Doubly reinforced beam]

Lecture GoalsLecture Goals
Doubly Reinforced beams
T Beams and L Beams

Analysis of Doubly ReinforcedAnalysis of Doubly Reinforced
SectionsSections
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 DoublyAnalysis of Doubly
Reinforced SectionsReinforced Sections
Effect of Compression Reinforcement on the Strength
and Behavior
( )12
2
sc
1
c
and
2
;C
ReinforcedDoubly
2
;
ReinforcedSingly
aa
a
dfAMCC
a
dfAMCC
ysn
ysn
<






−=′+=
⇒






−==
⇒

CompressionCompression
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 ProvidingReasons 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.
ρ < ρb

Change failure mode from compression
to tension. When ρ > ρbal addition of As
strengthens.
Effective reinforcement ratio = (ρ − ρ’)
Compression
zone
allows tension steel to
yield before crushing of
concrete.

Eases in Fabrication -
Use corner bars to hold & anchor stirrups.

Effect of CompressionEffect of Compression
Compare the strain distribution in two beams
with the same As

Effect of CompressionEffect of Compression
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 εs increases (εs2 >εs1).
1c
ss
1
11cc1c
ss
85.0
85.085.0
β
β
bf
fA
c
cbfbafCT
fAT
′
=
′=′==
=
1c
ssss
2
21css
2css
1cs
ss
85.0
85.0
85.0
β
β
bf
fAfA
c
cbffA
baffA
CCT
fAT
′
′′−
=
′+′′=
′+′′=
+′=
=

Doubly Reinforced BeamsDoubly Reinforced Beams
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

Rectangular SectionsRectangular Sections
Strain Compatibility Check
Assume εs’ using similar
triangles
( )
( )
s
s
0.003
'
'
*0.003
c d c
c d
c
ε
ε
′
= ⇒
−
−
′ =

Strain Compatibility
Using equilibrium and find a
( )
( )
( )
( )
( )
s s y
c s
c
s s y y
1 1 c 1 c
0.85
'
0.85 0.85
A A f
T C C a
f b
A A f d fa
c
f b f
ρ ρ
β β β
′−
′ ′= + ⇒ =
′
′− −
= = =
′ ′

The strain in the compression
steel is
( )
( )
s cu
1 c
y
1
0.85
1 0.003
'
d
c
f d
d f
ε ε
β
ρ ρ
′ ′ = − ÷
 
 ′ ′
= − ÷ ÷− 

Confirm
;
E
ys
s
y
ys εεεε ≥=≥′
f
( )
( )
y y1 c
s 3
y s
0.85
1 0.003
' E 29 x 10 ksi
f ff d
d f
β
ε
ρ ρ
 ′ ′
′ = − ≥ = ÷ ÷− 

Confirm
( )
( )
( )
( )
y1 c
y
1 c
y y
870.85
' 87
0.85 87
'
87
ff d
d f
f d
d f f
β
ρ ρ
β
ρ ρ
′ ′ −
− ≥
−
  ′ ′
− ≥  ÷ ÷ ÷ ÷−  

Find c
confirm that the tension steel has yielded
( )
s y c s y
ss s y
1
c 1
0.85
0.85
A f f ba A f
A A f
c a c
f b
β
β
′ ′+ =
′−
= ⇒ =
′
y
s cu y
sE
fd c
c
ε ε ε
− 
= ≥ = ÷
 

If the statement is true than
else the strain in the compression steel
( ) ( )n s s y s y
2
a
M A A f d A f d d
 ′ ′ ′= − − + − ÷
 
s sf Eε ′=

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
d
A E f b c
c
ε β
ε β
′= +
′′= +
′ ′= − + ÷
 

Rearrange the equation and find a quadratic equation
Solve the quadratic and find c.
( )
s y s s cu c 1
2
c 1 s s cu s y s s cu
1 0.85
0.85 0
d
A f A E f b c
c
f b c A E A f c A E d
ε β
β ε ε
′ ′= − + ÷
 
′ ′ ′⇒ + − − =

Find the fs’
Check the tension steel.
s s cu1 1 87 ksi
d d
f E
c c
ε
′ ′   ′= − = − ÷  ÷
   
y
s cu y
sE
fd c
c
ε ε ε
− 
= ≥ = ÷
 

Another option is to compute the stress in the
compression steel using an iterative method.
( )
( )
1 c3
s
y
0.85
29 x 10 1 0.003
'
f d
f
d f
β
ρ ρ
 ′ ′
′= − ÷ ÷− 

Go back and calculate the equilibrium with fs’
( )s y s s
c s
c
1
s
0.85
1 87 ksi
A f A f
T C C a
f b
a
c
d
f
c
β
′−
′ ′= + ⇒ =
′
=
′ 
′= − ÷
 
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 s
2
a
M A f A f d A f d d
 ′ ′ ′ ′ ′= − − + − ÷
 

Limitations on ReinforcementLimitations on Reinforcement
Ratio for Doubly ReinforcedRatio for Doubly Reinforced
beamsbeams
Lower limit on ρ
same as for single reinforce beams.
yy
c
min
2003
ff
f
≥
′
=ρ (ACI 10.5)

Example: Doubly ReinforcedExample: 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.

SectionSection
Compute the reinforcement coefficients, the
area of the bars #7 (0.6 in2
) and #5 (0.31 in2
)
( )
( )
( ) ( )
( ) ( )
2 2
s
2 2
s
2
s
2
s
4 0.6 in 2.4 in
2 0.31 in 0.62 in
2.4 in
0.0129
12 in. 15.5 in.
0.62 in
0.0033
12 in. 15.5 in.
A
A
A
bd
A
bd
ρ
ρ
= =
′ = =
= = =
′
′ = = =

SectionSection
Compute the effective reinforcement ratio and
minimum ρ
y
c
y
min
0.0129 0.0033 0.00957
200 200
0.00333
60000
3 3 4000
or 0.00316
60000
0.0129 0.00333 OK!
eff
f
f
f
ρ ρ ρ
ρ
ρ ρ
′= − = − =
= = =
= =
≥ ⇒ ≥

SectionSection
Compute the effective reinforcement ratio and
minimum ρ
( )
( )
( )( )( )
( )
1 c
y y
0.85 87
'
87
0.85 0.85 4 ksi 2.5 in. 87
0.0398
60 ksi 15.5 in. 87 60
f d
d f f
β
ρ ρ
  ′ ′
− ≥  ÷ ÷ ÷ ÷−  
  
≥ = ÷ ÷ ÷ −  
0.00957 0.0398≥ Compression steel has not
yielded.

SectionSection
Instead of iterating the equation use the quadratic
method
( )
( ) ( ) ( )
( )( ) ( ) ( )( )( )
( )( ) ( ) ( )
2
c 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
β ε ε′ ′ ′+ − − =
+
 + −
 
− =
− − =
− − =

SectionSection
Solve using the quadratic formula
( ) ( )
2
2
2.5969 3.8884 0
2.5969 2.5969 4 3.8884
2
3.6595 in.
c c
c
c
− − =
± − − −
=
=

SectionSection
Find the fs’
Check the tension steel.
s s cu
2.5 in.
1 1 87 ksi
3.659 in.
27.565 ksi
d
f E
c
ε
′   ′= − = − ÷  ÷
   
=
s
15.5 in. 3.659 in.
0.003 0.00971 0.00207
3.659 in.
ε
− 
= = ≥ ÷
 

SectionSection
Check to see if c works
( )( ) ( )( )
( ) ( ) ( )
2 2
s 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 f
c
f b
c
β
′ ′ −−
= =
=
The problem worked

SectionSection
Compute 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.
20.62 in 27.565 ksi
0.62 in 27.565 ksi 15.5 in. 2.5 in.
1991.9 k - in. 166 k - ft
n
a
M A f A f d A f d d
 ′ ′ ′ ′ ′= − − + − ÷
 
  
 ÷= − ÷
 ÷−   
+ −
= ⇒

SectionSection
If 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
φ
φ
= =
> ⇒ =
= =
=
From ACI (figure R9.3.2)or figure (pg 100 in your
text)
The resulting ultimate moment is

Analysis of FlangedAnalysis 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.

SectionsSections
Positive and Negative Moment Regions in a T-beam

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

SectionsSections
Effective Flange Width
Portions near the webs are more highly stressed than
areas away from the web.

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 forACI Code Provisions for
Estimating bEstimating beffeff
From ACI 318, Section 8.10.2
T Beam Flange:
eff
f w
actual
4
16
L
b
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
L
b b
h b
b b
≤ +
≤ +
≤ = +

From ACI 318, Section 8.10
Isolated T-Beams
weff
w
f
4
2
bb
b
h
≤
≥

Various Possible Geometries ofVarious Possible Geometries of
T-BeamsT-Beams
Single Tee
Twin Tee
Box

Doubly reinforced beam]

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