33_ch 08 Mechanical Design budynas_SM_ch08

# 33_ch 08 Mechanical Design budynas_SM_ch08 - 3 96 = − 393...

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236 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Bearing on cantilever: A b = 12(12) = 144 mm 2 F = 190 2 . 8 ± (144)(10 3 ) 2 . 343 ² = 4 . 17 kN Bending of cantilever: I = 1 12 (12)(50 3 12 3 ) = 1 . 233(10 5 )mm 4 I c = 1 . 233(10 5 ) 25 = 4932 F = M 151 = 4932(190) 2 . 8(151)(10 3 ) = 2 . 22 kN So F = 1 . 99 kN based on bearing on channel Ans. 8-49 F ± = 4 kN; M = 12(200) = 2400 N · m F ±± A = F ±± B = 2400 64 = 37 . 5 kN F A = F B = ³ (4) 2 + (37 . 5) 2 = 37 . 7 kN Ans. F O = 4 kN Ans. Bolt shear: A s = π (12) 2 4 = 113 mm 2 τ = 37 . 7(10) 3 113 = 334 MPa Ans. Bearing on member: A b = 12(8) = 96 mm 2 σ =− 37 . 7(10)
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Unformatted text preview: 3 96 = − 393 MPa Ans. Bending stress in plate: I = bh 3 12 − bd 3 12 − 2 ´ bd 3 12 + a 2 bd µ = 8(136) 3 12 − 8(12) 3 12 − 2 ± 8(12) 3 12 + (32) 2 (8)(12) ² = 1 . 48(10) 6 mm 4 Ans. M = 12(200) = 2400 N · m σ = Mc I = 2400(68) 1 . 48(10) 6 (10) 3 = 110 MPa Ans. a h a b d F' A ± 4 kN F" A ± 37.5 kN F" B ± 37.5 kN F' O ± 4 kN 32 32 A O B F' B ± 4 kN...
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## This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

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