Imágenes de páginas
PDF
EPUB

The polar moment of inertia of an area about any point in its plane is equal to the sum of the moments of inertia about any two axes at right angles, through the point, and in the plane.

Apply this, for example, to a circular section radius ». We have

[blocks in formation]

S

Moments of Inertia about Parallel Axes. When we speak simply of the moment of inertia of a section without mentioning any axis, the neutral axis is understood. This moment we denote by Io.

N

A

Fig. 272.

N

the I about SS.

Let now A (Fig. 272) be an area of any shape, NN its neutral axis, SS an S axis in the plane, parallel to NN and distant y, from it. We will calculate

Divide A into strips parallel to NN as usual. Then, taking one strip, distant y from NN,

I of strip about SS = area of strip × (1+1)2,

area (y2+2yyo+yo2),

area x2+2y, × area × y + area × y2,

[ocr errors]

=I of strip about NN+2y, x moment of strip about NN + area of strip xy2,

therefore, taking all the strips and adding up,

I of A about SS=I of A about NN+2y, × moment of A about

NN+Axy,

=I+Ay2.

The second term being zero, since NN is through the C. G. of A.

This expression is of great value, because knowing I we can at once deduce the I about any parallel axis. Or in some cases it is easier to find I about some axis parallel to NN than about NN itself, and the formula then enables us to deduce I..

For example, take a rectangle.

Then

I about one end = 1 Ah2,

distance y, between end and neutral axis:

... L= Ah2 – A

h

2

or we could have proceeded vice versâ.

=

h

2

Moment of Inertia of any Area.-The values of

r2 given on page 214, and the proposition just proved enable us to find the I of any practical section. Take for example

the trapezoidal section of page 358.

Find NN as explained

b

x

a

Fig. 273.

on page 358, and then divide up as shown by the dotted lines. We have then to add together the I's of two rectangles about their ends, two triangles about their bases, and two triangles about axes through their vertices parallel to their bases. These are

I of top rectangle =

I of bottom rectangle =

bc. c2,

xd. d2,

I of two top triangles=2{ ++c. xzb)e},

2

[ocr errors][merged small][subsumed]

.'. I。=} b&3+} xd3 + 11⁄2 ( x − b )c3 +1(a − x)d3.

[blocks in formation]

the values of c and d are found on page 358, and we substitute. The section is not an important one, so we will not conclude the calculation; enough has been done to show the method.

Beams of I Section.-We saw in the last chapter that the I shape of section was advantageous, but this is still more shown by the work of the present chapter. For we now see that the resistance offered by the stress on any slice of a section depends on its moment about the neutral axis, and thus varies as the square of the distance We thus get

E

of the slice from the axis, being bt. y2.

R

a still weightier reason for removing the bulk of the material as far as possible from the axis, i.e. for the adoption of the I section.

N

For example, consider a rectangular section, and let us vary its shape to the I shown in dotted lines, the shaded rectangle of material being moved to the new position also shaded, and the three equal rectangles similarly treated.

Fig. 274.

Then the shaded rectangle, area A, say, had in the rectangular section a moment of inertia Aa2/3 about NN, while in the new position its moment of inertia is something more than Aa2, and hence its power of resistance to bending is more than trebled. On the whole we have increased the M of I of the section by at least 4 (3 Aa2); although its sectional area, and hence the amount of metal in the beam, is unaltered.

The process of removing metal into the flanges is of course limited by other considerations. In the first place we must have a web strong enough to withstand a

And the web must

large portion if not all the shearing. not only be strong enough but also stiff, so that it may not give way under the compressive stresses which come on it, by buckling or sideway yielding; this effect is greater the greater the depth, and is in some cases, where the web is very deep, provided against by riveting angle bars at intervals to the web, as in Fig. 275, which shows what is

known as a solid

web bowstring girder, the vertical lines showing the stiffeners.

Fig. 275.

In Fig. 274 we show the increase of resistance effected while the depth remains fixed; but one way of increasing the distance of the metal from the neutral axis is by increasing the depth; in this case, however, we are limited by the necessity for stiffness, and thus the most economical depth is limited.

Again, the thinner the flanges the nearer is the metal in them to the outside edge, and hence the greater the moment of inertia, keeping their area constant. But again there is the necessity for stiffness in the compression flange to prevent buckling, and also, since the load is distributed over the width of the flange, it would, if too thin, bend as in Fig. 276.

Fig. 276.

Further, there are practical questions to be considered, such as strength to withstand forces which come on the beam during manufacture; liability to blows, which necessitate local strength-e.g. a thin flange might be broken by a slight accidental blow, though quite strong enough when in place. Another consideration applying to cast beams, is that below a certain thickness it is not possible to secure a sound casting.

These questions we cannot here deal with, but we have

shown what considerations govern the question; and the student will thus, when he has attained the necessary practical knowledge, know how to deal with them.

Moments of Inertia of I Beams—Approximate Methods.-The accurate calculation of the M. of I. of an I beam is sometimes rather long, and a quicker method is often of practical value.

The most usual method of approximation is to treat

[blocks in formation]

of the shape shown in (b),

the flanges being represented as lines; still using in the calculation, however, their actual areas.

Then NN is the centre line, and from (b)

[merged small][merged small][merged small][merged small][merged small][merged small][subsumed][merged small][ocr errors]

We will now see what error is made, by finding the accurate value; and for this purpose we will use a method of subtraction, and not, as on page 371, of addition.

The I is the difference between a rectangle 12" by 6", and two rectangles each 10" by 21′′; and these all have the same neutral axis NN.

I=(12×6) 122 - 2 × (10 × 21)102,
=864-4163=4473.

There is then a considerable discrepancy between the

« AnteriorContinuar »