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3 years ago

What are factor of safety for brittle and ductile material

Engineering

1 answer:

galben [10]3 years ago

5 0

Explanation:

Step1

Factor of safety is the number that is taken for the safe design of any component. It is the ratio of failure stress to the maximum allowable stress for the material.

Step2

It is an important parameter for design of any component. This factor of safety is taken according to the environment condition, type of material, strength, type of component etc.

Step3

Different material has different failure stress. So, ductile material fails under shear force. Ductile material’s FOS is based on yield stress as failure stress as after yield point ductile material tends to yield. Brittle material’s FOS is based on ultimate stress as failure stress.

The expression for factor of safety for ductile material is given as follows:

$FOS=\frac{\sigma_{yp}}{\sigma_{a}}$

Here, $\sigma_{f}$ is yield stress and $\sigma_{a}$ is allowable stress.

The expression for factor of safety for brittle material is given as follows:

$FOS=\frac{\sigma_{ut}}{\sigma_{a}}$

Here, $\sigma_{ut}$ is ultimate stress and $\sigma_{a}$ is allowable stress.

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Regeneration can only increase the efficiency of a Brayton cycle when working fluid leaving the turbine is hotter than the worki

storchak [24]

Answer:

True, Regeneration is the only process where increases the efficiency of a Brayton cycle when working fluid leaving the turbine is hotter than working fluid leaving the compressor.

Option: A

Explanation:

To increase the efficiency of brayton cycle there are three ways which includes inter-cooling, reheating and regeneration. Regeneration technique is used when a turbine exhaust fluids have higher temperature than the working fluid leaving the compressor of the turbine.

Thermal efficiency of a turbine is increased as the exhaust fluid having higher temperatures are used in heat exchanger where the fluids from the compressor enters and increases the temperature of the fluids leaving the compressor.

6 0

3 years ago

Read 2 more answers

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

Should aircraft wings have infinite stiffness?

Colt1911 [192]

Answer:

No, they need to be somewhat flexible so that forces such as turbulance don't shear the wing off.

3 0

3 years ago

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What was the first prototype of the artificial tree like?

olga_2 [115]

Answer:

An artificial Christmas tree is an artificial pine or fir tree manufactured for the specific purpose of use as a Christmas tree. The earliest artificial Christmas trees were wooden, tree-shaped pyramids or feather trees, both developed by Germans. Most modern trees are made of polyvinyl chloride (PVC) but many other types of trees have been and are available, including aluminum Christmas trees and fiber-optic illuminated Christmas trees.

Explanation:

7 0

4 years ago

A system consists of a disk rotating on a frictionless axle

kakasveta [241]

The system includes a disk rotating on a frictionless axle and a bit of clay transferring towards it, as proven withinside the determine above.

<h3>What is the angular momentum?</h3>

The angular momentum of the device earlier than and after the clay sticks can be the same.

Conservation of angular momentum the precept of conservation of angular momentum states that the whole angular momentum is usually conserved.

Li = Lf where;
li is the preliminary second of inertia
If is the very last second of inertia
wi is the preliminary angular velocity
wf is the very last angular velocity
Li is the preliminary angular momentum
Lf is the very last angular momentum

Thus, the angular momentum of the device earlier than and after the clay sticks can be the same.