The magnitude of the maximum stress at the tip of an internal crack ( σ m ) .

Question

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Answer 1

Question

Chapter 9.19, Problem 1QP

To determine

The magnitude of the maximum stress at the tip of an internal crack (σm) .

Expert Solution & Answer

Answer to Problem 1QP

The magnitude of the maximum stress at the tip of an internal crack is 3959.79 MPa_ .

Explanation of Solution

Given:

The radius of curvature of the crack tip, pt=1.9×10−4 mm

The crack length, a=3.8×10−2 mm

The tensile stress, σ0=140 MPa

Explanation:

Determine the maximum stress that occurs at the crack tip.

σm=2σ0(apt)1/2 (I)

Conclusion:

Substitute 3.8×10−2 mm for a, 140 MPa for σ0 , and 1.9×10−4 mm for pt in Equation (I).

σm=2(140 MPa)(3.8×10−2 mm1.9×10−4 mm)1/2=2(140)(14.142)=3959.79 MPa

Thus, the magnitude of the maximum stress at the tip of an internal crack is 3959.79 MPa_ .

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Answer 2

Question

Chapter 9.19, Problem 1QP

To determine

The magnitude of the maximum stress at the tip of an internal crack (σm) .

Expert Solution & Answer

Answer to Problem 1QP

The magnitude of the maximum stress at the tip of an internal crack is 3959.79 MPa_ .

Explanation of Solution

Given:

The radius of curvature of the crack tip, pt=1.9×10−4 mm

The crack length, a=3.8×10−2 mm

The tensile stress, σ0=140 MPa

Explanation:

Determine the maximum stress that occurs at the crack tip.

σm=2σ0(apt)1/2 (I)

Conclusion:

Substitute 3.8×10−2 mm for a, 140 MPa for σ0 , and 1.9×10−4 mm for pt in Equation (I).

σm=2(140 MPa)(3.8×10−2 mm1.9×10−4 mm)1/2=2(140)(14.142)=3959.79 MPa

Thus, the magnitude of the maximum stress at the tip of an internal crack is 3959.79 MPa_ .

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Answer 3

Question

Chapter 9.19, Problem 1QP

To determine

The magnitude of the maximum stress at the tip of an internal crack (σm) .

Expert Solution & Answer

Answer to Problem 1QP

The magnitude of the maximum stress at the tip of an internal crack is 3959.79 MPa_ .

Explanation of Solution

Given:

The radius of curvature of the crack tip, pt=1.9×10−4 mm

The crack length, a=3.8×10−2 mm

The tensile stress, σ0=140 MPa

Explanation:

Determine the maximum stress that occurs at the crack tip.

σm=2σ0(apt)1/2 (I)

Conclusion:

Substitute 3.8×10−2 mm for a, 140 MPa for σ0 , and 1.9×10−4 mm for pt in Equation (I).

σm=2(140 MPa)(3.8×10−2 mm1.9×10−4 mm)1/2=2(140)(14.142)=3959.79 MPa

Thus, the magnitude of the maximum stress at the tip of an internal crack is 3959.79 MPa_ .

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Answer 4

Question

Chapter 9.19, Problem 1QP

To determine

The magnitude of the maximum stress at the tip of an internal crack (σm) .

Expert Solution & Answer

Answer to Problem 1QP

The magnitude of the maximum stress at the tip of an internal crack is 3959.79 MPa_ .

Explanation of Solution

Given:

The radius of curvature of the crack tip, pt=1.9×10−4 mm

The crack length, a=3.8×10−2 mm

The tensile stress, σ0=140 MPa

Explanation:

Determine the maximum stress that occurs at the crack tip.

σm=2σ0(apt)1/2 (I)

Conclusion:

Substitute 3.8×10−2 mm for a, 140 MPa for σ0 , and 1.9×10−4 mm for pt in Equation (I).

σm=2(140 MPa)(3.8×10−2 mm1.9×10−4 mm)1/2=2(140)(14.142)=3959.79 MPa

Thus, the magnitude of the maximum stress at the tip of an internal crack is 3959.79 MPa_ .

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Answer 5

Chapter 9.19, Problem 1QP

To determine

The magnitude of the maximum stress at the tip of an internal crack (σm) .

Expert Solution & Answer

Answer to Problem 1QP

The magnitude of the maximum stress at the tip of an internal crack is 3959.79 MPa_ .

Explanation of Solution

Given:

The radius of curvature of the crack tip, pt=1.9×10−4 mm

The crack length, a=3.8×10−2 mm

The tensile stress, σ0=140 MPa

Explanation:

Determine the maximum stress that occurs at the crack tip.

σm=2σ0(apt)1/2 (I)

Conclusion:

Substitute 3.8×10−2 mm for a, 140 MPa for σ0 , and 1.9×10−4 mm for pt in Equation (I).

σm=2(140 MPa)(3.8×10−2 mm1.9×10−4 mm)1/2=2(140)(14.142)=3959.79 MPa

Thus, the magnitude of the maximum stress at the tip of an internal crack is 3959.79 MPa_ .

Answer 6

Chapter 9.19, Problem 1QP

To determine

The magnitude of the maximum stress at the tip of an internal crack (σm) .

Expert Solution & Answer

Answer to Problem 1QP

The magnitude of the maximum stress at the tip of an internal crack is 3959.79 MPa_ .

Explanation of Solution

Given:

The radius of curvature of the crack tip, pt=1.9×10−4 mm

The crack length, a=3.8×10−2 mm

The tensile stress, σ0=140 MPa

Explanation:

Determine the maximum stress that occurs at the crack tip.

σm=2σ0(apt)1/2 (I)

Conclusion:

Substitute 3.8×10−2 mm for a, 140 MPa for σ0 , and 1.9×10−4 mm for pt in Equation (I).

σm=2(140 MPa)(3.8×10−2 mm1.9×10−4 mm)1/2=2(140)(14.142)=3959.79 MPa

Thus, the magnitude of the maximum stress at the tip of an internal crack is 3959.79 MPa_ .

Answer 7

Chapter 9.19, Problem 1QP

To determine

The magnitude of the maximum stress at the tip of an internal crack (σm) .

Answer 8

Expert Solution & Answer

Answer to Problem 1QP

The magnitude of the maximum stress at the tip of an internal crack is 3959.79 MPa_ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

The radius of curvature of the crack tip, pt=1.9×10−4 mm

The crack length, a=3.8×10−2 mm

The tensile stress, σ0=140 MPa

Explanation:

Determine the maximum stress that occurs at the crack tip.

σm=2σ0(apt)1/2 (I)

Conclusion:

Substitute 3.8×10−2 mm for a, 140 MPa for σ0 , and 1.9×10−4 mm for pt in Equation (I).

σm=2(140 MPa)(3.8×10−2 mm1.9×10−4 mm)1/2=2(140)(14.142)=3959.79 MPa

Thus, the magnitude of the maximum stress at the tip of an internal crack is 3959.79 MPa_ .

Answer 12

Ch. 9.19 - Prob. 1QP Ch. 9.19 - Prob. 2QP Ch. 9.19 - Prob. 3QP Ch. 9.19 - Prob. 4QP Ch. 9.19 - Prob. 5QP Ch. 9.19 - Prob. 6QP Ch. 9.19 - Prob. 7QP Ch. 9.19 - Prob. 8QP Ch. 9.19 - Prob. 9QP Ch. 9.19 - Prob. 10QP

The magnitude of the maximum stress at the tip of an internal crack ( σ m ) .

The magnitude of the maximum stress at the tip of an internal crack (σm) .

Answer to Problem 1QP

Explanation of Solution

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Chapter 9 Solutions