An Encyclopædia of Architecture

of the weight of the body; but it may be easily imagined that the first thread, by being in the direction AD, will, besides the weight itself, have to sustain the effort of the power that draws it out of the vertical AB.

1245. If the direction of the horizontal force be prolonged till it meets the vertical, which would be in the first thread if it were not drawn away by the second, we shall have triangle ADB, whose sides will express the proportion of the weight to the forces of the two threads in the case of equilibrium being established; that is, supposing AB to express the weight, AD will express the effort of the thread attached to the point A, and BD that of the horizontal power which pulls the body away from the vertical AB.

1246. These different forces may also be found by transferring to the vertical DH (fig.520.) any length of line DF to represent the weight of the body. If from the point F the parallels FI, FG be drawn in the direction of the threads, their forces will be indicated by the lines ID, DG, so that the three sides of the triangle DGF, similar to the triangle ADB, will express the proportion of the weight to the two forces applied to the threads.

1247. Suppose the weight to be 30 lbs.: if from a scale of equal parts we set up 30 of those parts from D to F (fig. 519.), we shall find DG equal to 21, or the pounds of force of the horizontal line DE, and 35 for the oblique power ID.

1248. If the weight, instead of 30 lbs., were 100, we should find the value of the forces DG and ID by the proportions of 30: 21::100: x, where x expresses the force DG. The value resulting from this proportion is x = 30 = 70. The second proportion

21 x 100

35 x 100
30

=116-666.

30:35::100 y,where y represents the effort ID, whose value will be y = 1249. If the angle ADH formed by AD and DH be known, the same results may be obtained by taking DF for the radius, in which case IF=DG becomes the tangent, in this instance, of 35 degrees, and ID the secant; whence

DF: DI: IF:: radius: tang. 35 sec. 35.

If ID be taken for the radius, we have

ID: IF: FD:: radius: tang. 35 sin. 35.

1250. We have here to observe, that in conducting the operation above mentioned a figure DIFG has been formed, which is called the parallelogram of the forces, because the diagonal DF always expresses a compounded force, which will place in equilibrio the two others FI, FG, represented by the two contiguous sides IF, FG.

Fig. 522.

1251. Instead of two forces which draw, we may suppose two others which act by pushing from E to D (fig. 522.) and from A to D. If we take the vertical DF to express the weight, and we draw as before the parallels FG and FI in the E direction of the forces, the sides GD and DI of the parallelogram DGFI (fig. 519.) will express the forces with which the powers act relatively to DF to support the body: thus FI= GD the weight and two powers which support it will, in case of equilibrium, be represented by the three sides of a rectangular triangle DFI; so that if the weight be designated by H, the power which pushes from G to D by E, and that which acts from I to D by P, we shall have the proportion H: E: P:: DF: FI: ID, wherein, if we take DF for radius, it will be as radius is to the tangent of the angle FDI and to its secant. As a body in suspension is drawn away from the vertical line in which it hangs by a power higher than the body (fig. 520.), it follows that the oblique forces AB and BC each support, independent of any lateral efforts, a part of the weight of the body. In order to find the proportion of these parts to the total weight, take any distance BD on a vertical raised from the centre of the body B to express the weight, and complete the parallelogram DEBF, whose sides EB, BF will express the oblique forces of the powers A and C. These lines, being considered as the diagonals of the rectangular parallelograms LEIB, BHFM, may each be resolved into two forces, whereof one of them, vertical, sustains the body, and the other, horizontal, draws it away from the verticals AO, CQ. Hence IB will express the vertical force, or that part of the weight sustained by the power EB, and HB that sustained by the other power BF: as these two forces act in the same direction, when added together their sum will represent the weight DB. In short, IB being equal to HD, it follows that BH + BIBI + ID.

1252. As to the horizontal forces indicated by the lines LB and BM, as they are equal and opposite they destroy one another.

1253. It follows, from what has been said, that all oblique forces may be resolved into two others, one of which shall be vertical and the other horizontal, by taking their direction for the diagonal of a rectangular parallelogram.

1254. In respect of their ratio and value, those may be easily found by means of a scale if the diagram be drawn with accuracy; or by trigonometry, if we know the angles

ABD, DBC, which AB and BC form with the vertical BD, by taking successively for radius the diagonals BD, BE, and BF.

1255. In the accompanying diagram, the weight, instead of being suspended by strings acting by tension, is sustained by forces which are supposed to act by pushing. But as this arrangement makes no alteration in the system of forces, we may apply to this figure all that has been said with respect to the preceding one. The only difference is, that the parallelogram of the forces is below the weight instead of being above it. Thus ID+ IB=BD expresses the sum of the vertical forces which support the weight, and MB and BL the horizontal forces which counteract each other.

Fig. 523.

1256. In the two preceding figures the direction of the forces which act by tension or compression in supporting the weight form an acute angle. In those represented in fig. 521. and the figure at the side (524.), these directions make an obtuse angle; whence it follows that in fig. 521. the force C which draws the weight out of the vertical AG, instead of tending to support the weight B, increases its effect by its tendency to act in the same direction. In order to ascertain the amount of this effect upon BD in figs. 521. and 524., which represents the vertical action of the weight, describe the parallelogram BADF, for the purpose of determining the oblique forces BA, BF, and then take these sides for the diagonals of the two rectangles LAIB, BHFM, whose sides BI, BH will express the vertical forces, and LB, BM the horizontal

ones.

: B

Fig. 524.

1257. It must be observed that in fig. 521. the force AB acting upwards renders its vertical effect greater than the weight of a quantity ID, which serves as a compensation to the part BH, that the other force BF adds to the weight by drawing downwards. Similarly, the vertical effect of the force BE ( fig. 524.) exceeds the expression BD of the weight by a quantity DI, to counterpoise the effect BH of the other power BF, which acts downwards; so that in both cases we have BD only for the vertical effect of the weight. As to the horizontal effects LB and BM, they being equal and in opposite directions in both figures, they counteract each other.

1258. For the same reason that a force can be resolved into two others, those two others may be resolved into one, by making that one the diagonal of a parallelogram whose forces are represented by two contiguous sides. It is clear, then, that whatever x the number of forces which affect any point, they may be reduced into a single one. It is only necessary to discover the results of the forces two by two and to combine these results similarly two by two, till we come to the principal ones, which may be ultimately reduced to one, as we have seen above. By such a process we shall find that PY (fig. 525.) is the result of the forces PA, PB, PC, PD, which affect the point P.

Fig. 525.

1259. This method of resolving forces is often of great utility in the science of building, for the purpose of providing a force to resist several others acting in different directions but meeting in one point.

OF THE PROPERTIES OF THE LEVER.

1260. The lever is an inflexible rod, bar, or beam serving to raise weights, whilst it is supported at a point by a fulcrum, or prop, which is the centre of motion. To render the demonstrations relative to it easier and simpler, it is supposed to be void of gravity or weight. The different positions in which the power applied to it, and the weight to be affected, may be applied in respect of the fulcrum, have given rise to the distinction of three sorts of levers.

I. That represented in fig. 526., in which the fulcrum O is between the power applied P and the weight Q.

II. That represented in fig. 527., in which the weight Q is placed between the fulcrum

O and the power P, wherein it is to be remarked that the weight and the power act in contrary directions.

Fig. 526.

Fig. 527.

Fig. 528.

III. That represented in fig. 528., wherein the power P is placed between the weight and the fulcrum, in which case the power and the weight act in contrary directions.

1261. In considering the fulcrum of these three sorts of levers as a third species of power introduced for creating an equilibrium between the others, we must notice, 1st, That in which the directions of the weight and of the powers concur in the point R (fig. 529.). That in which they are parallel.

2d,

1262. In the first case, if from the point R (figs. 529. and 530.) we draw parallel to these directions

Om Rn, the ratio of these three forces, that is, the power, the weight, and the fulcrum, will be as the three sides of the triangle Om R, or its equal On R; thus we shall have P

Q: RmR: Rn: OR; and as the sides of a triangle are as the sines of their opposite angles, by taking OR as the radius we shall have

Fig. 529.

P: Q::sin. ORn ; sin. ORm.

Fig. 530.

And if from the point O two perpendiculars be let fall, OdOf, on the directions RQ, RP, Sin. ORn sin. ORm:: Od: Of;

from which two proportions we obtain

P: Q::Od: Of; whence P x Of=Q × Od.

This last expression gives equal products, which are called the momenta, moments, or quantities of motion of the force in respect of the fulcrum O. This property is the same for the straight as for the angular levers (figs. 529. and 530.). As this proportion exists, however large the angles mRO and ORn of the directions RQ, RP in respect to RO, it follows that when it becomes nothing, these directions become parallel without the proportion being changed; whence is derived the following general theorem, found in all works on mechanics: - If two forces applied to a straight or angular lever are in equilibrio, they are in an inverse ratio to the perpendiculars let fall from the fulcrum on their lines of direction: or in other words, In order that two forces applied to a straight or angular lever may be in equilibrio, their momenta in respect of the fulcrum must be equal.

1263. Since, in order to place the lever in equilibrio, it is sufficient to obtain equal momenta, it follows that if we could go on increasing or diminishing the force, we might place it at any distance we please from the fulcrum, or load it without destroying the equilibrium. This results from the formula Px Of=Q × Od,

one is straight and the other angular, and that the weight Q is 100 pounds, the arm DE of the lever 6 feet; its momentum will be 600. Then if we wish to ascertain at what distance Of a weight of 60 pounds must be placed so that it may be in equilibrio with the first, we shall have

1265. Reciprocally, to find the effect of a force P placed at the point C of the other arm of the lever at a known distance from the fulcrum, and marked Of, in order to counter

poise Q placed at the distance Of, we have the formula P=2x0d; and if we apply this formula to the numbers taken in the preceding example, the question will be, to find a force which placed at the distance of 10 feet from the fulcrum may be in equilibrio with a weight of 100 pounds at the end of the arm of a lever of 6 feet. We must in using the formula divide 600 by 10, and the quotient 60 will indicate the effect with which the force ought to act. If, instead of placing it in C, it is at B, 12 feet from the fulcrum, the force would be 600, which gives 50; and lastly, if we have to place it at a point 15 feet from the fulcrum, the force would be 600 = 40. Thus, in changing the situation of the force to a point more or less distant from the fulcrum, we must divide the momentum of the weight which is to be supported by the distance from the fulcrum taken perpendicularly to its direction.

OF THE CENTRE OF GRAVITY.

1266. The centre of gravity of a body is a certain point within it on which the body, if freely suspended, will rest in any position; whilst in other positions it will descend to the lowest place to which it can get. Not only do whole bodies tend by their weight to assume a vertical direction, but also all the parts whereof they consist; so that if we suspend any body, whatever be its form, by means of a string, it will assume such a position that the thread produced to the internal part of the body will form an axis round which all the parts will remain in equilibrium. Every time that the point of suspension of a body is changed, the direction of the thread produced exhibits a new axis of equilibrium. But it is to be remarked, that all these axes intersect each other in the same point situate in the centre of the mass of the body, supposing it composed of homogeneous parts but sometimes out of the mass of the body, as in the case of bodies much curved, this point is the centre of gravity. 1267. It is therefore easy to perceive that for a body to be in a state of rest its centre of gravity must be supported by a vertical force equal to the resultant of all the forces that affect it, but acting in a contrary direction. So in figs. 520. and 523., the weight supported by the forces AB and BC which draw or push, will be equally supported by a vertical force represented by the diagonal DB of the parallelogram which expresses the resultant of the forces.

There

1268. An acquaintance with the method of finding centres of gravity is indispensable in estimating the resistances, strains, and degree of stability of any part of an edifice. arise cases in which we may cast aside all consideration of the form of a body, especially too when it acts by weight, and suppose the whole figure collected in the centre of gravity. We may also, for the sake of simplifying operations, substitute a force for a weight.

OF THE CENTRE OF GRAVITY OF LINES.

1269. A straight line may be conceived to be composed of an infinite number of points, equally heavy, ranged in the same direction. After this definition, it is evident that if it be suspended by the middle, the two parts, being composed of the same number of equal points placed at equal distances from the point of suspension, will be necessarily in equilibrium; whence it follows that the centre of gravity of a right line is in the middle of its length.

1270. The points in a curve line not being in the same direction, the centre of its volume cannot be the same as its centre of gravity; that is to say, that a curve suspended by the middle cannot be supported in equilibrio but in two opposite situations; one when the branches of the curve are downwards, and the other when they are upwards, so that the curve may be in a vertical plane.

1271. If the curve is the arc of a circle ADB (fig. 533.), it is easy to see that from the uniformity of its curvature, its centre of gravity will be found in the right line DC drawn from the centre C to the middle D; moreover, if we draw the chord AB, the centre of gravity will be found between the points D and E.

Fig. 533.

1272. Let us suppose that through all the points of the line DE parallels to the chord AB be drawn, terminated on each side by the curve; and let us imagine that each of these

lines at its extremities bears corresponding points of the curve; then the line DE will be loaded with all these weights; and as the portions of the curve which answer to each parallel AB go on increasing as they approach D, the centre of gravity G will be nearer the point D than to the point E.

1273. To determine the position of this point upon the radius CD which divides the arc into two equal parts, we must use the following proportion: the length of the arc ABD is to the chord AB, as the radius CD is to the fourth term r, whose value is ABXCD

ABD

That is, in order to obtain upon the radius DC the distance CG of the centre of gravity from the centre of the arc of the circle, the chord AB must be multiplied by the radius CD and divided by the length of the arc ABD.

1274. When the circumference of the circle is entire, the axes of equilibrium being diameters, it is manifest that their intersection gives the centre of the curve as the centre of gravity. It is the same with all entire and symmetrical curves which have a centre, and with all combinations of right lines which form regular and symmetrical polygons.

OF THE CENTRE OF GRAVITY OF SURFACES.

1275. In order that a centre of gravity may be assigned to a surface, we must, as in the case of lines, imagine them to be material, that is, consisting of solid, homogeneous, and heavy particles.

1277. In order to find the centre of gravity of any triangle, bisect each of the sides, and from the points of bisection draw lines to the opposite angles; the point of intersection with each other of these lines will be the centre of gravity sought; for in the supposition that the surface of the triangle is composed of lines parallel to its sides, the lines AE, BF, and CD (fig. 537.) will be the axes of equilibrium, whose intersection at G gives

the centre of gravity.

We shall moreover find that this point is at one third of the distance from the base of each of the axes; so that, in fact, it is only necessary to draw a line from the point of bisection of one of the sides to the opposite angle, and to divide it into three equal parts, whereof that nearest the base determines the centre of gravity of the triangle.

Fig. 535.

Fig. 536.

h If g

Fig. 538.

1278. To find the centre of gravity of any irregular rectilinear surface, such as the pentagon, fig. 538., let it be divided into the three triangles, AED, ABC, ADC (fig. 538.), and by the preceding rule determine their centres of gravity F, G, H. Then draw the two lines NO, OP, which form a right angle surrounding the polygon. Multiply the area of each triangle by the distance of its centre of gravity on the line ON, indicated by Ff, Gg, Hh, and divide the sum of these products by the entire area of the pentagon, and this will give a mean distance through which an indefinite line IK parallel to ON is to be drawn. Conducting a similar operation in respect of the line OP, we obtain a new mean distance for drawing another line LO parallel to OP, which will intersect the first in the point M, the centre of gravity of the pentagon.

The centre of gravity of a sector of a circle AEBC (fig. 539.) must be upon the radius CE which divides the arc into two equal parts. To determine from the centre C, at

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An Encyclopædia of Architecture: Historical, Theoretical, and Practical