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hw4.sol.Fall.2010x

hw4.sol.Fall.2010x - ME 452 Machine Design II Solution to...

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ME 452: Machine Design II Solution to Homework Set 4. Moment arm should be 142 mm not 138 in this solution…… Problem 1. (Problem 6-29, page 351) The figure is a drawing of a 4- by 20-mm latch spring. A preload is obtained during the assembly by shimming under the bolts to obtain an estimated initial deflection of 2 mm(on the end of the spring). The latching operation itself requires an additional deflection of 4 mm. The material is ground high-carbon steel, bent then hardened and tempered to a minimum hardness of 490 Bhn. The radius of the bend is 4 mm. Estimate the yield strength to be 90% of the ultimate strength. (a) Find the maximum and minimum latching forces. (b) Is it likely the spring will achieve infinite life? SOLUTION: The solution of this problem will require the conversion of deflection information into load information so that stresses can be estimated. In addition, the sharp bend at the corner of the latch will introduce curvature effects in the bending stresses there, which have probably not been presented in previous courses on strength of materials. Since the beam is a cantilever loaded on the end, the critical plane will be at the wall where the bending moments are maximum. The beam is uniform thickness, with uniform material properties, and the curvature effects will occur where the bending moment is maximum (at the wall). So the critical points will be the top and bottom fibers of the beam in the plan at the wall. Consider the material properties first.

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The minimum ultimate tensile strength can be estimated using Equation (2-21), page 41, as follows: 3.4 3.4 490 1666 MPa ut B S H = = × = The problem statement suggests that the yield strength is 90 percent of the ultimate strength. Therefore 0.9 0.9 1666 1499.4 MPa y ut S S = = × = Because 1400 MPa ut S , then from Equation (6-8), page 282 we get: 700 MPa e S = From Table (6-2), page 288, for ground surface finish we have: 1.58 a = and 0.085 b = - Therefore from Equation (6-19), page 287, the surface condition modification factor is given by: 0.085 1.58 (1666) 0.841 b a ut k aS - = = × = Since A 95 for this part is not the same as a rotating beam test specimen, we need to adjust for equivalent diameter. For the given cross-section, the equivalent diameter is given by Equation (6-25), page 289 as: 1/ 2 0.808( ) e d hb = Therefore 1/2 0.808(4 20) 7.227 mm e d = × = Then the size factor is given by Equation (6-20), page 288 as: 0.107 0.107 1.24 1.24 (7.227) 1.003 b k d - - = = × = Note that we usually don’t use Marin factors greater than 1.0. In this case, since the calculation yields a Marin factor greater than 1.0., to be conservative we will use 1.00 b k = The loading factor from Equation (6-26), page 290 is given by (also see page 317 and 318 for discussion on combined loading): 1 c k =
Assuming that the operating temperature is 70 F ° the temperature factor from Table (6- 4), page 291, is given by: 1 d k = Assuming 50% reliability, the reliability factor from Table (6-5), page 293 is given by: 1 e k = Assuming there are no miscellaneous effects (see Pages 293 and 294), the miscellaneous- effects factor is given by: 1 f k = The endurance strength is then given by the Marin equation (see Equation 6-18, page 287) : e a b c d e f e S k k k k k k S = Then 0.841 1.00

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hw4.sol.Fall.2010x - ME 452 Machine Design II Solution to...

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