A rafter, like a post, is subject to a compressive force, and is liable to fail in three ways, namely: by flexure, by being crushed, or by crushing the material against which it presses. To render it entirely safe, therefore, it is requisite to ascertain the requirements for resisting failure in each of these three ways.

Of these it will be convenient to consider, first, that of the liability to being crushed. The rule for this is found in Art. 107. Let the rafter be of Georgia pine, then the value of C, Table I., will be 9500. The strain in the rafter (Art. 225) is 31,200 pounds. Now, taking the value of a, the factor of safety, at 10, we have, by Rule VI. (Art. 107.) -

A = 31200 x 10/9500 = 32.737;

or, 33 inches area of cross-section. This is the size of the rafter at its smallest section; for example, at any one of the joints where it is customary to reduce the area by cutting for the struts and rods.

Again: Let the liability of the rafter to flexure be now considered. For this we have a rule in Art. 114. The length of the rafter between unsupported points is nearly 9 2/8 feet, or 9 2/3 x 12 = 116 inches. Let the thickness of the rafter be taken at 6 inches. Then, by Rule XI. (Art. 114), we have -

227 The Rafter 114

r = l/t = 116/6 = 19 1/8

227 The Rafter 115

Then, 3/2 x .00109 x 373.8 = 0.611127 adding unity = 1.

1.611127

Substituting this, we have -

b = 31200x10x1.611127 = 502671.624 = 8.819;

9500x6 57000

or, to resist flexure the breadth is required to be 8.82, or, say, 9 inches; or, the rafter is to be 6x9 inches at the foot. The strain in the rafter at the upper end is only half that at the foot; the area of cross-section, therefore, at the head need not be more than half that which is required at the foot; but it is usual to make it there about 2/3 of the size at the foot. In this case it would be, therefore, 6x6 inches at the upper end.

Lastly, the requirement to resist crushing the surfaces against which the rafter presses is to be considered.

The fibres of timber yield much more readily when pressed together by a force acting at right angles to the direction of their length than when it acts in a line with their length.

The value of timber subjected to pressure in these two ways is shown in Arts. 94, 98. In Table I., the value per square inch of the first stated resistance is expressed by P, and the ultimate resistance of the other by C/a. The value of timber per square inch to safely resist crushing may be epressed by C/a, in which a is the factor of safety. Timber pressed in an oblique direction will resist a force exceeding that expressed by P, and less than that expressed by C/a.

When the angle of inclination at which the force acts is just 450, then the force will be an average between P and C/a.

And for any angle of inclination, the force will vary inversely as the angle; approaching P as the angle is enlarged, but approaching C/a as the angle is diminished. It will be equal to C/a when the angle becomes zero, and equal P

when the angle becomes 900. The resistance of timber per square inch to an oblique force is therefore expressed by -

M= P+Ao/ 90(C/a -P); (102.)

where Ao equals the complement of the angle of inclination. In a roof, A° is the acute angle formed by the rafter with a vertical line. If no convenient instrument be at hand to measure the angle, describe an arc upon the plan of the truss - thus: with C B (Fig. 84) for radius, describe the arc B g, and get the length of this arc in feet by stepping it off with a pair of dividers. Then -

Ao/90 = 0.63 2/3k/h;

where k equals the length of the arc, and h equals B C, the height of the roof. Therefore -

M = P+0.6 2/3 k/h(C/a-P) (103.)

equals the value of timber per square inch in a tie-beam, C and P being obtained from Table I., Art. 94. When C for the kind of wood in the tie-beam exceeds C set opposite the kind of wood in the rafter, then the latter is to be used in the rules instead of the former.

The value of M, equation (103.),, is the resistance per square inch of the surface pressed at the foot of the rafter. The resistance of the entire surface will therefore be MA, where A equals the area of the joint. Then, when the resistance equals the strain, we will have -

MA=S = A[p+0.6 2/3 k/h(C/a-p)]; from which we have -

A = S/P+0.63 2/3 k/h(C/a-P); (104.)

in which S is the strain to be resisted.

Now, the end of the rafter must be of sufficient size to afford a joint the area of which will not be less than that expressed by A in equation (104.).

For example, the strain to which the rafter, Fig. 84, is subject at its foot is ascertained to be (Art. 225) 31,200 pounds. For Georgia pine, the material of the tie-beam, P = 900 (Art. 94, Table I.), and C = 9500.

The length of the arc Bg is about 14.4 feet; the height B C is 13 feet. Let a, the factor of safety, be taken at 10, then we have (104.) -

A = 31200

900 + (0.63 2/3x114.4/13) (9500/10 - 900);

A = 31200 = 33.36;

900 + (0.705x50)

or, the superficial area of the bearing at the joint required to prevent crushing the tie-beam is 33 1/3 inches.

The results of the computations show that the rafter is required to be 6 inches thick, 9 inches wide at the foot, and 6 inches wide at the top. It is also ascertained that, in cutting for the bearing for the struts and boring for the suspension-rods, it is required that there shall be at least 33 inches area of cross-section left intact; and, farther, that the area of the surface of the joint against the tie-beam should not be less than 33 1/2 inches.