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3 years ago

Define the difference between elastic and plastic deformation in terms of the effect on the crystal lattice structure.

Engineering

1 answer:

grin007 [14]3 years ago

3 0

Answer:

Elastic Deformation:

Elastic deformation is due to the elasticity of the body i.e., the deformation disappears on removal of external forces and the lattice regain its original structure.
This type of deformation is temporary in nature.
This type of deformation depends mainly on the chemical bonding of the substance.
It occurs due to bending or stretching of chemical bond under applied stress where atoms do not slip pass over each other.
It is a reversible process.

Plastic Deformation:

Plastic deformation is due to the plasticity of the body i,e., the structure can be easily shaped or molded .
This type of deformation is permanent in nature.
This type of deformation occurs mainly due to breakage of chemical bonds(limited in number) between constituent atoms.
Atoms may slip pass each other causing dislocation of atoms , resulting in permanent deformation even after stress is removed.
It is an irreversible process.

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notka56 [123]

Answer:

L= 50000 lb

D = 5000 lb

Explanation:

To maintain a level flight the lift must equal the weight in magnitude.

We know the weight is of 50000 lb, so the lift must be the same.

L = W = 50000 lb

The L/D ratio is 10 so

10 = L/D

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3 years ago

An electric motor is to be supported by four identical mounts. Each mount can be treated as a linear prevent problems due requir

Artyom0805 [142]

GIVEN:

Amplitude, A = 0.1mm

Force, F =1 N

mass of motor, m = 120 kg

operating speed, N = 720 rpm

$\frac{A}{F}$ = $\frac{0.1\times 10^{-3}}{1} = 0.1\times 10^{-3}$

Formula Used:

$A = \frac{F}{\sqrt{(K_{t} - m\omega ^{2}) +(\zeta \omega ^{2})}}$

Solution:

Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K

We know that:

$\omega = \frac{2 \pi\times N}{60}$

so,

$\omega = \frac{2 \pi\times 720}{60}$ = 75.39 rad/s

Using the given formula:

Damping is negligible, so, $\zeta = 0$

$\frac{A}{F}$ will give the tranfer function

Therefore,

$\frac{A}{F}$ = $\frac{1}{\sqrt{(4K - 120\ ^{2})}}$

$0.1\times 10^{-3}$ = $\frac{1}{\sqrt{(4K - 120\ ^{2})}}$

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Answer:

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dybincka [34]

Answer:

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Explanation:

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