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ziro4ka [17]
3 years ago
5

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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Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa
Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air,  R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes  2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

(c) The next step is to find the total work needed in kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0
3 years ago
Challenge:
goldenfox [79]

Answer:

what?

Explanation:

8 0
3 years ago
A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur
anyanavicka [17]

Answer:

hello your question has some missing values below are the missing values

Mirror Radius (mm) Bending Failure Stress (MPa)

.603                                         225

.203                                         368

.162                                          442

answer : 191 mPa

Explanation:

<u>Determine the stress present at the time of fracture for the original plate</u>

Bending stress ∝ 1 / ( mirror radius )^n ------ ( 1 )

at 0.603  bending stress = 225

at 0.203  bending stress = 368

at 0.162  bending stress = 442

<u>applying equation 1   determine the value of n for several combinations</u>

 ( 225 / 368 ) = ( 0.203 / 0.603 )^n

hence : n = 0.452

also

 ( 368/442 ) = ( 0.162 / 0.203 ) ^n

hence : n = 0.821

also

( 225 / 442 ) = ( 0.162 / 0.603 ) ^n

hence : n = 0.514

Next determine the average value of n

n ( mean value ) =  ( 0.452 + 0.821 + 0.514 ) / 3 = 0.596

Calculate estimated stress present at the time of fracture for the original plate

= bending stress at x =  0.796 / bending stress at x = 0.603

= x / 225 = ( 0.603 / 0.796 ) ^ 0.596

therefore X ( stress present at the time of fracture of original plate )

     = 225 * 0.84747

     <em>=  191 mPa </em>

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Tom [10]

Answer:

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Explanation:

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Student A says hazardous waste can take the form of solid, liquid, or gas. Student B says hazardous waste can only take the form
lina2011 [118]

Answer:

Student A

Explanation:

hope this helps have a great day

4 0
3 years ago
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