Table 5.19 of Section 5.4.4 allows one to obtain an optimal composition close to 10% 40% 10% 40% If one uses the previous exact stress resultants,the calculation by computer of the optimal composition leads to the following result,which can be interpreted as described in Section 5.4.4. .1 45 1063 .71 Then one has for the minimum thickness of the laminate: tbicknes:e0.1063+3=2.17 mm 100 40% +10% 40% and for the two immediate neighboring laminates: tbickness:e=0.1068× (1800+1l900+1340 100 =2.18mm 15% 37.5% *10% 75/ 2003 by CRC Press LLC
Table 5.19 of Section 5.4.4 allows one to obtain an optimal composition close to If one uses the previous exact stress resultants, the calculation by computer of the optimal composition leads to the following result, which can be interpreted as described in Section 5.4.4. Then one has for the minimum thickness of the laminate: thickness: e = 0.1063 ¥ = 2.17 mm and for the two immediate neighboring laminates: thickness: e = 0.1068 ¥ = 2.18 mm ( ) 800 900 340 + + 100 ---------------------------------------------------- ( ) 800 900 340 + + 100 ---------------------------------------------------- TX846_Frame_C18a Page 367 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
thickness:.e=0.1096× 1800+|9001+1340 100 2=2.24mm 576 35% 15% 35e One notes a sensible difference between the initial composition estimated by the designer and the optimal composition.This difference in composition leads to a relative difference in thickness: 2.64-2.17 2.17 =21% which indicates a moderate sensibility concerning the effect of thickness and, thus,the mass.One foresees there a supplementary advantage:the possibility to reinforce the rigidity in given directions without penalizing very heavily the thickness.We can note this if we compare the moduli of elasticity obtained starting from the estimated design composition (Section 5.4.3)with the optimal composition,we obtain (Section 5.4.2,Tables 5.4 and 5.5)very different values noted below: estimated design composition optimum composition 30% 10% 20% 40% 30% 10% 20/ 40 Ex=55,333 MPa Ex=31,979 MPa G=16,315MPa Gixy=28,430 MPa 3.Bonding of the laminate:We represent here after the principal loadings deduced from the values of the stress resultants in Figure 18.5,in the immediate neighborhood of the border of titanium: 2003 by CRC Press LLC
thickness: e = 0.1096 ¥ = 2.24 mm One notes a sensible difference between the initial composition estimated by the designer and the optimal composition. This difference in composition leads to a relative difference in thickness: which indicates a moderate sensibility concerning the effect of thickness and, thus, the mass. One foresees there a supplementary advantage: the possibility to reinforce the rigidity in given directions without penalizing very heavily the thickness. We can note this if we compare the moduli of elasticity obtained starting from the estimated design composition (Section 5.4.3) with the optimal composition, we obtain (Section 5.4.2, Tables 5.4 and 5.5) very different values noted below: 3. Bonding of the laminate: We represent here after the principal loadings deduced from the values of the stress resultants in Figure 18.5, in the immediate neighborhood of the border of titanium: Ex = 55,333 MPa Ex = 31,979 MPa Gxy = 16,315 MPa Gxy = 28,430 MPa ( ) 800 900 340 + + 100 ---------------------------------------------------- 2.64 2.17 – 2.17 -------------------------- = 21% TX846_Frame_C18a Page 368 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
titanium- laminated 2.6 daN/mm A -24.6 729 629 41° 59 7 daN/mm One can for example,overestimate these loadings by substituting them with a fictitious distribution based on the most important component among them. Taking -59.7 daN/mm,one obtains then the simplified schematic below: titanium tg88 、laminated N=-59.7 daN/mm One must evaluate the width of bonding noted as e.For a millimeter of the border, this corresponds to a bonding surface of x 1 mm.For an average rupture criterion of shear of the adhesive,one can write (see Section 6.2.3): N e×1 0.2 X Tadhesive rupture then with t nupture=30 MPa: 597 #100mm 0.2×30 From this one obtains the following configuration such that e+e2+3=100 mm. t1 laminated 62 ta titanium 4.Bolting on the rest of the wing: ■“Pitch of bolts”:The tensile of bolting is assumed to be weak,then bolting strength is calculated based on shear.The bolt load transmitted by a bolt is denoted as AF and one has (cf.following figure): AF=NX pitchs 车XTboit rupture 2003 by CRC Press LLC
One can for example, overestimate these loadings by substituting them with a fictitious distribution based on the most important component among them. Taking –59.7 daN/mm, one obtains then the simplified schematic below: One must evaluate the width of bonding noted as . For a millimeter of the border, this corresponds to a bonding surface of ¥ 1 mm. For an average rupture criterion of shear of the adhesive, one can write (see Section 6.2.3): then with t rupture = 30 MPa: From this one obtains the following configuration such that 1 + 2 + 3 = 100 mm. 4. Bolting on the rest of the wing: “Pitch of bolts”: The tensile of bolting is assumed to be weak, then bolting strength is calculated based on shear. The bolt load transmitted by a bolt is denoted as DF, and one has (cf. following figure): N ¥ 1 ----------- £ 0.2 ¥ tadhesive rupture 597 0.2 30 ¥ ≥ ------------------- # 100 mm DF N pitch p∆2 4 = ¥ £ ---------- ¥ tbolt rupture TX846_Frame_C18a Page 369 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
where is the diameter of the bolt.One finds a pitch equal to 35 mm. titanium pitch △F border (intrados N=-59.7 daN/mm edge distance This value is a bit high.In practice,one takes pitch s5 for example,here: Pitch =30 mm. Thickness of the border:the bearing condition is written as: Nx pitch ☑titanium then: etitanium≥2.55mm Verification of the resistance of the border in the two zones denoted a in the previous figure:the stress resultant in this zone,noted as N,is such that: N×pitch=W'×(pitch-☑) then: N'=N-pitch pitch-=75.4 daN/mm The rupture stress being: Grupture =900 MPa and the minimum thickness 2.55 mm,one must verify N(daN/mm)e(daN/mm) e(mm) One effectively has 第0 2003 by CRC Press LLC
where ∆ is the diameter of the bolt. One finds a pitch equal to 35 mm. This value is a bit high. In practice, one takes pitch £ 5 ∆, for example, here: Pitch = 30 mm. Thickness of the border: the bearing condition is written as: then: Verification of the resistance of the border in the two zones denoted a in the previous figure: the stress resultant in this zone, noted as N¢, is such that: then: The rupture stress being: srupture = 900 MPa and the minimum thickness 2.55 mm, one must verify One effectively has N ¥ pitch ∆etitanium ---------------------- £ sbearing etitanium ≥ 2.55 mm N ¥ pitch = N¢ ¥ ( ) pitch – ∆ N¢ N pitch pitch – ∆ = = ----------------------- 75.4 daN/mm N¢( ) daN/mm e( ) mm --------------------------------- £ srupture( ) daN/mm 75.4 2.55 ---------- £ 90 TX846_Frame_C18a Page 370 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
Verification of the edge distance (see previous figure):One has to respect the following condition: △F 2 x edge distance×e -≤Ttitanium rupture then: edge distance≥7.8mm from which the configuration (partial)of the joint can be shown as in the following figure: 20 30 6.35mm 264mm 27 mm 18.1.7 Carbon Fiber Coated with Nickel Problem Statement: With the objective of enhancing the electrical and thermal conductivity of a laminated panel in carbon/epoxy,one uses a thin coat of nickel with a thickness e for the external coating of the carbon fibers by electrolytic plating process (see following figure). carbon HM>e=0.12d nickel 1.Calculate the longitudinal modulus of elasticity of a coated fiber. 2.Calculate the linear coefficient of thermal expansion in the direction of the coated fiber. Solution: 1.Hooke's law applied to a fiber with length subject to a load F(following figure)can be written as: F=EIS 2003 by CRC Press LLC
Verification of the edge distance (see previous figure): One has to respect the following condition: then: edge distance ≥ 7.8 mm from which the configuration (partial) of the joint can be shown as in the following figure: 18.1.7 Carbon Fiber Coated with Nickel Problem Statement: With the objective of enhancing the electrical and thermal conductivity of a laminated panel in carbon/epoxy, one uses a thin coat of nickel with a thickness e for the external coating of the carbon fibers by electrolytic plating process (see following figure). 1. Calculate the longitudinal modulus of elasticity of a coated fiber. 2. Calculate the linear coefficient of thermal expansion in the direction of the coated fiber. Solution: 1. Hooke’s law applied to a fiber with length subject to a load F (following figure) can be written as: DF 2 e ¥ dge distance ¥ e ----------------------------------------------------- £ ttitanium rupture F Ef s D = ------ TX846_Frame_C18a Page 371 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC