The formula of solution in this case is (Art. 81)

Let y be the unknown side of the square section of each rafter; then

Wherefore the equation (1) becomes, after obvious reductions,

The value of S is 1000, and from the data we find cos a = 8944. Substituting these and the values of W, 1, and cosec a, in the preceding equation (7 being taken in inches, and Win pounds avoirdupois), we get for the determination of y the equation

The integer part of the value of y in this equation, which is all that is required in practice, is readily found by trials. For on substituting 6 for y in the left-hand member of the equation (2,) the result is 204-78, instead of 226·4; the value of y therefore is nearly equal to 6. It is easily found to be 6.2.

4. Find the dimensions of each rafter in the last example, on the supposition that the rafters are braced by a stay attached to the middle point of each.

For illustration of the elementary principles, we will solve this example without the formula of Article 83.

Let AB, BC (fig. 39), represent the two rafters braced by the stay DF, attached to the centres D and F of the rafters. Then, because the weight W is distributed over the two rafters, each of the equal portions BF, FC, BD, DA, is strained by the load W distributed over its length. Hence (Art. 35) the load † W,

produces on each of these, on FC for instance, a strain which is equivalent to a weight W suspended from the centre of FC.

Wherefore, as the component of W in a direction perpendicular to the beam is W cos a, the moment of the strain on the centre of FC is (Art. 35),

Whence the tension per unit of area produced by flexion on the cross section of the middle of FC is (Art. 22)

y being the side of the square section of the beam.

Again the vertical pressure at B, arising from the loads distributed along BF and BD, is W + } W = { W. Resolving this along BA and BC, the component P along BC is

Cosec a = W cosec a.

Moreover, the vertical pressure at F, arising from the loads on BF and FC, is W + W = W. Resolving this along FC and FD, the component Q along FC is

Wherefore the total compression along the portion FC of the rafter BC is

P+Q

=

3

W (1) cosec a = W cosec a.

8

Hence the tension per unit of area produced on the cross section of FC at its middle by the compression along the rafter BC is

Let S be the pressure on each unit of area of the section under consideration; then, by (1) and (2),

Substituting in this the values of W, 1, &c. given in the preceding solutions, we get for the determination of y, the equation

2.81y = 56.6,

from which, by trials, y is easily found to be 4 inches nearly.

Hence, when the rafters are not supported by a stay, the side of the square section of each for safety must be 6 inches; but when braced, a square section whose side is 4 inches will be sufficient.

5. A roof formed by two rafters AB, BC, and a tie-beam AC, supports a load of 1 ton, the inclination of each rafter to the horizon being 28°; find the thrust along each rafter, the load being distributed along the rafters.

Ans. 1192.83 lbs.

6. In Example 5, determine the horizontal thrust on the tiebeam when an additional weight of 100 lbs. is placed on the top of the roof. Ans. 1147.24 lbs.

7. In Example 5, find the depth of each rectangular rafter in order that the load may be supported with safety, the breadth of each rafter being 8 inches, the length 16 feet, and the coefficient of safety 1,200. Ans. 4 inches nearly.

8. In the last example, find the depth of each rafter when the rafters are braced by a stay attached to their middle points. Ans. 2.01 inches.

9. Determine, in Example 3, the depth of the rectangular tie-beam, the uniformly distributed load upon it, including its own weight, being 180 lbs. per foot in length, and its breadth 5 inches, the coefficient of safety and other data the same as in Examples 1 and 3.

Obs. It will be a sufficiently close approximation to the true result to find the dimensions from the transverse strain alone. Ans. 17 inches.

Section 3.

THE TRUSSED ROOF.

85. In large buildings the trussed roof is always adopted. It consists of a combination of timbers, of iron, or of both, so arranged as to constitute an unyielding frame. It is named a truss, because the roof is trussed or tied together.

Let AC, CB, be two of the rafters of a roof, and AB the tie-beam (fig. 40), as in articles 79, 83. The feet of the rafters being joined together by the tie-beam, and resting on walls or supports, are in

capable of yielding in the directions of their length. The vertex C, therefore, where they meet, is a fixed point to which the centre D of the tie-beam is trussed or tied up by a beam DC, called the king-post.

The rafters of the truss are called the principal rafters. The principal rafters and consequently the trusses, are at about 10 or 12 feet apart; and between them are the common rafters, resting on horizontal beams called purlins. In order to prevent the deflection of the principal rafters, beams DE, DF, called struts, are fixed to the king-post at D, and inserted between the kingpost and the principal rafters. The struts being placed under the purlins at E and F, relieve the principal rafters CA, BC from all cross strain, except that which arises from their own weight between the points of support.

This is the simplest kind of truss.

86. There are other kinds of truss suited to different purposes; but the conditions are the same in all, viz., the establishing of fixed points to which the tie-beam is trussed. Thus in fig. 41 two points are substituted for a single one, and two suspending

rods are required. These are called queen-posts, and the truss in this case is called a queenpost truss.

Fig. 41.

THE STRAINS ON THE DIFFERENT PARTS OF A TRUSSED ROOF.

THE DISTRIBUTION OF THE LOAD.

87. We have remarked, in the introduction to the Resistance of Materials, that the reaction of three supports in a straight line are in general indeterminate. Hence, in the case of a trussed roof ACB (fig. 40), there are two hypotheses with respect to the pressures on A, E, and C, arising from the weight distributed on AC, and the corresponding pressures on B, F, and C. We will explain these hypotheses, designating one by "the method (A)," and the other by "the method (B)." The results obtained by both methods are nearly the same, so, for practical purposes, it is immaterial whether we adopt the one method or the other.

THE METHOD (A).

88. Let ACB (fig. 40) be a trussed roof; DE, DF, two struts parallel respectively to the rafters BC, CA, and D, the centre of the tie-beam AB.

Let W be the weight of the roof supported by the truss ACB; then W, which is distributed over the rafter AC, is supported at the points A, E, and C; and W, which is distributed over the other rafter, is supported at the points B, F, and C. We must keep in mind that the weight W, which is supported on each side, is transmitted to each rafter by the purlins at three different points, viz., at the vertex by the top purlin, at the strut by the middle purlin, and at the tie-beam by the bottom purlin.

Now, according to the first hypothesis, the vertical pressures