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Unformatted text preview: 4 Q6: Crosssection 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 σε 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.
 Three '11
 BillDaniel

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