PART IV.

PRINCIPLES OF CONSTRUCTION.

SECTION I.

THE MECHANICAL PRINCIPLES OF CONSTRUCTION. EQUILIBRIUM OF FORCES.-The following is a summary of the most important principles upon which the balance and stability of structures depend; and which is here given for the convenience of the young architect.

1. Composition and Resolution of Forces.-If two forces acting upon a point be represented in magnitude and direction by two straight lines drawn from the point, then the diagonal of the parallelogram described upon these lines will represent the resultant in magnitude and direction.

Let the forces act along lines ox, ov (fig. 18), and let them be represented by oA and оB. Complete the parallelogram OACB, then the resultant force will act in the direction of the diagonal oc and be equal to it in magnitude. These three forces are consequently in equilibrium if we suppose the resultant oc to act in an opposite direction in the same line. When we thus find a single force which is equivalent to two others the parallelogram of forces becomes a rule for the composition of forces.

Now the resultant oc may be found by trigonometry thus:

2

OC = √ OA2 + OB2 + 2 OA OB COS. AOB.

Its direction may be found by the formulæ,

Sin AOC sin AOB

ов
OC

ОА

sin BOC sin AOB

OC

When the two forces act on a point in directions which include a right angle, the case is one deserving

X

B

Fig. 18.

particular notice of the architect, as in the thrusts of two arches or frames at right angles to each other. The solution is then simply by Euc. I. 47.

where R represents resultant, P the force OA, and a the force OB. For example, if Þ be 15 lbs. and a be 8 lbs., then

R2 (15) + (8)2 = 225 +64
R2 =

therefore R = 17.

289,

The above proposition may be stated in another way, which, from its simplicity and constant use, is of great value in designing structures.

2. If three forces acting on a point be represented in magnitude and direction by the sides of a triangle taken in order, they will keep the point in equilibrium.

Let ABC be the triangle, and let P, Q and R be three forces proportional to the sides BC, CA, AB, and respectively parallel to them-P parallel to BC, Q parallel

D

B

Fig. 19.

R

to CA, and R parallel to AB. Then forces represented by BC, CA, and AB will keep the point at rest. The dotted lines show how the proposition may be proved by the parallelogram of forces, AB, AD, and CA being three forces in equilibrium.

The above proposition is called "The Triangle of Forces."

The same question may be solved by calculation. Let P, Q and R be the magnitudes of the three forces, and DA B, BAC, CAD the angles between their directions; then,

Sin BAC sin CAD sin DAB PR: Q.

Each of these forces is equal and opposite to the resultant of the other two; or if three forces as above keep a point in equilibrium, each force is proportional to the sine of the angle between the directions of the other two.

Another way of stating the same principle of great use in determining the strains upon frames, as roof trusses and the thrust of arches, is as follows:

If three forces acting on a point keep it in equilibrium, a triangle having its sides perpendicular to the directions of the forces may be drawn, the sides being respectively proportional to the forces perpendicular to them.

Note. Sometimes cases occur where it is easier to apply the latter method of finding the proportionate strains than that by parallel lines. This principle is applied in determining the weights for arch stones in equilibrium (see section on Stability).

3. If any number of forces act at a point and be represented in magnitude and direction by the sides of a polygon taken in order, they will be in equilibrium.

Let P, Q, R, S be forces acting at a point. From any point A draw AB to represent P in magnitude and direction, draw BC to represent Q, ci to represent R, and DE to represent s. Then EA will represent the

B

D

R

P

E

Fig. 20.

resultant in magnitude and direction, and these forces will be in equilibrium. This method is applicable to any number of forces, and is called "The Polygon of Forces" It is often necessary to determine the resultant of a system of forces acting at or through a point, or the thrusts of a polygonal frame of bars (see section on Stability).

Note.

It is not necessary that the forces should lie in one plane.

4. As we can compound two forces into one so we

can resolve one force into two others in given directions as ox, oy (fig. 18). For let oc represent a force. Draw a parallelogram OвCA having oc as a diagonal, then the force represented by oc is equal to two forces represented by Oв and OA respectively. Thus, if we assign the directions of the two components we can readily resolve any force into them. One case deserves special notice.

To resolve a force into two others at right angles to each other.

Let AX, AY (fig. 21) be at right angles, and let a denote the angle RAX, and R the force to be resolved, x and y the required components, then,

AX = AR cos a; AY = AR sin a; x2 + y2 = R2; (cos2 a + sin2 a) since sin2 a + cos2 a = 1.

Thus any force P may be resolved into two others, P'cos a and P sin a, which are rectangular components of the force P.

One very general rule of great importance may be stated here.

Y

=R2

a

To find the resolved part of a force Fig. 21. in any given direction, Multiply the expression for the force by the cosine of the angle between the given direction and that of the force.

Ex.—Let a weight of one ton be drawn up an incline, and let its inclination to the horizon be 4°, then the resolved part of the weight parallel to the incline will be 1 × sin 4°, or 156.25 lbs., which will be the necessary force required, neglecting friction. The resolved part of the weight, supported by the plane, will be 1 ton x cos 4° = 2234.5 lbs.

5. One very general case may be noticed. To find