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JOINT Erection of a 375-foot span over Osage river between St. Louis and Kan sas City, on line of St. Louis, Kansas City & Colorado Railroad, now part of the Chicago, Rcck Island Pacific Railroad System.

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table (C 150). In the column on the 12 inch line, is found 72.00 as the moment of inertia of one plate. The area of the section, Fig. 71, is: 2 Plates 12 in. by in. =12.00 square inches 4 Angles 5 in. by 3 in. by in. (C 111)=15.00 square inches Total 27.00 square inches Let it be required to obtain the radius of gyra tion Operation a is: Moment of inertia of 4 angles = 4x2.58 = 10.32 Moment of inertia of 2 plates = 2x72.00 = 144.00 Angles multiplied twice by distance : 4x3.75x5.25x5.25 = 413.39 Plate multiplied twice by distance : 2x6.00x0x0 (since axis coincides with a-a). 0.0 Operation c is: On (C 281) near bottom, and in second column, is found 2116. In first column is 46. Therefore 4.6 is the radius of gyration.

In regard to the it should be equal to or greater than Just to what distance back will be necessary to give this, is not known, and can be determined only by trial. A rough guess is to multiply the width of the plate by 0.7, and make it somewhere about that figure. In this case 12 X 0.7=8.4 inches—say make it inches. The computation will not be given, but this dis tance makes which shows the column to be safer about axis The distance might be cut down to inches, and a recomputation made. The result would probably be very close to 4.6. Usually two or three trials are necessary before it can be determined correctly. If the last computation comes within 1/10th of the other radius of gyration, and is greater, let it stand. Do not recompute.

47. Design of Compression Members. The strength of compression members also depends upon the length of the column, as is seen from the formula: Length in inches P=17.100-57x Radius of gyration Here P is the given load which is allowable on each square inch of column area. This is the formula used in (C 125) for fixed loads and for buildings. For live loads and for bridges, use 6/10ths of the above.

As an illustration of the use of this formula, suppose the column of Fig. 71 to be 30 feet ol long. Then, P=17,100-(57x30x12)÷4.6

=17,100-4,461 =12,639 pounds per square inch.

The area of the section was 27.00 square inches, and therefore the column could hold 12,639 X 27 =341,253 pounds.

The design of compression members of Z-bar columns and channel columns, is conveniently made from tables (C 127-142). The information on (C 125-126) should be carefully read before using them. These types are in very general use. In case other types are desired, a section must be assumed, and the radius of gyration worked out as shown above. The stress as given by the formula must then be divided into the total load on the column (or stress in the com pressive member); and this result—which is the required area—must be compared with the ac tual area of the assumed section. If the assumed section is less than the required, or differs greatly from it, another section must be as sumed and the operation repeated. This opera tion must be continued until the assumed and the required sections agree closely.

The student of Steel Construction will find it of great advantage if, in addition to the treat ment here given, he will familiarize himself with the principles of the "Strength of Materials" as specially presented in any of the standard text books on that subject.

As an example, let it be required to design a Z-bar column to carry the load of 341,253 pounds which is the strength of the 30-foot col umn of Fig. 71. This is equal to 341,253-÷-2,000 =170.67 tons. On (C 129), in the table of 10 inch Z-bars, on the 30-foot line, 172.2 is found, and this is the nearest value which can be found and still not be less than the 170.67. Looking up at the top of the column, it is seen that the Z-bars and plate must be inch thick. Also it is found that the area of the section is 32.7 square inches. By looking on (C 103, 130, and 248), the sizes of the Z-bars and the plates are found, and the section is determined to be: 4 Z-bars 5 in. by 31 in. by in.=27.84 square inches 1 Plate 7 in. by in. = 4.81 square inches 32.65 square inches In the table of 12-inch Z-bar columns on the same page of the handbook, is found 180.2 on the 30-foot length; and at the top of the column, the metal is found to be inch, and the area 31.2 square inches.

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