上海交通大学：《Design & Manufacturing II and Project》课程教学资源（讲义）Lecture 7 Springs

ME350 Mechanical Design and Manufacturing ll Springs

ME 350 Mechanical Design and Manufacturing II Springs

Basics Springs are elastic members which exert >Forces >Torques ·And absorb energy Springs are designed to provide a >Push >Pull,or >Twist It is often more economical to use stock spring

Basics • Springs are elastic members which exert ¾Forces ¾Torques • And absorb energy • Springs are designed to provide a ¾Push ¾Pull, or ¾Twist • It is often more economical to use stock spring

Types ·Torsion bar springs ·Vire springs >Helical springs =Round =Square or rectangular Generally not recommended ·Flat springs >Cantilever >Elliptical >Spring washers =Belleville springs Special-shaped springs

Types • Torsion bar springs • Wire springs ¾Helical springs =Round =Square or rectangular ™Generally not recommended • Flat springs ¾Cantilever ¾Elliptical ¾Spring washers =Belleville springs • Special-shaped springs

Types standard- variable pitch barrel constant rate variable rate hourglass conical (a)Helical compression springs.Push-wide load and deflection range-round or rectangular wire.Standard has constant coil diameter,pitch,and rate.Barrel,hourglass,and variable-pitch springs are used to minimize resonant surging and vibration.Conical springs can be made with minimum solid height and with constant or increasing rate. -a (b)Helical extension springs.Pull-wide (c)Drawbar springs.Pul/-uses comp- (d)Torsion springs.Twist- load and deflection range-round ression spring and drawbars to provide round or rectangular or rectangular wire,constant rate. extension pull with fail-safe,positive stop wire-constant rate

Types

Types Belleville wave slotted finger curved e) Spring washers.Push-Belleville has high loads and low deflections-choice of rates(constant,increasing,or decreasing) Wave has light loads,low deflection,uses limited radial space.Slotted has higher deflections than Belleville. Finger is used for axial loading of bearings.Curved is used to absorb axial end play. Volute spring.Push- (g)Beam springs.Push or Pull- (h)Power or motor springs. (i)Constant Force may have an inherently wide load but low deflection Twist-exerts torque over Pull-long high friction damping. range-rectangular or shaped many turns.Shown in,and deflection at low cantilever,or simply supported removed from,retainer. or zero rate

Types

Torsion Bar Springs Probably the simplest form of spring Common applications >Automotive suspension springs >Counterbalancing springs for car hoods and trunk lids >Doors Fixed end Bearirng Spline Torsion bar portion Generous radius d Bearing Spline (a) (b) Torsion bar with splined ends Rod with bent ends serving as torsion bar spring (type used in auto suspensions,etc.) (type used for auto hood and trunk counterbalancing.etc.l

Torsion Bar Springs • Probably the simplest form of spring • Common applications ¾Automotive suspension springs ¾Counterbalancing springs for car hoods and trunk lids ¾Doors

Torsion Bar Springs Shear stress T 16T T三 πd3 ·Angular deflection TL 32TL 0= 二 JG πd4G ·Spring rate JG πd4G k= L 32L ·T=F*D/2=torque ·r=wire radius ·J=πd4/32=polar moment of inertia A=wire cross-sectional area

Torsion Bar Springs • Shear stress 3 Tr T 16 J d τ π = = • Angular deflection 4 TL TL 32 JG dG θ π = = • Spring rate 4 32 JG dG k L L π = = • T = F ∗D/2 = torque • r = wire radius • J = π d 4/32 = polar moment of inertia • A = wire cross-sectional area

Stresses in Coil Springs Consider spring shown below and free body diagram of cut portion (b (a)

Stresses in Coil Springs • Consider spring shown below and free body diagram of cut portion

Stresses in Springs o Removed portion exerts force F and torsion T on remaining part Maximum stress in wire is: Tr F 十 max 士 A ·T=F*D/2=torque ·r=wire radius ·J=πd/32=polar moment of inertia A=wire cross-sectional area

Stresses in Springs • Removed portion exerts force F and torsion T on remaining part • Maximum stress in wire is: max Tr F J A τ =± + • T = F ∗D/2 = torque • r = wire radius • J = π d 4/32 = polar moment of inertia • A = wire cross-sectional area

Stresses in Springs ·Thus 8FD 4F max πd3 nd2 Positive sign indicates stress on inside fiber of spring. ·D=spring diameter ·d=wire diameter

Stresses in Springs • Thus max 3 2 8 4 FD F d d τ π π = + • D = spring diameter • d = wire diameter • Positive sign indicates stress on inside fiber of spring