determined as before. The point K, where LK and MK intersect, is the centre of the parallel forces P, Q, R, and S acting at the fixed points A, B, C, and D respectively. If the construction be repeated with the forces acting in any other direction, it will be found that the new resultant will act through the same point K.

In Fig. 180, the forces P, Q, R, and S have all the same sense, and therefore P', Q', R', and S' must have the same sense. But if the sense

of Q, say, were opposite to that of P, then the sense of Q' would be opposite to that of P'.

In applying the above method to the determination of the centre of a system of parallel forces, it is usually most convenient to take the two directions of the forces at right angles to one another.

96. Centres of Gravity or Centroids.-The particles of which any body is made up are attracted to the earth by forces which are proportional to the masses of these particles. For all practical purposes these forces may be considered to be parallel, and their resultant will pass through the centre of these parallel forces. In this case the centre of the parallel forces is called the centre of gravity or centroid of the body, and the determination of a centre of gravity resolves into finding the centre of a system of parallel forces.

The centre of gravity of a body may also be defined as that point from which if the body is suspended it will balance in any position.

When the term centre of gravity is applied to a line, the line is supposed to be an indefinitely thin wire; and when the centre of gravity of a surface is spoken of the surface is supposed to be an indefinitely thin sheet of material.

The following results, which are not difficult to prove, should be noted:

The centroid of a straight line is at its middle point.

The centroid of a triangle is at the intersection of its medians.

The centroid of a parallelogram is at the intersection of its diagonals.

If a plane figure is symmetrical about a straight line, the centroid of the figure is in that straight line.

To find the centroid of a line made up of a number of straight lines. At the centres of the straight lines apply parallel forces whose magnitudes are proportional to the lengths of these lines. The centre of these parallel forces is the centroid required.

To find the centroid of a curved line. Divide the line into a number of parts, preferably of equal length. At the centres of these parts apply parallel forces whose magnitudes are proportional to the lengths of the parts. The centre of these parallel forces is approximately the centroid required. Theoretically the approximation is closer the more numerous the parts into which the curved line is divided, but practically when the parts are very numerous or very short the drawing of the link polygon becomes less accurate.

To find the centroid of any irregular figure. Divide the figure into parts whose centroids and areas are known or easily found. At the centroids of these partial areas apply parallel forces whose magnitudes are proportional to these areas. The centre of these parallel forces is the centroid required. If the given figure has an irregular curved boundary line such as is shown in Fig. 181, divide the figure into a number of parallel strips as shown by the full straight lines. Draw the centre lines of these strips, shown dotted. The centroids of these strips may be taken at the middle points of their centre lines, and the areas of the strips may be taken as proportional to the lengths of their centre lines.

FIG. 181.

To find the centroid of a quadrilateral. Let ABCD (Fig. 182) be the quadrilateral. Draw the diagonals AC and BD, intersecting at O. Let OA be less than OC. Make CE equal to AO. Join BE and DE. G, the centroid of the triangle BDE is also the centroid of the quadrilateral ABCD.

To find the centroid of an arc of a circle. the arc of a circle of which O is the centre.

Let ABC (Fig. 183) be Draw OB at right angles

to the chord AC. Draw BD at right angles to OB and make BD equal to the arc BC. Join OD. OD at E. Draw EG parallel to centroid of the arc ABC.

Draw CE parallel to OB to meet
AC to meet OB at G. G is the

H

To find the centroid of a sector of a circle. Let OABC (Fig. 184) be the sector of a circle of which O is the centre. Draw OB at right angles to the chord AC. Draw BD at right angles to OB and make BD equal to the arc BC. Join OD. Find a point F in OC such that OF is two-thirds of OC. Draw FH parallel to OB to meet OD at H. Draw HG perpendicular to OB to meet OB at G. G is the centroid of the sector OABC.

To find the centroid of a figure considered as part of another figure. Frequently the addition of a simple figure to a given comparatively complicated one will make a simple figure. For example the figure ABCD (Fig. 185) only requires the addition of the triangle OCD to convert it into the sector of a circle whose centre is O. G, the centroid of the sector OAB is readily found and so is G2 the centroid of the triangle OCD.

FIG. 185.

Let G be the centroid of the figure ABCD. Then if at G. and G parallel forces P and Q be applied, the magnitudes of P and Q being proportional to the areas of OCD and ABCD, the centre of these parallel forces will be at G, the centroid of the sector OAB.

Hence, if parallel forces P and R be applied at G, and G1, but in opposite directions, the magnitudes of P and R being proportional to the areas of OCD and OAB, the centre of these parallel forces will be at G the centroid of the figure ABCD.

The area of a sector of a circle is equal to half the product of the arc and the radius.

97. Centre of Pressure and Centre of Stress.-If a plane figure be subjected to fluid pressure, the point in the plane of the figure at which the resultant of the pressure acts is called the centre of pressure. If the plane figure is a section of a bar or part of a structure which is subjected to stress, the point in the plane of the section at which the resultant of the stress acts is called the centre of

stress.

In what follows "pressure" will be taken to include "stress."

If the pressure be uniform over the figure, then the centre of pressure coincides with the centroid of the figure.

A general construction for determining the centre of pressure of any plane figure when the pressure varies uniformly in one direction is illustrated by Fig. 186. ABCD is a plane figure supposed to be vertical, and AB and CD are horizontal. AA, is the altitude of the figure, and the pressure is supposed to vary uniformly from an amount represented by AP at the level AB to an amount represented by A,Q at the level CD. AP and A,Q are horizontal.

Join QP and produce it to meet A,A at O. Draw any horizontal SRMN to cut the given figure. Draw the horizontal OF, and the verticals MM, and NN,. Through K, the middle point of MN, draw the vertical KF. Join FM, and FN, cutting MN at m and n. If this construction be repeated at a sufficient number of levels, and all points

corresponding to m be joined, also all points corresponding to n, a figure abnCDma is obtained and the centroid of this figure will be the centre of pressure of the original figure.

The proof is as follows. Suppose that the line MN is the centre line of a very narrow horizontal strip of the original figure, and let the width of this strip be denoted by w. The magnitude of the resultant pressure on this strip is equal to MN × w × RS, and it will act at K, the middle point of MN.

Since SRMN is parallel to QA,DC,

and, therefore, and

M,N,: mn :: OA, : OR,
MN: mn: A,Q: RS,
MN × RS = mn × A1Q,

MN × w × R$ = mn × w × A,Q,

that is, the resultant of the pressure on the strip of length mn when subjected to a pressure A,Q will have the same magnitude as the resultant of the pressure on the strip of length MN when subjected to a pressure RS, and it will act at the same point K which is also the middle point of mn.

It follows that the resultant of the pressure on the figure abnCDma when subjected to a uniform pressure A,Q will be the same as the resultant of the varying pressure on the original figure. But when the pressure on a plane figure is uniform, the centre of pressure is at its centroid. Therefore the centroid of the figure abnCDma is the centre of pressure of the original figure.

The figure abnCDma is called a modulus figure of the original figure.

A common and important practical case is that in which the given figure is a rectangle and the pressure on it varies uniformly in a direction parallel to a side of the rectangle.

In Fig. 187, ABCD is a rectangle on which the pressure varies uniformly from nothing at A to an amount DT at D. Applying the

construction just proved the student will have no difficulty in seeing that the modulus figure for the rectangle ABCD is the isosceles triangle OCD and that the centre of pressure for the rectangle ABCD is at C, on the vertical centre line of the rectangle and at a distance from AB equal to H where H is the height of the rectangle. Also the magnitude of R the resultant of the pressure on ABCD is equal to the area of the rectangle multiplied by the half of DT, where DT is the intensity of the pressure or the pressure per unit area at the level CD.

If the rectangle ABCD (Fig. 187) be divided into two rectangles by the line EF, then, C2, the centre of pressure of the rectangle ABFE is at a distance 3h from AB, where h is the height of the rectangle ABFE, and the magnitude of P the resultant of the pressure on ABFE is equal to the area of ABFE multiplied by the half of ES.

The centre of pressure of the rectangle CDEF will be at C, the centroid of the quadrilateral CDef and the magnitude of Q the resultant of the pressure on CDEF will be equal to the area of CDEF multiplied by half the sum of DT and ES. Q may however be determined both as regards magnitude and position by considering it as the equilibrant of the parallel forces P and R, the sense of R being opposite to that of P.

98. Masonry Dams.-A masonry dam is a wall for holding back the water at the end of a natural reservoir.

One form of dam section is shown in Fig. 188, AV being a vertical line. The principal problems connected with dams are: (1) the determination of the line of resistance when the reservoir is empty; (2) the determination of the line of resistance when the reservoir is full, and (3) the determination of the stresses at various horizontal sections of the dam.

Reservoir empty. When the reservoir is empty the stresses in the dam are those due to its weight. Referring to Fig. 188. Consider a portion of the dam lying between two vertical cross sections one foot apart, then the weight of a part of this between two horizontal section planes will be equal to the area, in square feet, of the cross section between these planes multiplied by the weight of a cubic foot of the material of the dam.

Let horizontal sections BC, EF, GH, and KL be taken. The resultant w of the weight of the top portion ADLK acts vertically through C, the centroid of ADLK. w, cuts KL at

A

D

P

Inner
Face

PA

K

W2 Outer
Face

WA

W

H

G

C2

W3

W

$3

TWA

FIG. 188.

c, which is the centre of pressure or centre of stress for the horizontal section KL.