Steel Design (Activate Learning with these NEW titles from Engineering!)
Steel Design (Activate Learning with these NEW titles from Engineering!)
6th Edition
ISBN: 9781337094740
Author: Segui, William T.
Publisher: Cengage Learning
Question
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Chapter 1, Problem 1.5.1P
To determine

(a)

The ultimate tensile stress of metal specimen.

Expert Solution
Check Mark

Answer to Problem 1.5.1P

120ksi.

Explanation of Solution

Given:

The diameter of metal specimen is 0.550inch.

The load at facture is 28,500pounds.

Concept Used:

Write the equation to calculate the ultimate tensile stress.

f=PA ...... (I)

Here, ultimate tensile stress is f, fractured load is P, and the cross-sectional area is A.

Calculation:

Calculate the cross-sectional area of specimen.

A=π4×d2 ...... (II)

Here, diameter of the specimen is d.

Substitute 0.550inch for d in the Equation (II).

A=π4×(0.550inch)2=0.95033inch24=0.2375inch2

Calculate the ultimate tensile stress.

Substitute 28,500pounds for P and 0.2375inch2 for A in the Equation (I).

f=28,500pounds0.2375inch2×(1klb1000pounds)=28.5klb0.2375inch2=120klb/inch2×(1ksi1klb/inch2)=120ksi

Conclusion:

Thus, the ultimate tensile stress on the metal specimen is 120ksi.

To determine

(b)

The elongation of the metal specimen.

Expert Solution
Check Mark

Answer to Problem 1.5.1P

13.3%.

Explanation of Solution

Given:

The original gage length is 2.03inches.

The change in gage length is 2.3inches.

Concept Used:

Write the equation to calculate the elongation.

e=LfL0L0×100 ...... (III)

Here, the elongation is e, the length of the specimen at fracture is Lf, and the original length is L0.

Calculation:

Calculate the elongation of the metal specimen.

Substitute 2.03inches for L0 and 2.3inches for Lf in Equation (III).

e=[(2.3inches)(2.03inches)(2.03inches)]×100=0.1330×100=13.3%

Conclusion:

Thus, the elongation of the metal specimen is 13.3%.

To determine

(c)

The reduction in the cross-sectional area of the metal specimen.

Expert Solution
Check Mark

Answer to Problem 1.5.1P

38.86%.

Explanation of Solution

Given:

The original diameter of metal specimen is 0.550inch.

The diameter after fracture load is 0.430inch.

Concept Used:

Write the equation to calculate reduction in cross-sectional area.

R=A0AfA0×100 ...... (IV)

Here, the reduction in cross-sectional area is R, original cross-sectional area is A0, and the cross-sectional area after fracture load is Af.

Calculation:

Calculate the cross-sectional area after fracture load.

Substitute 0.430inch for d in the Equation (II).

A=π4×(0.430inch)2=0.5808inch24=0.1452inch2

Calculate the reduction in the cross-sectional area.

Substitute 0.1452inch2 for Af and 0.2375inch2 for A0 in Equation (IV).

R=0.2375inch20.1452inch20.2375inch2×100=0.0923inch20.2375inch2×100=0.3886×100=38.86%

Conclusion:

Thus, the reduction in the cross-sectional area is 38.86%.

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