Subject :- Machine Design -I

Topic :- Spring

Difficulty :- Medium

$D=Mean \hspace{1mm} diameter \hspace{1mm} of \hspace{1mm}the\hspace{1mm} spring \hspace{1mm}coil $

$d=diameter \hspace{1mm}of \hspace{1mm}the\hspace{1mm} spring \hspace{1mm}wire$

$n=Number \hspace{1mm} of\hspace{1mm} active\hspace{1mm} coils$

$G=modulus\hspace{1mm} of \hspace{1mm}rigidity\hspace{1mm} for\hspace{1mm} the \hspace{1mm}spring\hspace{1mm} material$

$W=axial\hspace{1mm} load\hspace{1mm} on \hspace{1mm}the\hspace{1mm} spring $

$C=Spring\hspace{1mm} index=D/d$

$p=pitch\hspace{1mm} of\hspace{1mm} the\hspace{1mm} coils$

$\delta=deflection \hspace{1mm}of\hspace{1mm} the\hspace{1mm} spring \hspace{1mm}as\hspace{1mm} a \hspace{1mm}result\hspace{1mm} of\hspace{1mm} an \hspace{1mm}axial \hspace{1mm}load\hspace{1mm} W$

$Total \hspace{1mm}active \hspace{1mm}length\hspace{1mm} of \hspace{1mm} the \hspace{1mm} wire,$

$l=length \hspace{1mm}of\hspace{1mm}one\hspace{1mm} coil*no. \hspace{1mm} of \hspace{1mm} active\hspace{1mm} coils=\pi Dn$

$Let \hspace{1mm}\theta=angular\hspace{1mm} deflection\hspace{1mm} of\hspace{1mm} the \hspace{1mm}wire\hspace{1mm} when \hspace{1mm}acted \hspace{1mm}upon\hspace{1mm} by\hspace{1mm} the\hspace{1mm} torque \hspace{1mm}T$

$Axial \hspace{1mm} deflection \hspace{1mm}of \hspace{1mm}the \hspace{1mm}spring,$

$\delta=\theta*\frac{D}{2} ..... eqn .1$

$Also, $

$\frac{T}{J}=\frac{\tau}{D/2}=\frac{G\theta}{I}$

$\theta=\frac{TI}{JG}$

$J=Polar \hspace{1mm}moment \hspace{1mm} of \hspace{1mm}inertia\hspace{1mm}of\hspace{1mm}spring\hspace{1mm}wire$

$\theta=\frac{TI}{GJ}=\frac{W*\frac{D}{2}*\pi Dn}{\frac{\pi}{32}*d^4G} ....eqn .2$

$Substituting\hspace{1mm}this\hspace{1mm}value\hspace{1mm}in\hspace{1mm} eqn .1$

$\delta= \frac{16WD^2n}{Gd^4}*\frac{D}{2}=\frac{8WD^3n}{Gd^4}=\frac{8WC^3n}{Gd} $

$and\hspace{1mm}the\hspace{1mm}stiffness\hspace{1mm}of\hspace{1mm}the\hspace{1mm}spring\hspace{1mm}or\hspace{1mm}spring\hspace{1mm}rate,$

$\frac{W}{\delta}=\frac{Gd^4}{8D^3n}=\frac{Gd}{8C^3n}=constant$