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6 Balanced Sections, Tension-Controlled Sections, and Compression-Controlled or Brittle Sections

6 Balanced Sections, Tension-Controlled Sections, and Compression-Controlled or Brittle Sections

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C H A P T E R 3 Strength Analysis of Beams According to ACI Code



the structures of which they are part, and so on. The code (9.3) prescribes φ values or strength

reduction factors for most situations. Among these values are the following:

0.90 for tension-controlled beams and slabs

0.75 for shear and torsion in beams

0.65 or 0.75 for columns

0.65 or 0.75 to 0.9 for columns supporting very small axial loads

0.65 for bearing on concrete

The sizes of these factors are rather good indications of our knowledge of the subject in

question. For instance, calculated nominal moment capacities in reinforced concrete members

seem to be quite accurate, whereas computed bearing capacities are more questionable.

For ductile or tension-controlled beams and slabs where t ≥ 0.005, the value of φ for

bending is 0.90. Should t be less than 0.005, it is still possible to use the sections if t is

not less than certain values. This situation is shown in Figure 3.5, which is similar to Figure

R.9.3.2 in the ACI Commentary to the 2011 code.

Members subject to axial loads equal to or less than 0.10fc Ag may be used only when

is

no

lower than 0.004 (ACI Section 10.3.5). An important implication of this limit is that

t

reinforced concrete beams must have a tension strain of at least 0.004. Should the members be

subject to axial loads ≥ 0.10fc Ag, then t is not limited. When t values fall between 0.002 and

0.005, they are said to be in the transition range between tension-controlled and compressioncontrolled sections. In this range, φ values will fall between 0.65 or 0.70 and 0.90, as shown

in Figure 3.5. When t ≤ 0.002, the member is compression controlled, and the column φ

factors apply.

The procedure for determining φ values in the transition range is described later in this

section. You must clearly understand that the use of flexural members in this range is usually

uneconomical, and it is probably better, if the situation permits, to increase member depths

and/or decrease steel percentages until t is equal to or larger than 0.005. If this is done, not

only will φ values equal 0.9 but also steel percentages will not be so large as to cause crowding

of reinforcing bars. The net result will be slightly larger concrete sections, with consequent

smaller deflections. Furthermore, as you will learn in subsequent chapters, the bond of the

reinforcing to the concrete will be increased as compared to cases where higher percentages

of steel are used.



φ



(²t – 0.002) 150

3



0.90



Spiral 2011 code

0.75

0.65



(²t – 0.002) 250

3

lower bound on ²t for

members with factored axial

compressive load < 0.10 f 'c Ag



other

compression

controlled



transition



²t = 0.002

c⎢dt = 0.600



tension controlled



²t = 0.004 ²t = 0.005

c⎢dt = 3⎢7 c⎢dt = 0.375



F I G U R E 3.5 Variation of φ with net tensile strain

Grade 60 reinforcement and for prestressing steel.



t



and c/dt for



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3.7 Strength Reduction or φ Factors



We have computed values of steel percentages for different grades of concrete and steel

for which t will exactly equal 0.005 and present them in Appendix Tables A.7 and B.7 of this

textbook. It is desirable, under ordinary conditions, to design beams with steel percentages that

are no larger than these values, and we have shown them as suggested maximum percentages

to be used.

The horizontal axis of Figure 3.5 gives values also for c/dt ratios. If c/dt for a particular

flexural member is ≤ 0.375, the beam will be ductile, and if it is > 0.600, it will be brittle. In

between is the transition range. You may prefer to compute c/dt for a particular beam to check

its ductility rather than computing ρ or t . In the transition region, interpolation to determine

φ using c/dt instead of t , when 0.375 < c/dt < 0.600, can be performed using the equations

φ = 0.75 + 0.15



1

5



c/dt

3



for spiral members



φ = 0.65 + 0.25



1

5



c/dt

3



for other members



The equations for φ here and in Figure 3.5 are for the special case where fy = 60 ksi and for

prestressed concrete. For other cases, replace 0.002 with y = fy /Es . Figure 10.25 in Chapter

10 shows Figure 3.5 for the general case, where y is not assumed to be 0.002.

The resulting general equations in the range y < t < 0.005 are

0.15

(0.005 −



y)



0.25

(0.005 −



y)



φ = 0.75 + ( t −



y)



φ = 0.65 + ( t −



y)



and



for spiral members



for other members



The impact of the variable φ factor on moment capacity is shown in Figure 3.6. The two

curves show the moment capacity with and without the application of the φ factor. Point A

corresponds to a tensile strain, t , of 0.005 and ρ = 0.0181 (Appendix A, Table A.7). This

is the largest value of ρ for φ = 0.9. Above this value of ρ, φ decreases to as low as 0.65

as shown by point B, which corresponds to t of y . ACI 10.3.5 requires t not be less than

0.004 for flexural members with compressive axial loads less than 0.10 fm Ag. This situation

corresponds to point C in Figure 3.6. The only allowable range for ρ is below point C. From

the figure, it is clear that little moment capacity is gained in adding steel area above point A.

The variable φ factor provisions essentially permit a constant value of φMn when t is less



maximum ρ



0.4

0.35



M⎢f'cbd2



0.3

0.25



Mn⎢f'cbd2

φMn⎢f'cbd2



0.2



A



C



B



0.15

0.1

fc′ = 4000 psi

fy = 60,000 psi



0.05

0

0



0.005



0.01



0.015



0.02



ρ

F I G U R E 3.6 Moment capacity versus ρ.



0.025



0.03



0.035



73



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C H A P T E R 3 Strength Analysis of Beams According to ACI Code



than 0.005. It is important for the designer to know this because often actual bar selections

result in more steel area than theoretically required. If the slope between points A and C were

negative, the designer could not use a larger area. Knowing the slope is slightly positive, the

designer can use the larger bar area with confidence that the design capacity is not reduced.

For values of fy of 75 ksi and higher, the slope between point A and B in Figure 3.6

is actually negative. It is therefore especially important, when using high-strength reinforcing

steel, to verify your final design to be sure the bars you have selected do not result in a moment

capacity less than the design value.

Continuing our consideration of Figure 3.5, you can see that when t is less than 0.005,

the values of φ will vary along a straight line from their 0.90 value for ductile sections to 0.65

at balanced conditions where t is 0.002. Later you will learn that φ can equal 0.75 rather than

0.65 at this latter strain situation if spirally reinforced sections are being considered.



3.8



Minimum Percentage of Steel



A brief discussion of the modes of failure that occur for various reinforced beams was presented in Section 3.6. Sometimes, because of architectural or functional requirements, beam

dimensions are selected that are much larger than are required for bending alone. Such members

theoretically require very small amounts of reinforcing.

Actually, another mode of failure can occur in very lightly reinforced beams. If the

ultimate resisting moment of the section is less than its cracking moment, the section will fail

immediately when a crack occurs. This type of failure may occur without warning. To prevent

such a possibility, the ACI (10.5.1) specifies a certain minimum amount of reinforcing that

must be used at every section of flexural members where tensile reinforcing is required by

analysis, whether for positive or negative moments. In the following equations, bw represents

the web width of beams.

As min =



3 fc

b d

fy w



nor less than



200bw d

fy



In SI units, these expressions are



fc

4fy



(ACI Equation 10-3)



bw d and



1.4bw d

fy



, respectively.



The (200bw d )/ fy value was obtained by calculating the cracking moment of a plain

concrete section and equating it to the strength of a reinforced concrete section of the same

size, applying a safety factor of 2.5 and solving for the steel required. It has been found,

however, that when fc is higher than about 5000 psi, this value may not be sufficient. Thus, the

3 fc /fy bw d value is also required to be met, and it will actually control when fc is greater

than 4440 psi.

This ACI equation (10-3) for the minimum amount of flexural reinforcing can be written

as a percentage, as follows:

ρmin for flexure =



3 fc

200



fy

fy



Values of ρ min for flexure have been calculated by the authors and are shown for several grades of

concrete and steel in Appendix A, Table A.7 of this text. They are also included in Tables A.8 to

A.13. (For SI units, the appropriate tables are in Appendix B, Tables B.7 to B.9.)



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Courtesy of EFCO Corp.



3.9 Balanced Steel Percentage



Wastewater treatment plant, Fountain Hills, Arizona.



Section 10.5.3 of the code states that the preceding minimums do not have to be met if

the area of the tensile reinforcing furnished at every section is at least one-third greater than the

area required by moment. Furthermore, ACI Section 10.5.4 states that for slabs and footings

of uniform thickness, the minimum area of tensile reinforcing in the direction of the span is

that specified in ACI Section 7.12 for shrinkage and temperature steel which is much lower.

When slabs are overloaded in certain areas, there is a tendency for those loads to be distributed

laterally to other parts of the slab, thus substantially reducing the chances of sudden failure.

This explains why a reduction of the minimum reinforcing percentage is permitted in slabs of

uniform thickness. Supported slabs, such as slabs on grade, are not considered to be structural

slabs in this section unless they transmit vertical loads from other parts of the structure to the

underlying soil.



3.9



Balanced Steel Percentage



In this section, an expression is derived for ρ b , the percentage of steel required for a balanced

design. At ultimate load for such a beam, the concrete will theoretically fail (at a strain of

0.00300), and the steel will simultaneously yield (see Figure 3.7).

The neutral axis is located by the triangular strain relationships that follow, noting that

Es = 29 × 106 psi for the reinforcing bars:

c

0.00300

0.00300

=

=

d

0.00300 + (fy /Es )

0.003 + (fy /29 × 106 psi)

This expression is rearranged and simplified, giving

c=



87,000

d

87,000 + fy



75



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C H A P T E R 3 Strength Analysis of Beams According to ACI Code



0.003 in./in.



F I G U R E 3.7 Balanced conditions.



In Section 3.4 of this chapter, an expression was derived for depth of the compression

stress block, a, by equating the values of C and T. This value can be converted to the neutral

axis depth, c, by dividing it by β 1 :

a=



ρfy d



0.85fc

ρfy d

a

c=

=

β1

0.85β1 fc



Two expressions are now available for c, and they are equated to each other and solved for

the percentage of steel. This is the balanced percentage, ρ b :

ρfy d

0.85β1 fc



=



87,000

d

87,000 + fy



ρb =



or in SI units



0.85β1 fc

fy



0.85β1 fc

fy



87,000

87,000 + fy



600

600 + fy



Values of ρ b can easily be calculated for different values of fc and fy and tabulated for U.S.

customary units as shown in Appendix A, Table A.7. For SI units, it’s Appendix B, Table B.7.

Previous codes (1963–1999) limited flexural members to 75% of the balanced steel

ratio, ρ b . However, this approach was changed in the 2002 code to the new philosophy

explained in Section 3.7, whereby the member capacity is penalized by reducing the φ factor

when the strain in the reinforcing steel at ultimate is less than 0.005.



3.10



Example Problems



Examples 3.2 to 3.4 present the computation of the design moment capacities of three beams

using the ACI Code limitations. Remember that, according to the code (10.3.5), beams whose

axial load is less than 0.10fc Ay may not, when loaded to their nominal strengths, have net

tensile calculated strains less than 0.004.



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3.10 Example Problems



Example 3.2

Determine the ACI design moment capacity, φMn, of the beam shown in Figure 3.8 if fc = 4000 psi

and fy = 60,000 psi.

SOLUTION

Checking Steel Percentage

ρ=



4.00 in.2

As

=

= 0.0111

bd

(15 in.) (24 in.)



> ρmin = 0.0033

< ρmax = 0.0181

a=



As fy

0.85fc b



=



both from

Appendix A, Table A.7



(4.00 in.2 ) (60,000 psi)

= 4.71 in.

(0.85)(4000 psi) (15 in.)



β1 = 0.85 for 4000 psi concrete

c=



4.71 in.

a

=

= 5.54 in.

β1

0.85



Drawing Strain Diagram (Figure 3.9)

t



=



d−c

18.46 in.

(0.003) =

(0.003) = 0.0100

c

5.54 in.



> 0.005

∴ tension controlled

a

4.71 in.

= (4.00 in.2 ) (60 ksi) 24 in. −

Mn = As fy d −

2

2

= 5194.8 in-k = 432.9 ft-k

φMn = (0.9) (432.9 ft-k) = 389.6 ft-k



²c = 0.003

c = 5.54 in.

d = 24 in.



24 in.

27 in.



4 #9 bars

(4.00 in.2)



d – c = 18.46 in.



3 in.



F I G U R E 3.9 Neutral axis location for



Example 3.2.

15 in.

F I G U R E 3.8 Beam cross section for



Example 3.2.



77



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C H A P T E R 3 Strength Analysis of Beams According to ACI Code



Example 3.3

Determine the ACI design moment capacity, φMn , of the beam shown in Figure 3.10 if

fc = 4000 psi and fy = 60,000 psi.

SOLUTION

Checking Steel Percentage

ρ=



As

4.68 in.2

=

= 0.026 > ρmin = 0.0033

bd

(12 in.) (15 in.)



> ρ max = 0.0181 (from Appendix A, Table A.7). As a result, we know that

Computing Value of



t



will be < 0.005.



t



a=



As fy

0.85fc b



=



(4.68 in.2 ) (60,000 psi)

= 6.88 in.

(0.85) (4000 psi) (12 in.)



β1 = 0.85 for 4000 psi concrete

c=

t



=



a

6.88 in.

= 8.09 in.

=

β1

0.85

15 in. − 8.09 in.

d−c

(0.003) =

(0.003)

c

8.09 in.



= 0.00256 < 0.004

∴ Section is not ductile and may not be used as per ACI Section 10.3.5.



15 in.



3 #11 bars

(4.68 in.2)



18 in.



3 in.



12 in.



F I G U R E 3.10 Beam cross section for Example 3.3.



Example 3.4

Determine the ACI design moment capacity, φMn, for the beam of Figure 3.11 if fc = 4000 psi

and fy = 60,000 psi.

SOLUTION

Checking Steel Percentage

ρ=



As

3.00 in.2

=

= 0.020 > ρmin = 0.0033

bd

(10 in.) (15 in.)

but also < ρmax = 0.0181 (for t = 0.005)



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3.11 Computer Examples



Computing Value of



t



a=



As fy

0.85fc b



=



(3.00 in.2 ) (60,000 psi)

= 5.29 in.

(0.85) (4000 psi) (10 in.)



β1 = 0.85 for 4000 psi concrete

c=



t



=



5.29 in.

a

= 6.22 in.

=

β1

0.85



d−c

(0.003) =

c



15 in. − 6.22 in.

(0.003) = 0.00423 > 0.004 and < 0.005

6.22 in.

∴ Beam is in transition zone and



φ (from Figure 3.5) = 0.65 + (0.00423 − 0.002)

Mn = As fy d −



a

5.29 in.

= (3.00 in.2 ) (60 ksi) 15 in. −

2

2



250

3



= 0.836



= 2223.9 in-k = 185.3 ft-k



φMn = (0.836) (185.3 ft-k) = 154.9 ft-k



15 in.



3 #9 bars

(3.00 in.2)



18 in.



3 in.



10 in.



3.11



F I G U R E 3.11 Beam cross section for Example 3.4.



Computer Examples



Example 3.5

Repeat Example 3.2 using the Excel spreadsheet provided for Chapter 3.

SOLUTION

Open the Chapter 3 spreadsheet, and open the Rectangular Beam worksheet. Enter values only

in the cells highlighted yellow (only in the Excel spreadsheet, not the printed example). The final

result is φMn = 389.6 ft-k (same answer as Example 3.2).



79



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80



C H A P T E R 3 Strength Analysis of Beams According to ACI Code



Example 3.6

Repeat Example 3.3 using the Excel spreadsheet provided for Chapter 3.

SOLUTION

Open the Chapter 3 spreadsheet and the Rectangular Beam worksheet. Enter values only in

the cells highlighted yellow. The spreadsheet displays a message, ‘‘code violation . . . too much

steel.’’ This is an indication that the beam violates ACI Section 10.3.5 and is not ductile. This

beam is not allowed by the ACI Code.



Example 3.7

Repeat Example 3.4 using the Excel spreadsheet provided for Chapter 3.

SOLUTION

Open the Chapter 3 spreadsheet and the Rectangular Beam worksheet. Enter values only in the

cells highlighted yellow. The final result is φMn = 154.5 ft-k (nearly the same answer as Example

3.4). The φ factor is also nearly the same as Example 3.4 (0.0834 compared with 0.0836). The

difference is the result of the spreadsheet using the more general value for y of fy /Es = 0.00207

instead of the approximate value of 0.002 permitted by the code for Grade 60 reinforcing steel.

A difference of this magnitude is not important, as discussed in Section 1.25, ‘‘Calculation

Accuracy.’’



PROBLEMS

Problem 3.1 What are the advantages of the strength design

method as compared to the allowable stress or alternate design

method?



For Problems 3.7 to 3.9, determine the values of

for the sections shown.

Problem 3.7 (Ans. φMn = 379.1 ft-k)



Problem 3.2 What is the purpose of strength reduction

factors? Why are they smaller for columns than for beams?

Problem 3.3 What are the basic assumptions of the strength

design theory?

Problem 3.4 Why does the ACI Code specify that a certain

minimum percentage of reinforcing be used in beams?



24 in.



3 in.



Problem 3.5 Distinguish between tension-controlled and

compression-controlled beams.

Problem 3.6 Explain the purpose of the minimum cover

requirements for reinforcing specified by the ACI Code.



27 in.



4 #9 bars



12 in.



fy = 60,000 psi

f c' = 4,000 psi



t,



φ, and φMn



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Problems



Problem 3.8



Problem 3.10



12 in.

4 #10 bars

18 in.



21 in.



3 in.



3 #11 bars

18 in.



3 in.

14 in.



Problem 3.9 (Ans.

φMn = 1320.7 ft-k)



t



fy = 75,000 psi

f 'c = 5,000 psi



= 0.00408, φ = 0.797,



7 #11 bars



27 in.



30 in.



3 in.

20 in.



fy = 80,000 psi

f 'c = 6,000 psi



15 in.



fy = 60,000 psi

f c' = 4,000 psi



81



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Design of Rectangular Beams

and One-Way Slabs



C H A PT E R 4



4.1



Load Factors



Load factors are numbers, almost always larger than 1.0, that are used to increase the estimated

loads applied to structures. They are used for loads applied to all types of members, not just

beams and slabs. The loads are increased to attempt to account for the uncertainties involved

in estimating their magnitudes. How close can you estimate the largest wind or seismic loads

that will ever be applied to the building that you are now occupying? How much uncertainty

is present in your answer?

You should note that the load factors for dead loads are much smaller than the ones used

for live and environmental loads. Obviously, the reason is that we can estimate the magnitudes

of dead loads much more accurately than we can the magnitudes of those other loads. In this

regard, you will notice that the magnitudes of loads that remain in place for long periods of

time are much less variable than are those loads applied for brief periods, such as wind and

snow.

Section 9.2 of the code presents the load factors and combinations that are to be used for

reinforced concrete design. The required strength, U, or the load-carrying ability of a particular

reinforced concrete member, must at least equal the largest value obtained by substituting

into ACI Equations 9-1 to 9-7. The following equations conform to the requirements of the

International Building Code (IBC)1 as well as to the values required by ASCE/SEI 7-10.2

U = 1.4D

U = 1.2D + 1.6L + 0.5(Lr or S or R)

U = 1.2D + 1.6(Lr or S or R) + (L or 0.5W )

U

U

U

U



= 1.2D

= 1.2D

= 0.9D

= 0.9D



+

+

+

+



1.0W + L + 0.5(Lr or S or R)

1.0E + L + 0.2S

1.0W

1.0E



(ACI Equation 9-1)

(ACI Equation 9-2)

(ACI Equation 9-3)

(ACI

(ACI

(ACI

(ACI



Equation

Equation

Equation

Equation



9-4)

9-5)

9-6)

9-7)



In the preceding expressions, the following values are used:

U = the design or ultimate load the structure needs to be able to resist

D = dead load

L = live load



1 International



Code Council, 2012, International Building Code, Falls Church, Virginia 22041-3401.

Society of Civil Engineers, Minimum Design Loads for Buildings and Other Structures, ASCE 7-10 (Reston, VA:

American Society of Civil Engineers), p. 7.



2 American



82



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