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15_ch 05 Mechanical Design budynas_SM_ch05

15_ch 05 Mechanical Design budynas_SM_ch05 - max = S y n d...

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Chapter 5 129 5-17 Design decisions required: Material and condition Design factor Failure model Diameter of pin Using F = 416 lbf from Ex. 5-3 σ max = 32 M π d 3 d = 32 M πσ max 1 / 3 Decision 1: Select the same material and condition of Ex. 5-3 (AISI 1035 steel, S y = 81 000) . Decision 2: Since we prefer the pin to yield, set n d a little larger than 1. Further explana- tion will follow. Decision 3: Use the Distortion Energy static failure theory. Decision 4: Initially set n d = 1 σ
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Unformatted text preview: max = S y n d = S y 1 = 81 000 psi d = ³ 32(416)(15) π (81 000) ´ 1 / 3 = . 922 in Choose preferred size of d = 1 . 000 in F = π (1) 3 (81 000) 32(15) = 530 lbf n = 530 416 = 1 . 274 Set design factor to n d = 1 . 274 Adequacy Assessment: σ max = S y n d = 81 000 1 . 274 = 63 580 psi d = ³ 32(416)(15) π (63 580) ´ 1 / 3 = 1 . 000 in ( OK ) F = π (1) 3 (81 000) 32(15) = 530 lbf n = 530 416 = 1 . 274 (OK)...
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