commence with the primary truss or with the secondary trusses; but by commencing with the primary truss, the process is rendered more simple.

(1.) Primary Truss 1 2 3. Let W denote the weight of the roof, then W is distributed over each rafter, the resultants acting through the middle points of the rafters. Divide each of those resultants into two equal and parallel components, each equal to W, acting through the ends of the rafter; then W is to be considered as directly supported at 3, 4 W at 2, and 1 W + W W at 1; therefore the load at the joint 1 is

=

Let i be the inclination of the rafters to the horizon; then by the equations of Article 149

This is the pull upon the horizontal tie-rod of the primary truss, 23; and the thrust on each of the rafters 1 3, 1 2, is given by the equation

W cosec i
4

(2.)

(2.) Secondary Truss 1 4 3 5. The rafter 1 3 has the load W distributed over it; and reasoning as before, we are to leave two quarters of this out of the calculation, as being directly supported at 1 and 3, and to consider one-half, or W, as being the vertical load at the point 5. The truss is to be considered as consisting of a polygon of four pieces, 5 1, 1 4, 4 3, 3 5, two of which happen to be in the same straight line, and of the strut-brace, 5 4, which exerts obliquely upwards against 5, and obliquely downwards against 4, a thrust equal to the component perpendicular to the rafter of the load W; which thrust is given by the equation

Then we easily obtain the following values of the stresses on the rafter and ties, in which each stress is distinguished by having affixed to the letter R the numbers denoting the two joints between which it acts.

The difference between the thrusts on the two divisions of the rafter,

is the component along the rafter of the load at the point 5.

(3.) Smaller Secondary Trusses, 175, 56 3.-These trusses are similar in every respect to the larger secondary trusses, except that the load on each point is one-half, and consequently each of the stresses is reduced to one-half of the corresponding stress in the equations 3 and 4.

(4.) Resultant Stresses. The pull on the middle division of the great tie-rod 23 is simply that due to the primary truss, 1 2 3. The pull on the tie 47 is simply that due to the secondary truss 1 4 3. The pulls on the ties 5 7,5 6, are simply those due to the smaller secondary trusses, 157, 5 6 3. But agreeably to the Theorem of Art. 158, the pull on the tie 1 7 is the sum of those due to the larger secondary truss 1 4 3, and the smaller secondary truss 17 5. The pull on 6 4 is the sum of those due to the primary truss 1 2 3 and to the larger secondary truss 1 4 3. The pull on 6 3 is the sum of those due to the primary truss 1 2 3, to the larger secondary truss 143, and to the smaller secondary truss 5 6 3. The thrust on each of the four divisions of the rafter 1 3, is the sum of three thrusts, due respectively to the primary truss, the larger secondary truss, and one or other of the smaller secondary trusses.

Example II. Fig. 78 represents another form of truss common in roofs. Let W be the weight of the roof, as before, distributed over

7

Fig. 78.

the rafters 1 2, 13. 23 is the great tie-rod; 1 7, 6 5, 89, suspensionrods; 7 6,7 8, 5 4, 9 10, struts.

(1.) Primary Truss 1 2 3. The load at 1, as before, is to be taken a8 = W.

(2.) Secondary Trusses 7 6 3, 7 8 2. The load at 6 is to be held to consist of one-half of the load between 6 and 1, and one-half of the load between 6 and 3; that is, one-half of the load between 1 and 3, or W. The trusses are triangular, each consisting of two struts and a tie, and the stresses are to be found as in Article 149.

The suspension-rod 1 7 supports two-thirds of the load on 7 6 3, and two-thirds of the load on 7 8 2; that is, 4W W; and

=

this, together with W which rests directly on 1, makes up the load of W, already mentioned.

(3.) Smaller Secondary Trusses 34 5,9 10 2. Each of the points 4 and 10 sustains a load of W, from which the stresses on the bars of those smaller trusses can be determined.

One-half of the load on 4, that is ' W, hangs by the suspensionrod 6 5; and this, together with W, which rests directly on 6, makes up the load of W on that point, formerly mentioned. The same remarks apply to the suspension-rod 8 9.

(4.) Resultant Stresses. The pull between 5 and 9 is the sum of those due to the primary and larger secondary trusses; that between 5 and 3, and between 9 and 2, is the sum of the pulls due to the primary, larger secondary, and smaller secondary trusses.

The thrust on 1 6 is due to the primary truss alone; that on 64 to the primary and larger secondary truss; that on 4 3 to the primary, larger secondary, and smaller secondary trusses; and similarly for the divisions of the other rafter.

Example III. Suppose that instead of only three divisions, there are n divisions in each of the rafters 1 3, 12, of fig. 78; so that besides the middle suspension-rod 17, there are n-2 suspension-rods under each rafter, or 2 n - 4 in all; and n under each rafter, or 2 n centres of resistance; that is, the ridge-joint 1, and n each rafter; and the load directly supported on each of these

1 sloping struts 2 in all. There will thus be 2n-1

1 on

2 n

W

The total load on the ridge-joint, 1, will be as before, that

2

The total load on the upper joint of any secondary truss, distant

from the ridge-joint by m divisions of the rafter, will be,

1

W hung by

The stresses on the struts and tie of each truss, primary and secondary, being determined as in Article 149, are to be combined as in the preceding examples.

160. Compound Trusses-Several frames, without being distinguishable into primary and secondary, may be combined and con

nected in such a manner, that certain pieces are common to two or more of them, and require to have their stresses determined by the Theorem of Article 158.

Example I. In fig. 79, 89 represents part of the horizontal platform of a suspension bridge, supported and balanced by being hung from the top of a central pier, 1, by pairs of equally inclined rods or ropes, viz:-1 8 and 1 9; 1 6 and 1 7; 14 and 1 5; 1 2 and 1 3.

Fig. 79.

Here 8 19 is to be considered as a distinct triangular frame, consisting of a strut 8 9, and two ties 1 8 and 1 9, loaded with equal weights at 8 and 9, and supported at 1. Let x denote the height of the point of suspension I above the level of the loaded points, ya ye, the distance of those points on either side of the middle of the pier, P the load at each point, RR, the pull on each of the ties, 1 8, 1 9, T., the thrust between 8 and 9 along the platform. Then we have

and similar equations for each of the other distinct frames 6 17, 415, 2 1 3.

Then using a similar notation in each case, the thrust along the platform

and so on for as many pairs of divisions as the platform consists of. Example II. Fig. 80 represents the framework for supporting

one side of a timber bridge, resting on two piers at 1 and 4. It consists of four distinct trusses, viz.,

but all those trusses have the same tie-beam, 14; and the pull along that tie-beam is the sum of the pulls due to the four trusses. 161. Resistance of Frame at a Section.-THEOREM. If a frame be acted upon by any system of external forces, and if that frame be conceived to be completely divided into two parts by an ideal surface, the stresses along the bars which are intersected by that surface, balance the external forces which act on each of the two parts of the frame.

This theorem, which requires no demonstration, furnishes in some cases the most convenient method of determining the stresses along the pieces of a frame. The following consideration shows to

what extent its use is limited.

CASE 1. When the lines of resistance of the bars, and the lines of action of the external forces, are all in one plane, let the frame be supposed to be intersected anywhere by a plane at right angles to its own plane. Take the line of intersection of these two planes for an axis of co-ordinates; say for the axis of y, and any convenient point in it for the origin O; let the axis of x be perpendicular to this, and in the plane of the frame, and the axis of z perpendicular to both, and in the plane of sectio

The external forces applied to the part of the frame at one side of the plane of section (either may be chosen) being treated as in Article 59, give three data, viz., the total force along x = F.; the total force along y = F, and the moment of the couple acting round z = - M; and the bars which are cut by the plane of section must exert resistances capable of balancing those two forces and that couple. If not more than three bars are cut by the plane of section, there are not more than three unknown quantities, and three relations between them and given quantities, so that the problem is determinate; if more than three bars are cut by the plane of section, the problem is or may be indeterminate.

The formula to which this reasoning leads are as follows:-Let a be positive in a direction from the plane of section towards the part of the structure which is considered in determining F,, F„, and M; lety lie to the right of + when looking from z; let angles measured from Ox towards +y, that is, towards the right, he positive; and let the lines of resistance of the three bars cut by the plane of section make the angles i, i2, is, with x. Let n, ng, ng, be the perpendicular distances of those three lines of resistance from O, distances towards the

positive

right) of O 2 being considered as { negative }