This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.

14. Let it be required to design a Fink roof truss of 64 feet span and 1/4 pitch, the distance between trusses being 16 feet. The roof covering is taken as 12 pounds per square foot of roof surface, and the total snow and wind load will be taken as 30 pounds per square foot of horizontal projection. The weight of the steel in the roof truss will be computed from Merriman's formula (see Art. 4, p. 6). The total weight is now found to be:

Weight of truss, | 3 | X | 16 | X | 64 ( 11 | + | l | = | 5 580 pounds. | ||||||

4 | 10 | ||||||||||||||

Weight of roof cover, | 35.6 | X | 2 | X | 16 | x | 12 | = | 13 650 pounds. | ||||||

Weight of wind and snow 64 | X | 16 | X | 30 | = | 30 700 pounds. | |||||||||

Total | 49 930 pounds. |

Each apex load is therefore 49 930 ÷ 8 = 6 240 pounds. By multiplying this value by each of the stresses as given in Fig. 25, the stress in each member is computed as follows:

L0 U1 | = | 7.83 | X | 6 240 | = | 48 800 pounds |

L0 L1 | = | 7.00 | X | 6 240 | = | 43 700 " |

U1 L1 | = | 0.89 | X | 6 240 | = | 5 580 " |

U1 U2 | = | 7.38 | X | 6 240 | = | 46 000 " |

L1 U2 and U2 L3 | = | 1.00 | X | 6 240 | = | 6 240 " |

L1 L2 | = | 6.00 | X | 6 240 | = | 37 450 " |

U2 L2 | = | 1.79 | X | 6 240 | = | 11 150 " |

U2U3 | = | 6.93 | X | 6 240 | = | 43 200 |

L2 L5 | = | 4.00 | X | 6 240 | = | 24 950 |

L2 L3 | = | 2.00 | X | 6 240 | = | 12 475 " |

U3 L3 | = | 0.89 | X | 6 240 | = | 5 580 " |

L3 U4 | = | 3.00 | X | 6 240 | = | 18 725 " |

U3U4 | = | 6.48 | X | 6 240 | = | 40 500 " |

In the design of this truss, no material thinner than ¼-inch, and no angles smaller than 2½ by 2-inch, will be allowed.

Fig. 76 shows an outline diagram of the truss, with the stresses placed upon it. A positive sign signifies a tensile stress, and a negative sign signifies a compressive stress. The length of the top chords is

35.6 feet; and the length of each panel is ¼ of this, or 8.9 feet. The horizontal projection of one panel is ¼ of half the span, or 32 ÷ 4 = 8 feet.

Design of the Purlins. The distance between the trusses is 16 feet, and the distance between the purlins is 8.9 feet; therefore the load coming on one purlin is:

Fig. 76. Stresses in a Fink Truss.

Roof covering, 8.9 | X | 16 | X | 12 | = | 1 710 pounds |

Snow and wind, 8 | X | 16 | X | 30 | = | 3 840 |

Total | = | 5 550 pounds |

This should be resolved in two components, Vand H, perpendicular and parallel to the truss chord. These are determined by the proportions of similar triangles, as follows:

V | : 5 550 | = | 32 | : 35.6 |

V | = | 5 080 pounds. | ||

H | : 5 550 | = | 16 | : 35.6 |

H | = | 2 490 pounds. |

The bending moment caused by V is Mv = (5 080 X 16) ÷ 8 =

10 160 pound-feet. The bending moment caused by H is MH =

(2 490 X 16) ÷ 8 = 4 980 pound-feet. The stress caused by V is = Mc /I ; and the stress caused by H is MHC1/I' ; and there is also the condition that the sum of these two stresses shall not be greater than 15 000 pounds. Since the above formula involves the moment of inertia and half the depth of the beam, a beam must be chosen, and its moment of inertia and half-depth substituted in the above equation, and the equation solved. In case the sum of the stresses is in excess of 15 000 pounds, or very much smaller, a re-computation must be made, using a larger or a smaller beam.

A 15-inch 42-pound I-beam will be assumed, and will be examined to see if it fulfils the necessary conditions. The value of I and I* are taken from the Carnegie Handbook, p. 97. The value of c is - =

7½ in the first case, and 5.50/2 = 2.75 in the second case. The quantity 5.50 is the width of the flange of the I-beam. Substituting in the above formula, there results:

[10 160 X 12 X 7½] / 441.8 + [4 980 X 12 X 2.75] / 14.62 =13 320 pounds.

The above I-beam could be used; but in case the sheathing is laid closely and nailed tightly, we may consider it acting as a beam of a span of 16 feet, 8.9 feet deep, and of a thickness equal to that of the sheathing, which in this case will be assumed as 1½ inches. The sheathing will then take up the moment caused by the force H; and the purlin will take up the vertical bending moment alone. The

M c stress in the sheathing due to the force H isMHc/I. Here M =

4980 X 12;c = 8.9 X 12 ÷2;and l = 1.5 (8.9 X 12)3/12 Therefore

S = [4 980 X 12 X 8.9 X 12 X 12] / 2 X 1-5 (8.9 X 12)3

= 20.95 pounds per square inch, which is insignificant.

The vertical bending moment taken up by the purlin is 10 160 X 12 = 121 920 pound-inches, and this requires a section modulus of 121 920 ÷ 15 000 = 8.14. By consulting pages 101 and 102 of the Carnegie Handbook, the following is found to be true:

An 8-inch 11.25-pound channel is just too small. A 7-inch 17.25-pound channel gives the nearest section modulus. An 8-inch 13.75-pound channel would be lighter and stirrer. A 9-inch 13.25-pound channel would be still lighter and stiffer; and since it weighs less than any of the others, it will be more economical.

A 9-inch 13.25-pound channel will accordingly be used for the purlins.

On account of one half-panel load coming on the purlin at the ends and ridge of the truss, these purlins must theoretically be only one-half as strong as the other; but, on account of the fact that all purlins must be of the same height, these purlins are made of the lightest weight channel of the same height as the others. In this case it happens that the lightest weight 9-inch channel is required for the intermediate purlins as well as for the end ones. To illustrate the above, suppose that the purlins were required to be 10-inch 25-pound channels, then the end purlins would be made of 10-inch 15-pound channels.

In case sheathing is not used, then some other method must be employed to take up the bending moment due to the force H. The usual method of doing this is to bore holes in the center of the purlins at the middle point of their span, and to connect them with rods which run from one eave up over the ridge and down to the other eave (see Fig. 22).

Design of Tension Members. For Member L0 L1: The required net area is 43 700 ÷ 15 000 = 2.02 square inches. By consulting the Carnegie Handbook, p. 118, it is seen that two 3 by 3 by 5/10-inch angles give a gross area of 1.78 X 2 = 3.50 square inches. From this must be subtracted the rivet-hole made by a ¾-inch rivet. Since all rivet-holes are punched ⅛ inch larger in diam -ter than the rivet, the amount to be substracted from the above gross area is 5/16 X (¾ + ⅛) X 2 = 0.54, there being two rivet-holes taken out of the section. This gives a total net area of 3.56 - 0.54 = 3.02 square inches. As this is but slightly larger than there-quired net area, these angles will be used for this member. Since the stress in this member is the greatest stress in the bottom chord, and since the bottom chord is made of the same section up to the splice at L2 on account of economical construction, it being cheaper to run the same sized angle throughout than it would be to change the size of each panel and make a splice at each panel point, the size of angle as determined above will be used for the first two panels of the bottom chord at each end.

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