This rule applies to columns having a length of more than thirty times the diameter. Example.--Required to know the breaking weight of a column 14 feet high, 12 inches outside and 10.5 inches inside diameter. (The logarithms are found from the Table of Logarithms of Numbers, p. 35 of the Author's "Engineer's Pocket Remembrancer.") 88-8)128920-0(1452 tons nearly. 888 4012 3552 4600 4440 1600 Hence, allowing as the safe working strength of the column, we should have 8) 1452 181.5 tons' safe load. CHAPTER VI. IRON ROOFS. FROM its strength, durability, and incombustible nature, iron is a material eminently suitable for the construction of roof principals and purlins for warehouses, rail. way-stations, factories and store-houses; and, moreover, it admits of being arranged in such forms as will occupy a mínimum space. Roofs admit of being classified under four distinct heads : 1st. Roofs supported by triangular trusses or principals. 2nd. Roofs supported by arched trusses or principals. 3rd. Roofs supported by straight girder trusses or principals. 4th. Roofs supported by dome-shaped framework. The actual loads to which a roof may be subject consists of three elements, viz. :— The weight of the main principals, &c., The weight of the covering used, The weight of snow, ice, &c. for any particular case the two first elements will be constant, but the last must be taken for maximum cases. The load due to weather can of course only be determined from the nature of the climate in the locality for which the roof is required, by which also to a very great extent the material used for the covering will be determined, but conjointly with the purpose to which the structure is to be applied. Thus in some cases, such as for sheds in temperate climates, corrugated iron sheets, weighing 6 or 7 lbs. per square foot, may be used, whereas in other instances, as for exposed buildings in tropical climates, thick coverings of sand or concrete may be required to exclude the intense heat, and we have known of cases in which it has been stipulated that roof principals should be tested with a load of 100 lbs. per square foot of area. In such positions, as we are much exposed to wind, care must be taken to prevent the covering being blown off by the wind getting underneath. Obviously in all cases of roofs the load may be taken as uniformly distributed over the whole area of such roof, hence if it be required to find the load on any one principal it may be done by means of the following rules: RULE. To find the total load on any main principal in pounds, multiply the total load per square foot by the span of the principal in feet and by the distance between two contiguous principals in feet, the product will be the total distributed load in pounds on each principal. Example.-Let it be required to find the total load in pounds on a principal in a roof 50 feet span, the load per square foot being 30 lbs., and the distance between the principals 10 feet : 30 lbs. load per foot, 50 feet span of principal, 1500 10 feet distance between principals, 15000 lbs. total load on principal. This may, if required, be reduced to tons by dividing by 2240 thus: 2240) 15000 (6.69 (say) 6.7 tons. 13440 15600 13440 21600 20160 1440 To find the load per foot run of the main principals, we have RULE. To find the load in pounds per foot run on a main principal, multiply the load in pounds per square foot of roof by the distance in feet between two principals—the product will be the required load. Example.-Let the load per square foot be 25 lbs., and the distance between the principals 12 feet, then 25 lbs. load per square foot, 12 feet distance between principals, 300 lbs. load per foot run of principal. Let us take, as a general example, a light shed-roof of which each principal consists of two rafters meeting at the centre of the span, and having their ends tied by a horizontal tie-bar supported in the centre by a vertical tie-rod passing up to the crown of the roof. The two rafters will sustain the load, their ends being prevented from spreading by the tie-rod connecting them; the vertical rod only serves to prevent the tie-rod from sagging. This will in effect be the truss shown in Fig. 13, but in an inverted position. Let the span of the roof be 12 feet, its rise or height in the centre 2 feet 6 inches, the length of each rafter will be 6 feet 6 inches. Let the maximum load be 20 lbs. per square foot, and the distance of the principals apart 10 feet, then, by the above rule, the load on each principal will be found, 20 lbs. load per foot, 12 feet span, 240 10 feet distance between principals, 2400 lbs. load on principal. But the rafters being equal, each of them will carry one half of this load, or 1200 lbs., and the maximum strain on each rafter will be found from the rule as follows: RULE. To find the maximum strain on each rafter, multiply the load in pounds on the rafter by the length of the rafter in feet, and divide the product by the height of the roof in feet; the quotient will be the strain on the rafter (thrust) in pounds. This, the maximum strain on the rafter, occurs at its root, from which point the thrust diminishes towards |