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.
This problem may be solved, either by assuming the size of the rafters and computing their spacing, or by assuming the spacing and computing the size of the rafters. The latter method is the one most commonly used. The spacing of rafters is from 18 inches to 4 feet.
Fig. 20. Sheathing Laid Directly on Purlins.
The total weight per square foot which comes on the rafter is 12 + 10 + 4 = 2G pounds. Since each rafter carries a portion of the roof 10 by 2 feet, the total weight on one rafter is 10 X 2 X 26 = 520 pounds. The moment created by this weight is (520 X 10 X 12) ÷ 8 = 7 800 pound-inches. This should be equated to
S I /c
Allowing 1 000 pounds to the c square inch as the unit-stress on the extreme fibre, and noting that I ÷ C = bd3/12÷d/2=bd2/6 there results:
1000 bd2/6=7800 d =
The market widths of rafters are 1½, 2, 3, and 4 inches, 2 inches being the size usually employed. Substituting in the above formula, we have: d = =4.8 inches.
The rafters will be made 2 by 6 inches, since this is the nearest market size. If a 3-foot spacing of rafters was used, the required depth would be 5.92 inches, and a 2 by 6-inch would still be used. This spacing and this size of rafter would be the one to employ in the solution of the above problem.
2. Design the purlin for the roof of Problem 1, above, if the trusses are spaced 16 feet apart.
The rafters are spaced so close together that their own weight, the weight of the roof covering, and the snow load may be considered as uniformly distributed over the purlin. The total weight which comes upon one purlin is the weight of snow and roof covering on a space 16 feet long and 10 feet wide. This weight is:
Fig. 21. Use of Purlins Made of Tees.
Fig. 22. Use of Sag Rods to Prevent Sagging of Purlins at their Center.
Snow load = 10 X 16 X 12
1 920 pounds
Roof covering = 10 X 16 X 14
16 rafters 6 by 2-in., 10 feet long, at 3 lbs. per 144 cu. in.
4 640 pounds
The moment caused by this weight is:
(4 640 X 16 X 12) ÷ 8 = 111 360 pound-inches. The determination of the beam which will be used to withstand this bending moment is made by means of its section modulus. The formula M/S = I/c is used in the design of beams. The values of I and c are constant for any given beam, and therefore the value of I ÷ c for any particular beam is a constant, and this constant is called the section modulus. It is therefore evident that if we have a certain bending moment and a certain allowable unit-stress, we can obtain the value of the section modulus by dividing the moment by the allowable unit-stress. Then, looking into one of the steel handbooks, the beam can be determined which will have a section modulus equal to or slightly in excess of the value that has been obtained by dividing the bending moment by the unit-stress. This beam will be the beam which, with a unit-stress equal to the one assumed, will withstand the bending moment under consideration.
The handbooks issued by many of the steel companies are indispensable to the intelligent design of structural steel. That issued by the Carnegie Steel Company (edition of 1903) is one of the most convenient; and since it will be frequently referred to in this text, its purchase by the student is desired. This book may be obtained by addressing the Carnegie Steel Company at its offices in any of the larger cities. The cost to students has usually been 50 cents; to others, $2.00.
Assuming an allowable unit-stress of 18 000 pounds per square inch on the extreme fibre, the section modulus required to withstand the bending moment of 111 360 pound-inches is:
111 360/18 000=6.19
Looking in the Carnegie Handbook at column 11 on page 100, column 11 on page 102, and column 9 on page 104, it will be seen that any one of the following shapes will be sufficient:
One 5-inch 14.75-pound I-beam; One 7-inch 9.75 " channel; One 4 1/8 by 3 3/16 by 9/16 -inch Z-bar weighing 17.9 pounds per linear foot.
Instead of the 5-inch I-beam as given above, a 6-inch 12.25-pound I-beam with a section modulus of 7.3 could be used, and would be more economical, since it is less in weight; and it would also be stiffer, since its depth is greater and its section modulus is greater. A comparison of the above weights shows the channel to be the most economical, since its weight is considerably less than either of the other two shapes. Channels usually make the most economical purlins; and for this reason no other shapes are usually inspected, the channels being used in the first case without being compared with other sections. Inspection of column 11, page 110, Carnegie Handbook, shows that a 6 by 4 by ¾-inch angle could have been used for the purlin, since it gives a section modulus of 6.25. The weight of this angle, 23.6 pounds per linear foot, shows it to be far too uneconomical to employ.