Strain Analysis | Additional Study Material for Mechanical Engineering PDF Download

Introduction 

No matter what stresses are imposed on an elastic body, provided the material does not rupture, displacement at any point can have only one value. Therefore the displacement at any point can be completely given by the three single valued components u, v and w along the three co-ordinate axes x, y and z respectively. The normal and shear strains may be derived in terms of these displacements.

Normal strains

Consider an element AB of length δx (figure-2.3.2.1). If displacement of end A is u, that of end B is   . This gives an increase in length of   and therefore the strain in x-direction is    Similarly, strains in y and z directions are   .Therefore, we may write the three normal strain components as 

Shear strain

In the same way we may define the shear strains. For this purpose consider an element ABCD in x-y plane and let the displaced position of the element be A′B′C′D′ ( Figure-2.3.3.1). This gives shear strain in xy plane as    where α is the angle made by the displaced line B′C′ with the vertical and β is the angle made by the displaced line A′D′ with the horizontal. This gives

2.3.3.1F- Shear strain associated with the distortion of an infinitesimal element.

We may therefore write the three shear strain components as

Therefore, the complete strain matrix can be written as

Constitutive equation

The state of strain at a point can be completely described by the six strain components and the strain components in their turns can be completely defined by the displacement components u, v, and w. The constitutive equations relate stresses and strains and in linear elasticity we simply have   where E is  modulus of elasticity. It is also known that σ_x produces a strain of  in xdirection,  in y-direction and   in z-direction. Therefore we may write the generalized Hooke’s law as

It is also known that the shear stress    where G is the shear modulus and γ is shear strain. We may thus write the three strain components as 

In general each strain is dependent on each stress and we may write

For isotropic material

Rest of the elements in K matrix are zero.

On substitution, this reduces the general constitutive equation to equations for isotropic materials as given by the generalized Hooke’s law. Since the principal stress and strains axes coincide, we may write the principal strains in terms of principal stresses as

From the point of view of volume change or dilatation resulting from hydrostatic pressure we also have

where 

These equations allow the principal strain components to be defined in terms of principal stresses. For isotropic and homogeneous materials only two constants viz. E and ν are sufficient to relate the stresses and strains. The strain transformation follows the same set of rules as those used in stress transformation except that the shear strains are halved wherever they appear.

Relations between E, G and K

The largest maximum shear strain and shear stress can be given by 

Considering now the hydrostatic state of stress and strain we may write

 Substituting 

Elementary thermoelasticity

So far the state of strain at a point was considered entirely due to applied forces. Changes in temperature may also cause stresses if a thermal gradient or some external constraints exist. Provided that the materials remain linearly elastic, stress pattern due to thermal effect may be superimposed upon that due to applied forces and we may write

It is important to note that the shear strains are not affected directly by temperature changes. It is sometimes convenient to express stresses in terms of strains. This may be done using the relation    Substituting the above expressions for ε_x , ε_y and ε_z  we have.

and substituting   we have 

Combining this with   we  have 

Substituting   we may write the normal and shear stresses as

These equations are considered to be suitable in thermoelastic situations.