Chapter3d

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A. J. Clark School of Engineering Department of Civil and Environmental Engineering

Fifth EditionCHAPTER

3d

Reinforced Concrete Design

ENCE 355 - Introduction to Structural DesignDepartment of Civil and Environmental Engineering

University of Maryland, College Park

REINFORCED CONCRETE

BEAMS: T-BEAMS ANDDOUBLY REINFORCED BEAMS

Part I Concrete Design and Analysis

FALL 2002By

Dr Ibrahim Assakkaf

CHAPTER 3d. R/C BEAMS: T-BEAMS AND DOUBLY REINFORCED BEAMS Slide No. 1ENCE 355 Assakkaf

Doubly Reinforced BeamsIntroduction If a beam cross section is limited because

of architectural or other considerations, itmay happen that concrete cannot developthe compression force required to resistthe given bending moment.

In this case, reinforcing steel bars areadded in the compression zone, resultingin a so-called doubly reinforced beam ,that is one with compression as well astension reinforcement. (Fig. 1)

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Doubly Reinforced Beams

Introduction (contd)

d

b

h

d

Figure 1. Doubly Reinforced Beam

s A

s A


Doubly Reinforced BeamsIntroduction (contd) The use of compression reinforcement has

decreased markedly with the use ofstrength design methods, which accountfor the full strength potential of theconcrete on the compressive side of theneutral axis.

However, there are situations in whichcompressive reinforcement is used forreasons other than strength.

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Introduction (contd) It has been found that the inclusion of

some compression steel has the followingadvantages:

It will reduce the long-term deflections ofmembers.

It will set a minimum limit on bending loading It act as stirrup-support bars continuous

through out the beam span.


Doubly Reinforced BeamsIntroduction (contd) Another reason for placing reinforcement in

the compression zone is that when beamsspan more than two supports (continuousconstruction), both positive and negativemoments will exist as shown in Fig. 2.

In Fig. 2, positive moments exist at A andC; therefore, the main tensilereinforcement would be placed in thebottom of the beam.

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Introduction (contd)

+-

+ +

-MomentDiagram

Figure 2. Continuous Beam

w

AB

C



Introduction (contd) At B, however, a negative moment exists

and the bottom of the beam is incompression. The tensile reinforcement,therefore, must be placed near the top ofthe beam.

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Doubly Reinforced Beam Analysis

Condition I: Tension and CompressionSteel Both at Yield Stress The basic assumption for the analysis of

doubly reinforced beams are similar tothose for tensile reinforced beams.

The steel will behave elastically up to thepoint where the strain exceeds the yieldstrain y. As a limit = f y when thecompression strain

y.

If < y, the compression steel stress willbe = E s.

s f

s

s s f s


Condition I: Tension and CompressionSteel Both at Yield Stress (contd) If, in a doubly reinforced beam, the tensile

steel ratio is equal to or less than b, thestrength of the beam may be approximatedwithin acceptable limits by disregarding the

compression bars. The strength of such a beam will becontrolled be tensile yielding, and the leverarm of the resisting moment will be littleaffected by the presence of comp. bars.


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Condition I: Tension and CompressionSteel Both at Yield Stress (contd) If the tensile steel ratio is larger than b, a

somewhat elaborate analysis is required. In Fig. 3a, a rectangular beam cross

section is shown with compression steelplaced at distance from the compressionface and with tensile steel A s at theeffective depth d .


s Ad


Condition I: Tension and CompressionSteel Both at Yield Stress (contd)


Cross Section

(a)

Strain at UltimateMoment

(b)

Concrete-SteelCouple

(c)

Steel-SteelCouple

(d)

Figure 3

d

b

s A

s A

=

21 ad Z

c = 0.003

s

c a

c f 85.0

ab f N cC = 85.01

y sT f A N 11 =

s sC f A N =2

y sT f A N 22 =

s

d

d d Z =2

N.A

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Condition I: Tension and CompressionSteel Both at Yield Stress (contd) Notation for Doubly Reinforced Beam:


= total compression steel cross-sectional aread = effective depth of tension steel

= depth to centroid of compressive steel from compression fiber A s1 = amount of tension steel used by the concrete-steel couple A s2 = amount of tension steel used by the steel-steel couple A s = total tension steel cross-sectional area ( A s = A s1 + A s2) M n1 = nominal moment strength of the concrete-steel couple

M n2 = nominal moment strength of the steel-steel couple M n = nominal moment strength of the beams = unit strain at the centroid of the tension steel

= unit strain at the centroid of the compressive steel

s A

d

s


Condition I: Tension and CompressionSteel Both at Yield Stress (contd) Method of Analysis:

The total compression will now consist of twoforcesN C 1, the compression resisted by the concreteN C 2, the compression resisted by the steel

For analysis, the total resisting moment of thebeam will be assumed to consist of two parts or twointernal couples: The part due to the resistance ofthe compressive concrete and tensile steel and thepart due to the compressive steel and additionaltensile steel.


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Condition I: Tension and CompressionSteel Both at Yield Stress (contd) The total nominal capacity may be derived

as the sum of the two internal couples,neglecting the concrete that is displaced bythe compression steel.

The strength of the steel-steel couple isgiven by (see Fig. 3)


222 Z N M T n = (1)




Cross Section

(a)


(b)


(c)

Steel-SteelCouple

(d)

Figure 3

d

b

s A

s A

=

21 ad Z

c = 0.003

s

c a

c f 85.0

ab f N cC = 85.01

y sT f A N 11 =

s sC f A N =2

y sT f A N 22 =

s

d

d d Z =2

N.A

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The strength of the concrete-steel couple is

given by


( )2222

22 assuming

s s y s s sT C

y s y sn

A A f A f A N N

f f d d f A M

=====

Therefore,

( )d d f A M y sn =2 (2)

111 Z N M T n = (3)




( ) =

==

=+=

=

=

2

Therefore

then,since

assuming 2

1

1

2

2121

11

ad f A A M

A A A

A A

A A A A A A

f f a

d f A M

y s sn

s s s

s s

s s s s s s

y s y sn

(4)

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Condition I: Tension and CompressionSteel Both at Yield Stress (contd) Nominal Moment Capacity

From Eqs. 2 and 4, the nominal momentcapacity can be evaluated as


( ) ( )d d f Aad f A A

M M M

y s y s s

nnn

+=

+=

2

21 (5)


Condition I: Tension and CompressionSteel Both at Yield Stress (contd) Determination of the Location of Neutral

Axis:


( )

( )b f

f A

b f

f A Aa

f Aab f f A

N N N

ac

c

y s

c

y s s

y sc y s

C C T

=

=

+=+=

=

85.085.0

Therefore,

85.0

1

21

1

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Condition I: Tension and CompressionSteel Both at Yield Stress (contd) Location of Neutral Axis c


(6)( )

( )b f

f A Aac

b f

f A

b f

f A Aa

c

y s s

c

y s

c

y s s

==

=

=

11

1

85.0

85.085.0

(7)

NOTE: if 4,000 psi, then 1 = 0.85, otherwise see next slidec f


Condition I: Tension and CompressionSteel Both at Yield Stress (contd) The value of 1 may determined by

>

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ACI Code Ductility Requirements The ACI Code limitation on applies to

doubly reinforced beams as well as tosingly reinforced beams.

Steel ratio can be determined from

This value of shall not exceed 0.75 b asprovided in Table 1 (Table A-5, Textbook)


bd A s1= (9)



Recommended Design Values( ) psic f

y y

c

f f

f 2003 max = 0.75 b b (ksi)k

F y = 40,000 psi 3,000 0.0050 0.0278 0.0135 0.48284,000 0.0050 0.0372 0.0180 0.64385,000 0.0053 0.0436 0.0225 0.80476,000 0.0058 0.0490 0.0270 0.9657

F y = 50,000 psi 3,000 0.0040 0.0206 0.0108 0.48284,000 0.0040 0.0275 0.0144 0.64385,000 0.0042 0.0324 0.0180 0.80476,000 0.0046 0.0364 0.0216 0.9657

F y = 60,000 psi 3,000 0.0033 0.0161 0.0090 0.48284,000 0.0033 0.0214 0.0120 0.64385,000 0.0035 0.0252 0.0150 0.80476,000 0.0039 0.0283 0.0180 0.9657

F y = 75,000 psi 3,000 0.0027 0.0116 0.0072 0.48284,000 0.0027 0.0155 0.0096 0.64385,000 0.0028 0.0182 0.0120 0.80476,000 0.0031 0.0206 0.0144 0.9657

Table 1.Design Constants

(Table A-5 Text)

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Example 1Compute thepractical momentcapacity M n forthe beam havinga cross sectionas shown in thefigure. Use =3,000 psi and f y =

60,000 psi.


c f

21

2

02

11

10#2

stirrup3#

(typ)clear21

1

9#39#3


Example 1 (contd)Determine the values for and A s:

We assume that all the steel yields:


s AFrom Table 2 ( A-2, Textbook ),

2in54.2#102of area == s A2in0.6#96of area == s A

221

22

in46.354.20.6

in54.2

Therefore,

and

=====

==

s s s

s s

y s y s

A A A

A A

f f f f

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#3 #4 $5 #6 #7 #8 #9 #10 #111 0.11 0.20 0.31 0.44 0.60 0.79 1.00 1.27 1.562 0.22 0.40 0.62 0.88 1.20 1.58 2.00 2.54 3.123 0.33 0.60 0.93 1.32 1.80 2.37 3.00 3.81 4.684 0.44 0.80 1.24 1.76 2.40 3.16 4.00 5.08 6.245 0.55 1.00 1.55 2.20 3.00 3.95 5.00 6.35 7.806 0.66 1.20 1.86 2.64 3.60 4.74 6.00 7.62 9.367 0.77 1.40 2.17 3.08 4.20 5.53 7.00 8.89 10.928 0.88 1.60 2.48 3.52 4.80 6.32 8.00 10.16 12.489 0.99 1.80 2.79 3.96 5.40 7.11 9.00 11.43 14.04

10 1.10 2.00 3.10 4.40 6.00 7.90 10.00 12.70 15.60

Number of bars

Bar number Table 2. Areas of Multiple of Reinforcing Bars (in 2)

Table A-2 Textbook

Example 1 (contd)



Example 1 (contd)Doubly Reinforced Beam Analysis

Cross Section

(a)


(b)


(c)

Steel-SteelCouple

(d)

Figure 3

d

b

s A

s A

=

21 ad Z

c = 0.003

s

c a

c f 85.0

ab f N cC = 85.01

y sT f A N 11 =

s sC f A N =2

y sT f A N 22 =

s

d

d d Z =2

N.A

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Example 1 (contd)From Eq. 8:


( ) ( )

( ) ( )( ) k -in9.050,65.2206054.224.7

206046.3

2

21

=+=

+=

+=

d d f Aa

d f A A

M M M

y s y s s

nnn

kips-ft504.2kips-ft129.050,6

==n M


Example 1 (contd)The practical moment capacity is evaluated

as follows:

Check ductility according to ACI Code:

( ) kips-ft4542.5049.0 ==u M


( )( ) OK 0161.0)0157.0(Since

0157.0201146.3

max

1

=

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Condition II: Compression Steel BelowYield Stress The preceding equations are valid only if

the compression steel has yielded whenthe beam reaches its ultimate strength.

In many cases, however, such as for wide,shallow beams reinforced with higher-strength steels, the yielding of compressionsteel may not occur when the beamreaches its ultimate capacity.



Condition II: Compression Steel BelowYield Stress It is therefore necessary to to develop

more generally applicable equations toaccount for the possibility that thecompression reinforcement has not yieldedwhen the doubly reinforced beam fails inflexure.

The development of these equations willbe based on


y s

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Condition II: Compression Steel BelowYield Stress Development of the Equations for

Condition II Referring to Fig. 3,

But

and


( ) s sc y sC C T

A f ba f f A

N N N +=

+=85.0

21

ca 1= ( ) s s s E c

d c E f

s == 003.0

(11)

(12)

(13)


Cross Section

(a)


(b)


(c)

Steel-SteelCouple

(d)

Figure 3

Doubly Reinforced Beam AnalysisCondition II: Compression Steel BelowYield Stress

d

b

s A

s A

=

21 ad Z

c = 0.003

s

c a

c f 85.0

ab f N cC = 85.01

y sT f A N 11 =

s sC f A N =2

y sT f A N 22 =

s

d

d d Z =2

N.A

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Condition II: Compression Steel BelowYield Stress Substituting Eqs 12 and 13 into Eq. 11, yields

Multiplying by c , expanding, and rearranging, yield

If E s is taken as 29 10 3 ksi, Eq. 15 will take thefollowing form:


( ) ( ) s sc y s A E cd c

cb f f A += 003.085.0 1

( ) ( ) 0003.0003.085.0 21 =+ s s y s s sc A E d c f A A E cb f

(14)

(15)


Condition II: Compression Steel BelowYield Stress

The following quadratic equation can beused to find c when :


y s

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Example 2Compute the practicalmoment M n for abeam having a crosssection shown in thefigure. Use = 5,000psi and f y = 60,000 psi.


c f

02

11

21

2

stirrup3#

(typ)clear21

1

8#2

11#3See Textbook for completesolution for this example.

Chapter3d

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