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Case 10. Moments Of Inertia Of The Section Of An Unequal Sided Angle-iron.

Let the angle-iron be 7" x 3|" x \" as in Fig. 160.

To find the moments of inertia about the axes XOX and YO Y, through O the centre of area.

To find XOX, moments of area are taken about AB, to find YOY, they are taken round AC.

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Fig. 160.

66. Graphic method of obtaining the moment of inertia of an irregular section.

In this method an area is easily constructed which is proportional to the moment of inertia of the section. To find the moment of inertia about the axis XOX, through centre of area. For the upper portion of the section (Fig. 161) draw a line OY perpendicular to XOX, and at some whole number of units "b" from O, draw LM parallel to XOX. Draw any number of lines in the area, of which PQ is one parallel to XOX. Make LN=OP, or PN=OL = b. Join NQ meeting LM in M. Join OM meeting PQ in R. If this is done for all the lines such as PQ, then R will trace out a curve the area of which is proportional to the moment of inertia.

If A j is the total area of this new figure above and below XOX, then r 1.2 A

PQ PN

LM NL'

LM OL

PR OP'' multiplying, we get, since PN= OL = b,

PQ

PR OP2

Proof.

and

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Therefore b2. PR=PQ. OP2 (1).

Now if the coordinates of Q are x and y, and the coordinates of R are x1 and y, then

OP = y, PQ = x, PR = xl,

and equation (1) becomes b2xl = xy2 (2).

Considering an elementary strip of area at PQ of very small thickness dy, which is the same for both the Q and R curves; then moment of inertia of this strip about XOX is xdy. y2 and the moment of inertia of whole area

Ixx = jxy2dy = b2fx1 dy. From (2)
= b2Au

or moment of inertia about XOX is the area traced out by R, multiplied by b2.

EXERCISES.

1. The dimensions of the section of a cast-iron beam are:—Top flange 4 inches by 1| inches; bottom flange 12 inches by If inches; web 16 inches by 1| inches. Calculate the moment of inertia of the section about the neutral axis.

Ans. 2278-40.

2. Determine the moment of inertia and radius of gyration, about an axis through centre of area parallel to the flanges, of two tee-irons, each 8" x 3" x 1", placed flange to flange in cross shape.

Ans. 47 3.

3. The section of a plate girder is as follows:—Top flange, two plates each, 16 inches by | inch; bottom flange, three plates each, 14 inches by | inch; web plate 2 feet deep, % inch thick; angle-irons connecting flanges and web 4 inches by 4 inches by | inch. Find the moment of inertia of the section about neutral axis.

Ans. 8175.

4. Find the moment of inertia of the section of a circular tube 10 inches external diameter and 1 inch thick.

5. Show that in the case of a triangular area the moment of inertia about an axis through the centre of area is the same as if we consider one-third of the area concentrated at each of the middle points of the sides.

Note. This is useful in finding the moment of inertia of regular polygons which can be divided up into triangles.

6. Find the moment of inertia of a regular hexagon about an axis through its centre of area, and joining opposite angular points.

7. Find the moment of inertia of a rhomboidal or lozenge-shaped section about a diagonal as axis.

bh3

If b and h are diagonals, I= Tk

CHAPTER VII.

GIRDERS.

67. Modulus of section.

We have already proved that the relation between the bending moment and the stresses induced is given by the equation

y

i.e., the moment of the external forces is equal to the moment of internal stresses, or moment of resistance of the beam. The quantity

-is called the "modulus of the section," and is generally denoted by

the letter Z, and it is well to note that

Hence M=f=fZ.

y

Generally, it is necessary to consider the greatest tension and compression at any point of the cross section; then, if yt and yc are taken as the distances of the parts of the section furthest from the neutral axis, and if fu fc be the tensile and compressive stresses corresponding to the distances yt and yc,

M=^I or .£/
yt Vc

=ftZt orfcZc.

68. Graphic method of finding the " equivalent area" and "modulus of section."

Equivalent area. On the section of a beam we have the stress varying from zero at the neutral axis to a maximum at the outer fibre. If now we replace the section of the beam by a section which has spread over it the same amount of stress, but instead of being variable, is constant all over and equal to the stress on the outer fibre, we get a figure which is called the Equivalent Area or Modulus Figure.

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