Maurice Tutor

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  • MCS,PHD
    Argosy University/ Phoniex University/
    Nov-2005 - Oct-2011

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  • Professor
    Phoniex University
    Oct-2001 - Nov-2016

Category > Management Posted 17 Jan 2018 My Price 8.00

point of necking

Taking the logarithm of both sides of Equation 7.19 yields

width=

log σT = log K + n log ?T

Thus, a plot of log σT versus log ?T in the plastic region to the point of necking should yield a straight line having a slope of n and an intercept (at log σT = 0) of log K.

Using the appropriate data tabulated in Problem.

Problem

A cylindrical specimen of stainless steel having a diameter of 12.8 mm (0.505 in.) and a gauge length of 50.800mm(2.000 in.) is pulled in tension. Use the load–elongation characteristics tabulated below 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

 

Make a plot of log σT versus log ?T and determine the values of n and K. It will be necessary to convert engineering stresses and strains to true stresses and strains using Equations 7.18a and 7.18b.

width=

Answers

Maurice Tutor
(5)
Status NEW Posted 17 Jan 2018 07:01 PM My Price 8.00

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