mech design -4 deflection and stifness

8/11/2019 Mech Design -4 Deflection and Stifness

1/31

DEFLECTION AND

STIFFNESS


2/31

SPRINGRATES

Elasticity is that property of a material that enables it to regain

its original configuration after having been deformed.

A spring is a mechanical element that exerts a force when

deformed.

Two classification of springs are,

1. linear spring

2. non-linear spring


3/31

LINEARSPRING

Figure a shows a straight beam of length l simply supported at the

ends and loaded by the transverse force F.

The deflection y is linearly related to the force, as long as the elastic

limit of the material is not exceeded, as indicated by the graph.

This beam can be described as a linear spring.


4/31


5/31

NON-LINEARSTIFFENINGSPRING


6/31

NON-LINEARSOFTENINGSPRING

Figure (c) is an edge-view of a dish-shaped round disk. The forcenecessary to flatten the disk increases at first and then decreases asthe disk approaches a flat configuration, as shown by the graph.

Any mechanical element having such a characteristic is called anonlinear softening spring.


7/31

SPRINGCONSTANT

The general relationship between force(F) and

deflection() are,

F

F = k *

Where kis called as spring constant.


8/31


9/31

TENSION, COMPRESSIONANDTORSION

The angular deflection of a uniform round bar subjected

to a twisting moment T,

where is in radians.

The torsional spring rate as,


10/31

DEFLECTIONDUETOBENDING

The problem of bending of beams probably occurs more often thanany other loading problem in mechanical design.

Shafts, axles, cranks, levers, springs, brackets, and wheels, as wellas many other elements, must often be treated as beams in the design

and analysis of mechanical structures and systems.

The curvature of a beam subjected to a bending moment M is givenby

where is the radius of curvature.


11/31


The curvature of a plane curve is given by the equation

where the interpretation here is that y is the lateral deflection of the

beam at any point x along its length.

The slope of the beam at any point x is

1


12/31


For many problems in bending, the slope is very small, and for these

the denominator of Eq. (1) can be taken as unity.

Equation (1) can then be written as,

Differentiating Eq. (2) yields,

2


13/31


It is convenient to display these relations in a group as follows:

load

shear force

bending moment

slope

deflection


14/31

BEAMDEFLECTIONBYSUPERPOSITION

Superposition resolves the effect of combined loading on a structureby determining the effects of each load separately and adding theresults algebraically.

Superposition may be applied provided:

(1) Each effect is linearly related to the load that produces it,

(2) A load does not create a condition that affects the result of

another load,

(3) The deformations resulting from any specific load are not largeenough to appreciably alter the geometric relations of the parts of thestructural system.


15/31

STRAINENERGY

Strain energy principles are widely used for determining the

response of machines and structures to both static and dynamic

loads.

When load stretches the bar, strains are produced. The presence ofthese strains increase the energy level of the bar itself.

The energy absorbed by the bar during the loading process is called

strain energy.

1


16/31


17/31

STRAINENERGY

Let assume that the material of the bar shown in fig.1follows Hooke's law so that the load displacement curve

is a straight line.

Then the strain energy (U) stored in the bar (equal to thework (W) done by the load ) is


18/31

STRAINENERGY

If the member is deformed a distance y, and if the force-deflection

relationship is linear, this energy is equal to the product of theaverage force and the deflection,

This above equation is general in the sense that the force F can alsomean torque, or moment.

F y

F = k * y

y = F/k


19/31

STRAINENERGY

By substituting appropriate expressions for k, strain-energy formulas

for various simple loadings may be obtained.

For tension and compression and for torsion,


20/31

STRAINENERGY

To obtain an expression for the strain energy due to direct shear,

consider the element with one side fixed in Fig. a.

The force F places the element in pure shear, and the work done is

U = F/2.

Since the shear strain is = /l = /G = F/AG, we have

1


21/31


22/31

STRAINENERGY

Equation (2) is exact only when a beam is subject to pure bending. Even

when shear is present, Eq. (2) continues to give quite good results, except

for very short beams.

The strain energy due to shear loading of a beam is a complicated problem.

An approximate solution can be obtained by using Eq. (1) with a correctionfactor whose value depends upon the shape of the cross section.

If we use C for the correction factor and V for the shear force, then the

strain energy due to shear in bending is the integral of Eq. (1)


23/31

TABLEFORCORRECTIONFACTOR


24/31

CASTIGLIANOSTHEOREM

Castiglianos theorem states that when forces act on elastic systemssubject to small displacements, the displacement corresponding toany force, in the direction of the force, is equal to the partialderivative of the total strain energy with respect to that force.

The terms force and displacement in this statement are broadlyinterpreted to apply equally to moments and angular displacements.Mathematically, the theorem of Castigliano is i


25/31

CASTIGLIANOSTHEOREM

Axial and torsional deflections by using Castigliano'stheorem,


26/31

CASTIGLIANOSTHEOREM

Castiglianos theorem can be used to find the deflection at a point

even though no force or moment acts there. The procedure is:

1 . Set up the equation for the total strain energy U by includingthe energy due to a fictitious force or moment Qi acting at the pointwhose deflection is to be found.

2 . Find an expression for the desired deflection i , in thedirection of Qi , by taking the derivative of the total strain energywith respect to Qi .

3. Since Qi is a fictitious force, solve the expression obtained instep 2 by setting Qi equal to zero. Thus,


27/31

STATICALLYINDETERMINATEPROBLEMS

A system in which the laws of statics are not sufficient to

determine all the unknown forces or moments is said to

be statically indeterminate.

Problems of which this is true are solved by writing the

appropriate equations of static equilibrium and additionalequations pertaining to the deformation of the part. In all,

the number of equations must equal the number of

unknowns.


28/31


A simple example of a statically indeterminate problem is furnished by the

nested helical springs in Fig.


29/31


Only one equation of static equilibrium can be written. It is

which simply says that the total force F is resisted by a force F1 inspring 1 plus the force F2 in spring 2. Since there are two unknownsand only one equation, the system is statically indeterminate.

In the spring example, obtaining the necessary deformation equationwas very straightforward. However, for other situations, thedeformation relations may not be as easy. A more structuredapproach may be necessary.


30/31

PROCEDUREFORSTATICALLYINDETERMINATE

PROBLEMS

Procedure 1 :

1. Choose the redundant reaction(s). There may be alternative choices .

2. Write the equations of static equilibrium for the remaining reactions

in terms of the applied loads and the redundant reaction(s) of step 1.

3. Write the deflection equation(s) for the point(s) at the locations of the

redundant reaction(s) of step 1 in terms of the applied loads and the

redundant reaction(s) of step 1. Normally the deflection(s) is (are) zero.

If a redundant reaction is a moment, the corresponding deflectionequation is a rotational deflection equation.

4. The equations from steps 2 and 3 can now be solved to determine the

reactions


31/31

PROCEDUREFORSTATICALLYINDETERMINATE

PROBLEMS

Procedure 2 :

1. Write the equations of static equilibrium for the beam in

terms of the applied loads and unknown restraint reactions.

2 . Write the deflection equation for the beam in terms of the

applied loads and unknown restraint reactions.

3. Apply boundary conditions consistent with the restraints.

4. Solve the equations from steps 1 and 3.

mech design -4 deflection and stifness

Documents