4—nominal strength and detailing requirements 4 1 — column longitudinal reinforcement
recommendations for design of beam-column connections in monolithic reinforced concrete structures
... (22.2) 13 .8 (350) 17 .3 (44 0) 14 . 6 (370) 1. 000 (25 .4) 15 .8 (40 1) 19 .3 (49 1) 16 .1 ( 41 1) 1. 128 (28.6) 17 .8 (45 1) 21. 3 ( 5 41 ) 17 .8 (45 1) 10 1. 270 ( 31. 8) 20 .1 (502) 23.6 (592) 19 .6 (49 1) 11 1 . 41 0 ( 34. 9) ... h ≤ 6d b = ( 1. 128 in ) = 6.8 in in 1. 128 (28.6) 22.6 (572) 10 1. 270 ( 31. 8) 25 .4 (636) 11 1 . 41 0 ( 34. 9) 28.2 (698) 14 1. 693 (44 .5) 33.9 (890) *Based on size of longitudinal reinforcement ... requirements over those given in ACI 318 -02 are specified CHAPTER 4 NOMINAL STRENGTH AND DETAILING REQUIREMENTS 4. 1 Column longitudinal reinforcement Column longitudinal reinforcement passing through...
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Thermodynamics Kinetics of Dynamic Systems Part 4 pdf
... Homogeneous Flow Model and Fractional Flow Model, Proceedings of 2 010 IEEE Asia-Pacific Power and Energy Engineering Conference, ISBN978 -1 -42 44- 4 813 -5, Chengdu, China, March28- 21. 2 010 A Pecenko,; L.G.M ... Q.H., Ferre, D., and Pauchon, C. (19 98) Transient simulation of twophase flows in pipes Oil Gas Science Technology Vol 53, No.6, pp.8 01 811 ISSN 12 94- 447 5 Li, C J., Jia, W L., Wu, X.(2 010 ) Water Hammer ... tramsmission OGST, vol .17 , no.5, pp .11 17, ISSN1000-8 2 41 84 Thermodynamics – Kinetics of Dynamic Systems Hasan, A.R & Kabir, C.S. (19 92) Gas void fraction in two-phase up-flow in vertical and inclined annuli...
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MANAGING THE RISKS OF PAYMENT SYSTEMS CHAPTER 4 pot
... See U.C.C §§ 4A -40 1 and 4A-3 01( b) 99 Wire Transfers 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 Official Comment to U.C.C § 4A- 211 U.C.C § 4A- 211 (e) U.C.C § 4A- 211 (g) The rule ... U.C.C § 4A- 210 U.C.C § 4A- 210 (a) Official Comment to U.C.C § 4A- 211 U.C.C § 4A- 211 (b) U.C.C § 4A- 211 (a) Payment orders may be transmitted the same way under U.C.C § 4A -10 3(a) U.C.C § 4A- 211 (a) The ... instructions appear not to have been authentic U.C.C § 4A- 211 (c) U.C.C § 4A -10 6(a) refers to U.C.C § 1- 2 01( 27) Official Comment to U.C.C § 4A- 211 U.C.C § 4A- 211 (c) There is an exception, however, to the...
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Design and Optimization of Thermal Systems Episode 1 Part 4 pps
... liquids Radiative transport Condensation Fluidized bed h(W/m2 · K) 5 10 50 10 0 35 0—5 50 50 0—2 ,500 600 10 ,000 600 10 ,000 60 0—5 ,000 Convection has the advantage of heating the materials only up ... different sources, and constructive criticism Several such methods that may help to develop creative thinking are discussed by Alger and Hays (19 64) and by Lumsdaine and Lumsdaine (19 95) Techniques ... in thickness from D1 to D2 across a set of rollers, as shown in Figure 1. 10(d), mass conservation leads to the equation U1D1 U2 D2, where U1 is the speed before the rollers and U2 after, if the...
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Design and Optimization of Thermal Systems Episode 2 Part 4 doc
... and the p’s are the pressures Thus, h 4s h3 v3 (p4 – p3) 16 7.57 1. 0078 16 7.57 8. 64 17 6. 21 kJ/kg 10 (8.5 81 – 0.0073 84) 10 3 where the pressures are in MPa and 10 3 is used to obtain the work in kJ/kg ... valve –25°C Evaporator Entropy, s 2a T(°C) –25 P(kPa) 15 1.5 Saturated vapor 18 8.7 11 67 .1 Superheated vapor 30 11 67 .1 Saturated liquid –25 15 1.5 Liquid-vapor mixture FIGURE 5.3 Thermodynamic cycle ... then gives h4 16 7.57 10 .8 17 8.37 kJ/kg These values can now be used to obtain the water flow rate as 26. 24 kg/s The heat input is given by m (h1 – h4) 26. 24 (2 749 .0 – 17 8.37) kW 67 .45 MW The heat...
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Design and Optimization of Thermal Systems Episode 3 Part 4 ppsx
... equations, w1 0. 611 and w2 0.389 Therefore, C* (3.5/0. 611 )0. 611 ( 14 . 8/0.389)0.389 11 .965 Then, from Equation (10 .13 ) and Equation (10 . 14 ) , 3.5 m1 .4 0. 611 C* which gives m (0. 611 C*/3.5 )1/ 1 .4 1. 692 ... function, U(x1, x2, , xn), in unconstrained problems, which can be treated by geometric programming are U x2 / U U U x1x1/ x1 x1 550 10 5 x1 x1x2 x1 x1 x3.3 (10 .1) (10 .2) 12 0, 000 10 x1 x1 (10 .3) 2 ... Design and Optimization of Thermal Systems 0 .1 0.05 –0.05 –0 .1 –0 .15 –0.2 –0.25 –0.3 –0.35 –0 .4 –0 .45 10 1 –2 –3 4 –5 –6 –7 –8 –9 10 11 DP 10 –20 10 St –0.5 L2 L1 L2 10 0 0.5 (a) L1 (b) 0.026...
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Validation of Communications Systems with SDL phần 4 docx
... with SDL 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 51 52 else ; define test_failed True; echo Error found in test1.scn ||| test_res1.wri; ... Figure 4. 66, once you have executed the start transition, timer TIMER1 is started, and process proc1 is in state st1 system L_time block b1 process proc1 st1 SET (TIMER1) b1 proc1 NONE st1 x:= st1 ... script1.wri:’ in the simulator trace and into the file test res1.wri • Line 11 : source test1.scn: same as loading scenario test1.scn and redoing all • Line 14 : if (step = 41 ): step contains the current...
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Fundamentals of Database systems 3th edition PHẦN 4 pps
... Note Note Note 10 Note 11 Note 12 Note 13 Note 14 Note 15 Note 16 Note 17 Note 18 Note 19 Note Codd (19 85) specified 12 rules for determining whether a DBMS is relational Codd (19 90) presents ... connectives—(F1 and F2), (F1 or F2), not(F1), and not(F2)—depending on whether it is free or bound in F1 or F2 (if it occurs in either) Notice that in a formula of the form F = (F1 and F2) or F = (F1 ... go beyond the standard key, entity integrity, and referential integrity constraints imposed by the system 10 .4 Storage Organization in Oracle 10 .4. 1 Data Blocks 10 .4. 2 Extents 10 .4. 3 Segments A...
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FUNDAMENTALS OF DATABASE SYSTEMS Fourth Edition phần 4 pptx
... 12 345 6789 12 345 6789 12 345 6789 12 345 6789 6668 844 44 6668 844 44 45 345 345 3 45 345 345 3 45 345 345 3 45 345 345 3 45 345 345 3 33 344 5555 33 344 5555 33 344 5555 33 344 5555 33 344 5555 33 344 5555 33 344 5555 33 344 5555 FIGURE ... 33 344 5555 33 344 5555 9876 543 21 9876 543 21 33 344 5555 33 344 5555 9876 543 21 888665555 redundancy PLaCATION ENAME 12 345 6789 12 345 6789 6668 844 44 45 345 345 3 45 345 345 3 33 344 5555 33 344 5555 33 344 5555 33 344 5555 ... E 12 345 6789 33 344 5555 999887777 9876 543 21 6668 844 44 45 345 345 3 987987987 888665555 19 65- 01- 09 19 55 -12 -08 19 68-07 -19 1 9 41 -06-20 19 62-09 -15 19 72-07- 31 1969-03-29 19 37 -11 -10 ADDRESS DNUMBER 7 31 Fondren,Houston,TX...
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Analysis and Control of Linear Systems - Chapter 4 ppt
... are very similar to algorithms [4. 5] and [4. 4]: i i n * * A = {0}, A 1+ 1= E−1HA 1, ⇒ A = A = E−1HA [4. 18 ] n A = X , A i +1= H−1EA i , ⇒ A = A * = H−1EA * 2 2 [4. 19 ] The regularity of the beam ... the bound of algorithms [4. 4] and [4. 5]: L = X , L 1= L, L = L ∩ A−1L, L0 = {0}, L1 = L, L2 = L + AL, , L i +1= L ∩ A−1Li ⇒ L n = L* , Li +1 = L + ALi , ⇒ Ln = L* [4. 4] [4. 5] The concept of A-invariant ... ECC’93, p 11 19 -11 24, 19 93 [MAL 97] MALABRE M., MARTINEZ-GARCIA J.C., DEL MURO CUELLAR B., “On the fixed poles for disturbance rejection”, Automatica, vol 33, no 6, p 12 09 -12 11, 19 97 [MAR 94] MARTINEZ-GARCIA...
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the behavior of stock market prices eugene f fama the journal phần 4 pps
... 15 9 209 14 3 14 2 17 2 16 2 19 8 19 3 16 2 .1 153.7 17 2 .4 15 5.9 17 2.5 16 0 .4 15 9.3 2 21. 6 16 1.9 16 0 .1 2 24. 7 19 1 .4 205.2 15 7.6 1 64. 2 1 64. 0 19 3.9 16 0.0 16 8.6 19 0.6 17 2.8 228 .4 15 9.2 19 7.2 15 5.2 15 0.5 16 1 .4 ... 69 .4 73.3 81. 2 70.6 98.6 68.7 83.8 63 .4 66.7 68.2 70.3 84. 4 80.9 39 41 48 34 41 36 41 54 43 43 51 43 44 36 40 34 51 39 36 40 40 59 29 50 29 36 45 37 41 48 75.3 41 .6 - - 74. 6 38.6 39.0 43 .9 ... 80.5 1. 5 10 5.8 3.2 83.3 4. 7 12 8.0 -1. 0 -2.0 91. 0 72.5 1. 5 60.7 -0.7 10 8.8 4. 2 12 6.9 6 .1 14 9 .0 4. 0 96 .1 -1. 1 12 1 .1 -5 .1 95 .1 0.9 98.6 3 .4 98.9 -2.9 12 8 .1 -7 .1 139.3 1. 7 13 9 .4 4.6 13 7.5 1. 5 13 2.8...
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Mechanics of Microelectromechanical Systems - N.Lobontiu and E.Garcia Part 4 doc
... microhinges, microbridges 10 5 where the compliances of Eqs (2 .13 8) are calculated as: The superscripts (1) and (2) in Eqs (2 .13 9), (2. 14 0 ) and (2. 14 1 ) indicate the first component and the second component, ... compliance being: Chapter 80 Equations (2 . 41 ), (2 .43 ) and (2 .45 ) can be solved for the unknown amounts and by also employing Eqs (2 .42 ), (2 .44 ) and (2 .46 ) 2.2.2.2 Three Dimensional Force Detection ... in Eqs (2 .13 0) to (2 .1 34) , as shown in Fig 2.25 (b) Equation (2 .1 34) also serves at determining the corresponding axial compliance by simple inversion, so that the compliance of Eq (2 .13 2) Microcantilevers,...
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The Behavior of Structures Composed of Composite Materials Part 4 potx
... of 0 .12 7 mm (0.005") cited in Problem 6, calculate the elements of the [A], [B] and [D] matrices for the following laminates: (a) [ +45 / -45 / -45 / +45 ] (b) [ +45 / -45 / +45 / -45 ] (c) 81 10 11 12 13 (d) ... due to lateral loads given by Equations (3. 14 ) , (3 .15 ) and (3 .12 ) become: 92 where Equations (3 .16 ) and (3 .17 ) can be substituted into Equation (3 .18 ) with the result that: The above equations ... to be 0 .1" and neglecting Poisson’s Ratio, what are the new moduli in the and directions 15 The properties of graphite fibers and a polyimide matrix are as follows: GRAPHITE POLYIMIDE 16 17 (a)...
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Dynamics of Mechanical Systems 2009 Part 4 docx
... ( 10 n ) + ( 12 n ) × ( 15 n1 ) + (3n1 − 4n ) × ( −8n ) + O + 3n1 × (6n1 + 24n + 8n ) + 3n1 × (12 n + 16 n ) = 12 0n1 + 12 0n1 + 30n − 18 0n − 24n − 32n1 + 72n − 24n − 48 n or MQ = −32n1 − 42 n − 13 2n ... × F4 + OB × F5 + OO × F6 + OG × F7 = + 3n1 × ( 10 n ) + 4n × ( 15 n1 ) + 3n1 × ( −8n ) + 4n × (8n ) + + 12 n × (12 n1 + 16 n ) = 30n = 60n − 24n − 32n1 − 14 4 n + 19 2n1 or: MO = 16 0n1 − 30n − 16 8n ... validity of Eq (6.3 .17 ): ? 16 0n1 − 30n − 16 8n = −32n1 − 42 n − 13 2n + (12 n + 4n ) × (3n1 + 24n + 24n ) = −32n1 − 42 n − 13 2n − 36n + 288n1 + 12 n − 96n1 = 16 0n1 − 30n − 16 8n Equation (6.3 .17 ) is thus verified...
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The Design of Manufacturing Systems Part 4 pdf
... 4 Structural Control of Large-Scale Flexibly Automated Manufacturing Systems 4. 1 4. 2 Introduction The FMS Operational Model and the Manufacturing System Deadlock ... Policies — General Definitions 4. 4 Single-Unit RAS Admitting Polynomially Computable Optimal DAP Deadlock Detection in Single-Unit RAS • Single-Unit RAS Models with No Deadlock-Free Unsafe States 4. 5 ... Single-Unit RAS 4. 7 Additional Issues and Future Directions in FMS Structural Control Current strategic and technological trends in discrete-part manufacturing require extensive and flexible automation...
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Theory and Design of CNC Systems Part 4 potx
... 1) 1 0 2 0 2 2 2 a( j) a( j − 1) a( j − 2) 0 0 0 0 0 0 b( j) b ( j) b ( j) 0 0 0 0.25 1 0.75 2 4 4 4 2 4 2 2 2 2 0 2 11 2 21 10 11 12 6 6 6 4 6 4 4 4 4 2 4 31 41 13 14 15 16 17 18 19 20 6 6 4 ... X(k − 5) 10 10 10 10 10 10 10 10 10 10 10 10 10 10 11 10 12 10 13 10 Σ 80 Output of Output Adder: S(k) Pulse: Δ X0 (k) 10 20 30 40 50 10 50 10 50 10 50 10 40 30 20 10 0 80 As shown in Fig 4. 3, initial ... 4 4 6 6 4 6 6 6 4 6 6 6 6 6 6 6 6 6 6 6 4 6 6 6 51 6 6 51 21 22 23 24 2 2 2 4 2 4 4 4 4 6 4 41 31 25 26 27 28 0 0 0 2 0 2 2 2 2 4 2 21 2 11 29 30 31 32 0 0 0 0 0 0 0 0 0 0 0 2 0 1 0 11 2 21 1.25...
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Maintenance of Petroleum Systems B Episode 4 pot
... operations All settings and functions are the same as the 41 AF 49 UFC 3 -46 0-03 21 JANUARY 2003 Figure 5 .12 Combination Dual Rate-of-Flow Control Valve and Solenoid Valve ( 41 AF -10 ) 5 .16 .2 Combination ... rated for 22 71 liters per minute (600 gallons per minute) See Chapter for a detailed description 6 .4. 5 FSCV ( 41 AF-2C) 6 .4. 5 .1 The only difference between the 41 AF-2C (Figure 6 .4) and the 40 AF-2C (Figure ... open and the system pressure should drop 5 .16 .3 .1 .4 When the 51AF -4 closes, note the pressure gauge reading 51 UFC 3 -46 0-03 21 JANUARY 2003 5 .16 .3 .1. 5 If the pressure is more than psi, slowly...
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Modeling of Combustion Systems A Practical Approach 4 doc
... 3.0 4. 48 4. 89 5.29 5.70 6 .11 25 .44 25.03 24. 63 24. 22 23. 81 129. 61 132.88 13 5.93 13 8.78 14 1 .47 97.57 10 0. 84 10 3.88 10 6.73 10 9 .42 10 20 .1 1 018 .2 10 16 .4 10 14 . 7 10 13.2 11 17.6 11 19.0 11 20.3 11 21. 5 11 22.6 ... 15 3.65 15 6.30 15 8. 84 16 1.26 16 5.82 986.8 985 .1 983.6 982 .1 979.3 1 14 0 .4 1 14 1 .4 1 14 2 .4 1 14 3 .3 1 14 5 .1 0. 016 545 0. 016 5 61 0. 016 577 0. 016 592 0. 016 622 44 .733 42 .40 2 40 . 310 38 .42 0 35. 14 2 12 .0 13 .0 14 . 0 ... 20. 942 16 92.5 23.332 18 02.9 60.0 45 .3 292. 71 V hg 7. 815 12 08.0 8.3 54 12 33.5 9 .40 0 12 83.2 10 .42 5 13 32.3 11 .43 8 13 81. 5 12 .44 6 14 3 1. 3 13 .45 0 14 8 1. 8 14 . 452 15 33.2 15 .45 2 15 85.3 17 .44 8 16 92 .4 19 .44 1 1802.8...
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A methodology for validation of integrated systems models with an application to coastal-zone management in south-west sulawesi
... 10 5 10 5 10 6 10 9 10 9 11 0 References……………………………………………………………………… 11 1 Symbols…………………………………………………………………………… 12 3 Acronyms and abbreviations…………………… 12 7 Summary………………………………………………………………………… 12 9 ... of 19 91 to 20 01 is depicted in Fig .1. 7 It is noted that, due to the Asian economic crisis in 19 97, the inflation of the Indonesian Rupiah is remarkably high 1. 6 1 .4 1. 2 0.8 0.6 0 .4 0.2 19 91 1992 ... 6 .4 Recommendations……………………………………………………… 6 .4. 1 Other directions for the validation of integrated systems models… 6 .4. 2 Proper use of integrated systems models………………………… 10 3 10 3 10 3 1 04 1 04 10 5...
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DIA 4-HDSX of nguoi dan o DB NB
... ngày 20 tháng năm 2009 Địa lí Hoạt động sản xuất người dân đồng Nam Bộ (tt) Hoạt động 1: Thảo luận nhóm SGK /1 24 12 5 Dựa vào sách giáo khoa, đồ công nghiệp Việt Nam, hình ảnh nội dung mục 3, thu thập ... CHỮ KỲ DIỆU C H Ế B I Ế N 8 10 5 7 Đây ô chữ có chữ hoạt động sản xuất người dân lương thực thực phẩm, đem lại hiệu lớn TRÒ CHƠI: Ô CHỮ KỲ DIỆU C H Ế B I M Ê C Ô N G 8 10 Ế N 5 7 Đây ô chữ có chữ ... I M Ê C Ô N G Đ Ồ N G B Ằ N G 8 10 Ế N 5 7 Đây ô chữ có chữ vùng đất rộng lớn phẳng TRÒ CHƠI: Ô CHỮ KỲ DIỆU C M Ê C Ô Đ Ồ N G B Ằ N S Ô N G H Ế B I N G G Ế N 5 10 Đây ô chữ có chữ cái: Nét văn...
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