Aircraft Structures 1 2011 Part 10 ppt

346 Open and closed, thin-walled beams Derive expressions for the vertical and horizontal components of the deflection of the beam midway between the supports B and D. The wall thickness t and Young’s modulus E are constant throughout. Ans. u = 0.186W13/Ea3t, v = 0.177Wl3/Ea3t. P.9.3 A uniform beam of arbitrary, unsymmetrical cross-section and length 21 is built-in at one end and simply supported in the vertical direction at a point half-way along its length. This support, however, allows the beam to deflect freely in the horizontal x direction (Fig. P.9.3). For a vertical load W applied at the free end of the beam, calculate and draw the bending moment diagram, putting in the principal values. Ans. Mc = 0, MB = WI, MA = - W1/2. Linear distribution. Yt wl Fig. P.9.3 P.9.4 A beam, simply supported at each end, has a thin-walled cross-section shown in Fig. P.9.4. If a uniformly distributed loading of intensity w/unit length acts on the beam in the plane of the lower, horizontal flange, calculate the maximum ’t per unitlength Fig. P.9.4 Problems 347 direct stress due to bending of the beam and show diagrammatically the distribution of the stress at the section where the maximum occurs. The thickness t is to be taken as small in comparison with the other cross-sectional dimensions in calculating the section properties Ixx, Iyy and Ixy. Ans. uZ-;- = uz,3 = 13w12/384a2t, ui:l = w12/96a2t, uzT2 =-w12/48a2t P.9.5 A thin-walled cantilever with walls of constant thickness t has the cross- section shown in Fig. P.9.5. It is loaded by a vertical force W at the tip and a horizontal force 2W at the mid-section, both forces acting through the shear centre. Determine and sketch the distribution of direct stress, according to the basic theory of bending, along the length of the beam for the points 1 and 2 of the cross-section. The wall thickness t can be taken as very small in comparison with d in calculating the sectional properties I,,, Ixy etc. Ans. ui:] (mid-point) = -0.05 WZ/td2, uiF2 (mid-point) = -0.63 Wl/td2, uz,l (built-in end) = -1.85 Wl/td2 (built-in end) = 0.1 Wl/td2 Fig. P.9.5 P.9.6 A uniform cantilever of arbitrary cross-section and length I has section properties, JYX, Iyy and lYy with respect to the centroidal axes shown in Fig. P.9.6. It is loaded in the vertical (yz) plane with a uniformly distributed load of intensity wlunit length. The tip of the beam is hinged to a horizontal link which constrains it to move in the vertical direction only (provided that the actual deflections are small). Assuming that the link is rigid, and that there are no twisting effects: calculate: (a) the force in the link; (b) the deflection of the tip of the beam. Ans. (a) 3wZIxy/81xx; (b) wf/8EIxx. 348 Open and closed, thin-walled beams Fig. P.9.6 P.9.7 A thin-walled cantilever has a constant cross-section of uniform thickness with the dimensions shown in Fig. P.9.7. It is subjected to a system of point loads acting in the planes of the walls of the section in the directions shown. Calculate the direct stresses according to the basic theory of bending at the points 1 , 2 and 3 of the cross-section at the built-in end and half-way along the beam. Illustrate your answer by means of a suitable sketch. The thickness is to be taken as small in comparison with the other cross-sectional dimensions in calculating the section properties Ixx, IxY etc. Ans. At built-in end, u2,] = -11.4N/mm2, uz,2 = -18.9N/mmZ, a,,3 = 39.1 N/mmz Half-way, uz,l = -20.3 N/mm2, u2:2 = -1.1 N/mm2, (T,,~ = 15.4N/mm2 Fig. P.9.7 P.9.8 A uniform thin-walled beam has the open cross-section shown in Fig. P.9.8. The wall thickness t is constant. Calculate the position of the neutral axis and the maximum direct stress for a bending moment M, = 3.5Nm applied about the horizontal axis Cx. Take r = 5 mm, t = 0.64mm. Ans. a = 51.9", a,,,,, = 101 N/mm2. Problems 349 Fig. P.9.8 P.9.9 A beam has the singly symmetrical, thin-walled cross-section shown in Fig. P.9.9. The thickness t of the walls is constant throughout. Show that the distance of the shear centre from the web is given by p2 sin (Y cos (Y 1 + 6p + 2p3 sin2 (Y & = -d where p = d/h I I Fig. P.9.9 P.9.10 A beam has the singly symmetrical, thin-walled cross-section shown in Fig. P.9.10. Each wall of the section is flat and has the same length a and thickness t. Calculate the distance of the shear centre from the point 3. Ans. 5a cos a/8. 350 Open and closed, thin-walled beams a 4 Fig. P.9.10 1 3 5 P.9.11 Determine the position of the shear centre S for the thin-walled, open cross-section shown in Fig. P.9.11. The thickness t is constant. Ans. m/3. Fig. P.9.11 P.9.12 Figure P.9.12 shows the cross-section of a thin, singly symmetrical I-section. Show that the distance ts of the shear centre from the vertical web is given by - Es - 3PU - PI d - (1 + 12p) where p = d/h. The thickness tis taken to be neghgibly small in comparison with the other dimensions. Problems 351 Fig. P.9.12 P.9.13 A thin-walled beam has the cross-section shown in Fig. P.9.13. The thick- ness of each flange varies linearly from tl at the tip to t2 at the junction with the web. The web itself has a constant thickness t3. Calculate the distance Es from the web to the shear centre S. A~S. d2(2tl + l2)/[3d(tl + t.) + ht3] Fig. P.9.13 P.9.14 Figure P.9.14 shows the singly symmetrical cross-section of a thin-walled open section beam of constant wall thickness t, which has a narrow longitudinal slit at the corner 15. 4 Fig. P.9.14 352 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. [...]... panels 12 : 23, , 16 1 at the centre C of the circular cross-section and anticlockwise moments are taken as 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 14 5.8 0 -14 5.8... + @ @ Table 10 .1 0 0 0 @ 6x,/6r Boom Pz, (kN) 6y,lSz -10 0 0 .1 1 2 3 4 5 6 -13 3 -10 0 10 0 13 3 10 0 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) -10 0 10 -10 0 IO 5 6.7 5 5 6.7 5 -10 1.3 -17 7.3 -10 1.3 10 1.3 17 7.3 10 1.3 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... 3 81. 0 352.0 269.5 14 5.8 16 15 14 13 12 11 10 302.4 279.4 213 .9 11 5.7 0 0 -14 5.8 -269.5 -352.0 -3 81. 0 -11 5.7 - 213 .9 -279.4 -302.4 For this particular section Zxy = 0 since Cx (and C y ) is an axis of symmetry Further, My= 0 so that Eq.(9.6) reduces to in which + + Zxx = 2 x 216 .6 x 3 81. 02 4 x 216 .6 x 352.02 4 x 216 .6 x 26952 + 4 x 216 .7 x 14 5.82= 2.52 x lo8mm4 The solution is completed in Table 10 .2 10 .2.2... from Eq (10 .5) S,.,,"= -20 x lo3 - 53 320-6Yl 6Z + 53 320-6Y2 6Z in which, from Figs 10 .1 and 10 .2, we see that 6y1 - -10 0 - -0.05, sz 2 x 10 3 6Y2 - 10 0 - 6 , 2 x 10 3 - 0.05 Hence S,.:w, -20 x lo3 = + 53 320 x 0.05 + 53 320 x 0.05 = -14 668N The shear flow distribution in the web follows either from Eq (10 .6) or Eq (10 .7) and is (see Fig 10 .2(b)) 412 = 22.5' x 14 ' lo6 ([q150-s)ds+400 x 15 0 i.e 412 =... load P, in the rth boom is then given by Pqr = ai,rBr (10 .8) where B, is the cross-sectional area of the rth boom From Fig 10 .4(b) (10 .9) Further, from Fig 10 .4(c) or, substituting for Py,, from Eq (10 .9) (10 .10 ) The axial load Pr is then given by Pr = + p',r + pI,r )1' 2 (10 .11 ) or, alternatively (10 .12 ) 10 .1 Tapered beams 367 (a) z X ( b) (C) Fig 10 .4 Effect of taper on the analysis of open and closed... flow is 10 0 kN I Fig 10 .14 Alternative solution of Example 10 .5 380 Stress analysis of aircraft components constant between adjacent booms Suppose that the shear flow in the panel 21 is q Z l Then from symmetry and using the results of Table 10 .3 49 8 = 49 I O = q l 6 1 = q2 1 + q 21 q43 = q76 = 411 12 = q1 415 = 53.5 + 9 21 q32 = q S 7 ‘?loll = q1 516 = 30.3 q54 = 9 6 5 = 912 13 = q13 14 = 66.0 + q 21 The... there is a total of rn booms'in the section s = Sx,w + x m Px:r, sy = sy,,v + r=l n z (10 .13 ) Py,r r= 1 Substituting in Eqs (10 .13 ) for P.,,r and P,,,r from Eqs (10 .10 ) and (10 .9) we have sx, P:,r - s = sxw + x : r=l sz 1 sy = sy:w + pz,r I sY ' r= 1 6z (10 .14 ) Hence (10 .15 ) 368 Stress analysis of aircraft components Fig 10 .5 Modification of moment equation in shear of closed section beams due to boom load... distribution Example 10 .5 The fuselage of Example 10 .4 is subjected to a vertical shear load of 10 0 kN applied at a distance of 15 0mm from the vertical axis of symmetry as shown, for the idealized section, in Fig 10 .12 Calculate the distribution of shear flow in the section 10 .2 Fuselages 377 Fig 10 .12 Idealized fuselage section of Example 10 .5 As in Example 10 .4, Ixy = 0 and, since S, = 0, Eq (10 .17 ) reduces... 3 81. 0 352.0 269.5 14 5.8 0 -14 5.8 -269.5 -352.0 0 -30.3 -53.5 -66.0 -66.0 -53.5 -30.3 0 -32.8 -63 .1 -86.3 -98.8 -98.8 -86.3 -63 .1 -32.8 - 216 .7 216 .6 216 .6 (N/m) positive Clearly A12 = A23 = - - A I 6 ,= 4.56 x 10 5 /16 = 285OOmm’ Equation = (iv) then become 10 0 X lo3 X 15 0 2 x 28 500(-qb,, + - qbz - ’ ’ - qb1 61) 4-56 lo5q3,0 (v) Substituting the values of q b from Table 10 .3 in Eq (v), we obtain 10 0... the beam of Fig 10 .1, would simplify to S (10 .6) qs - - 3 tDy ds + B l y l ) Ixx ([ or S ([tDy ds + B2y2) (10 .7) 4s - - 22 Ixx Example 10 .1 Determine the shear flow distribution in the web of the tapered beam shown in Fig 10 .2, at a section midway along its length The web of the beam has a thickness of 't t' 400mm2 1 2 rnm 400 mm2 Section AA (a) Fig 10 .2 Tapered beam of Example 10 .1 10 .1 Tapered beams . (10 .5) 6Yl 6Y2 S,.,," = -20 x lo3 - 53 320- + 53 320- 6Z 6Z in which, from Figs 10 .1 and 10 .2, we see that - 0.05 6y1 - -10 0 6Y2 - 10 0 sz 2 x 10 3 6,. 2 x 10 3. by (10 .1) in which, for the direction of taper shown, Sy2 is negative. The axial load in flange 0 is given by Substituting for P,,l from Eqs (10 .1) we have (10 .2) Fig. 10 .1 Effect. -14 668N The shear flow distribution in the web follows either from Eq. (10 .6) or Eq. (10 .7) and is (see Fig. 10 .2(b)) 412 = 22.5 14 ''' x lo6 ([q150-s)ds+400 x 15 0