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.

In this case it is the metal which bears on the rivet or which the rivet bears on, which has to be considered; this is in compression and liable to failure, therefore, just as is the metal in a column or the compression side of a girder. The amount of stress which this metal will stand is determined by the ultimate compressive strength per square inch, and the area under compression, which area is the product of the diameter of the rivet and the thickness of the metal or in this case, the web of the girder.

If therefore t = thickness of metal d = diameter of rivet

C = ultimate compressive strength in pounds per square inch, then Vb = ultimate bearing value in pounds

= Cdt

The safe value usually used for power-driven rivets in building work is 18,000 pounds per square inch; for three-quarter-inch rivets, therefore, the bearing value becomes for a 5/16-in. web 18,000 X 3/4x5/16 = 4,219 pounds, and for hand-driven, rivets, 3,516 pounds.

The web of the beam in Fig. 191 is a case of bearing enclosed, that is, it is enclosed on both sides by other members, and therefore is stiffened against buckling under compression. The web of the girder is not enclosed, as it is free to buckle on one side. Most authorities allow a slightly greater bearing value, generally about 10 per cent for bearing on metal enclosed.

In designing such a connection as is illustrated in Fig. 191, the number of rivets through the web of the beam would be determined by the bearing value of one rivet, unless the thickness of this web was 5/8 in. or over, since for all thicknesses less than this the bearing value would be less than the double shear. The number of rivets through the web of the girder would be determined by the shearing value of one rivet for all thicknesses of webs of 5/16 in. and over, since for these thicknesses the bearing value is greater than single shear. Where two beams frame into a girder on opposite sides so that the rivets through the girder are common to both beams as shown in Fig. 192, these rivets are in bearing on the web of the girder for the combined load brought by both beams, in double shear for the combined loads, and in single shear for the load from each beam. If these loads were the same for each beam, single shear from the load from one beam would, of course, be equivalent to double shear for the load from both beams; if, however, the loads were greatly dissimilar the greatest load with the single shear value must be used. To illustrate this, suppose we have a 10-in. beam framed on one side of a 10-in. beam and an 8-in. beam framed opposite to it. Suppose the load brought by the 10-in. beam to the girder is 14,000 pounds, and that by the 8-in. beam 6,000 pounds. Now the web of a 10-in. 25-lb. beam is .31 inches thick, and the bearing value would therefore be .31 X 15,000 X .75 = 3,487 pounds, and for the total load this would require six rivets. To carry the load of 14,000 pounds in single shear at a value of 3,313 would require but five rivets, so that the bearing value and the total load from both beams would determine the number of rivets.

Fig. 191.

Fig. 192.

If, however, these beams were carried by a 12-in., 40-lb. beam "whose web is .46 inches thick, the bearing value would then be 5,175 pounds and this would require but four rivets; in this case the number would be determined by the greatest load and the single shear value of a rivet. Fig. 193 shows a single angle connection which would be determined by the rivet in single shear. It should be noticed that in designing connections a few rivets in excess of the actual number calculated should be used for connections; in general, 20 per cent should be added.

Fig. 193.

Fig. 194.

1. Suppose that certain rivets to be provided in a connection are in double shear. The rivets are all 7/8 in. in diameter. The outside plates are each 1/2 in. thick. What will be the thickness of the inside plate to make the rivet value equal to double shear?

2. Suppose a 6-in., 12.25-lb. I-beam that is 5 ft. long carries a load of 15,000 lb., uniformly distributed. How many rivets 3/4 in. in diameter, will be required for its connection to the beams at each end, allowing 6,000 lb. per square inch for shear on the rivets, and 12,000 pounds per square inch for bearing?

3. In the preceding problem, how many rivets 3/4 in. in diameter will be required to attach the connection angles to the 6 in. I-beam? In order to determine this, it will be necessary to first find the thickness of the web of the 6-in., 12.25-lb. I-beam. This can be found by referring to the tables on pages 30 and 31 of Part I. As the thickness of the webs is there given in decimals of an inch, these must be converted into the next smaller common fractions.

4. Given a 12-in., 3l 1/2-lb. I-beam 14 ft. long and a 10-in., 25-lb. I-beam 12 ft. long framed opposite to each other to a 15-in., 42-lb. I-beam. If these beams are each loaded to the safe capacity with a uniformly distributed load, what will be the number of 3/4-in. rivets required for the field connection to the girder, using 7,500 pounds for shear and 15,000 pounds for bearing?

5. In the above problem what will be the number of 3/4-in. shop rivets required on the end of each beam using 9,000 pounds for shear and 18,000 pounds for bearing?

6. Using the same values and loads as in problem 4, what will be the number of rivets required in each beam, if they do not frame opposite each other?

7. Give the lengths of field rivets and shop rivets required for each connection in each of the cases covered by problems 2, 3, 4, 5, 6.

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