This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: ME 452: Machine Design II Solution to Homework Set 3. Problem 1. (Problem 6-14, page 349) A rectangular bar is cut from an AISI 1020 cold-drawn steel flat. The bar is 2.5 inches wide by 3/8 inches thick and has a 0.5 inch diameter hole drilled through the center as depicted in Table A-15-1. The bar is concentrically loaded in push-pull fatigue by axial forces F a , uniformly distributed across the width. Using a design factor of n d = 2, estimate the largest force F a that can be applied ignoring column action. SOLUTION: In this problem the load on the plate is axial, giving rise to axial normal stresses. The critical point will be at the hole where the stress concentration exists, because the axial stress pattern gives rise to a uniform nominal stress field throughout the part. The stresses at the critical point are completely reversed, meaning that they vary between the same limits both tension and compression. This stress fluctuation pattern exactly matches that of the rotating beam test specimens, so the factor of safety for parts can be found by simply comparing the magnitude of the peak stresses in the part, a , to the magnitude of the peak stresses that would fail the part in 1,000,000 cycles (infinite life), S e , . That is, the design factor (or fatigue factor of safety) is given by: e f a S n = Our task for this problem is now one of estimating the endurance limit for the part, and the stress at the critical point. For AISI 1020 cold-drawn steel from Table (A-20), page 1040 we have S ut = 68 kpsi, and S y = 57 kpsi. From Equation (6-8), page 282, for S ut &lt; 200 kpsi 0.5 e ut S S = Therefore 0.5 68 34 kpsi e S = = From Table (6-2), page 288 for cold-drawn surface finish 2.7 a = and 0.265 b = - Then from Equation (6-19), page 287, the surface condition modification factor is given by: b a ut k aS = Therefore 0.265 2.7 (68) 0.8826 a k- = = From Equation (6-21), page 288, for axial loading there is no size adjustment factor, so: 1 b k = In this problem, there is axial loading ONLY, so the load factors listed in equation (6-26 on page 290 will be used. Note that alternately the general equations (6-55) and (6-56) on page 318 can be used for any loading conditions, or combinations of loads, and they are therefore a better approach. For axial loading, the loading factor is given by Equation (6-26), page 290 0.85 c k = Since no high-temperature information is given in the problem statement, assume that the operating temperature is room temperature. Therefore the temperature factor from Table (6-4), page 291, is given by: 1 d k = Since no reliability adjustments are given in the problem statement, assume the reliability of the fatigue data curve fits is adequate, which is 50% reliability. Therefore, 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 279) :...
View Full Document
- Fall '09
- Machine Design