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Saladsamurai
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HELP! Stress Strain Question
In part (a) of the problem, we found that due to a certain stress, the amount of
elastic strain that a material undergoes is
[itex]\epsilon_E=.0087[/itex]
and the amount of plastic strain is
[itex]\epsilon_{pl}=.0113[/itex].
The total strain is therefore
[itex]\epsilon_T=.02[/itex]We are then told that a sample of this material with original length
[itex]l_o=610 \ mm[/itex] undergoes that same stress involved in part (a).
What is the new length [itex]l_f[/itex]after the stress is removed ?So I believe the idea behind this is that we gain the elastic portion of the strain back, but the plastic elongation should be added onto the original length.
I wrote this quantitatively as:
[tex]l_f=l_0+\epsilon_{pl}\Delta l[/tex] (1)
To find the change in length we have:
[tex]\epsilon_T=\frac{\Delta l}{l_0} \Rightarrow \Delta l=\epsilon_Tl_0[/tex] (2)
Therefore (1) becomes:
[tex]l_f=l_0+\epsilon_{pl}(\epsilon_Tl_0)[/tex]
[tex]\Rightarrow l_f=l_0(1+\epsilon_T\epsilon_{pl})[/tex]
Plugging in numbers we have lf=.6101 mm
but the correct answer is .6167 which is waayyy off.
What am I missing here?
Homework Statement
In part (a) of the problem, we found that due to a certain stress, the amount of
elastic strain that a material undergoes is
[itex]\epsilon_E=.0087[/itex]
and the amount of plastic strain is
[itex]\epsilon_{pl}=.0113[/itex].
The total strain is therefore
[itex]\epsilon_T=.02[/itex]We are then told that a sample of this material with original length
[itex]l_o=610 \ mm[/itex] undergoes that same stress involved in part (a).
What is the new length [itex]l_f[/itex]after the stress is removed ?So I believe the idea behind this is that we gain the elastic portion of the strain back, but the plastic elongation should be added onto the original length.
I wrote this quantitatively as:
[tex]l_f=l_0+\epsilon_{pl}\Delta l[/tex] (1)
To find the change in length we have:
[tex]\epsilon_T=\frac{\Delta l}{l_0} \Rightarrow \Delta l=\epsilon_Tl_0[/tex] (2)
Therefore (1) becomes:
[tex]l_f=l_0+\epsilon_{pl}(\epsilon_Tl_0)[/tex]
[tex]\Rightarrow l_f=l_0(1+\epsilon_T\epsilon_{pl})[/tex]
Plugging in numbers we have lf=.6101 mm
but the correct answer is .6167 which is waayyy off.
What am I missing here?
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