1. Effect of preload in mechanical systems

2. Definition and examples
Basic example with springs
Conclusions
FE analysis

3. 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

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

5. Definition and examples
Basic example with springs
Conclusions
FE analysis

6. Fulfill your graphic when
this image is shown

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

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

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

10. 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

11. 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

12. 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

13. 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

14. 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

15. 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

16. 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

17. Definition and examples
Basic example with springs
Conclusions
FE analysis

18. 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

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

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

21. Example of FEA
No stress
Compression
tension 21

22. Example of FEA
da1 F0 e e
da2 22

23. Example of FEA Fe
da1 Fe F0 e e
da2 23

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

25. Example of FEA Fe
da1 Fe F0 e e
da2 25

26. Example of FEA
da1 F0 e e
da2 26

27. Example of FEA Fe
da1 F0 e e Fe
da2 27

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

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

30. 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

31. 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 = ( )*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