Belleville Spring DesignTechniques-Logic Diagram and Example

The logic diagram below is intended as a tool for belleville spring design. Hopefully it will be a guide and provide some inspiration for the designer. It is a template for one of many belleville spring design techniques, outlining the step-by-step procedure. No single logic diagram is the answer for all belleville spring problems. Hopefully this will help get you going in the right direction.

This logic diagram shows the basic steps and procedures which will determine if the belleville washer can be made to meet the specifications. The design that is derived is only an initial design and may not be the most economical one. What also needs to be considered are the manufacturing techniques required, the quantities involved, the tolerances needed, etc.

LINK 1: Load/deflection characteristics in belleville spring washers

LINK 2: Load vs. compressive stress for belleville spring washers

LINK 3: See chart "Design Stresses for Static Service" on this page

Belleville Spring Design Example

Problem: In a clutch, it is required to exert a minimum pressure of 204 lb. This pressure needs to be kept as constant as possible while the facing of the clutch wears down -

Belleville Spring Design Procedure

1. P₁=1.1 x 204=225 lb. Assume OD/ID=2. From LINK 1 choose load/deflection curve which gives approximately constant load, about 50% and 100% of deflection to flat. Choose curve with h/t=1.41

2. From LINK 1 the percent load at 50% deflection is 88%.

3. Calculate load at flat, P_F=225/0.88=255 lb.

4. Using LINK 2 estimated stress at flat is 200,000 psi.

5. Using the chart Design Stresses for Static Service, found on this page, yield point is 120% of tensile. From the chart showing tensile strength to hardness of spring steel found on this page the tensile will be 240,000 psi, with a hardness of Rc 48. Yield point without residual stress equals 1.2x240,000, 0r 288,000 psi .^.. S_max.

7. Calculate stock thickness, t.

bellevile spring washer

8. Calculate h and H. h=1.41t=1.41x0.054=0.076 in. H=h+t=0.076+0.054=0.130 in.

9. Referring to LINK 1 the load of 225 lb. will be reached at f₁=50% of maximum available deflection. f₁=0.5x0.076=0.038 in. deflection, or the load of 225 lb. will be reached at H₁=0.130-0.038=0.092 in. height. To allow for wear the spring should be preloaded at H₂=0.092-0.032=0.061 in. height.

This preload corresponds to a deflection f₂=0.130-0.061=0.069 in., which is 0.069/0.076 x 100=91% of h.

10. Because 91% exceeds the recommended 85%, then

11. Increase deflection range to 45% to 100%. From LINK 1 the percent load at 45% deflection is 0.835 and

P_F= 225/0.835 = 270 lb.

18. Repeating procedures 4, 5, 7, 8, and 9, and finding that

f₂/h x 100=85%

the calculated design values are-

OD=3 in.

ID=1.5 in.

t=0.055 in.

h=0.077 in.

f₁=0.034 in.

f₂=0.066 in.

P_F=270 lb.

S=200.000 psi

END OF DESIGN

*NOTE: THIS PROBLEM USES ONLY COMPRESSIVE STRESSES.

Check back for updated information related to Belleville Spring Design. Our team is constantly searching for items that will benefit you.

Got some info you want to share? Send it our way by going to our contact page.

Don't miss any updated information! Stay in touch by subscribing to Spring Makers Resource e-zine

Return from Belleville Spring Design to Belleville Spring Washers
Return to spring-makers-resource