978-0073397948 Chapter 3

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

Ec

Part a The solution is identical for grade 40 and grade 60 reinforcement

Part b

fy

fy1

Problem 3.1

Part c

Pu

Comments

1. There is no difference at fc = 1200 psi and elastic assumptions are used

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″

Ec

Part a The solution is identical for grade 40 and grade 60 reinforcement

P0.246The steel carries 25 percent of the load

Part b

fy

fy1

2/2

Problem 3.2

Part c

Pu

Comments

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o

f

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

0.00%

10.00%

20.00%

30.00%

40.00%

50.00%

60.00%

0.0000 0.0005 0.0010 0.0015 0.0020 0.0025 0.0030 0.0035

Pc/Ptotal

Ps/Ptotal

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

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.

PsfcnAst

 Ps105 kip 100

P13.7

Part (b): Compute the capacity of the column when the steel begins to yield εy

fy

Es



εy0.00207or 2/10 of one percent

2/2

Part (c): Compute the maximum load capacity of the section if loaded slowly

Part (d): If we reexamine the problem with a fast loading, then the concrete stress would be

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



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

PsfcnAst

 Ps91 kip 100

P14.6

Part (b): Compute the capacity of the column when the steel begins to yield εy

fy

Es



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

PsfsAst

 Ps570 kip 100

P27.5

Part (d): If we reexamine the problem with a fast loading as would occur in a building, then the

concrete stress would be

Note: the total increase is in the concrte contribution. 100

P24.3

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



3/3

Comments