12_ch 09 Mechanical Design budynas_SM_ch09

# 12_ch 09 Mechanical Design budynas_SM_ch09 - beads and then...

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250 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Had the weld bead gone around the corners, the situation would change. Here is a fol- lowup task analyzing an alternative weld pattern. 9-17 From Table 9-2 For the box A = 1 . 414 h ( b + d ) Subtracting b 1 from b and d 1 from d A = 1 . 414 h ( b b 1 + d d 1 ) I u = d 2 6 (3 b + d ) d 3 1 6 b 1 d 2 2 = 1 2 ( b b 1 ) d 2 + 1 6 ( d 3 d 3 1 ) length of bead l = 2( b b 1 + d d 1 ) fom = I u / hl 9-18 Computer programs will vary. 9-19 τ all = 12 800 psi . Use Fig. 9-17( a ) for general geometry, but employ
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Unformatted text preview: beads and then ±± beads. Horizontal parallel weld bead pattern b = 6 in d = 8 in From Table 9-2, category 3 A = 1 . 414 hb = 1 . 414( h )(6) = 8 . 48 h in 2 ¯ x = b / 2 = 6 / 2 = 3 in, ¯ y = d / 2 = 8 / 2 = 4 in I u = bd 2 2 = 6(8) 2 2 = 192 in 3 I = . 707 hI u = . 707( h )(192) = 135 . 7 h in 4 τ ± = 10 000 8 . 48 h = 1179 h psi 6" 8" b b 1 d d 1...
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