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Solid Mechanics - Statics

Chapter 9 Distributed Forces: Moments of

Inertia

Vector Mechanics for Engineers

F.P. Beer, E.R. Johnston, 6th edition

Second Moment Moment of Inertia

When a beam is subjected to a loading of a pure moment,

internal tension and compression stresses are formed on the

cross-section.

The distributed forces vary

linearly from the neutral axis

and expressed: F = k y A

The resultant of the distributed

forces is:

At the neutral axis the force equals

zero, above and below the neutral

axis the forces have opposite signs

Second Moment Moment of InertiaThe resultant of the distributed forces is:

The integral above is the first moment Qx of the cross-

section about the neutral axis. The integral over the whole

section is zero (compression = tension).

The moment of a finite F is:

M = yF = k y2A

The total moment of the forces over the

whole section:

The above integral is defined as the second moment or the

moment of inertia about the x-axis. It is denoted Ix and is

the product of y2 by the area, integrated for the entire

section. The moment of inertia is always positive.


We can define the moment of inertia about axis x or axis y

the following way:

These are also called: the rectangular moments of inertia.

If we choose a finite area A = (dx)(dy), the differential

moment of inertia would be:

About x-axis: dIx = y2dA , and

about y-axis: dIy = x2dA


Determine the moment of inertia

about axis-y, using a strip:

dA = (a x) dy

dIx = y2(a x) dy


about axis-x, using a strip:


of a rectangle, about the x-axis:

A general concept:

For the strip in the left diagram,

assign: b = dx, and h = y. Thus,

the differential second moment

about the x-axis becomes:

Example: moment of inertia rectangle

Polar Moment of Inertia

Definition: the polar moment of

inertia is about a point (not about

an axis):

Point O is the pole

From trigonometry we can write:

thus:

The moment of inertia of area A with

respect to axis x is Ix. If area A can be

concentrated into a thin strip parallel

to axis x, at a distance k from axis x,

and have the same moment of inertia:

Solve for kx:

kx is the radius of gyration

with respect to axis x.

Radius of Gyration of an area

Radius of Gyration of an area

Radius of gyration about a point, O

Thus we can

also write:

Same definition for axis y:

Radius of Gyration example

The radius of gyration of the

rectangle with respect to the base is

Note: the centroid is located at y = h/2

while the radius of gyration is located

at k = h /3 = h/1.732 . In other words,

y depends on the first moment, while k

depends on the second moment

Parallel-axis theoremMoment of inertia with respect

to axis a-a' is defined:

define: y = d + y', where y' is the

distance of dA from the centroid.

d distance of centroid from AA'

The first term is the moment of inertia about BB' , the

second term equals zero, thus,

the Parallel-axis Theorem:

Parallel-axis theorem The theorem can be extended to the radius of gyration:

Substituting k2A for I (moment of inertia about axis AA',

and k2A for I (moment of inertia about axis BB', the

theorem is expressed:

The theorem can be extended to the moment of

inertia about a point:

or:

Parallel-axis theorem - examples

Example 2 determine the

moment of inertia of a

triangle about axis BB', and

about axis DD'

Example 1 determine the moment

of inertia of a circle about line T

Ix = r4/4 about the diameter

Moment of

Inertia of

common

geometric

shapes

Moment

of Inertia

common

geometric

shapes

Properties of

rolled steel

shapes US

units.

Table 9.13

Properties of

rolled steel

shapes SI

units.

Table 9.13

Product of Inertia

Definition: the following integral is

defined as the product of inertia

about axes x and y:

The product of inertia can be positive,

negative or zero.

When one or both of the x and y axes are

axes of symmetry, the product of inertia

is zero, as seen in the channel example.

x'-y' is the centroid axes system,

thus the relations: x = x'+x and

y = y'+y.

by definition:

The first term is the product with respect to the centroid

axes. The second and third terms equal zero (first moment

about centroid axes), thus we get:

Parallel axis theorem for Product of Inertia

Principal axes and principal

moments of InertiaConsider an area A with known moments and product

of inertia

We want to determine the

moments of inertia with

respect to new axes x'-y'

which are rotated by an

angle of with respect to

the original axes.

We can write:

Principal axes and principal moments of Inertia

Moment of inertia with respect to the new axes:

Plugging in the values for the original moments of inertia:

Similarly we obtain:

trigonometry:


Now we can write:

Also, we can write:

This result could be expected from the relation

of the polar moment of inertia:


It can be shown that:

define:

The following

are parametric

circle equations

we can write:

Points A and B represent max.

and min. values for Ix'


Imin and Imax are 90 apart and

for them we define: principal

axes about point O

Also, at points A and B the

value of the product moment

of inertia Ix'y' is zero, and thus:


Summary:

For Ix'y' = 0 we get points A and B which correspond to

Imin and Imax which are 90 apart and are defined principal

moments of inertia, and are related to the principal axes

about point O

When Ix'y' is zero we obtain:

We get two values 2m which

are 180 apart, and thus two m which are 90 apart.

Mohr circle for the moment of inertiaMohrs circle (Otto Mohr, 1835-1918) is used to illustrate

the relations between the moment and the product of

inertia of a given area.

Mohr circle for the moment of inertiaIf the moment and the product of an area are known for a

set of rectangular axis, then with Mohr circle we can:

A. Principal axes and

principal moments

B. Moment and product

about any other axes

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