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

The strains in a roof-truss can be ascertained geometrically, as shown in Art. 224. To make a practical application of the results, in any particular case, it is requisite first to ascertain the load at the head of each brace, as represented by the line D a, Fig. 84. The load corresponding to any part of the roof is equal to the product of the superficial area of that particular part (measured horizontally) multiplied by the weight per square foot of the roof. Or, when M equals the weight per square foot, c the distance from centres at which the trusses are placed, and n the horizontal distance between the heads of the braces, then the total load at the head of a brace is represented by -

N= M c n. (101.)

The value of M is given in general terms in equation (96.). To show its actual value, let it be required to find the weight per square foot upon a root 52 feet span and 13 feet high at middle; or (Fig. 84), where A B equals half the space, or 26 feet, and CB 13 feet, then A C, the length of the rafter, will be 26.069, nearly. And where the weight of covering per square foot, on the inclination, is 12 pounds, the force of the wind against a vertical plane is 30 pounds; the weight of snow per foot horizontal is 20 pounds; the weight of the plastering forming the ceiling at the tie-beam is 9 pounds; and the load in the roof is nothing; - with these quantities substituted, equation (96.) becomes -

M = 29.069x 12 ( 2/52 + 0.0154) + 30 132/29.0692 + 20+9 +0;

M = (29.069 x 12 x 0.05386) + (30 x 0.2) + 20 + 9;

M=18.788 + 6 + 29=53.788; .

or, say, 53.8 pounds. Then if c, the distance from centres between trusses, is 10 feet, and;n, the distance between braces, is one third of A B, Fig. 84, or 26/3= 8 2/3, the total load at the head of a brace will be, as per equation (101.) -

N = 53.8 x 10 x 8 2/3 = 4663;

or, say, 4650 pounds. Now, by any decimally divided scale, make D a, Fig. 84, equal to 46 1/2 parts of the scale; this being the number of hundreds of pounds contained in the weight at D, as above. Then, by the same scale, the several lines in the figure drawn as before shown will be found to represent respectively the weights here set opposite to them, as follows:

Dd = d a = he = 23 1/4, and represents 2325 pounds; | |||

D a = d c = hf = F h = 46 1/2 | " | 4650 | " |

D c = Db = A l = Fg= 52 | " | 5200 | " |

Fe = Da + D d = 69 3/4 | " | 6975 | " |

Ff = 65 3/4 | " | 6575 | " |

Cj = 3Da = 139 1/2 | " | 13950 | " |

CK = 3 D b = 156 | " | 15600 | " |

C n = Ck + Fg+ D b + A 1= 312 | " | 31200 | " |

Cn=Ck+3 Db=6 Db=2Ck =312 | " | 31200 | " |

Nj = C o = 6 D a = 6x 46 1/2 = 279 | " | 27900 | " |

It should be observed here that the equality of the lines nj and Co is a coincidence dependent upon the relation which in this particular case the line CB happens to bear to the line A B; A B being equal to twice C B. And so of some other lines in the figure. If the inclination of the roof were made greater or less, the equality of the lines referred to would disappear. It should also be observed that the strains above found are not quite exact; they are, however, correct to within a fraction of a hundred pounds, which is a sufficiently near approximation for the purpose intended. From the results obtained above, we ascertain that the strain in the rafter, from F to C, is represented by CK, and is equal to 15,600 pounds; while the strain at the foot of the rafter, from A to D, is represented by Cn, and equals 31,200 pounds, or double that which is at the head of the rafter. We ascertain, also, that the maximum strain in the tie-beam, represented by n j, is 27,900 pounds; that that in the brace DE, represented by Dc, is 5200 pounds; and that that in the brace F B, represented by Ff, is 6575 pounds. The strain in the vertical rod D G is theoretically nothing. There is, however, a small strain in it, for it has to carry a part of the tie-beam and so much of the ceiling as depends for support upon that part. But the manner of locating the weights, adopted in this article, does not recognize any load located at the point G. This is an objection to this system, but it is not material.

For a recognition of weights at the tie-beam, see Arts. 205 to 211. The load at G may be found by obtaining the product of the surface carried into the weight per foot of the ceiling; or, say, 10 cn = 10 x 10 x 8 2/3 == 867 pounds. The load to be carried by the rod FE is shown at D d = he, which above is found to be 2325 pounds. To this is to be added 867 pounds for the ceiling at E, as before found for the ceiling at G; or, together, 3192 pounds. The central rod CB has to carry the two loads brought to B by the two braces footed there; and also the weight of the ceiling supported by B. The vertical strain from the brace F B is represented at Fh, and equals 4650 pounds; therefore, the total load on CB is 4650 + 4650 + 867 = 10,167 pounds.

226. - Roof-Timbers: the Tie-Beam. - The roof-timbers comprised in the truss shown in Fig. 84 are the rafters, tie-beam, two braces, and three rods. Of these, taking first the tie-beam, we have a piece subject to tension and sometimes to cross-strain (see Art. 682, Transverse Strains). In this case the tensile strain "only need be considered. For this a rule is given in Art. 117. In this rule, if the factor of safety be taken at 20, the result will be sufficiently large to allow for necessary cuttings at the joints. Therefore, if the beam be of Georgia pine, equation (16.), Art. 117, becomes -

A = 27900 x 20/16000 = 34 7/8;

or, say, 35 inches. This is ample to resist the tensile strain; but, to resist the transverse strains to which such a long piece of timber is subjected in the hands of the workman, it would be proper to make it, say, 6x9.

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