A cylindrical 1040 steel rod having a minimum tensile strength of 865 MPa (125,000 psi), a ductility of at least 10%EL, and a fi
1 answer:
Answer:
procedure attached below
The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw
Explanation:
<em>Given data:</em>
Minimum tensile strength = 865 MPa
Ductility = 10%EL
Desired Final diameter = 6.0 mm
20% cold worked 7.94 mm diameter 1040 steel stock
<u> Describe the procedure you would follow to obtain this material.</u>
assuming 1040 steel experiences cracking at 40%CW
<em>attached below is a detailed procedure of obtaining the material</em>
The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw
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Answer:
It will be equivalent to 338.95 N-m
Explanation:
We have to convert 250 lb-ft to N-m
We know that 1 lb = 4.45 N
So foe converting from lb to N we have to multiply with 4.45
So 250 lb = 250×4.45 =125 N
And we know that 1 feet = 0.3048 meter
Now we have to convert 250 lb-ft to N-m
So
So 250 lb-ft = 338.95 N-m
Answer:
a. 617.958kpa
b. 1.351kg/sec
Explanation:
Please see attachment for step by step guide.
