PART IV APPLICATIONS We have grouped in this last part of the book exercises and examples for applica- tions.These have various objectives and different degrees of difficulties.Leaving aside (except for special cases)the cases that are too academic,we will concern ourselves with applications of concrete nature,with an emphasis on the numerical aspect of the results.A few of these applications should be used as validation tests for numerical models. 2003 by CRC Press LLC
PART IV APPLICATIONS We have grouped in this last part of the book exercises and examples for applications. These have various objectives and different degrees of difficulties. Leaving aside (except for special cases) the cases that are too academic, we will concern ourselves with applications of concrete nature, with an emphasis on the numerical aspect of the results. A few of these applications should be used as validation tests for numerical models. TX846_Frame_C18a Page 341 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
18 APPLICATIONS 18.1 LEVEL 1 18.1.1 Simply Supported Sandwich Beam Problem Statement: 1.The following figure represents a beam made of duralumin that is supported at two points.It is subjected to a transverse load of F=50 daN.Calculate the deflection-denoted as A-of the beam under the action of the force F 0=10cm F=50 daN h=5mm 1) e=500 mm F=50 daN ep=2.5 mm 2) 瓶 ep ec=25 mm 安 2.We separate the beam of duralumin into two parts with equal thickness ep=2.5 mm,by imaginarily cutting the beam at its midplane.Each half is bonded to a parallel pipe made of polyurethane foam,making the skins of a sandwich beam having essentially the same mass as the initial beam (in neglecting the mass of the foam and the glue).The beam is resting on the same supports and is subjected to the same load F Calculate the deflection caused by F denoted by A'.Compare with the value of A found in Part 1.(Take the shear modulus of the foam to be:G=20 MPa.) Solution: 1.We will use the classical formula that gives the deflection at the center of the beam on two supports: F △= 48E1 with I= 12 2003 by CRC Press LLC
18 APPLICATIONS 18.1 LEVEL 1 18.1.1 Simply Supported Sandwich Beam Problem Statement: 1. The following figure represents a beam made of duralumin that is supported at two points. It is subjected to a transverse load of F = 50 daN. Calculate the deflection—denoted as D—of the beam under the action of the force F. 2. We separate the beam of duralumin into two parts with equal thickness ep = 2.5 mm, by imaginarily cutting the beam at its midplane. Each half is bonded to a parallel pipe made of polyurethane foam, making the skins of a sandwich beam having essentially the same mass as the initial beam (in neglecting the mass of the foam and the glue). The beam is resting on the same supports and is subjected to the same load F. Calculate the deflection caused by F, denoted by D¢. Compare with the value of D found in Part 1. (Take the shear modulus of the foam to be: Gc = 20 MPa.) Solution: 1. We will use the classical formula that gives the deflection at the center of the beam on two supports: D F 48EI ----------- with I bh3 12 = = -------- TX846_Frame_C18a Page 343 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
For duralumin (see Section 1.6):E=75,000 MPa.One finds △=16.7mm 2.Denoting by W the elastic energy due to flexure,one has' 高+高r heam with 1 (G两*G.(e.+2e,)×b Using Castigliano theorem,one has aF then: ∫尚=+∫高票 0≤x≤E/2:M=Fx/2;T=-F12 2≤x≤e:M=t-T=F2 r"高受×+nc- …+总-x-+× △'= Fe3 Fe k 8(ED+4(G两 Approximate calculation: (En=B,xe,xbxe+e2+E eb 2 12 then: (EI)=7090 MKS+7.8 MKS with E=60 MPa (cf.1.6) negligible TTo establish this relation,see Chapter 15,Equation 15.17. 2 See calculation of this coefficient in 18.2.1,and more precise calculation in 18.3.5. 2003 by CRC Press LLC
For duralumin (see Section 1.6): E = 75,000 MPa. One finds 2. Denoting by W the elastic energy due to flexure, one has1 with2 : Using Castigliano theorem, one has then: Approximate calculation: then: 1 To establish this relation, see Chapter 15, Equation 15.17. 2 See calculation of this coefficient in 18.2.1, and more precise calculation in 18.3.5. D = 16.7 mm W 1 2 -- M2 · Ò EI ----------- x 1 2 -- k · Ò GS ------------T 2 dx beam Ú d + beam Ú = k · Ò GS ------------ # 1 Gc ec + 2ep ( ) ¥ b --------------------------------------- D¢ ∂W ∂F = -------- D¢ M · Ò EI ----------- dM dF ------- x k · Ò GS ------------T dT dF ------ dx beam Ú d + beam Ú = 0 £ £ x /2: M = = Fx/2; T F– /2 /2 £ £ x : M F 2 = = --( ) – x ; T F /2 D¢ 1 · Ò EI ----------- Fx 2 ------ x 2 ¥ -- x F 2 --( ) – x ( ) – x 2 ----------------dxº /2 Ú d + 0 /2 ÚÓ ˛ Ì ˝ Ï ¸ = º k · Ò GS ------------- F 2 -- dx 2 – ¥ –------ F 2 -- dx 2 ¥ ------ /2 Ú + 0 /2 ÚÓ ˛ Ì ˝ Ï ¸ + D¢ F3 48 · Ò EI ----------------- F 4 ------ k · Ò GS = + ------------ · Ò EI # Ep ep b ec + ep ( )2 2 ---------------------- Ec ec 3 b 12 ¥ ¥ ¥ + ¥ ------- · Ò EI 7090 MKS 7.8 MKS negligible = = + with Ec 60 MPa ( ) cf. 1.6 TX846_Frame_C18a Page 344 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
one obtains for△'： △'=0.18mm+1.04mm bending shear moment △'=1.22mm Comparing with the deflection A found in Part 1 above: 14 Remarks: The sandwich configuration has allowed us to divide the deflection by 14 without significant augmentation of the mass:with adhesive film thickness 0.2 mm and a specific mass of 40 kg/m'for the foam,one obtains a total mass of the sandwich: m =700 g (duralumin)+50 g(foam)+48 g (adhesive) This corresponds to an increase of 14%with respect to the case of the full beam in Question 1. The deflection due to the shear energy term is close to 6 times more important than that due to the bending moment only.In the case of the full beam in question 1,this term is negligible.In effect one has: k=1.2 for a homogeneous beam of rectangular section,then: GS =8.27×108 (with G=29,000 MPa,Section 1.6).The contribution to the deflection A of the shear force is then: ∫会r乐=002mmxA 18.1.2 Poisson Coefficient of a Unidirectional Layer Problem Statement: Consider a unidirectional layer with thickness e as shown schematically in the figure below.The moduli of elasticity are denoted as E(longitudinal direction) and E,(transverse direction). 2003 by CRC Press LLC
one obtains for D¢: Comparing with the deflection D found in Part 1 above: Remarks: The sandwich configuration has allowed us to divide the deflection by 14 without significant augmentation of the mass: with adhesive film thickness 0.2 mm and a specific mass of 40 kg/m3 for the foam, one obtains a total mass of the sandwich: m = 700 g (duralumin) + 50 g (foam) + 48 g (adhesive) This corresponds to an increase of 14% with respect to the case of the full beam in Question 1. The deflection due to the shear energy term is close to 6 times more important than that due to the bending moment only. In the case of the full beam in question 1, this term is negligible. In effect one has: k = 1.2 for a homogeneous beam of rectangular section, then: (with G = 29,000 MPa, Section 1.6). The contribution to the deflection D of the shear force is then: 18.1.2 Poisson Coefficient of a Unidirectional Layer Problem Statement: Consider a unidirectional layer with thickness e as shown schematically in the figure below. The moduli of elasticity are denoted as E (longitudinal direction) and Et (transverse direction). D¢ = 0.18 mm 1.04 mm + bending shear moment D¢ = 1.22 mm D D¢ ----- 14 1 = ----- k GS ------ 8.27 10–8 = ¥ k GS ------ T dT dF ------dx Ú = 0.02 mm << D TX846_Frame_C18a Page 345 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
Show that two distinct Poisson coefficients ve,and vie are necessary to charac- terize the elastic behavior of this unidirectional layer.Numerical application:a layer of glass/epoxy.V=60%fiber volume fraction. Solution: Let the plate be subjected to two steps of loading as follows: b 1.A uniform stress oe along the e direction:the changes in lengths of the sides can be written as: b 2.A uniform stress o,along the t direction:for a relatively important elongation of the resin,one can only observe a weak shortening of the fibers along e.Using then another notation for the Poisson coefficient,the change in length can be written as: b a Now calculating the accumulated elastic energy under the two loadings above: When o is applied first,and then o,is applied, 1 W=soxaxexAb+-ic,×bxexha,.+axaxex△b, When o,is applied first,and then o is applied, W'=30×bxexha2+50×a×e×△b,+o,×bxe×△@ The final energy is the same: W=W' 2003 by CRC Press LLC
Show that two distinct Poisson coefficients nt and nt are necessary to characterize the elastic behavior of this unidirectional layer. Numerical application: a layer of glass/epoxy. Vf = 60% fiber volume fraction. Solution: Let the plate be subjected to two steps of loading as follows: 1. A uniform stress s along the direction: the changes in lengths of the sides can be written as: 2. A uniform stress st along the t direction: for a relatively important elongation of the resin, one can only observe a weak shortening of the fibers along . Using then another notation for the Poisson coefficient, the change in length can be written as: Now calculating the accumulated elastic energy under the two loadings above: When s is applied first, and then st is applied, When st is applied first, and then s is applied, The final energy is the same: Db1 b -------- s E ----- ; Da1 a --------- nt E = = –------s Db2 b -------- nt Et –------st ; Da2 a --------- st Et = = ---- W 1 2 --s a e Db1 1 2 = ¥ ¥ ¥ + --st ¥ b e ¥ ¥ Da2 + s ¥ a e ¥ ¥ Db2 W¢ 1 2 --st b e Da2 1 2 = ¥ ¥ ¥ + --s ¥ a e ¥ ¥ Db1 + st ¥ b e ¥ ¥ Da1 W W= ¢ TX846_Frame_C18a Page 346 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
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