Machinery Components Maintenance And Repair Episode 2 Part 10 ppt

... [W/mK] P01 10 50 16 20 6 8 < 2 6900 16 .2 1 .2 20 P05 25 49 16 15 8 12 < 2 7000 14 .2 1.8 20 P01–15 48 16 20 5 11 –* 7000 15.7 –* 20 P05 Total carbide: 94 Total metal: 6 –* 6100 17 .2 1.8 –* P10 Total ... a 1 ) 2 a 1 + a 2 ————— a 1 ≤ x < ———— (a 2 — a 1 ) 2 2 SF(x, a 1 , a 2 ) = { (A7.4b) 2( x – a 2 ) 2 a 1 + a 2 1– ————— ———— ≤ x < a 2 (a 2...

Ngày tải lên: 21/07/2014, 17:20

... stainless 21 24 22 25 23 26 24 27 25 29 austenitic stainless 14–17 15–18 17 21 22 25 26 29 high manganese 14 16 19 21 22 Aluminium pure, 100 0 series 20 0 24 0 20 0 23 0 20 0 22 5 190 22 0 – 20 00 to 7000 ... 115– 120 100 – 110 90 100 – – 60/70Cu–40/30Zn 25 –35 28 –35 27 –33 25 –30 – 90/95Cu 10/ 5Sn 15 25 20 25 23 –30 – – 60/90Cu–40/10Ni 6–15 8–18 12 22 – – Nickel pur...

Ngày tải lên: 21/07/2014, 17:20

... 3(s xx ′ 2 + s yy ′ 2 + s zz ′ 2 ) + 6(s 2 xy + s 2 yz + s 2 zy ) ≡ (s xx – s yy ) 2 + (s yy – s zz ) 2 + (s zz – s xx ) 2 + 6(s 2 xy + s 2 yz + s 2 zy ) = 6k 2 or 2Y 2 (A1 .26 b) which is the same ... s r 2 . So s r is also a stress invariant (it is known as the second stress invariant). From equation (A1 .25 ) s r 2 = s 2 xx + s 2 yy + s 2 zz +2( s 2...

Ngày tải lên: 21/07/2014, 17:20

... addition s r ′ 2 = s r 2 –3s m 2 =(s 1 2 + s 2 2 + s 3 2 ) – 3s m 2 (A1.5) After substituting for s m from equation (A1.1), 3s r ′ 2 =(s 1 – s 2 ) 2 +(s 2 – s 3 ) 2 +(s 3 – s 1 ) 2 (A1.6) The yield ... stresses: 1 s – 2 =— [ (s 1 – s 2 ) 2 +(s 2 – s 3 ) 2 +(s 3 – s 1 ) 2 ] = Y 2 or 3k 2 (A1.7) 2 330 Appendix 1 Fig. A1 .2 Geometrical representation...

Ngày tải lên: 21/07/2014, 17:20

... introduced of the Cutting edge design 25 5 Fig. 8 .24 Variation of chip shape and forces at w G =2. 6 mm and h B =0.4 mm Childs Part 2 28:3 :20 00 3:18 pm Page 25 5 Vyas, A. and Shaw, M. C. (1999) Mechanics ... conventional, (b) and (c) type I and II tools, K = 1 .2 Childs Part 2 28:3 :20 00 3:19 pm Page 26 2 where F* t , F* r and F* z are the specific cutting forc...

Ngày tải lên: 21/07/2014, 17:20

... the pinion at position 2. So 8, = Fpi, + Fp 12 and 5 2 = Fp 12 + Fp 22 . As the transmission error is the sum of 81 and S 2 T.E. = F(p n +2p, 2 + p 22 ) giving F. Once F ... frequency analysis of the input and of the output and the ratio (complete with phase) is the frequency transfer function required. 14 12 50 100 150 20 0 25 0 300...

Ngày tải lên: 05/08/2014, 09:20

... expanding a 467 sea gate landing and takeoff of seaplanes. 2. A platform to cover a large region of space and to display targets as soon as possible after they enter theat sea for landing and ... process [ CHEM ENG ] and production, in transportation and distribu- A nitrogen-fixation process used in Norway; em- tion, and in information processing and commu- ploys a very long (...

Ngày tải lên: 21/07/2014, 15:20

... l s 190 Advances in mechanics Childs Part 2 28:3 :20 00 3:13 pm Page 190 (x′ 6 – x′ C2 + t 1e,q3–4 sin h′ c ) 2 + (y′ 6 – y′ C2 + t 1e,q3–4 cos h′ c ) 2 = r 2 n (6.43) t 1eq3–4 is obtained as the ... axis by the amount l s : x′ = L 2 X (6 .20 a) where cos l s 0 sin l s 10 0 L 2 = L 21 L 22 = [ 010 ][ 0 cos a n sin a n ] –sin l s 0 cos l s 0 –sin a n cos a n } (6 .20 b)...

Ngày tải lên: 21/07/2014, 17:20

... 1.01a, ab+ = V 2 x 2 () C− 1 kN = M 2 x 2 () M− Ca b+ x 2 − () + ⎡ ⎣ ⎤ ⎦ 1 kN m⋅ = 024 6 810 10 5 0 5 Distance (m) Force (kN) V 1 x 1 () V 2 x 2 () x 1 x 2 , 024 6 810 60 40 20 0 20 Distance (m) Moment ... N B 2 W 2 − 0= N B 2 W 2 = N B 20 0.00 lb= Σ F x = 0; P μ ' s N B − 0= P 2 μ ' s N B = P 2 60.00 lb= Assume both crates A and B tip: Σ M...

Ngày tải lên: 21/07/2014, 17:20

... corporation definition (d) concern A10 n. injury; harm definition (b) damage 670 Questions Index PHOTOCOPIABLE © www.english-test.net Answers 322 TOEIC Vocabulary / Word by Meaning / Test # 322 (Answer Keys) A1 ... account A10 n. netting attached to a woman's hat definition (c) veil 6 72 Questions Index PHOTOCOPIABLE © www.english-test.net Answers 329 TOEIC Vocabulary / Word by Meanin...

Ngày tải lên: 21/07/2014, 22:20