Solution Manual for Design of Concrete Structures 15th Edition by Darwin Dolan Nilson
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Chap 3. Design of Concrete Structures and Fundamental Assumptions 3. 1. A 16 × 20 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. 2.15 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.
As := 6.0in2 16 " Ag := 16in .20in = 320 in2 Ac := Ag - As = 314 in2 f'c 4000psi fy 40000psi fy1 60000psi Ec 3600000psi Es 29000000psi n Es 8.1 E c Par t a
The solution is identical for grade 40 and grade 60 reinforcement fc = 1200psi P = fc Ac = n As434800 lbf Ps = fc n As =58000 lbf 20 " Ps
0.133 The steel carries 13.3 percent of the load
P
Part b εy fy 0.00138 εy1 f y 1 0.00207 Es Es For slow
loading fc 3000psi fc1 =3300psi P = Ac fc + As fy 1182000 lbf P1 = Ac fc1 + As fy1 =1396200 lbf Ps = As fy = 240000 lbf Ps1 = As fy1 = 360000 lbf P s P s 1 0.258 0.203 P1 P
Problem 3.1 Part c fc =3400psi Pu =Ac fc +As fy =1307600 lbf Ps =As fy =240000 lbf Ps =0.184 Pu Comments Ps1 =As fy1= 360000 lbf Ps1 =0.275 Pu
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.
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3. 2 The area of steel, expressed as a percentage of gross concrete area, for the column of Problem 3.1 is lower than would often be used in practice. Recalculate
the comparisons of Problem 3.1, using f y of 40 ksi and 60 ksi as before, but for a 16 × 20 in. column reinforced with eight No. 11 (No. 36) bars. Compare your results with those of Problem 3.1. As11 =1.56in 2 As =8 As11 =12.48 in2 Ag =16in 20in= 320 in2 Ac =Ag -As =307.52 in2 16" 4‐No. 11 (No. 36) As=12.48in2 20 " 4‐No. 11 (No. 36) f'c = 4000psi fy = 40000psi fy1= 60000psi Ec =3600000 psi Es =29000000 psi n Es 8.1 E c
Part a The solution is identical for grade 40 and grade 60 reinforcement
fc = 1200psi
P = fc (Ac + nA)s=48966lbf Ps = fc n As = 120640 lbf Ps
=0.246
The steel carries 25 percent of the load P Part b εy = fy 0.00138 εy1= f y 1 0.00207 Es Es
For slow
loading fc = 3000psi fc1 = 3300psi P = Ac fc + As fy
=1421760lbf
P1 = Ac fc1 + As fy1 =1763616 lbf
3.3. A square concrete column with dimensions 22 × 22 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. 3.3 . 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 0.0030. Determine the loads at strain increments of 0.0005 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 As = 10.12 in 2 Ac = 474 in3 fy = 60000 psi f'c = 4000 psi Strain fc (psi) Pc (kips) fs (psi) Ps (kips) Ptotal (kips) Pc/Pto tal Ps/Pto tal 0.000 0 0 0 0 0 0.00% 0.00% 0.005 1600 758 14500 147 905 83.8% 16.2% 0.001 2600 1232 29000 293 1526 80.8% 19.2% 0.0015 3100 1469 43500 440 1910 76.9% 23.1% 0.0020 3300 1564 58000 587 2151 72.7% 27.3% 0.0025 3400 1612 60000 607 2219 72.6% 27.4% 0.0030 3400 1612 60000 607 2219 72.6% 27.4% 1/2
3.4. A 20 × 24 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 Properties f'c = 4000p si Fy = 60000ps i f c = 1400 psi N = 8 E = 29000000psi Column Properties B = 20i n T = 24in Ag = b t =480 in2 Ast =6As11= 9.36 in2
Part (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)Ast.
Ac = Ag-Ast Ag 480 in2 Ast 9.36 in2 Ac 471 in2 P = fc[ Ag + ( n 1) Ast ] P 764 kip Concrete and steel contribution P c = fc ( Ag - Ast) Pc = 659 kip Pc 100 P 86.3 Ps = fc n Ast Ps 105 kip Ps 100 P 13.7 Part (b): Compute the capacity of the column when the steel fy
begins to yield εy Es εy 0.00207 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.
fs = εy Es fs = 60000 psi
From Figure 3.3
fc = 3100psi for slow loading
Since the problem is nonlinear, we must break out the concrete and steel areas. We can no longer use the elastic equation.
2/2 P = fc Ac + fs Ast P = 2021 kip Pc P =f A 10 0 32.6 Pc = 1459 kip P c c c Ps P =f A Ps = 562 kip 10 0 27.8 P s s st
Part (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.
fs = fy fs = 60000 psi
From Figure 3.3
fc = 3400psi for slow loading
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.
P= fc Ac+ fs Ast Pc =fc Ac Ps =fs Ast P = 2162 kip Pc = 1600 kip 10 0 Pc 74 P P562 kip 10 0 P s 26 s P
Part (d): If we reexamine the problem with a fast loading, then the concrete stress would be fc = 4000psi P = fc Ac + fs Ast P 2444 kip Pc = fc Ac Pc
Ps = fs Ast P =1883 kip 100 77 P c Ps = 562 kip 10 0 Ps 23 P
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 0.85 given in eq. 3.6.
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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 = 0.021 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 Properties Given Properties fc=4000psi fy Column Properties d 2 d = 24in A π g 4 Ast =ρ Ag
fc =1200psi n= 8 Es= 29000000psi =60000psi
ρ 0.021is the reinforcement ratio or the fraction of the section that is steel The total area of steel Ast is Ast 9.5 in
2
Part (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)Ast.
Ac = Ag -Ast Ag = 452 in2 Ast = 9.50 in2 Ac = 443 in2 P f A( n 1) A P= 623 kip Concrete and steel c g st contribution
P c = fc Ac +Ag Ast Pc = 531 kip Pc 100 P 85.4 Ps = fc n Ast Ps 91 kip Ps 100 P 14.6 Part (b): Compute the capacity of the column when the steel begins to yield εy fy Es εy 0.00207 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.
fs = εy Es fs 60000 psi
From Figure 1.16 fc 3100psi for slow
loadi ng
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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.
P= fc Ac + fs Ast P = 1943 kip Pc = fc Ac Pc = 1373 kip Ps= fs Ast Ps = 570 kip 100 Pc 27.4 P 100 Ps 29.3 P
Part (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. fs = fy fs = 60000 psi From Figure 3.3 fc =3400psi for slow loading 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. P = fc Ac + fs Ast P =2076kip Pc Pc = fc Ac Pc = 1506kip 10 0 72.5 P Ps = fs Ast Ps 570 kip 10 0 Ps 27.5 P
Part (d): If we reexamine the problem with a fast loading as would occur in a building, then the
concrete stress would be
fc = 4000psi P = fc Ac + fs Ast P = 2342 kip Pc = fc Ac Pc =1772 kip Pc Ps = fs Ast 100 P 75.7 Ps = 570
kip Ps
Note: the total increase is in the concrte
contribution. 100 24.3
P
Part (e): Determine the capacity for a slow loaded column with the steel changed to 6.5% Ast =0.065 Ag fs =fy From Figure 1.16 P= fc Ac +fs Ast Pc =fc Ac Ps =fs Ast Ast = 29.4 in2 fs = 60000 psi fc = 3400psi for slow loading P = 3270 kip Pc = 1506 kip
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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.
