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hw10sol.Spring07

# hw10sol.Spring07 - ME 452 Machine Design II Solution to...

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1 ME 452: Machine Design II Solution to Homework Set 10 Problem 1 (Problem 10-3 Page 543) (a) The helical compression spring has 12 total coils; i.e., 12 = t N The number of active coils for plain and ground ends, from Table 10-1, page 503, can be written as 1 = - a t N N Therefore, the number of active coils is 12 1 11 a N = - = The solid length for plain and ground ends, from Table 10-1, page 503, can be written as s t L d N = Therefore, the solid length of the compression spring is s L = 0.105 12 1.26 in × = The ultimate tensile strength for a spring material, see Equation (10-14), page 505, can be written as ut m A S d = For music wire, the intercept and the exponent, see Table 10-4, page 507, are m 201 kpsi.in A = and 0.145 m = Therefore, the ultimate tensile strength for the spring material is 0.145 201 278.7 kpsi (0.105) ut S = = The torsional yield strength of the music wire, see Table 10-6, page 508 and Example (10-1), page 509, can be written as 0.45 = sy ut S S Therefore, the torsional yield strength of the music wire is 0.45 278.7 125.4 kpsi sy S = × = The mean coil diameter of the spring, see Figure 10-1a, page 500 and Example (10-1), page 509 can be written as o D D d = -

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2 Therefore, the mean coil diameter of the spring is D 1.225 0.105 1.120 in = - = The spring index, see Equation (10-2), page 501, can be written as D C = d Therefore, the spring index is 1.120 C 10.67 0.105 = = The Bergstrasser factor, see Equation (10-6), page 501, can be written as 4 2 4 3 B C K C = - Therefore, the Bergstrasser factor is (4 10.67) 2 1.126 (4 10.67) 3 B K × + = = × - The spring force corresponding to the torsional yield stress, see Example (10-1), page 509 and see note below Equation (10-7), page 502, can be written as 3 sy B π d S F = 8 K D Note that this equation gives the force that will just yield the spring. It is a matter of user preference whether there should be a factor of safety used here (Using S sy /n s in stead of S sy ) or not. Since the problem does not specifically call for a factor of safety, none will be used. However, the book does suggest (see Equation (10-21), page 510) that as a rule of thumb, the factor of safety at the solid height should be greater than or equal to 1.2. To illustrate this recommendation, see the solution to the next problem in this homework set. So, the spring force corresponding to the torsional yield stress (with no factor of safety) is 3 3 (0.105) 125.4 10 45.2 lb 8 1.126 1.120 F π × × × = = × × The spring rate, see Equation (10-9), page 502, can be written as 4 3 a d G k = 8 D N where a N is the number of active coils. The modulus of rigidity for music wire with diameter 0.105 in d = , see Table 10-5, page 508, is 11.75 Mpsi G = Therefore, the spring rate for the helical compression spring is 4 6 3 (0.105) 11.75 10 11.55 lb/in 8 (1.120) 11 k × × = = × × The deflection caused by the above spring force, see Example (10-1), page 509, can be written as
3 F y k = Therefore, the deflection of the spring is 45.2 3.91 in 11.55 y = = The free length, see Example (10-1), page 509, can be written as 0 s L y L = + Therefore, the free length of the spring is 0 3.91 1.26 5.17 in L = + = (b) The force required to compress the spring to closure (i.e., the solid height or shut length) is the force calculated in part (a). The force required to compress the spring to closure is 45.2 lb F = (c) The spring rate of the helical spring was obtained; in part (a); i.e., the spring rate is

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hw10sol.Spring07 - ME 452 Machine Design II Solution to...

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