Question

Solution Initial Cross-sectional Area (in mm2) The Diameter of the Specimen (in mm) Final Length of the Specimen (in mm) Stress at the elastic load (in N/mm²) Young's Modulus of the Specimen in N/mm²) Final Area of the Specimen at Fracture (in mm) Final Diameter of the Specimen after Fracture (in mm) The following data are obtained from a tensile test of a copper specimen. The load at the yield point is 145 kN. Length of the specimen is 21 mm. - The yield strength is 73 kN/mm2. The percentage of elongation is 42 %. Determine the following (i) Diameter of the specimen, ii) Final length of the specimen, iii) Stress under an elastic load of 14 kN, iv) Young's Modulus if the elongation is 2.1 mm at 14 kN and (v) Final diameter if the percentage of reduction in area is 29%.

Answer 1

Given data:
Load at yield point: 145 kN
Length of specimen: 21 mm
Yield strength: 73 kN/mm²
Percentage of elongation: 42%
Elongation at 14 kN: 2.1 mm
Percentage of reduction in area: 29%

To determine the required values, we can use the formulas and equations related to tensile testing and material properties.

(i) Diameter of the specimen:
We can calculate the initial cross-sectional area using the yield strength and load at yield point:
Initial cross-sectional area = Load at yield point / Yield strength
Diameter of the specimen = sqrt(4 * Initial cross-sectional area / π)

(ii) Final length of the specimen:
The final length can be calculated using the percentage of elongation:
Final length = Length of specimen * (1 + Percentage of elongation / 100)

(iii) Stress under an elastic load of 14 kN:
Stress = Load / Cross-sectional area
Cross-sectional area = Load at yield point / Yield strength
Stress = Elastic load / Cross-sectional area

(iv) Young's Modulus if the elongation is 2.1 mm at 14 kN:
Young's Modulus = Stress / Strain
Strain = Elongation / Length of specimen
Stress = Elastic load / Cross-sectional area

(v) Final diameter if the percentage of reduction in area is 29%:
Final area of the specimen = Initial cross-sectional area * (1 - Percentage of reduction in area / 100)
Final diameter = sqrt(4 * Final area of the specimen / π)

Now, let's calculate the values using the given data:

(i) Diameter of the specimen:
Initial cross-sectional area = 145 kN / 73 kN/mm² = 1.986 kN/mm²
Diameter of the specimen = sqrt(4 * 1.986 / π) ≈ 2.819 mm

(ii) Final length of the specimen:
Final length = 21 mm * (1 + 42 / 100) = 21 mm * 1.42 ≈ 29.82 mm

(iii) Stress under an elastic load of 14 kN:
Cross-sectional area = 145 kN / 73 kN/mm² = 1.986 kN/mm²
Stress = 14 kN / 1.986 kN/mm² ≈ 7.05 N/mm²

(iv) Young's Modulus if the elongation is 2.1 mm at 14 kN:
Strain = 2.1 mm / 21 mm = 0.1
Stress = 14 kN / 1.986 kN/mm² ≈ 7.05 N/mm²
Young's Modulus = 7.05 N/mm² / 0.1 ≈ 70.5 N/mm²

(v) Final diameter if the percentage of reduction in area is 29%:
Final area of the specimen = 1.986 kN/mm² * (1 - 29 / 100) ≈ 1.410 kN/mm²
Final diameter = sqrt(4 * 1.410 / π) ≈ 2.375 mm

Therefore, the calculated values are:
(i) Diameter of the specimen: 2.819 mm
(ii) Final length of the specimen: 29.82 mm
(iii) Stress under an elastic load of 14 kN: 7.05 N/mm²
(iv) Young's Modulus if the elongation is 2.1 mm at 14 kN: 70.5 N/mm²
(v) Final diameter if the percentage of reduction in area is 29%: 2.375 mm