Excel tools to Check Beam Jacketting

Beam of 8 m

Output:Note : DRS = Doubly Reinforced Section, SRS = Singly Reinforced Section, NA = Not Applicable

Capacity-Demand Ratio, C-D-R Ratio= 1.05 Actual

Demand tensile Steel Reinforecement, Ast 3007.35 mm2

2864.140

Existing tensile Steel thro' bar pro. 20 3

Existing tensile Steel thro' bar pro 20 0

Existing tensile Steel Extra bar pro 20 0

Existing tensile Steel Reinforecement, Ast 942.48 mm2

Retrofitting Reqd.

Demand Compressive Steel Reinforecement, Asc 2296.93 mm2

2187.548

Existing Comp. Steel thro' bar pro. 20 3

Existing Comp. Steel thro' bar pro 20 0

Existing Comp. Steel Extra bar pro 20 0

Existing Comp. Steel Reinforecement, Asc 942.48 mm2

Retrofitting Reqd.

Stirrups Provided: Section Fails in Shear

Provide 8 Ф bar Spacing 100.00 mm @ C/C Leg = 2

Crack Width Calculation:Stress in tensile reinforced level fsb 242.19 N/mm

2

Spacing of reinforcement S 81.083 mm

acr = ((S / 2)^2+d'^2)^0.5 - Dia /2 = 46.669 mm

Ɛ1b = [fs*((a' - xu) / [(d-xu)*2*10 ^5] 0.0015366

Ɛ2b= bt*(a'-xu)*(D-xu) / [600000*Ast*(d - xu)] 0.0000328

Ɛmb = Ɛ1b-Ɛ2b 0.0015037

0.1734825

Stress in tensile reinforced level fst 0.000 N/mm2

Ɛ1t= fs*(a' - (-D)) / [(d- (-D))*2*10 ^5] 0.0000000

Ɛ2t= bt*(a'-(-D))*(D-(-D)) / [600000*Ast*(d - (-D))] 0.0000951

Ɛmt = Ɛ1t-Ɛ2t -0.0000951

-0.0125339

0.1609486 < 0.2

INPUT DATA:- Envelope for Load Combination

INPUT DATA:- Axial force, Torsion Moments, Bending moments and Shear force:

Design Shear force, Fy 155.390 KN

Design Torsional Moment for shear Tu=Mx 0.000 KNm

Moment M'uz 268.560 KN

Design Torsional Moment for moment Tu=Mx 0.000 KNm

Axial Force, Fx 0.000 KN T

Bending in another direction Muy 0.000 KNm

INPUT DATA:- Material Properties :

GF - LONG BEAM SPAN

RECTANGULAR BEAM CAPACITY CHECK

Wcrb = 3*acr*Ɛm/ [1+2*(acr - Cmin) / (D - xu)]

Wcrt = 3*acr x Ɛm / [1+2*(acr - Cmin) / (D - (-D))]

Wcr =Wcrb+Wcrt

Characteristic Strength of concrete, fck 20 N/mm2

Grade of Steel, fy 415 N/mm2

INPUT DATA:- Material Properties and Dimensions of Beam:

Thickness of flange, Df 125 mm

Total Depth of T- Beam, D 350 mm

Width of Flange, bf 230 mm

Width of web in compression fibre, bcw 230 mm

Width of web in tension, btw' 230 mm

Layer of bar in tension zone 1

Layer of bar in Compression zone 1

Cover, C 25 mm

Stirrups of Design purpose 8 mm

Dia of bar in Tension Reinforcement 20 mm

Dia of bar in Compression Reinforcement 20 mm

Spacer for vertical spacing 32 mm

Calculations of required Design Moments, Shear forces:

Resultant Moment of ( Muy and Muz') = M''uz 268.56 KNm

Design Moment Mu 268.56 KNm

Design shear force Vu 155.39 KN

Calculations of required Section properties:

Width of web in reiforcement level in tension, btw 230.00 mm

Width of Web, Average btw 230.00 mm

Effective Depth , d (Calculated) 307.00 mm

Width of small portion in the flange, bs 0.000 mm

Centroid of Section from Compression fibre Cc 161.111 mm

Centoid of section from tension fibre Ct 188.889 mm

Df/d (Calculated) 0.407

RESULT IN ACCORDANCE WITH DIFFERENT CONDITIONS:

1. Condition: If Neutral Axis lies in the flange? GO TO 2 (C)

Muf, lim= 0.36*fck*bf*Df*(d-0.416*Df) 52.79 KNm DRS-NA

-215.774 KNm DRS-NA

Neutral Axis lies in the Flange? NO DRS-NA

For Mu or Mu,lim Area of Steel, Ast 572.56 mm2

DRS-NA

For Tension Area of Steel, Ast2 -57.35 mm2

DRS-NA

for BM+Tension, total rqd. Steel, Ast 515.21 mm2

DRS-NA

Neutral Axis Ratio, Xu/d 0.40717 DRS-NA

Neutral Axis, Xu 125.000 mm DRS-NA

Due to Mu Compressive strain Ɛcc 0.00297 DRS-NA

Total Strain due to (Mu+Fx) Ɛcct 0.00297 < 0.0035 DRS-NA

for BM+compression, TrialTotal Rqd. Steel, Ast 515.210 mm2

DRS-NA

Test : Muf,lim > Mu for Neutral in flange

Due To Fx, Additional compressive stress fsc NA N/mm2

DRS-NA

Additional compressive stress fcc NA N/mm2

DRS-NA

Pu NA KN DRS-NA

Hit and trial, Extra Comprssve strain due to P, Ɛcc 0.00000000 NA DRS-NA

Test : Fx-Pu =0 0.000000 DRS-NA

for BM+compression, Extra Increased area Ast 0.000 mm2

DRS-NA

Required Ast 515.210 mm2

DRS-NA

2. Condition: If Neutral Axis lies in the webNeutral Axis lies in the Web YES

Neutral axis depth ratio α = Xu,max/d 0.4791

Maximum Neutral Axis for balanced design Xu,max 147.09 mm

yf,max 103.31 mm

Compressive force for straight portion, C1 129.33 KN

Compressive force for Parabolic portion, C2 114.96 KN

Compressive force Trap. Web, Cu = C1+C2 244.28 KN

Compressive force for Flange portion, C3 0.00 KN

Compressive force for Small Flange portion, C4 0.00 KN

Total Compressive force, C 244.28 KN

Moment of C1 about Neutral Axis, Integration I1 14945.94 KNmm

Moment C2 about Neutral Axis Integration I2 6038.76 KNmm

Moment f Cu about neutral axis Integration Iu 20984.70 KNmm

CG of Cu from Neutral axis ,Y 85.903 mm

CG of Cu from Extreme Compression fibre X 61.183 mm

For average web only X/d 0.19929

CG C from Extreme Compression fibre X 61.183 mm

60.05 KNm

Capacity Mu, lim (Only for trapezodal Web) 60.05 KNm

Capacity Mu, lim (trapezodal Web + Flange) 60.05 KNm

208.51 KNm DRS

245.812 mm

245.817 mm

676.591 mm2

676.591 mm2

-0.000689 DRS-NA

0.50839 DRS-NA

-60.049 DRS-NA

Neutral Axis Xu=(-b+√(b2-4*a*c))/2*a 147.661 mm DRS-NA

yf=0.15*xu+0.65*Df 103.399 mm DRS-NA

a= - 0.14976*fck*bw - 0.0050175*fck*(bf - bw)

b= 0.36*fck*bw*d + 0.00669*fck*(bf - bw)

c= 0.2899*fck*Df*(bf - bw)*(d-0.325*Df)-Mu

Capacity if average web +flange Muw, lim=

0.36*fck*bw*Xu,max*(d-0.416*Xu,max) + 0.446*fck*(bf-bw)*yf*(d-

Test, Mu > Mu,lim (Doubly or Singly), Mu - Mu,lim=

for average web Lever Arm z= jd

for (Trap. Web+ Flange) Lever Arm z= jd

For avg. web +flange balanced section Ast,max

For (web +Flange) balanced section Ast,max

Lever Arm z = jd 245.573 DRS-NA

677.263 mm2

DRS-NA

2 (C) Condition: Doubly Reinforced Section YES

Number of Layer of Comression bar 1 No.

Dia of Main reinforcement in compression 20 mm

Spacer for Vertical spacing 32 mm

Dia of Stirrups 8.00 mm

Clear cover, C 25.00 mm

Charactertistics strength of Concrete fck 20.00 N/mm2

Yield stress of steel fy 415.00 N/mm2

Helping calculation, d'' 52.00 mm

Effective depth from in compression side d' 43.00 mm

Effective depth at tension side d 307.00 mm

Neutral Axis depth ration Xu,max/d 0.479

Neutral Axis Depth Xu.max 147.09 mmm

Compression level strain Ɛsc Or Ɛcc 0.00248

Compressive level stress of steel fsc 344.720 N/mm2

Compressive level stress of concrete fcc=446*fck*(Ԑcc-250*Ԑcc^2) 8.920 N/mm2

Area of compressive stress Asc 2352.038 mm2

2187.548 mm2

For DRS Total Area of tension steel Ast 2864.140 mm2

for Tension, Area of Steel, Ast3 -2187.55 mm2


Neutral Axis, Xu 147.086 mm

Due to Mu Compressive strain Ɛcc 0.00350

Total Strain due to (Mu+P) Ɛcct 0.00350 < 0.0035 OK




Pu 0.000 KN

Hit and trial, Extra Comprssve strain due to P, Ɛcc 0.00000000 YES

Test : Fx-Pu =0 NA

Extra Increased area 0.000 mm2

Required Double reinforced Ast 676.591 mm2

Check for Shear forceDesign Shear force 155.39 KN

for Asc, Area of tension steel Ast2

For Df/d > 0.2 Ast

Shear stress 2.201 N/mm2

Percentage of tension steel, pt 1.335 %

Percentage of compressive steel, pc 1.335 %

Percentage of tension and compressive steel, pt 4.893 %

α=0.8*fck/6.89*pt 0.475

tc = 0.85*SQRT(0.8*fck)*(SQRT(1+5*α)-1)/6*α 0.999 N/mm2

1.00 x 0.999 = 0.999

< 2.201 Not OK

Shear Reinforcement is required.

Dia of of Vertical Stirrups 8.00 mm

Number of leg 2.00

Planned area to Provide Asv 100.53 mm2

Required spacing of stirrups Svreqd. 131.313 mm

Plan to Provide, spacing of stirrups Sv 100.000

Minimum required Asv 0.4*b*Sv/0.87*fy 25.481 mm2

Ф 4.028 mm

Rqd. Stirrups Ф 8.00 Leg 2.000 100.00 mm C/C

Permissible shear stress, K´tc =

Beam of GF - LONG BEAM SPAN 8 m

Output:Note : DRS = Doubly Reinforced Section, SRS = Singly Reinforced Section, NA = Not Applicable

Additional Tensile Steel reqd on jacketting, Ast 1493.81 mm2

Existing tensile Steel thro' bar pro. 20 3

Existing tensile Steel thro' bar pro 20 0

Jacket tensile Steel bar at 2nd Layer 25 5

Tensile Steel Reinft of Jacketted Beam, Ast 3396.85 mm2

OK.

Additional Comp. Steel reqd. on jacketting, Asc 1464.35 mm2

Existing Comp. Steel thro' bar pro. 20 3

Existing Comp. Steel thro' bar pro 20 0

Jacket Comp. Steel bar at 2nd Layer 25 3

Comp. Steel Reinft of Jacketted Beam, Asc 2415.10 mm2

OK.

Stirrups Provided: Section passed in Shear

Provide 10 Ф bar Spacing 75.00 mm @ C/C Leg = 2

Crack Width Calculation:Stress in tensile reinforced level fsb 236.52 N/mm

2

Spacing of reinforcement S 221.000 mm

acr = ((S / 2)^2+d'^2)^0.5 - Dia /2 = 49.497 mm

Ɛ1b = [fs*((a' - xu) / [(d-xu)*2*10 ^5] 0.0013898

Ɛ2b= bt*(a'-xu)*(D-xu) / [600000*Ast*(d - xu)] 0.0002685

Ɛmb = Ɛ1b-Ɛ2b 0.0011212

0.1509630

Stress in tensile reinforced level fst 0.000 N/mm2

Ɛ1t= fs*(a' - (-D)) / [(d- (-D))*2*10 ^5] 0.0000000

Ɛ2t= bt*(a'-(-D))*(D-(-D)) / [600000*Ast*(d - (-D))] 0.0006597

Ɛmt = Ɛ1t-Ɛ2t -0.0006597

-0.0944048

0.0565583 < 0.2

INPUT DATA:- Envelope for Load Combination

INPUT DATA:- Axial force, Torsion Moments, Bending moments and Shear force:

Design Shear force, Fy 155.390 KN

Design Torsional Moment for shear Tu=Mx 0.000 KNm

Moment M'uz 268.560 KN

Design Torsional Moment for moment Tu=Mx 0.000 KNm

Axial Force, Fx 0.000 KN T

Bending in another direction Muy 0.000 KNm

INPUT DATA:- Material Properties :

RECTANGULAR BEAM JACKETTING WITH STEEL BARS

Wcrb = 3*acr*Ɛm/ [1+2*(acr - Cmin) / (D - xu)]

Wcrt = 3*acr x Ɛm / [1+2*(acr - Cmin) / (D - (-D))]

Wcr =Wcrb+Wcrt

Characteristic Strength of concrete, fck 20 N/mm2

Grade of Steel, fy 415 N/mm2

INPUT DATA:- Material Properties and Dimensions of Beam:

Thickness of flange, Df 125 mm

Total Depth of T- Beam, D 650 mm 350 mm

Width of Flange, bf 430 mm 230 mm

Width of web in compression fibre, bcw 430 mm 230 mm

Width of web in tension, btw' 430 mm 230 mm

Layer of bar in tension zone 2

Layer of bar in Compression zone 2

Cover, C 25 mm

Stirrups of Design purpose 10 mm

Dia of bar in Tension Reinforcement 20 mm

Dia of bar in Compression Reinforcement 20 mm

Spacer for vertical spacing 32 mm

Jacketting thickness on Top 100 mm

Jacketting thickness on bottom 200 mm

Jacketting thickness on Sides 100 mm

Spacing of Jacketting Steel bar spacing 65 mm

Calculations of required Design Moments, Shear forces:

Resultant Moment of ( Muy and Muz') = M''uz 268.56 KNm

Design Moment Mu 268.56 KNm

Design shear force Vu 155.39 KN

Calculations of required Section properties:

Width of web in reiforcement level in tension, btw 430.00 mm

Width of Web, Average btw 430.00 mm

Effective Depth , d (Calculated) 579.00 mm

Width of small portion in the flange, bs 0.000 mm

Centroid of Section from Compression fibre Cc 303.801 mm

Centoid of section from tension fibre Ct 346.199 mm

Df/d (Calculated) 0.216

2. Condition: If Neutral Axis lies in the webNeutral Axis lies in the Web YES

Neutral axis depth ratio α = Xu,max/d 0.4791

Maximum Neutral Axis for balanced design Xu,max 277.40 mm

yf,max 122.86 mm


Compressive force for Parabolic portion, C2 405.34 KN

Compressive force Trap. Web, Cu = C1+C2 861.34 KN








CG of Cu from Extreme Compression fibre X 115.390 mm



399.32 KNm

Capacity Mu, lim (Only for trapezodal Web) 399.33 KNm

Capacity Mu, lim (trapezodal Web + Flange) 399.33 KNm

-130.77 KNm SRS

463.600 mm

463.610 mm

2385.652 mm2

2385.652 mm2

-0.001288

1.79258

-399.325

Neutral Axis Xu=(-b+√(b2-4*a*c))/2*a 278.487 mm

yf=0.15*xu+0.65*Df 123.023 mm

Lever Arm z = jd 463.149

2388.022 mm2

0.216 YES

Hit and Trial α = Xu/d 0.300000

9.37117

Neutral Axis for balanced design Xu 173.70 mm

yf,max 107.31 mm


Compressive force fo rParabolic portion, C2 253.81 KN

Compressive force Trap.Web, Cu = C1+C2 539.34 KN







Capacity if average web +flange Muw, lim=

0.36*fck*bw*Xu,max*(d-0.416*Xu,max) + 0.446*fck*(bf-






a= - 0.14976*fck*bw - 0.0050175*fck*(bf - bw)

b= 0.36*fck*bw*d + 0.00669*fck*(bf - bw)


For Df/d > 0.2 Ast

2 (A) IF Df/d > 0.2

α is chosen to make Test : C = T


CG of Cu from Extreme Comprssion fibre X 72.253 mm



272.51 KNm

Mu(Only for trapezodal Web) 273.31 KNm

Mu,lim (trapezodal Web + Flange) 273.31 KNm

4.75 KNm

506.741 mm

506.747 mm

1489.476 mm2

1493.81 mm2

For Tension Area of Steel, Ast2 0.00 mm2


529.97 KN

Neutral Axis, Xu 173.70 mm

Due to Mu Compressive strain Ɛcc 0.00219

Total Strain due to (Mu+P) Ɛcct 0.00219 < 0.0035 OK




Pu 0.000 KN

Hit and trial, Extra Comprssve strain due to P, Ɛcc 0.00000000 YES

Test : Fx-Pu =0 NA

Extra Increased area 0.000 mm2

Required Singly reinforced Ast 1493.810 mm2

-0.001288

1.79258

-268.560

Neutral Axis Xu=(-b+√(b2-4*a*c))/2*a 170.770 mm

yf=0.15*xu+0.65*Df 106.865 mm

Lever Arm z = jd 507.960

1464.349 mm2

Check for Shear forceDesign Shear force 155.39 KN

Shear stress 0.624 N/mm2

Percentage of tension steel, pt 0.631 %

Percentage of compressive steel, pc 0.379 %

average web +flange Muw= 0.36*fck*bw*Xu,max*(d-

0.416*Xu,max) + 0.446*fck*(bf-bw)*yf*(d-0.5*yf,max)






Force of Tension, T

a= - 0.14976*fck*bw - 0.0050175*fck*(bf - bw)

b= 0.36*fck*bw*d + 0.00669*fck*(bf - bw)


For Df/d > 0.2 Ast

Percentage of tension and compressive steel, pt 1.585 %

α=0.8*fck/6.89*pt 1.465

tc = 0.85*SQRT(0.8*fck)*(SQRT(1+5*α)-1)/6*α 0.729 N/mm2

1.00 x 0.729 = 0.729

> 0.624 OK

Shear Reinforcement is required.

Dia of of Vertical Stirrups 10.00 mm

Number of leg 2.00

Planned area to Provide Asv 157.08 mm2

Required spacing of stirrups Svreqd. -1255.245 mm

Plan to Provide, spacing of stirrups Sv 75.000

Minimum required Asv 0.4*b*Sv/0.87*fy 35.729 mm2

Ф 4.769 mm

Rqd. Stirrups Ф 10.00 Leg 2.000 75.00 mm C/C

Permissible shear stress,

K´tc =

Excel tools to Check Beam Jacketting

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