Foundations of Materials Science and Engineering
Foundations of Materials Science and Engineering
6th Edition
ISBN: 9781259696558
Author: SMITH
Publisher: MCG
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Chapter 7.10, Problem 45SEP

(a)

To determine

The critical length of crack if the vessel made of Al7178-T651.

(a)

Expert Solution
Check Mark

Answer to Problem 45SEP

The critical length of crack is 0.00216in.

Explanation of Solution

Write the expression for the circumferential stress or hoop stress in cylindrical vessel.

    σ=Prt                                                                                          (I)

Here, the internal pressure is P, the radius of vessel is r and the thickness of vessel is t.

Write the expression of typical fracture toughness is,

    KIC=Yσπa

    a=(KICYσπ)2                                                                              (II)

Here, fracture toughness is KIC, yield strength is Y, circumferential stress or hoop stress is σ and critical thickness of the vessel is a.

Write the expression of critical length of crack for the vessel is,

    C=2a                                                                                      (III)

Here, critical thickness of the vessel is a.

Conclusion:

Substitute 5000Psi for P, 18in for r and 0.25in for t in Equation (I).

    σ=(5000Psi)(18in)(0.25in)=(5000Psi)[1ksi1000Psi](18in)(0.25in)=360ksi

Refer table 7.1 “Typical fracture toughness values for selected engineering alloys”, to obtain the value of KIC as 21ksiin corresponding to the material Al7178-T651.

Substitute, 21ksiin for KIC, 1 for Y and 360ksi for σ in Equation (II).

    a=(21ksiin1×360(ksi)×π)2=(((21)2in)(360)2×π)=0.00108in

Substitute 0.00108in for a in Equation (III).

    C=2×0.00108in=0.00216in

Thus, the critical length of crack is 0.00216in.

(b)

To determine

The critical crack length if the vessel made of alloy steel (177pH).

(b)

Expert Solution
Check Mark

Answer to Problem 45SEP

The critical length of crack is 0.024in.

Explanation of Solution

Conclusion:

Refer table 7.1 “Typical fracture toughness values for selected engineering alloys”, to obtain the value of KIC as 70ksiin corresponding to the material alloy steel (177pH).

Substitute, 70ksiin for KIC, 1 for Y and 360ksi for σ in Equation (II).

    a=(70ksiin1×360(ksi)×π)2=(((70)2in)(360)2×π)=0.012in

Substitute 0.00108in for a in Equation (III).

    C=2×0.012in=0.024in

Thus, the critical length of crack is 0.024in.

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Chapter 7 Solutions

Foundations of Materials Science and Engineering

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