Lecture 6 - Flexure

Lecture 6 - Flexure

June 13, 2003CVEN 444

Lecture GoalsLecture Goals

Doubly Reinforced beamsT Beams and L Beams

Analysis of Doubly Analysis of Doubly Reinforced SectionsReinforced Sections

Effect of Compression Reinforcement on the Strength and Behavior

Less concrete is needed to resist the T and thereby moving the neutral axis (NA) up.

TC

fAT

ys

Analysis of Doubly Analysis of Doubly Reinforced SectionsReinforced Sections

Effect of Compression Reinforcement on the Strength and Behavior

12

2sc

1c

and

2; C

ReinforcedDoubly

2;

ReinforcedSingly

aa

adfAMCC

adfAMCC

ysn

ysn

Reasons for Providing Reasons for Providing Compression Compression ReinforcementReinforcement

Reduced sustained load deflections. Creep of concrete in compression

zone transfer load to compression steel reduced stress in concrete less creep less sustained load deflection

Reasons for Providing Reasons for Providing Compression ReinforcementCompression Reinforcement

Effective of compression reinforcement on sustained load deflections.


Increased Ductility

reduced stress block depth

increase in steel strain larger curvature are obtained.


Effect of compression reinforcement on strength and ductility of under reinforced beams.

b


Change failure mode from compression to tension. When bal addition of As strengthens.

Effective reinforcement ratio = (’)

Compression zone

allows tension steel to yield before crushing of concrete.


Eases in Fabrication - Use corner bars to hold & anchor

stirrups.

Effect of Compression Effect of Compression ReinforcementReinforcement

Compare the strain distribution in two beams with the same As

Effect of Compression Effect of Compression ReinforcementReinforcement

Section 1: Section 2:

Addition of A’s strengthens compression zone so that less concrete is needed to resist a given value of T. NA goes up (c2 <c1) and s increases (s2 >s1).

1c

ss1

11cc1c

ss

85.0

85.0 85.0

bf

fAc

cbfbafCT

fAT

1c

ssss2

21css

2css

1cs

ss

85.0

85.0

85.0

bf

fAfAc

cbffA

baffA

CCT

fAT

Doubly Reinforced Doubly Reinforced BeamsBeams

Under reinforced Failure ( Case 1 ) Compression and tension steel

yields ( Case 2 ) Only tension steel yields

Over reinforced Failure ( Case 3 ) Only compression steel yields ( Case 4 ) No yielding Concrete crushes

Four Possible Modes of Failure

Analysis of Doubly Analysis of Doubly Reinforced Rectangular Reinforced Rectangular SectionsSections

Strain Compatibility Check Assume s’ using similar triangles

s

s

0.003

'

'*0.003

c d c

c d

c


Strain Compatibility

Using equilibrium and find a

s s yc s

c

s s y y

1 1 c 1 c

0.85

'

0.85 0.85

A A fT C C a

f b

A A f d fac

f b f


Strain Compatibility The strain in the compression steel is

s cu

1 c

y

1

0.851 0.003

'

d

c

f d

d f


Strain Compatibility Confirm

;E ys

s

yys

f

y y1 cs 3

y s

0.851 0.003

' E 29 x 10 ksi

f ff d

d f

Analysis of Doubly Analysis of Doubly Reinforced Rectangular Reinforced Rectangular SectionsSectionsStrain Compatibility Confirm

y1 c

y

1 c

y y

870.85

' 87

0.85 87'

87

ff d

d f

f d

d f f

Analysis of Doubly Analysis of Doubly Reinforced Rectangular Reinforced Rectangular SectionsSectionsFind c

confirm that the tension steel has yielded

s y c s y

ss s y1

c 1

0.85

0.85

A f f ba A f

A A fc a c

f b

ys cu y

sE

fd c

c


If the statement is true than

else the strain in the compression steel

n s s y s y2

aM A A f d A f d d

s sf E


Return to the original equilibrium equation

s y s s c

s s s c 1

s s cu c 1

0.85

0.85

1 0.85

A f A f f ba

A E f b c

dA E f b c

c


Rearrange the equation and find a quadratic equation

Solve the quadratic and find c.

s y s s cu c 1

2c 1 s s cu s y s s cu

1 0.85

0.85 0

dA f A E f b c

c

f b c A E A f c A E d


Find the fs’

Check the tension steel.

s s cu1 1 87 ksid d

f Ec c

ys cu y

sE

fd c

c


Another option is to compute the stress in the compression steel using an iterative method.

1 c3s

y

0.8529 x 10 1 0.003

'

f df

d f

Analysis of Doubly Analysis of Doubly Reinforced Rectangular Reinforced Rectangular SectionsSectionsGo back and calculate the equilibrium with fs’

s y s s

c sc

1

s

0.85

1 87 ksi

A f A fT C C a

f b

ac

df

c

Iterate until the c value is adjusted for the fs’ until the stress converges.


Compute the moment capacity of the beam

n s y s s s s2

aM A f A f d A f d d

Limitations on Reinforcement Limitations on Reinforcement Ratio for Doubly Reinforced Ratio for Doubly Reinforced beamsbeams

Lower limit on

same as for single reinforce beams.

yy

cmin

200

3

ff

f

(ACI 10.5)

Example: Doubly Reinforced Example: Doubly Reinforced SectionSection

Given:

f’c= 4000 psi fy = 60 ksi

A’s = 2 #5 As = 4 #7

d’= 2.5 in. d = 15.5 in

h=18 in. b =12 in.

Calculate Mn for the section for the given compression steel.


Compute the reinforcement coefficients, the area of the bars #7 (0.6 in2) and #5 (0.31 in2)

2 2s

2 2s

2s

2s

4 0.6 in 2.4 in

2 0.31 in 0.62 in

2.4 in0.0129

12 in. 15.5 in.

0.62 in0.0033

12 in. 15.5 in.

A

A

A

bd

A

bd

Example: Doubly Reinforced Example: Doubly Reinforced SectionSectionCompute the effective reinforcement ratio and minimum

y

c

y

min

0.0129 0.0033 0.00957

200 2000.00333

60000

3 3 4000or 0.00316

60000

0.0129 0.00333 OK!

eff

f

f

f


Compute the effective reinforcement ratio and minimum

1 c

y y

0.85 87'

87

0.85 0.85 4 ksi 2.5 in. 870.0398

60 ksi 15.5 in. 87 60

f d

d f f

0.00957 0.0398 Compression steel has not yielded.

Example: Doubly Reinforced Example: Doubly Reinforced SectionSectionInstead of iterating the equation use the quadratic method

2c 1 s s cu s y s s cu

2

2 2

2

2

2

0.85 0

0.85 4 ksi 12 in. 0.85

0.62 in 29000 ksi 0.003 2.4 in 60 ksi

0.62 in 29000 ksi 0.003 2.5 in. 0

34.68 90.06 134.85 0

2.5969 3.8884 0

f b c A E A f c A E d

c

c

c c

c c

Example: Doubly Reinforced Example: Doubly Reinforced SectionSectionSolve using the quadratic formula

2

2

2.5969 3.8884 0

2.5969 2.5969 4 3.8884

23.6595 in.

c c

c

c


Find the fs’

Check the tension steel.

s s cu

2.5 in.1 1 87 ksi

3.659 in.

27.565 ksi

df E

c

s

15.5 in. 3.659 in.0.003 0.00971 0.00207

3.659 in.


Check to see if c works

2 2s y s s

c 1

2.4 in 60 ksi 0.62 in 27.565 ksi

0.85 0.85 4 ksi 0.85 12 in.

3.659 in.

A f A fc

f b

c

The problem worked

Example: Doubly Reinforced Example: Doubly Reinforced SectionSectionCompute the moment capacity of the beam

s y s s s s

2

2

2

2

2.4 in 60 ksi 0.85 3.659 in.15.5 in.

2 0.62 in 27.565 ksi

0.62 in 27.565 ksi 15.5 in. 2.5 in.

1991.9 k - in. 166 k - ft

n

aM A f A f d A f d d

Example: Doubly Reinforced Example: Doubly Reinforced SectionSectionIf you want to find the Mu for the problem

u u

3.66 in.0.236

15.5 in.

0.375 0.9

0.9 166 k - ft

149.4 k - ft

c

d

c

d

M M

From ACI (figure R9.3.2)or figure (pg 100 in your text)

The resulting ultimate moment is

Analysis of Flanged Analysis of Flanged SectionSection

Floor systems with slabs and beams are placed in monolithic pour.Slab acts as a top flange to the beam; T-beams, and Inverted L(Spandrel) Beams.

Analysis of Flanged Analysis of Flanged SectionsSectionsPositive and Negative Moment Regions in a T-beam

Analysis of Flanged Analysis of Flanged SectionsSections

If the neutral axis falls within the slab depth analyze the beam as a rectangular beam, otherwise as a T-beam.

Analysis of Flanged Analysis of Flanged SectionsSectionsEffective Flange Width

Portions near the webs are more highly stressed than areas away from the web.

Analysis of Flanged Analysis of Flanged SectionsSections

Effective width (beff)

beff is width that is stressed uniformly to give the same compression force actually developed in compression zone of width b(actual)

ACI Code Provisions for ACI Code Provisions for Estimating bEstimating beffeff

From ACI 318, Section 8.10.2

T Beam Flange:

eff

f w

actual

4

16

Lb

h b

b


From ACI 318, Section 8.10.3

Inverted L Shape Flange

eff w

f w

actual w

12

6

0.5* clear distance to next web

Lb b

h b

b b


From ACI 318, Section 8.10

Isolated T-Beams

weff

wf

42bb

bh

Various Possible Geometries Various Possible Geometries of T-Beamsof T-Beams

Single Tee

Twin Tee

Box

Documents

Lecture 6 - Flexure