This section is from the book "The American House Carpenter", by R. G. Hatfield. Also available from Amazon: The American House Carpenter.

The pressure in each timber may be obtained as shown in Fig. 84, where A B represents the axis of the tie-beam, A C the axis of the rafter, D E and F B the axes of the braces, and D G, FE, and CB the axes of the suspension-rods. In this design for a truss, the distance A B is divided into three equal parts, and the rods located at the two points of division, G and E. By this arrangement the rafter A C is supported at equidistant points, D and F. The point D supports the rafter for a distance extending half-way to A and half-way to F, and the point F sustains half-way to D and half-way to C. Also, the point C sustains half-way to F, and, on the other rafter, half-way to the corresponding point to F. And because these points of support are located at equal distances apart, therefore the load on each is the same in amount. On D G make Da equal by any decimally divided scale to the number of hundreds of pounds in the load on D, and draw the parallelogram abDc. Then, by the same scale, Db represents (Art. 71) the pressure in the axis of the rafter by the load at D; also, Dc the pressure in the brace D E. Draw cd horizontal; then D d is the vertical pressure exerted by the brace DE at E. The point F sustains, besides the common load represented by D a, also the vertical pressure exerted by the brace D E; therefore, make Fe equal to the sum of D a and Dd, and draw the parallelogram Fgef. Then Fg, measured by the scale, is the pressure in the axis of the rafter caused by the load at F, and Ff is the load in the axis of the brace FB. Draw fh horizontal; then Fh is the vertical pressure exerted by the brace FBat B. The point C, besides the common load represented by Da, sustains the vertical pressure Fh caused by, the brace FB, and a like amount from the corresponding brace on the opposite side. Therefore, make Cj equal to the sum of Da and twice Fh, and draw jk parallel to the opposite rafter. Then Ck is the pressure in the axis of the rafter at C. This is not the only pressure in the rafter, although it is the total pressure at its head C. At the point F, besides the pressure C k, there is Fg. At the point D, besides these two pressures, there is the pressure D b. At the foot, at A, there is still an additional pressure; for while the point D sustains the load halfway to F and half-way to A, the point A sustains the load half-way to D. This load is, in this case, just half the load at D. Therefore draw A m vertical, and equal, by the scale, to half of Da. Extend CA to l; draw ml horizontal. Then A l is the pressure in the rafter at A caused by the weight of the roof from A half-way to D. Now the total of the pressures in the rafter is equal to the sum of A /+ D b + Fg added to C k. Therefore make k n equal to the sum of A l+Db + Fg, and draw no parallel with the opposite rafter, and nj horizontal. Then Co, measured by the same scale, will be found equal to the total weight of the roof on both sides of CB. Since Da represents s, the portion of the weight borne by the point D, therefore Co, representing the whole weight of the roof, should equal six times Da, as it does, because D supports just one sixth of the whole load. Since C n is the total oblique thrust in the axis of the rafter at its foot, therefore nj is the horizontal thrust in the tie-beam at A.

Fig. 84.

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