Strain Energy

Strain Energy

Strain Energy of the member is defined as the internal work done in defoming the body by the action of externally applied forces. This energy in elastic bodies is known as elastic strain energy :

Strain Energy in uniaxial Loading

Fig .1

Let as consider an infinitesimal element of dimensions as shown in Fig .1. Let the element be subjected to normal stress s_x.

The forces acting on the face of this element is s_x. dy. dz

where

dydz = Area of the element due to the application of forces, the element deforms to an amount = Î_x dx

  Î_x = strain in the material in x – direction

       

Assuming the element material to be as linearly elastic the stress is directly proportional to strain as shown in Fig . 2.

Fig .2

\ From Fig .2 the force that acts on the element increases linearly from zero until it attains its full value.

Hence average force on the element is equal to ½ s_x . dy. dz.

\ Therefore the workdone by the above force

Force = average force x deformed length

           = ½ s_x. dydz . Î_x . dx

For a perfectly elastic body the above work done is the internal strain energy “du”.

where dv = dxdydz

   = Volume of the element

By rearranging the above equation we can write

The equation (4) represents the strain energy in elastic body per unit volume of the material its strain energy – density ‘u_o' .

From Hook's Law for elastic bodies, it may be recalled that

In the case of a rod of uniform cross – section subjected at its ends an equal and opposite forces of magnitude P as shown in the Fig .3.

Fig .3

ILLUSTRATIVE PROBLEMS

1. Three round bars having the same length ‘L' but different shapes are shown in fig below. The first bar has a diameter ‘d' over its entire length, the second had this diameter over one – fourth of its length, and the third has this diameter over one eighth of its length. All three bars are subjected to the same load P. Compare the amounts of strain energy stored in the bars, assuming the linear elastic behavior.

Solution :

From the above results it may be observed that the strain energy decreases as the volume of the bar increases.

2. Suppose a rod AB must acquire an elastic strain energy of 13.6 N.m using E = 200 GPa. Determine the required yield strength of steel. If the factor of safety w.r.t. permanent deformation is equal to 5.

Solution :

Factor of safety = 5

Therefore, the strain energy of the rod should be u = 5 [13.6] = 68 N.m

Strain Energy density

The volume of the rod is

Yield Strength :

As we know that the modulus of resilience is equal to the strain energy density when maximum stress is equal to s_x .

It is important to note that, since energy loads are not linearly related to the stress they produce, factor of safety associated with energy loads should be applied to the energy loads and not to the stresses.

STATICALLY INDETERMINATE PROBLEMS

 Statically indeterminate problems.
Members for which reaction forces and internal forces can be found out from static
equilibrium equations alone are called statically determinate members or structures.
Problems requiring deformation equations in addition to static equilibrium equations to
solve for unknown forces are called statically indeterminate problems.


















The reaction force at the support for the bar ABC in figure  can be determined
considering equilibrium equation in the vertical direction.




Now, consider the right side bar MNO in figure 1.22 which is rigidly fixed at both the ends.
From static equilibrium, we get only one equation with two unknown reaction forces R1 and
R2.

Hence, this equilibrium equation should be supplemented with a deflection equation which
was discussed in the preceding section to solve for unknowns.
If the bar MNO is separated from its supports and applied the forces , then
these forces cause the bar to undergo a deflection
R1,R2 and P
δMO that must be equal to zero.

δMN and δNO are the deflections of parts MN and NO respectively in the bar MNO.
Individually these deflections are not zero, but their sum must make it to be zero.
Equation 1.19 is called compatibility equation, which insists that the change in length of the
bar must be compatible with the boundary conditions.
Deflection of parts MN and NO due to load P can be obtained by assuming that the
material is within the elastic limit,

 

Substituting these deflections in equation

Combining both the  equations one can get,

 

From these reaction forces, the stresses acting on any section in the bar can be easily
determined.

STRESS, STRAIN AND DEFORMATION

Stress

Preamble

Engineering science is usually subdivided into number of topics such as

1.  Solid Mechanics

2.  Fluid Mechanics

3.  Heat Transfer

4.  Properties of materials and soon Although there are close links between them in terms of the physical principles involved and methods of analysis employed.

The solid mechanics as a subject may be defined as a branch of applied mechanics that deals with behaviours of solid bodies subjected to various types of loadings. This is usually subdivided into further two streams i.e Mechanics of rigid bodies or simply Mechanics and Mechanics of deformable solids.

The mechanics of deformable solids which is branch of applied mechanics is known by several names i.e. strength of materials, mechanics of materials etc.

Mechanics of rigid bodies:

The mechanics of rigid bodies is primarily concerned with the static and dynamic behaviour under external forces of engineering components and systems which are treated as infinitely strong and undeformable Primarily we deal here with the forces and motions associated with particles and rigid bodies.

Mechanics of deformable solids :

Mechanics of solids:

The mechanics of deformable solids is more concerned with the internal forces and associated changes in the geometry of the components involved. Of particular importance are the properties of the materials used, the strength of which will determine whether the components fail by breaking in service, and the stiffness of which will determine whether the amount of deformation they suffer is acceptable. Therefore, the subject of mechanics of materials or strength of materials is central to the whole activity of engineering design. Usually the objectives in analysis here will be the determination of the stresses, strains, and deflections produced by loads. Theoretical analyses and experimental results have an equal roles in this field.

Analysis of stress and strain :

Concept of stress : Let us introduce the concept of stress as we know that the main problem of engineering mechanics of material is the investigation of the internal resistance of the body, i.e. the nature of forces set up within a body to balance the effect of the externally applied forces.

The externally applied forces are termed as loads. These externally applied forces may be due to any one of the reason.

(i)   due to service conditions

(ii)  due to environment in which the component works

(iii)  through contact with other members

(iv)  due to fluid pressures

(v)   due to gravity or inertia forces.

As we know that in mechanics of deformable solids, externally applied forces acts on a body and body suffers a deformation. From equilibrium point of view, this action should be opposed or reacted by internal forces which are set up within the particles of material due to cohesion.

These internal forces give rise to a concept of stress. Therefore, let us define a stress Therefore, let us define a term stress

Stress:

Let us consider a rectangular bar of some cross – sectional area and subjected to some load or force (in Newtons )

Let us imagine that the same rectangular bar is assumed to be cut into two halves at section XX. The each portion of this rectangular bar is in equilibrium under the action of load P and the internal forces acting at the section XX has been shown

Now stress is defined as the force intensity or force per unit area. Here we use a symbol s to represent the stress.

Where A is the area of the X – section

Here we are using an assumption that the total force or total load carried by the rectangular bar is uniformly distributed over its cross – section.

But the stress distributions may be for from uniform, with local regions of high stress known as stress concentrations.

If the force carried by a component is not uniformly distributed over its cross – sectional area, A, we must consider a small area, ‘dA' which carries a small load dP, of the total force ‘P', Then definition of stress is

As a particular stress generally holds true only at a point, therefore it is defined mathematically as

Units :

The basic units of stress in S.I units i.e. (International system) are N / m² (or Pa)

MPa = 10⁶ Pa

GPa = 10⁹ Pa

KPa = 10³ Pa

Some times N / mm² units are also used, because this is an equivalent to MPa. While US customary unit is pound per square inch psi.

TYPES OF STRESSES :

only two basic stresses exists : (1) normal stress and (2) shear shear stress. Other stresses either are similar to these basic stresses or are a combination of these e.g. bending stress is a combination tensile, compressive and shear stresses. Torsional stress, as encountered in twisting of a shaft is a shearing stress.

Let us define the normal stresses and shear stresses in the following sections.

Normal stresses : We have defined stress as force per unit area. If the stresses are normal to the areas concerned, then these are termed as normal stresses. The normal stresses are generally denoted by a Greek letter ( s )

This is also known as uniaxial state of stress, because the stresses acts only in one direction however, such a state rarely exists, therefore we have biaxial and triaxial state of stresses where either the two mutually perpendicular normal stresses acts or three mutually perpendicular normal stresses acts as shown in the figures below :

Tensile or compressive stresses :

The normal stresses can be either tensile or compressive whether the stresses acts out of the area or into the area

Bearing Stress : When one object presses against another, it is referred to a bearing stress ( They are in fact the compressive stresses ).

Shear stresses :

Let us consider now the situation, where the cross – sectional area of a block of material is subject to a distribution of forces which are parallel, rather than normal, to the area concerned. Such forces are associated with a shearing of the material, and are referred to as shear forces. The resulting force interistes are known as shear stresses.

The resulting force intensities are known as shear stresses, the mean shear stress being equal to

Where P is the total force and A the area over which it acts.

As we know that the particular stress generally holds good only at a point therefore we can define shear stress at a point as

The greek symbol t ( tau ) ( suggesting tangential ) is used to denote shear stress.

However, it must be borne in mind that the stress ( resultant stress ) at any point in a body is basically resolved into two components s and t one acts perpendicular and other parallel to the area concerned, as it is clearly defined in the following figure.

The single shear takes place on the single plane and the shear area is the cross - sectional of the rivett, whereas the double shear takes place in the case of Butt joints of rivetts and the shear area is the twice of the X - sectional area of the rivett. 

1.5 Strain
The structural member and machine components undergo deformation as they are brought
under loads.
To ensure that the deformation is within the permissible limits and do not affect the
performance of the members, a detailed study on the deformation assumes significance.
A quantity called strain defines the deformation of the members and structures in a better
way than the deformation itself and is an indication on the state of the material.
Figure 1.12

Consider a rod of uniform cross section with initial length as shown in figure 1.12.
Application of a tensile load P at one end of the rod results in elongation of the rod by
L0
δ .
After elongation, the length of the rod is L. As the cross section of the rod is uniform, it is
appropriate to assume that the elongation is uniform throughout the volume of the rod. If
the tensile load is replaced by a compressive load, then the deformation of the rod will be a
contraction. The deformation per unit length of the rod along its axis is defined as the
normal strain. It is denoted by ε

Though the strain is a dimensionless quantity, units are often given in mm/mm, μm/m.

DEFORMATION IN AXIALLY LOADED MEMBER

Consider the rod of uniform cross section under tensile load P along its axis as shown in
figure 1.12.
Let that the initial length of the rod be L and the deflection due to load be δ . Using
equations 1.9 and 1.10,


Equation  is obtained under the assumption that the material is homogeneous and has
a uniform cross section.
Now, consider another rod of varying cross section with the same axial load P as shown in
figure 

Let us take an infinitesimal element of length dx in the rod that undergoes a deflection
dδ due to load P. The strain in the element is

The deflection of total length of the rod can be obtained by integrating above equation