Answer:
In ferrous metal iron present but on the other hand in the non ferrous material iron does not present.That is why there is a different heat treatment process for ferrous and nonferrous materials.
Ferrous materials contains iron is the main constitute.Like steel ,cast iron ,wrought iron .Steel and cast iron are the alloy element of iron ans carbon.Wrought iron is the purest from of iron.
Heat treatment process for ferrous materials :
1.Normalizing
2.Annealing
3.Quenching
4.Surface hardening
Heat treatment process for non ferrous materials :
Mostly annealing process is used for non ferrous materials.After annealing non ferrous will become soft.
When two metal plates are joined then they form a bimetallic structure.The bimetallic structure is used to find the relationship of thermal temperature and the mechanical displacement.
The use of bimetallic structure -In clock ,thermometers ,engines.
Explanation:
yes it has the answers to all repairs
<h2>Answer:</h2>
<h3>Required Answer is as follows :-</h3>
- 3/4 = (⅓ × m₁/Volume × V²)/(⅓ × m₂/Volume) × (V/2)²
- ¾ = m₁/Volume × (V)²/m₂/Volume × (V/2)²
- ¾ = m₁ × (V)²/m₂ × (V²/4)
<h3>Now,</h3>
- ∆m = m₂ - m₁
- ∆m = 16m₂/3 - m₁ = 13m₁/3
- Ratio = (13m₁/3)/ m₁
- Ratio = 13/3
Ratio = 13:3
<h3>Know More :-</h3>
Mass => It is used to measure it's resistance to its acceleration. SI unit of mass is Kg.
Answer:
a. 30°
b. 0.9MPa
Explanation:
The slip will occur along that direction for which the Schmid factor is maximum. The three possible slip directions are mentioned as 30°, 48°, 78°
The cosines for the possible λ values are given as
For 30°, cos 30 = 0.867
For 48°, cos 48 = 0.67
For 78°, cos 78 = 0.21
Among the three-calculated cosine values, the largest cos(λ) gives the favored slip direction
The maximum value of Schmid factor is 0.87. Thus, the most favored slip direction is 30° with the tensile axis.
The plastic deformation begins at a tensile stress of 2.5MPa. Also, the value of the angle between the slip plane normal and the tensile axis is mentioned as 65°
Thus, calculate the value of critical resolved shear stress for zinc:
From the expression for Schmid’s law:
τ = σ*cos(Φ)*cos(λ)
Substituting 2.5MPa for σ, 30° for λ and 65° for Φ
We obtain The critical resolved shear stress for zinc, τ = 0.9 MPa
