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

DISC SPRING EXAMPLES Machinery's Handbook, 31st Edition

363

9) Compressive stress σ 0 at point 0 due to maximum deflection, Equation (14): 3 π 0 σ = – 4 E · t · s · K 4 (1 – µ 2 ) · K 1 · D 2 a 4 · 206000 · 4 · 1.32 · 1 (1 – 0.3 2 ) · 0.6841 · 77.78 2 · = 3 π – · 2 = 160000 psi Because the compressive stress at point 0 does not exceed 1600 N/mm 2 , its current value satisfies the design requirement. 10) Tensile stress σ 2 at point 2 due to minimum deflection s = 0.9 mm, Equation (16): 0 σ = 1103 N/mm

s 2

` 4 E · K 4 · s · K 3 · t – K 2 · K 4 · h – (1 – µ 2 ) · K 1 · D 8

B

j

=

min 2 σ =

/ N 654 mm 2 2 a 4 · 206000 · 1 · 0.9 1.3589 · 4 – 1.2086 · 1 · 2.2 – (1 – 0.3 2 ) · 0.6841 · 77.78 2 11) Tensile stress σ 2 at point 2 due to maximum deflection s = 1.32 mm, Equation (16): = a k : D . 2 09

s 2

` 4 E · K 4 · s · K 3 · t – K 2 · K 4 · h – (1 – µ 2 ) · K 1 · D 8

B

j

=

max 2 σ =

a 2 a 4 · 206000 · 1 · 1.32 · 1.3589 · 4 – 1.2086) · 1 · 2.2 – (1 – 0.3 2 ) · 0.6841 · 77.78 2 Thus, σ 2 min = 654 N/mm 2 (94,850 psi) and σ 2 max = 1032 N/mm : . 2 132

D

k

/ N 1032 mm 2

=

2 (149,700 psi). 12) Tensile stress σ 3 at point 3 due to minimum deflection s = 0.9 mm, Equation (17):

s 2

+ K 3 · t

` 4 E · K 4 · s · K 4 · (2 K 3 – K 2 ) · h – (1 – µ 2 ) · K 1 · D 2 a · δ 8

B

j

=

min 3 σ =

. 2 09

+ 1.3589 · 4

: 4 · 206000 · 1 · 0.9 · 1 · (2 · 1.3589 – 1.2086) · 2.2 –

D

a

k

/ N 815 mm 2

=

(1 – 0.3 2 ) · 0.6841 · 77.78 2 · 1.95122 13) Tensile stress σ 3 at point 3 due to maximum deflection s = 1.32 mm, Equation (17):

s 2

+ K 3 · t

` 4 E · K 4 · s · K 4 · (2 K 3 – K 2 ) · h – (1 – µ 2 ) · K 1 · D 2 a · δ 8

B

j

=

max 3 σ =

. 2 132

+ 1.3589 · 4

: 4 · 206000 · 1 · 1.32 · 1 · (2 · 1.3589 – 1.2086) · 2.2 –

D

a

k

/ N 1149 mm 2

=

(1 – 0.3 2 ) · 0.6841 · 77.78 2 · 1.95122

Thus, σ 3 min = 815 N/mm

2 (118,200 psi) and σ

3 max = 1149 N/mm

2 (166,600 psi).

14) Functional tensile stress range at critical points 2 and 3. Point 2: σ 2 max − 0.5 σ 2 min = 1032 − 0.5 3 654 = 705 N/mm 2 Point 3: σ 3 max − 0.5 σ 3 min = 1149 − 0.5 3 815 = 741.5 N/mm 2 Because σ 3 max − 0.5 σ 3 min > σ 2 max − 0.5 σ 2 min , the tensile stresses at point 3 are used for fatigue life calculations.

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