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Unformatted text preview: Solution for Tutorial Two MECH2100, 2011 1 The University of Queensland School of Mechanical & Mining Engineering MECH2100  Machine Element Design Solution for Tutorial Two 2011 Q1: Nominal area A = (5010)25 = 1000 mm 2 ; nom = 250000/1000 = 250 MPa From Figure 4.40, K t = 2.43, for d/b = 10/50 = 0.2; max = 250 x 2.43 = 607.5 MPa Since the applied stress is more than the yield strength [400 MPa] of the material, the maximum stress experienced by the material is 400 MPa. Q2: The bar experiences a [compressive] residual stress. The value of the stress is 607.5  400 = 207.5 MPa. Q3: As a S R R = where S a is the gauge sensitivity (gauge factor) 300 10 100 06 . 2 1 1 6 = = = R R S a [microstrain] Q4: A 50 mm 25 mm 10 mm 400 MPa Solution for Tutorial Two MECH2100, 2011 2 Q5: Note: this is uniaxial stress, even though it varies with depth over the beam. I = 0.05 x 0.1 3 /12 = 4.16667 x 106 m 4 , = My/I = (2000 x 0.5) x 0.025/4.16667 = 6 MPa...
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 Three '11
 BillDaniel

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