Structural Concrete Textbook, second edition Volume 1 (PDF) fib Bulletins N° 51. As self compacting concrete, architectural concrete, fibre reinforced concrete,. A 16 × 20 in. Column is made of the same concrete and reinforced with the same six No. 29) bars as the column in Examples 1.1 and 1.2, except that a steel with yield strength f y= 40 ksi is used. The stress-strain diagram of this reinforcing steel is shown in Figure for f y = 40 ksi. For this column determine ( a) the axial load that will stress the concrete to 1200 psi; ( b) the load at which the steel starts yielding; ( c) the maximum load; and ( d) the share of the total load carried by the reinforcement at these three stages of loading. Compare results with those calculated in the examples for f y = 60 ksi, keeping in mind, in regard to relative economy, that the price per pound for reinforcing steels with 40 and 60 ksi yield points is about the same. EXAMPLE 1.1 A column made of the materials defined in Figure has a cross section of 16 × 20 in. And is reinforced by six No. 29) bars, disposed as shown in Figure. (See Tables A.1 and A.2 of Appendix A for bar diameters and areas and Section 2.14 for a description of bar size designations.) Determine the axial load that will stress the concrete to 1200 psi. The modular ratio nmay be assumed equal to 8. (In view of the scatter inherent in E c, it is customary and satisfactory to round off the value of nto the nearest integer.) Solution.One finds A g= 16 × 20 = 320 in 2, and from Appendix A, Table A.2, two No. 29) bars provide steel area A st = 6.00 in 2 or 1.88 percent of the gross area. The load on the column, from Eq. (1.8), is P= 1200[320 + (8 – 1)6.00] = 434,000 lb. Of this total load, the concrete is seen to carry P c= f cA c = f c( A g– A st) = 1200(320 – 6) = 377,000 lb, and the steel P s= f sA st = ( nf c) A st= 9600 × 6 = 57,600 lb, which is 13.3 percent of the total axial load. EXAMPLE 1.2 One may want to calculate the magnitude of the axial load that will produce a strain or unit shortening ε c = ε s= 0.0010 in the column of Example 1.1. At this strain the steel is seen to be still elastic, so that the steel stress fs= ε sE s= 0.001 × 29,000,000 = 29,000 psi. The concrete is in the inelastic range, so that its stress cannot be directly calculated, but it can be read from the stress-strain curve for the given value of strain. If the member has been loaded at a fast rate, curve bholds at the instant when the entire load is applied. The stress for ε = 0.001 can be read as f c= 3200 psi. Consequently, the total load can be obtained from (1.9) which applies in the inelastic as well as in the elastic range. Hence, P= 3200(320 – 6) + 29,000 × 6 = 1,005,000 + 174,000 = 1,179,000 lb. Of this total load, the steel is seen to carry 174,000 lb, or 14.7 percent. For slowly applied or sustained loading, curve crepresents the behavior of the concrete. Its stress at a strain of 0.001 can be read as f c = 2400 psi. Then P= 2400 × 314 + 29,000 × 6 = 754,000 + 174,000 = 928,000 lb. Of this total load, the steel is seen to carry 18.8 percent. FIGURE Typical stress-strain curves for reinforcing bars. FIGURE Concrete and steel stress-strain curves. Matlab convert to row echelon form. – nkjt, scoa, ciruvan, Kritner, Racil Hilan If this question can be reworded to fit the rules in the, please. The users who voted to close gave this specific reason: • 'Questions asking us to recommend or find a book, tool, software library, tutorial or other off-site resource are off-topic for Stack Overflow as they tend to attract opinionated answers and spam. 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