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# Beams Reinforced for Compression

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BEAMS REINFORCED FOR COMPRESSION Flexure Formulas.—It is frequently necessary to place steel in the compression as well as the tension side of a beam. When the size of a rectangular beam is limited, so that the concrete area is insufficient to carry the stress, steel may be used to take the surplus compression. In this case the concrete and steel act together, and the stress steel must be limited to such an amount as will not overtax the compressive strength of the concrete.

In this discussion, the following notation will be used, in addition to that employed for rectangular beams: A' = area of cross-section of compression steel; p'=ratio of compression steel area to effective area of beam, bd); d'= depth of center of gravity of compression steel below compression face of beans; stress in compression steel; C'= total compression on steel.

The same principles apply in this ease as in that of the beam reinforced for tension only, and the concrete is supposed to carry compression but no tension. It is easily seen that fS= f` (39) and that kd d) . (10) use of thee formulas in design will be illustrated by the following example: 14. A beam whose dimensions are b=12 in., d=22 in., d'=2 in., is to carry a bending moment of 1,100,000 in.-lbs. The safe unit stresses are 700 and 16,000 lb. in for concrete and steel respectively, n=15. Find the areas of steel required.

the given stresses (Table VII), k=.397 and j= .868. Formula (41) gives 115. labor of computation may be materially lessened by the use of tables, which may be made in several ways, of which the following seem most convenient for use.

Table XI. Transposing the terms of formula (43) we have in which is the ratio of steel for a beam with the same unit stresses and without compression steel.

Formula (45) may be put in the form Values of R, pi and f's, in terms of various values of f„ and for n=15, are given in Table _XI. This table may be used to find the areas of steel required when a beam of given dimensions must carry a bending moment too great to be resisted by tension reinforce ment only.

Table XII. Combining (41), (42) and (45) we have (1=ci from which 211 . . . . (4S) in which G is constant for definite values of unit stresses and steel ratios. In Table XII, values of p' and p are given directly for various values of and G when a =15 and To use this table in design, it is only necessary to find G by dividing the bending moment lI by for the proposed beam and take the required ratios of steel directly from the table.

Tables XIII and XIV. If Formulas (39) and (44) be combined a value for fs,'f in terms of p, p' and d' Id may be found. These values are tabulated in Table XIII.

If the values of from (40) be substituted in (48), it becomes Combining the above value of N with (39) we find that the value of N depends upon n, p' and d' d. In Table XIV, values of N are tabulated for various values of f, p' and d', cl when n=15.

These tables may be used in the investigation of beams of known dimensions and reinforcement, for the purpose of finding the safe resisting moment, or the unit stresses under given bending moment.

use of these tables will be best illustrated by a few examples.

15. Solve Problem 14 (p. 187) by the use of the tables. n=15 and fs=16,000 lb./in.2, Table XII may be used. d'/d = .09, G =11Z/bd' = 1100000 =190. From Table 12X22 X22 XII, with f=700, G=190 and d'/d=.09, we find directly that p = .0136 and p' = .0093, from which, and 16. A rectangular beam has the following dimensions; b=10 inches, d= 18 inches, d' = 1.5 inches, and is to carry a bending moment of 550,000 in.-lb. The safe unit stresses are 600 and 14,000 for concrete and steel respectively. n=15. Find the areas of steel required.

= .083. From Table XI for L=14,000, f = 600, and d' Id = .083, we find R= 102, p' = .008-1, = 7090 lb.,' in? substituting these values in (47) there results A' =1.88 ? and (46) 14000 17. A rectangular beam in which b= 10 inches, d=22 inches and d'=2 inches, is reinforced with 2.6 of steel in tension and the same amount in compression. The beam carries a bending moment of 850,000 in., lb. What are the maximum unit stresses upon the steel and concrete respectively? Solution.- 0118. For these values Table ZIII gives f,, A = 26.5 and Table XIV, N=280. Thcn formula (49) 18. A rectangular beam has b=10 inches, d=16 inches, inches, A'=2.4 A =2.25 If the safe unit stresses on steel and concrete are 16000 and 650 2 respectively, what is the safe resisting moment for the beam? Solution.—p'= p=iOXlG =.014, From Table ZIII, for these values 23.0 and Table XIV, N=306. If L=16,000, which is greater than is allowable. The safe moment will therefore be that which produces a stress of 650 lb./in.2 in the concrete. Substituting in