1/2. E c Part a The solution is identical for grade 40 and grade 60 reinforcement. P s f c n A s lbf. The steel carries 13.3 percent of the load

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1 1/ A in. column is made of the same concrete and reinforced with the same six No. 9 (No. 29) bars as the column in Examples 3.1 and 3.2, except t hat a steel with yield strength f y = 40 ksi is used. The stress-strain diagram of this reinforcing steel is shown in Fig for fy = 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. A s 6.0in 2 16" A g 16in20in 320in 2 A c A g A s 314in 2 f' c 4000psi f y 40000psi f y psi 20" E c psi E s psi E s n 8.1 E c art a The solution is identical for grade 40 and grade 60 reinforcement f c 1200psi f c A c na s lbf s f c na s lbf s art b The steel carries 13.3 percent of the load f y f y1 ε y ε E y s E s For slow loading f c 3000psi f c1 3300psi s A c f c A s f y s A s f y lbf lbf 1 s1 A c f c1 A s f y1 s A s f y lbf lbf

2 2/2 roblem 3.1 art c f c 3400psi u s A c f c A s f y s u A s f y lbf lbf u1 s1 A c f c A s f y1 s u A s f y lbf lbf Comments 1. There is no difference at fc = 1200 psi and elastic assumptions are used 2. As the strain increases, the steel with fy = 60,000 psi contributes more to the total load and the column has a higher total capacity 3. Grade 40 and Grade 40 have the same cost, therefore Grade 60 provides a 9% increase in capacity for no increase in cost.

3 1/2 3.2 The area of steel, expressed as a percentage of gross concrete area, for the column of roblem 3.1 is lower than would often be used in practice. Recalculate the comparisons of roblem 3.1, using f y of 40 ksi and 60 ksi as before, but for a in. column reinforced with eight No. 11 (No. 36) bars. Compare your results with those of roblem " A s in 2 A s 8A s in 2 20" 4 No. 11 (No. 36) A s =12.48in 2 A g 16in20in 320in 2 A c A g A s in 2 4 No. 11 (No. 36) f' c 4000psi f y 40000psi f y psi E c psi E s psi E s n 8.1 E c art a The solution is identical for grade 40 and grade 60 reinforcement f c 1200psi f c A c na s lbf s f c na s lbf s art b The steel carries 25 percent of the load f y f y1 ε y ε E y s E s For slow loading f c 3000psi f c1 3300psi s A c f c A s f y s A s f y lbf lbf 1 s1 A c f c1 A s f y1 s A s f y lbf lbf

4 2/2 roblem 3.2 art c f c 3400psi u s A c f c A s f y s u both cases A s f y lbf lbf u1 s1 A c f c A s f y1 s u A s f y lbf lbf Comments 1. There is no difference at fc = 1200 psi and elastic assumptions are used 2. There is a 16% capacity increase at nominal using Grade 60 reinforcement 3. The higher steel ratio produces a higher overall capacity compared to problem 3.1

5 3.3. A square concrete column with dimensions in. is reinforced with a total of eight No. 10 (No. 32) bars arranged uniformly around the column perimeter. Material strengths are f y = 60 ksi and f c = 4000 psi, with stressstrain curves as given by curves a and c of Fig Calculate the percentages of total load carried by the concrete and by the steel as load is gradually increased from 0 to failure, which is assumed to occur when the concrete strain reaches a limit value of Determine the loads at strain increments of up to the failure strain, and graph your results, plotting load percentages vs. strain. The modular ratio may be assumed at n = 8 for these materials. Using Concrete data from Figure 3.3 A s = in 2 A c = 474 in 3 fy = f' c = psi 4000 psi Strain f c (psi) c (kips) f s (psi) s (kips) total (kips) c / total s / total % 0.00% % 16.2% % 19.2% % 23.1% % 27.3% % 27.4% % 27.4% 90.00% 80.00% 70.00% 60.00% 50.00% 40.00% 30.00% 20.00% 10.00% 0.00% c/total s/total

6 1/ A in. column is made of the same concrete as used in Examples 3.1 and 3.2. It is reinforced with six No. 11 (No. 36) bars with f y = 60 ksi. For this column section, determine ( a ) the axial load that the section will carry at a concrete stress of 1400 psi; ( b ) the load on the section when the steel begins to yield; ( c ) the maximum load if the section is loaded slowly; and ( d ) the maximum load if the section is loaded rapidly. The area of one No. 11 (No. 36) bar is 1.56 in 2. Determine the percent of the load carried by the steel and the concrete for each combination. Reinforcement Areas Given roperties f' c 4000psi f y 60000psi f c 1400psi n 8 E s psi Column roperties b 20in t 24in A g bt 480 in 2 A st 6A s in 2 art (a) Compute the axial capacity of the section loaded below the elastic limit. Solution: The axial capacity is based on the gross area of the column plus the effective area of the steel. Since we count the holes where the steel is removed, the additional effective area of the steel is (n-1)a st. A c A g A st A g 480in 2 A st 9.36in 2 A c 471in 2 f c A g ( n 1) A st 764kip c f c A g A st c 659kip Concrete and steel contribution c s f c na st s 105kip s f y art (b): Compute the capacity of the column when the steel begins to yield ε y E s ε y or 2/10 of one percent Examining Figure 3.3, we are beyond the elastic portion of the concrete stress strain curve, but we are at the elastic limit of the steel. f s ε y E s f s 60000psi From Figure 3.3 f c 3100psi Since the problem is nonlinear, we must break out the concrete and steel areas. We can no longer use the elastic equation. for slow loading

7 2/2 f c A c f s A st 2021kip c c f c A c c 1459kip s s f s A st s 562kip art (c): Compute the maximum load capacity of the section if loaded slowly Examining Figure 3.3, we are beyond the elastic portion of the concrete stress strain curve and we are in the plastic range of the steel. f s f y f s 60000psi From Figure 3.3 f c 3400psi Since the problem is nonlinear, we must break out the concrete and steel areas. We can no longer use the elastic equation from 1.1. for slow loading f c A c f s A st 2162kip c s f c A c f s A st c s 1600kip c s 562kip art (d): If we reexamine the problem with a fast loading, then the concrete stress would be f c 4000psi c s f c A c f c A c f s A st f s A st 2444kip c c 1883kip s 562kip s As the concrete becomes non-linear, the steel picks up more load, but after the steel yields, the load goes to the conrete. 2. The slow loading is approximately 88% of the fast load scenario - This is slightly higher than the 0.85 given in eq. 3.6.

8 1/3 3.5 A 24 in. diameter column is made of the same concrete as used in Examples 3.1 and 3.2. The area of reinforcement equals 2.1 percent of the gross cross section (that is, A s = A g ) and f y = 60 ksi. For this column section, determine ( a ) the axial load the section will carry at a concrete stress of 1200 psi; ( b ) the load on the section when the steel begins to yield; ( c ) the maximum load if the section is loaded slowly; ( d ) the maximum load if the section is loaded rapidly; and ( e ) the maximum load if the reinforcement in the column is raised to 6.5 percent of the gross cross section and the column is loaded slowly. Comment on your answer, especially the percent of the load carried by the steel and the concrete for each combination. Reinforcement roperties Given roperties f' c 4000psi f y 60000psi f c 1200psi n 8 E s psi Column roperties d 24in A g π d2 ρ A st ρa g is the reinforcement ratio or the fraction of the section that is steel The total area of steel A st is A st 9.5in 2 art (a) Compute the axial capacity of the section loaded below the elastic limit. Solution: The axial capacity is based on the gross area of the column plus the effective area of the steel. Since we count the holes where the steel is removed, the additional effective area of the steel is (n-1)a st. A c A g A st A g 452in 2 A st 9.50in 2 A c 443in 2 f c A g ( n 1) A st 623kip c f c A g A st c 531kip Concrete and steel contribution c s f c na st s 91kip s f y art (b): Compute the capacity of the column when the steel begins to yield ε y E s ε y or 2/10 of one percent Examining Figure 1.16, we are beyond the elastic portion of the concrete stress strain curve, and we are at the elastic limit of the steel. ε y E s f s 60000psi From Figure 1.16 f c 3100psi f s for slow loading

9 2/3 Since the problem is nonlinear, we must break out the concrete and steel areas. We can no longer use the elastic equation from 1.1. f c A c f s A st 1943kip c c f c A c c 1373kip s s f s A st s 570kip art (c): Compute the maximum load capacity of the section if loaded slowly Examining Figure 1.16, we are beyond the elastic portion of the concrete stress strain curve and we are in the plastic range of the steel. f s f y f s 60000psi From Figure 3.3 f c 3400psi Since the problem is nonlinear, we must break out the concrete and steel areas. We can no longer use the elastic equation from 1.1. for slow loading f c A c f s A st 2076kip c f c A c c c 1506kip s f s A st s s 570kip art (d): If we reexamine the problem with a fast loading as would occur in a building, then the concrete stress would be f c 4000psi f c A c f s A st 2342kip c f c A c c 1772kip c s f s A st s 570kip s Note: the total increase is in the concrte contribution art (e): Determine the capacity for a slow loaded column with the steel changed to 6.5% A st 0.065A g A st 29.4in 2 f s f y f s 60000psi From Figure 1.16 f c 3400psi for slow loading f c A c f s A st 3270kip c f c A c c 1506kip c s s f s A st s 1764kip

10 3/3 Comments 1. As the concrete becomes non-linear, the steel picks up more load, but after the steel yields, the load goes to the conrete. 2. The slow loading is approximately 88% of the fast load scenario - This is slightly higher than the 85 percent given earlier.