Aircraft structures for engineering students - part 9 pps

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. 476 Structural constraint (a) Fig. 11.34 Distribution of axial constraint direct stress around the section. Equations (iii) to (viii) are, of course, valid for open section beams of any cross- section. Their application in a particular case is governed by the value of the torsion bending constant rR and the St. Venant torsion constant J [= (h + 2d)t3/3 for this example]. With this in mind we can proceed, as required by the example, to derive the direct stress and shear flow distributions. The former is obtained from Eqs (11.54) and (iv). Thus T sinh p(L - z) GJ coshpL = ARE - p or writing J = GJ/ErR and rearranging sinh p(L - z) cr = - {L TUR cosh pL GJTR In Eq. (ix) E, G, J and rR are constants for a particular beam, T is the applied torque, AR is a function of s and the hyperbolic term is a function of z. It follows that at a given section of the beam the direct stress is proportional to -2AR, and for the beam of this example the direct stress distribution has, from Fig. 11.31, the form shown in Figs 11.34(a) and (b). In addition, the value of cr at a particular value of s varies along the beam in the manner shown in Fig. 11.35. I- d& Fig. 11.35 Spanwise distribution of axial constraint direct stress. 11.5 Constraint of open section beams 477 Fig. 11.36 Calculation of axial constraint shear flows. Finally, the axial constraint shear flow, qr, is obtained from Eq. (11.57), namely At any section z, qr is proportional to ring to Fig. 11.36, 2AR = 240 - 2Afi so that in flange 12 2&t ds and is computed as follows. Refer- Hence [2A~tdS=t hd hd - h+d [ 4 2 (h+Zd)"] so that Similarly d38 h'd't dz3 4(h + 2d) qr,l = 0 and qr,2 = -E- " - 4(h hZd2t + 2d) 1 whence d38 h2d2t d38 h'd't qr:z=-E~ 4(h+2d)' qr~=E~4(,3+2d) Note that in the above d38/& is negative (Eq. (viii)). Also at the mid-point of the web where s2 = h/2, qr = 0. The distribution on the lower flange follows from antisymmetry and the distribution of qr around the section is of the form shown in Fig. 11.37. The spanwise variation of qr has the same form as the variation of Tr since d38 dz3 Tr = -ErR - 478 Structural constraint Fig. 11.37 Distribution of axial constraint shear flows. giving qr = “r 2ARtds from Eq. (11.57) rR 0 Hence for a given value of s, ($2A~tds), qr is proportional to Tr (see Fig. 11.33). 11.5.1 Distributed torque loading We now consider the more general case of a beam carrying a distributed torque load- ing. In Fig. 11.38 an element of a beam is subjected to a distributed torque of intensity T,(z), i.e. a torque per unit length. At the section z the torque comprises the St. Venant torque Tj plus the torque due to axial constraint Tr. At the section z + Sz the torque increases to T + ST(= Tj + STj + Tr + STr) so that for equilibrium of the beam element Tj + STj + Tr + STr + T~(z)SZ - Tj - Tr = 0 or -T~(z)SZ = STj + STr = ST . \\ = TJ + \ T+ 6T 6TJ + T,. + Fig. 11.38 Beam carrying a distributed torque loading. 11.5 Constraint of open section beams 479 Hence Now dTJ dTr dT dz dz &- T,(Z) = -+- - dB dz TJ = GJ- (Eq. (9.60)) and d30 Tr = -Er - dz3 (Eq. (1 1.58)) so that Eq. (1 1.63) becomes d46' d2B Er - - GJ - - T~(~) d# &2 - (1 1.63) (11.64) The solution of Eq. (11.64) is again of standard form in which the constants of integration are found from the boundary conditions of the particular beam under consideration. 11.5.2 Extension of the theory to allow for general Il_i-_l : systems of loading -" So far we have been concerned with open section beams subjected to torsion in which, due to constraint effects, axial stresses are induced. Since pure torsion can generate axial stresses it is logical to suppose that certain distributions of axial stress applied as external loads will cause twisting. The problem is to determine that component of an applied direct stress system which causes twisting. Figure 11.39 shows the profile of a thin-walled open section beam subjected to a general system of loads which produce longitudinal, transverse and -rotational A SY Fig. 11.39 Cross-section of an open section beam subjected to a general system of loads. 480 Structural constraint displacements of its cross-section. In the analysis we assume that the cross-section of the beam is undistorted by the loading and that displacements corresponding to the shear strains are negligible. In Fig. 11.39 the tangential displacement ut is given by Eq. (9.27), i.e. ut = pR8 + ucos$ + usin$ (1 1.65) Also, since shear strains are assumed to be negligible, Eq. (9.26) becomes (1 1.66) Substituting for vt in Eq. (1 1.66) from Eq. (1 1.65) and integrating from the origin for s to any point s around the cross-section, we have d8 du dv dz dz dz W, - WO = 2AR,o - -(x - XO) - - (y -yo) (11.67) where 240 = p~ ds. The direct stress at any point in the wall of the beam is given by Thus, from Eq. (1 1.67) Now AR,O = Ak + AR (Fig. 11.39) so that Eq. (1 1.68) may be rewritten d28 d2u d2v dz2 dz2 dz2 uZ = fi (z) - E - ~AR - E - x - E - (1 1.69) in which The axial load P on the section is given by where Jc denotes integration taken completely around the section. From Eq. (11.55) we see that sc 2A~t dr = 0. Also, if the origin of axes coincides with the centroid of the section Jc txds = Jc tyds = 0 and J tyds = 0 so that P = a2tds =fi(z)A (1 1.70) I in which A is the cross-sectional area of the material in the wall of the beam. The component of bending moment, M,, about the x axis is given by 11.5 Constraint of open section beams 481 Substituting for CT? from Eq. (1 1.69) we have d2 u d2v Mx =fi(z) IC tyds - 2ARtyds - EGIC txydS - Ic d? ty’ds We have seen in the derivation of Eqs (11.60) and (11.61) that Jc2ARtyds = 0. Also since tyrds=O! jcfxyds=Ixy, jcty2ds=I, (11.71) d2u d2v M, = -E- I - ED I,, dz2 x” Similarly (1 1.72) d2u d2v c2tx& = -Es I& - E- I . dz2 x’ Equations (1 1.71) and (11.72) are identical to Eqs (9.19) so that from Eqs (9.17) E- d2u = MxIxs -M y2 I xx, E- d2v = -My Iyy + My IxJ (1 1.73) dz2 Ixx Iyy - I,, dz2 Ixx Iyy - I$ The first differential, d2$/dz2, of the rate of twist in Eq. (1 1.69) may be isolated by multiplying throughout by 2ARt and integrating around the section. Thus lc~z2ARtds =fi(z) As before and jc 2ARt ds = 0, jc 2ARtx ds = 2ARty ds = 0 I or -=-Ic d2B 0,2A~t dS dz2 ErR Substituting in Eq. (1 1.69) from Eqs (1 1.70), (1 1.73) and (1 1.74), we obtain ( 1 1.74) 482 Structural constraint The second two terms on the right-hand side of Eq. (1 1.75) give the direct stress due to bending as predicted by elementary beam theory (see Eq. (9.6)); note that the above approach provides an alternative method of derivation of Eq. (9.6). Comparing the last term on the right-hand side of Eq. (1 1.75) with Eq. (1 1.54), we see that = Or ~AR Se Uz 2A~t ds rR It follows therefore that the external application of a direct stress system a, induces a self-equilibrating direct stress system Or. Also, the first differential of the rate of twist (d26/dzz) is related to the applied a, stress system through the term Jeaz2ARtds. Thus, if sc az2ARt ds is interpreted in terms of the applied loads at a particular section then a boundary condition exists (for d26/d2) which determines one of the constants in the solution of either Eq. (1 1.59) or Eq. (1 1.64). 11.5.3 Moment couple (bimoment) The units of sc a,2ARtds are force x (distance)2 or moment x distance. A simple physical representation of this expression would thus consist of two equal and opposite moments applied in parallel planes some distance apart. This combination has been termed a moment couple’ or a bimoment3 and is given the symbol Mr or B,. Equation (11.75) is then written (1 1.76) As a simple example of the determination of Mr consider the open section beam shown in Fig. 11.40 which is subjected to a series of concentrated loads PI, P2,. . . , Pk, . . . , Pn parallel to its longitudinal axis. The term azt ds in Jc a,2ARt ds may be regarded as a concentrated load acting at a point in the wall of the beam. Thus, Sc az2ARt ds becomes E;= Pk2Aw, and hence MXIYY - MYIXY M~~AR A IxxIyy - Ix:, >,+ ( IxxIyy - I2y )Y+T p (My Ixx - MxLy a,=-+ n Mr is determined for a range of other loading systems in Ref. 2. (1 1.77) Fig. 11.40 Open section beam subjected to concentrated loads parallel to its longitudinal axis. 11.5 Constraint of open section beams 483 - B (a) Fig. 11.41 Column of Example 11.3. Example 11.3 t l 100rnrn The column shown in Fig. 11.41(a) carries a vertical load of 100kN. Calculate the angle of twist at the top of the column and the distribution of direct stress at its base. E = 200 000 N/mm2 and GIE = 0.36. The centre of twist R of the column cross-section coincides with its shear centre at the mid-point of the web 23. The distribution of 2AR is obtained by the method detailed in Example 11.2 and is shown in Fig. 11.42. The torsion bending constant rR is given by Eq. (ii) of Example 11.2 and has the value 2.08 x 10'omm6. The St. Venant torsion constant J = Cst3/3 = 0.17 x 105mm4 so that dm (= ,LL in Eq. (iii) of Example 11.2) = 0.54 x Since no torque is applied to the Fig. 11.42 Distribution of area 2AR in the column of Example 11.3. [...]... (iii) -[ I 1 2 ; I -L 3 1 2J L k3 7 1-1 r 2 ; 2 I I 1 : -2 -I I I I -2 1 L I 1 : 5 1 k32 1 I Ll 21 2 I 1 1 1 [-; I ; I I I I k22 L-2 r-l -Li 4 - 2-I - 11 1 1 - 1 k33 I 1 I ZJ 506 Matrix methods of structural analysis u2 v2 u3 v 3 0 -1 0 0 0 ' 1 0 0 0 -1 21 = o 4 1- 1 1 1 0 I+ 21 = 0 1 2Jz 2Jz 2Jz 2 4 1 1 1 u2 1 0 0 -2 Jz 2J2 v2 2Jz 2.\/2 1 1 1 1 u3 = 0 -0 0 -2 Jz 2Jz 2 4 2J2 v =0 3 1 1 1 1 - -0 -1 1... displacement states we have, for the condition u1 = u l , u2 = 242, u3 = u3 Fx,1 = kau1 - kau2 Fx,2 = - k u l -k (ka -k kb)u2 - kbu3 Fx,3 = -kbU2 -k kbU3 Writing Eqs (12.11) in matrix form gives 1 { :} [- 4.1 { ::} : ka (12.11) -ka (12.12) k :-: = Comparison of Eq (12.12) with Eq (12.1) shows that the stiffness matrix [Kl of this two-spring system is [q= [ ka -k, 0 -ka ka+kb -kb 4.1 (12.13) Equation (12.13)... distribution at the built-in end if the cross-section of the beam remains undistorted by the loading and the shear modulus G and wall thickness t are each constant throughout the section Ans 91 2 = 399 2 .9/ R N/n~m, 92 3 = 71 1.3/R N/IIUTI: 431 = (1502.4 - 1 8 9 4 7 ~ 0 -4 ~ 2102.1 sinq5)/R N/mm 1 Fig P.11.3 P.11.4 A uniform, four-boom beam, built-in at one end, has the rectangular cross-section shown in Fig... Fx,2(= 0) and Fv,2( =- W ) Thus 1 1 1 +-2 Jz 2 4 u 1 v 1 1 0 -1 +- 2Jz 2J2 Inverting Eq (v) gives (vi) from which (vii) (viii) The reactions at nodes 1 and 3 are now obtained by substituting for u2 and v2 from Eq (vi) into Eq (iv) Thus O l 0 1 - -1 0 0 0 1 }:I:{ - 0 -1 - 12.6 Matrix analysis of space frames giving Fx,l = - Fx,2 - FY,2 = W Fy,l = 0 Fy,3 Fy>2= - W 1 Fy,3 = w Finally, the forces in the members... simpler form, namely x2 AP SYM P2 AE xu pu u2 [K ] - = V L -A2 -xp -xu ; x2 -xp -xu 2 -p -pu -pu -u 2 ; ; xp xu p2 pu (12.38) 2 u where A, p and u are the direction cosines between the x,y, z and X axes respectively The complete stiffness matrix for a space frame is assembled from the member stiffness matrices in a similar manner to that for the plane frame and the solution completed as before !ss... thin-walled cross-section shown in Fig P 11.10 If the thickness t is constant throughout and making the usual assumptions for a thin-walled cross-section, show that the torsion bending constant r R calculated about the shear centre S is TfE -+ 3 4 Fig P.11.10 P l l l l A uniform beam has the point-symmetric cross-section shown in Fig P 11.11 Making the usual assumptions for a thin-walled cross-section,... Hence and (as before) 7 - rences 1 Argyris, J H and Dunne, P C , The general theory of cylindrical and conical tubes under Parts I-IV, Feb 194 7, Part V, Sept and Nov torsion and bending loads, J Roy Aero SOC., 194 7, Part VI, May and June 194 9 2 Megson, T H G., Extension of the Wagner torsion bending theory to allow for general systems of loading, The Aeronautical Quarterly, Vol XXVI, Aug 197 5 3 Vlasov,... coordinates, i.e -Equation (12.23) does not change the basic relationship between Fx,i,Fx,j i and &as defined in Eq (12.22) From Fig 12.3 we see that - Fx,i Fx,i 8 + Fy,i 8 = cos sin = -Fx:isin 8 + Fy,icos 8 and - Fx,j=FxJcos8+ FY;,,sin8 - Fy,j =-Fx,jsin8+ Fy,jcos8 u Writing X for cos 8 an , for sin 8 we express the above equations in matrix form as -k I B 0 0 - p X: ] { ; y , j , or, in abbreviated form IF1... built-in end The shear modulus G is the same for all walls Ans q12= q56 = 46.6 N/mm, 95 2 = 180.8 N/mm, 1.4N/mm1 q43= 74.6N/mm7 = -6 30.1 mm, Y R = 0 (relative to mid-point of 52) 93 2 = q54 = XR 127 I27 ' Fig P.ll.l I 305mm I 254 m m q P.11.2 thin-walled two-cell beam with the singly symmetrica, cross-section shown in Fig P 1 1.2 is built-in at one end where the torque is 1 1 000 N m Assuming the cross-section... grouped toget her Finally, the internal forces in the springs may be determined from the forcedisplacement relationship of each spring Thus, if Sa is the force in the spring joining nodes 1 and 2 then Sa = ka(u 2- u , ) Similarly for the spring between nodes 2 and 3 Sb = kb(u3 - u 2 ) 12.4 Matrix analysis of pin-jointed frameworks The formation of stiffness matrices for pin-jointed frameworks and the subsequent . Constraint of open section beams 4 79 Hence Now dTJ dTr dT dz dz & ;- T,(Z) = -+ - - dB dz TJ = GJ- (Eq. (9. 60)) and d30 Tr = -Er - dz3 (Eq. (1 1.58)) so that Eq and integrating from the origin for s to any point s around the cross-section, we have d8 du dv dz dz dz W, - WO = 2AR,o - -( x - XO) - - (y -yo) (11.67) where 240 = p~. Aero. SOC., Parts I-IV, Feb. 194 7, Part V, Sept. and Nov. 194 7, Part VI, May and June 194 9. 2 Megson, T. H. G., Extension of the Wagner torsion bending theory to allow for general