- 2 A Spring Has A Free Length Of 1 497 In An Outside Diameter Of 0 430 In And A Wire Diameter Of 0 040 In Both Ends 1 (20.93 KiB) Viewed 52 times
- 2 A Spring Has A Free Length Of 1 497 In An Outside Diameter Of 0 430 In And A Wire Diameter Of 0 040 In Both Ends 2 (104.58 KiB) Viewed 52 times
- 2 A Spring Has A Free Length Of 1 497 In An Outside Diameter Of 0 430 In And A Wire Diameter Of 0 040 In Both Ends 3 (102.94 KiB) Viewed 52 times
2) A spring has a free length of 1.497 in., an outside diameter of 0.430 in., and a wire diameter of 0.040 in, both ends plain and pinned. Would the spring tend to buckle when compressed to 1.0 in.? Compute the pitch angle if the number of coils = 10. LP/DM
Solution Given- Step)) free length LF = 4.22 in D = 0.560 in d = 0.059m 18 d Dm = Dad : 0.501 0.059 Cet sicle elia Wire- olia Compress No. of Spring index Mean رل No Diel L = 4 in Step (2) Pitch where Na - no. So, P C Pitch emple Dm 8:491 P = L-2d Na of active coil Na = 18 Nt-2 18-22 16 4.22 9x0.059 6.2563 = temt IP F Dm +tent (-0.2563 TIX0.501 Step (3) 1 0.501 16 Step (4)
9.251 O KB Ans Step(5) for Buckling Critical Buckling load wa = for spring wire, property load wa= kx kB x L F K = ( Gid | Spring stiffness. 8³ Na by = 80 N/mm² modules 1.4946 sipaling 2= 0·059 x 25.4mm = 3 80 xto x 1.4986 = 1.529 N/mm. 8 x (8.4911³ x 16 Buckling facter depend on ration satio (LF) 4.22 남 2 = 7.53 0.56 from the chart KB = 0.045 (for hinged. end spring) So Wer = 1.529 x 0.045 x(4.22x 25.4 7.375 N "Where 1 in = 25.4mm] e f ks E When = 4.22-4 = 0.22 in F = 1.529 X (8₁22x25·4) = 8.544 N