Homework 10 Springs. Problem 17.23

Transcription

1 Homework 10 Springs Problem 17.2 An overflow vlve, shown in sketch b, hs piston imeter of 15 mm n slit length of 5 mm. The spring hs men coil imeter D 10 mm n wire imeter 2 mm. The vlve shoul open t 1 br (0.1 MP) pressure n be totlly open t br (0. MP) pressure when the spring is fully compresse. Clculte the number of ctive coils, the free length, n the pitch of the spring. The sher moulus for the spring mteril G 80 GP. The spring ens re squre n groun. Determine the mximum sher stress for this geometry. otes: The number of ctive coils is foun from Eq. (17.15). Equtions from Tble 17. re use for soli length, free length, pitch, n number of totl coils. The mximum sher stress is clculte from Eq. (17.10). The piston re is When the pressure is one br (0.1 MP), the vlve strts to open. The force ssocite with this pressure is ( m2)(0.1 MP) When the pressure is br (0. MP), the force is 5.0, n the spring eflects further 5 mm m. ote tht from Eq. (17.7), C D/ 10/2 5. Therefore, from Eq , If you use Eq. (17.17), you will get 22.19, more ccurte number. The stiffness of the spring, from Eq. (17.18) is Therefore, the spring From Tble 17., for squre n groun ens,

2 We know tht t 5.0, the spring is t the soli length of m. When this lo is remove, the eflection is δ P/k 5 /7060 /m m. Therefore the free length is Lf Ls + s ( m) m m, or 56.7 mm. From Tble 17., the pitch is The mximum sher stress is obtine from Eq. (17.11) fter the curvture sher correction fctor is clculte s Kw (4C 1)/(4C 4) /C 1.1, Problem A compression spring me of music wire is use for sttic loing. Wire imeter 1.5 mm, coil outsie imeter Do 12.1 mm, n there re eight ctive coils. Also, ssume tht the spring ens re squre n groun. Fin the following: () Spring rte n soli length (b) Gretest lo tht cn be pplie without cusing permnent set in excess of 2% (c) Spring free length with lo etermine in prt b tht cuses the spring to be soli () Whether buckling is problem. If it is, recommen wht you woul chnge in the reesign. otes: The spring rte is obtine from Eq. (17.18), n other equtions re tken from Tble 17.. The strength of the wire is obtine from Eq. (17.2) n t from Tble Whether or not buckling is problem is ssesse from Fig ote from Fig. 17.2(b), tht the spring imeter D D mm m. Therefore, C is, from Eq. 17.7, C D/ 10.6/ From Tble 17.1, G 79. GP. The spring rte is obtine from Eq. (17.18): From Tble 17., using squre n groun ens, the soli length is Ls t ( + 2) 1.5mm (8 + 2) 15mm From Tble 17.2 for music wire, Ap 2170 MP n m Therefore, from Eq. (17.2), S G (79.GP)(0.0015m) k C 1 + 8(7.067) (8) C A p ut m MP / m From Eq. (17.), Ssy 0.4Sut 0.4(2045 MP) MP

3 The trnsverse sher fctor is K (C + 0.5)/C 7.56/ , so tht Eq. (17.8) gives π τ P 8DK mx π S 8DK The eflection cuse by this lo is δ P k sy π (0.0015m) (2045.MP) 8(0.0106m)(1.071) m 18.2mm / m 95.5 Therefore, the free length is lf ls + δ 15 mm mm. mm. For this spring, δ/lf 18./. 0.55, n lf/d./ otice from Fig tht buckling is not problem for this spring. Problem 17.6 An 18-mm men imeter helicl compression spring hs 20 (ctive) coils, hs 2-mm wire imeter, n is me of chromium vnium. Determine the following: () The mximum lo-crrying cpcity for sfety fctor of 1.5 guring ginst yieling (b) The mximum eflection of the spring (c) The free length for squre n groun ens otes: The spring wire imeter is foun from the stress requirements, n Eq. (17.8) is use for the sher stress. The strength of the wire is obtine from Eq. (17.2) n t from Tble The solution ssumes tht 22 is the number of ctive coils, rther thn totl coils. Chrome vnium refers to the lloys for steel, so use the stiffness for steel wire, or G 11.5 Mpsi 79. GP. From Tble 17.2 for chrome vnium wire, Ap 2000 MP n m Therefore the strength of the wire in sher is, from Eqs. (17.2) n (17.), From (17.7), C D/ 18 mm/2 mm 9. The trnsverse sher fctor is K 9.5/ Therefore, from Eq. (17.8), the mximum sher stress is: Since the sfety fctor is 1.5, From Eq. (17.18),

4 From Tble 17. for squre n groun ens, Problem ls ( + 2) (0.002 m)(22) m lf ls + δt m m m. A helicl torsion spring, shown in sketch k, is me from hr-rwn steel with wire imeter of 2.2 mm n 8.5 turns. Dimensions re in millimeters. () Using sfety fctor of 2, fin the mximum force n the corresponing ngulr isplcement. (b) Wht woul the coil insie imeter be when the mximum lo is pplie? (c) We re skipping prt c. otes: The force is obtine from Eq. (17.40) fter the llowble stress in the wire is etermine. The ngulr isplcement is clculte from Eq. (17.4) once the number of ctive coils is foun from Eq. (17.45). The insie imeter uner mximum lo is obtine from Eq. (17.46). From Tble 17.1, for hr rwn wire, G 11.5 Mpsi 79. GP, n tke E 207 GP s for crbon steel from the insie front cover. From Tble 17.2 for hr rwn wire, Ap 1750 MP, m Therefore, the ultimte strength of the wire is obtine from Eq. (17.2) s: The llowble sher stress is clculte from Eq. (17.) s: Also, for this spring, from Eq. (17.7), C D/ 22 mm/2.2 mm 10, so from Eq. (17.40), Recognizing tht M P(0.05 m), Eq. (17.9) gives

5 From Eq. (17.44), b + e b l + l2 0.05m m πd π (0.022m) So tht from Eq. (17.42), θ rev 10.18MD E 10.18(12.5 )(0.05m)(0.022m)(8.84) 9 4 ( P)(0.0022m) 4 Therefore, uner the mximum lo, the number of coils is 0 +θrev From Eq. (17.45), 0.178