Problem 1 «A steel alloy has the following tensile properties and behavior in the plastic region:o Engineering strain, £g rs = 0.10 at the engineering tensile strength, TS; = 510 MPae...

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Answered Same DayDec 17, 2022

Answer To: Problem 1 «A steel alloy has the following tensile properties and behavior in the plastic...

Dr Shweta answered on Dec 17 2022
48 Votes
Ans 1 Actual strength of alloy is calculated as:
Given: Engineering strain ϵ = 0.10
Engineering t
ensile strength = TS = σ = 510Mpa
Strain-hardening exponent n = 0.11
Now, firstly we’ll calculate strength coefficient K
K = σ/ ϵn -------1
Putting Values in equation 1 we get,
K = 510 *10^6/0.100.11
on solving we get, 657.2Mpa
Now, at an engineering strain ϵ = 0.05, the true strength of alloy is
σ = K ϵn --------2
on solving we get,
σ = 472.67Mpa.
Sketch of stress-strain curve
Stress
Ans 2. Part 1 Angle between direction of tensile stress and direction of slip is calculated as:
As we know that
σ =τcrss/m ------[1]
Given: σ = 2.3 times of τcrss
So, on solving equation 1 we get,
m = 0.435
now as we know that
m = cosφcosλ
here φ = angle angle between the normal of...
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