A cylindrical specimen of stainless - Mechanical Engineering

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Free Chegg Question

A cylindrical specimen of stainless steel having a diameter of 12.8 mm (0.505 in.) and a gauge length of 50.800 mm (2.000 in.) is pulled in tension. Use the load–elongation characteristics shown in the following table to complete parts (a) through (f).

Load		Length
N	lbf	mm	in.
0	0	50.800	2.000
12,700	2,850	50.825	2.001
25,400	5,710	50.851	2.002
38,100	8,560	50.876	2.003
50,800	11,400	50.902	2.004
76,200	17,100	50.952	2.006
89,100	20,000	51.003	2.008
92,700	20,800	51.054	2.010
102,500	23,000	51.181	2.015
107,800	24,200	51.308	2.020
119,400	26,800	51.562	2.030
128,300	28,800	51.816	2.040
149,700	33,650	52.832	2.080
159,000	35,750	53.848	2.120
160,400	36,000	54.356	2.140
159,500	35,850	54.864	2.160
151,500	34,050	55.880	2.200
124,700	28,000	56.642	2.230
	Fracture

(a) Plot the data as engineering stress versus engineering strain.

(b) Compute the modulus of elasticity.

(c) Determine the yield strength at a strain offset of 0.002.

(d) Determine the tensile strength of this alloy.

(e) What is the approximate ductility, in percent elongation?

(f) Compute the modulus of resilience.

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Free Chegg Answer

Step 1 of 9

(a)

Calculate the engineering stress as follows:

Here, the area of the cylindrical specimen is A and the load applied on the specimen is .

Substitute for A.

Here, engineering stress is and the diameter for the specimen is.

Substitutefor and for .
Step 2 of 9

Calculate the engineering strain as follows:

Here, engineering strain is , instantaneous length of the specimen is , and the original length of the specimen is .

Substitute for and for .
Step 3 of 9

Similarly, calculate the engineering stress and strain for the other readings and tabulate as follows:
Step 4 of 9

Plot the graph between the stress and strain values (Graph-1):
Step 5 of 9

(b)

Calculate the modulus of elasticity as follows:

Here, modulus of elasticity is , the change in the stress is , and the change in the strain is .

Substitute for and for .

Therefore, the modulus of elasticity for the stainless steel is .
Step 6 of 9

(c)

Draw the graph of strain offset at 0.002 and the line intersects the Y axis at(Graph-2).

Therefore, the yield strength at the strain offset is .
Step 7 of 9

(d)

Refer to the graph 1, tensile strength of the alloy is maximum point of the curve and the peak point is .

Therefore, the tensile strength of the steel specimen is .
Step 8 of 9

(e)

Calculate the plastic strain or percent elongation for the specimen as follows:

Here, is the percent elongation, is the strain at fracture point, and is the strain in elastic region.

Substitute for and for .

Therefore, the percent elongation in specimen is .
Step 9 of 9

(f)

Calculate the modulus of resilience as follows:

Here, is the modulus of resilience, is the yield strength of the alloy, and is the modulus of elasticity.

Substitute for and for .

Therefore, the modulus of resilience is .