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navik [9.2K]
3 years ago
6

What’s the difference between engineering stress and strain and true stress and strain

Engineering
2 answers:
Nana76 [90]3 years ago
4 0

True strain and engineering strain? True stress is defined as the load divided by the cross-sectional area of the specimen at that instant and is a true indication of the internal pressures. ... Engineering stress is defined as the load divided by the initial cross-sectional area of the specimenAnswer:

Explanation:

svetoff [14.1K]3 years ago
4 0

Answer:

he value of true stress or true strain is calculated from this instantaneous change in cross section area. ... Conversely, since true strain is calculated by integrating the strain over the entire test, it is larger than engineering strain for the same value of displacement.

Explanation:

True strain and engineering strain? True stress is defined as the load divided by the cross-sectional area of the specimen at that instant and is a true indication of the internal pressures. ... Engineering stress is defined as the load divided by the initial cross-sectional area of the specimen.

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Levart [38]

Answer:

This is a for loop.

Explanation:

6 0
3 years ago
A medium-sized jet has a 3.8-mm-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to th
SCORPION-xisa [38]

Answer:

F_{thrust} ≅ 111 KN

Explanation:

Given that;

A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8

mass = 85,000 kg

drag co-efficient (C) = 0.37

(velocity (v)= 230 m/s

density (ρ) = 1.0 kg/m³

To calculate the thrust; we need to determine the relation of the drag force; which is given as:

F_{drag} = \frac{1}{2} × CρAv²

where;

ρ = density of air wind.

C = drag co-efficient

A = Area of the jet

v = velocity of the jet

From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0

SO, F_{drag}-F_{thrust} = 0

We can as well say:

F_{drag}= F_{thrust}

We can now replace F_{thrust} with F_{drag} in the above equation.

Therefore, F_{thrust} = \frac{1}{2} × CρAv²

The A which stands as the area of the jet is given by the formula:

A=\frac{\pi d^2}{4}

We can now have a new equation after substituting our A into the previous equation as:

F_{thrust} = \frac{1}{2} × Cρ (\frac{\pi d^2}{4})v^2

Substituting our data from above; we have:

F_{thrust} = \frac{1}{2} × (0.37)(1.0kg/m^3)(\frac{\pi(3.8m)^2 }{4})(230m/s)^2

F_{thrust} = \frac{1}{8}   (0.37)(1.0kg/m^3)({\pi(3.8m)^2 })(230m/s)^2

F_{thrust} = 110,990N

F_{thrust}  in N (newton) to KN (kilo-newton) will be:

F_{thrust} = (110,990N)*\frac{1KN}{1,000N}

F_{thrust} = 110.990 KN

F_{thrust} ≅ 111 KN

In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.

5 0
3 years ago
A well insulated turbine operates at steady state. Steam enters the turbine at 4 MPa with a specific enthalpy of 3015.4 kJ/kg an
Anarel [89]

Answer:

power developed by the turbine = 6927.415 kW

Explanation:

given data

pressure = 4 MPa

specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 = 2431.7 kJ/kg

velocity v2 = 90 m/s

mass flow rate = 11.95 kg/s

solution

we apply here  thermodynamic equation that

energy equation that is

h1 + \frac{v1}{2}  + q = h2 + \frac{v2}{2}  + w

put here value with

turbine is insulated so q = 0

so here

3015.4 *1000 + \frac{10^2}{2}  =  2431.7 * 1000 + \frac{90^2}{2}  + w

solve we get

w = 579700 J/kg = 579.7 kJ/kg

and

W = mass flow rate × w

W = 11.95 × 579.7

W = 6927.415 kW

power developed by the turbine = 6927.415 kW

7 0
3 years ago
For a turning operation, you have selected a high-speed steel (HSS) tool and turning a hot rolled free machining steel. Your dep
Alisiya [41]

Answer:

MRR = 1.984

Explanation:

Given that                              

Depth of cut ,d=0.105 in

Diameter D= 1 in

Speed V= 105 sfpm

feed f= 0.015 ipr

Now  the metal   removal  rate   given as

MRR= 12 f V d

d= depth of cut

V= Speed

f=Feed

MRR= Metal removal rate

By putting the values

MRR= 12 f V d

MRR = 12 x 0.015 x 105 x 0.105

MRR = 1.984

Therefore answer is -

1.944

8 0
3 years ago
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zheka24 [161]

Answer:

And Im still going with B..

7 0
3 years ago
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