It is then necessary to augment the thickness of the tube at the bond location. One starts from the relation: tha2πr2e placed in the form: GC NZGeec x ≤wit袖eKl (1-e)2π2 then: 22 ×(1-e) N2Geec One finds numerically: e>11.7mm we retain e =12 mm (one then has th a=1-8=0.987) ■Bond length In accordance with Section 6.2.3,the resistance condition is written as: 12 2.5 -0.2 schematic of fitting of tube 120 12 >44 M Taverage= 2πr2e -≤0.2 X Trupture then: e>44 mm 2003 by CRC Press LLC
It is then necessary to augment the thickness of the tube at the bond location. One starts from the relation: placed in the form: then: One finds numerically: e > 11.7 mm; we retain e = 12 mm (one then has th a = 1 - e = 0.987) Bond length In accordance with Section 6.2.3, the resistance condition is written as: then: a th a ---------- Mt 2pr 2 ¥ -------------- £ trupture GC 2GeeC ---------------- ( ) 1 – e ---------------- Mt 2pr 2 ¥ ----------- £ trupture with e << 1 GC 2GeeC --------------- trupture 2pr 2 Mt £ ¥ ----------- ¥ ( ) 1 – e t average Mt 2pr 2 = -------------- £ 0.2 ¥ trupture ≥ 44 mm TX846_Frame_C18a Page 357 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
3.Mass assessment: Mass of the shaft in carbon/epoxy Maminate=pX2元reXL with numerical values already cited: mlaminate =2.8 kg. If one takes a tubular shaft made of steel (Trupture=300 MPa)with a factor of safety that is 2 times less,say 3,and a minimum thickness of 2.5 mm,the resistance condition: M≤30OMPa 2πre3 leads to a radius of the tube of r≥43mm. From this we find a mass of:(pstee =7,800 kg/m2): msteel =10.5 kg The saving in mass of the composite solution over the steel solution is 73%.The real saving is higher because it takes into account the disappearance of the intermediate bearing and of one part of the universal joint. 18.1.5 Flywheel in Carbon/Epoxy Problem Statement: We show schematically in the figure below an inertia wheel made of carbon/ epoxy with 60%fiber volume fraction,with the indicated proportions for the orientation of the fibers. r(average b 5% +90% 5% 2003 by CRC Press LLC
3. Mass assessment: Mass of the shaft in carbon/epoxy mlaminate = r ¥ 2pre ¥ L with numerical values already cited: mlaminate = 2.8 kg. If one takes a tubular shaft made of steel (t rupture = 300 MPa) with a factor of safety that is 2 times less, say 3, and a minimum thickness of 2.5 mm, the resistance condition: leads to a radius of the tube of r ≥ 43 mm. From this we find a mass of: (rsteel = 7,800 kg/m3 ): msteel = 10.5 kg The saving in mass of the composite solution over the steel solution is 73%. The real saving is higher because it takes into account the disappearance of the intermediate bearing and of one part of the universal joint. 18.1.5 Flywheel in Carbon/Epoxy Problem Statement: We show schematically in the figure below an inertia wheel made of carbon/ epoxy with 60% fiber volume fraction, with the indicated proportions for the orientation of the fibers. Mt 2pr 2 e -------------- 300 3 £ -------- MPa TX846_Frame_C18a Page 358 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
1.Calculate the maximum kinetic energy that one can obtain with such a flywheel with a mass of 1 kg. 2.Compare the maximum kinetic energy that one can obtain with a steel flywheel with a mass of 1 kg.One will take:upure e=1,000 MPa. Solution: 1.The equilibrium of an element of the wheel (shown below)reveals inertia forces and cohesive forces. dm 62r(centrifugal load) ceb(cohesion load) 08 d0/2 One deduces from there the equilibrium equation along the radial direction: dm×w2r=2oeb2 Denoted by p the specific mass: pr de eba'r oebde p(ra)2=o Denoted by V=ro the circumferential speed,one obtains the maximum for the rupture strength of carbon/epoxy,as: 0 Numerical application:For the composition of the carbon/epoxy laminate indicated above,one reads in Section 5.4.2,Table 5.1: Orupture =1,059 MPa and with p 1,530 kg/m2(Table 3.4 of Section 3.3.3,or the calculation in Section 3.2.3 Vmax =832 m/s 2003 by CRC Press LLC
1. Calculate the maximum kinetic energy that one can obtain with such a flywheel with a mass of 1 kg. 2. Compare the maximum kinetic energy that one can obtain with a steel flywheel with a mass of 1 kg. One will take: srupture steel = 1,000 MPa. Solution: 1. The equilibrium of an element of the wheel (shown below) reveals inertia forces and cohesive forces. One deduces from there the equilibrium equation along the radial direction: Denoted by r the specific mass: Denoted by V = rw the circumferential speed, one obtains the maximum for the rupture strength of carbon/epoxy, as: Numerical application: For the composition of the carbon/epoxy laminate indicated above, one reads in Section 5.4.2, Table 5.1: srupture = 1,059 MPa and with r = 1,530 kg/m 3 (Table 3.4 of Section 3.3.3, or the calculation in Section 3.2.3): Vmax = 832 m/s dm w2 ¥ r 2seb dq 2 = ------ rr dq ebw2 r = sebdq r( ) rw 2 = s Vmax srupture r = --------------- TX846_Frame_C18a Page 359 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
from this the maximum kinetic energy obtained with 1kg of composite: then: WKinetic =346 kjoules 2.The maximum circumferential speed that one can obtain with a steel flywheel can be written as: Orupture steel Pstedl Therefore,the ratio of kinetic energies of composite/steel is Wkinetic carbon= Vmax cabon Orupt.carbon X pstee Wkinetic steel Vmux sed Orupt.steel Pcarbon with Paed =7800 kg/mand Cup e=1000 MPa,one obtains Wkinetic carbon =5.4 WKinetic steel With respect to the same mass,it appears then possible to accumulate 5 times more kinetic energy with a flywheel in carbon/epoxy composite. 18.1.6 Wing Tip Made of Carbon/Epoxy Problem Statement: Wing tip refers to a part of airplane wing as shown in Figure 18.1.It is made of a sandwich structure with carbon/epoxy skins (Figure 18.2)fixed to the rest of the wing by titanium borders as shown.Under the action of the aerodynamic forces (Figure 18.3),the wing tip is subjected to bending moments,torsional moments, and shear forces as shown in Figure 18.4(a). One can assume that the core of the sandwich structure transmits only shear forces,and the skins support the flexural moments.This is represented in Figure 18.4(b);the skins resist in their respective planes the in-plane stress resultants:N,N,and T Figure 18.5 shows the values of these stress resultants Recall the expression for the rotational kinetic energy of a mass m placed at a radiusr and rotating at a speed of @Wkinetie ==mr=mVcrconfer. 2003 by CRC Press LLC
from this the maximum kinetic energy obtained with 1kg of composite 7 : then: 2. The maximum circumferential speed that one can obtain with a steel flywheel can be written as: Therefore, the ratio of kinetic energies of composite/steel is with rsteel = 7800 kg/m 3 and srupt. steel = 1000 MPa, one obtains With respect to the same mass, it appears then possible to accumulate 5 times more kinetic energy with a flywheel in carbon/epoxy composite. 18.1.6 Wing Tip Made of Carbon/Epoxy Problem Statement: Wing tip refers to a part of airplane wing as shown in Figure 18.1. It is made of a sandwich structure with carbon/epoxy skins (Figure 18.2) fixed to the rest of the wing by titanium borders as shown. Under the action of the aerodynamic forces (Figure 18.3), the wing tip is subjected to bending moments, torsional moments, and shear forces as shown in Figure 18.4(a). One can assume that the core of the sandwich structure transmits only shear forces, and the skins support the flexural moments. This is represented in Figure 18.4(b); the skins resist in their respective planes the in-plane stress resultants: Nx, Ny, and Txy. Figure 18.5 shows the values of these stress resultants 7 Recall the expression for the rotational kinetic energy of a mass m placed at a radius r and rotating at a speed of w: . WKinetic 1 2 --Iw2 1 2 --mr2 w2 1 2 --mVcirconfer 2 == = WKinetic 1 2 -- 1 kg Vmax 2 = ¥ ¥ WKinetic = 346 kjoules Vmax. steel srupture steel rsteel = ----------------------- WKinetic carbon WKinetic steel ---------------------------- Vmax carbon 2 Vmax steel 2 ---------------------- srupt. carbon ¥ rsteel srupt. steel ¥ rcarbon = = ---------------------------------------- WKinetic carbon WKinetic steel ---------------------------- = 5.4 TX846_Frame_C18a Page 360 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
wing tip Figure 18.1 carbon/epoxy laminate titanium border(TA6V) duralumir Section AA bolting Figure 18.2 extrados support TI T intrado aerodynamic loads Figure 18.3 at a few points of the upper skin (or extrados). 1.According to Figures 18.4(a)and 18.4(b),deduce the elements of the stress resultants N,N,and Tsy from the knowledge of the moment resultants MyMy and Msy 2. Using a factor of safety of 2,define the carbon/epoxy skin that is suitable at the surrounding of the support made of titanium alloy (proportions, thickness,number of plies).One will use unidirectional plies with V= 60%fiber volume fraction. 2003 by CRC Press LLC
at a few points of the upper skin (or extrados). 1. According to Figures 18.4(a) and 18.4(b), deduce the elements of the stress resultants Nx, Ny, and Txy from the knowledge of the moment resultants Mx, My, and Mxy. 2. Using a factor of safety of 2, define the carbon/epoxy skin that is suitable at the surrounding of the support made of titanium alloy (proportions, thickness, number of plies). One will use unidirectional plies with Vf = 60% fiber volume fraction. Figure 18.1 Figure 18.2 Figure 18.3 TX846_Frame_C18a Page 361 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC