Given:

The diameter of metal specimen is 0.550 inch.

The load at facture is 28,500 pounds.

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 0.550 inch for d in the Equation (II).

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

Calculate the ultimate tensile stress.

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

f=28,500 pounds0.2375 inch2×(1 klb1000 pounds)=28.5 klb0.2375 inch2=120 klb/inch2×(1 ksi1 klb/inch2)=120 ksi

Conclusion:

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

Given:

The original gage length is 2.03 inches.

The change in gage length is 2.3 inches.

Concept Used:

Write the equation to calculate the elongation.

e=Lf−L0L0×100 ...... (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 2.03 inches for L0 and 2.3 inches for Lf in Equation (III).

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

Conclusion:

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

Given:

The original diameter of metal specimen is 0.550 inch.

The diameter after fracture load is 0.430 inch.

Concept Used:

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

R=A0−AfA0×100 ...... (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 0.430 inch for d in the Equation (II).

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

Calculate the reduction in the cross-sectional area.

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

R=0.2375 inch2−0.1452 inch20.2375 inch2×100=0.0923 inch20.2375 inch2×100=0.3886×100=38.86%

Conclusion:

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