hookes law and finite element analysis

Vadodara Institute of Engineering

Stress-Strain Relationship

Guided By: Prof. Prashant HazarePrepared by: Varia Niraj Chaturvedi Anupam

Patel Milan

D I R E C T S T R E SS WHE N A F ORCE IS AP PLI ED TO A N EL ASTI C B OD Y, THE

B ODY D EF ORMS. THE WAY I N WHICH THE B ODY DEF ORMS DEP EN DS UPON THE TYP E OF FORCE A P PLIE D T O IT.

Stress and strain

Compression force makes the body shorter.

A tensile force makes the body longer

AF

AreaForceStress

2/mN

Tensile and compressive forces are called DIRECT FORCESStress is the force per unit area upon which it acts.

….. Unit is Pascal (Pa) or

Note: Most of engineering fields used kPa, MPa, GPa.

( Simbol – Sigma)

LxStrain

DIRECT STRAIN , In each case, a force F produces a deformation x. In engineering, we usually change this force into stress and the deformation into strain and we define these as follows:Strain is the deformation per unit of the original length.

The symbol

Strain has no unit’s since it is a ratio of length to length. Most engineering materials do not stretch very mush before they become damages, so strain values are very small figures. It is quite normal to change small numbers in to the exponent for 10-6( micro strain).

called EPSILON

MODULUS OF ELASTICITY (E)

•Elastic materials always spring back into shape when released. They also obey HOOKE’s LAW.

•This is the law of spring which states that deformation is directly proportional to the force. F/x = stiffness = kN/m

•The stiffness is different for the different material and different sizes of the material. We may eliminate the size by using stress and strain instead of force and deformation:

•If F and x is refer to the direct stress and strain , then

AF Lx LA

xF

AxFL

hence

and

E

AxFL

•The stiffness is now in terms of stress and strain only and this constant is called the MODULUS of ELASTICITY (E)

• A graph of stress against strain will be straight line with gradient of E. The units of E are the same as the unit of stress.

ULTIMATE TENSILE STRESS•If a material is stretched until it breaks, the tensile stress has reached the absolute limit and this stress level is called the ultimate tensile stress.

STRESS STRAIN DIAGRAM


Elastic behaviour The curve is straight line trough out most of

the region Stress is proportional with strain Material to be linearly elastic Proportional limit

The upper limit to linear line The material still respond elastically The curve tend to bend and flatten out

Elastic limit Upon reaching this point, if load is remove,

the specimen still return to original shape


Yielding A Slight increase in stress above the elastic

limit will result in breakdown of the material and cause it to deform permanently.

This behavior is called yielding The stress that cause = YIELD STRESS@YIELD

POINT Plastic deformation Once yield point is reached, the specimen will

elongate (Strain) without any increase in load Material in this state = perfectly plastic

STRESS STRAIN DIAGRAMSTRAIN HARDENING

When yielding has ended, further load applied, resulting in a curve that rises continuously

Become flat when reached ULTIMATE STRESS The rise in the curve = STRAIN HARDENING While specimen is elongating, its cross sectional will decrease The decrease is fairly uniform

NECKING At the ultimate stress, the cross sectional area begins its

localised region of specimen it is caused by slip planes formed within material Actual strain produced by shear strain As a result, “neck” tend to form Smaller area can only carry lesser load, hence curve donward Specimen break at FRACTURE STRESS

Hooke’s law for plane stress• Materials that meet two important conditions: 1) The material is uniform throughout the body and has the same properties in all directions 2) The material follows Hooke’s law (ie is linearly elastic)• For e.g. the strain εx in the x direction due to the stress σx is equal to σx/E where E is the modulus of elasticity. But we also have a strain εx due to the stress σy and is equal to -v σy / E where v is the Poisson’s ratio.

Copyright 2005 by Nelson, a division of Thomson Canada Limited

FIG. 7-23Element of material in plane stress (z = 0)


FIG. 7-24 Element of material subjected to normal strains x, y, and z

Hooke’s law for plane stress

Special cases of Hooke’s law- Biaxial stress: σx = σy = 0 - Uniaxial stress: σy = 0 - Pure shear: σx = σy = 0, εx = εy = εz = 0 and γxy = τxy / G • Volume change: The change in volumecan be determined if the normal strainsIn the three perpendicular directions• Strain – Energy density in plane stress

Numerical


FIG. 7-29 Strain components x, y, and xyin the xy plane (plane strain)

Plane strain

• If the only deformations are those in the xy plane, then three strain components may exist – the normal strain εx in the x direction (fig 7-29b), the normal strain εy in the y direction (fig 7-29c) and the shear strain γxy (fig 7-29d). An element subjected to these strains (and only these strains) is said to be in a state of plane strain • It follows that an element in plane strain has no normal strain εz in the z direction and no shear strains γxz and γyz in the xz and yz planes respectively• The definition of plane strain is analogous to that for plane stress


FIG. 7-30 (1 of 2)Comparison of plane stress and plane strain


FIG. 7-30 (2 of 2)Comparison of plane stress and plane strain

DOUBLE SHEAR

Consider a pin joint with a support on both ends as shown. This is called CLEVIS and CLEVIS PIN By balance of force, the force in the two supports is F/2 eachThe area sheared is twice the cross section of the pinSo it takes twice as much force to break the pin as for a case of single shearDouble shear arrangements doubles the maximum force allowed in the pin

LOAD AND STRESS LIMIT

DESIGN CONSIDERATION Will help engineers with their important task in

Designing structural/machine that is SAFE and ECONOMICALLY perform for a specified function

DETERMINATION OF ULTIMATE STRENGTH An important element to be considered by a designer is

how the material that has been selected will behave under a load

This is determined by performing specific test (e.g. Tensile test)

ULTIMATE FORCE (PU)= The largest force that may be applied to the specimen is reached, and the specimen either breaks or begins to carry less load

ULTIMATE NORMAL STRESS (U) = ULTIMATE FORCE(PU) /AREA

ALLOWABLE LOAD / ALLOWABLE STRESSMax load that a structural member/machine component will be allowed to carry under normal conditions of utilization is considerably smaller than the ultimate loadThis smaller load = Allowable load / Working load / Design loadOnly a fraction of ultimate load capacity of the member is utilised when allowable load is appliedThe remaining portion of the load-carrying capacity of the member is kept in reserve to assure its safe performanceThe ratio of the ultimate load/allowable load is used to define FACTOR OF SAFETY

FACTOR OF SAFETY = ULTIMATE LOAD/ALLOWABLE LOAD@FACTOR OF SAFETY = ULTIMATE STRESS/ALLOWABLE STRESS

SELECTION OF F.S.

1. Variations that may occur in the properties of the member under considerations

2. The number of loading that may be expected during the life of the structural/machine

3. The type of loading that are planned for in the design, or that may occur in the future

4. The type of failure that may occur5. Uncertainty due to the methods of analysis6. Deterioration that may occur in the future because of poor

maintenance / because of unpreventable natural causes7. The importance of a given member to the integrity of the

whole structure

AXIAL FORCE & DEFLECTION OF BODY

Deformations of members under axial loading If the resulting axial stress does not exceed the

proportional limit of the material, Hooke’s Law may be applied

Then deformation (x / ) can be written as

AEFL

E

22Basics of Finite Element Analysis

Why FEM ?Modern mechanical design involves

complicated shapes, sometimes made of different materials.

Engineers need to use FEM to evaluate their designs.

23

Basics of Finite Element Analysis

FEA Applications Evaluate the stress or temperature

distribution in a mechanical component.Perform deflection analysis.Analyze the kinematics or dynamic

response.Perform vibration analysis.

24

Basics of Finite Element AnalysisConsider a cantilever beam shown.

Finite element analysis starts with an approximation of the region of interest into a number of meshes (triangular elements). Each mesh is connected to associated nodes (black dots) and thus becomes a finite element.

25

Basics of Finite Element Analysis

After approximating the object by finite elements, each node is associated with the unknowns to be solved.

For the cantilever beam the displacements in x and y would be the unknowns.

This implies that every node has two degrees of freedom and the solution process has to solve 2n degrees of freedom.

Once the displacements have been computed, the strains are derived by partial derivatives of the displacement function and then the stresses are computed from the strains.

Example problem

x

y

3

2 1

4

2

2The square block is in plane strain and is subjected to the following strains

2

2 3

2

3x

y

xy

xy

xy

x y

Compute the displacement field (i.e., displacement components u(x,y) and v(x,y)) within the block

Solution

Recall from definition

)3(

)2(3

)1(2

32

2

yxxv

yu

xyyv

xyxu

xy

y

x

Integrating (1) and (2)

)5()(),(

)4()(),(

23

12

xCxyyxv

yCyxyxu

Arbitrary function of ‘x’

Arbitrary function of ‘y’

Plug expressions in (4) and (5) into equation (3)

0)()(

)()(

)()(

)3(

21

322312

3223

12

32

xxC

yyC

yxxxCy

yyCx

yxx

xCxyy

yCyx

yxxv

yu

Function of ‘y’ Function of ‘x’

)constanta()()( 21 CxxC

yyC

Hence

Integrate to obtain

22

11

)()(

DCxxCDCyyC

D1 and D2 are two constants of

integration

Plug these back into equations (4) and (5)

23

12

),()5(

),()4(

DCxxyyxv

DCyyxyxu

How to find C, D1 and D2?

Use the 3 boundary conditions

0)0,2(0)0,0(0)0,0(

vvu

To obtain

000

2

1

DDC

Hence the solution is

3

2

),(

),(

xyyxv

yxyxu

x

y

3

2 1

4

2

2

31

Formulation of the Finite Element Method

The classical finite element analysis code (h version)The system equations for solid and structural mechanics problems are derived using the principle of virtual displacement and work (Bathe, 1982).

The method of weighted residuals (Galerkin Method)weighted residuals are used as one method of finite element formulation starting from the governing differential equation.

Potential Energy and Equilibrium; The Rayleigh-Ritz Method.Involves the construction of assumed displacement field. Uses the total potential energy for an elastic body

32


f B – Body forces (forces distributed over the volume of the body: gravitational forces, inertia, or magnetic)

f S – surface forces (pressure of one body on another, or hydrostatic pressure)

f i – Concentrated external forces

33

Formulation of the Finite Element MethodLet’s denote the displacements of any point (X, Y, Z) of the object from the unloaded configuration as UT

The displacement U causes the strains

and the corresponding stresses

The goal is to calculate displacement, strains, and stresses from the given external forces.

34Formulation of the Finite Element Method

Equilibrium condition and principle of virtual displacements

The left side represents the internal virtual work done, and the right side represents the external work done by the actual forces as they go through the virtual displacement. The above equation is used to generate finite element equations. And by approximating the object as an assemblage of discrete finite elements, these elements are interconnected at nodal points.

35


The equilibrium equation can be expressed using matrix notations for m elements.

where B(m) Represents the rows of the strain displacement matrix C(m) Elasticity matrix of element m H(m) Displacement interpolation matrix U Vector of the three global displacement

components at all nodes F Vector of the external concentrated forces applied to the nodes

36


The above equation can be rewritten as follows,

The above equation describes the static equilibrium problem. K is the stiffness matrix.

37

FEA - Flow Chart

REFERENCES

1. James M. Gere (2006) “ Mechanics of Materials”. 6th Edition, Thompson2. R.C. Hibbeler (2003) “ Mechanics of Materials”. 5th Edition, Prentice Hall3. Raymond Parnes (2001), “Solid Mechanics in Engineering”. John Willey

and Son