3 2 mathematical modeling of mechanical systems

... · PI3K]i V PIP2 Ns kass [PIP2 ]i [PTEN]cyto PIP2 Ns kdiss [PIP2 · PTEN]i PI3K·Rec(i)+PIP2 (i) → PI3K·Rec(i)+PIP3 (i) PI3K kcat [Rec·PI3K]i [PIP2 ]i PI3K PTEN·PIP2 (i)+PIP3 (i) → PTEN·PIP2 (i)+PIP2 ... (i)+PIP2 (i) PTEN kcat [Rec·PTEN]i [PIP3 ]i PTEN K M +[PIP3 ]i KM √D 3Ssite PIP2 (i)→PIP2 (j) √D 3Ssite PIP3 (i)→PIP3 (j) ′ ′ +[PIP2 ]i [PIP2 ]i − [PIP2 ] j + [PIP3 ]i − [PIP3 ] j + Table 3. 2 Physical ... 6 23 Springer, Berlin, pp 1 53 178 (20 03) 24 Rozi, A., Jia, Y.: A theoretical study of effects of cytosolic Ca2+ oscillations on activation of glycogen phosphorylase Biophys Chem., 106, 1 93 20 2...

Ngày tải lên: 27/06/2014, 10:20

... 6 .2 .3. 3 Point loads 28 9 29 0 2 93 29 9 30 3 30 4 30 5 30 7 30 7 31 1 31 2 31 2 31 3 31 3 31 3 31 4 31 6 31 6 31 6 31 7 31 8 31 9 31 9 32 2 32 2 32 4 32 4 32 5 32 6 xii ... 5 .3. 4.4 Uniform plate stretching 23 5 23 7 23 8 24 0 24 0 2 43 24 5 24 7 24 8 25 4 25 6 25 9 26 0 26 0 26 0 26 2 26 2 26 2 2 63 2 63 2 63 26 5 26 5 26 5 26 6 26 7 27 0 27 0 27 2 2 73 27 5 27 5 ... 177 180 186 188 189 190 190 190 1 93 196 196 20 0 20 0 20 5 20 7 20 7 20 9 21 0 21 0 21 1 2 13 2 13 21 4 21 7 21 8 22 0 22 0 22 1 22 2 22 2 22 4 22 9 23 1 23 1 23 2 x Contents 4.4.1 .2 Truncation stiffness for a free-free...

Ngày tải lên: 22/03/2014, 13:20

... (2. 8 .3) in the forms: A1 A ⋅ (B × C) = B1 C1 B1 = − A1 C1 A2 B2 C2 B2 A2 C2 A3 B1 B3 = C1 C3 A1 B3 A1 A3 = − C1 C3 B1 B2 C2 A2 B3 C1 C3 = A1 A3 B1 A2 C2 B2 C2 A2 B2 A3 C1 C3 = − B1 B3 A1 C3 A3 ... B = A1B1 + A2 B2 + A3 B3 (2. 8.1) and n1 A × B = A1 B1 n2 A2 B2 n3 A3 B3 (2. 8 .2) where the Ai and Bi (i = 1, 2, 3) are the ni components of A and B By comparing Eqs (2. 8.1) and (2. 8 .2) , we see ... Eq (2. 6 .21 ), C • A is: C ⋅ A = ( 4)(7 ) + (60)( 2) + ( 23 ) ( 4) = (2. 7 .26 ) C ⋅ B = ( 4)(1) + (60) (3) + ( 23 ) ( −8) = (2. 7 .27 ) Similarly, C · B is c From Eq (2. 7. 23 ) , B × A is: n1 B×A = n2 2 n3 −8...

Ngày tải lên: 10/08/2014, 20:20

... s1c2 c3 + c1s3 )n + ( − s1c2 s3 + c1c3 )n and n1 = c2N1 + c1s2N − s1s2N n = s2c3N1 + (c1c2c3 − s1s3 )N + ( s1c2c3 + c1s3 )N (4 .3. 14) n = − s2 s3N1 + ( − c1c2 s3 − s1c3 )N + ( − s1c2 s3 + c1c3 ... orientation of B in R (see Figure 4 .3. 7) Hence, the unit vectors are related by the expressions: N1 = c2n1 + s2c3n − s2 s3n N = − c1s2n1 + (c1c2c3 − s1s3 )n + ( − c1c2c3 − s1c3 )n (4 .3. 13) N = − s1s2n1 ... S V L = 22 .38 1n x − 1.571n y + 2. 721 n z ft sec (4.9.17) and S ( ) − 0.88n ) × (0. 433 n a L = −19 .36 n y + ( −5. 529 )n y × 0. 433 n y + 0 .25 n z [ +(6 .28 3n x − 0.88n z ) × (6 .28 3n x z y + 0 .25 n z )]...

Ngày tải lên: 10/08/2014, 20:20

... 19. 93 2. 258 2. 258 2. 258 2. 258 2. 258 2. 258 2. 258 2. 258 2. 258 2. 258 2. 201 2. 180 2. 127 2. 1 13 2. 089 2. 045 1.996 1.948 1.8 82 1. 833 9.91 13. 40 22 .36 24 .70 28 .55 35 .64 43. 39 50.87 60. 63 67.76 24 .07 23 . 59 ... 32 2. 1 2. 075 2. 125 2. 1 82 2. 23 7 20 1 /2" 20 3/ 4" 21 " 21 1/4" 20 .5000 20 .7500 21 .0000 21 .25 00 64.4 03 65.188 65.974 66.759 33 0.1 33 8 .2 34 6.4 35 4.7 2. 2 92 2 .34 8 2. 405 2. 4 63 21 1 /2" 21 3/ 4" 22 " 22 1/4" 21 .5000 ... 106 .2 101.6 96.1 90.8 86.6 80.5 6. 52 7.11 9.84 12. 88 14.58 15.74 19 .24 21 . 52 26.04 31 .5 36 .9 41.1 47.1 3. 34 3. 34 3. 34 3. 34 3. 34 3. 34 3. 34 3. 34 3. 34 3. 34 3. 34 3. 34 3. 34 3. 25 3. 24 3. 21 3. 17 3. 14 3. 13...

Ngày tải lên: 13/08/2014, 05:22

... W/m2 W/m2 K J/mol kJ/kg Mass [M] 1 –1 –1 1 1 1 Exponents Length Time θ [L] [θ] 3 3 3 2 2 –1 –1 –1 –1 1 1 1 1 2 2 2 Temperature [T] 2 2 2 2 2 2 2 2 –1 –1 –1 2 2 2 3 2 2 2 3 2 ... 4187 J 3. 968 Btu/hr 1.1 62 J/sec 0.1 124 Btu/ft3 4187 J/m3 2. 205 lb 0.0 027 8 g/sec-cm 0.6 72 lb/hr-ft 0.0 62 43 lb/ft3 34 13 Btu/hr 1 .34 1 hp 660.6 kcal/hr 31 7 .2 Btu/hr-ft2 176 .2 Btu/hr-ft2-°F 0.4 536 kg ... cal/cm2-sec = cal/cm-sec-°C = cal/g = cal/g-°C = centipoise = cm2/sec = ft = ft2/sec = g/cm3 = hp = 25 2.0 cal 1055 J 0.00890 cal/cm3 0. 037 3 MJ/m3 0.00 039 31 hp 0 .25 20 kcal/hr 0 .2 931 W 0 .2 93 MW 0.0 031 53...

Ngày tải lên: 13/08/2014, 05:22

... , 29 430 3, 21 23 , 20 10 ix àà, à, , , , 30 431 2, 21 23 , 20 10 àà, , ,  à, , 20 11, , 26 1, 20 11 àà, , , à, ààà : , 20 11, , 26 ... [21 ] , , , [22 ] à [19] à [ 23 ] [24 ] à àà à, [25 , 26 , 27 , 28 , 29 , 30 , 31 , 32 ], [2, 3, ... (3. 18) (3. 19) 3. 1 22 = 0, = ; = ; = > > > > () <  = > > > >  () : (3. 20 ) (3. 21 )  = 0 (,,) 3. 1.6 (3. 22 ) à, , = > > > > (, ) < > > > >0 : ; (3. 23 ) ...

Ngày tải lên: 09/09/2015, 17:54

... i 2 )k2 −(1 + i 2 )k2 (1 + iη1 )k1 0 0 + e1 0 −(1 + i 2 )k2 (1 + i 2 )k2 + (1 + i 3 )k3 +e2 1 0 (1 + i 2 )k2 −(1 + i 2 )k2 m1 + e3 −ω 1 m2 (1 + i 3 )k3 −(1 + i 2 )k2 (1 + i 2 )k2 H1 H2 = f1 f2 ... by:   2EI −EI  9l 3l2  d = F ( 32 )  −EI 2EI  θ M 3l2 3l3 And from a numerical point of view, the stiffness matrix is an interval matrix: [4 033 .68, 436 9.68] [−6554. 52, −6050. 52] [−6554. 52, −6050. 52] ... t3 is (t0 + δt )3 = t3 + 3 t.t2 + 3t0 δt2 + δt3 0 (38 ) The matrix equation (37 ) becomes : t3 [K](1 + iη) − ω t0 [M ] + δt 3. t2 [K](1 + iη) − ω [M ] 0 + 3t0 δt2 + δt3 [K](1 + iη) {H} = {F } (39 )...

Ngày tải lên: 12/01/2014, 21:45

... 18 Activity 3. 2: Identifying Sources of Information Exercise 1: Identifying Sources of Information ! Develop a list of sources of information Review the information ... the following table, list examples for each source of information You will discuss your results in a class discussion Sources Examples Artifacts Systems People ...

Ngày tải lên: 24/01/2014, 10:20

... Matrices 23 7 23 7 20 1 2 03 20 9 21 2 21 2 22 0 22 4 22 4 22 7 22 8 22 8 23 1 23 9 24 2 Contents 5.1 .3 Modal Decomposition Case I: Undamped System (C = 0) Case II: Proportional or Rayleigh Damping 5 .2 Continuous Systems ... Potential Energies of the Beam Exercise Problems xi 24 5 24 6 24 9 25 0 25 0 25 1 25 5 25 8 26 1 26 5 26 5 26 7 26 9 27 1 2 73 27 9 27 9 2 83 28 6 29 1 29 5 APPENDIX A: EQUIVALENT STIFFNESSES (SPRING CONSTANTS) OF BEAMS, ... 1 .3. 12, the total PE in the system is PEtot = 1 k1 x + k2 x1 2 (1 .3. 33) where from Figure 1 .3. 13, x1 = r3 θ (1 .3. 34) Substituting Equation 1 .3. 34 into Equation 1 .3. 33 and using Equation 1 .3. 23 , ...

Ngày tải lên: 08/04/2014, 11:49

... Information Scheduling General Outsource Process Systems Control C11 I 12 S 13 G14 O15 C21 I 22 S 23 G24 O25 C31 I 32 S 33 G34 O35 C41 I 42 S 43 G44 I 52 S 53 G45 Fig 10 Path through IBIS network In fact, ... not just one Refer to Figure 11 for an illustration of this process J MacGregor Smith 18 11 i 12 i1 i i2 i 22 11 i 21 12 21 i1 i2 i i 22 i3 32 31 Fig 11 Many explanations possible for i The paradox ... units of cycle time For the case of M > 2machine serial lines, the following formula had been derived (Enginarlar et al., 20 03a): e(1 Q)(eQ + e)(eQ + 2e) (2 Q) ì Q(2e 2eQ + eQ2 + Q 2) ...

Ngày tải lên: 08/04/2014, 12:26

... into the form Pe,sup δmin = − Γ|υ |2 δmin 2 = e−(δmin /2 ) ( 32 ) δmin 2 Pe δmin = erf c π = √ ∞ δmin /2 exp − x2 dx (33 ) So, the theoretical total probability of errors Pe can be ex∞ pressed as ... > δmin /2 The cdf of |υ |2 is F|υ |2 (u) = P(|υ |2 < u) or equivalently as F|υ |2 (u) = Γu/σ (1) = − exp(−u/σ ) For the upper bound Pe,sup (δmin ) = P(|υ |2 > (δmin /2) 2 ), the probability of errors ... theory to practice: an overview of MIMO space-time coded wireless systems, ” IEEE Journal on Selected Areas in Communications, vol 21 , no 3, pp 28 1 30 2, 20 03 [3] V Erceg, K V S Hari, M S Smith,...

Ngày tải lên: 22/06/2014, 19:20

... 6 .2 .3. 3 Point loads 28 9 29 0 2 93 29 9 30 3 30 4 30 5 30 7 30 7 31 1 31 2 31 2 31 3 31 3 31 3 31 4 31 6 31 6 31 6 31 7 31 8 31 9 31 9 32 2 32 2 32 4 32 4 32 5 32 6 xii ... 5 .3. 4.4 Uniform plate stretching 23 5 23 7 23 8 24 0 24 0 2 43 24 5 24 7 24 8 25 4 25 6 25 9 26 0 26 0 26 0 26 2 26 2 26 2 2 63 2 63 2 63 26 5 26 5 26 5 26 6 26 7 27 0 27 0 27 2 2 73 27 5 27 5 ... 177 180 186 188 189 190 190 190 1 93 196 196 20 0 20 0 20 5 20 7 20 7 20 9 21 0 21 0 21 1 2 13 2 13 21 4 21 7 21 8 22 0 22 0 22 1 22 2 22 2 22 4 22 9 23 1 23 1 23 2 x Contents 4.4.1 .2 Truncation stiffness for a free-free...

Ngày tải lên: 27/06/2014, 05:20

... 6 .2 .3. 3 Point loads 28 9 29 0 2 93 29 9 30 3 30 4 30 5 30 7 30 7 31 1 31 2 31 2 31 3 31 3 31 3 31 4 31 6 31 6 31 6 31 7 31 8 31 9 31 9 32 2 32 2 32 4 32 4 32 5 32 6 xii ... 5 .3. 4.4 Uniform plate stretching 23 5 23 7 23 8 24 0 24 0 2 43 24 5 24 7 24 8 25 4 25 6 25 9 26 0 26 0 26 0 26 2 26 2 26 2 2 63 2 63 2 63 26 5 26 5 26 5 26 6 26 7 27 0 27 0 27 2 2 73 27 5 27 5 ... 177 180 186 188 189 190 190 190 1 93 196 196 20 0 20 0 20 5 20 7 20 7 20 9 21 0 21 0 21 1 2 13 2 13 21 4 21 7 21 8 22 0 22 0 22 1 22 2 22 2 22 4 22 9 23 1 23 1 23 2 x Contents 4.4.1 .2 Truncation stiffness for a free-free...

Ngày tải lên: 27/06/2014, 08:20

... rad deg m2 cm2 m2 cm2 in .2 ft2 ft2 in .2 cm2 ft2 in .2 m2 Nm (or J) Nm (or J) in lb ft lb ft lb in lb N lb 3. 048 3. 048 1 .20 0 3. 105 3. 28 0 3. 937 1.000 1.018 8 .33 3 2. 540 2. 540 2. 587 3. 28 0 3. 937 1.000 ... 5.1 53 1.940 1.000 1.940 1.000 1 .35 5 1 .20 0 1. 129 8 .33 3 7 .37 5 8.850 1 .35 5 7 .37 5 1.818 1 .35 5 3. 030 2. 259 7.456 7 .37 5 4. 425 1 .34 1 5.500 3. 300 6.894 1.440 4.788 6.944 1.450 2. 088 3. 048 3. 048 1.097 1 .20 0 ... lb/in .2 (or psi) lb/ft2 m/sec cm/sec km/hr in./sec mi/hr (or mph) m/sec cm/sec km/hr 3. 048 3. 048 1 .20 0 2. 540 2. 540 8 .33 3 3. 28 0 3. 937 1.000 3. 28 0 3. 937 1.000 5 .28 0 1.609 6 .2 13 1.459 6.8 52 5.1 53 5.153...

Ngày tải lên: 27/06/2014, 15:20

... rad deg m2 cm2 m2 cm2 in .2 ft2 ft2 in .2 cm2 ft2 in .2 m2 Nm (or J) Nm (or J) in lb ft lb ft lb in lb N lb 3. 048 3. 048 1 .20 0 3. 105 3. 28 0 3. 937 1.000 1.018 8 .33 3 2. 540 2. 540 2. 587 3. 28 0 3. 937 1.000 ... 5.1 53 1.940 1.000 1.940 1.000 1 .35 5 1 .20 0 1. 129 8 .33 3 7 .37 5 8.850 1 .35 5 7 .37 5 1.818 1 .35 5 3. 030 2. 259 7.456 7 .37 5 4. 425 1 .34 1 5.500 3. 300 6.894 1.440 4.788 6.944 1.450 2. 088 3. 048 3. 048 1.097 1 .20 0 ... lb/in .2 (or psi) lb/ft2 m/sec cm/sec km/hr in./sec mi/hr (or mph) m/sec cm/sec km/hr 3. 048 3. 048 1 .20 0 2. 540 2. 540 8 .33 3 3. 28 0 3. 937 1.000 3. 28 0 3. 937 1.000 5 .28 0 1.609 6 .2 13 1.459 6.8 52 5.1 53 5.153...

Ngày tải lên: 10/08/2014, 20:20

... validity of Eq (6 .3. 17): ? 160n1 − 30 n − 168n = 32 n1 − 42n − 1 32 n + (12n + 4n ) × (3n1 + 24 n + 24 n ) = 32 n1 − 42n − 1 32 n − 36 n + 28 8n1 + 12n − 96n1 = 160n1 − 30 n − 168n Equation (6 .3. 17) is ... (l 2) 2 sin 2 − (l 2) 2 cos 2 N Z (5.6.14) Similarly, the velocity of acceleration of O3 is: ˙ ˙˙ vO3 = lθ1n11 + l θ2n 21 (5.6.15) ˙˙ 2 ˙˙ 2 aO3 = l θ1n11 − lθ1 n 13 + l θ2n 21 − lθ2n 23 ... in Eqs (5 .3. 26 ) and (5 .3. 27 ), we immediately obtain Eqs (5 .3. 22 ) and (5 .3. 23 ) Finally, observe that by differentiating in Eqs (5 .3. 22 ) and (5 .3. 23 ) we obtain Eqs (5 .3. 24 ) and (5 .3. 25 ) 5.4 Instant...

Ngày tải lên: 10/08/2014, 20:20

... I 22 + I 33 (7.7. 13) I II = I 22 I 33 − I 32 I 23 + I11I 33 − I 31 I 13 + I11I 22 − I 12 I 21 (7.7.14) I III = I11I 22 I 33 − I11I 32 I 23 + I 12 I 31 I 23 − I 12 I 21 I 33 + I 21 I 32 I 13 − I 31 I 13 ... as: 05 93_ C07_fm Page 20 4 Monday, May 6, 20 02 2: 42 PM 20 4 Dynamics of Mechanical Systems D = d11n1n1 + d12n1n + d13n1n + d21n 2n1 + d22n 2n + d23n 2n (7.4.6) + d31n 3n1 + d22n 3n + d33n 3n = dij ... operations: D ab = (a n 1 + a2n + a3n )(b1n1 + b2n + b3n ) = a1b1n1n1 + a1b2n1n + a1b3n1n + a2b1n 2n1 + a2b2n 2n + a2b3n 2n (7.4.5) + a3b1n 3n1 + a3b2n 2n + a3b3n 3n where the unit vector products...

Ngày tải lên: 10/08/2014, 20:20