The results shown below were obtained from a tensile test on a non-ferrous alloy with a 10 mm diameter and 50 mm gauge length - Leaving Cert Engineering - Question b - 2020
Question b
The results shown below were obtained from a tensile test on a non-ferrous alloy with a 10 mm diameter and 50 mm gauge length.
| Load (kN) | 15 | 25 | 40 | 60 |... show full transcript
Worked Solution & Example Answer:The results shown below were obtained from a tensile test on a non-ferrous alloy with a 10 mm diameter and 50 mm gauge length - Leaving Cert Engineering - Question b - 2020
Step 1
Using the graph paper supplied, plot the load-extension diagram.
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Answer
On the x-axis, label the extension in mm, ranging from 0 to 1.5 mm.
On the y-axis, label the load in kN, ranging from 0 to 120 kN.
Plot each pair of values from the table on the graph:
(0.06 mm, 15 kN)
(0.10 mm, 25 kN)
(0.16 mm, 40 kN)
(0.26 mm, 60 kN)
(0.38 mm, 80 kN)
(0.65 mm, 100 kN)
(0.90 mm, 107 kN)
(1.00 mm, 108 kN)
(1.20 mm, 95 kN)
(1.40 mm, 96 kN)
Connect the points to form a smooth curve.
Step 2
Indicate each of the following regions on your graph: Elastic region
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The Elastic Region is typically from the origin to the point where the curve begins to deviate linearly. Mark this region on the graph.
Step 3
Indicate each of the following regions on your graph: Plastic region
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The Plastic Region is where the material deforms permanently. Mark the section of the curve after the elastic region that continues to rise but begins to flatten.
Step 4
Indicate each of the following regions on your graph: Necking region
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The Necking Region occurs after the maximum load is reached, where the curve starts to drop. Mark this area on the graph.
Step 5
Determine the ultimate tensile strength (UTS) of the alloy in kN/mm².
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The Ultimate Tensile Strength (UTS) can be calculated using the formula:
extUTS=C.S.A.Max Load
The maximum load observed is 108 kN, and the circular cross-sectional area (C.S.A.) of the alloy is given by:
C.S.A.=π(2d)2=π(5)2=25πmm2
Using these values, we calculate:
UTS=25π mm2108 kN≈1.38 kN/mm2
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