2.9 Torsion of beams with open thin-walled cross-sections

Transcription

1 Kuva.17. thin-walled cross-section..9 Torsion of beams with open thin-walled cross-sections.9.1 Sectorial coordinate Consider a thin walled beam cross-section of arbitrary shape where the wall thickness is small as compared to the other measures of the cross-section shown in Figure.17. The origin of the global coordinate system x, y, z is located at the center of gravity of the cross-section, and the axial coordinate x coincides the beam axis. The coordinate s is chosen to coincide with the centerline of the cross-section. Then, an arbitrary point P o is chosen for the origin, and point for an estimation for the shear center or a pole. The sectorial coordinate ˆω is defined according to Figure.17 as a gradient dˆω = ±h (s)ds (.68) where the subscript refers to the pole and the superscript hat to the estimation of the sectorial coordinate, determined with respesct to the arbitrary point P o. The definition of the gradient of the sectorial coordinate can be expressed also as the projections in the direction of the global coordinate axes dˆω = (z z )dy + (y y )dz (.68a) The same definition can still be given as the length of the vector parallell to the beam axis, which is obtained as the vector product of of the position vector from point to point P o, and the tangent vector of the centerline d s on the cross-section plane d ˆω = ( r P r ) d s (.68b) where the vectors utilized are r P r = (y y ) j + (z z ) k ja d s = dy j + dz k. Or further, as the doubled area of the shaded triangle in the figure according to the definition of the vector product of two vectors. 1

2 Figure.18. The sectorial coordinate ˆω B (s). The sectorial coordinate can be determined by utilizing the definition and direct integration and using the point P o as the origin, when ˆω = P P o dˆω = ± P P o h (s)ds = P P o [(z z )dy (y y )dz] (.69) s an Example, we define the distribution of the sectorial coordinate for a U-profile shown in Figure.18, when point B is used as a pole, and point P o, the free end of the upper flange, as the origin of s coordinate. Sectorial coordinate at the interval 0 s 3b is s b 0 ds = b s, kun 0 s b s b ˆω B (s) = ˆω B (b) + b ds = b s, kun b s b s 3b ˆω B (b) b ds = 4b 3b s, kun b s 3b Consider next, how the distribution of the sectorial coordinate will change, if the pole is relocated, for example at point. Thus we can write very generally by applying directly the definition ˆω = P = P o [(z z )dy (y y )dz] P P o [(z z B + z B z )dy (y y B + y B y )dz] P P = [(z z B )dy (y y B )dz] [(z B z )dy (y B y )dz] P o P o = ˆω B (z B z )(y y o ) + (y B y )(z z o ) (.70) Let s define now the sectorial coordinate for the U-profile by using as the pole point C, at the corner of the upper flange and the web. Thus, z B z C = b ja y B y C = b

3 Figure.19. Sectorial coordinate ˆω C. and we get b s b 0 + ( b )s = 0, kun 0 s b b ˆω C = s b (s b) + ( b )b = 0, kun b s b 4b 3b s b b + ( b )(3b s) = b bs, kun b s 3b This distribution is shown in Figure xial displacement and stress distributions V.Z. Vlasov derived already in the 1950s the theory for the torsion of open thin-walled beams or girders. He assumed that no shear strain will take place at the centerline of the walls, or γ xs = 0 (.71) Consequently, no shear flow, similar to closed ring-type cross sections, doesn t appear. On the base of the definition of the shear strain it is obtained γ xs = u s + u s x = 0 = u s = u s x (.7) From this, it can be deduced that according to Figure.0, the displacement u s is and thus u s = h (s)ϕ(x) u s = h (s)ϕ(x) x = h (s)ϕ (x) (.73) If at point P o the axial displacemnet is zero, it is obtained by interating directly u = P P o u P s ds = h (s)ϕ (x)ds = ˆω (s)ϕ (x) (.74) P o 3

4 Figure.0. Rotation of the cross-section. The normal or axial strain is obtained according to the definition by differentiating ɛ x = u x = ˆω (s)ϕ (x) (.75) and the normal stress by adopting the one-dimensional linear elasticity as the material model σˆω = σ x = Eɛ x = E ˆω (s)ϕ (x) (.76) When the loading of the beam consists of pure torsion only, the equilibrium equation states N σˆω d = 0 = ˆω d = 0 S ω = 0 (.77) If the condition S ω = 0 is valid, the distribution of the sectorial coordinate is called normalized. Further, according to the equilibrium equations it is obtained M y σˆω zd = 0 = ˆω zd = 0 Iˆω z = 0 (.78) M z σˆω yd = 0 = ˆω yd = 0 Iˆω y = 0 These two conditions determine the location of the shear center of the cross-section. By inserting the expression of the sectorial coordinate, the location of the shear center will be calculated from the equations Iˆω y = ˆω yd = ˆω B d (z B z )( y d y o yd) + (y B y )( yzd z o yd) = IˆωB y (z B z )(I z y o S z ) + (y B y )(I yz z o S z ) (.79a) Correspondingly, the equation Iˆω z = IˆωB z (z B z )(I yz y o S y ) + (y B y )(I y z o S y ) (.79b) 4

5 is derived. When we will take still into account that the origin of the yz coordinates is located at the center of gravity of the cross-section, giving the conditions S y = S z = 0, it is obtained by requiring the disappearance of the two equations (.79a,b) the coordinates of the shear centre y = y B + I ziˆωb z I yz IˆωB y I y I z Iyz z = z B I yiˆωb y I yz IˆωB z I y I z I yz (.80) If the yz-coordinates are in addition principal coordinates of the cross-section, when I yz = 0, the simple expressions are obtained y = y B + Iˆω B z I y z = z B Iˆω B y I z (.81) Normalization of the sectorial coordinate is done by applying the condition S ω = ω d = 0 (.8) When we take into account that the surface integral in the definition of the sectorial coordinate can be split into pieces so that ω (s) = P P o dω = P P o dω Po When this is inserted into the expression (.8) it is obtained S ω = P o dω = ˆω (s) ˆω (s o ) (.83) ˆω d ˆω (s o ) = Sˆω ˆω (s o ) = 0 (.84) From this, the location of the origin P o of coordinate s can be determined ˆω (s o ) = Sˆω and finally, the expression for the sectorial coordinate is reached. ω (s) = ˆω (s) Sˆω (.85) Illustrative example The task is to determine the location of the shear center and the origin of s- coordinate for a U-profile, and then the distribution of the sectorial coordinate. In addition, we will determine the diagram of the sectorial static moment and the warping 5

6 Figure.1. Determining the shear center of a thin-walled cross-section. constant of the cross-section. The distance between the web of the cross-section and the centre of gravity is 3bt e = bt 1 b = e = 1 3 b The second moments or moments of inertia are correspondingly I y = 1 3 b3 t I z = 7 1 b3 t Next, we calculate the sectorial product moments I ωb y = ω B yd = ω B (s)y(s)tds s I ωb z = ω B zd = ω B (s)y(s)tds In Figure.1, the diagrams of ω B (s), y(s) ja z(s) are presented. When calculating line-integrals, various practical tables can be used. We get I ωb y = 1 bb ( b )t + b 6 [b ( b + b ) + b ( b + b )]t + b (b b ) b t = b4 t 4 I ωb z = 1 6 bb ( b 3 + b 3 )t + b (b + b ) b 3 t + b 6 [b ( b 3 b 3 ) b ( b 3 b 3 )]t = b4 t 3 Now, y, z are the principal coordinates, when the coordinates of the shear centre are obtained by using the definitions y = y B + I ω B z I y = b + b4 t/3 b 3 t/3 = 0 z = z B I ω B y I z = 5 6 b b4 t/4 7b 3 t/1 = 16 1 b Now, when the location of the shear centre is known, the sectorial coordinate is determined by using the shear centre as the pole. By applying the definition directly, we 6 s

7 Figure.. Sectorial coordinate ˆω. get s ˆω (s) = ω B (b) + 0 ω B (b) b ds = b s, kun 0 s b s b s b 3b ds = 13b b s, kun b s b 7 3b 13b ds = 14 b s, kun b s 3b which is depicted in Figure.. The integral due to the sectorial first moment over the whole cross-section takes the value Sˆω = ˆω d = ω (s)tds s = 1 b( b )t + 1 b( b b 14 )t + 1 b b( 14 8b 14 )t = t 6b3 7 Now, the final diagram of the sectorial coordinate is obtained by normalizing ω = ˆω Sˆω = ˆω 6b3 t/7 3bt and the sectorial coordinate diagram is finally = ˆω + 7 b 7 b b s, kun 0 s b ω (s) = 9 14 b + 3b s, kun b s b b b s, kun b s 3b It is presented in Figure.3. We determine still the distribution of the sectorial first moment of the cross section 7

8 Figure.3. Sectorial coordinate ω. S ω. It is obtained by direct integration S ω (s) = s 0 s 0 ω (s)tds 1 = 8 b3 t 1 8 b3 t + ( 7 b 1 bs)tds = 7 b ts 1 4 bts, kun 0 s b s b s b ( 9 14 b 3 13 bs)tds = 7 8 b3 t 9 14 b ts bts, kun b s b ( b 1 bs)tds = 39 8 b3 t b ts 1 4 bts, kun b s 3b Figure.4. Sectorial first momenti S ω (s). and it is depicted in Figure.4. The warping constant I ω is 3b I ω = ωd = ω(s)tds = 5 84 b5 t 0 8

9 Figure.5. Equilibrium of a wall element..9.4 Sectorial shear stresses Consider an element cut out of the wall of an open thin-walled girder under torsion without volume forces, presented in Figure.5. In figure, all the stress components parallel to the beam axis are presented. The equilibrium between them gives (σ ω t + (σ ωt) x The shear stresses take the form dx)ds σ ωtds + (τ ω t + (τ ωt) ds)dx τ ω tdx = 0 s (τ ω t) s of which we get by integrating = (σ ωt) x = Eω (s)ϕ (x)t(s) Consequently, s τ ω t τ ω (0) t(0) = Eϕ (x) ω (s)t(s)ds = Eϕ (x)s ω (s) }{{} 0 =0 τ ω (s) = Eϕ (x)s ω (s) 1 t(s) (.86) The shear stresses due to Saint Venant eli vapaan väännön leikkausjännitykset ovat puolestaan τ t (s) = ±Gϕ (x)t(s) (.87) and the total shear stress distribution is the sum of both of these Under torsion only, the stress resultants τ(s) = τ ω (s) + τ t (s) (.88) N = Q y = Q z = M y = M z 0 9

10 must disappear. Instead, the external torque loading M x has to be in equilibrium with the internal shear stress resultants. The external loading will generate in the beam shear stresses of Saint Venant and warping analyses. The torque resultant for warping stresses M ω is obtained by considering the equilibrium at the cross section plane and utilizing the expression (.86) when we get M ω = Performing once the integration by parts with sl τ ω h d = Eϕ (x) S ω (s)h (s)ds (.89) 0 dω (s) ds ds ω (s) ds = h (s) = ω (s) we get [ M ω = Eϕ (x)[ S ω (s)ω (s) }{{} =0 ] s=sl s=0 = Eϕ (x) ω(s)d = EI ω ϕ (x) ] sl ω (s)t(s)ω (s)ds 0 (.90) because S ω (0) = S ω (s l ) = 0. Thus, the shear stresses due to warping torsion are of the form τ ω (s) = M ωs ω (s) (.91) I ω t(s) Saint Venant shear stress resultant is correspondingly and shear stress distribution M t = GI t ϕ (x) τ t (s) = ± M tt(s) I t (.9) The total resultant, equilibrating the external loading, is thus the sum of these M x = M ω + M t = EI ω ϕ (x) + GI t ϕ (x) (.93) For the bimoment B, the definition is obtained on the base of equation (.76) B = σ ω ω d = Eϕ (x) ωd = EI ω ϕ (x) (.94) and from this further, the expression for the axial or warping stresses as expressed by the bimoment σ ω (s) = B I ω ω (s) (.95) 10

11 Figure.6. Equilibrium of a beam element. Finally, it is useful to observe that the connection between the bimoment and the torque due to warping torsion is M ω = db (.96) dx.9.4 Equilibrium of a beam element Consider generally an element of length dx cut out from a beam under torsional loading, shown in Figure.6. The equilibrium of this element states when we get M x + dm x dx dx M x + m t (x)dx = 0 ccording to (.93), the total torque is the sum dm x dx + m t = 0 (.97) M x = M ω + M t = EI ω ϕ (x) + GI t ϕ (x) and by inserting this into the equilibrium equation it is obtained EI ω ϕ (x) GI t ϕ (x) m t (x) = 0 (.98) Dividing the equation by the warping stiffness, the equation takes the form where ϕ (x) k l ϕ (x) = f(x) (.99) GIt k = l ja f(x) = m t(x) EI ω EI ω Equation (.99) can be expressed by utilizing the definition of the bimoment also in the form B (x) k l B(x) = m t(x) (.100) 11

12 Solving these equations is categorized into three different cases depending on the mutual value of the torsional stiffness due to Saint Venant and warping torsion. If the coefficient k > 10(...0), Saint Venant torsion is more dominant and the first term due to warping torsion can be dropped out of the equations. This corresponds to beams with massive solid and box-type cross-sections. The differential equation is simplified to the form GI t ϕ (x) = m t (x) (.101) If k < 1, Saint Venant torsion is more or less meaningless. Thus we have pure warping torsion, and it is often met with very thin-walled coldformed steel cross-sections. The differential equation is then of the form EI ω ϕ (x) = m t (x) (.10) The complete equation(.99) with both parts covers the range 1 < k < 10(...0). In this group, there are the hot-rolled steel profiles and thin-walled reinforced concrete girders..9.5 Solution for the differential equation The solution for the homogeneus equation ϕ (x) k l ϕ (x) = 0 (.103) is searched by a trial function ϕ(x) = exp(rx). Inserting this to the differential equation gives the characteristic equation of the problem Its roots (r 4 k l r ) exp(rx) = 0 r 1, = 0 r 3,4 = ± k l are all real numbers, and one of them is double root. Thus, the general solution of the homogeneous equation is ϕ(x) = c 1 + c x + c 3 exp ( k l x) + c 4 exp ( k l x) = c 1 + c x + c 3 c 4 Because the hyperbolic functions are defined by we can write further (exp ( k l x) exp ( k l x)) + c 3 + c 4 (exp ( k l x) + exp ( k l x)) sinh(x) = 1 (exp(x) exp( x)) cosh(x) = 1 (exp(x) + exp( x)) ϕ(x) = C 1 + C x + C 3 sinh ( k l x) + C 4 cosh ( k x) (.104) l 1

13 The complete solution for the differential equation is obtained by combining this with the particular solution corresponding to the cureent loading ϕ(x), or ϕ(x) = C 1 + C x + C 3 sinh ( k l x) + C 4 cosh ( k x) + ϕ(x) (.105) l The constants C 1, C, C 3 and C 4 are solved from the boundary conditions at each end of the beam. The stress rsultants can then be differentiated from the solution as M t = GI t ϕ k (x) = GI t [C + C 3 l cosh (k l x) + C k 4 l sinh (k l x) + ϕ (x)] B = EI ω ϕ (x) = GI t [C 3 sinh ( k l x) + C 4 cosh ( k l x) + ( l k ) ϕ (x)] M ω = B = EI ω ϕ k (x) = GI t [C 3 l cosh (k l x) + C k 4 l sinh (k l x) + ( l k ) ϕ (x)] (c) M x = M t + M ω = GI t [C + ϕ (x) ( l k ) ϕ (x)] (a) (b) (.106d).9.6 Some particular solutions When the beam is loaded by an evenly along the length distributed torque m o t, the inhomogeneous term on the right hand side of the differential equation is of the form ϕ (x) k l ϕ (x) = f(x) f(x) = mo t EI ω The particular solution can be searched by a trial function ϕ(x) = Cx. Inserting this into the differential equation results in the value for C as and the particular solution takes the form C = mo t GI t ϕ(x) = 1 m o t GI t x (.107) If the loading torque is distributed along linearly increasing function m o t x/l the inhomogeneous term in the differential equation is of the form f(x) = mo t x EI ω l and the particular solution is searched bu the trial function of ϕ(x) = Cx 3. Constant C takes the value C = mo t 6GI t l 13

14 and the particular solution is ϕ(x) = 1 6 m o t GI t l x3 (.108) concentrated point torque as loading Mt o at x = a corresponds to the particular solution 0, kun x a ϕ(x) = M t o l [ 1 GI t k sinh(k l (x a)) 1 l (x a)], kun x > a (.109) and a concentrated point bimoment B o at x = a correspondingly to the solution 0, kun x a ϕ(x) = B o [cosh( k GI t l (x a)) 1], kun x > a (.110).9.7 Boundary conditions The solution of a fourth order ordinary differential equation assumes four integration constants to be solved on the base of boundary conditions. Two conditions are usually adopted from each end. The most general restrictions are: 1. Rotation of the cross-section plane about the beam axis is prevented i.e. the angle of torsion is zero ϕ = 0. Warping of the cross section plane is prevented u ω ϕ = 0 = ϕ = 0 3. The known external torque is M o at one end of the beam ( can be also zero ) M x M ω + M t = EI ω ϕ + GI t ϕ = M o 4. The known external bimoment B o at one end of the beam ( can be also zero ) B = EI ω ϕ = B o The traditional suports in the practice: Fixed end with respect to the twisting ϕ = 0, ϕ = 0 (.111) Fully free end of the beam: M x = 0, B = 0 (.11) Simply supported end of the beam: ϕ = 0, B = 0 (.113) 14

15 Figure.7. Beam of the example. Illustrative example Consider a cantilever beam, shown in Figure.7, which is loaded by a uniformly distributed vertical load q along the beam axis. The load passes through the centre of gravity of the cross section. The geometry of the beam and cross-section is given by the mutual ratios of the measures l : b : t = 100 : 10 : 1. Poisson s ratio ν = 0.3. We have been considering the cross-section before and determined the location of the shear center and all the moment of inertia (second moments). The distibuted loading through the centre of gravity generates a distributed torque with the intensity of the load value times the distance between the shear centre and the centre of gravity of the cross-section, or The second moments are m o t = q( 1 3 b b) = 16 1 qb I ω = 5 84 b5 t = b6 I t = bt3 = b4 I z = 7 1 b3 t = 7 10 b4 When ν = 0.3, G = E/.6, the constant k is GIt k = l =.54 EI ω The full solution of the differential equation is the sum of the solution of the homogeneous equation and the particular solution ϕ(x) = C 1 + C x + C 3 sinh ( k l x) + C 4 cosh ( k l x) mo t GI t x From the boundary conditions which are in this case ϕ(0) = 0 = C 1 + C 4 = 0 ϕ (0) = 0 = C + k l C 3 = 0 M x (l) = 0 = GI t C + m o t l = 0 B(l) = 0 = GI t (C 3 sinh(k) + C 4 cosh(k)) ( l k ) m o t = 0 15

16 Kuva.8. Normal stress distributions. we get for the integration constants C 1, C, C 3 and C 4 the values C 1 = C 4 = mo t l C = k l C 3 = mo t l GI t k (k tanh(k) + cosh 1 (k)) kgi t the angle of rotation and the force quatities obtained by differentiating are thus ϕ(x) = mo t l [ x kgi t l sinh(k l x) + (tanh(k) + k 1 cosh 1 (k))(cosh( k ] l x) 1) k(x l ) M t (x) = m o t l [1 xl cosh(kl x) + (tanh(k) + k 1 cosh 1 (k)) sinh( kl ] x) [ M ω (x) = m o t l cosh( kl x) (tanh(k) + k 1 cosh 1 (k)) sinh( kl ] x) B(x) = mo t l [1 + k sinh( kl x) (k tanh(k) + cosh 1 (k)) cosh( kl ] x) k We determine at first the maximum values for the normal stresses due to bending and warping. Both the bimoment and the bending moment take the maximum values at the fixed end of the beam. The maximum value of the bending moment is M max z = M z (0) = 1 ql Correspondingly, the maximum value of the bimoment is B max = B(0) = mo t l [ ] 1 (k tanh(k) + cosh 1 (k)) = 0.580m o t l = 19.66qb 3 k The normal and warping stresses are calculated from the equation σ x = M z I z y(s) + B I ω ω(s) = q b y(s) 330 q b 3 ω(s) Figure.8 shows the normal stress distributions. 16

17 Kuva.9. Shear stress distribution. The shear stresses are also a combination of the stresses due to bending and due to torsion. The shear stresses due to bending are distributed according to the first moment of the cross sectioni S z, when the wall-thickness t is constant.the distribution of the first moment on the cross-section is s s S z (s) = y(s)tds = b tds = bts, kun 0 s b 0 0 s =S z (b) + y(s)tds = b t s + (s 3b )tds = b t 3bt s + t s, kun b s b =S z (b) + b s b y(s)tds = s b b b tds = t 3b + bt s, kun b s 3b The shear force takes its maximum value also at the fixed end of the beam, and it is The shear stresses due to this are Q max = Q y (0) = ql = 10qb τ = Q ys z (s) I z t = 1714 q b 4 S z(s) The torque due to warping M ω takes also its maximum value at the fixed end Mω max = m o t l = qb when the maximum shear stresses due to sectorial torsion are τ ω = M ωs ω (s) = 1800 q I ω t b 5 S ω(s) When the moment resultant due to Saint Venant torsion vanishes at the fixed end, the total shear stresses there consist only of the shear stresses due to bending and warping torsion, and are τ(s) = Q ys z (s) I z t Q ys z (s) I z t = 1714 q b 4 S z(s) 1800 q b 5 S ω(s) Figure.9 shows the distributions separately and as summed together. 17

18 Instead on the other sections of the beam, the shear stresses due to Saint Venant torsion do not vanish. The torque M t (x) reaches its maximum value at x = 0.467l M max t = 0.301m o t l =.93qb t this section the maximum shear stress values are τ t = ± M tt I t = 9.3 q b These stresses are distributed linearly over the wall-thickness and are zero on the centre line of the wall according to Vlasov s assumption. t the same section, the shear stresses due to warping torsion are constant over the wall-thickness, and are (M ω = 0.3m o t l) τ ω (0.467l) = M ωs ω (s) I ω t = 968 q b 5 S ω(s) and also due to bending τ(0.467l) = 0.533qlS z(s) I z t = q b 4 S z(s) Figure.30 shows the corresponding shear stress distributions. It can be observed that the shear stresses due to Saint Venant torsion have clearly higher values as compared to the other effects. For the maximum value of existing shear stresses we get the value τ max = ( ) q b = 94.7q b Kuva.30. Shear stress distributions. 18

19 .9.8 nalogy between the solution procedures of beam in combined tension and bending and alternatively of combined torsion of a thin-walled beam The formulation of the problem of combined torsion of a thin-walled beam is analogous to the analysis of the beam under combined bendig and tension/compression, and results in similar ordinary differential equation. Thus each quantity here has a corresponding one in the bending analysis. The table besides shows the analogy between these two theories. Bending EI z v Nv = q(x) deflection v Torsion EI ω ϕ GI t ϕ = m t (x) angle of torsion ϕ slope v warping displacement ϕ coordinate y sectorial coordinate ω first moment of the cross section S z (s) sectorial first moment of the crosssection S ω (s) moment of inertia (second moment) I z intensity of distributed loading q(x) concentarted load F = lim x 0 x q(x)dx warping constant I ω intensity of distributed torque m t (x) concentrated torque M = lim x 0 x m(x)dx axial normal force N normal force versus shear force Nv Saint Venant torsion stiffness GI t torque due to Saint Venant torsion M t = GI t ϕ bending moment M z = EI z v bimoment B = EI ω ϕ normal stress distribution σ x = M z I z y warping normal stress distribution σ ω = B I ω ω shear force Q = M z = EI z v warping torque M ω = B = EI ω ϕ shear stress distribution τ = Q y(x)s z (s) I z t vertical shear force component V = Nv + Q = Nv EI z v shear stress distribution due to sectorial torsion τ ω = M ω(x)s ω (s) I ω t total internal torque M x = GI t ϕ EI ω ϕ Taulukko.1 nalogy between various quantities and concept in bending and torsions. 19