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gtnhenbr [62]
3 years ago
9

A stainless-steel specimen from the same material characterized up above, was formed into a rectangular cross-section of dimensi

ons (3/4 in. x 1/8 in). The specimen is subject to a tensile force of 72,000 N.
(a) Determine the elastic and plastic strain values when the specimen is subject to the 72,000 N tensile load.
(b) If the original length is 24 in., what will be its final length after the 72,000 N tensile load is released
Engineering
1 answer:
AlexFokin [52]3 years ago
7 0

Our dates are in different Units, we will do a unique system. I will use International System Units.

F=72000N

\frac{3}{8}in= 19,05mm

\frac{1}{8}in=3,175mm

We can now calculate the stress with the force applied, that is

\sigma = \frac{F}{A} = \frac{72000}{(3,175)(19,05)*10^{-6}}

\sigma = 1190.40Mpa

A) The elastic straint is given by,

\epsilon_e = \frac{\sigma}{E}

Where E is the Elastic coefficient. This coefficient for Steel is 201Gpa or 207*10^3Mpa

\epsilon_e=\frac{1190.40}{207*10^3}

\epsilon_e =5.7507*10^{-3}

We know for the tables of Strain Steel that the Total Strain is \epsilon_f=0.01

Then the Plastic Strain is given by,

\epsilon_p=\epsilon_f-\epsilon_e

\epsilon_p= 0.01-5.7507*10^{-3}

\epsilon_p = 4.2492*10^{-3}

B) Now with the original length of 24in, i.e 609,6mm we have that

l_i = l_0 (1+\epsilon_p)

l_i = (609.6mm)(1+4.294*10^{-3})

l_i = 612.2176mm

<em>Therefore the final length after the 72000N will be 612.2176mm</em>

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79 f t^{3} is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

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V_{1}=100\ \mathrm{ft}^{3}

First has a natural water content of 25% = \frac{25}{100} = 0.25

Shrinkage limit, w_{1}=12 \%=\frac{12}{100}=0.12

G_{s}=2.70

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

V \propto[1+e]

\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}  ------> eq 1

e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}

The above equation is at S_{r}=1,

e_{1}=w_{1} \times G_{s}

Applying the given values, we get

e_{1}=0.25 \times 2.70=0.675

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e_{2}=w_{2} \times G_{s}

Applying the given values, we get

e_{2}=0.12 \times 2.70=0.324

Applying the found values in eq 1, we get

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V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}

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