This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.

It will be shown later that the strength of a beam depends partly on the form of its cross-section. The following discussion relates principally to cross-sections of beams, and the results reached (like shear and bending moment) will be made use of later in the subject of strength of beams.

47. Center of Gravity of an Area. The student probably knows what is meant by, and how to find, the center of gravity of any flat disk, as a piece of tin. Probably his way is to balance the piece of tin on a pencil point, the point of the tin at which it so balances being the center of gravity. (Really it is midway between the surfaces of the tin and over the balancing point.) The center of gravity of the piece of tin, is also that point of it through which the resultant force of gravity on the tin (that is, the weight of the piece) acts.

By "center of gravity" of a plane area of any shape we mean that point of it which corresponds to the center of gravity of a piece of tin when the latter is cut out in the shape of the area. The center of gravity of a quite irregular area can be found most readily by balancing a piece of tin or stiff paper cut in the shape of the area. But when an area is simple in shape, or consists of parts which are simple, the center of gravity of the whole can be found readily by computation, and such a method will now be described.

48. Principle of moments Applied to Areas. Let Fig. 21 represent a piece of tin which has been divided off into any number of parts in any way, the weight of the whole being "W", and that of the parts W1, W2, "W3, etc. Let C1 C2, C3, etc., be the centers of gravity of the parts, C that of the whole, and c1 c2, c3, etc., and c the distances from those centers of gravity respectively to some line (L L) in the plane of the sheet of tin. When the tin is lying in a horizontal position, the moment of the weight of the entire piece about L L is Wc, and the moments of the parts are W1C1, W1c2, etc. Since the weight of the whole is the resultant of the weights of the parts, the moment of the weight of the whole equals the sum of the moments of the weights of the parts; that is,

Wc=Wlcl+W2 c2+etc.

Now let A1, A2, etc. denote the areas of the parts of the pieces of tin, and A the area of the whole; then since the weights are proportional to the areas, we can replace the Ws in the preceding equation by corresponding A's, thus:

Ac=A1 c1+A2 c2 +etc.. (4)

If we call the product of an area and the distance of its center of gravity from some line in its plane, the "moment" of the area with respect to that line, then the preceding equation may be stated in words thus:

The moment of an area with respect to any line equals the algehraic sum of the moments of the parts of the area.

If all the centers of gravity are on one side of the line with respect to which moments are taken, then all the moments should be given the plus sign; but if some centers of gravity are on one side and some on the other side of the line, then the moments of the areas whose centers of gravity are on one side should be given the same sign, and the moments of the others the opposite sign. The foregoing is the principle of moments for areas, and it is the basis of all rules for finding the center of gravity of an area.

Fig. 21.

To find the center of gravity of an area which can be divided up into simple parts, we write the principle in forms of equations for two different lines as "axes of moments," and then solve the equations for the unknown distances of the center of gravity of the whole from the two lines. We explain further by means of specific examples.

Examples. 1. It is required to find the center of gravity of Fig. 22, a, the width being uniformly one inch.

The area can be divided into two rectangles. Let C. and

Fig. 22.

C3 be the centers of gravity of two such parts, and C the center of gravity of the whole. Also let a and b denote the distances of C from the two lines OL' and OL" respectively.

The areas of the parts are 6 and 3 square inches, and their arms with respect to OL' are 4 inches and ½ inch respectively, and with respect to OL" ½ inch and l½ inches. Hence the equations of moments with respect to OL' and OL" (the whole area being 9 square inches) are:

Hence, | 9 | X | a | = | 6 | X | 4 | + | 3 | X | ½ | = | 25.5, |

9 | X | b | = | 6 | X | ½ | + | 3 | X | 1½ | = | 7.5. | |

a | - | 25.5 | ÷ | 9 | = | 2.83 | inches, | ||||||

b | = | 7.5 | ÷ | 9 | = | 0.83 | " |

2. It is required to locate the center of gravity of Fig. 22, b, the width being uniformly one inch.

The figure can be divided up into three rectangles. Let Cl, C2 and C3 be the centers of gravity of such parts, C the center of gravity of the whole; and let a denote the (unknown) distance of C from the base. The areas of the parts are 4, 10 and 4 square inches, and their "arms " with respect to the base are 2, ½ and 2 inches respectively; hence the equation of moments with respect to the base (the entire area being 18 square inches) is:

18 x a = 4 x 2 + 10 x ½ + 4 x2 = 21. Hence, a = 21 ÷ 18 = 1.17 inches.

From the symmetry of the area it is plain that the center of gravity is midway between the sides.

1. Locate the center of gravity of

Fig. 23.

Ans. 2.6 inches above the base.

49. Center of Gravity of Built=up Sections. In Fig. 24 there are represented cross-sections of various kinds of rolled steel, called "shape steel," which is used extensively in steel construction. Manufacturers of this material publish "handbooks" giving full information in regard thereto, among other things, the position of the center of gravity of each cross section. with such a handbook

Fig. 23.

Fig. 24 available, it is therefore not necessary actually to compute the position of the center of gravity of any section, as we did in the preceding article; but sometimes several shapes are riveted together to make a "built-up" section (see Fig. 25), and then it may be necessary to compute the position of the center of gravity of the section. Example. It is desired to locate the center of gravity of the section of a built-up beam represented in Fig. 25. The beam consists of two channels and a plate, the area of the cross-section of a channel being 6.03 square inches.

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