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VikaD [51]
3 years ago
13

g A plane stress element has components sigma x = 160 MPa, tau xy = 100 MPa (CW). Determine the two values pf sigma y for which

the maximum shear stress is 73 MPa.

Engineering
1 answer:
Volgvan3 years ago
6 0

Answer:

The question is mentioned in the attachment.

Explanation:

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Do the coil resistances have any effect on the plots?
PolarNik [594]
Because of the skin depth effect, the current at high frequency tends to flow at very low depth from radius. Then at high frequency the effective cross section of the wire is narrower than at DC.

Fro example skin depth at 100 kHz is 0.206 mm (0.008”), a wire more thicker than AWG26 could be a waste of copper, better use a bunch of thin wire (Litz wire) to rise the Q factor.
8 0
2 years ago
A light bulb is switched on and within a few minutes its temperature becomes constant. Is it at equilibrium or steady state.
EleoNora [17]

Answer:

The temperature attains equilibrium with the surroundings.  

Explanation:

When the light bulb is lighted we know that it's temperature will go on increasing as the filament of the bulb has to  constantly dissipates energy during the time in which it is on. Now this energy is dissipated as heat as we know it, this heat energy is absorbed by the material of the bulb which is usually made up of glass, increasing it's temperature. Now we know that any object with temperature above absolute zero has to dissipate energy in form of radiations.

Thus we conclude that the bulb absorbs as well as dissipates it's absorbed thermal energy. we know that this rate is dependent on the temperature of the bulb thus it the temperature of the bulb does not change we can infer that an equilibrium has been reached in the above 2 processes i.e the rate of energy absorption equals the rate of energy dissipation.

Steady state is the condition when the condition does not change with time no matter whatever the surrounding conditions are.

6 0
3 years ago
A second inventor was driving down the highway in her Prius one day with her hand out the window. She happened to be driving thr
Eva8 [605]

Answer:

Explanation:

It wouldn't work because the wind energy she would be collecting would actually come from the car engine.

The relative wind velocity observed from a moving vehicle is the sum of the actual wind velocity and the velovity of the vehicle.

u' = u + v

While running a car will generate a rather high wind velocity, and increase the power generated by a wind turbine, the turbine would only be able to convert part of the wind energy into electricity while adding a lot of drag. In the end, it would generate less energy that what the drag casuses the car to waste to move the turbine.

Regenerative braking uses an electric generator connected to the wheel axle to recover part of the kinetic energy eliminated when one brakes the vehicle. Normal brakes dissipate this energy as heat, a regenerative brake uses it to recharge a batttery. Note that is is a fraction of the energy that is recovered, not all of it.

A "regenerative accelerator" makes no sense. Braking is taking kinetic energy out of the vehicle, while accelerating is adding kinetic energy to it. Cars accelerate using the power from their engines.

6 0
3 years ago
1. Fatigue equations are based solely on theoretical assumptions. Experimental data is only used to verify the theory. a. True.b
Rainbow [258]

Answer:

1.  b. False

2. b. False

3.  b. False

4.  b. False

5. a. True

6. a. True

7.  b. False

8.  b. False

9. a. True

Explanation:

1. The fatigue properties of a material  are determined by series of test.

2. For most steels there is a level of fatigue limit below which a component will survive an infinite number of cycles, for aluminum and titanium a fatigue limit can not be defined, as failure will eventually occur after enough experienced cycles.

3. Although there is a cyclic stress, there are also stresses complex circumstances involving tensile to compresive and constant stress, where the solution is given into the mean stress and the stress amplitude or stress range, which is double the stress amplitude.

4. Low‐cycle fatigue is defined as few thousand cycles and high cycle fatigue is around more than 10,000 cycles.

5. The number of cycles for failure on brittle materials are less and determined compared with the ductile materials.

6.  The bending fatigue could be handled with specific load requirements  for uniform bending or axial fatigue of the same section size where the material near the surface is subjected to the  maximum stress, as in torsional fatigue, which can be performed on  axial-type specially designed machines also, using the proper fixtures if  the maximum twist required is small, in which linear motion is changed to rotational motion.

7.  A SN-Curve for a given material, is a plot displayed on logarithmic scales of the magnitude of an alternating stress in relation to the number of cycles to failure

8. The strain life method measures the strain resistance of local stresses and strains around stress concentration that controls the fatigue life of the material. It is more accurate than determining fatigue performance as the stress-life method is for long life millions of cycles in elastic stresses, but an it gets an effective stress concentration in fatigue loading.

9. Linear Elastic Fracture Mechanics (LEFM) states that the material is isotropic and linear elastic so, when the stresses near the crack surpasses the material fracture toughness, the crack grows.

7 0
3 years ago
A transmission line (TL) of length L and conductance per unit length G' is connected to an ideal constant voltage generator V. T
Andru [333]

Answer:

The Current will decrease by a factor of 2

Explanation:

Given the conditions, it should be noted that the current in the circuit is determined by the LOAD. In other words, the amount of current generator will be producing depends upon the load connected to it.

Now, as the question says, the load is reduced to half its original value, we can write:

P1 = \sqrt{3} (V) (I1) Cos\alpha ----- (1)

P2 = \sqrt{3} (V) (I2) Cos\alpha\\

Since, P2 = P1/2,

P1/2 = \sqrt{3} (V) (I2) Cos\alpha ----- (2)\\

Dividing equations (1) and (2), we get,

P1 / (P1/2) = I1/ I2

I2 = I1 / 2\\

Hence, it is proved that the current in the transmission line will decrease by a factor of 2 when load is reduced to half.

7 0
2 years ago
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