Download - T-BEAM & Rectangular Beam(3sections)_final
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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
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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
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CL. NO. T - BEAM DESIGN CALCULATIONS
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.
Area of tension reinforcement ( Ast )
= +
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 :
Design the T-Beam for the following parameters
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
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CL. NO. T - BEAM DESIGN CALCULATIONS
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
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 = 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
Area of tension reinforcement ( Ast )
= ( 0.3616 fckbw xu + 0.4467 fck(bf- bw)yf) 1.15
= mm2
4.0 Example 4 :
Design the T-Beam for the following parameters
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
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CL. NO. T - BEAM DESIGN CALCULATIONS
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
Mu > Mulim Hence designed as doubly reinforced section.
Depth of N.A location
xu = xumax = 159.39 mm > Df N.A lies outside the flange. ( xu,max < 7Df/3 )
Area of tension reinforcement ( Ast )
= ( 0.3616 fckbw xu,max + 0.4467 fck(bf- bw)yf) 1.15
= mm2
Area of compression reinforcement ( Asc )
= esc = 0.0035 ( 1 - d' /xumax )
= 0.0035 ( 1 - 50/159.39 )
=
= mm2
fsc = N/mm2
fcc = N/mm2
5.0 Example 5 :
Design the T-Beam for the following parameters
bf = Df =
bw = d =
fck = N/mm2
fy = N/mm2
Mu =
Df/d = 90/450 = < 3xumax/ 7d
Limiting Moment of Resistance : ( Mulim )
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
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CL. NO. T - BEAM DESIGN CALCULATIONS
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
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 = 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
Area of tension reinforcement ( Ast )
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
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CL. NO. T - BEAM DESIGN CALCULATIONS
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 :
Design the T-Beam for the following parameters
bf = mm Df = mm
bw = mm d = mm
fck = N/mm2
fy = N/mm2
Mu = kNm d' = mm
Df/d = 90/450 = < 3xumax/7d
Limiting Moment of Resistance : ( Mulim )
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
Mu > Mulim Hence designed as doubly reinforced section.
(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
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CL. NO. T - BEAM DESIGN CALCULATIONS
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 )
Area of tension reinforcement ( Ast )
= ( 0.3616fckbw xumax + 0.4467 fck (bf- bw)Df ) 1.15
+
= mm2
Area of compression reinforcement ( Asc )
= 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')
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DATE
FERENCES
D
06-Oct-09
DATE
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DATE
FERENCES
D
06-Oct-09
DATE
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DATE
FERENCES
D
06-Oct-09
DATE
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DATE
FERENCES
D
06-Oct-09
DATE
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DATE
FERENCES
D
06-Oct-09
DATE
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DATE
FERENCES
D
06-Oct-09
DATE
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DATE
FERENCES
D
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DATE
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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
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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
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DOCUMENT. NO. DATE DESIGNED DATE
MP
Yugasoft
PROJECT
Design of T- Beam Technical NotesTITLE REVISION
CL. NO. DESIGN CALCULATIONS
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
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DOCUMENT. NO. DATE DESIGNED DATE
MP
Yugasoft
PROJECT
Design of T- Beam Technical NotesTITLE REVISION
CL. NO. DESIGN CALCULATIONS
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
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DOCUMENT. NO. DATE DESIGNED DATE
MP
Yugasoft
PROJECT
Design of T- Beam Technical NotesTITLE REVISION
CL. NO. DESIGN CALCULATIONS
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
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DOCUMENT. NO. DATE DESIGNED DATE
MP
Yugasoft
PROJECT
Design of T- Beam Technical NotesTITLE REVISION
CL. NO. DESIGN CALCULATIONS
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 )
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DOCUMENT. NO. DATE DESIGNED DATE
MP
Yugasoft
PROJECT
Design of T- Beam Technical NotesTITLE REVISION
CL. NO. DESIGN CALCULATIONS
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
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Page 22 of 23
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7/29/2019 T-BEAM & Rectangular Beam(3sections)_final
23/23
DOCUMENT. NO. DATE DESIGNED DATE
MP
Yugasoft
PROJECT
Design of T- Beam Technical NotesTITLE REVISION
CL. NO. DESIGN CALCULATIONS
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
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Af
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svy
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87.0
tt