Answer:
The hardenability increases with increasing austenite grain size, because the grain boundary area is decreasing. This means that the sites for the nucleation of ferrite and pearlite are being reduced in number, with the result that these transformations are slowed down, and the hardenability is therefore increased.
Answer:
In the steel: 815 kPa
In the aluminum: 270 kPa
Explanation:
The steel pipe will have a section of:
A1 = π/4 * (D^2 - d^2)
A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2
The aluminum core:
A2 = π/4 * d^2
A2 = π/4 * 0.7^2 = 0.3848 m^2
The parts will have a certain stiffness:
k = E * A/l
We don't know their length, so we can consider this as stiffness per unit of length
k = E * A
For the steel pipe:
E = 210 GPa (for steel)
k1 = 210*10^9 * 0.1178 = 2.47*10^10 N
For the aluminum:
E = 70 GPa
k2 = 70*10^9 * 0.3848 = 2.69*10^10 N
Hooke's law:
Δd = f / k
Since we are using stiffness per unit of length we use stretching per unit of length:
ε = f / k
When the force is distributed between both materials will stretch the same length:
f = f1 + f2
f1 / k1 = f2/ k2
Replacing:
f1 = f - f2
(f - f2) / k1 = f2 / k2
f/k1 - f2/k1 = f2/k2
f/k1 = f2 * (1/k2 + 1/k1)
f2 = (f/k1) / (1/k2 + 1/k1)
f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KN
f1 = 200 - 104 = 96 kN
Then we calculate the stresses:
σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPa
σ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa
Answer:
a) The mechanical force is -226.2 N
b) Using the coenergy the mechanical force is -226.2 N
Explanation:
a) Energy of the system:
If i = 2A and g = 10 cm
b) Using the coenergy of the system:
A storage room
common sense lol
