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nirvana33 [79]
3 years ago
5

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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There have been many attempts to manufacture and market plastic bicycles. All have been too flexible and soft. Which design-limi
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Answer:

The design-limiting property that insufficiently large is the elastic modulus (Young modulus)

Explanation:

Plastic usually have a relatively low elastic modulus, this couses the material to deform too much under stress. In the case of a bicycle, a little weight you put on it or little bumps will cause the bicycle to deform too much.

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What are the three elementary parts of a vibrating system?
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Answer:

the three part are mass, spring, damping

Explanation:

vibrating system consist of three elementary system namely

1) Mass - it is a rigid body due to which system experience vibration and kinetic energy due to vibration is directly proportional to velocity of the body.

2) Spring -  the part that has elasticity and help to hold mass

3) Damping - this part considered to have zero mass and  zero elasticity.

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Problem 4.079 SI A rigid tank whose volume is 3 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large
salantis [7]

Answer:

Q_{cv} = -1007.86kJ

Explanation:

Our values are,

State 1

V=3m^3\\P_1=1bar\\T_1 = 295K

We know moreover for the tables A-15 that

u_1 = 210.49kJ/kg\\h_i = 295.17kJkg

State 2

P_2 =6bar\\T_2 = 296K\\T_f = 320K

For tables we know at T=320K

u_2 = 228.42kJ/kg

We need to use the ideal gas equation to estimate the mass, so

m_1 = \frac{p_1V}{RT_1}

m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}

m_1 = 3.54kg

Using now for the final mass:

m_2 = \frac{p_2V}{RT_2}

m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}

m_2 = 19.59kg

We only need to apply a energy balance equation:

Q_{cv}+m_ih_i = m_2u_2-m_1u_1

Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i

Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)

Q_{cv} = -1007.86kJ

The negative value indidicates heat ransfer from the system

7 0
3 years ago
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