Aircraft structures for engineering students - part 7 ppt

352 Open and closed, thin-walled beams Calculate and sketch the distribution of shear flow due to a vertical shear force S, acting through the shear centre S and note the principal values. Show also that the distance & of the shear centre from the nose of the section is tS = 1/2( 1 + a/b). Am. q2 = q4 = 3bSY/2h(b + a), q3 = 3SY/2h. Parabolic distributions. P.9.15 Show that the position of the shear centre S with respect to the intersection of the web and lower flange of the thin-walled section shown in Fig. P.9.15, is given by 5's = -45a/97, 7s = 46a/97 Fig. P.9.15 P.9.16 Figure P.9.16 shows the regular hexagonal cross-section of a thin-walled beam of sides a and constant wall thickness t. The beam is subjected to a transverse shear force S, its line of action being along a side of the hexagon, as shown. Find the rate of twist of the beam in terms oft, a, S and the shear modulus G. Plot the shear flow distribution around the section, with values in terms of S and a. Fig. P.9.16 Problems 353 Ans. dO/dz = 0.192S/Gta2 (clockwise) q1 = -OXS/a, q4 = q6 = 0.13S/a, q2 = qs = -0.47S/a, q5 = 0.18S/a q3 = 47 = -0.17S/a Parabolic distributions, q positive clockwise. P.9.17 Figure P.9.17 shows the cross-section of a single cell, thin-walled beam with a horizontal axis of symmetry. The direct stresses are carried by the booms B1 to B4, while the walls are effective only in carrying shear stresses. Assuming that the basic theory of bending is applicable, calculate the position of the shear centre S. The shear modulus G is the same for all walls. Cell area = 135000mm2. Boom areas: B1 = B4 = 450mm , B2 = B3 = 550mm . 2 2 Wall Length (mm) Thickness (mm) 12, 34 23 41 500 580 200 0.8 1 .o 1.2 Ans. 197.2mm from vertical through booms 2 and 3. 100 mm 100 mm - - 1.0 mm 100 mm 0.8 mm 500 mm Fig. P.9.17 P.9.18 A thin-walled closed section beam of constant wall thickness t has the cross-section shown in Fig. P.9.18. Fig. P.9.18 354 Open and closed, thin-walled beams Assuming that the direct stresses are distributed according to the basic theory of bending, calculate and sketch the shear flow distribution for a vertical shear force S,, applied tangentially to the curved part of the beam. Ans. qol = S,,( 1.61 cos 8 - 0.80)/r q12 = Sy(0.57SS - 1.14rs - 0.33)/r P.9.19 A uniform thin-walled beam of constant wall thickness t has a cross- section in the shape of an isosceles triangle and is loaded with a vertical shear force Sy applied at the apex. Assuming that the distribution of shear stress is according to the basic theory of bending, calculate the distribution of shear flow over the cross-section. Illustrate your answer with a suitable sketch, marking in carefully with arrows the direction of the shear flows and noting the principal values. Ans. q12 = SY(33/d - h - 3d)/h(h + 2d) q23 = S,,(-6$ + 6h~2 - h2)/h2(h + 2d) 3 Fig. P.9.19 P.9.20 Find the position of the shear centre of the rectangular four boom beam section shown in Fig. P.9.20. The booms carry only direct stresses but the skin is fully effective in carrying both shear and direct stress. The area of each boom is lO0mm2. Ans. 142.5 mm from side 23. 3 I 14 I- 240 mm I Fig. P.9.20 Problems 355 I 250 mm P.9.21 A uniform, thin-walled, cantilever beam of closed rectangular cross- section has the dimensions shown in Fig. P.9.21. The shear modulus G of the top and bottom covers of the beam is 18 000 N/mm2 while that of the vertical webs is 26 000 N/m' . The beam is subjected to a uniformly distributed torque of 20 Nm/mm along its length. Calculate the maximum shear stress according to the Bredt-Batho theory of torsion. Calculate also, and sketch, the distribution of twist along the length of the cantilever assuming that axial constraint effects are negligible. Am. T~~ = 83.3N/mm2, 0 = 8.14 x lop9 t 2.1 mm 2.1 mm 1.2mm 1 11 11.2 mm Fig. P.9.21 P.9.22 A single cell, thin-walled beam with the double trapezoidal cross-section shown in Fig. P.9.22, is subjected to a constant torque T = 90 500 N m and is con- strained to twist about an axis through the point R. Assuming that the shear stresses are distributed according to the Bredt-Batho theory of torsion, calculate the distribu- tion of warping around the cross-section. Illustrate your answer clearly by means of a sketch and insert the principal values of the warping displacements. The shear modulus G = 27 500 N/mm2 and is constant throughout. AFZS. Wi = -Wg = -0.53m, W2 = -W5 = O.O5mm, W3 = -W4 = 0.38m. Linear distribution. 356 Open and closed, thin-walled beams 1.25 mm 3 1.25 mm r- 1.25 mm 500mm d+ 890 mm Fig. P.9.22 P.9.23 A uniform thin-walled beam is circular in cross-section and has a constant thickness of 2.5 mm. The beam is 2000 mm long, carrying end torques of 450 N m and, in the same sense, a distributed torque loading of 1 .O N m/mm. The loads are reacted by equal couples R at sections 500 mm distant from each end (Fig. P.9.23). Calculate the maximum shear stress in the beam and sketch the distribution of twist along its length. Take G = 30 000 N/mm2 and neglect axial constraint effects. Am. r,, = 24.2N/mm2, 8 = -0.85 .x 10-82rad, 0 < z < 500mm, 8 = 1.7 x 10-8(1450~ - z2/2) - 12.33 x rad, 500 < z < 1OOOmm Fig. P.9.23 P.9.24 A uniform closed section beam, of the thin-walled section shown in Fig. P.9.24, is subjected to a twisting couple of 4500Nm. The beam is constrained to twist about a longitudinal axis through the centre C of the semicircular arc 12. For the curved wall 12 the thickness is 2 mm and the shear modulus is 22 000 N/mm2. For the plane walls 23, 34 and 41, the corresponding figures are 1.6mm and 27 500 N/mm2. (Note: Gt = constant.) Calculate the rate of twist in radians/mm. Give a sketch illustrating the distribution of warping displacement in the cross-section and quote values at points 1 and 4. Problems 357 Fig. P.9.24 Am. de/& = 29.3 x rad/mm, w3 = -w4 = -0.19 mm, wz = - ~1 = -0.056m P.9.25 A uniform beam with the doubly symmetrical cross-section shown in Fig. P.9.25, has horizontal and vertical walls made of different materials which have shear moduli G, and Gb respectively. If for any material the ratio mass density/shear modulus is constant find the ratio of the wall thicknesses tu and tb, so that for a given torsional stiffness and given dimensions a, b the beam has minimum weight per unit span. Assume the Bredt-Batho theory of torsion is valid. If this thickness requirement is satisfied find the a/b ratio (previously regarded as fixed), which gives minimum weight for given torsional stiffness. Ans. tb/ta = Gu/Gb, b/a = 1. Fig. P.9.25 P.9.26 Figure P.9.26 shows the cross-section of a thin-walled beam in the form of a channel with lipped flanges. The lips are of constant thickness 1.27 mm while the flanges increase linearly in thickness from 1.27mm where they meet the lips to 2.54mm at their junctions with the web. The web has a constant thickness of 2.54 mm. The shear modulus G is 26 700 N/mmz throughout. The beam has an enforced axis of twist RR' and is supported in such a way that warping occurs freely but is zero at the mid-point of the web. If the beam carries a torque of 100Nm, calculate the maximum shear stress according to the St. Venant 358 Open and closed, thin-walled beams ~225 mml 50mm I 100 2.54 mm 1.27 mm Fig. P.9.26 theory of torsion for thin-walled sections. Ignore any effects of stress concentration at the corners. Find also the distribution of warping along the middle line of the section, illustrating your results by means of a sketch. hs. Tma = f 297.4N/m2, W1 = -5.48m = -Wg, w2 = 5.48mm = -w5, w3 = 17.98mm = -w4 P.9.27 The thin-walled section shown in Fig. P.9.27 is symmetrical about the x axis. The thickness to of the centre web 34 is constant, while the thickness of the other walls varies linearly from to at points 3 and 4 to zero at the open ends 1, 6, 7 and 8. Determine the St. Venant torsion constant J for the section and also the maximum value of the shear stress due to a torque T. If the section is constrained to twist about an axis through the origin 0, plot the relative warping displacements of the section per unit rate of twist. Problems 359 1 X 6 P.9.28. A uniform beam with the cross-section shown in Fig. P.9.28(a) is sup- ported and loaded as shown in Fig. P.9.28(b). If the direct and shear stresses are given by the basic theory of bending, the direct stresses being carried by the booms and the shear stresses by the walls, calculate the vertical deflection at the ends of the beam when the loads act through the shear centres of the end cross-sections, allowing for the effect of shear strains. 100 mrn 77- 100 mm t- ___ -t t2.5mm 2 t 1 100 mrn ___ A I 75 rnrn 100 mm A ! 75 rnrn t Fig. P.9.28(a) 360 Open and closed, thin-walled beams 4450 N v v n a $4450 N I- _- + A Fig. P.9.28(b) t/2 I ,i t/2 Take E = 69 000 N/mm2 and G = 26 700 N/mm2 Boom areas: 1, 3,4, 6 = 650mm2, 2, 5 = 1300mm2 Am. 3.4mm. P.9.29 A cantilever, length L, has a hollow cross-section in the form of a doubly symmetric wedge as shown in Fig. P.9.29. The chord line is of length c, wedge thickness is t, the length of a sloping side is a/2 and the wall thickness is constant and equal to to. Uniform pressure distributions of magnitudes shown act on the i 1.2po/unit area polunit area 4 + c/2 c/2 1 Fig. P.9.29 Problems 361 faces of the wedge. Find the vertical deflection of point A due to this given loading. If G=0.4E, t/c=0.05 and L =2c show that this deflection is approximately 5600p0c2/Et0. P.9.30 A rectangular section thin-walled beam of length L and breadth 3b, depth b and wall thickness t is built in at one end (Fig. P.9.30). The upper surface of the beam is subjected to a pressure which vanes linearly across the breadth from a value po at edge AB to zero at edge CD. Thus, at any given value of x the pressure is constant in the z direction. Find the vertical deflection of point A. Ans. po L' (9 L2/80 Eb2 + 1 609/2000G) / r. X , I Fig. P.9.30 t- 3b [...]... 0 @ 6x,/6r Boom Pz, (kN) 6y,lSz -1 00 0.1 1 2 3 4 5 6 -1 33 -1 00 100 133 100 0 -0 .1 -0 .1 0 0.1 -0 .05 -0 .05 -0 .05 0.05 0.05 0.05 Py,, a (kN) (kN) P, (kN) -1 0 0 10 -1 0 0 IO 5 6 .7 5 5 6 .7 5 -1 01.3 -1 77 .3 -1 01.3 101.3 177 .3 101.3 Px,r a @ @ 5, rlr 0.6 0 0.6 0.6 0 0.6 0.3 0.3 0.3 0.3 0.3 0.3 3 0 -3 -3 0 3 0 Px,Jr (m) (m) P?& (kNm) (kNm) -3 0 3 3 0 -3 370 Stress analysis of aircraft components From column... 300 = -3 3.2N/mm x 1200 x 300 = -7 7. 5N/mm x 900 x 300 = -1 10.7N/mm = -7 7. 5 N/mm (from symmetry) qb,56 = -3 3.2 N/mm (from symmetry) giving the distribution shown in Fig 10 .7 Taking moments about the centre of symmetry we have, from Eq (10.16) + 2 x 77 .5 x 600 x 300 + 110 .7 x 600 x 600 + 2 x 1200 x 600qs,0 -1 00 x lo3 x 600 = 2 x 33.2 x 600 x 300 Izll 33.2 10.1 Tapered beams 371 77 .5 110 .7 6 33.2 4 i 77 .5... 378 Stress analysis of aircraft components Table 1 03 Skin panel 1 2 3 4 5 6 7 8 1 16 15 14 13 12 11 10 2 3 4 5 6 7 8 9 16 15 14 13 12 11 10 9 Boom 2 3 4 5 6 1 8 1 16 15 14 13 12 11 10 B, (mm’) Y , (m) qb 216.6 216.6 216 .7 216.1 216.6 216.6 216.6 216.6 216.6 216.6 352.0 269.5 145.8 0 -1 45.8 -2 69.5 -3 52.0 381.0 352.0 269.5 145.8 0 -1 45.8 -2 69.5 -3 52.0 0 -3 0.3 -5 3.5 -6 6.0 -6 6.0 -5 3.5 -3 0.3 0 -3 2.8 -6 3.1... = -0 .364~~ 5.22 x lo8 Yr Hence Pi = P3 = -P4 = -P6 = -0 .364 X 295 X 900 = -9 6642N and P2 -Ps = -0 .364 x 295 x 1200 = -1 28 856N At a section 1.9 m from the built-in end M, = -1 00 x 2.1 = -2 lOkNm and the section dimensions are width = 1.22m, depth = 0.61 m so that I,, = 4 x 900 x 305’ + 2 x 1200 x 305’ = 5.58 x lo8mm4 and 0 z > r= -2 10 x lo6 = -0 . 376 ~~ 5.58 x lo8 Yr Hence Pi = P3 = -P4 = -P6 = -0 . 376 ... lie on the neutral axis of the section and are therefore unstressed; the calculation of boom areas B5 and B13 does not then arise  376 Stress analysis of aircraft components Table 1 02 1 2, 3, 4, 5, 6, 7, 8, 9 381.0 352.0 269.5 145.8 16 15 14 13 12 11 10 302.4 279 .4 213.9 115 .7 0 0 -1 45.8 -2 69.5 -3 52.0 -3 81.0 -1 15 .7 -2 13.9 -2 79 .4 -3 02.4 For this particular section Zxy = 0 since Cx (and C y ) is an... axial loads in the flanges are therefore PI1 = -Pz-2 -7 = 133.3 x 400 = 53 320 N The shear load resisted by the beam web is then, from Eq (10.5) S,.,,"= -2 0 x lo3 - 53 32 0-6 Yl 6Z + 53 32 0-6 Y2 6Z in which, from Figs 10.1 and 10.2, we see that 6y1 - -1 00 - -0 .05, sz 2 x 103 6Y2 - 100 - 6 , 2 x 103 - 0.05 Hence S,.:w, -2 0 x lo3 = + 53 320 x 0.05 + 53 320 x 0.05 = -1 4668N The shear flow distribution... -6 6.0 -6 6.0 -5 3.5 -3 0.3 0 -3 2.8 -6 3.1 -8 6.3 -9 8.8 -9 8.8 -8 6.3 -6 3.1 -3 2.8 - 216 .7 216.6 216.6 (N/m) positive Clearly A12 = A23 = - - A I 6 ,= 4.56 x 105/16 = 285OOmm’ Equation = (iv) then become 100 X lo3 X 150 2 x 28 500(-qb,, + - qbz - ’ ’ - qb161) 4-5 6 lo5q3,0 (v) Substituting the values of q b from Table 10.3 in Eq (v), we obtain 100 x lo3 x 150 = 2 x 28 500 (-2 62.4) + 2 x 4.56 x 105q3,0 from which... shear stress distribution is as shown in Fig 10.21 Fig 10.21 Shear stress (Wrnm’) distribution in wing section of Example 10 .7 10.3 Wings 3 87 10.3.3 Shear - I _ I * i i l P I U I _ Y _ - - - - - - - ~ ~ - - ~ - - ~ I - Initially we shall consider the general case of an N-cell wing section comprising booms and skin panels, the latter being capable of resisting both direct and shear stresses The... 10 .7 386 Stress analysis of aircraft components Choosing GREF = 27 600N/mm2 then, from Eq (10. 27) 24 200 t i p = - 1.22 = 1.07mm x 27 600 Similarly tf3 = r;4 = 1.07mm, t& = t i 6 = t& = 0.69mm Hence Similarly 61zi = 250, 613 = 6 = 72 5, % 233, 634 635 = 646 = 73 6, b56 = 368 Substituting the appropriate values of 6 in Eq (10.24) for each cell in turn gives the following For cell I For cell I1 de _- 1... -0 . 376 x 305 x 900 = -1 03212N and P2 = -Ps = -0 . 376 x 305 x 1200 = -1 376 16N Thus, there is an increase in compressive load of 103212 - 96 642 = 6 570 N in booms 1 and 3 and an increasein tensile load of 6 570 N in booms 4 and 6 between the two sections Also, the compressive load in boom 2 increases by 1 37 616 - 128 856 = 876 0N while the tensile load in boom 5 increases by 876 0N Therefore, the change in . (kNm) 1 -1 00 0.1 -0 .05 -1 0 5 -1 01.3 0.6 0.3 3 -3 2 -1 33 0 -0 .05 0 6 .7 -1 77 .3 0 0.3 0 0 3 -1 00 -0 .1 -0 .05 10 5 -1 01.3 0.6 0.3 -3 3 4 100 -0 .1 0.05 -1 0 5 101.3. = - or Thus qb.16 = qb.12 = 0 - 1.23 X X 900 X 300 = -3 3.2N/mm qb,23 = -3 3.2 - 1.23 x x 1200 x 300 = -7 7.5N/mm qb,34 = -7 7.5 - 1.23 x x 900 x 300 = -1 10.7N/mm. 10.2, we see that - 0.05 6y1 - -1 00 6Y2 - 100 sz 2 x 103 6,. 2 x 103 - -0 .05, - - Hence S,.:w, = -2 0 x lo3 + 53 320 x 0.05 + 53 320 x 0.05 = -1 4668N The shear