then: oe×a×e×△b2=o,×b×e×△a1 with the values obtained above for△b2and△a: axaxex-YG,xb=a,xbxex-wu oxa E, Numerical application:v=0.3,E=45,000 MPa,E,=12,000 MPa (see Section 3.3.3): Vt=0.3× 12,000 45,000 Vt=0.08 Remark:The same reasoning can be applied to all balanced laminates having midplane symmetry,by placing them in the symmetrical axes.'However,depending on the composition of the laminate,the Poisson coefficients in the two perpendicular directions vary in more important ranges: in absolute value and one with respect to the other. One can find in Section 5.4.2 in Table 5.14 the domain of evolution of the global Poisson coefficient vy of the glass/epoxy laminate,from which one can deduce the Poisson coefficient via using a formula analogous to the one above,as: VylEy Vx/Ex 18.1.3 Helicopter Blade This study has the objective of bringing out some important particularities related to the operating mode of the helicopter blade,notably the behavior due to normal load. Problem Statement: Consider a helicopter blade mounted on the rotor mast as shown schematically in the following figure. Or the orthotropic axes:see Chapter 12,Equation 12.9. 2003 by CRC Press LLC
then: with the values obtained above for Db2 and Da1: Numerical application: nt = 0.3, E = 45,000 MPa, Et = 12,000 MPa (see Section 3.3.3): Remark: The same reasoning can be applied to all balanced laminates having midplane symmetry, by placing them in the symmetrical axes. 3 However, depending on the composition of the laminate, the Poisson coefficients in the two perpendicular directions vary in more important ranges: in absolute value and one with respect to the other. One can find in Section 5.4.2 in Table 5.14 the domain of evolution of the global Poisson coefficient nxy of the glass/epoxy laminate, from which one can deduce the Poisson coefficient nyx using a formula analogous to the one above, as: 18.1.3 Helicopter Blade This study has the objective of bringing out some important particularities related to the operating mode of the helicopter blade, notably the behavior due to normal load. Problem Statement: Consider a helicopter blade mounted on the rotor mast as shown schematically in the following figure. 3 Or the orthotropic axes: see Chapter 12, Equation 12.9. s ¥ a e ¥ ¥ Db2 = st ¥ b e ¥ ¥ Da1 s a e nt Et ¥ ¥ ¥ –------ st ¥ b st b e nt E = ¥ ¥ ¥ –------ s ¥ a nt Et ------ nt E = ------ nt 0.3 12,000 45,000 = ¥ ------------------- nt = 0.08 nyx/Ey nxy = /Ex TX846_Frame_C18a Page 347 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
€/10 The characteristics of the rotor are as follows: Rotor with three blades;rotational speed:500 revolutions per minute. The mass per unit length of a blade at first approximation is assumed to have a constant value of 3.5 kg/m. ■e=5m;c=0.3m. The elementary lift of a segment dx of the blade (see figure above)is written as: dF:=p(cdx)C:v2 in which V is the relative velocity of air with respect to the profile of the blade. In addition,C()=0.35 (lift coefficient). p=1.3 kg/m'(specific mass of air in normal conditions). We will not concern ourselves with the drag and its consequences.One examines the helicopter as immobile with respect to the ground(stationary flight in immobile air).In neglecting the weight of the blade compared with the load application and in assuming infinite rigidity,the relative equilibrium configuration in uniform rotation is as follows: c0s0≠1 e small sin8≠e s(negligible) 2003 by CRC Press LLC
The characteristics of the rotor are as follows: Rotor with three blades; rotational speed: 500 revolutions per minute. The mass per unit length of a blade at first approximation is assumed to have a constant value of 3.5 kg/m. = 5 m; c = 0.3 m. The elementary lift of a segment dx of the blade (see figure above) is written as: in which V is the relative velocity of air with respect to the profile of the blade. In addition, Cz (7∞) = 0.35 (lift coefficient). r = 1.3 kg/m3 (specific mass of air in normal conditions). We will not concern ourselves with the drag and its consequences. One examines the helicopter as immobile with respect to the ground (stationary flight in immobile air). In neglecting the weight of the blade compared with the load application and in assuming infinite rigidity, the relative equilibrium configuration in uniform rotation is as follows: dFz 1 2 --r( ) cdx CzV2 = TX846_Frame_C18a Page 348 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
1.Justify the presence of the angle called"flapping angle"and calculate it. 2.Calculate the weight of the helicopter. 3.Calculate the normal force in the cross section of the blade and at the foot of the blade (attachment area). The spar of the blade'is made of unidirectional glass/epoxy with 60%fiber volume fraction "R"glass (e1700 MPa).The safety factor is 6.Calculate: 4.The longitudinal modulus of elasticity E of the unidirectional. 5.The cross section area for any x value of the spar,and its area at the foot of the blade. 6.The total mass of the spar of the blade. 7.The elongation of the blade assuming that only the spar of the blade is subject to loads. 8. The dimensions of the two axes to clamp the blade onto the rotor mast. Represent the attachment of the blade in a sketch. Solution: 1.The blade is subjected to two loads,in relative equilibrium: Distributed loads due to inertia,or centrifugal action,radial (that means in the horizontal plane in the figure,with supports that cut the rotor axis. Distributed loads due to lift,perpendicular to the direction of the blade (Ax in the figure). From this there is an intermediate equilibrium position characterized by the angle 0. Joint A does not transmit any couple.The moment of forces acting on the blade about the y axis is nil,then: d,xx=a.×xsinxa.×x e10 10 with: dF.=pc dx Cv-pe dx C(xcos0xpe dx C. dF.dm w'xcos0 m dx o'x (centrifugal load) then after the calculation: pccnm 4 3 03pcCixe 8 m See Section 7.2.3. 2003 by CRC Press LLC
1. Justify the presence of the angle called “flapping angle” q and calculate it. 2. Calculate the weight of the helicopter. 3. Calculate the normal force in the cross section of the blade and at the foot of the blade (attachment area). The spar of the blade4 is made of unidirectional glass/epoxy with 60% fiber volume fraction “R” glass (s rupture # 1700 MPa). The safety factor is 6. Calculate: 4. The longitudinal modulus of elasticity E of the unidirectional. 5. The cross section area for any x value of the spar, and its area at the foot of the blade. 6. The total mass of the spar of the blade. 7. The elongation of the blade assuming that only the spar of the blade is subject to loads. 8. The dimensions of the two axes to clamp the blade onto the rotor mast. Represent the attachment of the blade in a sketch. Solution: 1. The blade is subjected to two loads, in relative equilibrium: Distributed loads due to inertia, or centrifugal action, radial (that means in the horizontal plane in the figure, with supports that cut the rotor axis. Distributed loads due to lift, perpendicular to the direction of the blade (Ax in the figure). From this there is an intermediate equilibrium position characterized by the angle q. Joint A does not transmit any couple. The moment of forces acting on the blade about the y axis is nil, then: with: then after the calculation: 4 See Section 7.2.3. dFz ¥ x /10 Ú Fc x q # q dFc ¥ x /10 Ú d ¥ sin ¥ /10 Ú = dFz 1 2 --rc dx CzV2 1 2 --rc dx Cz( ) xcosq w¥ 2 # 1 2 --rc dx Cz x2 w2 = = dFc dm w2 x q # m dx w2 = cos x ( ) centrifugal load 1 2 --rcCzw2 4 4 /104 ( ) – 4 ------------------------------- q mw2 3 3 /103 ( ) – 3 = ------------------------------- q # 3 8 -- rcCz m ------------ ¥ TX846_Frame_C18a Page 349 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
or numerically: 0=0.073rad=4117 Remarks: One verifies that sin=0.073 0 and cos=0.997 1. ■ When the helicopter is not immobile,but has a horizontal velocity,for example vo.the relative velocity of air with respect to the blade varies between vo wx for the blade that is forward,and -+ox for a blade that is backward.If the incidence i does not vary,the lift varies in a cyclical manner,and there is vertical "flapping motion"of the blade.This is why a mechanism for cyclic variation of the incidence is necessary. We have not taken into account the drag to simplify the calculations.This can be considered similarly to the case of the lift.It then gives rise to an equilibrium position with a second small angle,called o,with respect to the radial direction from top view,as in the following figure.This is why a supplementary joint,or a drag joint,is necessary. 2.Weight of the helicopter:The lift and the weight balance themselves out. The lift of the blade is then: F.-Se ,cos0dpecn t10 then for the 3 rotor blades: Mg 3F. Mg =pcC:o'es numerically: Mg 2340 daN 2003 by CRC Press LLC
or numerically: Remarks: One verifies that sinq = 0.073 # q and cosq = 0.997 # 1. When the helicopter is not immobile, but has a horizontal velocity, for example v0, the relative velocity of air with respect to the blade varies between v0 + wx for the blade that is forward, and –v0 + wx for a blade that is backward. If the incidence i does not vary, the lift varies in a cyclical manner, and there is vertical “flapping motion” of the blade. This is why a mechanism for cyclic variation of the incidence is necessary. We have not taken into account the drag to simplify the calculations. This can be considered similarly to the case of the lift. It then gives rise to an equilibrium position with a second small angle, called j, with respect to the radial direction from top view, as in the following figure. This is why a supplementary joint, or a drag joint, is necessary. 2. Weight of the helicopter: The lift and the weight balance themselves out. The lift of the blade is then: then for the 3 rotor blades: numerically: q = = 0.073 rad 4∞11¢ Fz Fz q # dFz /10 Ú d cos /10 Ú 1 2 --rcCzw2 3 3 /103 ( ) – 3 = = ------------------------------- Mg = 3Fz Mg # 1 2 --rcCzw2 3 Mg = 2340 daN TX846_Frame_C18a Page 350 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
3.Normal load:It is denoted as M): N=m9(e2- 2 at the foot of the blade (x=//10): N(t/10)#12,000daN 4.Longitudinal modulus of elasticity: Using the relation of Section 3.3.1: Ee ErVr+EmVm with (Section 1.6):E,=86,000 MPa;E=4,000 MPa. E=53,200 MPa 5.Section of the spar of the blade made of glass/epoxy: The longitudinal rupture tensile stress of the unidirectional is Oe nupture 1700 MPa With a factor of safety of 6,the admissible stress at a section S(x)becomes 。==1700=283MPa S(x) 6 then: S(ax)=Na) S(x)= 20(-x m at the foot of the blade: S(/10)=4.24cm 2003 by CRC Press LLC
3. Normal load: It is denoted as N(x): at the foot of the blade (x = l/10): 4. Longitudinal modulus of elasticity: Using the relation of Section 3.3.1: with (Section 1.6): Ef = 86,000 MPa; Em = 4,000 MPa. 5. Section of the spar of the blade made of glass/epoxy: The longitudinal rupture tensile stress of the unidirectional is With a factor of safety of 6, the admissible stress at a section S(x) becomes then: at the foot of the blade: N x( ) Fc q # dFc x Ú d cos x Ú mw2 x dx x Ú = = N x( ) mw2 2 ----------- 2 x2 = ( ) – N( ) /10 # 12,000 daN E = EfVf + EmVm E = 53,200 MPa s rupture # 1700 MPa s N x( ) S x( ) ----------- 1700 6 === ----------- 283 MPa S x( ) N x( ) s = ----------- S x( ) mw2 2s ----------- 2 x2 = ( ) – S( ) /10 4.24 cm2 = TX846_Frame_C18a Page 351 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC