6.Mass of the spar (longeron)of the blade: mspar=Punidirec.S(x)dx =p x mo2 spar unidirect. Specific mass of the unidirectional layer (see Section 3.2.3): Punidirect.VrPr+Vmpm 1980 kg/m3 Then: mspar 2.38 kg 7.Elongation of the spar of the blade:The longitudinal constitutive relation is written as (see Section 3.1): Ex= N(x) Etx S(x)Et Elongation of a segment dx Ex(x)dx. For the whole spar of blade: ae=∫e,d 10 4f=0.9 Ee then: A0=2.4cm One has to reinforce the spar of the blade to diminish the elongation to resist the centrifugal force. 8.Clamped axes:For 2 axes in 30 NCD16 steel (rupture shear strength Tnup= 500 MPa;bearing strength obearng=1600 MPa);4 sheared sections;factor of safety =6: ■diameter::/10)/πo≤tup/6→p≥21.4mm ■length:N(e/10)/2bo≤earm/6→b≥10.5mm glass-resin ompound unidirectiona axⅪs foam resin-foam compound 2003 by CRC Press LLC
6. Mass of the spar (longeron) of the blade: Specific mass of the unidirectional layer (see Section 3.2.3): Then: 7. Elongation of the spar of the blade: The longitudinal constitutive relation is written as (see Section 3.1): Elongation of a segment dx : e x (x) dx. For the whole spar of blade: then: One has to reinforce the spar of the blade to diminish the elongation to resist the centrifugal force. 8. Clamped axes: For 2 axes in 30 NCD16 steel (rupture shear strength trupt = 500 MPa; bearing strength s bearing = 1600 MPa); 4 sheared sections; factor of safety = 6: diameter: N(/10)/pf2 £ trupt/6 Æ f ≥ 21.4 mm length: N(/10)/2hf £ sbearing/6 Æ h ≥ 10.5 mm mspar runidirect.S x( )dx /10 Ú = m spar r unidirect. mw2 s ----------- 1.7 6 -------3 = ¥ ¥ runidirect. Vf rf + Vmrm 1980 kg/m3 = = mspar = 2.38 kg ex sx Ex ----- N x( ) E ¥ S x( ) ---------------------- s E == = ---- D ex dx /10 Ú = D 0.9 s E = ------ D = 2.4 cm TX846_Frame_C18a Page 352 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
18.1.4 Transmission Shaft for Trucks Problem Statement: One proposes to replace the classical transmission shaft made of universal cardan joint and intermediary thrust bearing as shown below: 站部8 =回 with a solution consisting of a long shaft made of carbon/epoxy,with the following dimensions: 03120mm coupling plates (steel) L=2000mm The characteristics of the transmission shaft are as follows: Maximum torsional couple:M,300 m daN ■ Maximum rotation speed:N=4000 revolutions/minute The first resonant flexural frequency of a beam on two supports is given by: EI where m is the mass of the beam,and I is its flexure moment of inertia.It corresponds to a "critical speed"for a beam in rotation,which should not be reached during the operation The carbon/epoxy unidirectional has V=60%fiber volume fraction.The thickness of a cured ply is 0.125 mm. 1.Give the characteristics of a suitable shaft of carbon/epoxy composite. One will make use of the tables in Section 5.4.2 and will use a factor of safety of 6. 2.Study the adhesive fitting of the coupling plates to the shaft. 3.Carry out an assessment on the saving in weight with respect to the "shaft in steel"solution (not including the coupling plates). Solution: 1.Characteristics of the shaft:The shaft is assumed to be thin and hollow (thickness e is small compared with the average radius r as in the following figure). 2003 by CRC Press LLC
18.1.4 Transmission Shaft for Trucks Problem Statement: One proposes to replace the classical transmission shaft made of universal cardan joint and intermediary thrust bearing as shown below: with a solution consisting of a long shaft made of carbon/epoxy, with the following dimensions: The characteristics of the transmission shaft are as follows: Maximum torsional couple: Mt = 300 m daN Maximum rotation speed: N = 4000 revolutions/minute The first resonant flexural frequency of a beam on two supports is given by: where m is the mass of the beam, and I is its flexure moment of inertia. It corresponds to a “critical speed” for a beam in rotation, which should not be reached during the operation. The carbon/epoxy unidirectional has Vf = 60% fiber volume fraction. The thickness of a cured ply is 0.125 mm. 1. Give the characteristics of a suitable shaft of carbon/epoxy composite. One will make use of the tables in Section 5.4.2 and will use a factor of safety of 6. 2. Study the adhesive fitting of the coupling plates to the shaft. 3. Carry out an assessment on the saving in weight with respect to the “shaft in steel” solution (not including the coupling plates). Solution: 1. Characteristics of the shaft: The shaft is assumed to be thin and hollow (thickness e is small compared with the average radius r as in the following figure). f1 p 2 -- EI mL3 = --------- TX846_Frame_C18a Page 353 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
The shear stress t is as follows: M T= 2πre Taking into account the nature of the loading on the laminate making up the tube (pure shear),the composition of the tube requires e ■ Important percentages of unidirectionals in the direction of +45 (see Section 5.2.2). Minimum percentages in the order of 10%in other directions (see Section 5.2.3.6) This leads,for example,to the following distribution: 10% 40% 10% 40% In Section 5.4,one finds Table 5.3,which gives the maximum shear stress that can be applied to a laminate subject to pure shear,as a function of the proportions of the plies at0°，90°，+45°，-45°.One reads for the proportions above: Tmax 327 MPa from which the admissible stress taking into account a safety factor of 6: Tadmis.327/6 MPa One then has M,≤admis 2πre or numerically: re≥8760mm3 2003 by CRC Press LLC
The shear stress t is as follows: Taking into account the nature of the loading on the laminate making up the tube (pure shear), the composition of the tube requires Important percentages of unidirectionals in the direction of ±45∞ (see Section 5.2.2). Minimum percentages in the order of 10% in other directions (see Section 5.2.3.6). This leads, for example, to the following distribution: In Section 5.4, one finds Table 5.3, which gives the maximum shear stress that can be applied to a laminate subject to pure shear, as a function of the proportions of the plies at 0∞, 90∞, +45∞, -45∞. One reads for the proportions above: from which the admissible stress taking into account a safety factor of 6: One then has or numerically: t Mt 2pr 2 e = -------------- tmax = 327 MPa t admis. = 327/6 MPa Mt 2pr 2 e -------------- £ t admis. r 2 e 8 760 mm3 ≥ TX846_Frame_C18a Page 354 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
For a minimum specified radius r=60 mm,one obtains e≥2.43mm The corresponding number of plies of carbon/epoxy is 2.3#20 plies 0.125 giving a thickness of: e 2.5 mm One can verify that a number of 20 plies allows one to satisfy the following: (a)The required proportions 20y6 8(40%) 20 2(10% 8(40%) (b)The midplane symmetry,with the sequence: L90°101±45 Critical speed of such a shaft: = EI The longitudinal modulus E of the laminate (in the direction of the shaft) is (see Table 5.4 [longitudinal modulus]in Section 5.4.2) E=31,979MPa The specific density of the laminate is (see Section 3.2.3) Pum =Vrer+Vmpm with (Section 1.6):P=1750 kg/m;P=1,200 kg/m3.Then:Pum =1,530 kg/m3 (see also Table 3.4 in Section 3.3.3). 2003 by CRC Press LLC
For a minimum specified radius r = 60 mm, one obtains The corresponding number of plies of carbon/epoxy is giving a thickness of: One can verify that a number of 20 plies allows one to satisfy the following: (a) The required proportions (b) The midplane symmetry, with the sequence: Critical speed of such a shaft: The longitudinal modulus E of the laminate (in the direction of the shaft) is (see Table 5.4 [longitudinal modulus] in Section 5.4.2) E = 31,979 MPa The specific density of the laminate is (see Section 3.2.3) with (Section 1.6): rf = 1750 kg/m 3 ; rm = 1,200 kg/m3. Then: rlam = 1,530 kg/m 3 (see also Table 3.4 in Section 3.3.3). e ≥ 2.43 mm 2.43 0.125 ------------- # 20 plies e = 2.5 mm 90∞/0∞/ 454 ∞ ± s f1 p 2 -- EI mL2 = --------- rlam = Vf rf + Vmrm TX846_Frame_C18a Page 355 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
The second moment of inertia in flexure is I=e from which the first frequency is:f=76 Hz It corresponds to a critical speed of 4,562 rev/minute,superior to the maximum speed of rotation of the shaft.3 2.Bonded fittings: We will use the relation of Paragraph 6.2.3 (Figure 6.26)for simplification.This implies identical thicknesses for the tube making up the shaft and the coupling plate made of steel.The maximum shear stress has an order of magnitude of: tha ax M 心=0X*-th02z7 where e is the bond length,and G a=t N2Gee with G as the shear modulus of araldite,then G=1,700 MPa (Section 1.6). Glaminate=28430 MPa (Section 5.4.2;Table 5.5) ec bond thickness (Section 6.2.3):ec =0.2 mm. Thickness at bond location: If one conserves the thickness found for the tube,as e 2.5 mm,one obtains a=1×244.5 The resistance condition can then be written as: Tmaxs Trupure (15 MPa for araldite;see Section 6.2.3). Then: a× M, tha^2πr2 ≤Trupture 244.5、M, X ha2πr ≤Trupture numerically:th a 22.16 impossible (thx E]-1,1 [) 5 One also has to verify the absence of buckling due to torsion of the shaft,see annex 2 for this subject. 6For different thicknesses for the tube made of carbon/epoxy and for the coupling plate part, one can use the more general relation established in application 18.3.1.This also allows different shear moduli for each material. 2003 by CRC Press LLC
The second moment of inertia in flexure is I = p r 3 e from which the first frequency is: f1 = 76 Hz It corresponds to a critical speed of 4,562 rev/minute, superior to the maximum speed of rotation of the shaft. 5 2. Bonded fittings: We will use the relation of Paragraph 6.2.3 (Figure 6.26) for simplification. This implies identical thicknesses for the tube making up the shaft and the coupling plate made of steel. 6 The maximum shear stress has an order of magnitude of: where is the bond length, and with Gc as the shear modulus of araldite, then Gc = 1,700 MPa (Section 1.6). Glaminate = 28430 MPa (Section 5.4.2; Table 5.5) ec = bond thickness (Section 6.2.3): ec = # 0.2 mm. Thickness at bond location: If one conserves the thickness found for the tube, as e = 2.5 mm, one obtains a = l ¥ 244.5 The resistance condition can then be written as: t max £ t rupture (15 MPa for araldite; see Section 6.2.3). Then: numerically: th a ≥ 2.16 Æ impossible (th x Œ] – 1, + 1 [). 5 One also has to verify the absence of buckling due to torsion of the shaft, see annex 2 for this subject. 6 For different thicknesses for the tube made of carbon/epoxy and for the coupling plate part, one can use the more general relation established in application 18.3.1. This also allows different shear moduli for each material. tmax a th a ---------- ¥ t average a th a ---------- Mt 2pr 2 = = ¥ -------------- a Gc 2Geec = --------------- a th a ---------- Mt 2pr 2 ¥ -------------- £ trupture 244.5 th a ------------- Mt 2pr 2 ¥ ----------- £ trupture TX846_Frame_C18a Page 356 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC