Question

A bar of cross section area 200 mm² is axially pulled by a force P kN. If maximum stress induced in the bar is 30 MPa, determine P if elongation of 1.2 mm is observed over a guage length 3 m. Determine youngs modulus.​

Answer 1

$\large \underline \bold{Given}$:-

$\sf Cross \: section \: Area \: = \: 200 \: mm^{2} \: = \: 200\times 10^{-6} \: m^{2}$

$\: \: \: \: \: \:$$\sf pulling \: force \: = \: P \: kN \: = \: P\times 10^{3} \: N$

$\: \: \: \: \: \:$$\sf maximum \: stress \: = \: 30 \: MPa \: = \: 30\times 10^{6} \: Pa$

$\: \: \: \: \: \:$$\sf length \: of \: bar \: = \: 3 \: m$

$\sf elongation \: in \: length \: = \: 1.2 \: mm \: = \: 1.2\times 10^{-3} \: m$

$\large \underline \bold{To \: Find}$:-

$\sf determine \: \: P \: \: and \: \: also \: \: young \: \: modulus \: ?$

$\large \underline \bold{Usable \: Formula}$:-

1) $\qquad\qquad$$\large\boxed{\sf\pink{P \: = \: \dfrac{F}{A}}}$

2) $\qquad\qquad$$\large\boxed{\sf\pink{\gamma \: = \: \dfrac{\sigma}{\epsilon}}}$

$\sf where-$

$\qquad$$\sf F \: = \: pulling \: force$

$\qquad$$\sf P \: = \: inducing \: stress$

$\qquad$$\sf A \: = \: Cross \: Section \: Area$

$\qquad$$\sf \gamma \: = \: young \: \: modulus$

$\qquad$$\sf \sigma \: = \: longitudinal \: \: stress$

$\qquad$$\sf \epsilon \: = \: longitudinal \: \: de \: formity$

$\large \underline \bold{Solution}$:-

$\sf On \: using \: the \: above \: formula-$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$$\sf F \: = \: PA$

$\: \: \: \: \: \: \: \:$$\sf (P\times 10^{3}) \: = \: (30\times 10^{6})\times (2\times 10^{-4})$

$\: \: \: \: \: \:$$\sf (P\times 10^{3}) \: = \: (3\times 10^{7})\times (2\times 10^{-4})$

$\: \: \: \: \:$$\sf (P\times \cancel{10^{3}}) \: = \: (6\times \cancel{10^{3}})$

$\: \: \: \: \: \: \: \: \: \: \:$$\large\boxed{\sf\pink{P \: = \: 6}}$

$\sf So,$

$\qquad$$\sf F \: = \: 6\times 10^{3} \: N$

$\sf Now -$

$\qquad$$\sf \gamma \: = \: \dfrac{(\dfrac{F}{A})}{(\dfrac{\delta L}{L})}$

$\qquad$$\sf \gamma \: = \: \dfrac{FL}{A \: \delta L}$

$\: \: \: \: \: \: \: \: \:$$\sf \gamma \: = \: \dfrac{(\cancel{6}\times 10^{3})\times 3}{(\cancel{2}\times 10^{-4})\times (1.2\times 10^{-3})}$

$\qquad$$\sf \gamma \: = \: \dfrac{(3\times 10^{3})\times 3}{1.2\times 10^{-7}}$

$\qquad$$\sf \gamma \: = \: \dfrac{9\times 10^{3}}{1.2\times 10^{-7}}$

$\qquad$$\sf \gamma \: = \: \dfrac{3\times 10^{3}}{0.4\times 10^{-7}}$

$\qquad$$\sf \gamma \: = \: \dfrac{30\times 10^{3+7}}{4}$

$\qquad$$\sf \gamma \: = \: \dfrac{15\times 10^{10}}{2}$

$\: \: \: \: \: \:$$\large\boxed{\sf\pink{\gamma \: = \: 7.5\times 10^{10} \: N/m^{2}}}$

Answer 2

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⠀⠀⠀⠀⠀$\Large{\boxed{\underline{\overline{\mathfrak{\star \: AnSwer :- \: \star}}}}}$

$\large\underline{\tt{\red{ Formula}}}$

$\large{\tt\orange{P \: = \: \dfrac{F}{A}}}$

$\large{\tt\orange{\gamma \: = \: \dfrac{\sigma}{\epsilon}}}$

$\tt where$

$\tt F \: = \: pulling \: force$

$\tt P \: = \: inducing \: stress$

$\tt A \: = \: Cross \: Section \: Area$

$\tt \gamma \: = \: young \: \: modulus$

$\tt \sigma \: = \: longitudinal \: \: stress$

$\tt \epsilon \: = \: longitudinal \: \: de \: formity$

$\large \underline{\tt{\red{Solution}}}$

$\tt F \: = \: PA$

$\implies \tt (P\times 10^{3}) \: = \: (30\times 10^{6})\times (2\times 10^{-4})$

$\implies \tt (P\times 10^{3}) \: = \: (3\times 10^{7})\times (2\times 10^{-4})$

$\implies \tt (P\times \cancel{10^{3}}) \: = \: (6\times \cancel{10^{3}})$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\large\boxed{\tt\red{P \: = \: 6}}$

$\tt So,$

$\qquad \tt F \: = \: 6\times 10^{3} \: N$

Now−

$\tt \gamma \: = \: \dfrac{(\dfrac{F}{A})}{(\dfrac{\delta L}{L})}$

$\implies \tt \gamma \: = \: \dfrac{FL}{A \: \delta L}$

$\tt \gamma \: = \: \dfrac{(\cancel{6}\times 10^{3})\times 3}{(\cancel{2}\times 10^{-4})\times (1.2\times 10^{-3})}$

$\implies \tt \gamma \: = \: \dfrac{(3\times 10^{3})\times 3}{1.2\times 10^{-7}}$

$\implies \tt \gamma \: = \: \dfrac{9\times 10^{3}}{1.2\times 10^{-7}}$

$\implies \tt \gamma \: = \: \dfrac{3\times 10^{3}}{0.4\times 10^{-7}}$

$\implies \tt \gamma \: = \: \dfrac{30\times 10^{3+7}}{4}$

$\implies \tt \gamma \: = \: \dfrac{15\times 10^{10}}{2}$

$\qquad \qquad \large\underline{\tt\red{\gamma \: = \: 7.5\times 10^{10} \: N/m^{2}}}$

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