david roylance mechanics of materials Part 5 ppt

... s 2   =   0 .50 .51 .0 0 .50 .5 1.0 −0 .50 .50 .0   Then   = RAR −1  =   1.00.00.0 0.01.00.0 0.00.02.0     0 .50 .51 .0 0 .50 .5 1.0 −0 .50 .50 .0     1.00.00.0 0.01.00.0 0.00.00 .5   =    0.00 0.00 −0.02    Obviously, ... + ∂u ∂y dy 13 Transformation of Stresses and Strains David Roylance Department of Materials Science and Engineering Massachusetts Inst...

Ngày tải lên: 11/08/2014, 02:21

... 1982. 5 Statics of Bending: Shear and Bending Moment Diagrams David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, MA 02139 November 15, 2000 Introduction Beams ... using Eqn. 4, we have  ij = 1+ν E σ ij − ν E δ ij σ kk =   0.009 65 0.0377 0.0440 0.0377 0.02 85 0. 056 5 0.0440 0. 056 5 −0.009 15   These results a...

Ngày tải lên: 11/08/2014, 02:21

... material=steel 2 nodes=[67 ,58 ,55 ] 6 nodes= [50 ,41,40] . . . 7 nodes=[68,67,69] 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 ... x=0.98 456 2 y=0.167939 z=0 constraint=free 4 x=0.940634 y=0.3 358 41 z=0 5 x=1.07888 y=0.2 358 33 z=0 . . . 72 x=3.99602 y=3.01892 z=0 73 x=3.99602 y=3 .51 942...

Ngày tải lên: 11/08/2014, 02:21

... × 10 23 /mol Acceleration of gravity g =9.8 05 m/s 2 1 Matrix and Index Notation David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, ... ft-lb = 6. 250 ×10 18 ev Force 1 N = 10 5 d (dyne) = 0.2248 lbf = 0.1020 kg = 3 .59 7 oz = 1.124×10 −4 ton (2000lb) Length 1 m = 39.37 in = 3.281 ft =10 10 ˚ A Mass 1 kg = 2.2 05 lb...

Ngày tải lên: 11/08/2014, 02:21

... result can also be written 7 ATOMISTIC BASIS OF ELASTICITY David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, MA 02139 January 27, ... measure of the internal forces within the material. Much of Mechanics of Materials is concerned with elaborating this concept to include higher orders of dimensionality, worki...

Ngày tải lên: 11/08/2014, 02:21

... 124 3.6 2.3 1. 45 86 2 .5 22–33 boron 400 3 .5 1.0 2. 45 163 1.43 330–440 HS graphite 253 4 .5 1.1 1.80 140 2 .5 66–110 HM graphite 52 0 2.4 0.6 1. 85 281 1.3 220–660 Of course, these materials are not ... untransformed material just outside the neck. 5 INTRODUCTION TO COMPOSITE MATERIALS David Roylance Department of Materials Science and Engineering Massachusetts Inst...

Ngày tải lên: 11/08/2014, 02:21

... Advanced Strength of Materials, McGraw-Hill, New York, 1 952 12 The Kinematic Equations David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, ... constructed of carbon steel has a wall thickness of 0.0 75  , a diameter of 6  ,andalengthof30  . What are the hoop and axial stresses σ θ ,σ z when the cylinder ca...

Ngày tải lên: 11/08/2014, 02:21

... equilibrium: > eq2 := 0=(10* 15* 7 .5) -Rb*7 .5- Rc* 15; > # Now the three zero displacements at the supports: > eq3 := y(0)=0; > eq4 := y(7 .5) =0; > eq5 := y( 15) =0; > # set precision; ... c1:=-87.82;c2:=0;Ra:=28.11;Rb:=93.78;Rc:=28.11; > # the equation of the deflection curve is: > y(x); 33 4.686 x Heaviside(x) + 15. 63 (x - 7 .5) Heaviside(x - 7 .5) 34 + 4.686...

Ngày tải lên: 11/08/2014, 02:21

... compliance matrix: 0. 254 8E-10 -0. 355 4E-12 -0.1639E-16 0.7218D-10 0.7125D-19 -0.6022D-16 -0. 355 4E-12 0. 254 8E-10 -0.2 150 E- 15 0.3 253 D-18 -0.7218D-10 -0.1228D- 15 -0.1639E-16 -0.2 150 E- 15 0.2083E-09 -0.6022D-16 ... 0.0000D+00 -0.2798E+11 0.0000E+00 0.7 354 E+03 0.9876D+10 0.1377D+09 0.2 451 D+03 0.0000E+00 0.2798E+11 0.3473E+ 05 0.1377D+09 0.9876D+10 0.1 158 D+ 05 0.7 354 E+03...

Ngày tải lên: 11/08/2014, 02:21

... then 10 3        25. 0 −43.3 − 25. 043.30.00.00 −43.3 75. 043.3− 75. 00.00.00 − 25. 043. 350 .00.0− 25. 0 −43.30 43.3 − 75. 00.0 150 .0 −43.3 − 75. 00 0.00.0− 25. 0 −43.3 25. 043.30 0.00.0−43.3 − 75. 043.3 75. 00                     0 0 0 u 4 0 0              =              f 1 f 2 f 3 −1732 f 5 f 5              Note ... #...

Ngày tải lên: 11/08/2014, 02:21