4.4.1 The Rotating Disc - Engineering

1 Section solid mechanics Part II Kelly Rotating Discs The Rotating disc Consider a thin disc Rotating with constant angular velocity , Fig. Material particles are subjected to a centripetal acceleration 2 rar . The subscript r indicates an acceleration in the radial direction and the minus sign indicates that the particles are accelerating towards the centre of the disc . Figure : the Rotating disc The accelerations lead to an inertial force (per unit volume) 2 rFa which in turn leads to stresses in the disc .

2 The inertial force is an axisymmetric loading and so this is an axisymmetric problem. The axisymmetric equation of equilibrium is given by Adding in the acceleration term gives the corresponding equation of motion: 21 rrrrrrr , ( ) This equation can be expressed as 01 rrrrrbrr , ( ) where 2 rbr . Thus the dynamic Rotating disc problem has been converted into an equivalent static problem of a disc subjected to a known body force. Note that, in a general dynamic problem, and unlike here, one does not know what the accelerations are they have to be found as part of the solution procedure.

3 Using the strain-displacement relations and the plane stress Hooke s law then leads to the differential equation 22222111 rEurdrdurdrud ( ) This is Eqn. with a non-homogeneous term. The solution is derived in the Appendix to this section, : 2 rSection solid mechanics Part II Kelly 79232211811 rErCrCu ( ) As in , let 1/2 ECA and 12/1 ECC, and the full general solution is, using and , { Problem 1} 3222222222222222218112111811211183121131 812138121rCrrAEurCrAErCrAErCrArCrArrrr ( ) which reduce to when 0.

4 A solid disc For a solid disc , A in must be zero to ensure finite stresses and strains at 0 r. C is then obtained from the boundary condition 0)( brr , where b is the disc radius: 223161,0bCA ( ) The stresses and displacements are 22222222231183)(33183)(83)(rbrErurbrrbrr r ( ) Note that the displacement is zero at the disc centre, as it must be, but the strains (and hence stresses) do not have to be, and are not, zero there. Dimensionless stress and displacement are plotted in Fig.

5 For the case of . The maximum stress occurs at 0 r, where 2283)0()0(brr ( ) Section solid mechanics Part II Kelly 80 The disc expands by an amount 3241)(bEbu ( ) Figure : stresses and displacements in the solid Rotating disc A Hollow disc The boundary conditions for the hollow disc are 0)(,0)( barrrr ( ) where a and b are the inner and outer radii respectively.

6 It follows from that 2222223161,381baCbaA ( ) and the stresses and displacement are 2222222222222222222221131183)(33183)(83) (rbarbarErurbarbarrbarbarrr ( ) 2238b ubE32138 u rr Section solid mechanics Part II Kelly 81 which reduce to when 0 a. Dimensionless stress and displacement are plotted in Fig. for the case of and ba. The maximum stress occurs at the inner surface, where 222/31143)0(bab ( ) which is approximately twice the solid - disc maximum stress.

7 Figure : stresses and displacements in the hollow Rotating disc Problems 1. Derive the full solution equations for the thin Rotating disc , from the displacement solution Appendix: Solution to Eqn. As in , transform Eqn. using ter into 232221 teEudtud ( ) 1 0 1 2 br/ 2238b ubE32138 urr Section solid mechanics Part II Kelly 82 The homogeneous solution is given by Assume a particular solution of the form tpAeu3 which, from , gives tpeEu322181 ( ) Adding together the homogeneous and particular solutions and transforming back to r s then gives