Q7. A cylindrical rod of 1040 steel originally 15.2 mm (0.60 in.) in diameter is to be cold worked by drawing; the circular cros
s section will be maintained during deformation. A cold-worked tensile strength in excess of 840 MPa (122,000 psi) and a ductility of at least 12%EL are desired. Furthermore, the final diameter must be 10 mm (0.40 in.). Explain how this may be accomplished. (10 points)
1 answer:
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Answer:
Shearing strain will be 0.1039 radian
Explanation:
We have given change in length
Length of the pad L = 1.15 inch
We have to find the shearing strain
Shearing strain is given by
Shearing strain is always in radian so we have to change angle in radian
So
Answer:
(d) 2 pF
Explanation: the charge on capacitor is given by the expression
Q=CV
where Q=charge
C=capacitance
V=voltage across the plate of the capacitor
here we have given Q=500 pF, V=250 volt
using this formula C=
=500××
=2×
=2 pF
Answer:
1.44 mm
Explanation:
Compute the maximum allowable surface crack length using
where E is the modulus of elasticity,
is surface energy and
is tensile stress
Substituting the given values
The maximum allowable surface crack is 1.44 mm