Budynas-Nisbett:Shigley's I.Basics 1.Introduction to T©The McGraw-Hill Mechanical Engineering Mechanical Engineering Companies,2008 Design,Eighth Edition Design Mechanical Engineering Design the design factor in terms of a stress and a relevant strength.Thus Eg.(1-1)can be rewritten as loss-of-function strength S nd (1-3) allowable stress o(or t) The stress and strength terms in Eq.(1-3)must be of the same type and units.Also,the stress and strength must apply to the same critical location in the part. EXAMPLE 1-2 A rod with a cross-sectional area of A and loaded in tension with an axial force of P= 2000 lbf undergoes a stress of o=P/A.Using a material strength of 24 kpsi and a design factor of 3.0,determine the minimum diameter of a solid circular rod.Using Table A-17,select a preferred fractional diameter and determine the rod's factor of safery. Solution Since A =nd2/4.and o S/nd,then S24000P2000 0= nd 3=A=πd24 or, Answer d= 42000)31/2 =0.564in πS/ π(24000) From Table A-17,the next higher preferred size is in=0.625 in.Thus,according to the same equation developed earlier,the factor of safety n is π(24000)0.6252 Answer πSd2 n= =3.68 4P 4(2000) Thus rounding the diameter has increased the actual design factor. 1-12 Reliability In these days of greatly increasing numbers of liability lawsuits and the need to conform to regulations issued by governmental agencies such as EPA and OSHA,it is very important for the designer and the manufacturer to know the reliability of their product.The reliabil- ity method of design is one in which we obtain the distribution of stresses and the distribu- tion of strengths and then relate these two in order to achieve an acceptable success rate. The statistical measure of the probability that a mechanical element will not fail in use is called the reliability of that element.The reliability R can be expressed by a num- ber having the range 0<R 1.A reliability of R=0.90 means that there is a 90 per- cent chance that the part will perform its proper function without failure.The failure of 6 parts out of every 1000 manufactured might be considered an acceptable failure rate for a certain class of products.This represents a reliability of R=1-6 000=0.994 or 99.4 percent.Budynas−Nisbett: Shigley’s Mechanical Engineering Design, Eighth Edition I. Basics 1. Introduction to Mechanical Engineering Design 24 © The McGraw−Hill Companies, 2008 18 Mechanical Engineering Design the design factor in terms of a stress and a relevant strength. Thus Eq. (1–1) can be rewritten as nd = loss-of-function strength allowable stress = S σ(or τ ) (1–3) The stress and strength terms in Eq. (1–3) must be of the same type and units. Also, the stress and strength must apply to the same critical location in the part. EXAMPLE 1–2 A rod with a cross-sectional area of A and loaded in tension with an axial force of P 2000 lbf undergoes a stress of σ = P/A. Using a material strength of 24 kpsi and a design factor of 3.0, determine the minimum diameter of a solid circular rod. Using Table A–17, select a preferred fractional diameter and determine the rod’s factor of safety. Solution Since A = πd2 /4, and σ = S/nd , then σ = S nd = 24 000 3 = P A = 2 000 πd2/4 or, Answer d = 4Pnd πS 1/2 = 4(2000)3 π(24 000)1/2 = 0.564 in From Table A–17, the next higher preferred size is 5 8 in 0.625 in. Thus, according to the same equation developed earlier, the factor of safety n is Answer n = πSd2 4P = π(24 000)0.6252 4(2000) = 3.68 Thus rounding the diameter has increased the actual design factor. 1–12 Reliability In these days of greatly increasing numbers of liability lawsuits and the need to conform to regulations issued by governmental agencies such as EPA and OSHA, it is very important for the designer and the manufacturer to know the reliability of their product. The reliabil￾ity method of design is one in which we obtain the distribution of stresses and the distribu￾tion of strengths and then relate these two in order to achieve an acceptable success rate. The statistical measure of the probability that a mechanical element will not fail in use is called the reliability of that element. The reliability R can be expressed by a num￾ber having the range 0 ≤ R ≤ 1. A reliability of R = 0.90 means that there is a 90 per￾cent chance that the part will perform its proper function without failure. The failure of 6 parts out of every 1000 manufactured might be considered an acceptable failure rate for a certain class of products. This represents a reliability of R = 1 − 6 1000 = 0.994 or 99.4 percent.