PSI - Issue 71

Ritesh Kumar et al. / Procedia Structural Integrity 71 (2025) 364–371

366

In this analysis rotating disk is considered under a temperature field that changes along the radial direction. The thermal conduction equation without heat source is given by (Schmerr Jr., 2021): 0 r d dT K r dr dr     =         (5) After simplifying and rearranging the above heat conduction eq. (5), 2

  

dT dr B    +      

2 d T dr

A

CT D

+ =

(6)

where

r A rK =

dK

r dr = + = 0 and = 0 B K r

r

(7)

Case (I) Conditions at the boundary for Variable temperature field: It is assumed that an orthotropic annular rotating FG disk is subjected to a variable temperature field which has 100℃ at the outer radius (r = b) and 0 ℃ at the inner radius (r = a), solving the differential eq. (6) by considering the following boundary conditions: 0 0 T C = at r = a 0 100 T C = at r = b (8) Case (II) Conditions at the boundary for constant temperature field: In this case, it is assumed that the disk is subjected to uniform temperature ( =0℃ ) at each coordinate 2.2 Thermo-elastic Problem Considering FG polar orthotopic circular disk whose inner radius is ‘ a ’ and outer radius ‘ b ’ rotating at constant angular speed about its axis z which is perpendicular to the plane of the disk, as shown in Fig. 1. Equilibrium equation for a rotating disk is given by (Schmerr Jr., 2021): ( ) 2 2 0 r r d r r dr     − + = (10) In the above equation of equilibrium, a new variable (  ), called stress function is introduced to get the explicit solution. For the easiness of calculation of eq. (10) stress function is introduced here that will help to get the solution of radial stress, tangential stress, and radial displacement. Characteristics and specific properties of stress function depend on the challenges of the problem, the solution algorithm, and some assumptions taken for the solution. 0 0 T C = at r = a 0 0 T C = at r = b (9)



r 

=

(11)

r

Solving the eq. (10) and (11) we will get, 2 2 0 r d r dr     − + = Simplifying the above differential Eq (12): 2 2 r d r dr      = +

(12)

(13) By assuming the plane stress state problem, Hooke’s law for polar -orthotropic materials in plane stress state is given by * r 1 r r r r E E         = − + (14)