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:
(a) water height =408.66 in.
(b) mercury height=30.04 in.
Explanation:
Given: P=14.769 psi ( 1 psi= 6894.76 )
we know that
where
h=height.
Given that P=14.769 psi ⇒P= 101828.6 7
(a)
⇒101828.67=
=10.38 m
So water barometer will read 408.66 in. (1 m=39.37 in)
(b)
=13600
So 101828.67=
=0.763 m
So mercury barometer will read 30.04 in.