Fiber length,/(in.) T 5 e=10.2mm(0.4in.) 。=12.7mm(0.5in.) e=15.2mm(0.6in.) 4 Theoretical points taking l=12.7 mm Experimental points from three-point bending experiments 0 0 0 25 50 75 100 125 Fiber length,/(mm) FIGURE 3.7 Variation in the longitudinal strength of a unidirectional discontinuous fiber composite as a function of fiber length.(After Hancock,P.and Cuthbertson,R.C., J.Ma1er.Sci,5,762,1970.) for I>5lc,strengthening greater than 90%can be achieved even with discon- tinuous fibers.An example is shown in Figure 3.7. For l<l,there will be no fiber failure.Instead,the lamina fails primarily because of matrix tensile failure.Since the average tensile stress in the fiber is -the longitudinal tensile strength of the composite is given by Cuw =Ti dg Vf +omu(1 Vf), (3.18) where omu is the tensile strength of the matrix material. A simple method of determining the fiber-matrix interfacial shear strength is called a single fiber fragmentation test,which is based on the observation that fibers do not break if their length is less than the critical value.In this test,a single fiber is embedded along the centerline of a matrix tensile specimen (Figure 3.8).When the specimen is tested in axial tension,the tensile stress is transferred to the fiber by shear stress at the fiber-matrix interface.The embedded fiber breaks when the maximum tensile stress in the fiber reaches its tensile strength.With increased loading,the fiber breaks into successively shorter lengths until the fragmented lengths become so short that the maximum 2007 by Taylor Francis Group.LLC
for lf > 5lc, strengthening greater than 90% can be achieved even with discontinuous fibers. An example is shown in Figure 3.7. For lf < lc, there will be no fiber failure. Instead, the lamina fails primarily because of matrix tensile failure. Since the average tensile stress in the fiber is sf ¼ ti lf df , the longitudinal tensile strength of the composite is given by sLtu ¼ ti lf df vf þ smu(1 vf), (3:18) where smu is the tensile strength of the matrix material. A simple method of determining the fiber–matrix interfacial shear strength is called a single fiber fragmentation test, which is based on the observation that fibers do not break if their length is less than the critical value. In this test, a single fiber is embedded along the centerline of a matrix tensile specimen (Figure 3.8). When the specimen is tested in axial tension, the tensile stress is transferred to the fiber by shear stress at the fiber–matrix interface. The embedded fiber breaks when the maximum tensile stress in the fiber reaches its tensile strength. With increased loading, the fiber breaks into successively shorter lengths until the fragmented lengths become so short that the maximum Fiber length, lf (in.) lc=10.2 mm (0.4 in.) lc=12.7 mm (0.5 in.) lc=15.2 mm (0.6 in.) Theoretical points taking lc=12.7 mm Experimental points from three-point bending experiments Fiber length, lf (mm) Composite strength (104 psi) Composite strength (100 MPa) 0 1 2 3 4 5 8 7 6 5 4 3 2 1 0 0 25 50 75 100 125 FIGURE 3.7 Variation in the longitudinal strength of a unidirectional discontinuous fiber composite as a function of fiber length. (After Hancock, P. and Cuthbertson, R.C., J. Mater. Sci., 5, 762, 1970.) 2007 by Taylor & Francis Group, LLC.
Fiber embedded in matrix (a (b) Fragmented fiber FIGURE 3.8 Single fiber fragmentation test to determine fiber-matrix interfacial shear strength. tensile stress can no longer reach the fiber tensile strength.The fragmented fiber lengths at this point are theoretically equal to the critical fiber length,/How- ever,actual (measured)fragment lengths vary between //2 and /Assuming a uniform distribution for the fragment lengths and a mean value of equal to 0.75/,Equation 3.15 can be used to calculate the interfacial shear strengthm[3]: 3dgofu Tim 8i, (3.19) where is the mean fragment length. Equation 3.13 was obtained assuming that the interfacial shear stress r;is a constant.The analysis that followed Equation 3.13 was used to demonstrate the importance of critical fiber length in discontinuous fiber composites.How- ever,strictly speaking,this analysis is valid only if it can be shown that Ti is a constant.This will be true in the case of a ductile matrix that yields due to high shear stress in the interfacial zone before the fiber-matrix bond fails and then flows plastically with little or no strain hardening (i.e.,the matrix behaves as a perfectly plastic material with a constant yield strength as shown in Figure 3.9). When this occurs,the interfacial shear stress is equal to the shear yield strength of the matrix (which is approximately equal to half of its tensile yield strength) and remains constant at this value.If the fiber-matrix bond fails before matrix yielding,a frictional force may be generated at the interface,which transfers the load from the matrix to the fibers through slippage(sliding).In a polymer matrix composite,the source of this frictional force is the radial pressure on the fiber surface created by the shrinkage of the matrix as it cools down from the 2007 by Taylor Francis Group,LLC
tensile stress can no longer reach the fiber tensile strength. The fragmented fiber lengths at this point are theoretically equal to the critical fiber length, lc. However, actual (measured) fragment lengths vary between lc=2 and lc. Assuming a uniform distribution for the fragment lengths and a mean value of l equal to 0.75lc, Equation 3.15 can be used to calculate the interfacial shear strength tim [3]: tim ¼ 3dfsfu 8l , (3:19) where l is the mean fragment length. Equation 3.13 was obtained assuming that the interfacial shear stress ti is a constant. The analysis that followed Equation 3.13 was used to demonstrate the importance of critical fiber length in discontinuous fiber composites. However, strictly speaking, this analysis is valid only if it can be shown that ti is a constant. This will be true in the case of a ductile matrix that yields due to high shear stress in the interfacial zone before the fiber–matrix bond fails and then flows plastically with little or no strain hardening (i.e., the matrix behaves as a perfectly plastic material with a constant yield strength as shown in Figure 3.9). When this occurs, the interfacialshearstressis equal to the shear yield strength of the matrix (which is approximately equal to half of its tensile yield strength) and remains constant at this value. If the fiber–matrix bond fails before matrix yielding, a frictional force may be generated at the interface, which transfers the load from the matrix to the fibers through slippage (sliding). In a polymer matrix composite, the source of this frictional force is the radial pressure on the fiber surface created by the shrinkage of the matrix as it cools down from the Fiber embedded in matrix Fragmented fiber (a) (b) FIGURE 3.8 Single fiber fragmentation test to determine fiber–matrix interfacial shear strength. 2007 by Taylor & Francis Group, LLC
Stress (b)Elastic-strain hardening (a)Elastic-perfectly plastic 人于Elastic modulus Strain FIGURE 3.9 Stress-strain diagrams of(a)an elastic-perfectly plastic material and(b)an elastic-strain hardening material. curing temperature.In this case,the interfacial shear stress is equal to the product of the coefficient of sliding friction and the radial pressure. When the matrix is in the elastic state and the fiber-matrix bond is still unbroken,the interfacial shear stress is not a constant and varies with x. Assuming that the matrix has the same strain as the composite,Cox [4]used a simple shear lag analysis to derive the following expression for the fiber stress distribution along the length of a discontinuous fiber: coshB =E a0≤xs5 (3.20) where or=longitudinal fiber stress at a distance x from its end Er=fiber modulus s1=longitudinal strain in the composite B-1 2Gm Er In(R/r)' where Gm =matrix shear modulus r =fiber radius 2R=center-to-center distance from a fiber to its nearest neighbor 2007 by Taylor&Francis Group.LLC
curing temperature. In this case, the interfacial shear stress is equal to the product of the coefficient of sliding friction and the radial pressure. When the matrix is in the elastic state and the fiber–matrix bond is still unbroken, the interfacial shear stress is not a constant and varies with x. Assuming that the matrix has the same strain as the composite, Cox [4] used a simple shear lag analysis to derive the following expression for the fiber stress distribution along the length of a discontinuous fiber: sf ¼ Ef«1 1 cosh b lf 2 x cosh blf 2 2 6 6 4 3 7 7 5 for 0 x lf 2 , (3:20) where sf ¼ longitudinal fiber stress at a distance x from its end Ef ¼ fiber modulus «1 ¼ longitudinal strain in the composite b ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2Gm Efr2 f ln (R=rf) s , where Gm ¼ matrix shear modulus rf ¼ fiber radius 2R ¼ center-to-center distance from a fiber to its nearest neighbor (b) Elastic–strain hardening (a) Elastic–perfectly plastic Elastic modulus Strain Stress smy FIGURE 3.9 Stress–strain diagrams of (a) an elastic-perfectly plastic material and (b) an elastic-strain hardening material. 2007 by Taylor & Francis Group, LLC.
Increasing Bl Tensile stress in the fiber (a) X Fiber length=/ Increasing Interfacial 队 shear stress Increasing Bl (b) FIGURE 3.10(a)Normal stress distribution along the length of a discontinuous fiber according to Equation 3.20 and (b)shear stress distribution at the fiber-matrix interface according to Equation 3.21. Using Equations 3.11 and 3.20,shear stress at the fiber-matrix interface is obtained as: sinh B T=Ee Br (3.21) cosh 2 Equations 3.20 and 3.21 are plotted in Figure 3.10 for various values of Blr.It shows that the fiber stress builds up over a shorter load transfer length if Bl is high.This means that not only a high fiber length to diameter ratio (called the fiber aspect ratio)but also a high ratio of Gm/Er is desirable for strengthening a discontinuous fiber composite. Note that the stress distribution in Figure 3.5 or 3.10 does not take into account the interaction between fibers.Whenever a discontinuity due to fiber end occurs,a stress concentration must arise since the tensile stress normally assumed by the fiber without the discontinuity must be taken up by the surrounding fibers.As a result,the longitudinal stress distribution for each fiber may contain a number of peaks. 2007 by Taylor Francis Group,LLC
Using Equations 3.11 and 3.20, shear stress at the fiber–matrix interface is obtained as: t ¼ 1 2 Ef«1brf sinh b lf 2 x cosh blf 2 : (3:21) Equations 3.20 and 3.21 are plotted in Figure 3.10 for various values of blf. It shows that the fiber stress builds up over a shorter load transfer length if blf is high. This means that not only a high fiber length to diameter ratio (called the fiber aspect ratio) but also a high ratio of Gm=Ef is desirable for strengthening a discontinuous fiber composite. Note that the stress distribution in Figure 3.5 or 3.10 does not take into account the interaction between fibers. Whenever a discontinuity due to fiber end occurs, a stress concentration must arise since the tensile stress normally assumed by the fiber without the discontinuity must be taken up by the surrounding fibers. As a result, the longitudinal stress distribution for each fiber may contain a number of peaks. Increasing blf Tensile stress in the fiber Interfacial shear stress + − x Fiber length=lf Increasing blf Increasing blf (a) (b) FIGURE 3.10 (a) Normal stress distribution along the length of a discontinuous fiber according to Equation 3.20 and (b) shear stress distribution at the fiber–matrix interface according to Equation 3.21. 2007 by Taylor & Francis Group, LLC.
EXAMPLE 3.1 A unidirectional fiber composite contains 60 vol%of HMS-4 carbon fibers in an epoxy matrix.Using the fiber properties in Table 2.1 and matrix properties as Em=3.45 GPa and omy =138 MPa,determine the longitudinal tensile strength of the composite for the following cases: 1.The fibers are all continuous. 2.The fibers are 3.17 mm long and n is (i)4.11 MPa or (ii)41.1 MPa. SOLUTION Since HMS-4 carbon fibers are linearly elastic,their failure strain is 2480 MPa 8u= =0.0072. E345×103MPa Assuming that the matrix behaves in an elastic-perfectly plastic manner,its yield strain can be calculated as y== 138 MPa E3.45×10MPa=0.04. Thus,the fibers are expected to break before the matrix yields and the stress in the matrix at the instance of fiber failure is dm=Em8m=(3.45×103MPa)(0.0072)=24.84MPa. 1.Using Equation 3.9,we get u=(2480)(0.6)+(24.84)(1-0.6) =1488+9.94=1497.94MPa. 2.(i)When T;=4.11 MPa,the critical fiber length is 2480MPa（8×10-3mm)=2.414mm. (2)(4.11MPa) Since>l,we can use Equation 3.17 to calculate m=(2480)1- 2.414 (0.6)+(24.84)1-0.6) (2)(3.17) =921.43+9.94=931.37MPa. (ii)When Ti=41.1 MPa,le=0.2414 mm.Thus,lf >le. 2007 by Taylor Francis Group.LLC
EXAMPLE 3.1 A unidirectional fiber composite contains 60 vol% of HMS-4 carbon fibers in an epoxy matrix. Using the fiber properties in Table 2.1 and matrix properties as Em ¼ 3.45 GPa and smy ¼ 138 MPa, determine the longitudinal tensile strength of the composite for the following cases: 1. The fibers are all continuous. 2. The fibers are 3.17 mm long and ti is (i) 4.11 MPa or (ii) 41.1 MPa. SOLUTION Since HMS-4 carbon fibers are linearly elastic, their failure strain is «fu ¼ sfu Ef ¼ 2480 MPa 345 103 MPa ¼ 0:0072: Assuming that the matrix behaves in an elastic-perfectly plastic manner, its yield strain can be calculated as «my ¼ smy Em ¼ 138 MPa 3:45 103 MPa ¼ 0:04: Thus, the fibers are expected to break before the matrix yields and the stress in the matrix at the instance of fiber failure is s0 m ¼ Em«fu ¼ (3:45 103 MPa) (0:0072) ¼ 24:84 MPa: 1. Using Equation 3.9, we get sLtu ¼ (2480)(0:6) þ (24:84)(1 0:6) ¼ 1488 þ 9:94 ¼ 1497:94 MPa: 2. (i) When ti ¼ 4.11 MPa, the critical fiber length is lc ¼ 2480 MPa (2)(4:11 MPa) (8 103 mm) ¼ 2:414 mm: Since lf > lc, we can use Equation 3.17 to calculate sLtu ¼ (2480) 1 2:414 (2)(3:17) (0:6) þ (24:84)(1 0:6) ¼ 921:43 þ 9:94 ¼ 931:37 MPa: (ii) When ti ¼ 41.1 MPa, lc ¼ 0.2414 mm. Thus, lf > lc. 2007 by Taylor & Francis Group, LLC.