hw11sol.Fall.2010x

hw11sol.Fall.2010x - ME 452: Machine Design II Solution to...

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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. The problem statement asks for bolt load, F b , that loads the bolt to its proof strength, S p . From Table 8-9, see page 433, for SAE grade number 5 and bolt diameter range 0.25 - 1.0 in: S p = 85 kpsi, Since this problem asks for the bolt force that fails the bolt in the threads and shank, this means the factor of safety is 1.0. (a) If one thread of the nut takes all of the load, the shear area can be written as s o A dw 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 413, for 3/8” UNF threads, the number of threads per inch is 24 threads/in N = Therefore, the pitch of the thread, see page 410 and Figure (8-1), page 411, 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 of the load is 2 3 0.88 0.042 0.0435 in 8 s A = × × × = The shear stress in the thread is
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2 max b s F A τ = where F b is the axial force in the bolt. Since the strength of the nut material is a normal strength, and the stress is a shear stress, we use the von-Mises criteria to relate them. Then from equation (5-19), page 224, and equation (5-15), page 223, 2 2 max 3 3 P S σ = = + = Thus, the bolt load that fails the nut threads in shear is 3 P s b S A F = So the bolt load that shears the nut threads when one thread takes all of the load is 85,000(0.0435) 2135 lb 3 b 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 number of threads b F × The axial load for the shear failure when one thread of the nut takes all of the load was found in part (a) of this problem to be 2135 lb. Assuming a regular hexagonal nut, the thickness of the nut is H = 21/64 inches (See Appendix A-31, page 1055). Therefore, the number of threads in the nut, N th , is th N H N = × Therefore, the number of threads in the nut is given as 21 24 7.875 64 th N = × = Therefore, the axial load required for shear failure when all the nut threads equally share the load is 2135 7.875 16,811 lb b F = × = (c) If one thread of the bolt takes all of the load, the shear area can be written as s r i A d w p
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hw11sol.Fall.2010x - ME 452: Machine Design II Solution to...

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