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viva [34]
3 years ago
13

Define Plastic vs elastic deformation.

Engineering
1 answer:
Snowcat [4.5K]3 years ago
3 0

Answer:

Plastic deformation, irreversible or permanent. Deformation mode in which the material does not return to its original shape after removing the applied load. This happens because, in plastic deformation, the material undergoes irreversible thermodynamic changes by acquiring greater elastic potential energy.

Elastic deformation, reversible or non-permanent. the body regains its original shape by removing the force that causes the deformation. In this type of deformation, the solid, by varying its tension state and increasing its internal energy in the form of elastic potential energy, only goes through reversible thermodynamic changes.

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The metal control joints used to relieve stresses caused by expansion and contraction in large ceiling or wall expenses in inter
timurjin [86]

Answer:

It’s called Expansion Joints

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3 years ago
Clarifying the issues of a problem is the _____ step in the problem solving process.
ratelena [41]
The answer is 2nd Step because the first step is to define the problem and third is to define your goals
7 0
4 years ago
Calculate the viscosity(dynamic) and kinematic viscosity of airwhen
nikitadnepr [17]

Answer:

(a) dynamic viscosity = 1.812\times 10^{-5}Pa-sec

(b) kinematic viscosity = 1.4732\times 10^{-5}m^2/sec

Explanation:

We have given temperature T = 288.15 K

Density d=1.23kg/m^3

According to Sutherland's Formula  dynamic viscosity is given by

{\mu} = {\mu}_0 \frac {T_0+C} {T + C} \left (\frac {T} {T_0} \right )^{3/2}, here

μ = dynamic viscosity in (Pa·s) at input temperature T,

\mu _0= reference viscosity in(Pa·s) at reference temperature T0,

T = input temperature in kelvin,

T_0 = reference temperature in kelvin,

C = Sutherland's constant for the gaseous material in question here C =120

\mu _0=4\pi \times 10^{-7}

T_0 = 291.15

\mu =4\pi \times 10^{-7}\times \frac{291.15+120}{285.15+120}\times \left ( \frac{288.15}{291.15} \right )^{\frac{3}{2}}=1.812\times 10^{-5}Pa-swhen T = 288.15 K

For kinematic viscosity :

\nu = \frac {\mu} {\rho}

kinemic\ viscosity=\frac{1.812\times 10^{-5}}{1.23}=1.4732\times 10^{-5}m^2/sec

3 0
3 years ago
On a average work day more than work place firs are reorted​
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What is the question? It looks like a statement...
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3 years ago
For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about ____
7nadin3 [17]

Answer:

One

For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about _one___

<h3>what is supported mounted?</h3>
  • A structure that holds up or serves as a foundation for something else. Support is a synonym for mounting.

To learn more about it, refer

to brainly.com/question/25689052

#SPJ4

4 0
2 years ago
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