9_ch 09 Mechanical Design budynas_SM_ch09

# 9_ch 09 Mechanical Design budynas_SM_ch09 -...

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Chapter 9 247 Primary shear τ ± y = V A = F A = 20 000 17 h = 1176 h psi Secondary shear J u = ( b + d ) 3 6 = (6 + 6) 3 6 = 288 in 3 J = 0 . 707 h (288) = 203 . 6 h in 4 τ ±± x = τ ±± y = Mr y J = 20 000(6 . 25 + 3)(3) 203 . 6 h = 2726 h psi τ max = ± τ ±± 2 x + ( τ ±± y + τ ± y ) 2 = 1 h ² 2726 2 + (2726 + 1176) 2 = 4760 h psi Relate stress to strength τ max = τ all 4760 h = 12 800 h = 4760 12 800 = 0 . 372 in Decision: Specify 3 / 8 in leg size Speciﬁcations: Pattern: All-around square weld bead track Electrode: E6010 Type of weld: Fillet Weld bead length: 24 in Leg size: 3 / 8 in Attachment length: 12.25 in 9-15 This is a good analysis task to test the students’ understanding (1) Solicit information related to a priori decisions. (2) Solicit design variables b and d .
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