CHAPTER 5b. Flexure in Beams - assakkaf 5b REINFORCED CONCRETE ... Calculation of cb For Balanced...

1

Fifth EditionReinforced Concrete Design

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

CHAPTER

5b

REINFORCED CONCRETEA Fundamental Approach - Fifth Edition

FLEXURE IN BEAMS

ENCE 454 – Design of Concrete StructuresDepartment of Civil and Environmental Engineering

University of Maryland, College Park

SPRING 2004By

Dr . Ibrahim. Assakkaf

CHAPTER 5b. FLEXURE IN BEAMS Slide No. 1ENCE 454 ©Assakkaf

Reinforcement Ratio Limitations and Guidelines

Calculation of cb For Balanced Beam

N.A

cf ′85.0

a

b

td

Steel bars

abfC c′= 85.02a

ys fAT =

bc

003.0=cε

yε

2adjd −=

bcd −

Figure 12

Strain Stress

2



Calculation of cb For Balanced Beam

N.A

cf ′85.0

a

td

abfC c′= 85.02a

ys fAT =

bc

003.0=cε

yε

2adjd −=

bt cd −

Figure 13

Strain Stress



Calculation of cb For Balanced BeamFrom similar triangles of the strain diagram of Figure 13:

ty

b

yyc

c

t

b

dc

dc

ε

εεεε

+=

+=

+=

003.0003.0

or

003.0003.0 (15)

(16)

3



Calculation of cb For Balanced Beam– Or

– Since εy = fy/Es (where Es = 29 × 106 psi), hence,

yt

b

dc

ε+=

003.0003.0 (17)

61029003.0

003.0

003.0

003.0

×+

=+

=y

s

yt

b

fEfd

c(18)



Calculation of cb For Balanced BeamEq. 18 can rewritten to give

yt

b

fdc

+=

000,87000,87 (19)

cb = balanced neutral axis depthdt = effective depth to extreme tensile reinforcement layerfy = yield point of steel grade

4



Reinforcement Ratio ρb at Balanced Condition

With Eq. 19, and ρb = Ab/bdt

but

+

=⇒+

==

=

b

b

yb

y

b

b

b

t

b

b

bt

bcA

ffbAc

dc

bAd

000,87000,87ρ

000,87000,87

ρ

ρ

bcfabffACT bccyb 185.085.0 β′=′=⇒=

(20)

(21)

(22)



Reinforcement Ratio ρb at Balanced Condition

From Eq. 22, the ratio Ab/b is given by

Therefore, substituting into Eq. 21, gives y

bcb

fcf

bA 1β85.0 ′

= (23)

(24)

′+

=

′+

=

+

=

y

c

y

by

bc

yb

b

yb

ff

f

cfcf

fbcA

f

1

1

β85.0000,87

000,87

β85.0000,87

000,87000,87

000,87ρ

5



Steel Ratio Formula for Balanced Beam– Instead of using laborious techniques for

determining the balanced steel of beam, the following formula can be used to determine the steel ratio ρb at the balance condition:

+′

=000,87

000,87β85.0ρ 1

yy

cb ff

f (25)

where= compressive strength of concrete (psi)

fy = yield strength of steel (psi)β1 = factor that depends on as given by Eq. 3

cf ′

cf ′



Lower Limit for Steel Reinforcement– The ACI Code establishes a lower limit on

the amount of tension reinforcement. The code states that where tensile reinforcement is required , the steel area As shall not be less than that given by

dbf

dbff

A wy

wy

cs

2003min , ≥

′= (26)

Note that for rectangular beam bw = b

6



Steel Ratio– The steel ratio (sometimes called

reinforcement ratio) is given by

bdAs=ρ

b

h dN.A.

As

(27)

For ductile behavior, ACICode recommends that

bρ of 60% to50 from ranges ρmax



N.A

cf ′85.0

a

01 ′′

32 ′′

3 #8 bars

abfC c′= 85.02a

ys fAT =

bc

003.0

00207.0

Example 2 (cont’d)

2adjd −=

bcd −

Figure 14

Strain

7


Strain Limits Method for Analysis and Design

The nominal flexural strength of a concrete member is reached when the net compressive strain in the extreme fibers reaches the ACI Code-assumed limit of 0.003 in./in.It also stipulates that when the tensile strain in the extreme tension steel εt is sufficiently large at a value equal or greater than 0.005 in./in.; the behavior is fully ductile.



The reinforced concrete beam is characterized by ACI as– Tension-controlled

• Ductile mode of failure, with ample warning of failure as denoted by excessive cracking and deflection (preferred)

– Compression-controlled, or• Brittle mode of failure, with little warning of such

impending failure– Transitional

• Beams with small axial load and large moment.

8



ACI-318-02 Code Strain Limits– Compression-controlled limit:

– Tension-controlled limit:

– Transitional:

002.01029

000,606 =×

==s

yt E

fε

)63.0ρ/ρ (where 005.0 b ==tε

005.0002.0 ≤≤ tε



ACI-318-02 Code Strain Limits

Figure 14. Strain Limit Zones and variation of Strength Reduction Factor φ

9



Strain Limits in Terms of Neutral Axis Depth

N.A

b

d

Steel bars

dc 375.0=

/""003.0=cε

/""005.0=tε

cd −

/""003.0=cε

dc 6.0=

/""002.0=tε

d

Figure 15. Strain Limits (a) Tension Controlled; (b) Compression Controlled

(a) (b)

N.A



Variation of φ as a Function of Strain– Variation of the φ value for the range of strain

between εt = 0.002 and εt = 0.005 can be linearly interpolated to give the following expression:

[ ]

[ ] 90.06757.0φ70.0 :Sections Reinforced-Spirally

90.08348.0φ65.0 :Sections Tied

≤+=≤

≤+=≤

t

t

ε

ε (28)

(29)

10



Variation of φ as a Function of Neutral Axis Depth Ratio c/d

• Eq. 18 can be expressed in terms c/dt, where dt is the effective depth of the layer of reinforcement closest to the tensile face of the section as follows:

90.0/20.037.0φ70.0

:Sections Reinforced-Spirally

90.0/25.023.0φ65.0

:Sections Tied

≤

+=≤

≤

+=≤

t

t

dc

dc(30)

(31)



Flexural Members with Axial Load– The ACI Code stipulates that for flexural

members with axial load exceeding 0.10 f′cAg, the net tensile strain εt at nominal strength should not be less than

0.004– Therefore, in the transition zone of Figure 14,

the minimum strain value in flexural members for determining the φ value is 0.004.

11



Example 4– Determine the amount of steel

required to create a balanced condition for the beam shown, where = 4,000 psi. Assume A615 grade 60 steel that has a yield strength of 60 ksi and a modulus of elasticity = 29 × 106

psi. Also check the code requirement for ductile-type beam with the 3 #8 bars.

10 in.

25 in.23 in.

N.A.

3 #8 bars

cf ′



Example 4 (cont’d)1) Table (see in 0.79 bar 8 No.for Area 2=

( ) 2in 37.279.03 Therefore, ==sA

The strain at which the steel yields is

in./in. 00207.01029

000,606 =×

==s

yy E

fε

10 in.

25 in.23 in.

N.A.

In reference to the strain diagram of Fig. 16,and from similar triangles,

00207.0003.0bb cdc −

=

3 #8 bars

12



Bar Designation Diameter in

Area in2

Weight lb/ft

#3 [#10] 0.375 0.11 0.376 #4 [#13] 0.500 0.20 0.668 #5 [#16] 0.625 0.31 1.043 #6 [#19] 0.750 0.44 1.502 #7 [#22] 0.875 0.60 2.044 #8 [#25] 1.000 0.79 2.670 #9 [#29] 1.128 1.00 3.400

#10 [#32] 1.270 1.27 4.303 #11 [#36] 1.410 1.56 5.313 #14 [#43] 1.693 2.25 7.650 #18 [#57] 2.257 4.00 13.60

Table 1. ASTM Standard - English Reinforcing Bars

Note: Metric designations are in brackets



N.A

cf ′85.0

a

01 ′′

32 ′′

3 #8 bars

abfC c′= 85.02a

ys fAT =

bc

003.0

00207.0


2adjd −=

bcd −

Figure 16

Strain

13



10 in.

25 in.23 in.

N.A.

From which,00207.0

23003.0

bb cc −=

in. 6.13=bc

( ) in. 6.116.1385.0 psi 4,000 because 0.85

:4 and 3 Eqs. Using

1

1

====′=

cafc

ββ


3 #8 bars




10 in.

25 in.23 in.

N.A.

Therefore,

( )( )( ) kips 4.394106.11485.085.0 ==′= bafC bcb

ysbbb fATC ==

2in 57.660

4.394===

y

bsb f

TA

Hence, required steel for balanced condition = 6.57 in2

Or from Eq. 25, 3 #8 bars

02851.0000,87000,60

000,87000,60

(0.85))4000(85.0000,87

000,87β85.0ρ 1 =

+

=

+′

=yy

cb ff

f

14




10 in.

25 in.23 in.

N.A.

Also, we can check εt to make sure it isequal or greater than 0.005, or c/dt is equal or less than 0.375 (see Figure 14)

3 #8 bars

( )( ) 2in 56.6102302851.0ρρ ===⇒= dbAdbA

bsbsb

b ← Same answer

( )

OK 63.03614.002851.00103.0

ρρ

0103.0102337.2ρ

<==

===

b

s

dbA


Analysis of Singly Reinforced Rectangular Beams for Flexure

Variables that need to be found or answered include the following:

1. Find the strength φ Mn.2. Check the adequacy of a given beam, or3. Find an allowable load that the beam can

carry.The flow chart of Figure 17 can be used for the analysis of a reinforced concrete beam.

15


Figure 17.Flowchart for Analysis ofSingly Reinforced RectangularBeams in Bending



Example 5For the beam cross-section shown in the figure, determine whether the failure of the beam will be initiated by crushing of concrete or yielding of steel:

= 4000 psi, fy = 60,000 psi, and As = 10 in2cf ′

12 in.

27 in. 24 in.

As

(to be discussed in class)

16



Example 6Compute the nominal moment strength of the beam section shown in the figure:

= 6000 psi, fy = 60,000 psi.cf ′

8 ′′

8 ′′

01 ′′

51 ′′5.2 ′′

7 No. 5

(to be discussed in class)

CHAPTER 5b. Flexure in Beams - assakkaf 5b REINFORCED CONCRETE ... Calculation of cb For Balanced...

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