1) It is necessary to select a ceramic material to be stressed using a three-point loading scheme. The specimen must have a circular cross section, a radius of 3.8 mm (0.15 in.), and must not experience fracture or a deflection of more than 0.021 mm (0.00085 in.) at its center when a load of 445 N (100 lbf) is applied. If the distance between support points is 50.8 mm (2 in.),
which of the materials in Table 12.5 are candidates? The magnitude of the center-point deflection may be computed using the equation
supplied. Choose a ceramic material from the table with proper justification.
Tabulation of Flexural Strength (Modulus of Rupture) and
Modulus of Elasticity for Ten Common Ceramic Materials
Modulus of
Flexural Strength Elasticity
Material MPa ksi GPa 106 psi
Silicon nitride (Si3N4) 250–1000 35–145 304 44
Zirconiaa (ZrO2) 800–1500 115–215 205 30
Silicon carbide (SiC) 100–820 15–120 345 50
Aluminum oxide (Al2O3) 275–700 40–100 393 57
Glass-ceramic (Pyroceram) 247 36 120 17
Mullite (3Al2O3-2SiO2) 185 27 145 21
Spinel (MgAl2O4) 110–245 16–35.5 260 38
Magnesium oxide (MgO) 105b 15b 225 33
Fused silica (SiO2) 110 16 73 11
Soda-lime glass 69 10 69 10
2) A circular specimen of MgO is loaded using a three-point bending mode. Compute the minimum possible radius of the specimen without fracture, given that the applied load is 5560 N (1250 the flexural strength is 105 MPa (15,000 psi), and the separation between load points is 45 mm (1.75 in.).
3) A three-point bending test is performed on a spinel specimen having a rectangular cross section of height d 3.8 mm (0.15 in.) and width b 9 mm (0.35 in.); the distance between support points is 25 mm (1.0 in.).
(a) Compute the flexural strength if the load at fracture is 350 N (80 lbf).
(b) The point of maximum deflection occurs at the center of the specimen and is described by where E is the modulus of elasticity and I is the cross-sectional moment of inertia. Compute deflection at center at a load of 310 N (70 lbf)
4) A continuous and aligned fiber-reinforced composite is to be produced consisting of 45 vol% aramid fibers and 55 vol% of a polycarbonate matrix; mechanical characteristics of these two materials are as follows:
Modulus Tensile
of Elasticity Strength
[GPa (psi)] [MPa (psi)]
Aramid fiber 131 (19 106) 3600 (520,000)
Polycarbonate 2.4 (3.5 105) 65 (9425)
Also, the stress on the polycarbonate matrix when the aramid fibers fail is 35 MPa (5075 psi).
(i) For this composite, compute
(a) the longitudinal tensile strength, and
(b) the longitudinal modulus of elasticity
(ii) Assume that the composite described in
Problem 16.8 has a cross-sectional area of
480 mm2 (0.75 in.2) and is subjected to a longitudinal
load of 53,400 N (12,000 lbf).
(a) Calculate the fiber–matrix load ratio.
(b) Calculate the actual loads carried by
both fiber and matrix phases.
(c) Compute the magnitude of the stress on
each of the fiber and matrix phases.
(d) What strain is experienced by the composite?
5) Calculate the number-average molecular weight of a random poly(isobutylene-isoprene) copolymer in which the fraction of isobutylene repeat units is 0.25; assume that this concentration corresponds to a degree of polymerization of 1500.
6) A random poly(styrene-butadiene) copolymer has a number-average molecular weight of 420,000 g/mol and a degree of polymerization of 6000. Compute the fraction of styrene and butadiene repeat units in this copolymer.
7) (a) Determine the ratio of butadiene to acrylonitrile repeat units in a copolymer having a number-average molecular weight of 250,000 g/mol and a degree of polymerization of 4640.
(b) Which type(s) of copolymer(s) will this copolymer be, considering the following possibilities: random, alternating, graft, and block? Why?
8) A continuous and aligned glass fiber-reinforced composite consists of 40 vol% of glass fibers having a modulus of elasticity of 69 GPa ( psi) and 60 vol% of a polyester resin that, when hardened, displays a modulus of 3.4 GPa ( psi).
(a) Compute the modulus of elasticity of this composite in the longitudinal direction.
(b) If the cross-sectional area is 250 mm2 (0.4 in.2) and a stress of 50 MPa (7250 psi) is applied in this longitudinal direction, compute the magnitude of the load carried by each of the fiber and matrix phases.
(c) Determine the strain that is sustained by each phase when the stress in part (b) is applied.
The modulus of elasticity of the composite is calculated using Equation 16.10a: 30 GPa 14.3 106 psi2
To solve this portion of the problem, first find the ratio of fiber load to matrix load, using Equation 16.11; thus, or
In addition, the total force sustained by the composite may be computed
from the applied stress and total composite cross-sectional area according to
However, this total load is just the sum of the loads carried by fiber and matrix phases; that is, Substitution for from the above yields or whereas Thus, the fiber phase supports the vast majority of the applied load.
(c) The stress for both fiber and matrix phases must first be calculated.Then, by using the elastic modulus for each (from part a), the strain values may be determined. For stress calculations, phase cross-sectional areas are necessary: and Thus,
Finally, strains are computed as Therefore, strains for both matrix and fiber phases are identical, which
they should be, according to Equation 16.8 in the previous development. f sf Ef 116.4 MPa 69 103 MPa
1.69 10 3 m sm Em 5.73 MPa 3.4 103 MPa 1.69 10 3 sf Ff Af 11,640 N 100 mm2 116.4 MPa 116,875 psi2 sm Fm Am
860 N 150 mm2 5.73 MPa 1833 psi2
Af VfAc 10.421250 mm22 100 mm2 10.16 in.22 Am VmAc 10.621250 mm22 150 mm2 10.24 in.22
Ff Fc Fm 12,500 N 860 N 11,640 N 12700 lbf2 Fm 860 N 1200 lbf2 13.5 Fm Fm 12,500 N Ff Fc Ff Fm 12,500
