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Flywheel Problem

Flywheel Problem. The Second Tutorial. Finding acceleration and velocity. y. r = xe x +ye y = re r e r =cos θ e x +sin θ e y. z. e θ. e r. r. ye y. v = d/dt(re r ). a=d/dt(d/dt(re r )). x. xe x. Finding acceleration and velocity. v= ω r e θ.

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Flywheel Problem

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  1. Flywheel Problem The Second Tutorial

  2. Finding acceleration and velocity y r = xex+yey = rer er=cosθex+sinθey z eθ er r yey v = d/dt(rer) a=d/dt(d/dt(rer)) x xex

  3. Finding acceleration and velocity v=ωr eθ v=d/dt(rer)=r*d/dt (cosθex+sinθey) v= r*(-sinθ*(dθ/dt) ex+cos θ*(dθ/dt)ey ) (1) Now a=dv/dt= d/dt(ωr eθ) = ωr*d/dt(eθ) • Now recall that: • eθ= (cosθex - sin θey) • ω=(dθ/dt) This becomes: a=- ω2r er Substitution into (1) gives the form: v=ωr eθ (2)

  4. 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

  5. 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

  6. 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))

  7. Stress-Strain relations Simplifying we find that: εrr=d/dθ(ur), εθθ=ur/r, εzz=d/dz (uz) εθr= (1/2)(d/dz(uθ) +r-1d/dθ(uz)-uθ/r) = 0 εθz=(1/2)(d/dz(uθ)+r-1d/dθ(uz)) = 0 εrz=(1/2)(d/dr(uz)+d/dz(ur)) = 0

  8. Stress-Strain relations Using the governing Stress-Strain relation: σij= (E/(1+ν) [εij + (v/(1-2v))(Σ εkk)δij] σθz = σrz = σrθ=0 σzz=(E/(1-v2))(εrr+vεθθ) = 0 (from assumption) σrr=(E/(1-v2))(εrr+vεθθ) σθθ=(E/(1-v2))(vεrr+εθθ)

  9. 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)

  10. 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

  11. 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

  12. Stress field • After solving for “A” and “B” substitute to • obtain final Stresses σrr=ρω2(3+v)/8[(R2)(1-(r/R)2)] σθθ=(R2ρω2(3+v)/8 )[1-(r/R)2((1+3v)/(v+3))] • When r=0 you get the maximum stress σrr=σθθ=(R2ρω2(3+v))/8

  13. Product bound Plug in max stresses to mises yield condition to obtain: σ(mises)= σ(max) σ(max)<σ(yield) ωR< 2(2σ(yield)/(ρ(3+v))).5

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