《纺织复合材料》课程参考文献（Composite materials design and applications）18 APPLICATIONS

M h(shear forces Qx and Qy taken up by nida) a b Figure 18.4 -22 daN/mm -28 daN/mm -40 daN/mm -10 daN/mm -45 daN/mm 0 daN/mm -8 daN/mm -13 daN/mm -17 daN/mm] -5 daN/mm -14 daN/mm -3 daN/mm -5 daN/mm -9.4 daN/mm -3.4 daN/mm -10 daN/mm -12 daN/mm -3 daN/mm Figure 18.5 3.The skin is bonded on the edge of the titanium (Figure 18.2).Provide the dimensions of the bonded surface by using an average shear stress in the adhesive (araldite:t rupture=30 MPa). 4.The border of the titanium is bolted to the rest of the wing (Figure 18.2) Determine the dimensional characteristics of the joint:"pitch"of the bolts, thickness,foot,with the following data: Bolts:30 NCD 16 steel:=6.35 mm,adjusted,negligible tensile loading, Grupture =1,100 MPa;Trupture =660 MPa;Ciearing =1600 MPa TA6V titanium alloy:Orupture=900 MPa;Tupure=450 MPa;bearing=1100 MPa Duralumin:Gnupture =420 MPa;Cpearing =550 MPa Solution: 1.The moment resultants M,M Msy(and M,not shown in Figure 18.4a) are taken up by the laminated skins.One then has in the upper skin (Figure 18.4b),b being the mean distance separating the two skins: N=告N=兴=% 2003 by CRC Press LLC

3. The skin is bonded on the edge of the titanium (Figure 18.2). Provide the dimensions of the bonded surface by using an average shear stress in the adhesive (araldite: t rupture = 30 MPa). 4. The border of the titanium is bolted to the rest of the wing (Figure 18.2). Determine the dimensional characteristics of the joint: “pitch” of the bolts, thickness, foot, with the following data:
Bolts: 30 NCD 16 steel: ∆ = 6.35 mm, adjusted, negligible tensile loading, srupture = 1,100 MPa; trupture = 660 MPa; sbearing = 1600 MPa
TA6V titanium alloy: srupture = 900 MPa; trupture = 450 MPa; sbearing = 1100 MPa
Duralumin: srupture = 420 MPa; sbearing = 550 MPa Solution: 1. The moment resultants Mx , My, Mxy (and Myx, not shown in Figure 18.4a) are taken up by the laminated skins. One then has in the upper skin (Figure 18.4b), h being the mean distance separating the two skins: Figure 18.4 Figure 18.5 Nx My h ------; Ny –Mx h ---------; Txy Mxy h == = –-------- TX846_Frame_C18a Page 362 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC

Remark:The moment resultants,that means the moments per unit width of the skin-1mm in practice-have units of daN x mm/mm.The stress resultants NN Ty have units of daN/mm. 2.Looking at the most loaded region of the skin,we can represent the principal directions and stresses by constructing Mohr's circle (shown in the following figure).Then we note that there must be a nonnegligible proportion of the fibers at +45.However,the laminate has to be able to resist compressions along the axes x and y.The estimation of the propor- tions can be done following the method presented in Section 5.4.3.One then obtains the following composition°： 307% 20% 30%+X0u0 20% -45 daN/mm (N) -17 daN/mm -40 daN/mm (N -17 daN/mm T -59.7 -45 -40 -25.3 daN/mm -25.3 daN/mm -59.7 daN/mm 4 Let o,o,te be the stresses along the principal axes l,t of one of the plies for the state of loading above,the thickness e of the laminate (unknown a priori) such that one finds the limit of the Hill-Tsai criterion of failure.One then has 6+ 601+ -1 rupture Tet rupture The calculation to estimate these proportions is shown in detail in the example of Section 5.4.3,where one has used the same values of the resultants with a factor of safety of 2,as: N.=-800N/mm:N--900N/mm;Tg=-340N/mm. See Section 5.3.2 and also Chapter 14. 2003 by CRC Press LLC

Remark: The moment resultants, that means the moments per unit width of the skin – 1mm in practice – have units of daN ¥ mm/mm. The stress resultants Nx, Ny, Txy have units of daN/mm. 2. Looking at the most loaded region of the skin, we can represent the principal directions and stresses by constructing Mohr’s circle (shown in the following figure). Then we note that there must be a nonnegligible proportion of the fibers at ±45∞. However, the laminate has to be able to resist compressions along the axes x and y. The estimation of the propor￾tions can be done following the method presented in Section 5.4.3. One then obtains the following composition8 : Let s
, st , t
t be the stresses along the principal axes l, t of one of the plies for the state of loading above, the thickness e of the laminate (unknown a priori) such that one finds the limit of the Hill-Tsai criterion of failure.9 One then has 8 The calculation to estimate these proportions is shown in detail in the example of Section 5.4.3, where one has used the same values of the resultants with a factor of safety of 2, as: Nx = -800 N/mm; Ny = -900 N/mm; Txy = -340 N/mm. 9 See Section 5.3.2 and also Chapter 14. s
2 s
rupture 2 ------------------ st 2 st rupture 2 ----------------- s
st s
rupture 2 ------------------ t
t 2 t
t rupture 2 + – + ------------------ = 1 TX846_Frame_C18a Page 363 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC

If one multiplies the two sides by the square of the thickness e: (oe)2 (oe)(oe)(o;e),(te) =e [1] Ot rupture Or rupture O rupture Ter rupture one will obtain the values (oe),(o,e),(te e),by multiplying the global stresses Ty with the thickness e,as (oxe),(c e),(Te),which are just the stress resultants defined previously: Nx=(xe);N=(oe);Tty=(txye) Units:the rupture resistances are given in MPa(or N/mm)in Appendix 1.As a consequence: Nx=-400MPa×mm N,=-450MPa×mm Txy=-170MPa×mm with a factor of safety of 2,one then has Ng=-800MPa×mm Ng=-900MPa×mm Tg=-340MPa×mm We use the Plates in annex 1 which show the stresses o,o,ter in each ply for an applied stress resultant of unit value (1 MPa,for example): (a)Plies at0°： ■Loading N=-800MPa×mm only: For the proportions defined in the previous question,one reads on Plate 1: 0=2.4 (oee)=2.4×-800=-1920MPa×mm 0=0.0 (oe)=0 Te,=0 (t,e)=0 ■Loading N'=-900MPa×mm only: One reads from Plate 5: 0=-0.54 (oee)=-0.54×-900=486MPa×mm 0,=0.12 (o,e)=0.12×-900=-108MPa×mm Tir =0 (Tere)=0 2003 by CRC Press LLC

If one multiplies the two sides by the square of the thickness e: [1] one will obtain the values (s
e), (st e), (t
t e), by multiplying the global stresses sx, sy, txy with the thickness e, as (sx e), (sy e), (txy e), which are just the stress resultants defined previously: Units: the rupture resistances are given in MPa (or N/mm2 ) in Appendix 1. As a consequence: with a factor of safety of 2, one then has We use the Plates in annex 1 which show the stresses s
, st , t
t in each ply for an applied stress resultant of unit value (1 MPa, for example): (a) Plies at 0∞:
Loading = -800 MPa ¥ mm only: For the proportions defined in the previous question, one reads on Plate 1:
Loading = -900 MPa ¥ mm only: One reads from Plate 5: ( ) s
e 2 s
rupture 2 ------------------ ( ) ste 2 st rupture 2 ----------------- ( ) s
e ( ) ste s
rupture 2 -------------------------- ( ) t
te 2 t
t rupture 2 + – + ------------------ e 2 = Nx == = ( ) sxe ; Ny ( ) sye ; Txy ( ) txye Nx = –400 MPa mm ¥ Ny = –450 MPa mm ¥ Txy = –170 MPa mm ¥ Nx ¢ = –800 MPa mm ¥ Ny ¢ = –900 MPa mm ¥ Txy¢ = –340 MPa mm ¥ Nx ¢ s
= 2.4 st = 0.0 t
t = 0 ˛ Ô ˝ Ô ¸ ( ) s
e = 2.4 800 ¥ – = –1920 MPa mm ¥ ( ) ste = 0 Ó ( ) t
t e = 0 Ô Ì Ô Ï Æ Ny ¢ s
= –0.54 st = 0.12 t
t = 0 ˛ Ô ˝ Ô ¸ ( ) s
e = –0.54 ¥ –900 = 486 MPa mm ¥ ( ) ste = 0.12 900 ¥ – = –108 MPa mm ¥ Ó ( ) t
t e = 0 Ô Ì Ô Ï Æ TX846_Frame_C18a Page 364 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC

■Loading Ty1=-34oMPa×mm only: One reads from Plate 9: 0e=0 (ee)=0 0,=0 (o,e)=0 tm=0.26 L(te,e)=0.26×-340=-89MPa×mm The superposition of the three loadings will then give the plies at 0a total state of stress of (oee)=-1920+486=-1434MPa×mm (o,e)=-108MPa×mm (te)=-89MPa×mm Then we can write the Hill-Tsai criterion in the modified form written above (relation denoted as [1))in which one notes the denominator with values of the rupture strengths indicated at the beginning of annex 1: e2= 14342,10821434×108,892 11302 1412 11302 63407 e =2.02mm (0°) One resumes the previous calculation as follows: plies at 0o (ore) (o,e） (Tere) N -1920 0 0 N 486 -108 0 T 0 0 -89 e=2.02mm total -1434 -108 -89 (MPa×mm) (b) Plies at 90:One repeats the same calculation procedure by using the Plates 2,6,and 10.This leads to the following analogous table,with a thickness e calculated as previously (this is the minimum thickness of the laminate below which there will be rupture of the 90 plies). plies at90° (Gre) (oe) (tre) N 432 -96 0 N -2160 0 0 T 0 0 89 e=2.16mm total -1728 -96 89 (MPa x mm) 2003 by CRC Press LLC

Loading l = -340 MPa ¥ mm only: One reads from Plate 9: The superposition of the three loadings will then give the plies at 0∞a total state of stress of Then we can write the Hill-Tsai criterion in the modified form written above (relation denoted as [1]) in which one notes the denominator with values of the rupture strengths indicated at the beginning of annex 1: One resumes the previous calculation as follows: (b) Plies at 90∞: One repeats the same calculation procedure by using the Plates 2, 6, and 10. This leads to the following analogous table, with a thickness e calculated as previously (this is the minimum thickness of the laminate below which there will be rupture of the 90∞ plies). plies at 0∞ (s
e) (ste) (t
te) e = 2.02 mm -1920 0 0 486 -108 0 0 0 -89 total (MPa ¥ mm) -1434 -108 -89 plies at 90∞ (s
e) (ste) (t
te) e = 2.16 mm 432 -96 0 -2160 0 0 0 0 89 total (MPa ¥ mm) -1728 -96 89 Txy ¢ s
= 0 st = 0 t
t = 0.26 ˛ Ô ˝ Ô ¸ ( ) s
e = 0 ( ) ste = 0 Ó ( ) t
t e = 0.26 ¥ –340 = –89 MPa mm ¥ Ô Ì Ô Ï Æ ( ) s
e = = –1920 486 + –1434 MPa mm ¥ ( ) ste = –108 MPa mm ¥ ( ) t
te = –89 MPa mm ¥ e 2 14342 11302 -------------- 1082 1412 ---------- 1434 ¥ 108 11302 --------------------------- 892 632 = = + – + ------- 4.07 e ( ) 0∞ = 2.02 mm Nx ¢ Ny ¢ Txy¢ Nx ¢ Ny ¢ Txy ¢ TX846_Frame_C18a Page 365 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC

(c)Plies at +45:Using Plates 3,7,and 11 one obtains: plies at45° (ore) (o,e) (tue) N -752 -48 72 N -846 -54 -81 T -1384 55 0 e=2.64mm total -2982 -47 -9 (MPa×mm) (d)Plies at -45:Using Plates 4,8,12 one obtains: plies at-45° (cre) (o,e) (te) N -752 -48 -72 N -846 -54 81 T 1384 -55 0 e=1.13mm total -214 -157 9 (MPa×mm) Then the theoretical thickness to keep here is the largest out of the four thicknesses found,as: e 2.64 mm (rupture of the plies at +45). The thickness of each ply is 0.13 mm.It takes 2.64/0.13 =20 plies minimum from which is obtained the following composition allowing for midplane symmetry: 630y%】 4(20%) 20 6(30% 4(20%) Remark:Optimal composition of the laminate:For the complex loading considered here,one can directly obtain the composition leading to the minimum thickness by using the tables in Section 5.4.4.One then uses the reduced stress resultants,deduced from the resultants taken into account above,to obtain N.=-800/(800l+1900l+l340)=-39% N=-900/(1800l+1900l+1340l)=-44% 7gy=-17% 2003 by CRC Press LLC

(c) Plies at +45∞: Using Plates 3, 7, and 11 one obtains: (d) Plies at -45∞: Using Plates 4, 8, 12 one obtains: Then the theoretical thickness to keep here is the largest out of the four thicknesses found, as: e = 2.64 mm (rupture of the plies at +45∞). The thickness of each ply is 0.13 mm. It takes 2.64/0.13 = 20 plies minimum from which is obtained the following composition allowing for midplane symmetry: Remark: Optimal composition of the laminate: For the complex loading considered here, one can directly obtain the composition leading to the minimum thickness by using the tables in Section 5.4.4. One then uses the reduced stress resultants, deduced from the resultants taken into account above, to obtain plies at 45∞ (s
e) (ste) (t
te) e = 2.64 mm -752 –48 72 –846 -54 -81 -1384 55 0 total (MPa ¥ mm) –2982 –47 -9 plies at -45∞ (s
e) (ste) (t
te) e = 1.13 mm -752 –48 -72 –846 -54 81 1384 -55 0 total (MPa ¥ mm) –214 –157 9 Nx ¢ Ny ¢ Txy ¢ Nx ¢ Ny ¢ Txy ¢ Nx = = –800/( ) 800 900 340 + + –39% Ny = = –900/( ) 800 900 340 + + –44% Txy = –17% TX846_Frame_C18a Page 366 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC