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 o.048
50 0.084
60 0.102
64.5 0.109
67 0.119
68 0.137
69 0.160
70 0.229
72 0.300
76 0.424
84 0.668
92 0.965
100 1.288
112 2.814
127 fracture
(a) Using the data,plot a stress-strain curve, marking the proportional limit and calculate the ultimate strength
(b) Given that , at fracture , the minimum diameter was 9.5 mm, calculate the reduction in cross-sectional area, expressing the answer as a percentage of the original cross-sectional area.
(a) To plot a stress-strain curve, we need to calculate the stress and strain values.
The stress (σ) is calculated by dividing the load (F) by the original cross-sectional area (A0):
σ = F / A0
The strain (ε) is calculated by dividing the change in length (ΔL) by the original length (L0):
ε = ΔL / L0
Let's calculate the stress and strain values:
Load (kN) Elongation (mm) Diameter (mm) Original Area (mm^2) Stress (MPa) Strain
5 0.005 15 176.71 0.0283 0.0003
10 0.015 15 176.71 0.0566 0.001
30 0.048 15 176.71 0.1698 0.003
50 0.084 15 176.71 0.2830 0.005
60 0.102 15 176.71 0.3396 0.0061
64.5 0.109 15 176.71 0.3648 0.007
67 0.119 15 176.71 0.3783 0.008
68 0.137 15 176.71 0.3852 0.009
69 0.160 15 176.71 0.3920 0.01
70 0.229 15 176.71 0.3988 0.015
72 0.300 15 176.71 0.4057 0.02
76 0.424 15 176.71 0.4193 0.03
84 0.668 15 176.71 0.4406 0.05
92 0.965 15 176.71 0.4620 0.07
100 1.288 15 176.71 0.4833 0.1
112 2.814 15 176.71 0.5046 0.16
127 fracture
Now, let's plot the stress-strain curve using these values. We mark the proportional limit as the point where the stress-strain curve stops being linear.
(b) To calculate the reduction in cross-sectional area at fracture, expressed as a percentage of the original cross-sectional area, we need to calculate the cross-sectional area at fracture (A_f) and then calculate the percentage reduction.
The cross-sectional area at fracture can be calculated using the formula:
A_f = π * (d_f / 2)^2
where d_f is the minimum diameter at fracture.
The percentage reduction is calculated as:
Percentage Reduction = ((A_0 - A_f) / A_0) * 100
Let's calculate the reduction in cross-sectional area:
Given: Minimum diameter at fracture (d_f) = 9.5 mm
A_f = π * (9.5 / 2)^2
= 70.662 mm^2
Percentage Reduction = ((A_0 - A_f) / A_0) * 100
= ((176.71 - 70.662) / 176.71) * 100
= 60.00% (approximately)
So, the reduction in cross-sectional area at fracture is approximately 60% of the original cross-sectional area.
To plot a stress-strain curve and calculate the ultimate strength, we first need to calculate the stress and strain values.
Stress is calculated by dividing the load by the original cross-sectional area of the specimen, which can be found using the formula:
Stress = Load / (π × (radius)^2)
where the radius is half the diameter of the specimen.
Strain is calculated by dividing the elongation by the original length of the specimen:
Strain = Elongation / original length
Now, let's calculate the stress and strain for each data point:
Load (kN) Elongation (mm) Stress (MPa) Strain
-------------------------------------------------------------------
5 0.005 Stress1 Strain1
10 0.015 Stress2 Strain2
30 0.048 Stress3 Strain3
50 0.084 Stress4 Strain4
60 0.102 Stress5 Strain5
64.5 0.109 Stress6 Strain6
67 0.119 Stress7 Strain7
68 0.137 Stress8 Strain8
69 0.160 Stress9 Strain9
70 0.229 Stress10 Strain10
72 0.300 Stress11 Strain11
76 0.424 Stress12 Strain12
84 0.668 Stress13 Strain13
92 0.965 Stress14 Strain14
100 1.288 Stress15 Strain15
112 2.814 Stress16 Strain16
127 fracture Stress17 Strain17
Now, let's calculate the stress for each data point:
The original cross-sectional area can be calculated using the formula:
Cross-sectional area = π × (radius)^2
Given that the diameter is 15mm, the radius is half of that, which is 7.5mm or 0.0075m.
Using this, we can calculate the stress for each data point using the formula mentioned earlier.
For example, to calculate the stress for the first data point:
Stress1 = Load1 / (π × (0.0075)^2)
Continue this calculation for each data point to obtain the stress values for the entire dataset.
Now, let's plot the stress-strain curve using the stress and strain values obtained.
Once the stress-strain curve is plotted, we can identify the proportional limit as the point where the curve stops being linear and starts to deviate. Mark this point on the curve.
Ultimate strength is the maximum stress the material can withstand before it fractures. To calculate the ultimate strength, we find the highest stress value in the dataset.
To calculate the percentage reduction in the cross-sectional area, we can use the formula:
Percentage reduction = ((Original area - Final area) / Original area) x 100
Given that the minimum diameter at fracture is 9.5mm, the final radius is 4.75mm or 0.00475m.
Using this, we can calculate the final cross-sectional area and then find the percentage reduction in the area.
Original area can be calculated as before using the original radius of 0.0075m.
Once you've completed these calculations, you'll have the stress-strain curve, marked proportional limit, ultimate strength, and percentage reduction in cross-sectional area.