1. Cooling Pipes: Force Analysis Thermal forces Disc deflections Manufacturing tolerance forces Glue joint analysis Friction forces

2. Thermal forces Temperature decrease 40 K radial force [N] tangential force [N] torque [Nmm] primaryinner 0.10.0518 middle 0.10.234 outer 0.1 22 secondarymiddle 0.73.80 outer 0.73.80

3. Thermal Deflections Maximum deflection of the disc is 31  m Thermal deflection of the cooling pipe between two cooling blocks if one end would be free:  L = L ·  ·  T = 65 · 15 ·10 -6 · 40 = 39  m A manufacturing tolerance of this value would result in the same forces

4. Manufacturing tolerance Analysis of pieces of cooling pipe between two adjacent blocks. One end gets a prescribed deflection, other end: –all six DOFs fixed –all DOFs fixed except one rotational DOF Random effect, the average disc deflection is zero, deflections because of standard deviation Stresses in the glue layer between insert and disc surface

5. Manufacturing tolerance forces One end displaced 0.1 mm in tangential direction all 6 DOFs fixed1 rot DOF free rad [N] tang [N] torque [Nmm] rad [N] tang [N] primaryinner 01.635.700.6 middle 01.129.500.4 outer 01.937.600.7 secondarymiddle 0.317.814704.6 outer 0.19.594.302.4

6. Actual situation Holes in cooling blocks have diameter 2.2 mm, screws to attach cooling blocks to inserts are M1.6 This means that the clearance between screw and hole is between 0 and 0.6 mm Below this tolerance, the rotation can be set free. In a worst case situation the rotation has to be fixed if the manufacturing tolerance > 0.3 mm  2.2  1.6

7. Forces Manufacturing Tolerance [mm] 0.10.20.30.5 highest force [N] 4.69.213.849.4 highest torque [Nmm] ---294 (Highest forces and torques occur at the inserts of the secondary middle cooling blocks)

8. Disc Deflection Mean total force = 0, because of the random direction of the manufacturing tolerances, assume a normal distribution of the forces Standard Deviation of the total tangential force  total = F ·  n Per insert: 4.2 ·  total /n = 4.2F/  n (at Z = 4.2, the chance of exceeding this value is 0.1% in 18 discs) Take mean tangential force of the secondary cooling blocks and the number n as the overall amount of secondary cooling blocks

9. Disc Deflection n = 92Manufacturing tolerance [mm] 0.10.20.30.40.5 force per insert [N] 1.53.14.610.616.6 disc deflection [  m] 24.248.472.6167262 Same deflection as thermal deflection (31.2  m) will be reached at a manufacturing tolerance of 130  m Maximum deflection of disc (200  m) will be reached at a manufacturing tolerance of about 400  m

10. Analytical glue joint analysis glue with Filling in all properties:  max = 0.4 MPa (with 0.5 mm manufacturing tolerance) Force

11. Analytical glue joint analysis glue Torque with Filling in all properties:  max = 5.7 MPa (with 0.5 mm manufacturing tolerance)

12. Considerations The calculated values for the shear stress are heavily depending on the boundaries of the glue joint By making a well-finished glue joint, stresses can be brought down Tests have to be done on the glue joint to get a good feeling of the glue strength

13. Numerical glue joint analysis 19.8 MPa

14. Numerical glue joint analysis 8.9 MPa

15. Numerical glue joint analysis 11.3 MPa

16. Prestressed screws Cooling blocks are attached to insert with M1.6 Max. prestressing force F s =  max · A If  max = 150 MPa and A = 1.27 mm², then F s = 191 N per screw Friction force F f = 2·F s ·  = 2·191·0.1 = 38.2 N If manufacturing tolerance is 0.4 mm, then the force and the torque together give a force of 31.6 + 147/22.8 = 38.0 N

17. Concluding remarks Keeping the manufacturing tolerance at 0.3 mm will be safe because of the low forces and zero moments If 0.3 mm cannot be reached, 0.4 mm is the absolute maximum with respect to disc deflections Glue joint analyses give very different results, because of the uncertatinties of the boundary geometry 0.4 mm is a limit if there wouldn’t be a positioning pin between cooling block and insert; the positioning pin can take a lot of the friction force away