Chapter 14. Flexural Reinforced Concrete Members
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Principles of Structural Design: Wood, Steel, and Concrete
The American Concrete Institute (ACI), which is a primary agency in the United States that prepares the national standards for structural concrete, provides the empirical relations for the modulus
of elasticity based on the compression strength, fc′.
Although the stress–strain curves have different slopes for concrete of different strengths, the
following two characteristics are common to all concretes:
1. The maximum compression strength, fc′ in all concrete is attained at a strain level of
approximately 0.002 in./in.
2.The point of rupture of all curves lies in the strain range of 0.003–0.004 in./in. Thus, it is
assumed that concrete fails at a strain level of 0.003 in./in.
Design Strength of Concrete
To understand the development and distribution of stress in concrete, let us consider a simple rectangular beam section with steel bars at bottom (in the tensile zone), which is loaded by an increasing
transverse load.
The tensile strength of concrete being small, the concrete will crack at bottom soon at a low
transverse load. The stress at this level is known as the modulus of rupture and the bending moment
is referred to as the cracking moment. Beyond this level, the tensile stress will be handled by the
steel bars and the compression stress by the concrete section above the neutral axis. Concrete being
a brittle (not a ductile) material, the distribution of stress within the compression zone could be considered linear only up to a moderate load level when the stress attained by concrete is less than 1/2
fc′, as shown in Figure 14.1. In this case, the stress and strain bear a direct proportional relationship.
As the transverse load increases further, the strain distribution will remain linear (Figure 14.2b)
but the stress distribution will acquire a curvilinear shape similar to the shape of the stress–strain
curve. As the steel bars reach the yield level, the distribution of strain and stress at this load will be
as shown in Figure 14.2b and 14.2c.
єc
(a)
(b)
fc = єc Ec
f s = єs · Es
єs
(c)
Figure 14.1 Stress–strain distribution at moderate loads: (a) section, (b) strain, and (c) stress.
fc΄
єc
0.85fc΄
a
c
(a)
(b)
fy/E
fy
(c)
fy
(d)
Figure 14.2 Stress–strain distribution at ultimate load: (a) section, (b) strain, (c) stress, and (d) equivalent
stress.
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For simplification, Whitney (1942) proposed a fictitious but an equivalent rectangular stress
distribution of intensity 0.85 fc′, as shown in Figure 14.2d. This has since been adopted by the ACI.
The property of this rectangular block of depth a is such that the centroid of this rectangular block
is the same as the centroid of actual curved shape and that the area under the two diagrams in
Figure 14.2c and d are the same. Thus, for the design purpose, the ultimate compression of concrete
is taken to be 0.85 fc′, uniformly distributed over the depth, a.
Strength of Reinforcing Steel
The steel bars used for reinforcing are round, deformed bars with some forms of patterned ribbed
projections onto their surfaces. The bar sizes are designated from #3 through #18. For #3 to #8
sizes, the designation represents the bar diameter in one-eighths of an inch, i.e., #5 bar has a diameter of 5/8 in. The #9, #10, and #11 sizes have diameters that provide areas equal to the areas of
the 1 in. × 1 in. square bar, 1 in.ì1 in. square bar, and 1ẳ in.ì1ẳ in. square bar, respectively.
Sizes #14 and #18 are available only by special order. They have diameters equal to the areas of
1½ in.ì1ẵ in. square and 2in.ì2in. square bar, respectively. The diameter, area, and unit weight
per foot for various sizes of bars are given in Appendix D.1.
The most useful properties of reinforcing steel are the yield stress, f y and the modulus of elasticity, E. A large percentage of reinforcing steel bars are not made from new steel but are rolled from
melted, reclaimed steel. These are available in different grades. Grade 40, Grade 50, and Grade 60
are common where Grade 40 means the steel having an yield stress of 40 ksi and so on. The modulus
of elasticity of reinforcing steel of different grades varies over a very small range. It is adopted as
29,000 ksi for all grades of steel.
Concrete structures are composed of the beams, columns, or column–beam types of structures
where they are subjected to flexure, compression, or the combination of flexure and compression.
The theory and design of simple beams and columns have been presented in the book.
LRFD Basis of Concrete Design
Until mid-1950, concrete structures were designed by the elastic or working stress design (WSD)
method. The structures were proportioned so that the stresses in concrete and steel did not exceed
a fraction of the ultimate strength, known as the allowable or permissible stresses. It was assumed
that the stress within the compression portion of concrete was linearly distributed. However, beyond
a moderate load when the stress level is only about one-half the compressive strength of concrete,
the stress distribution in concrete section is not linear.
In 1956, the ACI introduced a more rational method wherein the members were designed for
a nonlinear distribution of stress and the full strength level was to be explored. This method was
called the ultimate strength design (USD) method. Since then, the name has been changed to the
strength design method.
The same approach is known as the load resistance factor design (LRFD) method in steel and
wood structures. Thus, the concrete structures were the first ones to adopt the LFRD method of
design in the United States.
The ACI Publication #318, revised numerous times, contains the codes and standards for concrete buildings. The ACI codes 318-56 of 1956 for the first time included the codes and standards
for the ultimate strength design in an appendix to the code. The ACI 318-63 code provided an equal
status to the WSD and ultimate strength methods bringing both of them within the main body of
the code. The ACI 318-02 code made the ultimate strength design, with a changed name of the
strength design as the mandatory method of design. The ACI 318-08 code contains the latest design
provisions.
In the strength design method, the service loads are amplified using the load factors. The member’s strength at failure known as the theoretical or the nominal capacity is somewhat reduced by a
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strength reduction factor to represent the usable strength of the member. The amplified loads must
not exceed the usable strength of member, namely,
Amplified loads on member ≤ usable strength of member
(14.1)
Depending upon the types of structure, the loads are the compression forces, shear forces, or
bending moments.
Reinforced Concrete Beams
A concrete beam is a composite structure where a group of steel bars are embedded into the tension
zone of the section to support the tensile component of the flexural stress. The areas of the group
of bars are given in Appendix D.2. The minimum widths of beam that can accommodate a specified number of bars in a single layer are indicated in Appendix D.3. These tables are very helpful
in designs.
Equation 14.1 in the case of beams takes the following form similar to wood and steel structures:
Mu ≤ φM n
(14.2)
where
Mu is the maximum moment due to application of the factored loads
Mn is the nominal or theoretical capacity of the member
ϕ is the strength reduction (resistance) factor for flexure
According to the flexure theory, Mn = FbS where Fb is the ultimate bending stress and S is the section modulus of the section. The application of this formula is straightforward for a homogeneous
section for which the section modulus or the moment of inertia could be directly found. However,
for a composite concrete–steel section and a nonlinear stress distribution, the flexure formula presents a problem. A different approach termed as the internal couple method is followed for concrete
beams.
In the internal couple method, two forces act on the beam cross section represented by a compressive force, C acting on one side of the neutral axis (above the neutral axis in a simply supported
beam) and a tensile force, T acting on the other side. Since the forces acting on any cross section of
the beam must be in equilibrium, C must be equal and opposite of T, thus representing a couple. The
magnitude of this internal couple is the force (C or T) times the distance Z between the two forces
called the moment arm. This internal couple must be equal and opposite to the bending moment
acting at the section due to the external loads. This is a very general and a convenient method for
determining the nominal moment, Mn in concrete structures.
Derivation of the Beam Relations
The stress distribution across a beam cross section at the ultimate load is shown in Figure 14.3 representing the concrete stress by a rectangular block according to the “Design Strength of Concrete”
section.
The ratio of stress block and depth to the neutral axis is defined by a factor β1 as follows:
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β1 =
a
c
(14.3)
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0.85fc΄
b
d
c
a/2
a
h
d
C
Z
fy
T
Figure 14.3 Internal forces and couple on a section.
Sufficient test data are available to evaluate β1. According to the ACI
1. For fc′ ≤ 4000 psi β1 = 0.85
(14.4a)
2. For fc′ > 4000 psi but ≤ 8000 psi
f ′ − 4000
(0.05)
β1 = 0.85 − c
1000
(14.4b)
3. For f ′ > 8000 psi β1 = 0.65
(14.4c)
With reference to Figure 14.3, since force = (stress) (area)
C = (0.85 fc′)(ab)
T = f y As
(a)
(b)
Since, C = T,
(0.85 fc′)(ab) = f y As
(c)
or
a=
As f y
0.85 fc′b
(d)
a=
ρf y d
0.85 fc′
(14.5)
or
where
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ρ = steel ratio =
As
bd
(14.6)
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Since moment = (force) (moment arm)
a
a
M n = T d − = f y As d −
2
2
(e)
Substituting a from Equation 14.5 and As from Equation 14.6 into (e)
ρf y
M n = ρf ybd 2 1 −
1.7 f c′
(f)
ρf y
Mu
= ρf y 1 −
φbd 2
1.7 fc′
(14.7)
Substituting (f) into Equation 14.2
Equation 14.7 is a very useful relation to analyze and design a beam.
_
If we arbitrarily define the expression to the right side of Equation 14.7 by K called the coefficient
of resistance, then Equation 14.7 becomes
Mu = φbd 2 K
(14.8)
ρf y
K = ρf y 1 −
1.7 f c′
(14.9)
where
_
_
The coefficient K depends on (1) ρ, (2) f y, and (3) fc′ . The values of K for different combinations of
ρ, f y, and fc′ are listed in Appendices D.4 through D.10.
In place of Equation 14.7, these tables can be directly used in beam computations.
The Strain Diagram and Modes of Failure
The strain diagrams in Figures 14.1 and 14.2 show a straight line variation of the concrete compression strain εC to the steel tensile strain, εS; the line passes through the neutral axis. Concrete
can have a maximum strain of 0.003 and the strain at which steel yields is εy = f y /E. When the
strain diagram is such that the maximum concrete strain of 0.003 and the steel yield strain of εy
are attained at the same time, it is said to be a balanced section, as shown by the solid line labeled
I in Figure 14.4.
In this case, the amount of steel and the amount of concrete balance each other out and both of
these will reach the failing level (will attain the maximum strains) simultaneously. If a beam has
more steel than the balanced condition, then the concrete will reach a strain level of 0.003 before
the steel attains the yield strain of εy. This is shown by condition II in Figure 14.4. The neutral axis
moves down in this case.
The failure will be initiated by crushing of concrete which will be sudden since concrete is
brittle. This, mode of failure in compression, is undesirable because a structure will fail suddenly
without any warning.
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0.003
III
NA at under reinforced condition
Balanced
neutral axis (NA)
I
II
єy =
NA at over
reinforced condition
fy
E
Figure 14.4 Strain stages in beam.
If a beam has lesser steel than the balanced condition, then steel will attain its yield strain
before the concrete could reach the maximum strain level of 0.003. This is shown by condition III
in Figure 14.4. The neutral axis moves up in this case. The failure will be initiated by the yielding of the steel, which will be gradual because of the ductility of steel. This is a tensile mode of
failure, which is more desirable because at least there is an adequate warning of an impending
failure. The ACI recommends the tensile mode of failure or the under-reinforcement design for a
concrete structure.
Balanced and Recommended Steel Percentages
To ensure the under-reinforcement conditions, the percent of steel should be less than the balanced
steel percentage, ρb which is the percentage of steel required for the balanced condition.
From Figure 14.4, for the balanced condition,
0.003 f y E
=
c
d −c
(a)
By substituting c = a/β1 from Equation 14.3 and a = ρf y d 0.85 fc′ from Equation 14.5 and E =29 × 106
psi in Equation (a), the following expression for the balanced steel is obtained:
0.85β1 fc′ 870, 000
ρb =
f y 87, 000 + f y
(14.10)
The values for the balanced steel ratio, ρb calculated for different values of fc′ and f y are tabulated
in Appendix D.11. Although a tensile mode of failure ensues when the percent of steel is less than
the balanced steel, the ACI code defines a section as tension controlled only when the tensile strain
in steel εt is equal to or greater than 0.005 as the concrete reaches to its strain limit of 0.003. The
strain range between εy = ( f y /E) and 0.005 is regarded as the transition zone.
The values of the percentage of steel for which εt is equal to 0.005 are also listed in Appendix
D.11 for different grades of steel and concrete. It is recommended to design beams with the percentage of steel, which is not larger than these listed values for εt of 0.005.
If a larger percentage of steel is used than for εt = 0.005, to be in the transition region, the strength
reduction factor ϕ should be adjusted, as discussed in the “Strength Reduction Factor for Concrete”
section.
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Minimum Percentage of Steel
Just as the maximum amount of steel is prescribed to ensure the tensile mode of failure, a minimum
limit is also set to ensure that the steel is not too small so as to cause the failure by rupture (cracking)
of the concrete in the tension zone. The ACI recommends the higher of the following two values for
the minimum steel in flexure members.
( As ) min =
3 fc′
bd
fy
(14.11)
200
bd
fy
(14.12)
or
( As )min =
where
b is the width of beam
d is the effective depth of beam
The values of ρmin which is (As)min /bd, are also listed in Appendix D.11, where higher of the
values from Equations 14.10 and 14.11 have been tabulated.
The minimum amount steel for slabs is controlled by shrinkage and temperature requirements,
as discussed in the “Specifications for Slab” section.
Strength Reduction Factor for Concrete
In Equations 14.2 and 14.7, a strength reduction factor ϕ is applied to account for all kinds of uncertainties involved in strength of materials, design and analysis, and workmanship. The values of the
factor recommended by the ACI are listed in Table 14.1.
For the transition region between the compression-controlled and the tension-controlled stages
when εt is between εy (assumed to be 0.002) and 0.005 as discussed above, the value of ϕ is interpolated between 0.65 and 0.9 by the following relation
250 *
φ = 0.65 + (ε t − 0.002)
3
(14.13)
The values† of εt for different percentages of steel are also indicated in Appendices D.4 through D.10. When it is not listed in
these tables, it means that εt is larger than 0.005.
Table 14.1
Strength Reduction Factors
Specifications for Beams
The ACI specifications for beams are as follows:
1.Width-to-depth ratio: There is no code requirement for b/d
ratio. From experience, the desirable b/d ratio lies between
1/2 and 2/3.
Structural System
1. Tension-controlled beams
and slabs
2. Compression-controlled
columns
Spiral
Tied
3. Shear and torsion
4. Bearing on concrete
ϕ
0.9
0.70
0.65
0.75
0.65
* For spiral reinforcement this is ϕ = 0.70 + (εt − 0.002)(250/3).
† ε is calculated by the formula
ε t = 0.00255 fc′β1 ρf y − 0.003.
t
(
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)
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Hook
Hanger
Effective
depth
Stirrup (see Chapter 15)
Main reinforcement
Clear
cover
Clear cover
Bar spacing
Figure 14.5 Specifications sketch of beam.
2.Selection of steel: After a required reinforcement area is
computed, Appendix D.2 is used to select the number of bars Table 14.2
First Estimate of Beam
that provide for the necessary area.
3.Minimum beam width required to accommodate multiples Weight
of various size bars are given in Appendix D.3. This is an Design Moment,
Estimated
Mu, ft-k
Weight, lbs/ft
useful design aid as demonstrated in the example.
4.The reinforcement is located at certain distance from the ≤200
300
surface of the concrete called the cover. The cover require- >200 but ≤300
350
400
ments in the ACI code are extensive. For beams, girders, and >300 but ≤400
450
columns that are not exposed to weather or are not in contact >400 but ≤500
500
with the ground, the minimum clear distance from the bot- >500
tom of the steel to the concrete surface is 1½ in. There is a
minimum cover requirement of 1½ in. from the outermost
longitudinal bars to the edge toward the width of the beam, as shown in Figure 14.5.
5.Bar spacing: The clear spacing between the bars in a single layer should not be less than
any of the following:
• 1 in.
• The bar diameter
1ìmaximum aggregate size
6.Bars placement: If the bars are placed in more than one layer, those in the upper layers
are required to be placed directly over the bars in the lower layers and the clear distance
between the layers must not be less than 1 in.
7.Concrete weight: Concrete is a heavy material. The weight of the beam is significant. An
estimated weight should be included. If it is found to be appreciably less than the weight
of the section designed, then the design should to be revised. For a good estimation of
concrete weight, Table 14.2 could be used as a guide.
Analysis of Beams
Analysis relates to determining of the factored or service moment or the load capacity of a beam of
known dimensions and known reinforcement.
The procedure of analysis follows:
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1. Calculate the steel ratio from Equation 14.6
ρ=
As
bd
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2.Calculate (As)min from Equations 14.11 and 14.12 or use Appendix D.11.
Compare this to the As of beam to ensure that it is more than the minimum.
3.For known ρ, read εt from Appendices D.4 through D.10. If no value is given, then εt = 0.005.
If εt < 0.005, determine ϕ _from Equation 14.13.
4.For known ρ, compute K from Equation 14.9 or read the value from Appendices D.4
through D.10.
5.Calculate Mu from Equation 14.7
Mu = φbd 2 K
6.Break down into the loads if required.
Example 14.1
The loads on a beam section are shown in Figure 14.6. Whether the beam is adequate to support
the loads. fc ′ = 4,000 psi and fy = 60,000 psi.
Solution
A. Design loads and moments
1. Weight of beam/ft = (12/12) × (20/12) × 1 × 150 = 250 lb/ft or 0.25 k/ft
2. Factored dead load, wu = 1.2 (1.25) = 1.5 k/ft
3. Factored live load, Pu = 1.6 (15) = 24 k
4. Design moment due to dead load = wuL2/8 = 1.5(20)2/8 = 75 ft-k
5. Design moment due to live load = PuL/4 = 24(20)/4 = 120 ft-k
6. Total design moment, Mu = 195 ft-k
7. As = 3.16 in.2 (from Appendix D.2 for 4 bars of #8)
8. ρ = As/bd = 3.16/12×17 = 0.0155
9. ρmin = 0.0033 (from Appendix D.11) < 0.0155 OK
10. ε_t ≥ 0.005 (value not listed in Appendix D.9), ϕ = 0.9
11. K = 0.8029 ksi (for ρ = 0.0155 from Appendix D.9)
12. Mu = φbd 2K
= (0.9)(12)(17) (0.8029) = 2506 in.-k or 209 ft-k > 195 ft-k OK
2
LL = 15 k
DL = 1 k/ft
(excluding weight)
20 ft
12 in.
17 in.
20 in.
4–#8
Figure 14.6 Loads and section of Example 14.1.
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Design of Beams
In wood beam design in Chapter 7 and steel beam design in Chapter 11, beams were designed for
bending moment capacity and checked for shear and deflection.
In concrete beams, shear is handled independently, as discussed in Chapter 15. For deflection,
the ACI stipulates that when certain depth requirements are met, deflection will not interfere with
the use or cause a damage to the structure. These limiting values are given in Table 14.3 for normal
weight (120–150 lb/ft3) concrete and Grade 60 steel. For other grade concrete and steel, the adjustments are made as indicated in the footnotes to the Table 14.3.
When the minimum depth requirement is met, deflection needs not be computed. For members
of lesser thickness than those listed in Table 14.3, the deflections should be computed to check for
safe limits. This book assumes that the minimum depth requirement is satisfied.
The beam design falls into the two categories discussed below.
Design for Reinforcement Only
When a beam section has been fixed from the architectural or any other consideration, only the
amount steel has to be selected. The procedure is as follows:
1. Determine the design moment, Mu including the beam weight for various critical load
combinations.
_
2.Using d = h – 3, and ϕ = 0.9, calculate the required K from Equation 14.8 expressed as
K=
Mu
φbd 2
_
3.From Appendices D.4 through D.10, find the value of ρ corresponding to K of step 2.
From the same appendices,
confirm that εt ≥ 0.005. If εt < 0.005, reduce ϕ by Equation
_
14.13 and recompute K and find the corresponding ρ.
4.Compute the required steel area As from Equation 14.6
As = ρbd
Table 14.3
Minimum Thickness of Beams and Slabs, for Normal Weight
Concrete and Grade 60 Steel
Minimum Thickness, h, in.
Member
Beam
Slab (one-way)
Simply Supported
Cantilever
One End
Continuous
Both Ends
Continuous
L/16
L/20
L/18.5
L/24
L/21
L/28
L/8
L/10
Notes: L is the span in inches.
For lightweight concrete of unit weight 90–120 lb/ft3, the table values should be
multiplied by (1.65 – 0.005Wc) but not less than 1.09, where Wc is the
unit weight in lb/ft3.
For other than Grade 60 steel, the table value should be multiplied by
(0.4 + fy /100), where fy in ksi.
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5.Check for minimum steel (ρ)min from Appendix D.11.
6.Select the bar size and the number of bars from Appendix D.2. From Appendix D.3, check
whether the selected steel (size and number) can fit into width of the beam, preferably in a
single layer. They can be arranged in two layers. Check to confirm that the actual depth is
at least equal to h − 3.
7.Sketch the design.
Example 14.2
Design a rectangular reinforced beam to carry a service dead of 1.6 k/ft and a live load of 1.5 k/ft
on a span of 20 ft. The architectural consideration requires the width to be 10 in. and depth to be
24 in. Use fc ′ = 3,000 psi and fy = 60,000 psi.
Solution
1. Weight of beam/ft = (10/12)×(24/12)×1×150 = 250 lb/ft or 0.25 k/ft
2. wu = 1.2 (1.6 + 0.25) + 1.6 (1.5) = 4.62 k/ft
3. Mu = wuL2/8 = 4.62(20)2/8 = 231 ft-k or 2772 in.-k
4. d = 24 − 3 = 21 in.
_
5. K = 2772/(0.9)(10)(21)2 = 0.698 ksi
6. ρ = 0.0139 εt = 0.0048 (from Appendix D.6)
7. From Equation
14.13, ϕ = 0.65 + (0.0048 − 0.002)(250/3) = 0.88
_
8. Revised K = 2772/(0.88)(10)(21)2 = 0.714 ksi
9. Revised ρ = 0.0143 (from Appendix D.6)
10. As = ρbd = (0.0143)(10)(21) = 3 in.2
11. ρ(min) = 0.0033 (from Appendix D.11) < 0.0143 OK
12. Selection of steel
Bar Size
#6
#7
#9
No. of Bars
As, from
Appendix D.2
Minimum Width in One
Layer from Appendix D.3
7
5
3
3.08
3.0
3.0
15 NG
12.5 NG
9.5 OK
Select 3 bars of #9.
13. Beam section is shown in Figure 14.7.
Design of Beam Section and Reinforcement
The design comprises of determining the beam dimensions and
selecting the amount of steel. The procedure is as follows:
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1. Determine the design moment, Mu including the beam weight
for various critical load combinations.
2.Select steel ratio ρ corresponding to εt = 0.005 from
Appendix D.11.
_
3.From Appendices D.4 through D.10, find K for the steel ratio
of step 2.
4.For b/d ratio of 1/2 and 2/3, find two values of d from the
following expression
10 in.
Stirrups
(not designed)
21 in.
3-#9
3 in.
Figure 14.7 Beam section
of Example 14.2.
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