Hw11sol.Spring07 - ME 452 Machine Design II Solution to Homework Set 11 Problem 1 Calculate the axial load that will stress an SAE 3/8 UNF Grade 5

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1 ME 452: Machine Design II Solution to Homework Set 11 Problem 1. Calculate the axial load that will stress an SAE 3/8” UNF Grade 5 bolt to its proof strength assuming: a. Shear failure of the nut threads when one thread takes the entire load. b. Shear failure of the nut threads when all the threads equally share the load. c. Shear failure of the bolt threads when one thread takes the entire load. d. Shear failure of the bolt threads when all of the threads equally share the load. e. Tensile failure of the bolt on area A t . Solution The given bolt is specified to be an SAE 3/8” UNF grade 5 bolt. From Table 8-9, see page 418, for SAE grade number 5 and bolt diameter range 0.25 - 1.0 in: The minimum proof strength = 85 kpsi, the minimum yield strength = 92 kpsi, and the minimum tensile strength = 120 kpsi. (a) If one thread of the nut takes all the load, the shear area can be written as s o Ad w p π = The constant o w is an area reduction factor that defines the percentage of the pitch occupied by metal at the major diameter. For UNF threads, the value of o w can be taken to be 0.88 (see class notes). From Table (8-2), page 399, for 3/8” UNF threads, the number of threads per inch is given as 24 threads/in N = Therefore, the pitch of the thread, see page 396, can be written as 1 p N = Therefore, the pitch of the thread is given as 1 0.042 in/thread 24 p == Therefore, the shear area for the nut when one thread takes all the load is given as 2 3 0.88 0.042 0.0435 in 8 s A =×× × = The axial load required for shearing can be written as max s FA τ =
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2 where max τ is the shear strength of the material. The shear strength of the material can be approximated using the von-Mises criterian to be 57.7% of its proof strength. Therefore, the shear strength of the material can be estimated as max 0.577 p S = Therefore, the shear strength of the material is given as max 0.577 85 49.05 kpsi = ×= Therefore, the axial load required for shearing if one thread of the nut takes all the load is given as 3 49.05 10 0.0435 2134 lb F × = (b) When all the nut threads equally share the load, the axial load required for shear failure can be written as = the axial load for one thread
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This note was uploaded on 02/09/2012 for the course ME 352 taught by Professor Staff during the Fall '08 term at Purdue University-West Lafayette.

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Hw11sol.Spring07 - ME 452 Machine Design II Solution to Homework Set 11 Problem 1 Calculate the axial load that will stress an SAE 3/8 UNF Grade 5

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