Effect of preload in mechanical systems

Definition and examples Basic example with springs Conclusions FE analysis

Such a system is hyperstatic Principle A mechanical system is under preload when an internal load exists inside the system, having no external loads. Such a system is hyperstatic Studying the internal stress requires to set the balance of loads the balance of displacements the relationship between loads and displacements for each preloaded component 3

Angular bearings Bolt assemblies Main industrial uses 01 02 4 Bâti Arbre Bâti Bolt assemblies 4

Definition and examples Basic example with springs Conclusions FE analysis

Fulfill your graphic when this image is shown

Without preload: L = L01+L02 Example with springs Without preload: L = L01+L02 Fe Initial equilibrium Position L01 L02 7

Load-length relation: Example with springs Without preload: L = L01+L02 Fe Load-length relation: Fe = K2 da = F2 F2 da Fe abruption 8

Load-length relation: Example with springs Without preload: L = L01+L02 Fe Load-length relation: Fe = K1 da = -F1 -F1 da Fe abruption 9

Equilibrium of the washer Example with springs With preload: L = L01+L02 - e Fe Preload length F2 e Equilibrium of the washer F1 + F2 = 0 Thus : Abs(F1) = Abs(F2) Preload length F1 ? e 10

With preload: L = L01+L02 - e Abs(F2) Abs(F1) e F0 da02 da01 Example with springs With preload: L = L01+L02 - e Abs(F1) Preload length Abs(F2) e Preload force F0 da02 da01 Equations: F1 - F2 = 0 da01 + da02 = e F1 = K1 da01 F2 = K2 da02 e 11

Equilibrium of displacement Example with springs Equations: F1 - F2 = 0 da01 + da02 = e F1 = K1 da01 F2 = K2 da02 Load balance Equilibrium of displacement Load-displacement relation for each component

With preload and positiveFe Example with springs With preload and positiveFe FALSE! Abs(F1) Abs(F2) F0 F2 F1 F2 = F0+ Fe F1 = F0 - Fe No load balance Fe + F1 - F2 ≠ 0 No displacement equilibrium Fe e 13

With preload and positiveFe Example with springs With preload and positiveFe Abs(F1) Abs(F2) Fe + F1 - F2 = 0 Thus Fe = F2 - F1 e F2 F1 da2 da1 Equations: Fe + F1 - F2 = 0 da1 + da2 = e F1 = K1 da1 F2 = K2 da2 da = da2 - da02 da > 0 Fe e 14

With preload and negativeFe Example with springs With preload and negativeFe Abs(F1) Abs(F2) Fe + F1 - F2 = 0 D’où Fe = F2 - F1 e F2 F1 da2 da1 Equations: Fe + F1 - F2 = 0 da1 + da2 = e F1 = K1 da1 F2 = K2 da2 da = da2 - da02 da < 0 Fe e 15

Rigidity increased in the preloaded zone Example with springs Study of the global sttifness Abs(F1) Abs(F2) Rigidity increased in the preloaded zone Fe da 16

Definition and examples Basic example with springs Conclusions FE analysis

Preload => hyperstaticsystem Conclusions Preload => hyperstaticsystem No clearance Stiffness increased (A.2 page 1) Load balance Equilibrium of displacements Load-displacements relations (A.3 page 1) 18

Definition and examples Basic example with springs Conclusions FE analysis Back to the preliminary exercise

Example of FEA No preload preloaded e e/2 F0 e 20

Example of FEA No stress Compression tension 21

Example of FEA da1 F0 e e da2 22

Example of FEA Fe da1 Fe F0 e e da2 23

Example of FEA Fe da1 = 0 Fe F0 e e da2 24

Example of FEA Fe da1 Fe F0 e e da2 25

Example of FEA da1 F0 e e da2 26

Example of FEA Fe da1 F0 e e Fe da2 27

Example of FEA Fe F0 e e da1 Fe da2 = 0 28

Definition and examples Basic example with springs Conclusions FE analysis Back to the preliminary exercise

Initial equilibrium (middle) 2 compression springs : free length 39 mm ; Rate 0,5 KgF/mm A Washer is in contact between the 2 springs Available axial space : 66 mm (including the washer) Determine the required axial force to obtain an axial displacement of the washer of 5 mm 10 mm Initial equilibrium (middle) 30

With preload: L = L01+L02 - e Abs(F2) Abs(F1) 12 mm F0 e Example with springs With preload: L = L01+L02 - e In Preloaded area F = (K1 + K2)*d F = (0.5 + 0.5)*5 F = 5 kgF Abs(F1) Abs(F2) 12 mm F0 Out of Preloaded area F = K2* (d0 + d) F = (0.5)*(6 + 10) F = 8 kgF 5 10 e 31