Tutorial one solution

# Tutorial one solution - 4 Q6 Cross-section of the rope =...

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Solution for Tutorial One MECH2100, 2011 1 The University of Queensland School of Mechanical & Mining Engineering MECH2100 - Machine Element Design Solution for Tutorial One 2011 Q1: There are four assessment items in this course. The computer-based assessment is worth [15]%. The two design project assignments are worth, project one [20]% and project two [15]%. The final exam is worth [50]%. Q2: QFD stands for [Quality] [Function] [Deployment]. It is usually implemented as a spreadsheet table where the row headings correspond to the product [Attributes/Attribute] as defined by the client and the column headings correspond the engineering [Characteristics/characteristic]. Q3: Steel: Density, kg/m 3 7700 Aluminum: Density, kg/m 3 2800 Steel: Young's Modulus (Modulus of Elasticity), GPa 207 Aluminum: young's Modulus, GPa 70 Poisson's ratio of metals 0.33 Poisson's ratio of Ceramics 0.25 Poisson's ratio of Polymers 0.40 Q4: I = π ( R 0 4 - R i 4 )/4 or ( D 0 4 - D i 4 )/64 = 0.3645 m 4 Q5: I x = (0.8 x 1.8 3 – 0.6 x 1.6 3 )/12 = 0.184 m

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Unformatted text preview: 4 Q6: Cross-section of the rope = A (m 2 ), Density of steel = 7700 kg/M 3 (from Q3). The maximum stress is at the top of the rope and is calculated as: σ max = ρ gh = 7700 x 9.81 x 600 = 45.32 x 10 6 Pa = 45.32 MPa Q7: Eq. 3.11 Su = 500H B =500x111 = 55500 [psi] Solution for Tutorial One MECH2100, 2011 2 Q8: Eq. 3.12 Sy = 1.05Su – 30,000 = 28275 [psi] Q9: [1] The yield strength can be found from the σ-ε curve as Sy = 250 MPa. [2] The maximum stresses ( σ ) are caused by moment W(L) and equals to Mc/I, where σ = S y = My/I = ( WL )( d /2)/( bd 3 /12 ) = 6( WL )/( bd 2 ) 250 x 10 6 = 6 W x 0.15/(0.05 x 0.025 2 ) Hence W = 8680.55 N Q10: Moment arm to centroid = 25 +4 mm = 29 mm Stressing is 2 2 3 05 . 016 . 6 15000 029 . 05 . 016 . 15000 6 029 . ) 12 / ( ) 2 / ( 029 . × × × ± × = × ± = ± = ± bd P bd P bd d P bd P I My A P = 18.75 + 65.25 MPa = 84 MPa on the right side, and = 18.75 – 65.25 MPa = -46.5 on the left side. W L 25 50 x...
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## This note was uploaded on 11/26/2011 for the course MECH 2100 taught by Professor Billdaniel during the Three '11 term at Queensland.

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Tutorial one solution - 4 Q6 Cross-section of the rope =...

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