Tee Beam Book Example.xlsx

0.600

CLEAR ROAD

0.300 0.150 0.300 0.600

1.350

G1

3.000

Cross Section At Mid-Span All Dimensions are in 'm'

CLEAR ROADWAY 7.5

0.150 x0.300

G1

G2

3.000

d-Span

n 'm'

G3

3.000

DESIGN OF RCC T - GIRDER USING STAAD RESULTS :

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

BASIC DESIGN DATA Effective span Angle of skew Clear carriage way Spacing of main girder c/c Spacing of cross girder c/c Width of crash barier Thk of deck slab Thk of wearing coat Length of cantilever Cantilever slab thk at fixed end Cantilever slab thk at free end No of main girder Depth of main girder Web thk of main girder ( at center ) Web thk of main girder ( at support ) Top haunch Bottom haunch Bottom bulb No of cross girder Depth of cross girder Web thk of cross girder

22

Leff Ang Bcw Spmg Spcg Wkerb Df Wc Lcan Dcan1 Dcan2 Nomg Dmg bwmc bwms Thw x Thh Bhw x Bhh Bbw x Bbh Nocg Dcg bwcg

21.40 0.000 7.50 3.000 5.35 0.325 0.210 0.070 1.050 0.310 0.160 3 1.800 0.300 0.600 0.150 0.150 0.600 3 1.200 0.300

Grade of concrete

Cgrade

30

23 24

Grade of reinforcement Clear cover

Sgrade cov

500 0.040

25

Unit weight of concrete

wcon

2.500

26

Stress in concrete (compression)

fc

830

27 28

Stress in steel (tension) Modular ratio

ft m

20000 10

D RESULTS :

x x x

0.300 0.300 0.300

m degree m m m m m m m m m m m m m m m m m m m N/mm2 N/mm2 m t/m3 t/m2 t/m2

OUTER GIRDER SECTION

L/2 of span

DATA Moment (kN.m) D(m) Df Overall Depth of Structure bw (m) Effective Span Spacing of girder (m) Clear Clover Effective Cover (mm) Dia of Bar (mm) Stirrups (mm) Grade of concrete Fck Grade of reinforcement Fy Design Ultimate Stress Fyd Design Bond Stress Fbd k b1 lo

2400.34 1.8 0.21 2.2 0.3 21.40 3.175 45 250 32 10 30 500 434.7826086957 2.7 36 1.2 21.40 Output

Effective Depth (mm) Df/d Ratio

0.1076923077

Effective width Beff.1

2.38

Beff (m)

5.06

Bf (m)

2.5

Xumax

0.864

Df (m)

0.21

1950

Mu (N-mm)

13406635620

Mu (KN-m)

13406.63562

Ma (KN-m)

2400.34

Mu > Ma

OK

Ast (Required)

2853.0831262543

Ast(Provided) Lb.net (mm)

9847.04 1449.2753623188 966.1835748792

Length of Bearing (mm)

GIRDER 3L/8 of span

L/4 of span

2145.24 1.8 0.21 2.2 0.3 21.40 3.175 45 240 32 10 30 500 434.7826086957 2.7 36 1.2 21.40

1400.16 1.8 0.21 2.2 0.3 21.40 3.175 45 220 32 10 30 500 434.7826086957 2.7 36 1.2 21.40

1960

1980

0.1071428571

0.1060606061

2.38

2.38

5.06

5.06

2.5

2.5

0.864

0.864

0.21

0.21

13478644620

13622662620

13478.64462

13622.66262

2145.24

1400.16

put

OK

OK

2534.5787004901

1633.522895758

9847.04 1449.2753623188 966.1835748792

9847.04 1449.2753623188 966.1835748792

INNER GIRDER SECTION

L/2 of span

DATA Moment (kN.m) D(m) Df Overall Depth of Structure bw (m) Effective Span Spacing of girder (m) Clear Clover Effective Cover (mm) Dia of Bar (mm) Stirrups (mm) Grade of concrete Fck Grade of reinforcement Fy Design Ultimate Stress Fyd Design Bond Stress Fbd k b1 lo

3L/8 of span

5623.49 1.8 0.21 2.2 0.3 21.40 3 45 250 32 10 30 500 434.7826086957 2.7 36 1.2 21.40

4814.21 1.8 0.21 2.2 0.3 21.40 3 45 240 32 10 30 500 434.7826086957 2.7 36 1.2 21.40

Output Effective Depth (mm) Df/d Ratio

1950

1960

0.1076923077

0.1071428571

Effective width Beff.1

2.38

2.38

Beff (m)

5.06

5.06

Bf (m)

2.5

2.5

Xumax

0.864

0.864

Df (m)

0.21

0.21

Mu (N-mm)

12667687200

12735727200

Mu (KN-m)

12667.6872

12735.7272

Ma (KN-m)

5623.49

4814.21

Mu > Ma

OK

OK

Ast (Required)

6763.1482455596

5742.8083554637

Ast(Provided)

8616.16 1449.2753623188 966.1835748792

8616.16 1449.2753623188 966.1835748792

Lb.net (mm) Length of Bearing (mm)

L/4 of span 3925.46 1.8 0.21 2.2 0.3 21.40 3 45 220 32 10 30 500 434.7826086957 2.7 36 1.2 21.40

1980 0.1060606061 2.38 5.06 2.5 0.864 0.21 12871807200 12871.8072 3925.46 OK 4619.7604564391 8616.16 1449.2753623188 966.1835748792

DESIGN OF CANTILEVER SLAB At Intermediate Section Dead Laod Moment Summary of moments (DL+ SIDL) Sl. No. 1 2 3 4 Total

Items. Crash barrier Wearing coat Kerb Deck slab

Loads (kN) 8.125 1.65 4.05 6.16875 Total Load

= 8.125 = 1.650 = 4.050 = 6.169 = 19.99

Live Load Moment :Class A Loading

As Per IRC:6:2000, only Class A and Class B Vehicles can come to Cantilever Portio

Effective width (beff) = 1.2a + b1 Using IRC 112:2011, AN Where, a is the distance of load cg from support. b1 is tyre width + twice the thickness of wearing coat Effective width (beff) As beff <1.2m, hence overlaping won't oc Therefore, beff Impact factor

As Per IRC 112:2011

Therefore, live load /m width including impact Moment at the face of support due to live load =109.615*0.325 Design bending moment (DL + SIDL + live load ) Main Reinforcement Required effective depth Required effective depth Depth provided Spacing Required Ast Provided Ast

=(49.37/0.36*30*0.48*(1-0.416*0.48))^0.5

Dcan1-(Cover+ S=1000 *π/4×(16)^2

=(0.5*Fck)/Fy*(1-(√1-(4.6M///Fckbd^2))*bd) Provide 40mm Clear Cover and 16mm dia bars @ spa

Required Ast(446.02)

<

Distribution Reinforcement

Design bending moment [0.2 x (DL + SIDL)BM + 0.3 x live load BM]

Design B.M Required Ast Provide 12mm

=(0.5*Fck)/Fy*(1-(√1-(4.6M///Fckbd^2))*bd) @ 175 c/c at top & bottom

Provided Ast

Provided Ast(646.19) >Req

Hence Ok Check For Shear in Cantilever Portion Dead Load Shear Total Shear 19.99+1.5×(57/0.780) (dead load shear + liveload shear) Design Shear 1.5*129.6091346 Ved As Per Clause 10.3.2(2) of IRC 112:2011 Page No.88 Shear Resistance of a Structure is given by VRdc=[0.2k Subject to a Minimum of VRdc=(Vmin+0.15)bwd K=1+√(200/d) K=1+√(200/410)

<2.0 Where d= depth in mm 1.6984302958 <2.0 Hence OK

Vmin=0.031k3/2fck1/2 Vmin =0.031*(1.6942)^3/2*(30)^0.5 Vmin 1

=

0.3758321 0.3758321

=1058.22/(1000*410)

VRdc=(0.12*1.6942*(80*0.00255*30)^(1/3)*1000*410 VRdc(min)=Vmin*b*d = 0.374429*1000*410

Ved > Hence OK Also IRC 112:2011, Clause-10.3.2(5) Specified the Following Criteria VRdc(min

Ved ≤ 0.5bwdvfcd V So

= = 0.5bwdvfcd = = > Hence OK

0.6(1-Fck/310) 0.5 1199845.2 1199.8452 Ved

Lever arm (m) 0.8875 0.3625 0.7500 0.4691489362 Total B.M

Moment at the face of support (kN-m) 7.211 0.598 3.038 2.894 13.74

me to Cantilever Portion.

sing IRC 112:2011, ANNEXURE B.3.2 PAGE NO.279

a b1 beff nce overlaping won't occurs.

ng impact 109.615*0.325

48*(1-0.416*0.48))^0.5

Dcan1-(Cover+d/2) 1000 *π/4×(16)^2

bd^2))*bd) 16mm dia bars @ spacing of 190 mm c/c

= = =

0.33 0.39 0.780

m m m

=

0.780

m

=

1.5

= 109.62 kN/m = 35.63 kN/m =

49.37

kN/m

= = = =

109.08 109.08 mm 262.00 mm 207.60 mm

= 446.02 mm2/m = 1058.09 mm2/m

Provided Ast(1058.09)

+ 0.3 x live load BM]

=

Design B.M

20.56

kN-m

= 30.84094

bd^2))*bd)

= 275.57 mm2/m = 646.19 mm2/m

ded Ast(646.19) >Required Ast (275.57) Hence Ok

= 19.99 kN 129.6091346154 kN

194.4137019231 kN 112:2011 Page No.88

Where d= depth in mm

=

0.0025807035 153367.01439 N 153.36701439 kN 154091.17607 N 154.09117607 kN

ollowing Criteria

6(1-Fck/310) N kN

Near Expansion Joint

5.7 t

0.21

1.050 m 0.25 0.5 Tyre size

Class A

Dead Laod Moment Summary of moments (DL+ SIDL) Sl. No. 1 2 3 4 Total

Items. Crash barrier Wearing coat Kerb Deck slab

Loads (kN) 8.125 1.65 4.125 2.1 Total Load

Live Load Moment :Class A Loading

As Per IRC:6:200, only Class A and Class B Vehicles can come to Cantilever Portion

Effective width (beff) = 1.2a + b1 Where, a is the distance of load cg from support. b1 is tyre width + twice the thickness of wearing coat Effective width (beff) As beff < 1.2m, hence overlaping won't oc Therefore, beff Impact factor

As Per IRC 112:2011

Therefore, live load /m width including impact Moment at the face of support due to live load Design bending moment (DL + SIDL + live load ) Main Reinforcement Required effective depth Required effective depth

=(47.06/0.36*30*0.48*(1-0.416*0.48))^0.

Depth provided Spacing

S=1000 *π/4×(16)^2

=(0.5*Fck)/Fy*(1-(√1-(4.6M///Fckbd^2))*bd) Provide 40mm Clear Cover and 16mm dia bars @ spac

Required Ast Provided Ast

Required Ast(424.61) Distribution Reinforcement

Design bending moment [0.2 x (DL + SIDL)BM + 0.3 x live loa

Required Ast Provide 12mm

=(0.5*Fck)/Fy*(1-(√1-(4.6M///Fckbd^2))*bd) @ 225 c/c at top & bottom

Provided Ast

Provided Ast(502.59) >Requ

Check For Shear in Cantilever Portion Dead Load Shear Total Shear 16+1.5×(57/0.790) (dead load shear + liveload shear) Design Shear 1.5*124.2278481 As Per Clause 10.3.2(2) of IRC 112:2011 Page No.88 Shear Resistance of a Structure is given by VRdc=[0.2k VRdc=(Vmin+0.15)bwd

K=1+√(200/d)

<2.0

K=1+√(200/410)

1.6984302958 Hence OK

Vmin=0.031k fck 3/2

1/2

Vmin =0.031*(1.6942)^3/2*(30)^0.5 Vmin 1

=

=893.50/(1000*410)

VRdc=(0.12*1.6984302*(80*0.00255*30)^(1/3)*1000*410 VRdc(min)=Vmin*b*d = 0.37583*1000*410

VRdc(min

> Hence OK

Also IRC 112:2011, Clause-10.3.2(5) Specified the Following Criteria Ved ≤ 0.5bwdvfcd V So

0.5bwdvfcd

= = = = > Hence OK

N) = 8.125 = 1.650 = 4.125 = 2.100 = 16.00

Lever arm (m) 0.8875 0.3625 0.7500 0.4691489362 Total B.M

Moment at the face of support (kN-m) 7.211 0.598 3.094 0.985 11.89

n come to Cantilever Portion Using IRC 112:2011, ANNEXURE B.3.2 PAGE NO.279

port. earing coat

a b1 beff m, hence overlaping won't occurs.

= 0.33 m = 0.40 m = 0.790 m = 0.790 m =

1.5

ncluding impact =108.228*0.325

= 108.23 kN/m = 35.17 kN/m = 47.06 kN/m

*30*0.48*(1-0.416*0.48))^0.5

= 106.51 = 106.51 mm

Dcan1-(Cover+d/2) S=1000 *π/4×(16)^2

Fckbd^2))*bd) r and 16mm dia bars @ spacing of 225 mm c/c

= 262.00 mm = 207.60 mm = 424.61 mm2/m

= 893.50 mm2/m Provided Ast(893.50)

<

L + SIDL)BM + 0.3 x live load BM] Design B.M

Fckbd^2))*bd)

19.96 kN-m = 29.94695 = 267.44 mm2/m = 502.59 mm2/m

rovided Ast(502.59) >Required Ast (267.44) Hence Ok

= 16.00 kN 124.2278481013 kN Ved 186.3417721519 kN IRC 112:2011 Page No.88

Where d= depth in mm <2.0

0.3758321368 0.3758321368 =

3)*1000*410

Ved

d the Following Criteria

0.6(1-Fck/310) 0.5 1199845.1613 N 1199.8451613 kN Ved

0.002179261 144970.6659 N 144.9706659 kN 154091.1761 N 154.0911761 kN

SHEAR RESISTANCE CHECK Let us Provide 1 bent up bar at 645 mm interval Maximum Spacing of Bent up bars as Per Clause-16.5.2(8) of IRC 112:2011 Sbmax=0.6d(1+cot∝)

=

0.6*1950*2

=

Now =

1*(∏/4)*30*30

=

706.5

=

1*(∏/4)*28*28

=

615.44

As per clause 10.3.3.3 of IRC:112-2011, Now

z

S = =(0.9d) = =0.8fy = = =

0.645 m 1755 mm 24 N/mm2 bw

300

65246.29 N 65.24629 kN

= =1×300×0.9×1950×0.6×0.36×30×2/2 = 3411720 = 3411.72 > 65.2463 Hence OK = =

56836.77 N 56.83677 kN

<

3411.72

Hence OK Design shear resistance of member without shear reinforcement is given by VRdc =

=0 vmin=

K=

=

1.32026 Hence OK

vmin= (VRdc)min

0.25758 = = 150683.5 N = 150.6835 kN ( )O.G

Now

( )I.G (VRdc)O.G

(VRdc)I.G

= =

0.00842 < 0.00736 <

= =

249893.3 N 249.8933 kN > Hence OK

(VRdc)min

= =

239120.8 N 239.1208 kN > Hence OK

(VRdc)min

IRC 112:2011 2.34 > 0.645 Hence OK mm2

(Outer Girder)

mm2

(Inner Girder)

mm

nt is given by

< Hence OK

2

0.02 0.02

SHEAR REINFORCEMENT DISTRIBUTION ON OUTER GIRDER Outer Girder Total Shear Design Shear

= =

458.69 kN 688.035 kN

(VRdc)O.G Shear resisted by girder without shear reinforcement Hence design shear for which shear reinforcement will be provided Ved

As per clause -10.3.3.3(2)& clause-16.5.2(3) of IRC :112-2011 only 50% of the shear will be re Let’s provide 4-legged 8mmф vertical stirrups

4 legged 8 mmф

Asw =

200.96 mm2 Using clause-10.3.3.2 of IRC:112-2011, Vrd.max=

=

Vrd.max=

1022690.584 N 1022.690584 kN > 219.0708477 Hence OK As per same clause ,spacing of vertical stirrups given by S=

= 643.965 mm = 321.982 mm Provide 4-legged 8mm ф vertical stirrups@ 300mm c/c starting/end of girders. As per cl-16.5.2 of IRC:112-2011, Min. Shear reinforcement ratio is

w.min =

=

0.00079

0.00208 >

0.00079

Provided shear reinforcement ratio is (w)=

=

Hence OK

UTER GIRDER

SHEAR REINFORCEMENT DIS Outer Girder Total Shear Design Shear

=

249.893 kN

= 438.142 kN 50% of the shear will be resisted by the bent up bars =

of girders.

Shear resisted by girder without shear rei Hence design shear for which shear reinf

As per clause -10.3.3.3(2)& clause-16.5.2

219.071 kN

Let’s provide 4-legged 8mmф vertical sti Asw =

200.96 Using clause-10.3.3.2 of IRC:112-2011, 0.59952

Vrd.max= Vrd.max=

924466.0645 924.4660645 Hence OK As per same clause ,spacing of vertical st S=

Provide 4-legged 8mm ф vertical stirrups As per cl-16.5.2 of IRC:112-2011, Min. Shear reinforcement ratio is

Provided shear reinforcement ratio is (w)=

R REINFORCEMENT DISTRIBUTION ON INNER GIRDER

r Total Shear

= =

412.36 kN 618.54 kN

(VRdc)O.G = d by girder without shear reinforcement 239.121 kN n shear for which shear reinforcement will be provided Ved = 379.419 kN e -10.3.3.3(2)& clause-16.5.2(3) of IRC :112-2011 only 50% of the shear will be resisted by the bent up bars

=

e 4-legged 8mmф vertical stirrups

4 legged 8 mmф

mm2 e-10.3.3.2 of IRC:112-2011, =

0.54194

N kN > 189.71 Hence OK clause ,spacing of vertical stirrups given by =

743.631 mm

gged 8mm ф vertical stirrups@ 200mm c/c starting/end of girders. .5.2 of IRC:112-2011,

einforcement ratio is

w.min =

=

0.00079

ear reinforcement ratio is =

0.0009 > 0.00079 Hence OK

189.71 kN

y the bent up bars

Design of Cross Girder(Intermediate) Maximum L.L Moment Maximum D.L Moment Total Moment

= = =

84.25 kN-m 586.23 kN-m 670.48 kN-m

Asuuming Effective Depth

=

1700 mm

Ast

=(792.9*10^6)/0.87*500.1700*(1-0.416*(250/1700)) Ast(Req) =

965.7477 mm2

Provide 4 bars of 20mm dia No. of bars Dia of bars

= =

4 20 mm

Ast(Prov) =

1256.6368 mm2

Ast(Prov) > Ast(Req) Hence OK Side Face Reinforcement (As Per clause 31.4 of IS456:2000) 0.1% of web area on either face with Spacing not more than 450 mm Ast(Req) = 360 mm2 Provide 4-12mm dia bars each face uniformly as side reinforcement. Shear Check for Intermediate Cross Girder Clause-10.3.2 of IRC-112:2011 K vmin=

= = = = = =

1.3429972 0.2642622 0 1256.6368 mm2 300 mm 0.002464

VRdc

= =

147753.34 N 147.75334 kN

(VRdc)min

=

134773.7 N

vmin=

Ast Bw VRdc = VRdc (VRdc)min

=

(VRdc)min VRdc

Dead Load Shear from STAAD Live Load Shear from STAAD Design Shear Ved Provide 8 mm dia and 4 legged Stirrups

Asw

= 134.7737 kN (VRdc)min > Hence OK = 241.042 kN = 34.634 kN =

413.514 kN 4 legged 8 mmф

=

Min. Shear reinforcement ratio is

201.06189 mm2 w.min =

=

0.0007887

Provided shear reinforcement ratio is (w)=

=

0.002681 > 0.0007887 Hence OK

As Per Clause 16.5.2 of IRC:112-2011(6, 7, 8, 9) Specifies that Smin

= = =

dg+10

40

Whichever is Largest

2øs

Maximum Longitudinal Spacing Smax 0.75d(1+cotα) =

=

Maximum Longitudinal Spacing for Bent-up bars Sbmax 0.6d(1+cotα) = =

1275 mm

1020 mm

Design of Cross Girder(End) Maximum L.L Moment Maximum D.L Moment Total Moment

= = =

Asuuming Effective Depth

=

Ast

=(792.9*10^6)/0.87*500.1700*(1-0.416*(250/1700)) Ast(Req)

=

Provide 4 bars of 20mm dia No. of bars Dia of bars

= = Ast(Prov) =

Ast(Prov) > Hence OK Side Face Reinforcement (As Per clause 31.4 of IS456:2000) 0.1% of web area on either face with Spacing not more than 450 mm Ast(Req) = Provide 4-12mm dia bars each face uniformly as side reinforcement. Shear Check for End Cross Girder Clause-10.3.2 of IRC-112:2011 K vmin=

vmin=

Ast Bw

= = = = = =

VRdc = VRdc (VRdc)min

VRdc

= =

(VRdc)min

=

=

(VRdc)min VRdc

Dead Load Shear from STAAD Live Load Shear from STAAD Design Shear Ved Provide 8 mm dia and 4 legged Stirrups

Asw

= > Hence OK = = =

=

Min. Shear reinforcement ratio is

w.min =

= Provided shear reinforcement ratio is (w)=

=

0.0026808 > Hence OK

As Per Clause 16.5.2 of IRC:112-2011(6, 7, 8, 9) Specifies that Smin

= = =

dg+10

40

Whichever is Largest

2øs

Maximum Longitudinal Spacing Smax 0.75d(1+cotα) =

=

Maximum Longitudinal Spacing for Bent-up bars Sbmax 0.6d(1+cotα) = =

25.98 kN-m 24.63 kN-m 50.61 kN-m 1700 mm

6*(250/1700)) 72.897762 mm2

4 20 mm 1256.6368 mm2 Ast(Req)

nce OK 1.4 of IS456:2000) pacing not more than 450 mm

360 mm2 mly as side reinforcement.

rder

1.3429972 0.2642622 0 1256.6368 mm2 300 mm 0.002464

147753.34 N 147.75334 kN 134773.7 N

134.7737 kN (VRdc)min

nce OK 74.242 kN 25.615 kN 149.7855 kN 4 legged 8 mmф 201.06189 mm2

0.0007887 40.249224 0.0007887

nce OK

) Specifies that

Whichever is Largest

1275 mm

1020 mm