The allowance usually made for the load on floors is— Example 24.-It is required to construct a floor to a warehouse to be carried by steel-rolled girders. The span between the side walls is 20 feet; the main girders are to rest on these walls and to be placed 10 feet apart. On these girders are to rest smaller girders running transversely with them and placed 3 feet apart; on these latter a 4" timber floor is laid. The greatest live load estimated to come on the floor is 3 cwts. per square foot distributed over its surface. Determine suitable sections for the girders. The working load on each main girder is, therefore, 33 tons. The breaking load = 4 × 33 = 132 tons. The breaking load for a span of 1 foot = 132 × 20 = 2,640 tons. Referring to the Table we find the nearest section which corresponds with this is No. 2, or a steel joist 18′′ × 7′′. Next consider the cross-beams : The breaking load on cross-girder=5 × 4 = 20 tons. The breaking load for a span of 1 foot = 20 x 10 = 200 tons. By referring to the Table we find a suitable girder is No. 11, which weighs 16 lbs. per foot. If the cross-girders be continuous over two or more spans their strength is increased in the proportion explained in the chapter on Bending moments (see Arts. 96 to 98). Example 25.-A rolled iron girder of No. 3 section rests on two abutments 15 feet apart. Determine (1) what weight placed at the centre will break it; (2) what weight placed 5 feet from one abutment will break it. (1) From the Table it is seen that the distributed breaking weight in tons: = 1218 81.2 tons. 81.2 = 40.6 tons. The central weight, therefore, (2) In the second case, the maximum bending moment occurs at the point of application of the weight. 10 If W = required weight, the bending moment = W foot = = 3 tons, or - 40 W inch-tons. This must be equal to the moment of resistance of the section, or Example 26.-In the last example, if two weights of 10 and 15 tons be placed on the girder at two points 18 inches at each side of the centre, what is the maximum tensile or compressive stress per square inch on the fibres ? The maximum bending moment occurs at the point of application of the 15 tons, and is equal to 78 foot-tons, or 936 inch-tons. If f=stress in tons per square inch on the flanges, I moment of inertia of section = 731. = Example 27.-Two rolled-steel girders placed side by side, span an opening of 16 feet. They are required to support a brick wall 20 feet high and 173 inches in thickness. A cross-beam is also suspended from their centres, which imposes on them an extra weight of 10 tons. Determine a suitable section for the girders. The weight of a cubic foot of brick work is about 100 lbs. X Total weight of brickwork = 20 × 16 x 1.45 × 100 lbs. 20.7 tons, Estimated weight of girders between abutments = 1.3 Total distributed load on two girders = 22.0 29 Each girder, therefore, is loaded with a distributed weight of 11 tons, and a concentrated central weight of 5 tons. The maximum bending moment occurs at the centre of the girder, and is equal to 42 foot-tons, or 504 inch-tons. If the steel be strained to one-fourth of its breaking weight, the moment of resistance of the section must be equal to four times the maximum bending moment, or = 504 × 4 = 2,016. In looking down the column of the moments of resistance in the Table, we find that girders 4, 5, and 6 give results nearest to what is required, and as No. 4 or a 15" x 5′′ girder is the lightest, it is the most economical to use. This girder will give a margin of strength, as the estimated weight of the two girders, or 1-3 tons, is considerably in excess of their actual weight. If iron girders be used instead of steel, it will be necessary to use No. 2 section, which weighs 81.6 lbs. per foot. If the girders have a bearing on each abutment of 15 inches, we have Weight of steel girders = 37 × 50=1,850 lbs., 148. Approximate Method of Calculating the Strength of Rolled Girders. The method of determining the strengths of rolled girders, which we have been considering, involves the determination of their moments of inertia and resistance; the calculation of these quantities is somewhat tedious, and where very great accuracy is not necessary, a shorter and much simpler rule may be adopted, which we will now proceed to explain. By this latter method the aid which the web supplies in resisting the horizontal stresses is left out of consideration, so that the strength of the girder as thus found is somewhat less than its real strength. This discrepancy is somewhat modified by taking as the effective depth of the girder its extreme depth over all, instead of the depth between the centres of gravity of the flanges. = Let S horizontal flange-stress at any section of the girder, d= total depth of the girder. Then, in all cases, M We have, therefore, the following approximate rule for determining the flange-stress at any portion of a H-girder with a thin web. The flange-stress at any section of a rolled girder is equal to the bending moment at this section, divided by the depth of the girder. Applying this approximate rule to example 26, we have And, as the sectional area of the flange = 5 square inches, we get stress per square inch on flanges= 58.5 = 11.7 tons, instead of 10.24 tons as previously found. Example 28.-What must be the distance between the supports so that the girder No. 5 in the Table will break by its own weight -(1) in steel; (2) in iron? (1) In the case of steel W = 53 lbs. Substituting this in the above equation, we get No. TABLE XXV.-STRENGTHS OF ROLLED GIRDERS IN WROUGHT IRON 2 18.0 7.10 71 94 24.67 81.6 84 1208 316 0 606 64 82 19 15627 64.5 731 4 15.0 5.06 50 80 14 88 48 6 50 510 5 14.0 5.87 50 81 15.7651.5 53 6 12.0 6-23 73 87 18:45 60 2 62 7 10.0 6.16 66 70 14.27 467 48 8 9.0 3.75 50 50 7.74253 26 9 8.0 402 42 56 74 24.3 25 10 7.0 3.70 32 46 5:36 17.5 18 11 6.25 3:38 26 50 4.76 15.6 16 12 6.0 3:09 39 50 5.06 16.5 17 1 20 0 8.26 76 97 29 67 97.2 100 1825 3650 5475 2433 3650 494 404 221.5 89.0 396 74.2 371 42.4 242 31.2 200 |