15_ch 09 Mechanical Design budynas_SM_ch09

15_ch 09 Mechanical Design budynas_SM_ch09 - associated...

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Chapter 9 253 6 4.8 7.2 A B G 1" 7.5" 9-22 h = 0 . 6cm, b = 6cm, d = 12 cm . Table 9-3, category 5: A = 0 . 707 h ( b + 2 d ) = 0 . 707(0 . 6)[6 + 2(12)] = 12 . 7cm 2 ¯ y = d 2 b + 2 d = 12 2 6 + 2(12) = 4 . 8cm I u = 2 d 3 3 2 d 2 ¯ y + ( b + 2 d ) ¯ y 2 = 2 ( 12 ) 3 3 2 ( 12 2 )( 4 . 8 ) + [6 + 2 ( 12 ) ]4 . 8 2 = 461 cm 3 I = 0 . 707 hI u = 0 . 707 ( 0 . 6 )( 461 ) = 196 cm 4 τ ± = F A = 7 . 5 ( 10 3 ) 12 . 7 ( 10 2 ) = 5 . 91 MPa M = 7 . 5 ( 120 ) = 900 N · m c A = 7 . 2cm , c B = 4 . 8cm The critical location is at A. τ ±± A = Mc A I = 900(7 . 2) 196 = 33 . 1MPa τ max = ± τ ± 2 + τ ±± 2 = (5 . 91 2 + 33 . 1 2 ) 1 / 2 = 33 . 6MPa n = τ all τ max = 120 33 . 6 = 3 . 57 Ans. 9-23 The largest possible weld size is 1 / 16 in. This is a small weld and thus difficult to accom- plish. The bracket’s load-carrying capability is not known. There are geometry problems
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Unformatted text preview: associated with sheet metal folding, load-placement and location of the center of twist. This is not available to us. We will identify the strongest possible weldment. Use a rectangular, weld-all-around pattern – Table 9-2, category 6: A = 1 . 414 h ( b + d ) = 1 . 414(1 / 16)(1 + 7 . 5) = . 751 in 2 ¯ x = b / 2 = . 5 in ¯ y = d 2 = 7 . 5 2 = 3 . 75 in...
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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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