19_ch 09 Mechanical Design budynas_SM_ch09

19_ch 09 Mechanical Design budynas_SM_ch09 - E6010...

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Chapter 9 257 (d) From Table 9-2, category 6: A = 1 . 414(0 . 25)(0 . 25 + 2 . 5) = 0 . 972 in 2 I u = d 2 6 (3 b + d ) = 2 . 5 2 6 [3(0 . 25) + 2 . 5] = 3 . 39 in 3 The second area moment about an axis through G and parallel to z is I = 0 . 707 hI u = 0 . 707(0 . 25)(3 . 39) = 0 . 599 in 4 Ans. (e) Refer to Fig. P.9-26 b . The shear stress due to F y is τ 1 = F y A = 600 0 . 972 = 617 psi The shear stress along the throat due to F x is τ 2 = F x A = 1039 0 . 972 = 1069 psi The resultant of τ 1 and τ 2 is in the throat plane τ ± = ( τ 2 1 + τ 2 2 ) 1 / 2 = (617 2 + 1069 2 ) 1 / 2 = 1234 psi The bending of the throat gives τ ±± = Mc I = 439(1 . 25) 0 . 599 = 916 psi The maximum shear stress is τ max = ( τ ± 2 + τ ±± 2 ) 1 / 2 = (1234 2 + 916 2 ) 1 / 2 = 1537 psi Ans. (f) Materials: 1018 HR Member: S y = 32 kpsi, S ut = 58 kpsi (Table A-20)
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Unformatted text preview: E6010 Electrode: S y = 50 kpsi (Table 9-3) n = S sy τ max = . 577 S y τ max = . 577(32) 1 . 537 = 12 . Ans. (g) Bending in the attachment near the base. The cross-sectional area is approximately equal to bh. A 1 . = bh = . 25(2 . 5) = . 625 in 2 τ xy = F x A 1 = 1039 . 625 = 1662 psi I c = bd 2 6 = . 25(2 . 5) 2 6 = . 260 in 3 At location A σ y = F y A 1 + M I / c σ y = 600 . 625 + 439 . 260 = 2648 psi...
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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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