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sertanlavr [38]
3 years ago
9

Calculate the biaxial stresses σ1 and σ2 for the biaxial stress case, where ε1 = .0020 and ε2 = –.0010 are determined experiment

ally on an aluminum member of elastic constants, E = 71 GPa and v = 0.35. Also, determine the value for the maximum shear stress.
Engineering
1 answer:
Llana [10]3 years ago
3 0

Answer:

i) σ1 = 133.5 MPa

  σ2 = -2427 MPa

ii) 78.89 MPa

Explanation:

Given data:

ε1 = 0.0020 and ε2 = –0.0010

E = 71 GPa

v = 0.35

<u>i) Determine the biaxial stresses  σ1 and σ2 using the relations below</u>

ε1 = σ1 / E - v (σ2 / E)   -----( 1 )

ε2 = σ2 / E - v (σ1 / E)  -------( 2 )

resolving equations 1 and 2

σ1 = E / 1 - v^2 {  ε1 + vε2 } ---- ( 3 )

σ2 = E / 1 - v^2 {  ε2 + vε1 } ----- ( 4 )

input the given data into equation 3 and equation 4

σ1 = 133.5 MPa

σ2 = -2427 MPa

<u>ii) Calculate the value of the maximum shear stress ( Zmax )</u>

Zmax = ( σ1 - σ2 ) / 2

         = 133.5 - ( - 2427 ) / 2

         = 78.89 MPa

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8 0
4 years ago
An equilibrium mixture of 3 kmol of CO, 2.5 kmol of O2, and 8 kmol of N2 is heated to 2600 K at a pressure of 5 atm. Determine t
garri49 [273]

Answer:

x_{CO}=0.0203\\x_{O_2}=0.0926\\x_{CO_2}=0.227\\x_{N_2}=0.660

Explanation:

Hello,

In this case, we consider the reaction:

CO(g)+\frac{1}{2} O_2(g)\rightleftharpoons CO_2

For which the law of mass action is expressed as:

Kp=\frac{n_{CO_2}}{n_{CO}*n_{O_2}^{1/2}} (\frac{P}{n_{Tot}} )^{1-1-1/2}

Whereas the exponents are referred to the stoichiometric coefficients in the chemical reaction. Moreover, in table A-28 (Cengel's thermodynamics) the natural logarithm of the undergoing reaction at 2600 K is 2.801, thus:

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In such a way, in terms of the change x the equilibrium goes:

16.46=\frac{x}{(3kmol-x)*(2.5kmol-0.5x)^{0.5}} (\frac{5}{13.5kmol-0.5x} )^{-0.5}

Hence, solving for x:

x=2.754kmol

Thus, the moles at equilibrium:

n_{CO}=3-2.754=0.246kmol\\n_{O_2}=2.5-0.5(2.754)=1.123kmol\\n_{CO_2}=x=2.754kmol\\n_{N_2}=8kmol

Finally the compositions:

x_{CO}=\frac{0.246}{0.246+1.123+2.754+8} =0.0203\\\\x_{O_2}=\frac{1.123}{0.246+1.123+2.754+8} =0.0926\\\\x_{CO_2}=\frac{2.754}{0.246+1.123+2.754+8} =0.227\\\\x_{N_2}=\frac{8}{0.246+1.123+2.754+8} =0.660

Best regards.

7 0
4 years ago
A specimen of some metal having a rectangular cross section 10.4 mm × 12.8 mm is pulled in tension with a force of 15900 N, whic
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Answer:

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Explanation:

We apply Hooke's Law as follows :

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Where l\ and w are the cross-sectional dimensions.

=>We equate the two stress equations:

E\varepsilon=\frac{F}{lw}\\\\\varepsilon=\frac{F}{Elw}

#We substitute our values in the equation as:

\varepsilon=\frac{15900\ N}{(79\times 10^9\ N/m^2)(10.4\ m\times 10^{-3}\times 12.8\ m\times 10^{-3})}}\\\\\\=0.00151

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8 0
3 years ago
A steam turbine has isentropic efficiency of 0.8. Isentropically, it is supposed to deliver work of 100 kW. What is the actual w
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Part 2

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Part 3

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