lesson 6 - design website - home...either increase the concrete strength, or doubly reinforced the...

11

Lesson 6Lesson 6Doubly Reinforced BeamsDoubly Reinforced Beams

22

ContentsContents

Introduction to Doubly reinforced concrete Introduction to Doubly reinforced concrete beamsbeams Worked exampleWorked example Design methodologyDesign methodology

33

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

44

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==

55

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

66

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

77

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

88

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

.

. .

.

.

99

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

1111

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’’))))

1212

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.