document.xls Slab design 4.5m-ANAND 18:10:52 7.0) Design of slab 7.0 Analysis & Design of Precast slab above Road Beam at RP1 to RP4 7.1 General Details : Width of slab Panel = 5.000 m Width of slab Panel without ledg (5-0.287-0.288) = 5.000 m Length of slab Panel = 4.630 m Bearing below slab on Road beam @ North side = 0.000 m Bearing below slab on Retaining wall = 0.000 m Deck slab depth = 0.45 m Effective Deck slab depth (450- 40- 0.5*20) = 400.0 mm Effective Span of simply supported slab = (4.63-0-0+0.4-0) = 4.500 m Wearing coat thickness = 0.106 m Concrete density = 25 Wearing coat material density = 24 Concrete Grade Slab fck = 30 Mpa Reinforcement fy = 500 Mpa Fill height = 1m 7.2 Loading & Design forces: 7.2.1) Dead Loads Uniformly distributed load for per metre width for dead load due to self weigth of slab and SIDL due to wear moment and Shear force at distance d from face of support are tabulated below. Crash barrier is present at b slab are supported on beam on one side andon retaining wall on other side. Hence crash barrier load will not slab. Table - 1 ; Design UDL and BM & SF for Dead Load & SIDL Load type Max Midspan moment (kNm) Result Result Result Dead Load (0.45x25) = 11.25 11.25x4.5^2 / 8 28 11.25x4.5 / 2 25 18 kN/m 3 kN/m 3 Max Uniformly distributed load (kN/m) Max shear force @ centre of support (kN) Max Shear force @ distance d from face of support
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document.xls Slab design 4.5m-ANAND 16:44:13
7.0) Design of slab
7.0)Analysis & Design of Precast slab above Road Beam at RP1 to RP4
7.1)General Details :Width of slab Panel = 5.000 m
Width of slab Panel without ledge (5-0.287-0.288) = 5.000 m
Length of slab Panel = 4.630 m
Bearing below slab on Road beam @ North side = 0.000 m
Bearing below slab on Retaining wall = 0.000 m
Deck slab depth = 0.45 m
Effective Deck slab depth (450- 40- 0.5*20) = 400.0 mm
Effective Span of simply supported slab = (4.63-0-0+0.4-0) = 4.500 m
Wearing coat thickness = 0.106 m
Concrete density = 25
Wearing coat material density = 24
Concrete Grade Slab fck = 30 Mpa
Reinforcement fy = 500 MpaFill height = 1 m
7.2)Loading & Design forces:
7.2.1) Dead Loads
Uniformly distributed load for per metre width for dead load due to self weigth of slab and SIDL due to wearing coat and its Bending
moment and Shear force at distance d from face of support are tabulated below. Crash barrier is present at both end of slab where
slab are supported on beam on one side andon retaining wall on other side. Hence crash barrier load will not cause any bending in
slab.
Table - 1 ; Design UDL and BM & SF for Dead Load & SIDL
Load type Max Midspan moment (kNm)
Result Result ResultDead Load (0.45x25) = 11.25 11.25x4.5^2 / 8 28 11.25x4.5 / 2 25 18
7.2.2) Live LoadsLive Load for different vehicles (i.e. IRC 70R Wheeled, IRC 70R Tracked, IRC Class A & 40T boggie) are placed symmetrically on span to get maximum bending moment & nearer to support to get maximum shear force are tabulated below: -
a) IRC 70R TRACKED VEHICLE :-Load of tracked vehicle = 70 TonsLength of tracked vehicle = 4.57 mDispersion in longitudnal direction =. 4.57 + 2* 1 6.57 mVehicle width in transverse direction 2.90 mDispersion in transverse direction =. 2.9 + 2* 1 4.90 mImpact factor = 25.0 %Load intensity per square m= 70*9.81*(1+25/100) / ( 6.57*4.9 ) 26.7 KN/m2
document.xls Slab design 4.5m-ANAND 16:44:13
i) Bending moment calculation
4.57
27 kN/m 27 kN/m 1.2 5.00
0.85
4.500Va = 60 kN 4.5 Vb = 60 kN
4.50067 kNm
ii) Shear Force calculationMaximum shear force is obatined for the same laoding condition
43 kN
0.625 0.6254.500
Va = 43 kNVb = 43 kN
B) IRC 40T Boggie Load :-IRC 40T Boggie Load consists of 2 axles of 20T each.Axle No. 1 = 20 Tons Axle No. 2 = 20 Tons
Impact factor = 25.0 % 1.2
Total Live Load on the slab panel = 20+20= 40 TonsLoad on one side wheels including impact factor = 40 Tons/2 x 1.250 x 9.81 = 245 kNi) Bending moment calculation
1
20T 20T
document.xls Slab design 4.5m-ANAND 16:44:13
Wheel contact surface in longitunal direction 0.31 mWheel spacing in longitunal direction 1.22 mWheel contact surface in transverse direction 0.86 mWheel spacing in transverse direction 1.93 mDispersion due to fill of 1m = 2* 1m in each direction 2.00 mTotal diespersion in longitudnal direction =0.31+1.22 +2 3.53 mTotal diespersion in transverse direction =0.86+1.93 +2 4.79 mLoad intensity per square m= 25*9.81*(1+40/100) / ( 3.53*4.79 ) 29.0 KN/m2
7.2.2) Live LoadsLive Load for different vehicles (i.e. IRC 70R Wheeled, IRC 70R Tracked, IRC Class A & 40T boggie) are placed symmetrically on span to get maximum bending moment & nearer to support to get maximum shear force are tabulated below: -
a) IRC 70R TRACKED VEHICLE :-Load of tracked vehicle = 70 TonsLength of tracked vehicle = 4.57 mDispersion in longitudnal direction =. 4.57 + 2* 1 6.57 mVehicle width in transverse direction 2.90 mDispersion in transverse direction =. 2.9 + 2* 1 4.90 mImpact factor = 21.3 %Load intensity per square m= 70*9.81*(1+21.25/100) / ( 6.57*4.9 ) 25.9 KN/m2
document.xls Slab design 6m 16:44:13
i) Bending moment calculation
4.57
26 kN/m 26 kN/m 1.2 20.00
0.85
6.000Va = 78 kN 6 Vb = 78 kN
6.000116 kNm
ii) Shear Force calculationAs dispersion width is greater than slab panel span. The SF force vehicle position as used for BM calcuation is used.
56 kN
41 kN
0.825 0.8256.000
Va = 56 kNVb = 56 kN
B) IRC 40T Boggie Load :-IRC 40T Boggie Load consists of 2 axles of 20T each.Axle No. 1 = 20 Tons Axle No. 2 = 20 Tons
Impact factor = 25.0 % 1.2
Total Live Load on the slab panel = 20+20= 40 TonsLoad on one side wheels including impact factor = 40 Tons/2 x 1.250 x 9.81 = 245 kNi) Bending moment calculation
1
20T 20T
document.xls Slab design 6m 16:44:13
Wheel contact surface in longitunal direction 0.31 mWheel spacing in longitunal direction 1.22 mWheel contact surface in transverse direction 0.86 mWheel spacing in transverse direction 1.93 mDispersion due to fill of 1m = 2* 1m in each direction 2.00 mTotal diespersion in longitudnal direction =0.31+1.22 +2 3.53 mTotal diespersion in transverse direction =0.86+1.93 +2 4.79 mLoad intensity per square m= 25*9.81*(1+40/100) / ( 3.53*4.79 ) 29.0 KN/m2
0.00
0.86
29 kN/m 29 kN/m 1.93 6.0
0.31
6.000Va = 51 kN 3.53 Vb = 51 kN
20.000108 kNm
ii) Shear Force calculationTo obtain maximum shear force, the load is so placed that the dispersion just touches the support, thus bringing the concentratedload nearer to the support.
Load case 1 : DL + SIDL + LL = Bending Moment : 67+11+108+116= 303 kNmShear force @ deff: 33+6+52+68= 158 kN
7.3) Distribution Moment Calculation for Lateral distribution of LoadsFor solid slabs spanning in one direction, distributing bars shall be provided at right angles to the main tensile bars to provide forlateral distribution of loads.
0.2*187+0.3*116 = 72 kNm
Maximum Midspan
Moment (kNm)
Max Shear force @
distance d from face of support (kN)
Distribution moment = MD = 0.2 x (DL & SIDL moment) + 0.3 x (LL moment)
MD =
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1.1 Moment & Reaction for Rectangular Plate - Interpolation - (Chamber Wall C1) - Water PressureChamber design _ Moment, Shear force & Direct tension calcualation
Note: Co-efficient are taken from Moody"s chart.
0where ka = 1.000
γ = Density of soil = 10.000h = Height at which pressure is acting
Earth pressure upto bottom slab = 1.000 x 5.55 x 10 = 55.5Phi
a = Half width 1.25 kab = Height 5.55a/b 1/4.44
a/ b = 1/4.00Mx - Horizontal moment coefficient My - Vertical moment coefficient
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1.2 Moment & Reaction co-efficient for Rectangular Plate of (Chamber Wall C1) - Water PressureChamber design _ Moment, Shear force & Direct tension calcualation
Note: Co-efficient are taken from Moody"s chart.
0where ka = 1.000
γ = Density of soil = 10.000h = Height at which pressure is acting
Earth pressure upto bottom slab = 1.000 x 5.55 x 10 = 55.5Phi
a = Half width 1.25 kab = Height 5.55a/b 1/4.44
Mx - Horizontal moment coefficient My - Vertical moment coefficienty/b x/a 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1
Assume angle of friction of sol ɸ = Co-efficient of active pressure = (1-sinɸ)/(1+sinɸ) =
kN/m3
kN/m2
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document.xls Chamber des - Page 14/20
1.3 Design of tank - (Chamber wall C1) (LC - 100 - Water Pressure) :-
Grade of concrete = 20 MpaGrade of steel = 415 MpaModular ratio = 13.33Permissible comp. flex. stress in concrete = 7.0 Mpa IS 456-2000, Table 21Permissible tensile. stress in steel Liquid f = 150 Mpa IS 3370 Part II, Table 2
Remote = 190 MpaLiquid face Remote face
n1 = 0.384 0.329j = 0.872 0.890Q = 1.17 1.03 0 0
Design of sectionA. Design of section for flexural
Location Tension Moment Direct b D clear d Total Ast Remark. Pt %
Face Tension cover (for flex.)(for dir tension.) dia @ Spacing Prov.
Grade of concrete = 20Height of wall above Footing level = #REF! mDepth of foundation top below Road lvl = #REF! mTotal height of wall H = #REF! mSaturated Density of soil = 18Angle of internal friction ' phi' = 30 DegreeCoeff. Of friction m = 0.5Surcharge load SL = 0*18 = 0 No surchargeOver burden load SL = 0*18 = 0 No OverburdenConcrete density = 24
Coefficient of earth pressure Ka = #REF!Dynamic increament in Ka due to EQ, ' Ka1
#REF! #REF! = #REF!
#REF! = #REF!
= #REF! = 0.07.4.2) Design of Abutment @ Base : -7.4.2.1)Summary of forces @ Base of Abutment: Stem depth = #REF! m
Grade of concrete = #REF!Height of wall above ground = #REF! mDepth of foundation below ground = #REF! mTotal height of wall H = #REF! m
Saturated Density of soil = #REF!Angle of internal friction = #REF! DegreeCoeff. Of friction m = #REF!
Concrete density = #REF!Horizontal Siesmic coeff. = #REF!Vertical Siesmic coeff. = #REF!Coefficient of earth pressure Ka = #REF!Dynamic increament in Ka due to EQ, ' Ka1
#REF! #REF! = #REF!Dynamic decreament in Kp due to EQ, ' Kp1 = #REF!
#REF! #REF!
= #REF! = #REF!
= 0 = 0.0
7.6.2) Design of Abutment @ Base : -7.6.2.1) Summary of forces @ Base of Abutment: