Tutorial Two solution

# Tutorial Two solution - Solution for Tutorial Two MECH2100...

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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 = (50-10)25 = 1000 mm 2 ; σ nom = 250000/1000 = 250 MPa From Figure 4.40, K t = 2.43, for d/b = 10/50 = 0.2; σ 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 ε a S R R = ∆ where S a is the gauge sensitivity (gauge factor) με ε 300 10 100 06 . 2 1 1 6 = = ∆ = R R S a [microstrain] Q4: A 50 mm 25 mm 10 mm σ ε 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 10-6 m 4 , σ Α = My/I = (2000 x 0.5) x 0.025/4.16667 = 6 MPa με σ ε...
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Tutorial Two solution - Solution for Tutorial Two MECH2100...

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