Job: ISBT Flyover on NH37 Note No:
Bracket for Future Prestressing
Height of Lighting Post (Approximate Height) = 11.50m
Thickness = 0.006m
Sectional Area of the Lighting Post = 0.01m^2
Total Volume = 0.0575 m^3
Density of Material = 7850kg/m^3
Weight of Lighting Post = 451.4 kg
Increasing by 10%to cater for the services and other joinings
Weight of Lighting Post = 496.5 kg
say = 500kg
Debabrata 1/14/2015
Job: ISBT Flyover on NH37 Note No:
Bracket for Future Prestressing
The following calculations are carried out following the design procedures of " Concrete Bridge Practice : Analysis, Design and Economics " by Dr. V.K.Raina. )
Basic Inputs
1. Distance of point load from face of support a = 1264 mm
2. Overall depth of section at face of support h = 3300 mm
3. Clear Cover c = 40 mm
4. Dia of main reinforcement = 25 mm
5. Effective Depth of Section (with double layer reinf.) d' = 3222.5 mm
6. Design Depth of Section ( d = 0.8 d' ) d = 2578 mm
7. Depth of section at edge s = 1850 mm
8. Width of Cantilever / Corbel / Bracket b = 1200 mm
9. Grade of Concrete = 40
10. Grade of Reinforcement = 415 Mpa
11. Ultimate Shear Value = 2850 kN
12. Actual Horizontal Load = 0 kN
Conditions
Condition I
For the section to be designed as a corbel " a / d' " shall be less than 1.
Hence a / d' = 1264 / 3222.5 = 0.392 < 1.0
As the ratio " a / d' " is less than 1 so the section is to be designed as a corbel
Condition II
For the section to be designed as a corbel " s / d' " shall be greater than 0.5.
Hence s / d' = 1850 / 3222 = 0.57 > 0.5 Proceed with the design
Loadings
Ultimate Shear Value 2850.00 kN (Refer to separate calculations)
= 1.7 x actual horizontal force in working load condition = 0 kN
but not less than
= = 0.2 x 2850 = 570.00 kN
570.00 kN
Now the design for the corbel is carried out as per the following steps. STEP I
Check for Nominal Shear Strength
where; = 2850.00 kN = 2850000 N bd = 1200 x 2578 sqmm = 3093600 Therefore; = 2850000 / 3093600 = 0.92
Now; = 28 - day standard cylinder strength of concrete used. = 0.80 times the standard cube strength
= 0.15 x 0.8 x 40 = 4.80 Ensured Ensured fsy Vu Hu Vertical Load (Vu) Horizontal Load (Hu) H u 0.2 * V u
Maximum Horizontal Load " H u "
Ensure V u / b d < = 0.15 fc' V u mm2 V u / b d V u / b d N / mm2 fc' 0.15 fc' N / mm2 Hence V u / b d < = 0.15 fc' is
Page-STUP Consultants P Ltd. Designed ByDebabrata
'file:///var/www/apps/conversion/tmp/scratch_6/257217810.xls'#$Corbel Date: 1/14/2015
Job: ISBT Flyover on NH37 Note No:
STEP II
=
where; = yeild stress value of the reinforcement used.
= 415
Type of Surface
1 Concrete placed monolithically across interface. 1.40
2
2 Concrete placed against hardened concrete but with roughened surface 1.00
3 Concrete anchored to structural steel 0.70
4 Concrete placed against hardened concrete but surface not roughened 0.60
= 1
(for Concrete placed against hardened concrete but with roughened surface)
= 2850 x 1000 0.85 x 415 x 1 = 8079 STEP III = = 570.00 kN = 570 x 1000 0.85 x 415 = 1616 STEP IV = = [ ( 2850 x 1000 x 1264 ) + ( 570 x 1000 ) x ( 3300 - 3222.5 ) ] 0.85 x 415 x 2578 = 4010 STEP V >= >= >= = 4009.92 + 1615.88 = 5625.79 = 2 / 3 x 8079.38 + 1615.88 = 7002.13 = 0.04 x 0.8 x 40 / 415 x 1200 x 3222. = 11927.13 Hence = maximum of ( 5625.79 , 7002.13 , 11927.13 )
Calculation of Shear Friction Reinforcement " A v f "
A vf V u 0.85 fsy fsy N/ mm2 Type of Surface ( 1 / 2 / 3 / 4 ) A vf mm2 A vf mm2
Calculation for Direct Tension Reinforcement " A t ". A t H u 0.85 fsy H u A t mm2 mm2 Calculation for Flexural Tension Reinforcement " A f "
A f [ V u a + H u ( h - d' ) ] 0.85 fsy d
mm2 Total Primary Tensile Reinforcement " A s "
A s ( A f + A t ) Provide the largest of these three
magnitudes as Total Primary Tensile Reinforcement As. ( 2 / 3 A vf + A t ) ( 0.04 fc' / fsy ) b d' ( A f + A t ) mm2 mm2 ( 2 / 3 A vf + A t ) mm2 mm2 ( 0.04 fc' / fsy ) b d' mm2 mm2 A s
Job: ISBT Flyover on NH37 Note No:
Page-STUP Consultants P Ltd. Designed ByDebabrata
'file:///var/www/apps/conversion/tmp/scratch_6/257217810.xls'#$Corbel Date: 1/14/2015
Job: ISBT Flyover on NH37 Note No:
STEP VI >= >= Therefore; = 0.5 x 4009.92 and = 0.333 x 8079.38 = 2004.96 = ### Hence; = maximum of ( 2004.96 , 2690.43 ) = 2690 STEP VII = Now; V c = 10 bd in kgs = 10 x 120 x 257.8
where b & d are in cms = 309360 kg = 3093.6 kN
Pitch = 225 mm
= 0.5 x ( 2850 - 3093.6 ) x 225 x 1000 415 x 2578
= -25.62 Though no steel is rerquire we provide
Reinforcement Details some R/F for extra R/F
= 11927
Diameter of Primary Steel = 25 mm
Area of one bar = 490.87
No. of Bars provided = 26 nos.
= 26 * 490.88
= 12763 > R/F is adequate
So we provide 26 nos. 25 mm diameter bars as main reinforcement
= 2690.43
2 / 3 d' = (2 / 3 ) x 3222.5
= 2149.00 mm
Diameter of Stirrup Bar = 10 mm
Area of one bar = 78.54
No. of layers of Stirrups = 5 nos.
Spacing of Stirrups = 430 mm
No. of legs of stirrups = 8
= 78.54 x 5 x 8
= 3142 > R/F is adequate
So we provide 8 legged 10 mm diameter bars as horizontal stirrups in 5 layers
= -25.62
Diameter of Stirrup Bar = 10 mm
Area of one bar = 78.54
Pitch = 225 mm
No. of legs of stirrups = 4
= 78.54 x 4
= 314.16 > R/F is adequate
The total sectional area of the stirrups is " A h " (closed ties) to be provided horizontally, one below the other, and next to " As "
A h 0.5 * A f Provide the largest of these two magnitudes as Ah. 0.333 * A vf 0.5 * A f 0.333*A vf mm2 mm2 A h mm2 The total steel in vertical stirrups is " A v "
A v 0.50 * ( V u - V c ) * p f sy d
A v
A v mm2
Total Primary Tensile Reinforcement " A s "
A s mm2
mm2
A s provided
mm2 A
s
Horizontal Steel ( Closed Stirrups) " A h "
A h mm2
The stirrups shall be provided below As and within a depth of " 2 / 3 d' " below As.
mm2
A h provided
mm2 A
h
Vertical Steel ( Closed Stirrups) " A v "
A v mm2
mm2
A v provided
mm2 A
Job: ISBT Flyover on NH37 Note No: