It is important because now a days we all need help from engineers
Answer:
See explanation below
Explanation:
Hypo-eutectoid steel has less than 0,8% of C in its composition.
It is composed by pearlite and α-ferrite, whereas Hyper-eutectoid steel has between 0.8% and 2% of C, composed by pearlite and cementite.
Ferrite has a higher tensile strength than cementite but cementite is harder.
Considering that hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains whereas hypereutectoid steel contains a higher amount of cementite, the following properties are obtainable:
Hypo-eutectoid steel has higher yield strength than Hyper-eutectoid steel
Hypo-eutectoid steel is more ductile than Hyper-eutectoid steel
Hyper-eutectoid steel is harder than Hyper-eutectoid steel
Hypo-eutectoid steel has more tensile strength than Hyper-eutectoid steel.
When making a knife or axe blade, I would choose Hyper-eutectoid steel alloy because
1. It is harder
2. It has low cost
3. It is lighter
When making a die to press powders or stamp a softer metals, I will choose hypo-eutectoid steel alloy because
1. It is ductile
2. It has high tensile strength
3. It is durable
Answer:
The percentage of the remaining alloy would become solid is 20%
Explanation:
Melting point of Cu = 1085°C
Melting point of Ni = 1455°C
At 1200°C, there is a 30% liquid and 70% solid, the weight percentage of Ni in alloy is the same that percentage of solid, then, that weight percentage is 70%.
The Ni-Cu alloy with 60% Ni and 40% Cu, and if we have the temperature of alloy > temperature of Ni > temperature of Cu, we have the follow:
60% Ni (liquid) and 40% Cu (liquid) at temperature of alloy
At solid phase with a temperature of alloy and 50% solid Cu and 50% liquid Ni, we have the follow:
40% Cu + 10% Ni in liquid phase and 50% of Ni is in solid phase.
The percentage of remaining alloy in solid is equal to
Solid = (10/50) * 100 = 20%
Answer:
i) 796.18 N/mm^2
ii) 1111.11 N/mm^2
Explanation:
Initial diameter ( D ) = 12 mm
Gage Length = 50 mm
maximum load ( P ) = 90 KN
Fractures at = 70 KN
minimum diameter at fracture = 10mm
<u>Calculate the engineering stress at Maximum load and the True fracture stress</u>
<em>i) Engineering stress at maximum load = P/ A </em>
= P / 
= 90 * 10^3 / ( 3.14 * 12^2 ) / 4
= 90,000 / 113.04 = 796.18 N/mm^2
<em>ii) True Fracture stress = P/A </em>
= 90 * 10^3 / ( 3.24 * 10^2) / 4
= 90000 / 81 = 1111.11 N/mm^2
Pretty sure it’s Cenozoic
