hw6sol

hw6sol - ME 363 1. (a) Homework #6 σ fm Κ ε n +1 = ε n...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ME 363 1. (a) Homework #6 σ fm Κ ε n +1 = ε n +1 = ε = ln 1.22 Αo 0.5 = 0.713 = 2 ln Α 0.35 45,000 ( 0.1713) 0.713 1.22 = 34, 240 psi ⎡⎛ ⎣⎝ σ d = σ fm ⎢⎜ 1 + μ ⎞ Αo 2 ⎤ +α ln α ⎟ Α1 3 ⎥ ⎠ ⎦ ⎡ ⎛ 0.1 ⎞ 0.52 2 ⎤ = 34, 240 ⎢ ⎜ 1 + + ( 0.1) ⎥ ⎟ ln 2 ⎣ ⎝ 0.1 ⎠ 0.35 3 ⎦ ⎛ ⎞ 12 0 π × = 0.1 ⎟ α= ⎜ 2 180 ⎝ ⎠ = 34, 240 × 1.49 = 51,133 psi Fd = A f × σ d = (b) π 4 ( 0.35) 2 × 51,133 = 4,920 lbs P= Fd × V 4920 × 2 = = 17.9 Hp 550 550 2. UTS = 600 MPa = 87,000 psi l = 2 ( 23.5 + 2 ) + 2 = 13 in Pmax = 0.7 (UTS ) × t × l 5 × 13 32 = 123,731 lbs ( = 549kN ) = 0.7 (87,000 ) × 3. (a) Ri ⎛ RY ⎞ ⎛ RY = 4 ⎜ i ⎟ - 3⎜ i Rf ⎝ ET ⎠ ⎝ ET 3 3 ⎞ ⎟ +1 ⎠ ⎛ 10 × 120 ⎞ ⎛ 10 × 120 ⎞ = 4⎜ ⎟ - 3⎜ ⎟ +1 3 3 ⎝ 70 × 10 × 2 ⎠ ⎝ 70 × 10 × 2 ⎠ = 4 ( 0.0085) - 3 ( 0.0085) + 1 3 = 0.975 Rf = Ri 10 = = 10.26 0.975 0.975 ∴ D f = 2 × R f = 20.52 mm (b) R f = 20 Ri ⎛ RY ⎞ ⎛ RY ⎞ = 4 ⎜ i ⎟ - 3⎜ i ⎟ + 1 20 ⎝ ET ⎠ ⎝ ET ⎠ ⎛ R × 120 ⎞ ⎛ Ri × 120 ⎞ Ri 4⎜ i 3 +1 = 0 ⎟ - 3⎜ ⎟− 70 × 10 × 2 ⎠ 70 × 103 × 2 ⎠ 20 ⎝ ⎝ 19.5 Ri = 3 3 (c) 50 −1 T r 20 50 = −1 2 r ∴ r = 4.5 = Rf 4. (a) UTS = 320 MPa = 46, 410 psi 1 ⎛ 12 ⎞ Fmax = π × 9 × × 46, 410 ⎜ − 0.7 ⎟ 8 ⎝9 ⎠ = 103,883lbs (b) 103,883 = 415,534lbs 0.25 415,532 Pressure = = 8, 400 psi π (122 − 92 ) 4 Force = ...
View Full Document

Ask a homework question - tutors are online