The application of these principles to the theory and practice of framework, will now be illustrated and described. The pressure or strain exerted by any given weight, acting upon or being hung from the apex of two inclined beams, as a b, a c, fig. 460, increases as the inclination. To find what this pressure is, proceed as follows : Find the centre lines of the beams, and from the point a of their intersection at the apex drop a perpendicular a b. Let the amount of the pressure of the ball a be supposed to be 8 cwt. Then, from any scale of equal parts, make a b equal to eight of these parts. From b draw b c parallel to the opposite beam a d, and b d parallel to a c. Measure off the distance a c on the compasses, and apply it to the same scale of equal parts as that from which the distance a b was taken, and it will be found to be 5 ½ and as the beams are equally inclined, the pressure on a d will be same as on a c; so that by doubling 5 ½ we obtain the amount of united pressure which the two beams sustain, namely, 11, 8 being that at the point a. To simplify the constructions, and to prevent errors in the measurements, it is better to construct all problems with simple lines, as in fig. 459. The various points of intersection will thus be more clearly seen. The united pressure of two beams, varying in their inclination, is illustrated by fig. 461, in which the inclination is less than in fig. 460; by proceeding as in fig. 460, making a b equal 8, the distance a c will be 4 ¾, which, doubled, will give the united pressure, 9 ½, in place of 11, as in fig. 460. Where two beams are placed at different inclinations - exemplified in the weaving-shed roof illustrated in the Chapter upon Roofs - the pressure on each beam can be found by the same process as now described. Thus, let a b, a c, fig. 462, be the two beams; and the pressure exerted at the apex a, by the ball, or by that suspended from it, as in the diagram, be equal to 10 cwt. From the point a of intersection of the central lines of the two beams drop the perpendicular a d, and make a d equal to 10, from any scale of equal parts. From the point d draw, parallel to ab, a line cutting a c in the point e. From d, parallel to a c, draw d f. Measure a e, it will equal 7 ¼; measure a f, it will equal 5½; the united pressure of the two beams will thus be 12 ¾, the pressure at point a being 10.
The student should project a series of diagrams with different degrees of inclination given to the beams, and marking in the figures, as in fig. 462. If we have two beams acting in two different directions, as the beams a b,a c, fig. 463, and if we know the force or pressure acting on these, we can find the direction in which a third beam should be placed to meet these two, and also ascertain the pressure which this third piece sustains, by proceeding thus : Let a b, b c, fig. 463, be the two inclined beams, the pressure a b 6, and on c b 11 cwt., and the direction in which these pressures act shown by the arrows; continue these directions, indefinitely, to e and d, and make e equal to 6 and b d equal to 11, from any scale of equal parts; from e, parallel to d b, draw a line, and from d another parallel to b e, intersecting in the point f, thus completing the parallelogram b e f d. Draw the diagonal b f. The line b f will be the direction in which a third piece should be placed to balance or equal the two pieces a b, c b; and the amount of pressure which that piece, as f g, has to sustain will be found by measuring the diagonal f b from the scale of equal parts, and which will be 14 ½. This is a problem in what we have shown to be called the " composition " of forces, in which a third force is found to counterbalance or resist two other forces acting in different directions. As stated in a preceding paragraph, when treating of Roofs generally, the tendency of two inclined beams a b,a c, fig. 464, acted upon by pressures or forces, represented by the balls and arrows, is to force the parts or walls f g upon which the feet of the beams rest apart, in the direction of the arrows d e. The amount of the thrust or force thus exercised may be ascertained thus: From the point a drop' a perpendicular, as a h, and make a h equal to the weight a, say 10 cwt.; from h, parallel to the two beams a b, a c, draw lines cutting the beams in the points i and j; join i and j by a line cutting a h. The distance k i taken from the scale of equal parts will give the pressure tending to force out the wall c f, and distance k j that to force out the wall b g. Where the inclination of the beam is unequal, as in fig. 462, the pressures tending to force the walls b and c outwards will also be unequal. To find these construct the diagram, as already there described, and then from the point / draw at right angles to a d the line f h, cutting a d in h, and from the point & the line eg. Measure g e from the scale of equal parts, this will give the pressure tending to force out the wall c = 4; the distance h f = 3 ¾ will be the pressure on the wall 6.
The pressure of beams upon walls, as in fig. 464, is made up of two kinds, one horizontal, the other vertical. To make the pressure of the beams on the walls as vertical as possible - which is the direction in which they are best calculated to sustain a pressure - the member of a truss, known as a tie beam, connecting the feet of the two beams, as the line b c, fig. 464, is used. The angle at which there is the least oblique pressure exerted by two inclined beams uniformly-loaded on their surface is found to be 35.16. The pressure tending to tear asunder or rend the tie beam c b, fig. 462, will be equal to f h + g e, or 3 ¾ + 4 = 7 ¾. The amount of horizontal thrust or pressure upon the wall c will, as already stated, be equal to the distance g e; upon wall b, distance f h. The distance f a will be the measure (5 ½ cwt.) of the pressure upon the point b, the distance a e that on the point c. The direction of the pressure on these points being a b, a c. But in the case of roofs, the pressure is not exerted when on the apex of the inclined beam, as in figs. 462 and 464; but the roof covering being spread over the surface of the beams, from top a to bottom b, fig. 464, the direction of the pressure on the beams is out of the lines of their length. To find the direction proceed thus : Let a b, fig. 465, be one of the beams, and c its centre of gravity; from a drop a perpendicular a d, and parallel to this, through c, draw a line c e f, and from a, a line at right angles to a d, cutting e f in the point e, join e b; e b is the direction in which the thrust or pressure acts upon the beam a b. To ascertain the amount of pressure exerted on the wall; from e set off from a scale of equal parts a distance e g equal to the weight sustained by the beams, and from g, parallel to a e, draw the line g h; the distance g h gives the pressure. To ascertain the amount of horizontal and vertical thrusts or pressures at the foot of a beam, as at the point i, produce the line a i to the point j, and make i j equal to the weight placed on the beam, as, say, 5. From j, parallel to d a and e a, draw lines j l, j k, and from i draw i k; the distance i l gives the amount of horizontal, the distance i k the amount of vertical pressure, at the foot of the beam at i.