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jeka94
3 years ago
5

A strain gauge with a 4 mm gauge length gives a displacement reading of 1.5 um. Calculate the stress at the location of the stra

in gauge if the material is a) structural steel, and b) PMMA.
Engineering
1 answer:
Art [367]3 years ago
3 0

Answer:

1)  75Mpa

2) 1.125 MPa

Explanation:

given data:

gauge length = 4 mm

displacement = 1.5\mu m = 1.5\times 10^{-3} m

a) structural steel

Young modulus for steel is 200 GPa = 200\times 10^3 MPa

we know that

E  =\frac{stress}{strain}

Stress = 200\times 10^3 \times \frac{1.5\times 10^{-3}}{4}

          = 75Mpa

b) PMMA

Young's modulus = 3GPa = 3\times10^3 MPa

stress = 3\times \frac{1.5\times10^{-3}}{4}

stress = 1.125 MPa

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What do you own that might not be manufactured?
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3 years ago
Calculate the unit cell edge length for an 80 wt% Ag−20 wt% Pd alloy. All of the palladium is in solid solution, the crystal str
alisha [4.7K]

Answer:

λ^3 = 4.37

Explanation:

first let us to calculate the average density of the alloy

for simplicity of calculation assume a 100g alloy

80g --> Ag

20g --> Pd

ρ_avg= 100/(20/ρ_Pd+80/ρ_avg)

         = 100*10^-3/(20/11.9*10^6+80/10.44*10^6)

         = 10744.62 kg/m^3

now Ag forms FCC and Pd is the impurity in one unit cell there is 4 atoms of Ag since Pd is the impurity we can not how many atom of Pd in one unit cell let us calculate

total no of unit cell in 100g of allow = 80 g/4*107.87*1.66*10^-27

                                                          = 1.12*10^23 unit cells

mass of Pd in 1 unit cell = 20/1.12*10^23

Now,

                      ρ_avg= mass of unit cell/volume of unit cell

                      ρ_avg= (4*107.87*1.66*10^-27+20/1.12*10^23)/λ^3

                          λ^3 = 4.37

6 0
4 years ago
Which of the following terms describes the path from an electrical source to a switch or plug?
mart [117]

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transmitter hope thus helped!

Explanation:

8 0
3 years ago
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Use Excel, MatLab or a similar program to plot carrier thermal velocity as a function of temperature for both electrons and hole
Sidana [21]

Answer:

Solution 2:

The thermal energy of any particle is given as = \frac{1}{2}mv^{2}

where m= Effective mass of the particle,

v= Velocity of the particle

The average kinetic energy is given as =  \frac{3}{2}KT,

where K= Boltzmann Constant

T= Temperautre of the particle

At equilibrium,

1/2 mv² =  3/2 KT

Hence, v= \sqrt{\frac{3KT}{m}}

As, effective mass ratio is calculated with respect to rest mass of electron,

melectron = 0.11 * 9.11 *10-31 kg

mhole =  0.35 *9.11 *10-31 kg

K = 1.38 × 10-23 m2 kg s-2 K-1

Plotting, over a range of temperature of 0 K to 400 K, we obtain the attached Graphs (attached).

Solution 3:

The above two curves plotted are not identical as seen. From the same value of Temperature, and under identical conditons, the Thermal velocity solely depends on effective mass ratio. And it is inversely proportional to effective mass ratio. More the effective mass ratio, less the thermal velocity and flatter the slope of the curve and vice versa. As, the mass ratio of holes is more than that of electrons, the curve of electrons has a steeper slope than that of holes. Hence, the curves are not just identical.

8 0
3 years ago
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