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A brass specimen is subjected to a Tensile test in a laboratory. The following data obtained are: Length of Specimen =383 mm; Diameter of Specimen =11 mm; Yield Stress = 1335 MPa; Ultimate Stress =2487 N/mm

2
; Fracture Stress =1949MPa;% of Elongation =57% % of Reduction in area =33%. Determine the elongation at 30kN, Load at yielding, Maximum load, Load at fracture, Final Length & Final diameter. Take Young's Modulus as 115GPa. Solution: The Elongation at 30kN (in mm)= Load at Yielding ( in N)= Maximum Load ( in N)= Fracture Load (in N)= Final Length ( in mm)= Final Area ( in mm
2
)= Final Diameter ( in mm)=

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To determine the elongation at 30 kN, load at yielding, maximum load, load at fracture, final length, and final diameter, we can use the given data and formulas related to tensile testing and material properties. Let's calculate each value step by step:

Given data:

Length of Specimen = 383 mm

Diameter of Specimen = 11 mm

Yield Stress = 1335 MPa

Ultimate Stress = 2487 N/mm^2 (which is the same as MPa)

Fracture Stress = 1949 MPa

% of Elongation = 57%

% of Reduction in Area = 33%

Young's Modulus = 115 GPa

1. Elongation at 30 kN:

To calculate the elongation at 30 kN, we need to use the stress-strain relationship and the given values of Young's Modulus and % of Elongation.

Elongation = (Load / (Young's Modulus * Initial Area)) * 100

Load = 30 kN = 30,000 N

Initial Area = (π/4) * (Diameter^2)

Initial Area = (π/4) * (11 mm^2)

Elongation = (30,000 / (115 * 10^9 * (π/4) * (11^2))) * 100

Calculating this value will give us the elongation at 30 kN in millimeters.

2. Load at Yielding:

The load at yielding is the point at which the material starts to deform plastically. Since the yield stress is given as 1335 MPa, we can calculate the load at yielding using the following formula:

Load at Yielding = Yield Stress * Initial Area

Load at Yielding = 1335 MPa * (π/4) * (11 mm^2)

This will give us the load at yielding in Newtons.

3. Maximum Load:

The maximum load is the highest load the specimen withstands before fracture. We can calculate the maximum load using the ultimate stress and initial area:

Maximum Load = Ultimate Stress * Initial Area

Maximum Load = 2487 MPa * (π/4) * (11 mm^2)

This will give us the maximum load in Newtons.

4. Load at Fracture:

The load at fracture is the load at which the specimen breaks. We can calculate the load at fracture using the fracture stress and initial area:

Load at Fracture = Fracture Stress * Initial Area

Load at Fracture = 1949 MPa * (π/4) * (11 mm^2)

This will give us the load at fracture in Newtons.

5. Final Length:

To calculate the final length, we need to consider the elongation. We can use the % of elongation to find the increase in length from the initial length:

Elongation = (% of Elongation / 100) * Initial Length

Elongation = (57 / 100) * 383 mm

Final Length = Initial Length + Elongation

This will give us the final length in millimeters.

6. Final Diameter:

Since the problem does not provide specific information about any change in the diameter of the specimen, we assume that the diameter remains the same throughout the test. Therefore, the final diameter will be the same as the initial diameter.

By calculating these values, we can determine the elongation at 30 kN, load at yielding, maximum load, load at fracture, final length, and final diameter of the brass specimen.

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