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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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How much power is needed to operate a Carnot heat pump if the pump receives heat 10°C and delivers 50 kW of heat at 40°C? at A)

mariarad [96]

Answer:

Power needed to pump=4.79 KW.

Explanation:

Given that: $T_{1}=283K,T_{2}=313K,Q_{H}=50KW$

We know that coefficient of performance of heat pump

COP= $\dfrac{T_{H}}{T_{H}-T_{L}}$

So COP= $\dfrac{313}{313-283}$

COP=10.43

COP= $\frac{Q_{H}}{W_{in}}$

10.43 = $\frac{50}{W_{in}}$

$W_{in}$ =4.79 KW

So power needed to pump=4.79 KW.

3 0

3 years ago

In an apartment the interior air temperature is 20°C and exterior air temperatures is 5°C. The wall has inner and outer surface

ELEN [110]

Answer:

20 W/m², 20 W/m², -20 W/m²

Yes, the wall is under steady-state conditions.

Explanation:

Air temperature in room = 20°C

Air temperature outside = 5°C

Wall inner temperature = 16°C

Wall outer temperature = 6°C

Inner heat transfer coefficient = 5 W/m²K

Outer heat transfer coefficient = 20 W/m²K

Heat flux = Concerned heat transfer coefficient × (Difference of the temperatures of the concerned bodies)

q = hΔT

Heat flux from the interior air to the wall = heat transfer coefficient of interior air × (Temperature difference between interior air and exterior wall)

⇒ Heat flux from the interior air to the wall = 5 (20-6) = 20 W/m²

Heat flux from the wall to the exterior air = heat transfer coefficient of exterior air × (Temperature difference between wall and exterior air)

⇒Heat flux from the wall to the exterior air = 20 (6-5) = 20 W/m²

Heat flux from the wall to the interior air = heat transfer coefficient of interior air × (Temperature difference between wall and interior air)

⇒Heat flux from the wall and interior air = 5 (16-20) = -20 W/m²

Here the magnitude of the heat flux are same so the wall is under steady-state conditions.

7 0

3 years ago

What should you use to keep battery terminals from corroding

Ray Of Light [21]

Answer: Apply battery-terminal grease to the terminals to help prevent corrosion. It's available at any auto parts store and usually comes in a little ketchup-like packet. Another great option is AMSOIL Heavy-Duty Metal Protector. It creates a protective coating on terminals that wards off corrosion.

Explanation:

3 0

3 years ago

What is the stress concentration factor of a shaft in torsion, where D=1.25 in. and d=1 in. and the fillet radius is, r=0.2 in.a

klio [65]

Answer:

Concentration factor will be 1.2

So option (C) will be correct answer

Explanation:

We have given outer diameter D = 1.25 in

And inner diameter d = 1 in and fillet ratio r = 0.2 in

So $\frac{r}{d}$ ratio will be $=\frac{0.2}{1}=0.2$