Civil Engineering Online Notes: Strength Of Materials

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Monday, October 14, 2013

MOMENT OF INERTIA

Moment of inertia, also called mass moment of inertia or the angular mass, (SI units kg m²) is a measure of an object’s resistance to changes in its rotation rate. It is the rotational analog of mass. That is, it is the inertia of a rigid rotating body with respect to its rotation. The moment of inertia plays much the same role in rotational dynamics as mass does in basic dynamics, determining the relationship between angular momentum and angular velocity, torque and angular acceleration, and several other quantities. While a simple scalar treatment of the moment of inertia suffices for many situations, a more advanced tensor treatment allows the analysis of such complicated systems as spinning tops and gyroscope motion.

The symbol I and sometimes J are usually used to refer to the moment of inertia.

The moment of inertia of an object about a given axis describes how difficult it is to change its angular motion about that axis. For example, consider two discs (A and B) of the same mass. Disc A has a larger radius than disc B. Assuming that there is uniform thickness and mass distribution, it requires more effort to accelerate disc A (change its angular velocity) because its mass is distributed further from its axis of rotation: mass that is further out from that axis must, for a given angular velocity, move more quickly than mass closer in. In this case, disc A has a larger moment of inertia than disc B.

The moment of inertia has two forms, a scalar form I (used when the axis of rotation is known) and a more general tensor form that does not require knowing the axis of rotation. The scalar moment of inertia I (often called simply the "moment of inertia") allows a succinct analysis of many simple problems in rotational dynamics, such as objects rolling down inclines and the behavior of pulleys. For instance, while a block of any shape will slide down a frictionless decline at the same rate, rolling objects may descend at different rates, depending on their moments of inertia. A hoop will descend more slowly than a solid disk of equal mass and radius because more of its mass is located far from the axis of rotation, and thus needs to move faster if the hoop rolls at the same angular velocity. However, for (more complicated) problems in which the axis of rotation can change, the scalar treatment is inadequate, and the tensor treatment must be used (although shortcuts are possible in special situations). Examples requiring such a treatment include gyroscopes, tops, and even satellites, all objects whose alignment can change.

The moment of inertia can also be called the mass moment of inertia (especially by mechanical engineers) to avoid confusion with the second moment of area, which is sometimes called the moment of inertia (especially by structural engineers) and denoted by the same symbol I. The easiest way to differentiate these quantities is through their units. In addition, the moment of inertia should not be confused with the polar moment of inertia, which is a measure of an object’s ability to resist torsion (twisting).

Definition

A simple definition of the moment of inertia of any object, be it a point mass or a 3D-structure, is given by:

where

‘dm’ is the mass of an infinitesimally small part of the body

and r is the (perpendicular) distance of the point mass to the axis of rotation.

Detailed Analysis

The (scalar) moment of inertia of a point mass rotating about a known axis is defined by

The moment of inertia is additive. Thus, for a rigid body consisting of N point masses m_iwith distances r_i to the rotation axis, the total moment of inertia equals the sum of the point-mass moments of inertia:

For a solid body described by a continuous mass density function ?(r), the moment of inertia about a known axis can be calculated by integrating the square of the distance (weighted by the mass density) from a point in the body to the rotation axis:

where

V is the volume occupied by the object.

? is the spatial density function of the object, and

are coordinates of a point inside the body.

Diagram for the calculation of a disk’s moment of inertia. Here k is 1/2 and r is the radius used in determining the moment.

Based on dimensional analysis alone, the moment of inertia of a non-point object must take the form:

where

M is the mass

R is the radius of the object from the center of mass (in some cases, the length of the object is used instead.)

k is a dimensionless constant called the inertia constant that varies with the object in consideration.

Inertial constants are used to account for the differences in the placement of the mass from the center of rotation. Examples include:

k = 1, thin ring or thin-walled cylinder around its center,

k = 2/5, solid sphere around its center

k = 1/2, solid cylinder or disk around its center.

Parallel axis theorem

Once the moment of inertia has been calculated for rotations about the center of mass of a rigid body, one can conveniently recalculate the moment of inertia for all parallel rotation axes as well, without having to resort to the formal definition. If the axis of rotation is displaced by a distance R from the center of mass axis of rotation (e.g. spinning a disc about a point on its periphery, rather than through its center,) the displaced and center-moment of inertia are related as follows:

This theorem is also known as the parallel axes rule and is a special case of Steiner’s parallel-axis theorem.

Perpendicular Axis Theorem

The perpendicular axis theorem for planar objects can be demonstrated by looking at the contribution to the three axis moments of inertia from an arbitrary mass element. From the point mass moment, the contributions to each of the axis moments of inertia are

Composite bodies

If a body can be decomposed (either physically or conceptually) into several constituent parts, then the moment of inertia of the body about a given axis is obtained by summing the moments of inertia of each constituent part around the same given axis.

Common Moments of Inertia

THEORY OF SIMPLE BENDING

Labels: Strength Of Materials

THEORY OF SIMPLE BENDING

Assumptions:

Plane sections of the beam, originally plane, remain plane.
The material of the beam is homogeneous and obeys Hooke’s law.
The moduli of elasticity for tension and compression are equal.
The beam is initially straight and of constant cross-section.
The plane of loading must contain a principle axis of the beam cross-section and the loads must be perpendicular to the longitudinal axis of the beam.

Flexure Formula:

Where M= bending moment

I = moment of inertia of the section about the bending axis.

=fibre stress at a distance ‘y’ from the centroidal/neutral axis.

E = Young’s Modulus of the material of the beam.

R = radius of curvature of the bent beam.

If y is replaced by c, the distance to remotest element, then

Where, Z= section modulus and is given by

Pure and Non-uniform Bending:

Pure bending refers to flexure of a beam under constant bending moment, which means that the shear force is zero.

i.e

In contrast, non-uniform bending refers to flexure in the presence of shear forces, which means that the bending moment changes as we move along the axis of the beam.

Examples of beams in pure bending:

(a) Beam subjected to positive bending moments (figure 1)

Figure 1

(b) Beam with central region in pure bending (figure 2)

Figure 2

Normal Strains:

As a result of bending, somewhere between the top and bottom of the beam is a surface in which the longitudinal fibres do not change in length. This surface is called the neutral surface of the beam and its intersection with any cross-sectional plane is called the neutral axis of the cross-section.

All the longitudinal fibres other than those in the neutral surface either lengthen or shorten, thereby creating longitudinal strains

Where k = curvature = 1/R

This equation shows longitudinal strains are proportional to the curvature and that they vary linearly with the distance y from the neutral surface. This equation is derived from the geometry of the deformed beam and is independent of the properties of the material. The equation is valid irrespective of the stress-strain diagram of the material.

Transverse Strain:

The axial strains

are accompanied by lateral or transverse strains due to the effect of Poisson’s ratio. Positive strains are accompanied by negative transverse strains

Where

is the Poisson’s Ratio.

As a result of these strains, the shape of the cross-section change. For example, let us study the case of a beam of rectangular cross-section subjected to pure bending so as to induce tension at the top and compression at the bottom. The sides of the rectangular cross-section become inclined to each other. The top surface becomes saddle shaped. If the longitudinal curvature in the xy plane is considered positive, then the transverse curvature in the yz plane is negative. All planes of the beam that were initially parallel to neutral surface develop antiplastic curvature.

Applicability of Flexure Formula:

The normal stresses determined from flexure formula concern pure bending, which means no shear forces act on the cross-section. In case of non-uniform bending the presence of shear forces produces warping or put of place distortion of the cross-section, thus, a section that is plane before bending is no longer plane after bending. Warping due to shear greatly complicates the behaviour of the beam, but more elaborate analysis shows that the normal stresses calculated from the flexure formula are not significantly altered by the presence of the shear stresses and the associated warping. Thus use of the theory of pure bending for calculating normal stresses in cases of non-uniform bending is considered justified.

Ultimate strength in bending:

When a beam is loaded upto failure, then using bending moment at failure, flexural stress is calculated and is called Modulus of Rupture. it is used to compare the ultimate strength of beams of various sizes and materials

SHEAR CENTRE -WITH EXAMPLES

Labels: Strength Of Materials

SHEAR CENTRE -WITH EXAMPLES

If a beam is subjected to bending moments and shear force in a plane, other than the plane of geometry, which passes through the centroid of the section, then bending moment will be accompanied by twisting. In order to avoid twisting and cause bending only, the transverse forces must act through a point which may not coincide with the centroid, but will depend upon the shape of the section and such a point is termed as shear centre.

Figure 1

Consider a channel section as shown in figure 1. Now we shall find the position of the plane through which the vertical loads must act so as to produce simple bending, with the x-axis as neutral axis.

It may be assumed that the vertical shearing force, F at the section is taken up by the web alone. In the flanges, there will be horizontal shear stresses which will be denoted by q.

Let us consider an element ‘abcd’ cut from the lower flange by two adjacent cross-sections

apart and by a vertical plane parallel to the web and at distance ’u’ (which is variable) from the free end of the lower flange. The difference in tensile forces T and

must be equal to the shear force on the side ‘ad’ of the element. Assuming a uniform distribution of shear stress (since the thickness is small) over the thickness, we have,

The integration being carried out over the portion ‘ab’ of the flange.

The stress per unit length of the centre line of the section,

Therefore, it is seen that q is proportional to u.

The maximum value of

At the junction of the flange and web, the distribution of the shear stress is complicated, so we may assume that the equation

holds good for u = 0 and u = b.

The average shear stress

The longitudinal shear force in the top and bottom of the flange clip_image011

The couple about the z-axis of these shear forces clip_image012

Let us assume that the vertical shear force F acts through point ‘o’, the shear centre at a distance c from O on the centre line of the web.

The twisting of this section is avoided if

which gives the position of the shear centre.

Note: the shear centre for cross-sectional areas having one axis of symmetry, is always located on the axis of symmetry. In the case of the I-beam which is symmetrical about both the x-axis and y-axis, the shear centre coincides with the centroid of the section. The exact location of the shear centre for unsymmetrical sections are complicated and can be located by inspection.

EXAMPLE – 1