Tài liệu shi20396 chương 16 pdf

... psi P 1 P 2 P 1 P 2 P P 1 P 2 ␣ F 200 125 275 P 2 P 1 125 shi20396_ ch16.qxd 8/28/03 4:01 PM Page 416 Chapter 16 417 Eq. (16- 14): P 2 = P 1 exp(− f φ) = 168 0 exp(−0.942) = 655 lbf T = (P 1 − P 2 ) D 2 = (168 0 − 655) 16 2 = 8200 lbf ·in ... 1803/2 = 901 lbf. (c) Eq. (16- 22): p = 2P bD p| θ=0° = 2P 1 3 (16) = 2 (168 0) 3 (16) = 70 psi Ans. As it should be p| θ=270° = 2P 2 3 (16)...

Ngày tải lên: 27/01/2014, 05:20

... = 5.18 ◦ 1 -16 (a) σ = 600 20(6) = 5MPa (b) I = 1 12 8(24) 3 = 9 216 mm 4 (c) I = π 64 32 4 (10 −1 ) 4 = 5.147 cm 4 (d) τ = 16( 16) π(25 3 )(10 −3 ) 3 = 5.215(10 6 ) N/m 2 = 5.215 MPa shi20396_ ch01.qxd ... = 0.25, x i = 0, x f = 0.2 F i = 10 lbf; F f = 10 + 2.5(0.2) = 10.5 lbf Ans. x y F fN N ␪ T r shi20396_ ch01.qxd 6/5/03 12:11 PM Page 2 Chapter 1 3 From Eq. (1) of Prob. 1-6 N = F −...

Ngày tải lên: 22/01/2014, 15:20

... 1.8 log(0.029 317) =−0 .166 531, a 3 = 38 (34 110) −0 .166 531 = 216. 10 kpsi N 3 =  32 216. 1  −1/0 .166 531 = 95 740 cycles Ans. 7-31 Using Miner’s method a = [0.9(100)] 2 50 = 162 kpsi b =− 1 3 log 0.9(100) 50 =−0.085 ... a = 10 2.2095 = 162 .0 b = log 50/90 3 =−0.085 09 (S f ) ax = 162 −0.085 09 10 3 ≤ N ≤ 10 6 in kpsi Ans. Check: 10 3 (S f ) ax = 162 (10 3 ) −0.085 09 = 90 kpsi 1...

Ngày tải lên: 22/01/2014, 15:20

... 14 R =  14 2 + 9 2 = 16. 64 σ 1 = 4 + 16. 64 = 20.64 σ 2 = 4 − 16. 64 =−12.64 φ p = 1 2  90 + tan −1 14 9  = 73.63 ◦ cw τ 1 = R = 16. 64, φ s = 73.63 ◦ − 45 ◦ = 28.63 ◦ cw 4 x 16. 64 4 28.63Њ 12.64 20.64 73.63Њ x 2␾ s (Ϫ10, ... φ s = 61 ◦ − 45 ◦ = 16 ◦ cw x 14 14 9.434 16 x 23.43 4.57 61Њ 2␾ s (9, 8 cw ) C R D ␶ 2 ␶ 1 ␴ 1 ␴ 2 2␾ p (19, 8 ccw ) y x ␴ ␶ cw ␶ ccw x 17 17 9.22 24.7Њ x...

Ngày tải lên: 22/01/2014, 15:20

... 33.67 τ  = 884 0.186 = 4753 psi, vol = 0.186 2 2 16 = 0.277 in 3 τ  = 2210 0.186 = 11 882 psi fom  = I u hl = 256 0.186 (16) = 86.0 eff = I (h 2 /2)l = 33.67 (0.186 2 /2 )16 = 121.7 Conclusions: To meet allowable ... 3000 h = 861 3000 = 0.287 → 5 /16 " Decision: Use 5 /16 in fillet welds Ans. 100 16 3 562 266.7 266.7 462 F F B B A R x A R y A 60Њ y x shi20396_ ch09.qxd 8/19/...

Ngày tải lên: 27/01/2014, 05:20

... (11-18) when  a f F D C 10  3 x D < x 0 R = 1 The smallest 02-series roller has a C 10 = 16. 8kN for a basic load rating.  0.427 16. 8  3 (900) < ? > 0.02 0.0148 < 0.02 ∴ R = 1 Spotting this early ... kN, (x) 1 = 115(2000) (16) 10 6 = 13.8 This establishes point 1 on the R = 0.90 line. 1 0 10 2 10 18 1 2 39.6 100 1 10 13.8 72 1 100 x 2 log x F A B log F R ϭ 0.90 R ϭ 0.2...

Ngày tải lên: 27/01/2014, 05:20

Ngày tải lên: 15/12/2013, 01:16

... 1 2 )(1 2 ) 2(1 2 )(1.5 2 )  = 4167 psi Ans. (σ t ) i =−p =− 4167 psi, (σ r ) i =− 4167 psi σ  i = [(− 4167 ) 2 − (− 4167 )(− 4167 ) + (− 4167 ) 2 ] 1/2 = 4167 psi Ans. (σ t ) o = 4167  1.5 2 + 1 2 1.5 2 − ... (−5860)(−3 516) + (−3 516) 2 ] 1/2 = 5110 psi Ans. Outer member: Eq. (4-58) (σ t ) o = 3 516  1.5 2 + 1 2 1.5 2 − 1 2  = 9142 psi (σ r ) o =−p =−3 516 psi Eq. (6-13) σ  o = [9...

Ngày tải lên: 22/01/2014, 15:20

... σ. shi20396_ ch02.qxd 7/21/03 3:28 PM Page 18 Chapter 2 19 2 -16 Fraction discarded is α +β . The area under the PDF was unity. Having discarded α +β fraction, the ordinates to the truncated PDF ... 0.008 696 0.017292 130 10 1300 169 000 130 0.014 493 0. 0166 12 140 8 1120 156 800 140 0.011 594 0.013185 150 5 750 112 500 150 0.007 246 0.008647 160 2 320 51200 160 0.002 899 0.004685...

Ngày tải lên: 22/01/2014, 15:20

... 708 2 /500) 500 −1 = 1.7808 C x = 1.7808/43. 416 = 0.041 02, ¯y = ln 43. 416 − ln(1 + 0.041 02 2 ) = 3.7691 ␥ (10 Ϫ3 ) ␶ (MPa) 0 50 100 150 200 250 300 01234567 shi20396_ ch03.qxd 8/18/03 10:18 AM Page ... 1.06(10 8 )in Ans. 2024 T4 aluminum: E W = 10.3(10 6 ) 0.098 = 1.05(10 8 )in Ans. Ti-6Al-4V titanium: E W = 16. 5(10 6 ) 0 .16 = 1.03(10 8 )in Ans. Gray cast iron: E W = 14.5(10 6 ) 0....

Ngày tải lên: 22/01/2014, 15:20