Introduction to Elasticity Part 1 docx

... kJ/mol 10 −6 , ◦ C 1 Pb 14 (2) 327 5.4 29 Al 69 (10 ) 660 10 .5 22 Cu 11 7 (17 ) 10 84 13 .5 17 Fe 207 (30) 15 38 15 .3 12 W 407 (59) 3 410 32 4.2 The system will generally have sufficient thermal energy to ... parameter w/AE corresponding to λ = 3 is read from the graph to be 1. 11. For a jumper weight of 15 0 lb, this corresponds to A =1. 35 in 2 , or a cord diameter o...

Ngày tải lên: 11/08/2014, 09:20

... 2.0 2.49 34.3 1. 8 22–33 aramid 12 4 3.6 2.3 1. 45 86 2.5 22–33 boron 400 3.5 1. 0 2.45 16 3 1. 43 330–440 HS graphite 253 4.5 1. 1 1. 80 14 0 2.5 66 11 0 HM graphite 520 2.4 0.6 1. 85 2 81 1.3 220–660 Of ... lattice energy U. 16 type r 0 (pm) K (GPa) A n U(kJ/mol) U expt LiF 2 01. 4 6. 710 e+ 01 1.750 -10 14 NaCl 282.0 2.400e+ 01 1.750 -764 KBr 329.8 1. 480e+ 01 1.750 -663 T...

Ngày tải lên: 11/08/2014, 09:20

... type system MPa Nickel fcc {11 1} 11 0 5.7 Copper fcc {11 1} 11 0 0.98 Gold fcc {11 1} 11 0 0.90 Silver fcc {11 1} 11 0 0.60 Magnesium hcp {11 01} 0 01 0. 81 NaCl cubic {11 0} 11 0 0.75 Slip occurs when ... systems; using Miller indices, these are the {11 1} planes and the 11 0 directions. There are 4 independent nonparallel (11 1) planes, and 3 independent [11 0] directions in each p...

Ngày tải lên: 11/08/2014, 09:21

... class="bi x0 y1 w2 h2" alt=""

Ngày tải lên: 04/05/2014, 14:43

... delivering 10 0 hp (horsepower) at 18 00 rpm (revolutions per minute) to the drive shaft, and we wish to compute the shearing stress. From Eqn. 8, the torque on the shaft is T = W ω = 10 0 hp  1 1.3 41 10 −3  N·m s·hp 18 00 rev min 2π rad rev  1 60  min s = ... element’s volume is ∆V V = a  b  c  − abc abc = a (1 +  x ) b (1 +  y ) c (1 +  z ) − abc abc = (1+  x ) (...

Ngày tải lên: 11/08/2014, 09:20

... ’sigma_prime’=map(evalf,evalm(aa&*sigma&*transpose(aa))); [ 1. 3.232 1. 598] sigma_prime = [3.232 8.830 3.366] [1. 598 3.366 1. 170] 10 Figure 4: Finite displacements.  x = O 1 B 1 − OB OB = O 1 B 1 − 1 =  1+ 2 ∂u ∂x +  ∂u ∂x  2 +  ∂v ∂x  2 1 Using ... matrices: > a1:=array (1 3 ,1 3,[[cos(psi),sin(psi),0],[-sin(psi),cos(psi),0],[0,0 > ,1] ]); [cos(psi...

Ngày tải lên: 11/08/2014, 09:21

... relative to the x-y axes is then S = RA 1 R 1 SA =   .8830 × 10 10 − .19 70 × 10 10 − .12 22 × 10 −9 − .19 71 × 10 10 .2072 × 10 −9 −.83 71 × 10 10 − .12 22 × 10 −9 −.8369 × 10 10 −.2905 × 10 −9   Note ... (material) direction is: S =   1/ E 1 −ν 21 /E 2 0 −ν 12 /E 1 1/E 2 0 0 01/ G 12   =   .12 20 × 10 10 −.3050 × 10 11 0 −.30...

Ngày tải lên: 11/08/2014, 09:21

... equation: q(x)=R a x 1 +R b x−7.5 1 +R c x 15  1 − 10 x 0 V (x)=−  q(x)dx = −R a x 0 − R b x − 7.5 0 − R c x − 15  0 +10 x 1 M(x)=−  V(x)dx = R a x 1 + R b x − 7.5 1 + R c x − 15  1 − 10 2 x 2 EIy  (x)=  M(x)dx ... in Eqn. 5 to obtain ∂M ∂P = 1 2 x 1 −x− L 2  1 Then δ P = 1 EI  L  P 2 x 1 − Px − L 2  1  1 2 x 1 −...

Ngày tải lên: 11/08/2014, 09:21

... 0.2548E -10 -0. 215 0E -15 0.3253D -18 -0.7 218 D -10 -0 .12 28D -15 -0 .16 39E -16 -0. 215 0E -15 0.2083E-09 -0.6022D -16 -0 .12 28D -15 0.2228D -19 0.7 218 E -10 0 .10 84E -18 -0.6022E -16 0.3058D-09 -0.4265D -11 -0 .19 67D -15 0.6 214 E-22 ... strains to the x-y stresses is then the transformed compliance matrix in the x-y direction: 4 Sbar :=   .8828 10 10 − .19 68 10 10 −...

Ngày tải lên: 11/08/2014, 09:21

... =           k (1) 11 k (1) 12 k (1) 13 k (1) 14 00 k (1) 21 k (1) 22 k (1) 23 k (1) 24 00 k (1) 31 k (1) 32 k (1) 33 + k (2) 11 k (1) 34 + k (2) 12 k (2) 13 k (2) 14 k (1) 41 k (1) 42 k (1) 43 + k (2) 21 k (1) 44 + k (2) 22 k (2) 23 k (2) 24 00 k (2) 31 k (2) 32 k (2) 33 k (2) 34 00 ... added into the global matrix according to...

Ngày tải lên: 11/08/2014, 09:21