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The following data were collected from a 20-mm-diameter test specimen of a ductile cast iron ( l 0 = 40.00 mm): Load (N) ? l (mm) 0 0.0000 25,000 0.0185 50,000 0.0370 75,000 0.0555 90,000 0.20 105,000...

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The following data were collected from a 20-mm-diameter test specimen of a ductile cast iron (l0= 40.00 mm):

Load (N)

?l (mm)

0

0.0000

25,000

0.0185

50,000

0.0370

75,000

0.0555

90,000

0.20

105,000

0.60

120,000

1.56

131,000

4.00 (maximum load)

125,000

7.52 (fracture)

After fracture, the total length was 47.42 mm and the diameter was 18.35 mm. Plot the engineering stress–strain curve and the true stress–strain curve. Since the instantaneous cross-sectional area of the specimen is unknown past the point of necking, truncate the true stress–true strain data at the point that corresponds to the ultimate tensile strength. Use of a software graphing package is recommended. Calculate

(a) the 0.2% offset yield strength;

(b) the tensile strength;

(c) the modulus of elasticity, using a linear fit to the appropriate data;

(d) the % elongation;

(e) the % reduction in area;

(f) the engineering stress at fracture; and

(g) the modulus of resilience.

Answered Same Day Dec 24, 2021

Solution

Robert answered on Dec 24 2021
122 Votes
The following data were collected from a 20-mm-diameter test specimen of a ductile cast iron (l0= 40.00 mm):
Load (N) ?l (mm)
0 0.0000
25,000 0.0185
50,000 0.0370
75,000 0.0555
90,000 0.20
105,000 0.60
120,000 1.56
131,000 4.00 (maximum load)
125,000 7.52 (fracture)
After fracture, the total length was 47.42 mm and the diameter was 18.35 mm. Plot the engineering stress–strain
curve and the true stress–strain curve. Since the instantaneous cross-sectional area of the specimen is unknown
past the point of necking, truncate the true stress–true...
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