t-beam & rectangular beam(3sections)_final

7/29/2019 T-BEAM & Rectangular Beam(3sections)_final

1/23

esign of Beams

rade of concrete M 25 =

ear Cover (cc) 32 mm =

rade of Steel Fe 415 pt & pc,max =

Moment Shear

Mu Vu Leff bw D bf Df d Pt Pc

kN m kN mm mm mm mm mm mm mm mm mm kN m % % No dia

200.00

Size of beam

450 7501 B1 Simply 10000

Support 750.00 230.00

800.00

1.127 0.000 6 2532 40.0 710.0 782.62 Singly 3.306

25

6 25

Dia

of

bar

cc

pt,min

Type of

Section

Mu,limMu/bd

2

N/mm2

16

161200 100

450

16

pt,limit

450 100

% of Steel

Requried

T

Reinforc.o

Beam Section

Design Forces

Type of

Beam

40.0 710.0 0.9813.527Singly

215.00 100Support

Span

900.00 640.0 Doubly 3.967

0.000

1.346 0.160

32

32 710.0

1339.49

782.62

TITLE

1.194%

0.205%

4.000%

d'

Eff.

depth

Type1

PROJECT

DESIGN OF T & RECTANGULAR BEAMS


2/23

1.0 Example 1 :Design the T-Beam for the following parameters :

bf = Df =

bw = d =

fck = N/mm2

fy = N/mm2

Mu =

Df/d = 100/300 =

3xumax/7d < < xumax/d

Mulim = 0.3616 fckbw xumax (d - 0.416xumax) + 0.4467 fck(bf- bw) yf(d - yf/2 )

xumax = 0.5313 d = 0.5313 x 300 = mm

yf = 0.15 xumax + 0.65 Df = 88.9085 mm

Mulim = kNm

Mu < Mulim Hence designed as singly reinforced section.

Location of N.A

When xu = Df , Moment of Resistance Mu1 = 0.3616 fckbfDf(d -0.416 Df)

Mu1 = kNmMu < Mu1, N.A lies within the flange .( xu < Df)

Hence designed as a rectangular beam of width b =bf

Method - 1 :

Area of tension reinforcement ( Ast )

Ast = = mm2

Method - 2 :

Depth of N.A (xu) Mu = 0.3616 fckbfxu (d - 0.416xu)

xu = = 87.33 mm < Df

Ast = 0.3616 x fckx bf x xu x 1.15 = mm2

2.0 Example 2 :

Design the T-Beam for the following parameters

0.2277

3489.45

CL. NO. T - BEAM DESIGN CALCULATIONS

250

kNm

200

mm 100 mm

300 mm

DATE CHECKE

RE

Yugasoft

PROJECT

TITLE REVISION

DESIGNE

MP

DOCUMENT. NO.

1200

20

200

mm

DATE

0.33

Df/d

fy

0.5313

3488.26

256.8583

224.2498

159.39

2

6.411202.1

dbf

Md

fck

u

dbdbf

M4.611

f

f0.5f2

fck

u

y

ck


3/23


DATE CHECKE

RE

Yugasoft

PROJECT

TITLE REVISION

DESIGNE

MP

DOCUMENT. NO. DATE

bf = mm Df =bw = mm d = mm

fck = N/mm2

fy = N/mm2

Mu = kNm d' = mm

Df/d = 100/200 = 0.5 > xumax/d

Mulim = 0.3616 fckbfxumax (d - 0.416xumax) xumax = 0.456 d

= kNm =

Mu > Mulim Hence designed as doubly reinforced section.

Depth & location of N.A :

xu = xumax = 91.2 mm < Df N.A lies within the flange.


= +

fy(d-d')

= mm2

Area of compression reinforcement ( Asc )

= esc = 0.0035 ( 1 - d'/xumax )

= 0.0035 ( 1 - 30/91.2 )

== mm

2fsc = N/mm

2

fcc = N/mm2

3.0 Example 3 :


bf = Df =

bw = d =

fck = N/mm2

fy = N/mm2

Mu =

1200 100

310 kNm

50020

200 200

30135

1200 mm 100

(Mu - Mulim) x 1.15

(0.456)

91.2mm

Mu - Mulim

(fsc-fcc)(d-d')

25 415

Ast

1911.48

0.3616 x fckx bf x xumax x 1.15

fy

102.56

200 mm

Asc

395.13

8.92

0.00235

mm

300 mm

mm

128.266


4/23


DATE CHECKE

RE

Yugasoft

PROJECT

TITLE REVISION

DESIGNE

MP

DOCUMENT. NO. DATE

Df/d = 100/300 =

3xumax/7d < Df/d < xumax/d

Mulim = 0.3616 fckbw xumax (d - 0.416xumax) + 0.4467 fck(bf- bw) yf(d - yf/2 )

xumax = 0.4791 d = 0.4719 x 300 = mm

yf = 0.15 xumax + 0.65 Df = mm

Mulim = kNm


Location of N.A

When xu = Df , Moment of Resistance Mu1 = 0.3616 fckbfDf(d -0.416 Df)

Mu1 = kNm

Mu > Mu1, N.A lies outside the flange . xu > Df

When xu = 7Df/ 3 ,

Moment of Resistance Mu2 = 0.3616 fck bf (7Df/3) d-0.416 (7Df/3)

Mu2 = kNm > 310 kNm

Mu < Mu2 Non Uniform stress block in flange. xu < 7Df/3

( Rectangular cum parabolic stress block )

Depth of N.A ( Xu)

Mu = 0.3616 x f ckx bw x xu x (d-0.416xu)+ 0.4467 x fckx (bf- bw) x yfx (d - yf/ 2)

a = 0.416 + 0.0225 k k = (21/34) * ((bf/bw) -1) = 3.088

b = - (1+ 0.3 k - 0.195 (Df/d) )

c = M_k + 0.4225 k (Df/d)2

- 1.3 k (Df/d)M_k = =

xu/d = =

xu = 142.8956 mm < 143.73 mm (xumax)

yf = 0.15 xu + 0.65 Df = mm < Df Hence O.K


= ( 0.3616 fckbw xu + 0.4467 fck(bf- bw)yf) 1.15

= mm2

4.0 Example 4 :


bf = mm Df =

bw = mm d = mm

fck = N/mm2

fy = N/mm2

Mu = kNm d' = mm

Df/d = 100/300 =

0.33

(0.4791)

310.581

86.5595

0.3616 x fckx bw x d1.905Mu

Ast

86.434

(0.2053)

1200 100

50

200 300

20 250

258

0.33

mm

143.73

364.798

280.3123

0.476319

fy

3390.729

a

acbb

2

42


5/23


DATE CHECKE

RE

Yugasoft

PROJECT

TITLE REVISION

DESIGNE

MP

DOCUMENT. NO. DATE

3xumax/7d < Df/d < xumax/d

Limiting Moment of Resistance : ( Mulim )

Mulim = 0.3616 fckbw xumax (d - 0.416xumax) + 0.4467 fck(bf- bw) yf(d - yf/2)

xumax = 0.5313 d = 0.5313 x 200 = mm

yf = 0.15 xumax + 0.65 Df = mm

= kNm


Depth of N.A location

xu = xumax = 159.39 mm > Df N.A lies outside the flange. ( xu,max < 7Df/3 )


= ( 0.3616 fckbw xu,max + 0.4467 fck(bf- bw)yf) 1.15

= mm2


= esc = 0.0035 ( 1 - d' /xumax )

= 0.0035 ( 1 - 50/159.39 )

=

= mm2

fsc = N/mm2

fcc = N/mm2

5.0 Example 5 :


bf = Df =

bw = d =

fck = N/mm2

fy = N/mm2

Mu =

Df/d = 90/450 = < 3xumax/ 7d


Ast

4735.32

Asc Mu - Mulim

1200 mm

8.92

0.2277 0.5313

253.858

fy

(fsc-fcc)(d-d')

0.0024

21.906 217.39

90 mm

200 mm 450 mm

(0.2053 )

30 415

654.5 kNm

0.2

88.9085

159.39


6/23


DATE CHECKE

RE

Yugasoft

PROJECT

TITLE REVISION

DESIGNE

MP

DOCUMENT. NO. DATE

Mulim = 0.3616 fckbw xumax (d - 0.416xumax) + 0.4467 fck(bf- bw) Df(d - Df/2)xumax = 0.4791 d = 0.4719 x 450 = mm

Mulim = kNm


Location of N.A

When xu = Df, Moment of Resistance Mu1 = 0.3616 fckbfDf(d -0.416 Df)

Mu1 = kNm

Mu > Mu1, N.A lies outside the flange . xu > Df

When xu = 7Df/3 ,

Moment of Resistance Mu2 = 0.3616 fckbw (7Df/3) x (d - 0.416 (7Df/3))

+ 0.4467 x fckx (bf- bw) x Dfx (d - Df/2)

= kNm > 654.5 kNm ( Mu)

Mu > Mu2 Uniform stress block in the flange. xu > 7Df/3


Method - 1:

Ast =

= mm2

Method - 2:

Mu,flange = 0.4467 x fckx (bf- bw) x Dfx (d - Df/ 2)

= kNm

Mu,web = Mu - Muflange

= kNm

Ast =

= mm2

Method - 3:

653.69

483.35

4612.731

657

488.467

166.034

215.6

4612.7308

)2/(

15.16.411

5.02

fy

uflange

w

w

uweb

cky

ck

Ddf

Mdb

db

M

ff

f

y

fwfckw

ff

w

f

w

u

cky

ck

f

Dbbfdb

d

D

d

D

b

bf

dbM

fff ck

15.1)(4467.021

2

4467.06.411

5.02


7/23


DATE CHECKE

RE

Yugasoft

PROJECT

TITLE REVISION

DESIGNE

MP

DOCUMENT. NO. DATE

Depth of N.A ( Xu)

Mu = 0.3616 f ckbw xu (d - 0.416xu)+ 0.4467 fck(bf- bw) Df (d - Df/ 2)

xu =

xu = mm > 7Df/3 ( 210mm )

Uniform stress block in the flange & xu > 7Df/3

= + 0.4467 fck(bf- bw) Df) 1.15

= mm2

6.0 Example 6 :


bf = mm Df = mm

bw = mm d = mm

fck = N/mm2

fy = N/mm2

Mu = kNm d' = mm

Df/d = 90/450 = < 3xumax/7d


Mulim = 0.3616 fckbw xumax (d - 0.416xumax)+ 0.4467 fck(bf- bw) Df (d - Df/2 )

xumax = 0.479 d = 0.479 x 200 = mm

Mulim = kNm


(0.2053)

215.6

700 50

0.2

90

200 450

30 415

211.403

Ast

4613.16

657.01

fy

( 0.3616 fckbw xu

1200

d

D

d

D

b

b

dbf

Md

ff

w

f

wck

u 2134

21

3616.0664.111202.1

2


8/23


DATE CHECKE

RE

Yugasoft

PROJECT

TITLE REVISION

DESIGNE

MP

DOCUMENT. NO. DATE

Depth of N.A location

xu = xumax = 215.6 mm > Df N.A lies outside the flange. ( xu,max > 7Df/3 )


= ( 0.3616fckbw xumax + 0.4467 fck (bf- bw)Df ) 1.15

+

= mm2


= esc = 0.0035 ( 1 - d'/xumax )

= 0.0035 ( 1 - 50/215.6 )

=

= mm2

fsc = N/mm2

fcc = N/mm2

Asc

4936.236Ast

Ast

fy

0.002688

319.32 349.95

13.38

Asc Mu-Mulim

(fsc-fcc)(d-d')

(Mu-Mulim) X 1.15

fy(d-d')


9/23

DATE

FERENCES

D

06-Oct-09

DATE


10/23

DATE

FERENCES

D

06-Oct-09

DATE


11/23

DATE

FERENCES

D

06-Oct-09

DATE


12/23

DATE

FERENCES

D

06-Oct-09

DATE


13/23

DATE

FERENCES

D

06-Oct-09

DATE


14/23

DATE

FERENCES

D

06-Oct-09

DATE


15/23

DATE

FERENCES

D

06-Oct-09

DATE


16/23

GeneralThe slab portion deforming with the rectangular portion of the beam increases the moment

resisting capacity of the beam where it is in compression.where it is in tension along with

the beam,such as at supports in continuous beams,it cracks and only tensile reinforcement

provided are effective in tension and compressive force is taken by web(rectangular portion)

reducing the beam to a rectangular one.

Structurally,flanged beams may be of two types- T beams & L beams.If slab is

extending on both sides of a beam it is called a T-beam;but if the slab is only on one side

it is an L-beam.

Which portion of the slab will work with the beam depends on span of beam,

width of slab on adjacent sides,breadth of beam,thickness of slab and slab reinforcement at the

junction of beam and slab.

For ready reference the code defines the effective width of flange (slab) as follows:

a)

i) For T-beam, bf = l0 + bw + 6Df

6

ii) For L-beam, bf = l0 + bw + 3Df

12

bf = Effective width of flange for T or L-beam

l0 = distance between sections of zero moments in a beam

bw = breadth of web,and

Df = Thickness of slab

b) If the beam is an isolated one,i.e,only one beam has been provided in the floor/roofsystem,the effectve width of flange shall be as follows:

i) For T-beam, bf = + bw

ii) For L-beam, bf = + bw

Note: lo = 0.7 leff

In case,if the main reinforcement of a slab is parallel to the beam,the transverse reinforcement

of slab at the junction shall not be less than 60% of the main reinforcement at midspan of the

slab extending into it upto atleast quarter span of beam,to have monolithic behaviour.

Location of Neutral Axis

For a Singly Reinforced Section: (Mu


17/23

DOCUMENT. NO. DATE DESIGNED DATE

MP

Yugasoft

PROJECT

Design of T- Beam Technical NotesTITLE REVISION

CL. NO. DESIGN CALCULATIONS

DATEDATE CHECKED

REFERENCES

415

xumax(TABLE B of SP16)

Case 1: When the Neutral Axis (N.A) lies within the flange (xu


18/23


MP

Yugasoft

PROJECT



DATEDATE CHECKED

REFERENCES

=

= Pt,lim x bf x d +

100

=

415

3 x xumax

Case 2: When the Neutral Axis (N.A) lies outside the flange

& 3/7 xu lies within the flange

Rectangular cum Parabolic portion of stress block is acting on the flange.Non Uniform

stress block is acting on the flange.

3 x xumax

=

+

Cuw = 0.361 x fckx bw x xumax yfmax = 0.15 xumax +0.65 DfCuf = 0.446 x fckx (bf - bw) x yf

Cu = 0.361 x fckx bw x xumax + 0.446 x fckx (bf- bw) x yfTu = 0.87 x fy x Ast,w + 0.87 x fy x Ast,f

Mulim = Muw + Muw

Mulim = Cuw x zw + Cufxzf= 0.361 x fckx bw x xumax x (d-0.446xumax)+ 0.446 x fck x (bf - bw) x yf

.

Ast

Df

d 7d

fy

Mu0.87 x fy x (d - 0.416 xu )

Asc

Mu -Mulim(fsc - fcc) x (d - d')

For Doubly Reinforced Section

Ast Mu -Mulim0.87 x fy x (d - d')

250 500

0.2276 0.2053 0.19547

Page 18 of 23


19/23


MP

Yugasoft

PROJECT



DATEDATE CHECKED

REFERENCES

Mulim = 0.361 x fckx bw x xu x (d-0.446xumax) + 0.446 x fck x (bf - bw) x yf

= 0.361 x xumax/d x (1- 0.416 x xumax/d)

fckx bw x d + 0.223 x (bf/bw-1) x (yf/d) x (2-(yf/d))

Mur = Cuw x zw + Cufxzf= 0.361 x fckx bw x xu x (d-0.446xumax)+ 0.446 x fck x (bf - bw) x yf x (d-yf/2)

Equating Tu = Cu

0.87 x fy x Ast = 0.361 x fckx bw x xu + 0.446 x fckx (bf- bw) x yf

= 0.361 x fckx bw x xu +0.446 fck (bf - bw)yf

0.87 x fy

0.361 x fckx bw x xumax + 0.446 x fckx (bf- bw) x yf= 0.87 x fy 0.87 x fy

+

= 0.87 x fy 0.87 x fy x (d - yf/2)

a = 0.416 + 0.0225 k

b = 1+ 0.3 k- 0.195 (Df/d)

c = M_k + 0.4225 k (Df/d)2 - 1.3 k (Df/d)

xu/d =

k = (21/34) * ((bf/bw) -1)

M_k =

Case3: When the Neutrl Axis (N.A) lies outside the flange

& 3/7 xu lies outside the flange

Uniform stress block in flange.

=

Mulim

Ast,lim

Ast,lim

Mu_flange

Mu

Ast

Mu_web

0.3616 x fckx bw x d

a

acbb

2

42

Page 19 of 23


20/23


MP

Yugasoft

PROJECT



DATEDATE CHECKED

REFERENCES

+

3 x xumax

Cuw = 0.361 x fckx bw x xumaxC

uf= 0.446 x f

ck

x (bf- b

w)

x Df

Cu = 0.361 x fckx bw x xumax + 0.446 x fckx (bf- bw) x DfTu = 0.87 x fy x Ast,w + 0.87 x fy x Ast,f

Equating Cu = Tu ie Cuw + Cuf = Tuw + Tuf

xu = 0.87 x fy x Ast- 0.446 x fck(b f-bw) x Df

0.361 fckbw

or

xu =

Mulim = Muw + Muw

Mulim = Cuw x zw + Cufxzf= 0.361 x fckx bw x xumax x (d-0.446xumax)+ 0.446 x fck x (bf - bw) x Df

.

Mulim = 0.361 x fckx bw x xu x (d-0.446xumax) + 0.446 x fckx (bf- bw) x Df

= 0.361 x xumax/d x (1- 0.416 x xumax/d) +

fckx bw x d 0.223 x (bf/bw-1) x (Df/d) x (2-(Df/d))

Mur = Cuw x zw + Cufxzf= 0.361 x fckx bw x xu x (d-0.446xumax)+ 0.446 x fckx (bf- bw) x Dfx (d-Df/2)

Astf =

0.87 x fy

Astw =

Ast =

Equating Tu = Cu

0.87 x fy x Ast = 0.361 x fckx bw x xu + 0.446 x fckx (bf- bw) x yf

Df

d d

Mulim

0.446 fck(bf-bw)Df

0.87 x fy x Astw

0.361 x fckx bw

w

ff

w

f

w

u

cky

ck db

d

D

d

D

b

bf

db

M

ff

fck

.021223.0

6.411

5.02

dbdbf

)M(4.611

f

f0.5w2

ck

u

y

ck

w

uflangeM

Page 20 of 23


21/23


MP

Yugasoft

PROJECT



DATEDATE CHECKED

REFERENCES

0.361 x fckx bw x xumax + 0.446 x fckx (bf - bw) x Df= 0.87 x fy 0.87 x fy

+

= 0.87 x fy 0.87 x fy x (d - Df/2)

Determination of fsc corresponding to strain ( sc )


22/23


MP

Yugasoft

PROJECT



DATEDATE CHECKED

REFERENCES

Eqn of Parabola

@ x = 0.002 - esc , y = gcc

y = x2

or

Note

If the amount of tension reinforcement required to resist the Design moment Mu is less

than the minimum reinforcement,then the minimum reinforcement (Ast,min) is provided for

creep,shrinkage,thermal and other environmental requirements irrespective of the

strength requirement.

More over for singly reinforced sections Pt,provided < pt,limit (under Reinforced Section)

For doubly Reinforced section,the area of tension and compression reinforcement Cl 26.5.1.1 b)

provided should be less than the maximum reinforcement to avoid practical difficulty 26.5.1.2 Note

in placing and compacting concrete. IS 456:2000

3 Transverse or Shear Reinforcement

Nominal Shear Stress tv = Cl 40.1

Under no circumstances even with shear reinforcement the nominal shear stress

in beamstv shall exceed maximum shear stress (tc,max) given in Table 20 of Cl 40.2.3IS 456:2000 for different grades of concrete. IS 456:2000

Grade of concrete M15 Table 20

IS 456:2000

a) If tv > tc,maxThe section is to be redesigned by changing the value of b and d

b)If tv < tc given in Table 19 of IS 456:2000Minimum shear reinforcement shall be provided in accordance with 26.5.1.6

Design shear strength of concrete

1

Table 19 Design Shear Strength of Concrete tc N/mm2

M15

0.28

0.35

0.46

0.54

0.6

0.64

0.68

0.71

0.71

0.71

0.71

0.71

0.71

M25 M30 M35

0.446fck

(0.002)2

Vu/ bd (N/mm2)

M40

tc,max2.5 2.8 3.1 3.5 3.7 4

N/mm2

M20

Grade of concrete

pt M20 M25 M30 M35 M40

0.25 0.36 0.36 0.37

0.15 0.28 0.29 0.29 0.29 0.30.37 0.38

0.5 0.51

0.59 0.6

0.50 0.48

0.75 0.56 0.57 0.59

0.49 0.5

1.25 0.67 0.7 0.71

1.00 0.62 0.64 0.66 0.67 0.68

0.73 0.74

0.78 0.791.50 0.72

1.75 0.75 0.78 0.8

0.74 0.76

2.25 0.81 0.85 0.88

0.82 0.84

0.86 0.88

0.9 0.92

0.93 0.95

0.96 0.98

2.50 0.82

2.75 0.82 0.9 0.94

0.88 0.91

0.99 1.013.00 0.82 0.92 0.96

2.00 0.79 0.82 0.84

2

2)002.0(

)002.0(

446.0sc

ck

cc

fg e

2

002.01446.0

scckcc fg

e

t 6/)151(8.085.0 ckc ft

ck

p

f

89.6

8.0

Page 22 of 23


23/23


MP

Yugasoft

PROJECT



DATEDATE CHECKED

REFERENCES

c) tv > tc given in Table 19 of IS 456:2000, Shear reinforcement

shall be provided to carry a stress equal to tv - tc

Spacing of Shear Reinforcement (Sv)

Cl 26.5.1.6

When tv < tc Sv = IS 456:2000

tv > tc Sv = Cl 40.4 a

IS 456:2000

Asv - Total Cross Sectional area of stirrup legs(mm2)

fy - Characteristic strength of the stirrups 415 N/mm2

Maximum Spacing of Shear Reinforcement (Sv,max)

Cl 26.5.1.5 of IS 456:2000 stipulates that the maximum spacing of Shear reinforcement

measured along the axis of the member shall not be more than 0.75d or 300mm

whichever is less,where d is the effective depth of the section.

Provide Shear Reinforcement at a spacing in such a way that it should not exceed

the minimum of Sv and Sv,max

4 Deflection Check

The Vertical deflection limits may generally be assumed to be satisfied ,provided Cl 23.2.1

that the span to depth ratio is not greater than the value obtained as below IS 456:2000

Allowable L/d = Basic Value x Span Factor x MFt x MFc

a) Basic Values of L/d

Cantilever Simply Supported Continuous Cl 23.2.1 a

IS 456:2000

b) Span Factor = 1 Cl 23.2.1 b

IS 456:2000

c) Modification factor for Tension Reinforcement (MFt)

Cl 23.2.2c

MFt = 2 & Fig 4

IS 456:2000

Service stress in steel fs = 0.58f y

Ast-Provided

d) Modification factor for Compression Reinforcement (MFc)

MFc = 1 + pc 1.5 Cl 23.2.1 d

3 + pc & Fig 5

pc - % of Compression Reinforcement IS 456:2000

Note

For Spans (Cantilever ) above 10m the deflection calculation should be made.

e) Reduction factors for ratios of span to effective depth for flanged beams (F3) 0.8

F3 = 0.8 + 2/7 (bw / bf- 0.3)

The final span/depth ratio allowed is (Basic ratio) x (F1) x (F2) x (F3)

7 20 26

Ast-Required

10

Span (m)

1

0.225 + 0.00322 fs + 0.625 log10 (pt)

b

Af svy

4.0

87.0

b

Af

cv

svy

)(

87.0

tt

t-beam & rectangular beam(3sections)_final

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