52. Compound Girders

Where spans occur too great to admit the use of rolled beams, compound girders are employed, made up of plates and angle irons.

The single web or "plate girder" is the most economical, the most accessible for painting, for inspection, and for connecting the floor beams. But where a thick wall is to be supported and lateral stiffness is required, the double web or "box girder" is used.

If girders have no lateral bracing, the flanges should not be less than one twentieth the span.

53. Webs

The webs should be of such thickness that there shall be no tendency to buckle and the vertical shearing stress per square inch shall not exceed 6000 pounds. This stress is greatest near the bearings, and is obtained by dividing one half the load upon the girder by the web section.

54. Buckling

Vertical angle irons are riveted to webs to prevent buckling, at intervals of not more than the depth of the web, and should always be used at the bearings and where concentrated loads occur; it is good practice to use stiffners when the thickness of web is less than one sixtieth (1/60) the depth.

55. Flanges

The flange embraces all the metal in top and bottom of girder, exclusive of web plates.

56. Deflection

For deflection the depth of girder should be about one twentieth (1/20) the span.

57. Rivets In Girders

Rivets In Girders should not be spaced closer than two and one half (2 1/2) inches between centres or three times the diameter, nor farther apart than sixteen times the thickness of plate connected, and should be closer near bearings.

Rivets five eighths (5/8), three quarters (J), and seven eighths (7/8) of an inch in diameter are commonly used.

If the webs are connected to the flanges by single angles as in box girders, the rivets will be in single shear; but if a pair of angles are placed each side of the web as in a plate girder, the rivets will be in double shear.

In long girders it is often necessary to increase the thickness of webs near ends, to give a greater bearing surface for the rivets.

58. Strain On Flanges Of Girders

To calculate a girder accurately we should allow for the rivet holes; but by taking the safe strength of the iron at 14,000 pounds per square inch strain on the flanges, and using a formula for 12,000 pounds, and disregarding the rivet holes, we can compute it with sufficient accuracy for all practical purposes in buildings.

If 12,000 pounds strain per square inch is desired, add one sixth (1/6) to the total area of flanges for loss by rivet holes.

Proceeding on the first assumption, we have the following rule for the strength :

Safe loads in tons = 12 X area of one flange X height of web in inches 3 X span in feet.

To find area of one flange in inches, divide (3 X weight in tons X span in feet) by (12 X height of web in inches).

Example

Given a brick wall 12 inches wide, 29 feet 1 1/2 inches high, the span 24 feet, and girder 20 inches in height: what should be the area of each flange, the load being 40.2 tons, at 115 pounds per square foot of 12-inch wall?

Ans

Area of one flange = (3x40.2x24)//(12x20)= 12.00 square inches.

To divide the area of 12.06 square inches into a proper working flange, we should require a Plate 10 X 3/4 in. = 7.5 inches.

Two angles 4 X 3 X 3/8 in. = 4.96 "

Total, 12.46 inches - an excess of 0.4 in.

59. Shearing

A sufficient area having been provided in the top and bottom flanges to resist the compressive and tensile strains, there will be needed in the web sufficient metal to resist the shearing strain. This strain is, theoretically, nothing at the middle of a girder uniformly loaded, but from thence increases by equal increments to each support, where it is equal to one half of the whole load.

For example: In the case considered, the girder of 24 feet span carries 80,400 pounds uniformly distributed over its whole length. Taking half of the load over half the beam, at the centre the shearing strain is nothing; at 4 feet from the centre it is equal to 1/3 of half the load, or 13,400 pounds; at 8 feet it is 26,800 pounds; at 12 feet, or at the supports, it is 40,200 pounds, or half the whole load.

The thickness of web may be determined as follows: Let G = shearing stress;

A = area of web in inches at point of stress; k = effective resistance of wrought iron to shearing; t == thickness of web; d = height of web. Then A = td, and G = ktd, or t= G/dk , or t= 40,200/(20x6000) = .335, or nearly 3/8 inch.

As the least practicable thickness of plates is 1/4 inch, it is not worth while to compute that of the web at the 4 and 8 feet stations.