HW 7 Solution

HW 7 Solution - P avg 5.24 kW = P avg T avg ϖ ⋅ =...

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Unformatted text preview: P avg 5.24 kW = P avg T avg ϖ ⋅ = Average power P min kW = P min T min ϖ ⋅ = Minimum power P max 10.5 kW = P max T max ϖ ⋅ = Maximum power T avg 1000 N m ⋅ = T avg T max T min + 2 = Average torque Use equations (9.1) to calculate the power values. Solution: See Mathcad file P0916a. T max 2000 N ⋅ m ⋅ = Maximum torque T min N ⋅ m ⋅ = Minimum torque ϖ 5.236 rad sec = ϖ 50 rpm ⋅ = Shaft speed Given: N newton = rpm 2 π ⋅ rad ⋅ min 1- ⋅ = Units: Statement: What are the maximum, minimum, and average power values for the shaft shown in Fig P9-2 for the data in row a of Table P9-1 if the shaft speed is 50 rpm? PROBLEM 9-16a M a 48.45 N ⋅ m ⋅ = 2. Calculate the mean and alternating components of torque. T m T max T min + 2 = T m 1000 N m ⋅ = T a T max T min- 2 = T a 1000 N m ⋅ = 3. Calculate the unmodified endurance limit. S' e 0.5 S ut ⋅ = S' e 372.5 MPa = 4. Determine the endurance limit modification factors for a rotating round shaft. Load C load 1 = Size C size d ( ) 1.189 d mm 0.097- ⋅ = Surface A 4.51 = b' 0.265- = (machined) C surf A S ut MPa b' ⋅ = C surf 0.782 = Temperature C temp 1 = Reliability C reliab 0.814 = ( R = 99%) 5. Determine the modified endurance limit as a function of the unknown diameter, d . PROBLEM 9-2a Statement: A simply supported shaft is shown in Figure P9-2. A constant magnitude distributed load p is applied as the shaft rotates subject to a time-varying torque that varies from T min to T max . For the data in row a of Table P9-1, find the diameter of shaft required to obtain a safety factor of 2 in fatigue loading if the shaft is stee S ut = 745 MPa and S y = 427 MPa. The dimensions are in cm, the distributed force in N/cm, and the torque is in N-m. Assume no stress concentrations are present. Units: N newton = MPa 10 6 Pa ⋅ = Given: Distance between bearings L 20 cm ⋅ = Distance to p a 16 cm ⋅ = Distance to end of p b 18 cm ⋅ = Tensile strength S ut 745 MPa ⋅ = Applied distributed load p 1000 N ⋅ cm 1- ⋅ = Yield strength S y 427 MPa ⋅ = Minimum torque T min N ⋅ m ⋅ = Design safety factor N fd 2 = Maximum torque T max 2000 N ⋅ m ⋅ = Assumptions: The finish is machined, reliability is 99%, and the shaft is at room temperature. Solution: See Figure P9-2 and Mathcad file P0902a. 1. The maximum moment in the shaft occurs between a and b . See the appendix to this problem below for the determination of M a . S x z , ( ) if x z ≥ 1 , , ( ) = 6. For a Mathcad solution, define a step function S. This function will have a value of zero when x is less than z , and a value of one when it is greater than or equal to z . x m ⋅ 0.005 L ⋅ , L .. = 5. Define the range for x R 2 1700 N = R 2 p b a- ( ) ⋅ R 1- = R 1 300 N = R 1 p 2 L ⋅ L a- ( ) 2 L b- ( ) 2- ⋅ = M R 1 L ⋅ p 2 L a- ( ) 2 ⋅- p 2 L b- ( ) 2 ⋅ + V R 1 p L a- ( ) ⋅- p L b- ( ) ⋅ + R 2 + At x = L + , V = M = 0 4. 4....
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This note was uploaded on 07/23/2011 for the course EML 4501 taught by Professor Staff during the Spring '11 term at UNF.

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HW 7 Solution - P avg 5.24 kW = P avg T avg ϖ ⋅ =...

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