Lecture7 True

8/13/2019 Lecture7 True

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Lecture 7 - Flexure

September 18, 2001

CVEN 444


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L ecture Goals

Basic Concepts Rectangular Beams Non-uniform beams Balanced Beams


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F lexural Stress

Plane sections remain plane ( not true for deep

beams h > 4b) The strain in the reinforcement is equal to the

strain in the concrete at the same level, i.e. es = ecat same level.

Stress in concrete & reinforcement may becalculated from the strains using s-e curves forconcrete & steel.

Basic Assumptions in F lexure Theory


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F lexural Stress

Tensile strength of concrete is neglected for

calculation of flexural strength. Concrete is assumed to fail in compression, when

ec (concrete strain) = ecu (limit state) = 0.003 Compressive s-e relationship for concrete may

be assumed to be any shape that results in anacceptable prediction of strength.

Additional Assumptions for design (f or simplif ication)


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F lexural Stress

The concrete may exceed the ec at the outside edge of thecompressive zone.


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F lexural Stress

The compressive force is modeled as C c = k 1k 3fc b*c atthe location x = k 2*c


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F lexural Stress

The compressive coefficientsof the stress block at given forthe following shapes.

k 3 is ratio of maximum stressat f c in the compressive zone

of a beam to the cylinderstrength, f c (0.85 is a typicalvalue for common concrete)


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F lexural Stress

The compressive zone is modeled with a equivalentstress block.


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F lexural Stress

The equivalent rectangular concrete stress distribution haswhat is known as a b 1 coefficient is proportion of averagestress distribution covers.

65.0 4000

4000*15.085.0

psi4000 for85.0

c1

c1

-- f

f

b

b


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F lexural Stress

Requirements for analysis of reinforced concrete beams

[1] Stress-strain Compatibility

-Stress at a point in membermust correspond to strain at a

point

[2] Equilibrium

Internal forces balance externalforces


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F lexural Stress Example of rectangular reinforced concrete beam .

(1) Setup equilibrium.

- n

css

x

M2T 0

85.0

CT 0

ad M

ab f f A

F


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(2) Find flexural capacity.

-2

M85.0

85.0

ys

n

c

ys

c

ss

ad f A

Tjd b f

f Aa

ab f C f AT


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(3) Need to confirm es>ey

ycs

1

s

yy

e e e

b

s e

-c

cd

ac

E


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F lexural Stress - Example

Given that the beam with concreterated at f c =6000 psi and the steel israted at f s =60,000 psi. d= 12.5 in.

(a) Determine the area of the steelfor a balanced system.

(b) Determine the moment capacityof the beam. M n

(c) Determine the NA.

For a non-rectangular beam


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Three possibi l i ties in I nelastic

Behavior

Compression Failure - (over-reinforced beam)

Tension Failure - (under-reinforced beam) Balanced Failure - (balanced reinforcement)


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I nelastic Behavior

Compression Failure

The concrete will crush before the steel yields. Thisis a sudden failure.

The beam is known as anover-r einf orced beam .


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I nelastic Behavior

Tension Failure

The reinforcement yields before the concrete crushes.The concrete crushes is asecondary compression

failure.The beam is known as anunder-r einf orced beam .


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I nelastic Behavior

Which type of failure is the most desirable?

The under-r einforced beam is the most desirable.

f s = f y

es >> ey

You want ductility systemdeflects and still carries load.


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L imitations on Reinforcement

Ratio,For rectangular cross-sections, r:

[1] Upper Limit on r

ACI 10.3.3

This will ensure steel yields; es (1.8 to 2.0) ey @failure

bdA

s r

bal75.0 r r


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Ratio,It is best to keep r = ( 0.4 to 0.5 ) r bal to allow enoughspace to place reinforcement and to limit cracking and

deflection.[2] Lower Limit on r ACI 10.5.1

ACI Eqn. 10 - 3

f c & f y are in psi

d b f d b f

f A w

yw

y

cs(min) *

200 *

3


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Ratio,[2] Lower Limit on r ACI 10.5.1

Lower limit used to avoid Piano Wire beams.

Very small A s ( M n < M cr )es is huge (large deflections)

when beam cracks (M n > M cr ) beam fails right away because M n < M cr

yy

cmin200

3

f f

f r


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Additional Requirements for L ower

L imit onTemperature and Shrinkage reinforcement instructural slabs and footings (ACI 7.12) place

perpendicular to direction of flexural reinforcement. GR 40 or GR 50 Bars: A s (T&S) = 0.0020 A g

GR 60 or Welded Wire Fabric (WWF):

As (T&S) = 0.0018 A g

Ag - Gross area of the concrete


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Determine Balanced Reinforcement

Ratio, bal

y1 b1 by b

y b

by b

003.0

003.0ca

003.0

003.0c

003.0003.0c

cc003.0003.0

e b b

e

e

e -

d d

d

d

The equation can be rewritten to find c b


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Determine Balanced Reinforcement

Ratio, bal The equation can be rewritten to find r bal

yy

1cs(bal) bal

yy

1cs(bal)

s

s

yy

1c

y

bcs(bal)

ys(bal) bc

000,87000,87

*85.0

bd

000,87

000,87*

bd85.0

*003.0

003.0*

d b85.0 ba85.0

ba85.0

f f f A

f f

f A

E E

f f

f f

A

f A f T C

b r

b

e b


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Example - Balanced Reinforcement

Ratio, bal Given:

b =12 in. d = 15.5 in. h = 18 in.

f c= 4000 psi

f y= 60 ksi (4 #7 bar)

Determine A s(bal), A s(allow)

As(min)


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Example

Given:

f c = 3 ksi & f y = 40 ksi

Determine:

(1) the balanced steel reinforcement.

(2) the maximum reinforcement areaallowed

(3) position of the NA and depth ofthe compressive stress block.

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