(Part A) Machinerys Handbook 31st Edition Pages 1-1484

Machinery's Handbook, 31st Edition

362 DISC SPRING EXAMPLES 3) The number of springs in the stack is found from Equation (1): n = L 0 / ( t + h ) = 40/(4 + 2.2) = 40 ∕ 6.2 = 6.45. Rounding n to the nearest integer gives n = 6. The actual length of unloaded spring stack is L 0 = 6.2 3 6 = 37.2 mm (1.465 inch) and it satisfies the L 0 < 40 mm condition. 4) Calculating the cone angle α from Equation (7) and actual outside diameter D a from Equation (9) gives: . . . . . . ( . ) atan atan sin atan cos atan D D 80 41 2 22 # 011282 64 = h 80 2 4 # 011282 6 1 + h @ 011282 7778 mm 3062 in a a ° α = − = = − = a a ^ ^ ^ k h k @ 6 6

5) Calculating constant K 1 from Equation (11): . 195122 δ = =

d D

. 195122 195122 1 . −

2

a

k

. 06841

K

=

=

. . 195122 1 195122 1

. 195122 2

1

− + −

π

;

E

·

ln

h

^

6) Calculating minimum and maximum forces, F min and F max from Equation (10): Based on the design requirements, the disc spring stack is deflecting by 5.5 mm (0.217 in) under minimum load, and each individual disc spring is deflecting by 5.5 / 6 @ 0.92 mm (0.036 in). A single disc spring deflection s min = 0.9 mm (0.035 in) is used to cal- culate F min . Under maximum load, the disc spring stack is permitted maximum deflection of 8 mm (0.315 in), and each individual disc spring deflects by 8 / 6 @ 1.33 mm (0.0524 in). A disc spring deflection s max = 1.32 mm (0.052 in) will be used to calculate F max . If disc springs are made of AISI 6150 alloy steel, then modulus of elasticity E = 206,000 N/mm 2 (30 3 10 6 psi) and Poisson’s ratio μ = 0.3. . . . . . . . . ( ) . . . . . . . . ( ) F F N F F N 1 03 06841 7778 4 206000 22 2 09 22 09 4 409 14390 3235 1 03 06841 7778 4 206000 22 2 132 22 132 4 4 132 20050 4510 lbf lbf min min max 2 2 3 2 2 3 = − − − + = = − − − + = ^ ^ ^ ^ ^ ^ a a ^ ^ h h h h h h k k h h : : D D · · · · · ·

max

7) Calculating constant K 2 , Equation (18): . 41 80 195122 d D δ = = =

. 195122 6 195122 195122 1 1 . . ln ln c ^ π − − ^ h h

ln π δ δ δ ln 6 1 1 − − a ·

m

k

. 12086

K

=

=

=

2

· 8) Calculating constant K 3 (Equation (19)):

3

1

. 195122 3 1 95122 1 1 3589 . . ln π − = ^ ^ h h · ·

h

ln π δ δ − ^ · ·

K

=

=

3

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