The following data was obtained from a tensile test of steel. The test specimen was 15mm in diameter and 50 mm in length.
Load(kN)/elongation mm)
5 /0.005
10 /0.015
30 /0.048
50 /0.084
60 /0.102
64.5 /0.109
67 /0.119
68 /0.137
69 /0.16
70 /0.229
72 /0.3
76 /0.424
84 /0.668
92 /0.965
100 /1.288
112 /2.814
127 /fracture
Given that at fracture, the minimum diameter was 9.6mm. calculate the reduction in cross-sectional area. Expressing the answer as a percentage of the original cross-sectional area.
You don't need all that load and elongation data to answer the question. At fracture, the diameter was reduced by a factor 9.6/15 = 0.64. The area was reduced by the square of that factor, 0.4096
The area at that time was 40.96% of the initial area.
To calculate the reduction in cross-sectional area, we need to find the difference between the original cross-sectional area and the cross-sectional area at fracture, and then express it as a percentage of the original cross-sectional area.
First, let's find the original cross-sectional area. The original cross-sectional area (A1) of a circular specimen can be calculated using the formula:
A1 = π * (d1/2)^2
Given that the diameter (d1) of the specimen is 15mm, we can substitute this value into the formula:
A1 = π * (15/2)^2
A1 = π * 7.5^2
A1 ≈ π * 56.25
A1 ≈ 176.71 mm^2 (rounded to two decimal places)
Next, let's find the cross-sectional area at fracture (A2). We know that the minimum diameter at fracture (d2) was 9.6mm. So, we can calculate A2 using the same formula:
A2 = π * (d2/2)^2
A2 = π * (9.6/2)^2
A2 ≈ π * 4.8^2
A2 ≈ π * 23.04
A2 ≈ 72.16 mm^2 (rounded to two decimal places)
Now, we can calculate the reduction in cross-sectional area (ΔA) by subtracting A2 from A1:
ΔA = A1 - A2
ΔA ≈ 176.71 - 72.16
ΔA ≈ 104.55 mm^2 (rounded to two decimal places)
Finally, to express the reduction in cross-sectional area as a percentage of the original cross-sectional area, we can use the following formula:
Percentage reduction = (ΔA / A1) * 100
Applying the values:
Percentage reduction ≈ (104.55 / 176.71) * 100
Percentage reduction ≈ 0.59 * 100
Percentage reduction ≈ 59% (rounded to the nearest whole number)
Therefore, the reduction in cross-sectional area is approximately 59% of the original cross-sectional area.