This section is from the book "Building Construction", by R. Scott Burn. Also available from Amazon: Building Construction.

Fig. 458.

The resistance of a horizontal beam to cross strain increases, as we have said, as the shortness of the distance between the supports or bearings, as a b, fig. 458; the horizontal position being the weakest in which a beam can be placed, the vertical the strongest. The point at which the pressure or cross strain is applied to a beam, influences also its power to resist strain; and it follows from what we have said that the nearer this pressure is applied to any one of its bearings the greater will its power to resist strains be. Hence, as already stated, a beam will support twice the weight if that weight be uniformly distributed over its surface, which it would do if applied at its centre. Hence, also, the varying pressure as the distance from the bearings at which the pressure is applied is increased or diminished. At the point c, fig. 458, as we have shown, the pressure is just twice the amount which it would be if the same weight was distributed over the whole surface of the beam, as illustrated by the dotted line. The proportion of pressure to that borne at the centre c, which a given weight exercises at any other point may be ascertained thus: let the distance between the bearings a and b be 30 feet, and the pressure applied at the point d, say 10 feet from the point a, and 20 feet from b; then multiply these two together, which gives 200; next square the distance a c, or c b, which is 15 x 15 = 225; so that the proportion of pressure which the same weight exercises at the two points d and c is as 200 to 225. If the pressure was applied at e, 5 feet from the support or bearing b, then e b x e a or 5 x 25 = 125; and a c2 or c b2 = 225, gives the proportion which the pressure at the point e exercises, namely 125, as compared with that at c, namely 225. To find the strain to which the beam is subjected at the points of support a b, by any given weight pressing at a point as d: let the weight be 450 lbs., and the distance a d, as above stated, 10 feet; then multiply the distance bd = 20 by the weight 450, and divide by the span or distance between a b as 30, the quotient will be the pressure at the point b. To find that at a, multiply the distance ad =10 by the weight 450, and divide by 30 (the span a b) the quotient is the pressure at the point a. The pressure sustained at other points, as e and f while that at e and its distance from a is known, may be thus ascertained. Let the distance e b be 5, and of 7 feet. To find the pressure at the pointy, multiply weight at e (450) by the distance ad =10, then multiply the product by the distance b e = 5, and divide by the distance or span ab = 30. To find the pressure at the point e, multiply 450 (the weight at d) by b e = 5, and the product by distance a f= 7, and divide by span a b = 30; the quotient will be the pressure at the point f = 10. As the pressure thus decreases from the centre c, the reason will now be seen for the statement previously made, namely, that the depth of a beam may decrease from the centre towards the ends. We have said that a beam is the weakest when laid horizontally, as a b at fig. 457, and strongest vertically, as c d; its strength, when in an inclined position, will evidently be in proportion to its angle of inclination from the vertical line c d. The proportion may be ascertained geometrically, as in fig. 457, where the distances a b c d are laid down from a scale of equal parts. From c as a centre, with c a or c b as radius, describe the semi-circle a d b. From the points where the inclined beams as c e, c f, or c g cut this, drop perpendiculars cutting c b in points, as at h, i, and j; then the strength of c f will diminish in the proportion as c i is greater than c h, or cj greater than c i. By adding two inclined beams, as shown by the dotted lines a d, b d, to the vertical beam d c, to the horizontal one a b, we obtain what is called technically a " truss," and an arrangement which is the strongest possible one we can get. Theoretically the truss is perfect without the vertical piece, when the pressure is vertical, as by the arrow k; in this case the pressure is transmitted along the inclined beams d a, d b, in the direction of the arrows l m, and is a pressure or strain of compression; this has a tendency to bulge or force the walls a b out in the direction of the arrows n o; but this is overcome by the beam a b, technically called a "tie beam," the strain or pressure upon which is that of tension acting in the direction of the arrows p q tending to pull its fibres asunder. But in the case of a roof, in which the inclined beams (or rafters), as d a, d b, are loaded along their whole surface, the pressure is changed from the purely vertical one as at k, and has therefore to be provided for. To do so, the vertical piece d c is added, the strain upon which is that of tension in the direction of the arrow r. The junction of the foot of this member (technically called a " king post," or if of wrought-iron, which may be substituted for timber, as the strain is tensional, a "king bolt" or "king rod") affords a butting place from which other members spring to relieve the pressure upon the inclined pieces (rafters) a d, ab. . This is illustrated by the inclined piece c l (which is called a "strut" or " brace"), and the strain upon which is that of "compression" in the direction of the arrow s.

58. Having thus briefly stated the strains and pressures to which beams placed vertically and horizontally are subjected, we now turn our attention to those which affect beams inclined to the horizontal, these forming, as the student now knows, the most important members of trussed framework. Two points require to be ascertained in determining the strains to which the parts of trussed framework are subjected; first, the direction in which the pressure is communicated to the part; and, second, the amount or value of the pressure. These can be ascertained by calculation; but the simplest, and that perhaps most generally employed, is by geometrical construction, this being based upon the well-known problem, the "parallelogram of forces," by means of which we can find two pressures or forces, which, acting in two different directions, can counterbalance or be equal to one pressure or force acting in a certain direction, this being called the uresolution of forces or pressures;" or we can, on the other hand, or converse of the above, find the value and direction of one pressure or force, which will counterbalance or be equal to two forces or pressures acting in two directions, this latter process being called the "composition of forces;" thus, by means of construction, we can find the strains to which different parts of trussed framework are subjected. The "parallelogram of forces" may be illustrated by the following diagram : let a, fig. 459, be supposed to be a ball discharged from a cannon in the direction b c, as indicated by the arrow; and d another ball, propelled in the direction b e; both would travel if acted upon singly, in these directions. But suppose a single ball, as b, to be acted upon by two forces, one tending to send it forward in the direction b c, the other in the direction b e, the ball would follow neither of these directions, but would take another and different course, and would go off in the direction of the diagonal, as at g, towards f. This diagonal evidently represents the amount or result of the two forces, b c and b e, as it is made up of these, and is therefore called the " resultant" of these two compound forces, or "component forces" b c and b e. Should the force or pressure sending the ball b, fig. 459, in the direction b c, be equal to that sending it in the direction b e, the diagonal would be b h, or of a square, as b c h i; but if the force sending the ball b, in the direction b e, was twice that sending it in the direction b c, the diagonal would be that of a rectangle, the length of which would be equal to twice the breadth; as, for example, b f is the diagonal of the rectangle chef. The finding of the amount of the pressure of the diagonal ball g, which will balance, so to say, the two pressures a and d, i.e., one pressure acting in one direction, as b f, to be equal to two acting in two different directions, b c, b e, is called the "composition of forces," or pressures; while the converse of this finding, the amount of two pressures, as b c, b e, which shall be equal to one pressure, b f, is called the "resolution of forces." These, as we have said, can be found geometrically. Thus, by taking the measurement, or setting off, in the first instance, in the construction of the problem, the distance b c from any scale of equal parts (see Chapter I. - Drawing), to represent the force of b in any given weight, as lbs. or tons, and in like manner b e, from the same scale, and finishing the parallelogram, of which these will be the two sides, and drawing the diagonal b f by measuring this, the amount of the pressure or force of g will be known. Thus b c, from a certain scale of equal parts, we find to be equal to 8½ - say 8½ cwts. - be 14¼ , and the diagonal 19¾, which is the component of the two, b c, b e, and balances these, so to say.

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