Illustrative Problem 7.1 A laminar composite of polyester reinforced with continuous glass fibers is loaded under the isostrain situation.The modulus of elasticity for the glass fibers is 69GPa and for polyester it is 3.4GPa.The volume fraction of glass fibers is 40%. (a)Find the modulus for the composite. (b)When a stress of 50MPa is applied to the composite with a cross-sectional area of 250 mm2,find the stresses carried by the fibers and the matrix. (c)Find the strains of the fibers and the matrix. PDF文件使用"pdfFactory Pro”试用版本创建ww,fineprint..con.cn
A laminar composite of polyester reinforced with continuous glass fibers is loaded under the isostrain situation. The modulus of elasticity for the glass fibers is 69GPa and for polyester it is 3.4GPa. The volume fraction of glass fibers is 40%. (a) Find the modulus for the composite. (b) When a stress of 50MPa is applied to the composite with a cross-sectional area of 250 mm2 , find the stresses carried by the fibers and the matrix. (c) Find the strains of the fibers and the matrix. Illustrative Problem 7.1 PDF 文件使用 "pdfFactory Pro" 试用版本创建 ÿwww.fineprint.com.cn
Solution (a)Ec=Emvm+Ervr =3.4×0.6+69×0.4=30GPa (b)First,find the ratio F/F based on the formula: F E/i F F1- 69GPa×0.4 =13.5 Fm 3.4GPa×0.6 F=13.5Fm PDF文件使用"pdfFactory Pro'”试用版本创建wm,fineprint.com,cn
(a) Ec = Emvm + Efvf = 3.4 ×0.6 + 69 × 0.4 = 30 GPa Solution (b) First, find the ratio Ff /Fm based on the formula: m m f f m f E v E v F F = f m m f F F F F 13.5 13.5 3.4 GPa 0.6 69 GPa 0.4 = = ´ ´ = PDF 文件使用 "pdfFactory Pro" 试用版本创建 www.fineprint.com.cn
The load carried by the composite (F)is. F=Aco=250mm2×50MPa=12,500N The load (Fc)is carried by both the fibers and the matrix acting to share the burden: 13.5Fm+Fm=12,500N F=860N F/=F。-Fm=12,500N-860N=11,640N PDF文件使用"pdfFactory Pro”试用版本创建ww,fineprint..com,cn
The load carried by the composite (Fc ) is: Fc = Acs = 250 mm2 ´ 50 MPa = 12,500 N The load (Fc) is carried by both the fibers and the matrix acting to share the burden: 13.5 Fm + Fm = 12,500 N Fm = 860 N Ff = Fc - Fm = 12,500 N - 860 N = 11,640 N PDF 文件使用 "pdfFactory Pro" 试用版本创建 www.fineprint.com.cn
The cross-sectional areas of the fibers and the matrix are: Am=vmAc=(0.6)(250mm2)=150mm2 4y=vA.=(0.4)(250mm2)=100mm2 n- 860W Om=- =5.73MPa Am 50mm F 11,640N =116.4MPa A 100mm PDF文件使用"pdfFactory Pro'”试用版本创建wm,fineprint.com,cn
The cross-sectional areas of the fibers and the matrix are： Am = vmAc = (0.6)(250 mm2 ) = 150 mm2 Af = vfAc = (0.4)(250 mm2 ) = 100 mm2 MPa mm N A F MPa mm N A F f f f m m m 116.4 100 11,640 5.73 150 860 2 2 = = = = = = s s PDF 文件使用 "pdfFactory Pro" 试用版本创建 www.fineprint.com.cn
Solution (C) 5.73MPa ￡m= =1.69×10-3 Em 3.4×103MPa 116.4MPa 69×103MPa =1.69×10 PDF文件使用"pdfFactory Pro”试用版本创建w,fineprint..com.cn
3 3 3 3 1.69 10 69 10 116.4 1.69 10 3.4 10 5.73 - - = ´ ´ = = = ´ ´ = = MPa MPa E MPa MPa E f f f m m m s e s e Solution (C) PDF 文件使用 "pdfFactory Pro" 试用版本创建 ÿwww.fineprint.com.cn