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hw10sol.Fall.2010x

# hw10sol.Fall.2010x - 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 561) A helical compression spring is wound using 2.5 mm diameter music wire. The spring has an outside diameter of 31 mm with plain ground ends, and 14 total coils. (a) What should the free length be to ensure that when the spring is compressed solid the torsional stress does not exceed the yield strength, that is, that it is solid-safe? (b) What force is needed to compress this spring to closure? (c) Estimate the spring rate. (d) Is there a possibility that the spring might buckle in service? SOLUTION: (a) The helical compression spring has 14 total coils; i.e., N t = 14 The number of active coils for plain and ground ends, from Table 10-1, page 521, can be written as 1 = - a t N N Therefore, the number of active coils is 14 1 13 a N = - = The solid length for plain and ground ends, from Table 10-1, page 521, can be written as s t L d N = Therefore, the solid length of the compression spring is s L = 2.5 14 35 mm × = The ultimate tensile strength for a spring material, see Equation (10-14), page 523, can be written as ut m A S d = For music wire, the intercept and the exponent, see Table 10-4, page 525, are m 2211 MPa.mm A = and 0.145 m = Therefore, the ultimate tensile strength for the spring material is 0.145 2211 1936 MPa (2.5) ut S = = No setting operation was discussed in the problem statements, so we will assume none was performed on the spring. Therefore, the torsional yield strength of the music wire, see Table 10-6, page 526 and Example (10-1), page 527, can be written as

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2 0.45 = sy ut S S Therefore, the torsional yield strength of the music wire is 0.45 1936 871.2 MPa sy S = × = The mean coil diameter of the spring, see Figure 10-1a, page 518 and Example (10-1), page 527 can be written as o D D d = - Therefore, the mean coil diameter of the spring is D 31 2.5 28.5 mm = - = The spring index, see Equation (10-1), page 519, can be written as D C = d Therefore, the spring index is 28.5 C 11.4 2.5 = = The Bergstrasser factor, see Equation (10-5), page 519, is 4 2 4 3 B C K C = - Therefore, the Bergstrasser factor is (4 11.4) 2 1.117 (4 11.4) 3 B K × + = = × - The spring force corresponding to the torsional yield stress, see Example (10-1), page 527 and Equation (10-7), page 520, 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 instead 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 528) 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 (2.5) 871.2 167.9 N 8 1.117 28.5 F π × × = = × × The spring rate, see Equation (10-9), page 520, can be written as 4 3 a d G k = 8 D N
3 where N a is the number of active coils. The modulus of rigidity for music wire with diameter d = 2.5 mm (0.0984 in), see Table 10-5, page 526, is 81 GPa = 81,000 MPa G =

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