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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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Answer:

(a) Increases

(b) Increases

(c) Increases

(d) Increases

(e) Decreases

Explanation:

The tensile modulus of a semi-crystalline polymer depends on the given factors as:

(a) Molecular Weight:

It increases with the increase in the molecular weight of the polymer.

(b) Degree of crystallinity:

Tensile strength of the semi-crystalline polymer increases with the increase in the degree of crystallinity of the polymer.

(c) Deformation by drawing:

The deformation by drawing in the polymer results in the finely oriented chain structure of the polymer with the greater inter chain secondary bonding structure resulting in the increase in the tensile strength of the polymer.

(d) Annealing of an undeformed material:

This also results in an increase in the tensile strength of the material.

(e) Annealing of  a drawn material:

A semi crystalline material which is drawn when annealed results in the decreased tensile strength of the material.

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3 years ago
During a medical evaluation, the doctor can __________.
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An ideal Otto cycle has a compression ratio of 9.2 and uses air as the working fluid. At the beginning of the compression proces
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Answer:

(a) The amount of heat transferred to the air, q_{out} is 215.5077 kJ/kg

(b) The net work output, W_{net}, is 308.07 kJ/kg

(c) The thermal efficiency is 58.8%

(d) The Mean Effective Pressure, MEP, is 393.209 kPa

Explanation:

(a) The assumptions made are;

c_p = 1.005 kJ/(kg·K), c_v = 0.718 kJ/(kg·K), R = 0.287 kJ/(kg·K),

Process 1 to 2 is isentropic compression, therefore;

T_{2}= T_{1}\left (\dfrac{v_{1}}{v_{2}}  \right )^{k-1} = 300.15\times 9.2^{0.4} = 729.21 \, K

From;

\dfrac{p_{1}\times v_{1}}{T_{1}} = \dfrac{p_{2}\times v_{2}}{T_{2} }

We have;

p_{2} = \dfrac{p_{1}\times v_{1}\times T_{2}}{T_{1} \times v_{2}} = \dfrac{98\times 9.2\times 729.21}{300.15 } = 2190.43 \, kPa

Process 2 to 3 is reversible constant volume heating, therefore;

\dfrac{p_3}{T_3} =\dfrac{p_2}{T_2}

p₃ = 2 × p₂ = 2 × 2190.43 = 4380.86 kPa

T_3 = \dfrac{p_3 \times T_2}{p_2} =\dfrac{4380.86  \times 729.21}{2190.43} = 1458.42 \, K

Process 3 to 4 is isentropic expansion, therefore;

T_{3}= T_{4}\left (\dfrac{v_{4}}{v_{3}}  \right )^{k-1}

1458.42= T_{4} \times \left (9.2 \right )^{0.4}

T_4 = \dfrac{1458.42}{(9.2)^{0.4}}  = 600.3 \, K

q_{out} = m \times c_v \times (T_4 - T_1) = 0.718  \times (600.3 - 300.15) = 215.5077 \, kJ/kg

The amount of heat transferred to the air, q_{out} = 215.5077 kJ/kg

(b) The net work output, W_{net}, is found as follows;

W_{net} = q_{in} - q_{out}

q_{in} = m \times c_v \times (T_3 - T_2) = 0.718  \times (1458.42 - 729.21) = 523.574 \, kJ/kg

\therefore W_{net} = 523.574 - 215.5077 = 308.07 \, kJ/kg

(c) The thermal efficiency is given by the relation;

\eta_{th} = \dfrac{W_{net}}{q_{in}} \times 100=  \dfrac{308.07}{523.574} \times 100= 58.8\%

(d) From the general gas equation, we have;

V_{1} = \dfrac{m\times R\times T_{1}}{p_{1}} = \dfrac{1\times 0.287\times 300.15}{98} =0.897\, m^{3}/kg

The Mean Effective Pressure, MEP, is given as follows;

MEP =\dfrac{W_{net}}{V_1 - V_2} = \dfrac{W_{net}}{V_1 \times (1- 1/r)}= \dfrac{308.07}{0.897\times (1- 1/9.2)} = 393.209 \, kPa

The Mean Effective Pressure, MEP = 393.209 kPa.

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