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Anna71 [15]
3 years ago
6

Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 43.1 deg

rees and 47.9 degrees, respectively , with the tensile axis. If the critical resolved shear stress is 20.7 MPa (3000 psi), will an applied stress of 45 MPa (6500 psi) cause the single crystal to yield? If not, what stress will be necessary?
Engineering
1 answer:
Marysya12 [62]3 years ago
4 0

Answer:

resolved shear stress = 22.0 MPa

so we can say that here single crystal will yield  because critical resolved shear stress i.e 20.7 MPa is less than resolved shear stress i.e 22.0 MPa

Explanation:

given data

angles φ  = 43.1 degrees

angles λ = 47.9 degrees

shear stress =  20.7 MPa (3000 psi)

stress σ = 45 MPa (6500 psi)

solution

we have given shear stress so first we calculate here resolved shear stress that is express as

resolved shear stress = σ cosφ cosλ    .................1

here σ is stress and φ and λ are angles given

so put here value we get

resolved shear stress = σ cosφ cosλ

resolved shear stress = 45 cos(43.1) cos(47.9)

resolved shear stress = 22.0 MPa

so we can say that here single crystal will yield  because critical resolved shear stress i.e 20.7 MPa is less than resolved shear stress i.e 22.0 MPa

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adell [148]

Answer:

<em>a) 4.51 lbf-s^2/ft</em>

<em>b) 65.8 kg</em>

<em>c) 645 N</em>

<em>d) 23.8 lb</em>

<em>e) 65.8 kg</em>

<em></em>

Explanation:

Weight of the man on Earth = 145 lb

a) Mass in slug is...

32.174 pound = 1 slug

145 pound = x slug

x = 145/32.174 = <em>4.51 lbf-s^2/ft</em>

b) Mass in kg is...

2.205 pounds = 1 kg

145 pounds = x kg

x = 145/2.205 = <em>65.8 kg</em>

c) Weight in Newton = mg

where

m is mass in kg

g is acceleration due to gravity on Earth = 9.81 m/s^2

Weight in Newton = 65.8 x 9.81 = <em>645 N</em>

d) If on the moon with acceleration due to gravity of 5.30 ft/s^2,

1 m/s^2 = 3.2808 ft/s^2

x m/s^2 = 5.30 ft/s^2

x = 5.30/3.2808 = 1.6155 m/s^2

weight in Newton = mg = 65.8 x 1.6155 = 106

weight in pounds = 106/4.448 = <em>23.8 lb</em>

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mass on the moon = <em>65.8 kg</em>

3 0
3 years ago
A weighted, frictionless piston-cylinder device initially contains 5.25 kg of R134a as saturated vapor at 500 kPa. The container
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Answer:

-6.326 KJ/K

Explanation:

A) the entropy change is defined as:

delta S_{12}=\int\limits^2_1  \, \frac{dQ}{T}

In an isobaric process heat (Q) is defined as:

Q= m*Cp*dT

Replacing in the equation for entropy  

delta S_{12}=\int\limits^2_1 \frac{m*Cp*dT}{T}

m is the mass and Cp is the specific heat of R134a. We can considerer these values as constants so the expression for entropy would be:  

delta S_{12}= m*Cp*\int\limits^2_1 \frac{ dT }{T}  

Solving the integral we get the expression to estimate the entropy change in the system  

delta S_{12}= m*Cp *ln(\frac{T_{2}}{T_{1}})

The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is 0.85\frac{kJ}{Kg*K}

We can get the temperature at the beginning knowing that is saturated vapor at 500 KPa. Consulting the thermodynamic tables, we get that temperature of saturation at this pressure is: 288.86 K

The temperature in the final state we can get it from the heat expression, since we know how much heat was lost in the process (-976.71 kJ). By convention when heat is released by the system a negative sign is used to express it.

Q= m*Cp*dT

With dt=T_{2}-T_{1} clearing for T2 we get:

T_{2}=\frac{Q}{m*Cp}+T1= \frac{-976.71kJ}{5.25Kg*0.85\frac{kJ}{Kg*K}}+288.86 K =69.98 K

Now we can estimate the entropy change in the system

delta S_{12}= m*Cp*ln(\frac{T_{2}}{T_{1}})= 5.25Kg*0.85\frac{kJ}{Kg*K}*ln(\frac{69.98}{288.861})= -6.326\frac{kJ}{K}

The entropy change in the system is negative because we are going from a state with a lot of disorder (high temperature) to one more organize (less temperature. This was done increasing the entropy of the surroundings.  

b) see picture.

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Answer:

Plumber and pipefitters

Explanation:

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2 years ago
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Xelga [282]

Answer:

a)-True

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3 0
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Answer:

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Explanation:

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