david roylance mechanics of materials Part 1 docx

... ET m ∆H α L GPa (Mpsi) ◦ C 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 ... ELASTICITY David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, MA 0 213 9 January 21, 2000 Introduction Thi...

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... 0.2 012 0 .19 89 0 .19 65 1. 2 0 .19 42 0 .19 19 0 .18 95 0 .18 72 0 .18 49 0 .18 26 0 .18 04 0 .17 81 0 .17 58 0 .17 36 1. 3 0 .17 14 0 .16 91 0 .16 69 0 .16 47 0 .16 26 0 .16 04 0 .15 82 0 .15 61 0 .15 39 0 .15 18 0.4 0 .14 97 0 .14 76 0 .14 56 ... 0 .14 76 0 .14 56 0 .14 35 0 .14 15 0 .13 94 0 .13 74 0 .13 54 0 .13 34 0 .13 15 1. 5 0 .12 95 0 .12 76 0 .12 57 0 .12 38 0 .12 19 0 .12 0...

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... 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 course, these materials ... 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 The column labeled U expt lists experiment...

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... Strength of Materials, McGraw-Hill, New York, 19 52 12 The Kinematic Equations David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, MA 0 213 9 Septemb ... Simeon Denis Poisson, (17 81 18 40). 6 SHEAR AND TORSION David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technolo...

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... + ∂u ∂y dy 13 Transformation of Stresses and Strains David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, MA 0 213 9 May 14 , 20 01 Introduction One of the most ... 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 the series expansion √ 1+ x =1+ x/2+x...

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... 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 that this ... E 2 =4,G 12 =2.8(allGPa)and ν 12 =0.25. The compliance matrix S in the 1- 2 (material) direction is: S =   1/ E 1 −ν 21 /E 2 0 −ν 12 /E 1 1/E 2 0...

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... 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 ... 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 −x− L...

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... 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 ... strain 14 Sbar :=   .8828 10 10 − .19 68 10 10 − .12 22 10 −9 − .19 69 10 10 .20 71 10 −9 −.8370 10 10 − .12 22 10 −9 −....

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... =           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 + ... Displacements Node # DOF 1 DOF 2 DOF 3 DOF 4 DOF 5 DOF 6 1 000000 2 0. 013 333 -0.03 219 0000 3 0.02 -0.084...

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... load=load_case _1 8 nodes=[68,58,67] 9 nodes=[57,58,68] load=load_case _1 10 nodes=[57, 51, 58] 11 nodes=[52, 51, 57] load=load_case _1 12 nodes=[37,39,52] load=load_case _1 13 nodes=[39, 51, 52] . . . 11 6 nodes=[2,3 ,1] material ... nodes=[2,3 ,1] material properties steel E=2.05e +11 nu=0.33 t =1 distributed loads load_case _1 color=red direction=GlobalY values= (1, 1) (3 ,1) 13...

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