.8 Warping function for the section in Figure 3.7 5 4 3 1 O 6 C S 2 7 8 F IGURE 3.9 Thin-walled cross section 9 3.4. CROSS SECTIONS WITH OPEN AND CLOSED PARTS 71 y coordinate (mm) z coordinate (mm) Node 1 2 3 4 5 6 7 8 9 180 160 150 140 80 60 30 20 0 90 90 60 30 0 30 40 90 90 TABLE 3.5 Nodal coordinates of the cross section in Figure 3.9 Node m 6 4 4 3 8 7 3 2 1 8 Node n 5 4 3 6 7 6 2 8 2 9 !m (mm2 ) !n (mm2 ) ;8:96 ;82:66 ;82:66 ;233:4 ;233:4 101:6 101:6 ;8:96 ;351:6 552:2 552:2 ;8:96 101:6 248:3 248:3 ;351:6 ;581:3 248:3 ;351:6 478:0 TABLE 3.6 Numerical values of the principal warping function sketched in Figure 3.10 72 III. THIN-WALLED ELASTIC BEAMS OF CLOSED CROSS SECTION 5 6 4 S 7 3 1 2 8 F IGURE 3.10 Principal warping function for the section in Figure 3.9 9 3.4. CROSS SECTIONS WITH OPEN AND CLOSED PARTS Thickness Property Area (mm2 ) 1 mm Iy (mm4 ) Iz (mm4 ) Iyz (mm4 ) J (mm4 ) I! (mm6 ) Area (mm2 ) 5 mm Iy (mm4 ) Iz (mm4 ) Iyz (mm4 ) J (mm4 ) I! (mm6 ) Area (mm2 ) 10 mm Iy (mm4 ) Iz (mm4 ) Iyz (mm4 ) J (mm4 ) I! (mm6 ) Area (mm2 ) 12 mm Iy (mm4 ) Iz (mm4 ) Iyz (mm4 ) J (mm4 ) I! (mm6 ) 73 Linear Theory BEAMSTRESS Difference (%) 524 455200 1143708 5270 749424 18:3 106 2619 2:276 106 5:719 106 26352 3:747 106 91:5 106 5237 4:552 106 11:44 106 52703 7:49 106 183 106 6286 5:46 106 13:72 106 63244 8:99 106 219:6 106 522 453976 1136910 4843 754965 18:88 106 2571 2:25 106 5:555 106 16740 3:892 106 116:1 106 5047 4:485 106 10:81 106 19487 8:11 106 321 106 6010 5:38 106 12:84 106 18361 9:90 106 437 106 0:38 0:27 0:60 8:80 0:73 3:07 1:87 1:15 2:95 57:40 3:73 21:2 3:8 1:5 5:8 170 7:6 43 4:6 1:5 6:9 244 9:2 50 TABLE 3.7 Properties of the cross section in Figure 3.9