by dr. attaullah shah swedish college of engineering and technology wah cantt
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By Dr. Attaullah Shah Swedish College of Engineering and Technology Wah Cantt. Reinforced Concrete Design-4 Design of doubly reinforced beams. Due to size limitations compression reinforcement may be required in addition to anchor bars to support stirrups. - PowerPoint PPT PresentationTRANSCRIPT
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By
Dr. Attaullah Shah
Swedish College of Engineering and Technology Wah Cantt.
Reinforced Concrete Design-4
Design of doubly reinforced beams
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− Due to size limitations compression reinforcement may be required in addition to anchor bars to support stirrups.
− If the applied ultimate moment is more than the factored nominal capacity allowed by maximum steel ratio, the additional steel may be required in compression and tension to support the excess moment.
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Analysis of Doubly Reinforced SectionsAnalysis of Doubly Reinforced Sections
Effect of Compression Reinforcement on the Strength and BehaviorLess concrete is needed to resist the T and thereby moving the neutral axis (NA) up.
TC
fAT
ys
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Analysis of Doubly Reinforced Analysis of Doubly Reinforced SectionsSections
Effect of Compression Reinforcement on the Strength and Behavior
12
2sc
1c
and2
; C
ReinforcedDoubly 2
;
ReinforcedSingly
aa
adfAMCC
adfAMCC
ysn
ysn
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Reasons for Providing Compression ReinforcementReasons for Providing Compression Reinforcement
− 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
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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
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Analysis of Doubly Reinforced Analysis of Doubly Reinforced Rectangular SectionsRectangular Sections
Strain Compatibility Check Assume s’ using similar triangles
s
s
0.003'
'*0.003
c d c
c dc
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Analysis of Doubly Reinforced Analysis of Doubly Reinforced Rectangular SectionsRectangular SectionsStrain 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 facf b f
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Analysis of Doubly Reinforced Analysis of Doubly Reinforced Rectangular SectionsRectangular Sections
Strain Compatibility The strain in the compression steel is
s cu
1 c
y
1
0.851 0.003
'
dc
f dd f
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Analysis of Doubly Reinforced Analysis of Doubly Reinforced Rectangular SectionsRectangular Sections
Strain Compatibility Confirm
;E ys
s
yys
f
y y1 cs 3
y s
0.851 0.003
' E 29 x 10 ksif ff d
d f
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Analysis of Doubly Reinforced Analysis of Doubly Reinforced Rectangular SectionsRectangular Sections
Strain Compatibility Confirm
y1 c
y
1 c
y y
870.85' 87
0.85 87'
87
ff dd f
f dd f f
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Analysis of Doubly Reinforced Analysis of Doubly Reinforced Rectangular SectionsRectangular Sections
Find 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
sEfd c
c
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Analysis of Doubly Reinforced Analysis of Doubly Reinforced Rectangular SectionsRectangular Sections
If the statement is true than
else the strain in the compression steel
n s s y s y2aM A A f d A f d d
s sf E
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Analysis of Doubly Reinforced Analysis of Doubly Reinforced Rectangular SectionsRectangular Sections
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 cc
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Analysis of Doubly Reinforced Analysis of Doubly Reinforced Rectangular SectionsRectangular Sections
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 cc
f b c A E A f c A E d
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Analysis of Doubly Reinforced Analysis of Doubly Reinforced Rectangular SectionsRectangular Sections
Find the fs’
Check the tension steel.
s s cu1 1 87 ksid df Ec c
ys cu y
sEfd c
c
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Analysis of Doubly Reinforced Analysis of Doubly Reinforced Rectangular SectionsRectangular SectionsAnother 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 d
fd f
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Analysis of Doubly Reinforced Analysis of Doubly Reinforced Rectangular SectionsRectangular SectionsGo back and calculate the equilibrium with fs’
s y s sc s
c
1
s
0.85
1 87 ksi
A f A fT C C a
f ba
c
dfc
Iterate until the c value is adjusted for the fs’ until the stress converges.
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Analysis of Doubly Reinforced Analysis of Doubly Reinforced Rectangular SectionsRectangular Sections
Compute the moment capacity of the beam
n s y s s s s2aM A f A f d A f d d
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Limitations on Reinforcement Ratio Limitations on Reinforcement Ratio for Doubly Reinforced beamsfor Doubly Reinforced beams
Lower limit on
same as for single reinforce beams.
yy
cmin
200 3
fff
(ACI 10.5)
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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.
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Example: Doubly Reinforced Example: Doubly Reinforced SectionSection
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 in 0.012912 in. 15.5 in.
0.62 in 0.003312 in. 15.5 in.
A
A
Abd
Abd
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Example: Doubly Reinforced Example: Doubly Reinforced SectionSection
Compute the effective reinforcement ratio and minimum
y
c
y
min
0.0129 0.0033 0.00957
200 200 0.00333 60000
3 3 4000or 0.0031660000
0.0129 0.00333 OK!
eff
f
ff
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Example: Doubly Reinforced Example: Doubly Reinforced SectionSectionCompute the effective reinforcement ratio and minimum
1 c
y y
0.85 87' 87
0.85 0.85 4 ksi 2.5 in. 87 0.039860 ksi 15.5 in. 87 60
f dd f f
0.00957 0.0398 Compression steel has not yielded.
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Example: Doubly Reinforced Example: Doubly Reinforced SectionSection
Instead 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
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Example: Doubly Reinforced Example: Doubly Reinforced SectionSection
Solve using the quadratic formula
2
2
2.5969 3.8884 0
2.5969 2.5969 4 3.88842
3.6595 in.
c c
c
c
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Example: Doubly Reinforced Example: Doubly Reinforced SectionSection
Find the fs’
Check the tension steel.
s s cu2.5 in.1 1 87 ksi
3.659 in.27.565 ksi
df Ec
s15.5 in. 3.659 in. 0.003 0.00971 0.00207
3.659 in.
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Example: Doubly Reinforced Example: Doubly Reinforced SectionSectionCheck 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
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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
na
M A f A f d A f d d
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Example: Doubly Reinforced Example: Doubly Reinforced SectionSectionIf you want to find the Mu for the problem
u u
3.66 in. 0.23615.5 in.
0.375 0.9
0.9 166 k - ft
149.4 k - ft
cd
cd
M M
From ACI (figure R9.3.2)or figure (pg 100 in your text)
The resulting ultimate moment is