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Derive an expression for deflection of helical spring of circular wire.

Subject :- Machine Design -I

Topic :- Spring

Difficulty :- Medium

1 Answer
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D=Meandiameterofthespringcoil

d=diameterofthespringwire

n=Numberofactivecoils

G=modulusofrigidityforthespringmaterial

W=axialloadonthespring

C=Springindex=D/d

p=pitchofthecoils

δ=deflectionofthespringasaresultofanaxialloadW

Totalactivelengthofthewire,

l=lengthofonecoilno.ofactivecoils=πDn

Letθ=angulardeflectionofthewirewhenacteduponbythetorqueT

Axialdeflectionofthespring,

δ=θD2.....eqn.1

Also,

TJ=τD/2=GθI

θ=TIJG

J=Polarmomentofinertiaofspringwire

θ=TIGJ=WD2πDnπ32d4G....eqn.2

Substitutingthisvalueineqn.1

δ=16WD2nGd4D2=8WD3nGd4=8WC3nGd

andthestiffnessofthespringorspringrate,

Wδ=Gd48D3n=Gd8C3n=constant

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