The center of gravity of any irregular plane figure can be determined by applying the following principle: The static moment of any plane figure with regard to a line in its planethat is, the product of its area A by the distance D of its center of gravity from that line-is equal to the algebraic sum of the static moments of the separate parts into which the figure may be divided, with regard to the same axis, or

AD=aidi+a2d2, etc.,

in which, a1 az, etc., denote the areas of the subdivided parts of the figure, and dı, d2, etc. are the distances of their respective centers of gravity from the reference line. Solving this equa

tion for the value of D,

aidi+a2d2+etc.

D=

A

The figure whose center of gravity is required is divided into separate parts whose centers of gravity are easily ascertained, usually into rectangles

or triangles. A suitable axis is then assumed with reference to which the expressions aıdı, a2d2, etc. are found, and their sum is divided by A = ai + a2+ etc., the quotient giving D. The center of gravity of the whole figure lies, therefore, on a line parallel to the assumed axis and distant D from it. In a similar manner, another line containing the center of gravity is obtained, the intersection of the two lines giving

its exact position.

EXAMPLE 1.-Find the center of gravity of the cross-section

of the dam shown in Fig. 5.

SOLUTION.-Divide the section into the rectangles ABC'I and HEFG and the triangle CDC', and assume the lines X-X and Y-Y as reference lines.

Then,

From the illustration, a1 = 100×20=2,000, y1 =70, x=20; 69 X 46

2

and x3=43.

= 1,587, y2=43, x2=45.33; as=86×20=1720, yз= 10,

Substituting these values,

2,000×70+1,587X43+1,720X10

y=

42.48

2,000+1,587+1,720

2,000×20+1,587×45.33+1,720×43

and

x=

-= 35.03

5,307

EXAMPLE 2.-Find the center of gravity of the bridge chord section shown in Fig. 6.

SOLUTION. The center of gravity is on the line YY, which is an axis of symmetry. To find the distance y, divide the section into angles and plates and take moments about XX.

αχ εχζε

-X

are:

The areas and centers of gravity of the angles might be located by the preceding principles or taken from a manufacturer's handbook. They for the 4"X4"X" angle, area=3.75 sq. in. and distance from center of gravity to back of angle =1.18 in.; for the 3"X3"X" angle, area=3.25 sq. in. and distance from center of gravity to back of angle=1.06 in. Hence, the moment of the bottom angles is 2X3.75X1.18=8.85 and that of the top angles is 2X3.25 (15-1.06) 90.61. The moment of the two web plates is 2X15X1X7.5=112.5, and that of the cover-plate, 24XX 15.25 183.00. The sum of the moments is, 8.85+90.61+112.5 +183.00 394.96. The sum of the areas is 2 X 3.25 + 2 × 3.75 +24X+2X15X}=41 sq. in. Then, y=394.96÷41-9.63 in.

FIG. 6

Center of Gravity of Solids.-For a solid having three axes of symmetry, all perpendicular to each other, like a sphere, cube, right parallelopiped, etc., the point of intersection of these axes is the center of gravity.

For a cone or pyramid, draw a line from the apex to the center of gravity of the base; the required center of gravity is one-fourth the length of this line from the base, measured on the line.

For two bodies, the larger weighing W lb., and the smaller P lb., the center of gravity will lie on the line joining the centers of gravity of the two bodies and at a distance from the Pa P+W'

larger body equal to

where a is the distance between

the centers of gravity of the two bodies.

For any number of bodies, first find the center of gravity of two of them, and consider them as one weight whose center of gravity is at the point just found. Find the center of gravity of this combined weight and a third body. So continue for the rest of the bodies, and the last center of gravity will be the center of gravity of the whole system of bodies.

To find the center of gravity mechanically, suspend the object from a point near its edge and mark on it the direction of a plumb-line from that point; then suspend it from another point and again mark the direction of a plumb-line. The intersection of these two lines will be directly over the center of gravity.

MOMENT OF INERTIA

The moment of inertia of a plane surface about a given axis is the sum of the products obtained by multiplying each of the elementary areas, into which the surface may be conceived to be divided, by the square of its distance from the axis.

The moment of inertia is usually designated by the letter I. The value of the moment of inertia used in calculating the strength of beams and columns is usually taken about the neutral axis of the figure, which, with the exception of reinforced-concrete sections, is passing through the center of gravity of the figure.

Formulas for the values of I about an axis passing through the center of gravity of the section are given for various forms of sections in the accompanying table. For any other section,

it can be computed by means of the following principles:

Principle I.-The moment of inertia of a section about any axis is equal to the algebraic sum of the moments of inertia about the same axis, of the separate parts of which the figure may be conceived to consist.

Principle II.-The moment of inertia of any figure about an axis not passing through the center of gravity, is equal to the moment of inertia about a parallel axis through the center of gravity, plus the product of the entire area of the section by the square of the distance between the two axes.

EXAMPLE 1.-Find the moment of inertia about the neutral axis XX of the Bethlehem I column section having dimensions as shown in Fig. 1.

SOLUTION.-Conceive the section to consist of the square ABCD minus twice the rectangle abcd. Then, by applying principle I and the formulas of the table for moments of inertia, 124 2X5.75X10.53

I=

618.6

NOTE. This result can be obtained directly by the I beam formula, given in the same table.

EXAMPLE 2.-Find the moment of inertia of the section shown in Fig. 2 about the neutral axis parallel to the coverplate.

SOLUTION.-The neutral axis passes through the center of gravity, which has been found to be 9.63 in. from the back of the