Machinery's Handbook, 31st Edition
284 FORMULAS FOR COLUMNS Factor of Safety for Machine Columns: When the conditions of loading and the physical qualities of the material used are accurately known, a factor of safety as low as 1.25 is sometimes used when minimum weight is important. Usually, however, a factor of safety of 2 to 2.5 is applied for steady loads. The factor of safety represents the ratio of the critical load P cr to the working load. Application of Euler and Johnson Formulas: To determine whether the Euler or Johnson formula is applicable in any particular case, it is necessary to determine the value of the quantity Q ÷ r 2 . If Q ÷ r 2 is greater than 2, then the Euler Formula (1) should be used; if Q ÷ r 2 is less than 2, then the J. B. Johnson formula is applicable. Most compression members in machine design are in the range of proportions covered by the Johnson formula. For this reason a good procedure is to design machine elements on the basis of the Johnson formula and then as a check calculate Q ÷ r 2 to determine whether the Johnson formula applies or the Euler formula should have been used. Example 1, Compression Member Design: A rectangular machine member 24 inches long and 1 ∕ 2 3 1 inch in cross section is to carry a compressive load of 4000 pounds along its axis. What is the factor of safety for this load if the material is machinery steel having a yield point of 40,000 psi, the load is steady, and each end of the rod has a ball connection so that n = 1? Solution: From Formula (3) . . , , , . Q 1 3 1416 3 1416 30 000 000 40 000 24 24 00778 # # # # # = = (The values 40,000 and 30,000,000 were obtained from the table Strength Data for Iron and Steel on page 429 .) The radius of gyration r for a rectangular section (page 242 ) is 0.289 3 the dimension in the direction of bending. In columns, bending is most apt to occur in the direction in which the section is the weakest, the 1 ∕ 2 -inch dimension in this example. Hence, least ra- dius of gyration r = 0.289 3 1 ∕ 2 = 0.145 inch. . . . r Q 0145 00778 370 2 2 = = ^ h which is more than 2, so the Euler formula will be used.
. 370 40 000 2 1 1 5400 5400 4000 1 35 pounds so that the factor of safety is cr y 2 # # ' = = = = , . P Q s Ar
Example 2, Compression Member Design: In the preceding example, the column formu las were used to check the adequacy of a column of known dimensions. The more usual problem involves determining what the dimensions should be to resist a specified load. For example: A 24-inch long bar of rectangular cross section with width w twice its depth d is to carry a load of 4000 pounds. What must the width and depth be if a factor of safety of 1.35 is to be used? Solution: First determine the critical load P cr :
. 4000 1 35 5400 working load factor of safety pounds # # =
P
= =
cr
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