4_ch 09 Mechanical Design budynas_SM_ch09

4_ch 09 Mechanical Design budynas_SM_ch09 -...

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242 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 9-8 Primary shear τ ± = 0 (why?) Secondary shear Table 9-1: J u = 2 π r 3 = 2 π (4) 3 = 402 cm 3 J = 0 . 707 hJ u = 0 . 707(0 . 5)(402) = 142 cm 4 M = 200 F N · m( F in kN) τ ±± = Mr 2 J = (200 F )(4) 2(142) = 2 . 82 F (2 welds) F = τ all τ ±± = 140 2 . 82 = 49 . 2kN Ans. 9-9 Rank fom ± = J u lh = a 3 / 12 ah = a 2 12 h = 0 . 0833 ± a 2 h ² 5 fom ± = a (3 a 2 + a 2 ) 6(2 a ) h = a 2 3 h = 0 . 3333 ± a 2 h ² 1 fom ± = (2 a ) 4 6 a 2 a 2 12( a + a )2 = 5 a 2 24 h = 0 . 2083 ± a 2 h ² 4 fom ± = 1 3 ³ 8 a 3 + 6 a 3 + a 3 12 a 4 2 a + a ´ = 11 36 a 2 h = 0 . 3056 ± a 2 h ² 2 fom ± = a ) 3 6 h 1 4 a = 8 a 3 24 = a 2 3 h = 0 . 3333 ± a 2 h ² 1 fom ± = 2 π ( a / 2) 3 π = a 3 4 = a 2 4 h = 0 . 25 ± a 2 h ² 3 These rankings apply to fillet weld patterns in torsion that have a square area a × a in which to place weld metal. The object is to place as much metal as possible to the border.
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