Download Solution Manual for Design of Concrete Structures 15th Edition by Darwin Dolan Nilson

Solution Manual for Design of Concrete Structures 15th Edition by Darwin Dolan Nilson Link full download: http://testbankcollection.com/download/solution-manual-fordesign-of-concrete-structures-15th-edition-by-darwin-dolan-nilson/ 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.

16 "

As := 6.0in2

Ac := Ag - As = 314 in2 f'c 4000psi

fy 40000psi fy1 60000psi

Ec Es 3600000psi 29000000psi Es n 8.1 E c Par The solution is identical for grade 40 and grade 60 t a reinforcement fc = 1200psi P = fc Ac = n As434800 lbf Ps = fc n As =58000 lbf P s 0.133 The steel carries 13.3 percent of the load P

20 "

Ag := 16in .20in = 320 in2

Part b f

ε

y 1

fy

y

0.00138

Es For slow loading fc 3000psi P = Ac fc + As fy 1182000 lbf Ps = As fy = 240000 lbf P s

P 0.203

εy1

0.00207 Es

fc1 =3300psi P1 = Ac fc1 + As fy1 =1396200 lbf Ps1 = As fy1 = 360000 lbf P s 1 0.258 P

1

Problem 3.1 Part c fc =3400psi Pu =Ac fc +As fy =1307600 lbf Ps =As fy =240000 lbf Ps =0.184 Pu

Ps1 =As fy1= Ps1 =0.275 Pu

360000 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.

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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. 16" As11 =1.56in2 As =8 As11 =12.48 in

320 in2

Ac =Ag -As =307.52 in2

20 "

Ag =16in 20in=

4‐No. 11 (No. 36) As=12.48in2

2

4‐No. 11 (No. 36)

f'c = fy = fy1= 4000psi 40000psi 60000psi Ec =3600000 Es =29000000 psi psi Es n 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 P s The steel carries 25 =0.246 percent of the load P Part b f y fy 1 ε y = 0.00138 εy1= 0.00207 Es Es

For slow loading fc = 3000psi P = Ac fc + As fy =1421760lbf

fc1 = 3300psi P1 = Ac fc1 + As fy1 =1763616 lbf

Ps = As fy = 499200 lbf

Ps1 = As fy1 = 748800 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 in2 Ac = 474 in3 fy = 60000 psi f'c = 4000 psi Strain fc fs Ps (psi) Pc (kips) (psi) (kips) 0.000 0 0 0 0 0.005 1600 758 14500 147 0.001 2600 1232 29000 293 0.0015 3100 1469 43500 440 0.0020 3300 1564 58000 587 0.0025 3400 1612 60000 607 0.0030 3400 1612 60000 607

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Ptotal Pc/Pto Ps/Pto (kips) tal tal 0 0.00% 0.00% 905 83.8% 16.2% 1526 80.8% 19.2% 1910 76.9% 23.1% 2151 72.7% 27.3% 2219 72.6% 27.4% 2219 72.6% 27.4%

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 = bt

=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 480 Ast 9.36 Ac 471 Ag-Ast in2 in2 in2 P 764 Concrete and P = fc[ Ag + ( n 1) Ast ] kip steel contribution Pc Pc = 659 P f A A c = c ( g - st) kip 100 86.3 P Ps Ps 105 100 Ps = fc n Ast kip 13.7 P Part (b): Compute the capacity of the column when the steel fy

begins to yield εy or 2/10 of one εy 0.00207 percent

Es

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 From Figure 3.3 Since the problem is nonlinear, we must break out the concrete and steel areas. We can no longer use the elastic equation.

fs = 60000 psi fc = 3100psi for slow loading

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P = fc Ac + fs Ast

P = 2021 kip 10 Pc 0 P 32.6

P =f A c c c

Pc = 1459 kip 10 0

P =f A s s st

Ps = 562 kip

Ps 27.8 P

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 = f y

fs = 60000 psi fc = 3400psi for slow loading From Figure 3.3 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. P= fc Ac+ Pc =fc Ac

fs Ast

Ps =fs Ast

P = 2162 kip Pc = 1600 kip

P562 kip s

10 Pc 0 P 74 P 10 s 0 26 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 P2444 kip Pc = fc Ac

Pc

Ps = fs Ast

P =1883 kip c Ps = 562 kip

10 0

77

P 10 Ps 0 P 23

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

Es= 29000000psi

fy =60000psi ρ 0.021is the reinforcement ratio or the fraction of the section that is steel d 2 The total area of steel Ast is Ast 9.5 in2

fc=4000psi Column Properties d = 24in

fc =1200psi n= 8

Aπ g 4

Ast =ρ Ag 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 = Ac = 2 2 Ag -Ast 452 in 9.50 in 443 in2 P= Concrete and P f A( n 1) A 623 kip steel c g st contribution

P

f

A

c = c Ac + g

A

st

Pc = 531 kip

Pc 100 85.4

P

Ps Ps = fc n Ast

Ps 91 kip

100

14.6 fy Part (b): Compute the capacity of the column when the steel begins to yield εy or 2/10 of one Es εy 0.00207 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. P

fs = εy Es From Figure 1.16

fs 60000 psi 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 100 Pc 27.4 the elastic equation from 1.1. P P = 100 Ps 29.3 P= fc Ac + fs Ast 1943 kip P Pc = Pc = fc Ac 1373 kip Ps = Ps= fs Ast 570 kip 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 = fs = f y 60000 psi fc for slow =3400psi loading From Figure 3.3 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. P = fc Ac + fs Ast

P =2076kip Pc = 1506kip

10 Pc Pc = fc Ac 0 72.5 P 10 Ps Ps = fs Ast Ps 570 kip 0 P 27.5 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 Pc = fc Ac Ps = fs Ast

P = 2342 kip Pc =1772 kip Ps = 570

Pc 100 P

75.7

kip Note: the total increase is in the concrte contribution.

10 0

Ps 24.3

P Part (e): Determine the capacity for a slow loaded column with the steel changed to 6.5% Ast =0.065 Ag Ast = 29.4 in2 fs =fy fs = 60000 psi From Figure 1.16 for slow P= fc Ac +fs Ast fc = 3400psi loading Pc =fc Ac P = 3270 kip Pc = 1506 Ps =fs Ast kip

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.