For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
his material elongate when a true stress of 414 MPa (60050 psi) is applied if the original length is 520 mm (20.47 in.)? Assume a value of 0.22 for the strain-hardening exponent, n.
1 answer:
Answer:
24.348mm
Explanation:
NB: I'll be attaching pictures so as to depict missing mathematical expressions or special characters which are not easily found on keyboards
K = d / €^n
Note : d represents the greek alphabet epsilion.
K = 345 / 0.02⁰.²² = 816mPa
The true strain based upon the stress of 414mPa =
€= (€/k)^1/n = (414/816)¹/⁰.²² = 0.04576
However the true relationship between true strain and length is given by
€ = ln(Li/Lo)
Making Li the subject of formula by rearranging,
Li = Lo.e^€
Li = 520e⁰.⁰⁴⁵⁷⁶
Li = 544.348mm
The amount of elongation can be calculated from
Change in L = Li - Lo = 544.348 - 520 change in L = 24.348mm.
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Work energy theorem:
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If the distance is doubled and the potential difference is halved, then
Explanation:
As can be seen from the relationship between kinetic energy and the potential difference, the distance between the plates has no effect on the relation between kinetic energy and the potential difference. Since the charge of the second particle is equal to that of the first one, the new kinetic energy would be half of the first kinetic energy.