Winter 2023 MECH 321-Properties and Failure of Materials Problem # 1: In an aligned and continuous glass fiber-reinforced nylon 6,6 composite, the fibers are to carry 94% of a load applied in the longitudinal direction. a) Using the data provided, determine the volume fraction of fibers that will be required. b) What will be the tensile strength of this composite? Assume that the matrix stress at fiber failure is 30 MPa (4350 psi). c) Explain the role of fibre and matrix in a composite material. Modulus of Elasticity [GPa (psi)] Glass fiber 72.5 (10.5 × 10°) 3.0 (4.35 × 103) 3400 (490,000) 76 (11,000) Nylon 6,6 Tensile Strength [MPa (psi)] Solution: a) Given some data for an aligned and continuous glass-fiber-reinforced nylon 6,6 composite, we are asked to compute the volume fraction of fibers that are required such that the fibers carry 94% of a load applied in the longitudinal direction. From Equation 16.11 Now, using values for Ffand Fm from the problem statement 0.06 0.94 = = 15.67 (3.0 GPa)(1 -Vf) (72.5 GPa)V f And, solving for Vf yields, Vf= 0.393 b) We are now asked for the tensile strength of this composite. From Equation 16.17, "¿ = 0m(1-Vj) + o;vý = (30 MPa)(1 - 0.393) + (3400 MPa)(0.393) = 1354 MPa (196,400 psi) Since values for of* (3400 MPa) and om' (30 MPa) are given in the problem statement. c) The primary function of the fibers is to carry the loads along their longitudinal directions. I.e., the fibers are the load carrying members in the composite material. The primary functions of the matrix are to transfer stresses between the reinforcing fibers (hold fibers together) and protect the fibers from mechanical and/or environmental damages. The fiber provides the reinforcement for the composite material whereas the matrix holds the fiber together thus protecting the alignment of the fiber in the composite. 1
Winter 2023 MECH 321-Properties and Failure of Materials Problem # 2: Compute the longitudinal tensile strength of an aligned glass fiber-epoxy matrix composite in which the average fiber diameter and length are 0.010 mm (4 x 10-4 in.) and 2.5 mm (0.10 in.), respectively, and the volume fraction of fibers is 0.40. Assume that (1) the fiber-matrix bond strength is 75 MPa (10,900 psi), (2) the fracture strength of the fibers is 3500 MPa (508,000 psi), and (3) the matrix stress at fiber failure is 8.0 MPa (1160 psi). Solution: It is first necessary to compute the value of the critical fiber length using Equation 16.3. If the fiber length is much greater than lc, then we may determine G* using Equation 16.17, otherwise, use of either Equations 16.18 or 16.19 is necessary. Thus, 20 (3500 MPa)(0.010 mm) = 0.233 mm (0.0093 in.) 2(75 MPa) Inasmuch as l > lc (2.5 mm > 0.233 mm), but since l is not much greater than l c , then use of Equation 16.18 is necessary. Therefore, od-[1- |- "(1 - V) = (3500 MPa)(0.40) 1 - (2)(2.5 mm) 0.233 mm