36_ch 04 Mechanical Design budynas_SM_ch04

# 36_ch 04 Mechanical Design budynas_SM_ch04 - / 2 F sin (100...

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Chapter 4 105 4-69 Must use Eq. (4-33) A = 80(60) 40(60) = 2400 mm 2 R = (25 + 40)(80)(60) (25 + 20 + 30)(40)(60) 2400 = 55 mm Section is equivalent to the “T” section of Table 3-4 r n = 60(20) + 20(60) 60 ln[(25 + 20) / 25] + 20 ln[(80 + 25) / (25 + 20)] = 45 . 9654 mm e = R r n = 9 . 035 mm I z = 1 12 (60)(20 3 ) + 60(20)(30 10) 2 + 2 ± 1 12 (10)(60 3 ) + 10(60)(50 30) 2 ² = 1 . 36(10 6 )mm 4 For 0 x 100 mm M =− Fx , M F =− x ; V = F , V F = 1 For θ π/ 2 F r = F cos θ , F r F = cos θ ; F θ = F sin θ , F θ F = sin θ M = F (100 + 55 sin θ ), M F = (100 + 55 sin θ ) Use Eq. (5-34), integrate from 0 to π/ 2, double the results and add straight part δ = 2 E ³ 1 I ´ 100 0 Fx 2 dx + ´ 100 0 (1) F (1) dx 2400( G / E ) + ´ π/ 2 0 F (100 + 55 sin θ ) 2 2400(9 . 035) d θ + ´ π/ 2 0 F sin 2 θ (55) 2400 d θ ´ π/ 2 0 F (100 + 55 sin θ ) 2400 sin θ d θ ´
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Unformatted text preview: / 2 F sin (100 + 55 sin ) 2400 d + / 2 (1) F cos 2 (55) 2400( G / E ) d Substitute I = 1 . 36(10 3 ) mm 2 , F = 30(10 3 ) N, E = 207(10 3 ) N/mm 2 , G = 79(10 3 ) N/mm 2 = 2 207(10 3 ) 30(10 3 ) 100 3 3(1 . 36)(10 6 ) + 207 79 100 2400 + 2 . 908(10 4 ) 2400(9 . 035) + 55 2400 4 2 2400 (143 . 197) + 207 79 55 2400 4 = . 476 mm Ans. F F F r M 100 mm x y z 30 mm 50 mm Straight section...
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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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