Lecture 8. Stress Strain in Multi-dimension

Transcription

1 Lecture 8. Stress Strain in Multi-dimension Module. General Field Equations

2 General Field Equations [] Equilibrium Equations in Elastic bodies xx x y z yx zx f x 0, etc [2] Kinematics xx u x x,etc. [3] Constitutive Relation [Stress-Strain Relation] xx [ xx ( yy zz )], etc (for isotropic case) E

3 General Field Equations [] Derivation of Equilibrium Equations in Elastic bodies xx y z zx - Consider Force Equilibrium condition in the x direction. z z yx zx zx x dz yx y y yx xx x dy x xx dx xx yx zx f x 0 x y z xy yy zx f y x y z 0 xz x y z yz zz f z 0 2

4 General Field Equations [2] Kinematics xx yy zz u x u y uz z x y 2 2 u ux x y u u z y yz y z y xy xy yz u x u z z x 2zx zx 3

5 General Field Equations [3] Constitutive Relation [Stress-Strain Relation] ) For linear, elastic isotropic materials xx [ xx ( yy zz )] ( T T0 ) E zz [ zz ( xx yy )] ( T T0 ) E yy [ yy ( zz xx )] ( T T0 ) E xy xz xy xz yz G G yz G (unless stated otherwise, thermal expansion will not be explicitly considered from now on.) E = Young s modulus ( Esteel 200GPa ) ν = Poisson s ratio ( steel 0.3 ) G=Shear modulus= (will be shown) 4

6 General Field Equations 2) For General Anisotropic Materials, ( Normal Stress may cause shearing deformation) xx s s2 s3 s4 s5 s6 xx 2 yy s2 s22 s23 s24 s25 s 26 2 yy 3 zz s3 s32 s33 s34 s35 s 36 3 zz 4 yz s4 s42 s43 s44 s45 s 46 4 yz s s s s s s 5 zx zx s 6 xy 6 s62 s63 s64 s65 s66 6 xy Symbolically, we write ε S σ * Remark: There will be 36 Coefficients for the stress-strain relation, but because of symmetry (e.g., c ij = d ji ), one can show that only 2 independent coefficients are independent. 5

7 General Field Equations 2) For General Anisotropic Materials (continued), ε S σ Instead of, One can use: σ Cε. C c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c c * Remark: There will be 36 Coefficients for the stress-strain relation, but because of symmetry (e.g., s ij = s ji ), one can show that only 2 independent coefficients are independent. 6

8 General Field Equations 3) Orthotropic Materials (e.g. Wood, Composites) y x (9 coefficients) z * No coupling between normal and shearing components in the material axes. 76

9 General Field Equations 3) Cubic Materials (continued) s s s xx xx 2 yy 2 zz yz s 44 yz c c c xx xx 2 yy 2 zz c c c yy 2 xy yy 2 zz c c c zz 2 xx 2 yy zz yz c44 yz zx c44 xy c 44 xy zx (unit 0^0 Pa) c c2 c44 Aluminum Copper Tungsten (3 coefficients) (3 coefficients) c ( s s ) / [( s s )( s 2 s )] c s / [( s s )( s 2 s )] c / s

10 General Field Equations * Proof of G = E/2(+ v) in the isotropic case - Approach : Consider Pure Shear Case and Use Stress/Strain Transformation 9

11 General Field Equations * Proof of G = E/2(+ v) in the isotropic case (continued) [for Path ] [for Path ] () Path 2) Stress in the -2 axes(45 degrees from x-y axes) (2) Path 3) Strain in the -2 axes : (3) Path 4) Strain back in the x-y axes from strain in the -2 axes [from shear strain in by Path = shear strain by Path 2-3-4] (4) 0

12 General Field Equations Revisit the case of Cubic Materials y x * Anisotropy Factor = c / [( c c ) / 2] 44 2 shear stiffness in xy direction o shear stiffness in 45 direction c c c xx xx 2 yy 2 zz c c c yy 2 xy yy 2 zz c c c zz 2 xx 2 yy zz yz c44 yz zx c44 xy c 44 xy zx (3 coefficients)

13 Module 2. Problems in Non-Cartesian Coordinates

14 Problems in Non-Cartesian Coordinates Some Remarks [Note] Material Behavior is independent from the selected coordinate system or orientation when isotropic materials are considered. [] For any Cartesian Coordinates (x, y, z ), the same stress-strain relation holds: x' x' [ x' x' ( y' y' z ' z ' )] E z ' z ' [ z ' z ' ( x' x' y' y' )] E y ' y ' [ y ' y ' ( z ' z ' x' x' )] E xy ' ' xz ' ' xy ' ' xz ' ' G G yz ' ' G yz ' ' [2] For Cylindrical (or Spherical) Coordinates (i.e. orthogonal coordinate systems), the stress-strain relation is the same as that for the Cartesian coordinate system. rr [ rr ( zz )] E zz [ zz ( rr )] E [ ( rr zz )] E r z rz r z rz G G G

15 Problems in Non-Cartesian Coordinates In Cylindrical Coordinates * Equilibrium equation * Kinematics (strain-displacement relation) rr u r r u r v zz r r w z r ur v v r r r rz w r u z r r r w v z 2

16 Problems in Non-Cartesian Coordinates Example : Stress on the cylindrical surface of a cylindrical shell [Analysis] t (0) Circular Symmetry: stress independent of. F p z r F: axial force, p=pressure, t<<r F (0) No loading variation along z axis: stress independent of z (far away from the F- loading point) (0) We will calculate all six components,,,,,,. zz zr z r rr 3

17 Problems in Non-Cartesian Coordinates Example : Continued [A] For zz, zr, z, consider the following infinitesimal element with n=e z F p t F z r ) t zr =0 because of symmetry and compatibility 2) t z =0. If not, a torsional moment will be produced. 3) Force Equilibrium in z: F= zz * 2prt F, 0 zz 2 rt zr z p 4

18 Problems in Non-Cartesian Coordinates Example : continued [B] For, and r r, consider the following infinitesimal element with n=e 2) To calculate, consider the force equilibrium in the y direction for the upper part as: 0) We know z = t z =0 ) r = t r =0 because of symmetry and compatibility 5

19 Stress and Strain Transformation Problems in Non-Cartesian Coordinates Example : continued [C] About rr stress, note that At the inside surface, At the outside surface, 0 Thus, p 0 rr rr p r Compared with p, t 0 rr rr rr t F p F z r 6

20 Stress and Strain Transformation Yielding in multi-dimension Module 3. Yielding in Multi-dimension

21 Stress and Strain Transformation Yielding in multi-dimension Yielding Criteria for multi-dimensional stress state? Nonzero stress: 0, 0, 0 (No shear stress) zz 0, 0, 0 rr 2 2 Crandall, An Introduction to the Mechanics of solid, Mc Graw-Hill, 999 -D Uniaxial Tensile Test Crandall, An Introduction to the Mechanics If Y, yielding occurs for -D cases. xx of solid, McGraw-Hill, 999 (Fig & 29) (Yielding of thin-walled tube under combined stress (materials: isotropic metals))

22 Stress and Strain Transformation Yielding in multi-dimension Von-Mises and Tresca Yielding Criteria [] The von-mises Criterion (expressed in three-dimensional principal stresses) Equivalently, in terms of general stress components / 2 ( ) ( ) ( ) 3t 3t 3t xx yy yy zz zz xx xy yz zx Y [2] The Tresca or the maximum shear stress criterion For plane problems, The von-mises criterion represents a 45- degree rotated ellipse with respect to and 2. max min Y t max = 2 2 (choose and among, and ) max min 2 3 2

23 Stress and Strain Transformation Yielding in multi-dimension Mechanics involved in Plastic Deformation Plastic deformation depends on the motion of individual stress is the dominant agent in the migration of the dislocations. Thus, hydrostatic state of stress would not tend to move the dislocations / 2 ( 2) ( 2 3) ( 3 ) Y Crandall, An Introduction to the Mechanics of solid, McGraw-Hill, 999 (Fig. 5.27) Elastic Case Plastic Case 3

24 Stress and Strain Transformation Yielding in multi-dimension Example 2: The load P that will cause the onset of yielding? Crandall, An Introduction to the Mechanics of solid, McGraw-Hill, 999 Example 7.7 (Fig. 7.25) A circular rod of radius r (Yield Stress: Y) 5 possible locations for the most critically-stressed points: A, B, etc. 4

25 Stress and Strain Transformation Yielding in multi-dimension Example 2: Continued Crandall, An Introduction to the Mechanics of solid, McGraw-Hill, 999 Example 7.7 (Fig. 7.25) Consider the free-body diagram. For A For C For C 2 For B For B 2 5

26 Stress and Strain Transformation Yielding in multi-dimension Example 2: Continued For B (2) Principal Stress t PLr 2PLr 5 PLr R 2 I0 I0 2 I0 () Stress Distribution Crandall, An Introduction to the Mechanics of solid, McGraw-Hill, 999 Example 7.7 (Fig. 7.25&26) 3PLr 5PLr PLr 2I 2I I PLr PLr 4 PLr 3 0 2I 2I I I 0 I 0 I xx 2I 0 Remark: The Max Bending and Twist Stress Occur on the top of the beam. However, shear stress by P on the top is zero. I 0 Radius r x 3 x 2 x I xx p r 2 4 p r I I I I 4 2 x x x x 0 x x

27 Stress and Strain Transformation Yielding in multi-dimension Example 2: Continued (consider FBD s) For B * Substituting the results into the Tresca criterion, ( PLr PLr, 4, 0) 2 3 Izz Izz t Y 2 2 max min max = * Substituting the results into the Mises criterion, / 2 ( 2) ( 2 3) ( 3 ) Y 7

28 Stress and Strain Transformation Yielding in multi-dimension Example 2: Continued (consider FBD s) Mr b 4PLr zz I I 0 0 Crandall, An Introduction to the Mechanics of solid, McGraw-Hill, 999 Example 7.7 (Fig. 7.25&26) For B 2 Mr t 3PLr t xz (2 I ) 2I 2 I * We will find: c r; c r 2 9PLr PLr I 2I 0 0 c 4PLr 3PLr z( zz, zx) (, ) Ixx Izz 4PLr 3PLr I 0 =(, ) I 2I 0 0 From von Mises: P From Tresca: P IY IY Lr Lr I0 I0 R 5PLr 2PLr, c 2I I 0 0 8

29 Stress and Strain Transformation Yielding in multi-dimension Example 2: Continued For B For B 2 Yielding starts first at B 2 by von Mises. From von Mises: P Lr 0 From Tresca: P IY Lr IY From von Mises: P Lr 0 From Tresca: P IY Lr IY Yielding starts simultaneously at B and B 2 by Tresca. (More conservative Criterion) 9