lesson 6 - design website - home...either increase the concrete strength, or doubly reinforced the...
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Lesson 6Lesson 6Doubly Reinforced BeamsDoubly Reinforced Beams
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ContentsContents
Introduction to Doubly reinforced concrete Introduction to Doubly reinforced concrete beamsbeams Worked exampleWorked example Design methodologyDesign methodology
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Introduction to Doubly Introduction to Doubly RR’’fedfed beamsbeams
M M –– moment calculated from formulae moment calculated from formulae MuMu –– ultult. Moment = 0.156fcubd. Moment = 0.156fcubd22
If M > If M > MuMu Concrete can no more withstand the actual Concrete can no more withstand the actual
moment.moment. Either increase the concrete strength,Either increase the concrete strength, Or doubly reinforced the beamOr doubly reinforced the beam
Providing compression steel will resist the Providing compression steel will resist the moment in excess of moment in excess of MuMu
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Stress Stress –– Strain blocksStrain blocks
b
d
x=d/2 s = 0.9x
Fst
Fccz
0.45f
stCross section Strains Stresses
As
As
' d '
sc
cu Fsc==
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Area of reinforcementArea of reinforcement
Compression steel Compression steel reinfreinf. Is cal. Using. Is cal. Using
Tension steel (kTension steel (k’’=0.156)=0.156)
)(95.0 ''
ddfMM
Ay
us
d’: depth to the compression steel from the top surface
s
ys A
zfMA '
95.0
)9.025.0(5.0
'Kdz
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AssumptionAssumption
Compression steel has yielded i.e. steel Compression steel has yielded i.e. steel stress = 0.95 stress = 0.95 fyfy, but this will only apply if, but this will only apply if
If If dd’’/x/x>0.43 >0.43 -- compression steel will be at a compression steel will be at a stress less than yield, in which case this stress less than yield, in which case this stress can be found from the stress can be found from the elastoelasto--plastic plastic stress strain curve shown abovestress strain curve shown above
dx
dd
x d z' '
.
0.43 OR 0.215 where 0 45
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Worked exampleWorked example A simply supported rectangular beam of 9m A simply supported rectangular beam of 9m
span carries a characteristic dead (span carries a characteristic dead (ggkk ) load ) load (inc. Self wt. of beam ), and imposed ((inc. Self wt. of beam ), and imposed (qqkk ) ) loads of 6 loads of 6 kN/mkN/m and 8 and 8 kN/mkN/m respectively. respectively. Check whether it is a SRB or DRBCheck whether it is a SRB or DRB
Doubly Reinforced Beam??
b =225 mm
h=400 mm q = 8kN/mg = 6kN/m
kk
9m
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Worked exampleWorked example The beam dimensions : The beam dimensions : bb= 225mm, = 225mm, hh= 400mm, = 400mm, fcufcu
=30MPa and =30MPa and ffyy =460N/mm2 calculate the area of =460N/mm2 calculate the area of reinforcement required.reinforcement required.
Since M>Since M>MuMu, doubly , doubly rr’’fedfed
Ultimate load ( ) = . .= . . kN / m
Design Moment ( ) = kNm
Assuming bar diameter = Effective depth = - / - cover
(say 40mm to main steel inc. link )=
Ultimate Moment ( ) = .156
.156 30
Nmm kNm
w g q
M wl
mmd h
mm
M f bd
k k
u cu
14 1614 6 16 8 212
8212 9
8214 65
2525 2
348
0
0 225 348
127 5 10127 5
2 2
2
2
6
.
. .
.
.
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Worked exampleWorked example
Compression steel calculationCompression steel calculationd mm
z d K
z mm
x d z
dx
'
'
'
/ /
(0. (0. . ))
(0. (0. .. ))
( ). .
. .
cover
Compression steel has yielded
2 40 20 2 50
5 25 0 9
348 5 25 01560 9
270
0 45348 270
0 45173
50173
0 289 0 43
1010
Worked exampleWorked example
Choose appropriate reinforcement Choose appropriate reinforcement diametersdiameters
26
'' 669
)50348(46095.01015.87
)(95.0mm
ddfMMA
y
us
66927046095.0
105.12795.0
6'
s
y
us A
zfMA
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Design methodologyDesign methodology
a). Calculate K from M/fcubd2a). Calculate K from M/fcubd2 K` = 0.402(bb K` = 0.402(bb –– 0.4) 0.4) –– 0.18(bb 0.18(bb –– 0.4)20.4)2 If K < KIf K < K’’, compression steel is not required, beam , compression steel is not required, beam
shall be designed as a singly reinforced one.shall be designed as a singly reinforced one. If K > KIf K > K’’, compression steel is required., compression steel is required.
b). Calculate x = (bb b). Calculate x = (bb –– 0.4)d0.4)d If If dd’’/x/x < 0.37, the compression steel has yielded and < 0.37, the compression steel has yielded and
fscfsc = 0.95fy= 0.95fy If If dd’’/x/x > 0.37, calculate the steel compressive strain > 0.37, calculate the steel compressive strain
esc and hence esc and hence fscfsc c). Calculate the area of compression steel, c). Calculate the area of compression steel, AsAs’’ = (K = (K –– KK’’)fcubd2/(fsc(d )fcubd2/(fsc(d –– dd’’))))
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Design methodologyDesign methodology
d). Calculate the area of tension steel fromd). Calculate the area of tension steel from As = KAs = K’’fcubd2/(0.95fyZ) + Ascfcubd2/(0.95fyZ) + Asc’’fsc/0.95fyfsc/0.95fy Where Z = d Where Z = d –– 0.9x/20.9x/2
e). Linkse). Links The links should pass round the corner bars and each The links should pass round the corner bars and each
alternate bar.alternate bar. The link size should be at least oneThe link size should be at least one--quarter the size of quarter the size of
the largest compression bar.the largest compression bar. The spacing of links should not be greater than twelve The spacing of links should not be greater than twelve
times the size of the smallest compression bars.times the size of the smallest compression bars. NO compression bar should be more than 150mm NO compression bar should be more than 150mm
from a restrained bar.from a restrained bar.