hw3sol

# hw3sol - Homework#3 1(a Solution rt = t 0.015 = = 0.437 t c...

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Unformatted text preview: Homework #3 1. (a) Solution rt = t 0.015 = = 0.437 t c 0.0343 rt cos α 1 − rt sin α tan φ = = 0.437 cos10o = 0.466 1 − 0.437 ⋅ sin10o ∴φ = 25o (b) Fs = Fc cos φ − Ft sin φ = 300 ⋅ cos 25o − 125 ⋅ sin 25o = 219.1 bs ( = 974.6 N ) (c) τs = Fs Fs ⋅ sin φ = As w⋅t w= 2 = 0.0393 in 25.4 219.1 × sin 25o = 0.0393 × 0.015 = 157,003 b 2 ( = 1081 MPa ) in 2. (a) HP = Fc ⋅ Vc 33, 000 450 × 165 = = 2.25 Hp 33,000 Hp 2.25 E= = = 0.84 HPm 2.68 ∴ 84% (b) MRR = 12Vcf r ⋅ t = 12 × 165 × 0.01 × 0.1 = 1.94 in HPs = HP 2.25 = = 1.14 HP 3 in MRR 1.98 3 min. min 3. rt = 0.1 = 0.5 0.2 r cos α 0.5 × 10o = tanφ = t 1 − rt sin α 1 − 0.5 × sin10o = 0.539 ∴φ = 28.33o Fs = Fc cos φ = 500cos 28.3o − 200sin 28.3o = 345.2 N shear energy U s = total energy U = = Fs ⋅ Vs Fc ⋅ Vc Fs Fc cos α cos (φ − α ) 345.2 cos10o = ⋅ 500 cos ( 28.3 − 10o ) = 0.72 ∴ 72% ...
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