Lecture 2 - Flexure - Brief.pdf

7/28/2019 Lecture 2 - Flexure - Brief.pdf

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Reinforced Concrete Design

Lecture no. 2 - Flexure


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Flexure in Beams and Slabs

Beams and slabs are subjected primarily

to flexure (bending) and shear.

At any section within the beam, the

internal resisting moment is necessary toequilibrate the bending moments causedby external loads.


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Fig. 1. One-way flexure (MacGregor 1997, Fig. 4-1)


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Continuous one-way slab


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Fig. 2. Internal forces in a beam (MacGregor 1997, Fig. 4-3)


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Basic Assumptions in Flexure Theory

1) Plane section remains plane.

2) The strain in the reinforcement is equalto the strain in the concrete at the same

level (perfect bond).3) The stresses in the concrete and

reinforcement can be computed from the

strains using the stress-strain curves.


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Basic Assumptions in Flexure Theory

(contd)

4) The tensile strength of concrete isneglected.

5) Concrete is assumed to fail when the

compressive strain reaches a limitingvalue, for example, a value of 0.003.


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Plane Section Remains Plane

Fig. 3. Assumed linear strain distribution (Notes 1990, Fig. 6-5)


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Fig. 4. Cracking of reinforced concrete beam (MacGregor 1992, p. 79)

BMD

SFD

+


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Elastic Stresses, Cracked Section

E.N.A.

c

d

nAs

b

fc

c/3

d - c/3 M

fsT

Fig. 6. Elastic stresses and strains in cracked section (at service loads)

kd

jd= d-kd/3

kd/3


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Static Test on Under-reinforced Beam


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Failure of Under-reinforced Beam

Concrete fails atstrain = 0.003


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Analysis for Ultimate Moment

Capacity of Beam Section

1) Stress and strain compatibility: stress-strain relationships are used.

2) Equilibrium: internal moments must

balance the bending moment due toapplied load.

To compute the moment capacity of thebeam, two requirements must be satisfied:


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Tension, Compression, and

Balanced Failures

Flexural failures may occur in three different ways:

1. Tension failure. Reinforcement yields before concretestrain reaches its limiting value. (Under-reinforced)

2. Compression failure. Concrete strain reaches itslimiting value before steel yields. (Over-reinforced)3. Balanced failure. Concrete reaches its limiting value

and steel yields at the time of failure.

Failure mode depends on the reinforcement ratio, sA

bd =


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Tension, Compression, and

Balanced Failures (contd)

,s >,y ,s =,y ,s =,y

0 0 0,u = 0.003 ,u = 0.003 ,u = 0.003 ,u = 0.003

,s =,y

Balanced sectionStrength controlled bytension in reinforcement

Strength controlled bycompression in concrete

Pure

(underreinforced)

(overreinforced)

compression

Fig. 7. Strain distribution in concrete beam (Notes 1990, p. 6-21)

Balanced Failure


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Equivalent Rectangular Stress

Block in Concrete

ACI permits the use of equivalentrectangular concrete stress distribution forultimate strength calculations.


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Block in Concrete (contd) Uniform concrete stress:

Depth of stress block:

'0.85

cf

1a c= Distance from the fiber of maximum

strain to the neutral axis

Strain Equivalent rectangular stress block

T = As fss

Actual Stress Distribution

Fig. 9. Equivalent rectangular stress block (MacGregor 1997, Fig. 4-17)


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Block in Concrete (contd)

The factor is a function of compressivestrength of concrete as follows:1

'

'

'

1

'

0.85 for 280 ksc

2800.85 0.05 for 280 560 ksc

70

0.65 for 560 ksc

c

cc

c

f

ff

f

=


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Balanced Failureb

d

cb

Asb =Dbbd

,s =,y = fy/Es

,u = 0.003

ab =$1cb

0.85 fc

ab/2

Cb = 0.85 fcbab

Tb = Asbfy

N.A.

1. From similar triangles,

0.003 0.003

b

y

c d

=

+

1 1

0.003

0.003b b

ya c d = = +

(1)


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Balanced Failure (contd)

The balanced reinforcement ratio is then:

2. Force equilibrium,

'0.85

sb y cf f ba=

Substituting Eq. (1) into (2) gives:

(2)

'1 0.003

0.850.003

csb

y y

fA bd

f

= +

'

10.003

0.850.003

sb cb

y y

A f

bd f

= = +

(3)

(4)

T C=


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Maximum Reinforcement in Design

To ensure a ductile behavior, the maximum reinforcement ratio is given by:

Note: ACI defines a section as being tension-controlled if the net

tensile strain in the layer of steel farthest from the compressionface of the beam equals or exceeds 0.005 in tension.

max0.75

b = (5)


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P.N.A.

c

b

0.85 fc

C

AsT = Asfy

0.85 fc

a/2

C = 0.85 fcba

d - a/2

T = Asfy

c a

b

P.N.A.

As

As

d h

b

,c

,s

c

fc

C

T = Asfy

P.N.A.

a =$1c

0.85 fc

a/2

C

d - a/2

T

Fig. 10. Stresses and strains at nominal flexural strength (Nawy 1996, p. 93)

Under-reinforced Section (Tension Failure)

Exceed yield strain of steel !s y

>


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'0.85

s s cf f ba=

T C=From force equilibrium ,

'0.85

s y

c

fa

f b=

Because ,s yf f=

Therefore, the nominal (theoretical) flexural capacity is:

(7)

(8)

(9b)

Or

2n s y

aM A f d

=

'

0.852

n c

aM f ba d

=

(9a)


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Finally, the actual flexural capacity of the section is:

n

0.9 for tension-controlled section0.7 for compression-controlled section

(10)

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Lecture 2 - Flexure - Brief.pdf