Question

A hollow cast-iron cylinder 4m long, 300mm outer diameter, and thickness of metal 50mm is subjected to a central load on the top when standing straight. The stress produced is 75000kN/m2. Assume Young's modulus for cast iron as 1.5x 108 kN/m^2 and find (i) magnitude of the load, (ii) longitudinal strain produced, and (iii) total decrease in length. ​

Answer 1

$\Large\bf{\color{indigo}GiVeN,}$

• A hollow cast-iron cylinder,

• Length (L) = 4 m

• Outer diameter (D) = 300 mm = 0.3m

• Thickness of metal (t) = 50 mm = 0.05m

• Stress $\bf{(\sigma)}\:=\:75000\:k.N/m^2\:$

• Young's modulus for cast iron (E) = $\bf{1.5\times{10^8}\:k.N/m^2\:}$

$\Large\bf{\color{cyan}To\:FiNd,}$

1. Magnitude of the load.
2. Longitudinal strain.
3. Total decrease in length.

$\Large\bf{\color{coral}CaLcUlAtIoN,}$

$\red\checkmark\:\:\bf{Diameter\:of\:the\:cylinder\:=\:D\:-\:2t\:}$

$:\longmapsto\:\:\bf{Diameter\:of\:the\:cylinder\:=\:0.3\:-\:(2\times{0.05})\:}$

$:\longmapsto\:\:\bf{Diameter\:of\:the\:cylinder\:=\:0.3\:-\:0.1\:}$

$:\longmapsto\:\:\bf\blue{Diameter\:of\:the\:cylinder\:=\:0.2\:m}$

(i) Magnitude of the load (Force) ;-

$\bf\pink{We\:know\:that,}$

$\red\bigstar\:\:\bf\orange{Stress\:(\sigma)\:=\:\dfrac{Force}{Area\:of\:cross\:section}\:}$

$\longmapsto\:\:\bf{Force\:=\:\sigma\times {Area\:of\:cross\:section}\:}$

$\longmapsto\:\:\bf{Force\:=\:\sigma\times{\dfrac{\pi}{4}}\:(D^2\:-\:d^2)\:}$

$\longmapsto\:\:\bf{Force\:=\:75000\times{\dfrac{\frac{22}{7}}{4}}\:\Big(\:(0.3)^2\:-\:(0.2)^2\:\Big)\:}$

$\longmapsto\:\:\bf{Force\:=\:75000\times{\dfrac{22}{28}}\:\Big(\:0.09\:-\:0.04\:\Big)\:}$

$\longmapsto\:\:\bf{Force\:=\:75000\times{\dfrac{22}{28}}\times{0.05}\:}$

$\longmapsto\:\:\bf{Force\:=\:\dfrac{82500}{28}\:}$

$\longmapsto\:\:\bf\blue{Magnitude\:of\:the\:load\:=\:2946.4\:k.N\:}$

(ii) Longitudinal Strain ;-

$\bf\red{We\:know\:that,}$

$\orange\bigstar\:\:\bf\purple{Young's\:modulus\:(E)\:=\:\dfrac{Stress\:(\sigma)}{Strain\:(e)}\:}$

$\longmapsto\:\:\bf{e\:=\:\dfrac{\sigma}{E}\:}$

$\longmapsto\:\:\bf{e\:=\:\dfrac{75000}{1.5\times{10^8}}\:}$

$\longmapsto\:\:\bf{e\:=\:\dfrac{750000}{15}\times{10^{-8}}\:}$

$\longmapsto\:\:\bf{e\:=\:50000\times{10^{-8}}\:}$

$\longmapsto\:\:\bf{\color{darkkhaki}Longitudinal\:strain\:=\:0.0005\:}$

(iii) Total decrease in length (dl) ;-

$\bf\blue{We\:know\:that,}$

$\green\bigstar\:\:\bf\pink{Strain\:(\sigma)\:=\:\dfrac{Change\:in\:length}{Original\:length}\:}$

$\longmapsto\:\:\bf{Change\:in\:length\:(dl)\:=\:Strain\times {Original\:length}\:}$

$\longmapsto\:\:\bf{Change\:in\:length\:(dl)\:=\:0.0005\times {4}\:}$

$\longmapsto\:\:\bf\orange{Change\:in\:length\:(dl)\:=\:0.002\:m\:}$

$\longmapsto\:\:\bf\green{Change\:in\:length\:(dl)\:=\:2\:mm\:}$