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
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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
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76
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.
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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).
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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
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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
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