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:
2.77mpa
Explanation:
compressive strength = 20 MPa. We are to find the estimated flexure strength
We calculate the estimated flexural strength R as
R = 0.62√fc
Where fc is the compressive strength and it is in Mpa
When we substitute 20 for gc
Flexure strength is
0.62x√20
= 0.62x4.472
= 2.77Mpa
The estimated flexure strength is therefore 2.77Mpa
Answer:
A range from 5,000 miles to 7,500 miles, on average. Some recommended intervals may be shorter or longer.
Hazard is a possible source of danger
Answer:
a) -8 lb / ft^3
b) -70.4 lb / ft^3
c) 54.4 lb / ft^3
Explanation:
Given:
- Diameter of pipe D = 12 in
- Shear stress t = 2.0 lb/ft^2
- y = 62.4 lb / ft^3
Find pressure gradient dP / dx when:
a) x is in horizontal flow direction
b) Vertical flow up
c) vertical flow down
Solution:
- dP / dx as function of shear stress and radial distance r:
(dP - y*L*sin(Q))/ L = 2*t / r
dP / L - y*sin(Q) = 2*t / r
Where dP / L = - dP/dx,
dP / dx = -2*t / r - y*sin(Q)
Where r = D /2 ,
dP / dx = -4*t / D - y*sin(Q)
a) Horizontal Pipe Q = 0
Hence, dP / dx = -4*2 / 1 - 62.4*sin(0)
dP / dx = -8 + 0
dP/dx = -8 lb / ft^3
b) Vertical pipe flow up Q = pi/2
Hence, dP / dx = -4*2 / 1 - 62.4*sin(pi/2)
dP / dx = 8 - 62.4
dP/dx = -70.4 lb / ft^3
c) Vertical flow down Q = -pi/2
Hence, dP / dx = -4*2 / 1 - 62.4*sin(-pi/2)
dP / dx = -8 + 62.4
dP/dx = 54.4 lb / ft^3