Aircraft Structures 3E Episode 13 pps

466 Structural constraint Fig. 11.23 (a) Torsion of I-section beam; (b) plan view of beam showing undistorted shape of flanges. Fig. 11.24. Obviously the beam still twists along its length but the rate of twist is no longer constant and the resistance to torsion is provided by the St. Venant shear stres- ses (unrestrained warping) plus the resistance of the flanges to bending. The total torque may therefore be written T = TJ + Tr, where TJ = GJ d8/dz from the uncon- strained torsion of open sections but in which d8/dz is not constant, and Tr is obtained from a consideration of the bending of the flanges. It will be instructive to derive an expression for Tr for the I-section beam of Fig. 11.25 before we turn our attention to the case of a beam of arbitrary section. Suppose that at any section z the angle of twist of the I-beam is 8. Then the lateral displacement u of the lower flange is h U=8- 2 Fig. 11.24 Bending effect of axial constraint on flanges of I-section beam subjected to torsion. 11.5 Constraint of open section beams 467 tY z Fig. 11 25 Torsion of I-section beam fully built-in at one end. and the bending moment MF in the plane of the flange is given by d2u MF = -EIF - dz2 (see Section 9.1 for sign convention) where I, is the second moment of area of theflange cross-section about they axis. It is assumed here that displacements produced by shear are negligible so that the lateral deflection of the flange is completely due to the self-equilibrating direct stress system c7r set up by the bending of the flange. We shall not, however, assume that the shear stresses in the flange are negligible. The shear S, in the flange is then d3 u dMF - SF = -EIFT dz dz or substituting for u in terms of 8 and h h d38 2 dz3 SF = -EIF - - Similarly, there is a shear force in the top flange of the same magnitude but opposite in direction. Together they form a couple which represents the second part Tr of the total torque, thus and the expression for the total torque may be written de h2 d38 dz 2 dz T = GJ EIF - 7 The insight into the physical aspects of the problem gained in the above will be found helpful in the development of the general theory for the arbitrary section beam shown in Fig. 11.26. 468 Structural constraint t’ Fig. 11.26 Torsion of an open section beam fully built-in at one end. The theory, originally developed by Wagner and Kappus, is most generally known as the Wagner torsion bending theory. It assumes that the beam is long compared with its cross-sectional dimensions, that the cross-section remains undistorted by the loading and that the shear strain T~~ of the middle plane of the beam is negligible although the stresses producing the shear strain are not. From similar assumptions is derived, in Section 9.6, an expression for the primary warping w of the beam, viz. In the presence of axial constraint, de/dz is no longer constant so that the longitudi- nal strain aw/az is not zero and direct (also shear) stresses are induced. Thus (11.54) The or stress system must be self-equilibrating since the applied load is a pure torque. Therefore, at any section the resultant end load is zero and IC or,tds = 0 ( IC denotes integration around the beam section 1 or, from Eq. (11.54) and observing that d29/d? is a function of z only 2A~t ds = 0 (11.55) The limits of integration of Eq. (1 1.55) present some difficulty in that AR is zero when w is zero at an unknown value of s. Let 2.4~ = 2A~,o - 2Ak where AR.0 is the area swept out from s = 0 and A; is the value of AR!O at w = 0 (see Fig. 11.27). Then in Eq. (1 1.55) 11.5 Constraint of open section beams 469 Fig. 11.27 Computation of swept area AR. and giving (1 1.56) The axial constraint shear flow system, qr, is in equilibrium with the self- equilibrating direct stress system. Thus, from Eq. (9.22) aqr d0-r -+t-=o as dz I- dqr - a0r dS dZ Hence Substituting for mr from Eq. (1 1.54) and noting that qr = 0 when s = 0, we have d3 B qr = 2AREt 2 ds or Now or, from Eq. (1 1.57) (1 1.57) 470 Structural constraint The integral in this equation is evaluated by substituting p~ = (d/ds)(2A~) and integrating by parts. Thus At each open edge of the beam qr, and therefore $ 2ARtds, is zero so that the integral reduces to - Jc 4Ait ds, giving (1 1.58) where rR = sc 4Ait ds, the torsion-bending constant, and is purely a function of the geometry of the cross-section. The total torque T, which is the sum of the St. Venant torque and the Wagner torsion bending torque, is then written de d3 8 dz dz T = GJ- - ErR 7 (1 1.59) (Note: Compare Eq. (1 1.59) with the expression derived for the I-section beam.) tD of the beam wall so that rR, for a beam with n booms, may be generally written In the expression for rR the thickness t is actually the direct stress carrying thickness n where B, is the cross-sectional area of the rth boom. The calculation of rR enables the second order differential equation in dO/dz (Eq. (1 1.59)) to be solved. The constraint shear flows, qr, follow from Eqs (1 1.57) and (1 1.56) and the longitudinal constraint stresses from Eq. (1 1.54). However, before illustrating the complete method of solution with examples we shall examine the calculation of rR. So far we have referred the swept area AR, and hence rR, to the centre of twist of the beam without locating its position. This may be accomplished as follows. At any section of the beam the resultant of the qr shear flows is a pure torque (as is the resultant of the St. Venant shear stresses) so that in Fig. 11.28 IC qr sin $ds = Sy = 0 Fig. 11.28 Determination of the position of the centre of twist. 11.5 Constraint of open section beams 471 Therefore, from Eq. (1 1.57) Now dY d sin$= , -(UR)=PR ds ds and the above expression may be integrated by parts, thus The first term on the right-hand side vanishes as si 2ARt ds is zero at each open edge of the beam, leaving y2A~t ds = 0 Again integrating by parts The integral in the first term on the right-hand side of the above equation may be recognized, from Chapter 9, as being directly proportional to the shear flow produced in a singly symmetrical open section beam supporting a shear load Sy. Its value is therefore zero at each open edge of the beam. Hence Similarly, for the horizontal component S, to be zero (11.60) (1 1.61) Equations (11.60) and (11.61) hold if the centre of twist coincides with the shear centre of the cross-section. To summarize, the centre of twist of a section of an open section beam carrying a pure torque is the shear centre of the section. We are now in a position to calculate rR. This may be done by evaluating sc 4Ait ds in which 2AR is given by Eq. (1 1 S6). In general, the calculation may be lengthy unless the section has flat sides in which case a convenient analogy shortens the work considerably. For the flat-sided section in Fig. 11.29(a) we first plot the area 2AR:o swept out from the point 1 where we choose s = 0 (Fig. 11.29(b)). The swept area AR,O increases linearly from zero at 1 to (1/2)pI2dl2 at 2 and so on. Note that move- ment along side 23 produces no increment of 2kfR,o as p23 = 0. Further, we adopt a sign convention for p such that p is positive if movement in the positive s direction of the foot of p along the tangent causes anticlockwise rotation about R. The increment of 2AR.0 from side 34 is therefore negative. 472 Structural constraint 2AR10 cs 1-34 5 6 di2 d23 Etc. (a) (b) Fig. 11.29 Computation of torsion bending constant rR: (a) dimensions of flat-sided open section beam; (b) variation of 2AR,o around beam section. In the derivation of Eq. (1 1.56) we showed that Suppose now that the line 1’2’3‘. . .6‘ is a wire of varying density such that the weight of each element 6s’ is tSs. Thus the weight of length 1‘2’ is tdI2 etc. They coordinate of the centre of gravity of the ‘wire’ is then Comparing this expression with the previous one for 2AL, y and J are clearly analo- gous to 2AR,o and 2Ak respectively. Further Expanding and substituting 2Ak IC t ds for jc 2AR,ot ds gives rR = (2A~,o)~tds - (2AL)’ tds (1 1.62) Thus, in Eq. (1 1.62), rR is analogous to the moment of inertia of the ‘wire’ about an axis through its centre of gravity parallel to the s axis. I I Example 11.2 An open section beam of length L has the section shown in Fig. 11.30. The beam is My built-in at one end and carries a pure torque T. Derive expressions for the direct stress and shear flow distributions produced by the axial constraint (the or and qr systems) and the rate of twist of the beam. 11.5 Constraint of open section beams 473 2 Fig. 11.30 Section of axially constrained open section beam under torsion. The beam is loaded by a pure torque so that the axis of twist passes through the shear centre S(R) of each section. We shall take the origin for s at the point 1 and initially plot 2AR-o against s to determine rR (see Fig. 11.31). The position of the centre of gravity, (2A’,), of the wire 1’2’3‘4’ is found by taking moments about the s axis. Thus t(2d+h)2A’, = td - + th - + rd - (3 (7) (hqd) from which hd(h + d) 2(h + 2d) 3A‘, = 474 Structural constraint rR follows from the moment of inertia of the ‘wire’ about an axis through its centre of gravity. Hence which simplifies to td3h2 2h+d rR =- 12 (-) h+2d Equation (1 1.59), that is de d3 8 dz dz T=GJ EE~RT may now be solved for dO/dz. Rearranging and writing p2 = GJ/ErR we have d38 2d8 T dz3 z=-p The solution of Eq. (iii) is of standard form, i.e. (ii) (iii) de T dz GJ - + Acoshpz + Bsinhpz The constants A and B are found from the boundary conditions. dO/dz = 0 at the built-in end. load. Therefore, from Eq. (1 1.54), d’O/d2 = 0 at the free end. (1) At the built-in end the warping w = 0 and since w = -2ARd8/dz then (2) At the free end gr = 0, as there is no constraint and no externally applied direct From (1) From (2) so that or A = -T/GJ B = (T/GJ) tanh pL d0 T - = - (1 - coshpz + tanhpLsinh pz) dz GJ dz GJ cosh pL 1 The first term in Eq. (iv) is seen to be the rate of twist derived from the St. Venant torsion theory. The hyperbolic second term is therefore the modification introduced by the axial constraint. Equation (iv) may be integrated to find the distribution of angle of twist 8, the appropriate boundary condition being 8 = 0 at the built-in end. Thus (4 1 sinh p(L - z) sinh pL - GJ p cosh pL p cosh pL 11.5 Constraint of open section beams 475 z z=L Fig. 11.32 Stiffening effect of axial constraint. and the angle of twist, at the free end of the beam is ei=:E==(1-7) TL tanh pL Plotting 6 against z (Fig. 1 1.32) illustrates the stiffening effect of axial constraint on the beam. The decrease in the effect of axial constraint towards the free end of the beam is shown by an examination of the variation of the St. Venant ( TJ) and Wagner (Tr) torques along the beam. From Eq. (iv) and d38 coshp(L - z) dz3 cosh pL Tr = -ErR - = T (vii) (viii) Tj and Tr are now plotted against z as fractions of the total torque T (Fig. 11.33). At the built-in end the entire torque is carried by the Wagner stresses, but although the constraint effect diminishes towards the free end it does not disappear entirely. This is due to the fact that the axial constraint shear flow, qr, does not vanish at z = L, for at this section (and all other sections) d38/dz3 is not zero. Fig. 11.33 Distribution of St. Venant and torsion-bending torques along the length of the open section beam shown in Fig. 11.30. [...]... are expressed in terms of displacements or forces Generally, as we have previously noted, actual structures must be idealized to some extent before they become amenable to analysis Examples of some simple idealizations and their effect on structural analysis have been presented in Chapter 9 for aircraft structures Outside the realms of aeronautical engineering the representation of a truss girder by... is the continuum structure; in this category are dams, plates, shells and, obviously, aircraft fuselage and wing skins A method, extending the matrix technique for skeletal structures, of representing continua by any desired number of elements connected at their nodes was developed by Clough et aL2 at the Boeing Aircraft Company and the University of Berkeley in California The elements may be of any... system is [q= [ ka -k, 0 -ka ka+kb -kb 4.1 (12 .13) Equation (12 .13) is a symmetric matrix of order 3 x 3 It is important to note that the order of a stiffness matrix may be predicted from a knowledge of the number of nodal forces and displacements For example, Eq (12.7) is a 2 x 2 matrix connecting two nodal forces with two nodal displacements; Eq (12 .13) is a 3 x 3 matrix relating three nodal forces... the realms of aeronautical engineering the representation of a truss girder by a pin-jointed framework is a well known example of the 12.1 Notation idealization of what are known as ‘skeletal’ structures Such structures are assumed to consist of a number of elements joined at points called nodes The behaviour of each element may be determined by basic methods of structural analysis and hence the behaviour... relative twist between the free ends is given by TI 9- 2p1( 10cosh2p l - 1) where p2 = GJ/EI' and G = shear modulus (constant throughout) Fig P.lt.14 Matrix methods of structura I analysis Actual aircraft structures consist of numerous components generally arranged in an irregular manner These components are usually continuous and therefore, theoretically, possess an infinite number of degrees of freedom... thefinite element method, as it is known, in greater detail later Initially, we shall develop the matrix stiffness method of solution for simple skeletal and beam structures The fundamentals of matrix algebra are assumed Generally we shall consider structures subjected to forces, Fr,l, Fy,l, z , l ,Fy,2,F,.,2, F Fr,2, , F,.+ Fy,,, F,,,, at nodes 1, 2, ., n at which the displacements are u l , V I , wl,... thickness t Show that the torsion bending constant about the shear centre S is t' Fig P.11.12 P.11 .13 A thin-walled, I-section beam, of constant wall thickness t, is mounted as a cantilever with its web horizontal At the tip, a downward force is applied in the plane of one of the flanges, as shown in Fig P 11 .13 Assuming the necessary results of the elementary theory of bending, the St Venant theory of... actual structure Standard methods of structural analysis, some of which we have discussed in the preceding chapters, are inadequate for coping with the necessary degree of complexity in such idealized structures It was this situation which led, in the late 1940s and early 1950s, to the development of matrix methods of analysis and at the same time to the emergence of high-speed, electronic, digital... This procedure has been discussed to some extent in Chapter 9 where we noted that the greater the simplification introduced by the idealization the less complex but more inaccurate became the analysis In aircraft design, where structural weight is of paramount importance, an accurate knowledge of component loads and stresses is essential so that at some stage in the design these must be calculated as accurately... remarks in the preceding paragraph and by reference to Eq (12.2) we could have deduced at the outset of the analysis that the stiffness matrix for the two-spring assembly would be of the form kll kl2 k13 [lul = k21 k22 k31 k32 [ (12.14) The element kl of this matrix relates the force at node 1 I the displacemen at node 1 and so on Hence, remembering the stiffness matrix for the single spring (Eq (12.7))