Flywheel Problem

The Second Tutorial

Finding acceleration and velocity

xex

yey

r

r = xex+yey = rer

er=cosθex+sinθey

ereθ

x

y

z

v = d/dt(rer)

a=d/dt(d/dt(rer))

Finding acceleration and velocity

v= r*(-sinθ*(dθ/dt) ex+cos θ*(dθ/dt)ey ) (1)

v=d/dt(rer)=r*d/dt (cosθex+sinθey)

Now recall that:

• eθ= (cosθex - sin θey)

• ω=(dθ/dt)

Substitution into (1) gives the form:

v=ωr eθ

Now a=dv/dt= d/dt(ωr eθ ) = ωr*d/dt(eθ)

(2)

v=ωr eθ

This becomes: a=- ω2r er

Kinetic Energy

K=(1/2)I*ω2 where I = ∫r2dm I= ∫ ρ r2dV = ∫∫∫ρ r2 dtdrdθ = (2πρtR4)/4

dθ=2πrdr ∫dt=t, thickness

Substitution gives:

K=(1/4) πρtR4ω2

Material consideration

Yield- At high speeds centrifugal forces cause tensile strength Material needs to be designed resistant to these

stresses Young’s modulus and Possion’s ratio must be

accounted for in design

Stress-Strain relations

• Assuming plane stress t<<R

• Use Strain-Displacement relationships given:ur=u(r), uz, uθ=rωt

εrr=d/dθ(ur), εθθ=(1/r)d/dθ(uθ) + ur/r, εzz=d/dz (uz)

εθr= (1/2)(d/dz(uθ) +r-1d/dθ(uz)-uθ/r)εθz=(1/2)(d/dz(uθ)+r-1d/dθ(uz))εrz=(1/2)(d/dr(uz)+d/dz(ur))

Stress-Strain relationsSimplifying we find that:

εθr= (1/2)(d/dz(uθ) +r-1d/dθ(uz)-uθ/r) = 0εθz=(1/2)(d/dz(uθ)+r-1d/dθ(uz)) = 0εrz=(1/2)(d/dr(uz)+d/dz(ur)) = 0

εrr=d/dθ(ur), εθθ=ur/r, εzz=d/dz (uz)

Stress-Strain relations

σθz = σrz = σrθ = 0

Using the governing Stress-Strain relation:

σij= (E/(1+ν) [εij + (v/(1-2v))(Σ εkk)δij]

σzz=(E/(1-v2))(εrr+vεθθ) = 0 (from assumption)

σθθ=(E/(1-v2))(vεrr+εθθ)

σrr=(E/(1-v2))(εrr+vεθθ)

Equation of Motion for ur

Since looking for ur use the following Equation of Motion:

d/dr(σrr)+r-1d/dθ(σrθ)+d/dz(σrz)+(1/r)(σrr- σθθ)=ρd2/dt2(ur)

(ignores body force ρbr)

Using the stresses derived earlier we can simplify to the following:

d/dr(σrr)+(1/r)(σrr- σθθ)=ρd2/dt2(ur)= -ρω2r

d2/dr2(ur)+(1/r)(d/dr(ur))- (ur)/r2= -[ρω2r(1-v2)]/E (3)

Equation of Motion for ur

d2/dr2(ur)+(1/r)(d/dr(ur))- (ur)/r2= -[ρω2r(1-v2)]/E (3)

This equation will solve for ur. But first re-write using reverse chain rule:

d/dr[(1/r)(d/dr(urr))]= -[ρω2r(1-v2)]/E

Integrating yields the solution:

ur=Ar+Br-1-[((1-v2)/8E)ρω2r3

Stress field Using the stress-strain and the strain-

displacement relations derived earlier

σrr=(E/(1-v2))(εrr+vεθθ) σθθ=(E/(1-v2))(vεrr+εθθ)

εrr=d/dθ(ur), εθθ=ur/r, εzz=d/dz (uz)

Plug into new equation for ur setting boundary conditions:

σr(R)=0 (no mass rotating and pulling on outside)ur(0)=0 no displacement in the r direction at center

Stress field After solving for “A” and “B” substitute to obtain final Stresses

σrr=ρω2(3+v)/8[(R2)(1-(r/R)2)]σθθ=(R2 ρω2(3+v)/8 )[1-(r/R)2((1+3v)/(v+3))]

When r=0 you get the maximum stress

σrr=σθθ=(R2 ρω2(3+v))/8

Product bound

Plug in max stresses to mises yield condition toobtain:

σ(mises)= σ(max)

σ(max)<σ(yield)

ωR< 2(2σ(yield)/(ρ(3+v))).5