Callister Composite Questions Solutions PDF

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CHAPTER 16

COMPOSITES

PROBLEM SOLUTIONS

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Influence of Fiber Orientation and Concentration 16.8 A continuous and aligned fiber-reinforced composite is to be produced consisting of 30 vol% aramid fibers and 70 vol% of a polycarbonate matrix; mechanical characteristics of these two materials are as follows: Modulus of Elasticity [GPa (psi)]

Tensile Strength [MPa (psi)]

131 (19 × 106)

3600 (520,000)

Aramid fiber

5

65 (9425)

2.4 (3.5 × 10 )

Polycarbonate

Also, the stress on the polycarbonate matrix when the aramid fibers fail is 45 MPa (6500 psi). For this composite, compute (a) the longitudinal tensile strength, and (b) the longitudinal modulus of elasticity Solution This problem calls for us to compute the longitudinal tensile strength and elastic modulus of an aramid fiber-reinforced polycarbonate composite. (a) The longitudinal tensile strength is determined using Equation 16.17 as ∗ = σ ' (1 − V σ cl m f

)+

σ ∗f V f

= (45 MPa)(0.70) + (3600)(0.30)

= 1100 MPa (160,000 psi)

(b) The longitudinal elastic modulus is computed using Equation 16.10a as

Ecl = EmVm + E f V f

= (2.4 GPa)(0.70) + (131 GPa)(0.30)

= 41 GPa (5.95 × 10 6 psi)

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16.9 Is it possible to produce a continuous and oriented aramid fiber-epoxy matrix composite having longitudinal and transverse moduli of elasticity of 57.1 GPa (8.28 × 106 psi) and 4.12 GPa (6 × 105 psi), respectively? Why or why not? Assume that the modulus of elasticity of the epoxy is 2.4 GPa (3.50× 105 psi). Solution This problem asks for us to determine if it is possible to produce a continuous and oriented aramid fiberepoxy matrix composite having longitudinal and transverse moduli of elasticity of 57.1 GPa and 4.12 GPa, respectively, given that the modulus of elasticity for the epoxy is 2.4 GPa. Also, from Table 16.4 the value of E for aramid fibers is 131 GPa. The approach to solving this problem is to calculate values of Vf for both longitudinal and transverse cases using the data and Equations 16.10b and 16.16; if the two Vf values are the same then this composite is possible. For the longitudinal modulus Ecl (using Equation 16.10b),

Ecl = Em (1 − V fl ) + E f V fl

57.1 GPa = (2.4 GPa)(1 − V fl ) + (131 GPa)V fl Solving this expression for Vfl (i.e., the volume fraction of fibers for the longitudinal case) yields Vfl = 0.425. Now, repeating this procedure for the transverse modulus Ect (using Equation 16.16)

Ect =

4.12 GPa =

EmE f

(1 − V ft ) E f

+ V ft Em

(2.4 GPa)(131 GPa) GPa) + V ft (2.4 GPa)

(1 − V ft ) (131

Solving this expression for Vft (i.e., the volume fraction of fibers for the transverse case), leads to Vft = 0.425. Thus, since Vfl and Vft are equal, the proposed composite is possible.

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16.10 For a continuous and oriented fiber-reinforced composite, the moduli of elasticity in the longitudinal and transverse directions are 19.7 and 3.66 GPa (2.8 × 106 and 5.3 × 105 psi), respectively. If the volume fraction of fibers is 0.25, determine the moduli of elasticity of fiber and matrix phases. Solution This problem asks for us to compute the elastic moduli of fiber and matrix phases for a continuous and oriented fiber-reinforced composite. We can write expressions for the longitudinal and transverse elastic moduli using Equations 16.10b and 16.16, as

Ecl = Em (1 − V f

)

+ Ef Vf

19.7 GPa = Em (1 − 0.25) + E f (0.25)

And

Ect =

EmE f

(1

3.66 GPa =

− V f ) E f + V f Em Em E f (1 − 0.25) E f + 0.25Em

Solving these two expressions simultaneously for Em and Ef leads to

Em = 2.79 GPa (4.04 × 10 5 psi)

E f = 70.4 GPa (10.2 × 10 6 psi)

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16.11 (a) Verify that Equation 16.11, the expression for the fiber load–matrix load ratio (Ff/Fm), is valid. (b) What is the F f/Fc ratio in terms of Ef, Em, and Vf? Solution (a) In order to show that the relationship in Equation 16.11 is valid, we begin with Equation 16.4—i.e.,

Fc = Fm + F f

which may be manipulated to the form

Fc Fm

= 1 +

Ff Fm

or

Ff Fm

Fc

=

Fm

− 1

For elastic deformation, combining Equations 6.1 and 6.5

σ =

F = εE A

or

F = AεE We may write expressions for Fc and Fm of the above form as

Fc = AcεEc Fm = AmεEm which, when substituted into the above expression for Ff/Fm, gives

Ff Fm

=

AcεEc − 1 AmεEm

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But, Vm = Am/Ac, which, upon rearrangement gives

Ac

1 Vm

=

Am

which, when substituted into the previous expression leads to

Ff

=

Fm

Ec EmVm

− 1

Also, from Equation 16.10a, Ec = EmVm + EfVf, which, when substituted for Ec into the previous expression, yields

Ff Fm

=

EmVm + E f V f − 1 EmVm

=

EmVm + E f V f − EmVm Ef Vf = EmVm EmVm

the desired result. (b) This portion of the problem asks that we establish an expression for Ff/Fc. We determine this ratio in a similar manner. Now Fc = Ff + Fm (Equation 16.4), or division by Fc leads to

1 =

Ff Fc

+

Fm Fc

which, upon rearrangement, gives

Ff

F = 1− m Fc Fc

Now, substitution of the expressions in part (a) for Fm and Fc that resulted from combining Equations 6.1 and 6.5 results in

Ff Fc

= 1−

AmεEm AcεEc

= 1−

AmEm Ac Ec

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Since the volume fraction of fibers is equal to Vm = Am/Ac, then the above equation may be written in the form

Ff Fc

V E = 1− m m Ec

And, finally substitution of Equation 16.10(a) for Ec into the above equation leads to the desired result as follows:

Ff Fc

= 1−

VmEm VmEm + V f E f

VmEm + V f E f − VmEm VmEm + V f E f

=

=

=

Vf Ef VmEm + V f E f Vf Ef

(1 − V f ) Em + V f E f

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16.12 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). Modulus of Elasticity [GPa (psi)]

Tensile Strength [MPa (psi)]

72.5 (10.5 × 106)

3400 (490,000)

Glass fiber

5

3.0 (4.35 × 10 )

Nylon 6,6

76 (11,000)

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

Ff Fm

=

Ef Vf EmVm

=

Ef Vf Em (1 − V f

)

Now, using values for Ff and Fm from the problem statement

Ff

=

Fm

0.94 = 15.67 0.06

And when we substitute the given values for Ef and Em into the first equation leads to

Ff Fm

= 15.67 =

(72.5 GPa)V f (3.0 GPa)(1 − 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,

∗ = σ ' (1 − V σ cl f m

)

+ σ f∗ V f

= (30 MPa)(1 − 0.393) + (3400 MPa)(0.393) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

= 1354 MPa (196,400 psi) ' (30 MPa) are given in the problem statement. since values for σ ∗ (3400 MPa) and σ m f

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16.13 Assume that the composite described in Problem 16.8 has a cross-sectional area of 320 mm2 (0.50 in.2) and is subjected to a longitudinal load of 44,500 N (10,000 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? Solution The problem stipulates that the cross-sectional area of a composite, Ac, is 320 mm2 (0.50 in.2), and the longitudinal load, Fc, is 44,500 N (10,000 lbf) for the composite described in Problem 16.8. Also, for this composite Vf = 0.3 Vm = 0.7 Ef = 131 GPa Em = 2.4 GPa (a) First, we are asked to calculate the Ff/Fm ratio. According to Equation 16.11

Ff Fm

=

EfVf (131 GPa)(0.30) = = 23.4 EmVm (2.4 GPa)(0.70)

Or, Ff = 23.4Fm (b) Now, the actual loads carried by both phases are called for. From Equation 16.4

F f + Fm = Fc = 44,500 N

23.4Fm + Fm = 44,500 N

which leads to

Fm = 1824 N (410 lb f ) F f = Fc − Fm = 44, 500 N − 1824 N = 42, 676 N (9590 lb f )

(c) To compute the stress on each of the phases, it is first necessary to know the cross-sectional areas of both fiber and matrix. These are determined as

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A f = V f Ac = (0.30)(320 mm 2 ) = 96 mm2 (0.15 in.2 ) Am = Vm Ac = (0.70)(320 mm2 ) = 224 mm2 (0.35 in.2 )

Now, the stresses are determined using Equation 6.1 as

σf =

σm =

Ff Af

=

42,676 N = 445 × 10 6 N/m2 = 445 MPa (63,930 psi) (96 mm2 )(1 m /1000 mm) 2

Fm 1824 N = 8.14 × 10 6 N/m 2 = 8.14 MPa (1170 psi) = (224 mm2 )(1 m /1000 mm) 2 Am

(d) The strain on the composite is the same as the strain on each of the matrix and fiber phases; applying Equation 6.5 to both matrix and fiber phases leads to

εm =

εf =

8.14 MPa σm = 3.39 × 10 -3 = Em 2.4 × 10 3 MPa σf Ef

=

445 MPa = 3.39 × 10 -3 131 × 10 3 MPa

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16.14 A continuous and aligned fiber-reinforced composite having a cross-sectional area of 1130 mm2 (1.75 in.2) is subjected to an external tensile load. If the stresses sustained by the fiber and matrix phases are 156 MPa (22,600 psi) and 2.75 MPa (400 psi), respectively, the force sustained by the fiber phase is 74,000 N (16,600 lbf) and the total longitudinal strain is 1.25 × 10-3, determine (a) the force sustained by the matrix phase (b) the modulus of elasticity of the composite material in the longitudinal direction, and (c) the moduli of elasticity for fiber and matrix phases. Solution (a) For this portion of the problem we are asked to calculate the force sustained by the matrix phase. It is first necessary to compute the volume fraction of the matrix phase, Vm. This may be accomplished by first determining Vf and then Vm from Vm = 1 – Vf. The value of Vf may be calculated since, from the definition of stress (Equation 6.1), and realizing Vf = Af/Ac as

σf =

Ff Af

=

Ff V f Ac

Or, solving for Vf

Vf =

Ff σ f Ac

=

(156

74,000 N = 0.420 × 10 6 N / m 2 )(1130 mm2 )(1 m /1000 mm) 2

Also

Vm = 1 − V f = 1 − 0.420 = 0.580 And, an expression for σm analogous to the one for σf above is

σm =

Fm Am

=

Fm Vm Ac

From which

Fm = Vmσ m Ac = (0.580)(2.75 × 10 6 N/m2 )(1.13 × 10 -3 m 2 ) = 1802 N (406 lb f )

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(b) We are now asked to calculate the modulus of elasticity in the longitudinal direction. This is possible Fm + F f σ (from Equation 6.1). Thus realizing that Ec = c (from Equation 6.5) and that σ c = Ac ε

Fm + F f σc = ε

Ec =

=

(1.25

Ac ε

=

Fm + F f εAc

1802 N + 74,000 N = 53.7 ×10 9 N/m 2 = 53.7 GPa (7.77 × 10 6 psi) × 10−3)(1130 mm2 )(1 m /1000 mm) 2

(c) Finally, it is necessary to determine the moduli of elasticity for the fiber and matrix phases. This is possible assuming Equation 6.5 for the matrix phase—i.e.,

Em =

σm εm

and, since this is an isostrain state, εm = εc = 1.25 × 10-3. Thus

Em =

2.75 × 10 6 N / m2 σm = 2.2 × 10 9 N/m 2 = 1.25 × 10−3 εc

= 2.2 GPa (3.2 × 10 5 psi)

The elastic modulus for the fiber phase may be computed in an analogous manner:

Ef =

σf εf

=

σf εc

=

156 × 10 6 N / m 2 = 1.248 × 1011 N/m2 1.25 × 10−3

= 124.8 GPa (18.1 × 10 6 psi)

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The Fiber Phase The Matrix Phase 16.19 For a polymer-matrix fiber-reinforced composite, (a) List three functions of the matrix phase. (b) Compare the desired mechanical characteristics of matrix and fiber phases. (c) Cite two reasons why there must be a strong bond between fiber and matrix at their interface. Solution (a) For polymer-matrix fiber-reinforced composites, three functions of the polymer-matrix phase are: (1) to bind the fibers together so that the applied stress is distributed among the fibers; (2) to protect the surface of the fibers from being damaged; and (3) to separate the fibers and inhibit crack propagation. (b) The matrix phase must be ductile and is usually relatively soft, whereas the fiber phase must be stiff and strong. (c) There must be a strong interfacial bond between fiber and matrix in order to: (1) maximize the stress transmittance between matrix and fiber phases; and (2) minimize fiber pull-out, and the probability of failure.

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16.20 (a) What is the distinction between matrix and dispersed phases in a composite material? (b) Contrast the mech...