Problem 68. Find the moment of inertia of the section of a box girder, shown in Fig. 77, with respect to its horizontal gravity axis. The moment of inertia of one of the angle sections with respect to its own horizontal gravity axis, is 31.92 in. to the 4th power.

Problem 69. Find the moment of inertia of the column section, shown in Fig. 78, with respect to the two gravity axes gx gy The column is built up

and

"

1.78

"

%" -g'

-14"

FIG. 76

of one central plate, two outside plates, and four Z-bars. The legs

of the Z-bars are equal, and have a length of 31⁄2 in. The moment of

48"

%

-18

"

1.82"

FIG. 77

inertia of each Z-bar

section with respect to its own horizontal and vertical gravity axis is 42.12 and 15.44 in. to the 4th power, respectively.

Problem 70. Find the moment of inertia of the section shown in Fig. 79, with respect to the horizontal and vertical gravity axes 9x This section

and gy.

is made up of plates moment of inertia of each angle section with respect to both its own and angles. The horizontal and vertical gravity axes is 28.15 in. to the 4th power.

51. Moment of Inertia by Graphical Method. —It will often be necessary to find the moment of inertia of a plane section whose bounding curve is of a complicated form, as

in the case when it is necessary to compute the strength of rails or deck beams. The graphical method given below may be used for such cases.

Let the section be a rail section, Fig. 80. Draw two lines, AB and CD, parallel to the required gravity axis, at any distance, l, apart. The present section is symmetrical with respect to the y-axis, so that it will only be necessary to consider the part on one side of that axis, say the part to the right. Suppose the section divided into strips parallel to AB and CD, and let z denote the length of one of these strips, and dy its width. For each value of x there is a length x' found, such that

Then for every point P on the original section whose coördinates are x and y, there will be a point P' on the transformed section whose coördinates are x' and y. Suppose all these points, P', constructed, and a boundary line drawn through them. Let F denote the area of the orig

inal curve, and F" the area of the transformed curve, also

let

N=SydF= Syxdy = 1Sx'dy = 1F".

But SydF=ÿF (Art. 24), where y is the distance of the center of gravity of F from the line AB. It follows that

This locates the center of gravity.

The moment of inertia will be found by substituting for each x, x', and for each x', x'', such that

Every point, P', now goes over into a point P", forming a new transformed boundary. Call the area of this last

curve F". Since x" = 2', and x' = x24,

ין

=

y2
72°

-g'

= Sy2dF = Sy2xdy = l2 Sx"dy = P2F"'',

giving the moment of inertia of the original section with respect to the line AB.

gx

To determine I it is simply necessary to use the formula I = I'- Fd2. This gives

gx

The areas of the sections are measured by
planimeter.

means of

An 52. Moment of Inertia by Use of Simpson's Rule. approximate value for the moment of inertia of irregular sections, such as rail sections, may be obtained by the use of Simpson's Rule. Let the irregular area be the rail sec

tion (Fig. 43) and let it be required to find the moment of inertia of the section with respect to the base of the rail. We may write

...

29

I = EyA = 3%A0+gi4+34, ... ty ng where the A's represent the areas and the y's the distance from the center of the A's to the base of the rail. In this

1

case yo, y1 = 2, y2=4, etc., and A。= 2.95, A1=1.95, A=.61, etc. (see Problem 33).

A more exact summation of the terms would be given by adding by means of Simpson's Formula.

This gives

where uo, u, ug, etc., have the values given in Problem 33. The student should compare the result obtained by this method with that obtained by the method of direct addition given above. Compare the value obtained with that resulting when Durand's Rule (Art. 28) is used. Use both methods to find the moment of inertia of the sections in Problem 33 and Problem 34.

53. Least Moment of Inertia of Area. In considering the strength of columns and struts it is necessary to know the least moment of inertia of a cross section, since bending will take place about an axis of its cross section having such least moment of inertia. It was shown in Art. 42 that, if the moment of inertia of the area with respect to two rectangular axes in its plane is known, the moment of inertia with respect to any other axis,