Machinery's Handbook, 31st Edition
328
Spring Design
75 to 85 Percent of Inside Diameter
2 Pounds (Initial Tension) 9 Pounds ( P )
” 3 4
1 ( CL ) ” 7 16
( F ) ” 5 8
2 (to Inside Diameter of Hooks) ” 1 16
Fig. 17. Extension Spring Design Example
Step 3: The number of active coils (say 6) . . AC f F
. 01062 0625 586 = = =
This result should be reduced by 1 to allow for deflection of 2 hooks (see notes 1 and 2 that follow these calculations.) Therefore, a quick answer is: 5 coils of 0.0625 inch diameter wire. However, the design procedure should be completed by carrying out the following steps: Step 4: The body length = ( TC + 1) 3 d = (5 + 1) 3 0.0625 = 3 ∕ 8 inch. Step 5: The length from the body to inside hook . . . . 2 1 4375 0 375 0 531 0531 inch = − = − = 0625 0531 = 85 percent = This length is satisfactory, see Note 3 following this procedure. Step 6: . . d OD 1 00625 075 1 11 The spring index = − = − = Step 7: The initial tension stress is , , S P S IT 7 it = This stress is satisfactory, as checked against curve in Fig. 16. Step 8: The curvature correction factor K = 1.12 (Fig. 13). . . FL Body ID 2 Percentageof ID = 50 200 2 14 340pounds per square inch # # = = Step 9: The total stress = (50,200 + 14,340) 3 1.12 = 72.285 pounds per square inch This result is less than 106,250 pounds per square inch permitted by the middle curve for 0.0625 inch wire in Fig. 3 and therefore is a safe working stress that permits some addi tional deflection that is usually necessary for assembly purposes.
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