Definition of stress and strain Sunday, March 1, :04 PM

Transcription

1 Definition of stress and strain Sunday, March 1, :04 PM To solve any problem in elasticity we must satisfy 1. Compatibility of displacements (ie single value displacements and their derivatives in the body) 2. Displacement-to-strain relationships 3. Strain-to-stress relationships (constitutive equations) 4. Equilibrium of stresses at any point 5. Boundary conditions Definition of continuous displacement due to body deformation (excludes free-body motion) Continuous deformation in the body. There is no discontinuity (for example, cracks) in the body. Definition of strain Definition of stress at a point Stress at point A: 3D-stress at a point ELASTICITY Page 1

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3 Relationship between stress and strain in elastic materials: Generalized Hooke's Law Sunday, March 1, :11 PM Stress induces strain In the linear-elastic range, the six components of stress are expressed as linear functions of the six components of strain The 36 constants C are the material-dependent elastic constants that relate strain to stress. For a LINEAR or NON-LINEAR ELASTIC material (Green's Solid) This is to assure that ENERGY IS NOT LOST IN DEFORMATION. This leaves 21 constants C that relate stress to strain. For ISOTROPIC materials, we can use symmetry to reduce these constants to two (2) independent elastic constants called the Lame' constants λ and G. We CANNOT measure the Lame' constants directly. We can measure several engineering constants and relate them to the Lame' constants. In practice, the elastic constants for an Isotropic material are: K--Bulk's modulus ELASTICITY Page 3

4 E--Elastic Modulus G--Shear Modulus ν--poison's ratio We only need to obtain by testing two (2) elastic constants to fully define an isotropic material. We obtain these constants by the study of some special states of stress Simple Tension Test--Measure E and ν Hydrostatic Pressure Test--Measure K Shear Modulus, Pure Shear Test--Measure G In what follows, we will investigate the Tension and the Hydrostatic states of stress in more detail ELASTICITY Page 4

5 Engineering Tests and relationship to Lame's Constants--Simple Tension Test--Measuring Young's Modulus E and Poisson's Ratio Sunday, March 1, :55 PM DEFINE the Elastic Modulus as: SOLVE for the relationship between E and the Lame's constants Subtract Eqn(2) and Eqn(3) to get Substitute back into Equation(2) to get Substitute into Equation(1) to get ELASTICITY Page 5

6 DEFINE Poisson's Ratio as RELATE to Lame's Constants Now, we can solve for λ and G in terms of G and ν ELASTICITY Page 6

7 Hydrostatic Pressure Test--Measuring Bulk Modulus K Monday, March 2, :25 PM Unit Cube under Hydrostatic Pressure Change in volume of unit cube Define DILATATION as Define BULLK MODULUS as K = hydrostatic pressure / dilatation Relate K to Lame's Constants ELASTICITY Page 7

8 Note that with a PURE SHEAR test we can find G which is one of the Lame's constants. ELASTICITY Page 8

9 Stress-Strain Relationships for elastic, isotropic material Monday, March 2, :51 PM Usually, E, ν, and G are the constants most often used to relate stresses and strains. In 3-Dimentional elasticity, the stress is given in terms of strain as: The strain is given in terms of stress as: For PLANE STRESS conditions, all STRESS occurs in the x-y plane. Any stress in the z-direction is zero Strain exists in the z plane. ELASTICITY Page 9

10 For PLANE STRAIN conditions, all STRAIN occurs in the x-y plane. Any strain in the z-direction is zero In case of initial strains due to temperature, ELASTICITY Page 10

11 Strain-Displacement Relationships Tuesday, March 3, :30 AM The easiest visualization of strain is shown here, where the deformations in the x and y directions due to axial and shear forces are shown de-coupled from each other. u = displacement in the x-direction v = displacement in the y-direction For a more detailed analysis, consider that a displacement field describes how a body deforms as well as how it displaces. A point P originally located at coordinate (x,y) moves to P', and after deformation is located at coordinate (x+u, y+v) The displacement components u and v are functions of location (x,y) of point P. The displacement components u and v of every single point in the body are called the DISPLACEENT FIELD Displacement Vector Position Vector ELASTICITY Page 11

12 One can also develop the relation of the shear strain to the displacement derivatives. In summary, ELASTICITY Page 12

13 In Matrix operator format: ELASTICITY Page 13

14 Compatibility of Displacements Tuesday, March 3, :37 AM When the body is deformed, it does not crack, does not break, no kinks appear in bending, and the material particles do not interpenetrate. The compatibility condition requires that the displacements must be continuous and single-valued functions of position. Continuous, smooth displacements At point A, two different values for the displacement At point B, kink, not a smooth derivative Arbitrarily assumed expressions for strain may NOT satisfy the compatibility equations,. The strains must be related to each other as: COMPATIBILITY EQUATIONS Arbitrarily assumed displacement fields are certain to SATISFY compatibility equations IF they are single-valued and continuous ELASTICITY Page 14

15 A big reason for the use of the displacement-based Fes which assume polynomials as displacement fields is the ease with which compatibility is satisfied SO, by assuming a polynomial as a displacement field COMPATIBILITY IS SATISFIED ELASTICITY Page 15

16 Equilibrium Tuesday, March 3, :41 AM In 2-D In 3-D ELASTICITY Page 16

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18 In matrix arrangement and assuming the body forces F are zero:.. ELASTICITY Page 18

19 Boundary Conditions Tuesday, March 3, :55 AM Prescription of DISPLACEMENT or STRESSES on the sides (2D) or surface (3D) of a body. The surface traction creates normal and shear stresses around the surface element. Define =Surface Tractions The surface tractions are in EQUILIBRIBUM with the surface element On EACH surface we have: 1) Let's work with the TOP surface ELASTICITY Page 19

20 2) Let's work with the side surface ELASTICITY Page 20

21 Exact vs approximate solutions Tuesday, March 3, :49 AM If we start with a compatible displacement field Find strains using Find stresses using If these stresses satisfy at every point of the volume and at all boundary tractions Then You have obtained the exact solution of the mathematical model Let's take this very simple problem. Here, we know that we have a linear displacement function u(x) CLOSED-FORM solutions in elasticity usually start with an assumption on the stress field, not the displacement field that we used for this very simple problem. We can find stress functions in a few simple geometries/loadings/b.cs. In the next problem, We introduce an Airy-Stress function to find a closed-form solution to a uniformly-loaded cantilever. FINITE ELEMENT solutions in solid mechanics are based on ASSUMED displacement fields u(x,y) and v(x,y). These displacement fields are continuous and have smooth derivatives. As such COMPATIBILITY is immediately satisfied. We can use our ASSUMED displacement fields to find strains. We can use strains and stress-strain relationships (constitutive models of the material behavior) to find stresses. These stresses DO NOT SATISFY EQUILIBRRIUM at every point in the body as expected from closed-form solutions. Equilibrium and BCs are satisfied in an integral (average) sense. Equilibrium IS SATISFIED at the NODES ELASTICITY Page 21

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23 Using Airy-Stress Functions to find exact stress distribution in uniformly-loaded cantilever Tuesday, March 3, :00 AM We can find the exact stress distribution in this beam due to the distributed load P First we must find a function (usually a polynomial or a sine/cosine) that satisfies (1) Equilibrium (2) Compatibility (3) Boundary Conditions. In this problem, we are given such a function. A. Determine whether the following COMPATIBLE stress field is possible within an elastic, uniformly loaded cantilever beam Answer The field is given as compatible. It must also satisfy EQUILIBRIUM and BOUNDARY CONDITIONS For EQUILIVBRIUM Therefore, stress field satisfies equilibrium For BOUNDARY CONDITIONS ELASTICITY Page 23

24 We expect that Boundary Conditions are satisfied if we can show that: Checking TOP surface Checking Bottom surface ELASTICITY Page 24

25 Checking Left surface ELASTICITY Page 25

26 B. Given P=10KN/m, L=2m, h=100mm, t=40mm, ν, E=200GPa Calculate the magnitude and direction of the principal stresses at POINT Q Answer ELASTICITY Page 26

27 C. Compare these answers to the ones we get from Elementary Beam Mechanics ELASTICITY Page 27

28 Using Mohr's Circle or Eigenvalue analysis ELASTICITY Page 28