Aircraft Structures 3E Episode 9 pptx

306 Open and closed, thin-walled beams 1 9a 6a Fig. 9.26 Closed section beam of Example 9.6. Eq. (9.35) simplifies to S q 2 t s - p Yd.S+%,O Ixx 0 3 in which I,, = 2 [ 1; t ( $SI>’ dsl + 1: f ( As2>’ ds2] Evaluating this expression gives I,, = 1 152a3t. for the wall 41 The basic shear flow distribution qb is obtained from the first term in Eq. (i). Thus, In the wall 12 which gives (iii) The qb distributions in the walls 23 and 34 follow from symmetry. Hence from Eq. (9.48) giving qs,o = - ” (58.7;) 1 1 52a3 9.5 Torsion of closed section beams 307 Taking moments about the point 2 we have or Spa sin e 1 loa (- ?s: + 58.7~~) dsl SY(& + 9a) = 1152a3 0 We may replace sin 6’ by sin(O1 - 0,) = sin el cos O2 - cos O1 sin Q2 where sin O1 = 15/17, cosO2 = 8/10, cosQ1 = 8/17 and sine2 = 6/10. Substituting these values and integrating Eq. (v) gives & = -3.35a which means that the shear centre is inside the beam section. A closed section beam subjected to a pure torque T as shown in Fig. 9.27 does not, in the absence of an axial constraint, develop a direct stress system. It follows that the equilibrium conditions of Eqs (9.22) and (9.23) reduce to dq/ds = 0 and dq/dz = 0 respectively. These relationships may only be satisfied simultaneously by a constant value of q. We deduce, therefore, that the application of a pure torque to a closed section beam results in the development of a constant shear flow in the beam wall. However, the shear stress 7 may vary around the cross-section since we allow the wall thickness t to be a function of s. The relationship between the applied torque and this constant shear flow is simply derived by considering the torsional equilibrium of the section shown in Fig. 9.28. The torque produced by the shear flow acting on an element 6s of the beam wall is pq6s. Hence or, since q is constant and fpds = 2A (as before) T = 2Aq X z (9.49) Fig. 9.27 Torsion of a closed section beam 308 Open and closed, thin-walled beams 't Fig. 9.28 Determination of the shear flow distribution in a closed section beam subjected to torsion. Note that the origin 0 of the axes in Fig. 9.28 may be positioned in or outside the cross-section of the beam since the moment of the internal shear flows (whose resul- tant is a pure torque) is the same about any point in their plane. For an origin outside the cross-section the term $p ds will involve the summation of positive and negative areas. The sign of an area is determined by the sign ofp which itself is associated with the sign convention for torque as follows. If the movement of the foot of p along the tangent at any point in the positive direction of s leads to an anticlockwise rotation of p about the origin of axes, p is positive. The positive direction of s is in the positive direction of q which is anticlockwise (corresponding to a positive torque). Thus, in Fig. 9.29 a generator OA, rotating about 0, will initially sweep out a negative area since PA is negative. At B, however, pB is positive so that the area swept out by the generator has changed sign (at the point where the tangent passes through 0 and p = 0). Positive and negative areas cancel each other out as they overlap so that as the generator moves completely around the section, starting and returning to A say, the resultant area is that enclosed by the profile of the beam. Fig. 9.29 Sign convention for swept areas. 9.5 Torsion of closed section beams 309 The theory of the torsion of closed section beams is known as the Bredt-Batho rlteory and Eq. (9.49) is often referred to as the Bredt-Batho formula. 9.5.1 Displacements associated with the Bredt-Batho shear flow The relationship between q and shear strain established in Eq. (9.39), namely q=Gt (E -+- 2) is valid for the pure torsion case where q is constant. Differentiating this expression with respect to z we have or (9.50) In the absence of direct stresses the longitudinal strain div/az( = E,) is zero so that Hence from Eq. (9.27) d28 d’u d% dz2 dz2 dz2 p- + -cos + + -sin @ = 0 (9.51) For Eq. (9.51) to hold for all points around the section wall, in other words for all values of + d2 8 d2u d2v dz- 7=0, &2-0, dz2 - It follows that 8 = Az + B, u = Cz + D, v = Ez + F, where A, B, C, D, E and F are unknown constants. Thus 8, w and v are all linear functions of z. Equation (9.42), relating the rate of twist to the variable shear flow qs developed in a shear loaded closed section beam, is also valid for the case qs = q = constant. Hence d6’ which becomes, on substituting for q from Eq. (9.49) (9.52) The warping distribution produced by a varying shear flow, as defined by Eq. (9.45) for axes having their origin at the centre of twist, is also applicable to the case of a 3 10 Open and closed, thin-walled beams a t constant shear flow. Thus Replacing q from Eq. (9.49) we have (9.53) where The sign of the warping displacement in Eq. (9.53) is governed by the sign of the applied torque T and the signs of the parameters So, and Aos. Having specified initially that a positive torque is anticlockwise, the signs of So, and Aos are fixed in that So, is positive when s is positive, i.e. s is taken as positive in an anticlockwise sense, and Aos is positive when, as before, p (see Fig. 9.29) is positive. We have noted that the longitudinal strain E, is zero in a closed section beam sub- jected to a pure torque. This means that all sections of the beam must possess identical warping distributions. In other words longitudinal generators of the beam surface remain unchanged in length although subjected to axial displacement. Example 9.7 Determine the warping distribution in the doubly symmetrical rectangular, closed section beam, shown in Fig. 9.30, when subjected to an anticlockwise torque T. From symmetry the centre of twist R will coincide with the mid-point of the cross- section and points of zero warping will lie on the axes of symmetry at the mid-points of the sides. We shall therefore take the origin for s at the mid-point of side 14 and measure s in the positive, anticlockwise, sense around the section. Assuming the shear modulus G to be constant we rewrite Eq. (9.53) in the form 2 3t 54 I t l4 Fig. 9.30 Torsion of a rectangular section beam. 9.5 Torsion of closed section beams 31 1 where In Eq. (i) wo=O, S=2 -+- and A=ab (: t) From 0 to 1,0 < s1 < b/2 and Note that Sos and Aos are both positive. Substitution for So, and Aos from Eq. (ii) in Eq. (i) shows that the warping distribu- tion in the wall 01, wol, is linear. Also which gives IV] =-( ) T ba 8abG tb (iii) The remainder of the warping distribution may be deduced from symmetry and the fact that the warping must be zero at points where the axes of symmetry and the walls of the cross-section intersect. It follows that w2 = -w1 = -w3 = w4 giving the distribution shown in Fig. 9.31. Note that the warping distribution will take the form shown in Fig. 9.31 as long as Tis positive and b/tb > a/t,. If either of these conditions is reversed w1 and w3 will become negative and H:., and w4 positive. In the case when b/tb = a/t, the warping is zero at all points in the cross-section. Fig. 9.31 Warping distribution in the rectangular section beam of Example 9.7. 3 12 Open and closed, thin-walled beams 2 t a 1 Fig. 9.32 Arbitrary origin for s. Suppose now that the origin for s is chosen arbitrarily at, say, point 1. Then, from Fig. 9.32, So, in the wall 12 = q/t, and Aos = $sIb/2 = Slb/4 and both are positive. Substituting in Eq. (i) and setting wo = 0 so that +vi2 varies linearly from zero at 1 to I T ba U wp=-2 -+- 2abG (tb tu) [ 2(b/tb + u/t,)t, 4 at 2. Thus or .;= ( ) T ba 4abG tb Similarly The warping distribution therefore varies linearly from a value -T(b/rb - a/tu)/4abG at 2 to zero at 3. The remaining distribution follows from symmetry so that the complete distribution takes the form shown in Fig. 9.33. Comparing Figs 9.31 and 9.33 it can be seen that the form of the warping distribu- tion is the same but that in the latter case the complete distribution has been displaced axially. The actual value of the warping at the origin for s is found using Eq. (9.46). Thus (vii) 9.5 Torsion of closed section beams 3 13 4 Fig. 9.33 Warping distribution produced by selecting an arbitrary origin for s. Substituting in Eq. (vii) for wi2 and )vi3 from Eqs (iv) and (vi) respectively and evaluating gives (viii) Subtracting this value from the values of w:(= 0) and d’(= -T(b/tb - a/tU)/4abG) we have as before. Note that setting wo = 0 in Eq. (i) implies that wo, the actual value of warping at the origin for s, has been added to all warping displacements. This value must therefore be subtracted from the calculated warping displacements (i.e. those based on an arbitrary choice of origin) to obtain true values. It is instructive at this stage to examine the mechanics of warping to see how it arises. Suppose that each end of the rectangular section beam of Example 9.7 rotates through opposite angles 8 giving a total angle of twist 28 along its length L. The corner 1 at one end of the beam is displaced by amounts a8/2 vertically and b8/2 horizontally as shown in Fig. 9.34. Consider now the displacements of the web and cover of the beam due to rotation. From Figs 9.34 and 9.35(a) and (b) it can be seen that the angles of rotation of the web and the cover are, respectively 4b = (ae/2)/(~/2) = ae/L and 4, = (b8/2)/(L/2) = bB/L The axial displacements of the corner 1 in the web and cover are then b a8 a be 2L’ 2L respectively, as shown in Figs 9.35(a) and (b). In addition to displacements produced by twisting, the webs and covers are subjected to shear strains ’yb and corresponding to __ __ 314 Open and closed, thin-walled beams Fig. 9.34 Twisting of a rectangular section beam. the shear stress system given by Eq. (9.49). Due to yb the axial displacement of corner 1 in the web is ybb/2 in the positive z direction while in the cover the displacement is yaa/2 in the negative z direction. Note that the shear strains yb and ya correspond to the shear stress system produced by a positive anticlockwise torque. Clearly the total axial displacement of the point 1 in the web and cover must be the same so that ba0 babe a -+^lYb-=- ^la- 2L 2 2L 2 from which The shear strains are obtained from Eq. (9.49) and are T T ^la==' yb =- I_ 7- a012 Fig. 9.35 Displacement due to rotation (b) Cover Displacement due to shear strain (a) Web Displacements due to twist and shear strain. 9.5 Torsion of closed section beams 31 5 whence TL The total angle of twist from end to end of the beam is 28, therefore 28 T /2a 2b\ or as in Eq. (9.52). corner 1 gives the warping wl at 1. Thus Substituting for 8 in either of the expressions for the axial displacement of the ab TL fa b\ T a i.e. wl=-( ) T ba 8abG tb as before. It can be seen that the warping of the cross-section is produced by a com- bination of the displacements caused by twisting and the displacements due to the shear strains; these shear strains correspond to the shear stresses whose values are fixed by statics. The angle of twist must therefore be such as to ensure compatibility of displacement between the webs and covers. 9.5.2 Condition for zero warping at a section The geometry of the cross-section of a closed section beam subjected to torsion may be such that no warping of the cross-section occurs. From Eq. (9.53) we see that this condition arises when or Differentiating Eq. (9.54) with respect to s gives (9.54) [...]... manner to that of a closed section beam From Fig 9. 16 (9. 63) 3 18 Open and closed, thin-walled beams Referring the tangential displacement wt to the centre of twist R of the cross-section we have, from Eq (9. 28) (9. 64) Substituting for dwt/dz in Eq (9. 63) gives from which (9. 65) On the mid-line of the section wall rzs= 0 (see Eq (9. 57)) so that, from Eq (9. 65) Integrating this expression with respect... idealize the panel of Fig 9. 47(a) into a combination of direct stress carrying booms and shear stress only carrying skin as shown in Fig 9. 47(b) In Fig 9. 47(a) the direct stress carrying thickness tD of the skin is equal to Fig 9. 46 Idealization of a wing section 9. 8 Structural idealization 3 29 (a) Actual (b) Idealized Fig 9. 47 Idealization of a panel its actual thickness t while in Fig 9. 47(b) tD = 0 Suppose... Fig 9. 54 (9. 79) in which A is the area endosed by the web and the lines joining the ends of the web to the point 0 This result may be used to determine the distance of the line of action of the resultant shear force from any point From Fig 9. 54 Se = 2Aq12 from which Fig 9. 54 Moment produced by a constant shear flow 338 Open and closed, thin-walled beams Substituting for q12from Eq (9. 78) gives 9. 9.3... section properties contained in Eqs (9. 6) and (9. 7) refer to the direct stress carrying area of the beam 9. 9 Effect of idealization 341 n N P Fig 9. 57 Alternative solution to Example 9. 13 section In cases where the shear force is not constant over the unit length of beam the method is approximate We shall illustrate the method by applying it to Example 9. 13 In Fig 9. 51 the shear load of 4.8 kN is applied... crosssection of the beam remains undistorted by the loading Example 9. 9 Determine the shear flow distribution in the beam section shown in Fig 9. 42, when it is subjected to a shear load in its vertical plane of symmetry The thickness of the walls of the section is 2 mm throughout Y I i t s4 C X 1 8 4 s5 - 5 9 200 mm Fig 9. 42 Beam section of Example 9. 9 - _ - I 1OOmm 324 Open and closed, thin-walled beams The... becomes dB rzs= 2Gn -, (9. 57) r,,, = 0 dz Eq (3.28) becomes (9. 58) and Eq (3. 29) is J = C - 1 st3 or J = 3 3 Lt t3ds (9. 59) In Eq (9. 59) the second expression for the torsion constant is used if the cross-section has a variable wall thickness Finally, the rate of twist is expressed in terms of the applied torque by Eq (3.12), viz T=GJ- d0 dz (9. 60) The shear stress distribution and the maximum shear stress... section beam is based on the equilibrium equation (9. 22) The thickness r in this 9. 9 Effect of idealization 333 Table 9. 1 0 0 1 2 3 4 5 6 7 8 9 0 uz(N/mm2) B [mm2) Boom +660 +600 +420 +228 + 25 -204 - 396 -502 -540 640 600 600 600 620 640 640 8 50 640 278 x 216 x 106 x 31 x 0.4 x 27 x lo6 IO6 IO6 lo6 lo6 35.6 32.3 22.6 12.3 1.3 -11.0 -21.4 -27.0 - 29. 0 106 100 x 106 214 x IO6 187 x IO6 equation refers... beam with booms; (b) equilibrium of boom element 334 Open and closed, thin-walled beams area Br are q1 and q2 Then, from Fig 9. 50(b) which simplifies to (9. 73) Substituting for czin Eq (9. 73) from Eq (9. 6) we have or, using the relationships of Eqs (9. 11) and (9. 12) Equation (9. 74) gives the change in shear flow induced by a boom which itself is subjected to a direct load (czBr).Each time a boom is... is usually ignored in the thin-walled sections common to aircraft structures Equation (9. 66) may be rewritten in the form (9. 67) or, in terms of the applied torque W, 4s : = -2A (see Eq (9. 60)) (9. 68) in which A R = pR ds is the area swept out by a generator, rotating about the centre of twist, from the point of zero warping, as shown in Fig 9. 37 The sign of w,, for a given direction of torque, depends... remaining warping distribution follows from symmetry and the complete distribution is shown in Fig 9. 39 In unsymmetrical section beams the position of the point of zero 4 Fig 9. 39 Warping distribution in channel section of Example 9. 8 9. 6 Torsion of open section beams 321 i , -t 8.04mm 2.5mm 1.5 mrn 4 s3 M Fig 9. 40 Determination of points of zero warping warping is not known but may be found using the method . rzs = 2Gn -, r,,, = 0 Eq. (3.28) becomes (9. 57) (9. 58) and Eq. (3. 29) is 1 (9. 59) st3 J=C- or J=- t3ds In Eq. (9. 59) the second expression for the torsion constant is. 9. 29 Sign convention for swept areas. 9. 5 Torsion of closed section beams 3 09 The theory of the torsion of closed section beams is known as the Bredt-Batho rlteory and Eq. (9. 49) . thin-walled sections common to aircraft structures. Equation (9. 66) may be rewritten in the form or, in terms of the applied torque W, = -2A (see Eq. (9. 60)) (9. 67) (9. 68) in which AR =