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Chapter 14. Flexural Reinforced Concrete Members

# Chapter 14. Flexural Reinforced Concrete Members

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278

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

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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279

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

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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Principles of Structural Design: Wood, Steel, and Concrete

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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Flexural Reinforced Concrete Members

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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Principles of Structural Design: Wood, Steel, and Concrete

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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Flexural Reinforced Concrete Members

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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Principles of Structural Design: Wood, Steel, and Concrete

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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Flexural Reinforced Concrete Members

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:

73397_C014.indd 285

1. Calculate the steel ratio from Equation 14.6

ρ=

As

bd

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Principles of Structural Design: Wood, Steel, and Concrete

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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Flexural Reinforced Concrete Members

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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Principles of Structural Design: Wood, Steel, and Concrete

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:

73397_C014.indd 288

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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