A copper rod with initial length l_{0} is pulled by a force. The instantaneous length of the rod is given by \(l = {l_0}\left( {1 + 2{e^{4t}}} \right)\)** **where t represents time. True strain rate at t = 0 is:

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BHEL ET Mechanical Held on May 2019

Option 2 : \(\frac{8}{3}\)

CT 1: Ratio and Proportion

2536

10 Questions
16 Marks
30 Mins

__Concept:__

**The relation between true strain (ϵ _{T}) and engineering strain (ϵ) is: **

\(ϵ{_T} = \ln \left( {1 +ϵ} \right) = \;ln\left( {\frac{{{l_f}}}{{{l_i}}}} \right)\)

__Calculation:__

__Given:__

\(l_f = {l_0}\left( {1 + 2{e^{4t}}} \right)\)

\(ϵ{_T} = \ln \left( {1 +ϵ} \right) = \;ln\left( {\frac{{{l_f}}}{{{l_i}}}} \right)\)

l_{i} = l_{0}

l_{f} = final length

\(ϵ{_T} = \ln \left( {\frac{{{l_0}\left( {1 + 2{e^{4t}}} \right)}}{{{l_0}}}} \right)\)

\(ϵ{_T} = ln\left( {1 + 2{e^{4t}}} \right)\)

\(strain\;rate\;\left( {{ϵ_T}} \right) = \;\frac{{d{ϵ_T}}}{{dt}} = \frac{{2{e^{4t}}.\left( 4 \right)}}{{1 + 2{e^{4t}}}} = \frac{{8{e^{4t}}}}{{1 + 2{e^{4t}}}}\)

\({\left| {{ϵ_T}} \right|_{t = 0}} = \frac{{8{e^0}}}{{1 + 2{e^0}}} = \frac{8}{{1 + 2}} = \frac{8}{3}\)