2_ch 09 Mechanical Design budynas_SM_ch09

# 2_ch 09 Mechanical Design budynas_SM_ch09 - . 5)] = min[15,...

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240 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Secondary shear Table 9-1 J u = d ( 3 b 2 + d 2 ) 6 = 2[ ( 3 )( 2 2 ) + 2 2 ] 6 = 5 . 333 in 3 J = 0 . 707 hJ u = 0 . 707 ( 5 / 16 )( 5 . 333 ) = 1 . 18 in 4 τ ±± x = τ ±± y = Mr y J = 7 F ( 1 ) 1 . 18 = 5 . 93 F kpsi Maximum shear τ max = ± τ ±± 2 x + ( τ ± y + τ ±± y ) 2 = F ² 5 . 93 2 + (1 . 13 + 5 . 93) 2 = 9 . 22 F kpsi F = τ all 9 . 22 = 20 9 . 22 = 2 . 17 kip Ans. (1) (b) For E7010 from Table 9-6, τ all = 21 kpsi Table A-20: HR 1020 Bar: S ut = 55 kpsi, S y = 30 kpsi HR 1015 Support: S ut = 50 kpsi, S y = 27 . 5 kpsi Table 9-5, E7010 Electrode: S ut = 70 kpsi, S y = 57 kpsi The support controls the design. Table 9-4: τ all = min[0 . 30(50), 0 . 40(27
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Unformatted text preview: . 5)] = min[15, 11] = 11 kpsi The allowable load from Eq. (1) is F = τ all 9 . 22 = 11 9 . 22 = 1 . 19 kip Ans. 9-6 b = d = 2 in Primary shear τ ± y = V A = F 1 . 414(5 / 16)(2 + 2) = . 566 F Secondary shear Table 9-1: J u = ( b + d ) 3 6 = (2 + 2) 3 6 = 10 . 67 in 3 J = . 707 h J u = . 707(5 / 16)(10 . 67) = 2 . 36 in 4 τ ±± x = τ ±± y = Mr y J = (7 F )(1) 2 . 36 = 2 . 97 F F 7"...
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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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