Counterfort Wall Design

Provided spacing = Provided tensile reinforcement , Asc = (Asc x Calculated Spacing)/Provided spacing

150 2

2094.47 mm Use tensile steel of 

20

mm dia.@

150 mm c/c K

Design of Rear of  Rear Counterfort:

F Net downward pressure on heel slab at E = Net downward pressure on heel slab at D =

285.93  kN/m

2

16.08  kN/m

2

G 9.00

B 1.7

C 5 .0 0

1.5 E

D

The magnitude of reaction of  reaction transferred to each counterfort at F and G Reaction at F = 714.82 kN/m 80.47 3 5 0 .3 2 Reaction at G = 40.21 kN/m Thus at any depth(h) from top, the horizontal e/p acting 8.56 h = 15.65 Therefore, Horizontal pressure at top of front of front counterfort = 134.01 kN/m Shear force at top of front of  front counterfort = 1048.59 kN/m Counterfort act as a T‐Beamand the depth available is more than required from actual desing consideration. Even as a rectangular beam the depth required is given by; M

Rb

Required depth of Stem of Stem slab =

=

551.20

1232.42 mm Ok , Provided depth is greater

Available depth(GK) calculate by: Tanθ =FG/AB

 

0.436 30 Degree 4.450 m

θ =

GK = M   

Required tension steel, Ast = Required Number of reinforcement of reinforcement of  Providing the reinforcemt of

1.75

st



 j



d 

=

4589.98 mm2

32 mm dia. @ layer each layer contained

7 Nos. 4 bar

Curtailment of reinforcement of  reinforcement in counterfort;

Since the earth pressure and other forces in the counterfort decreases towards the top,So it is possible to curtail reinforcement h

Hs^2 * (n - m) 

total number of bar of  bar provided at bottom, n = total number of bar of bar to be curtail, m =

n

Therefore, h =

7 .8 0 5

7 6

m below top Plus the development length

Shear Force Check;

Effective shear force in a case of beam of beam of varying of varying depth is determined by; V = V'‐M/d tan θ where, d= de/Cos θ

773298.3 773298.3 N 5138.42 mm

VFG = V' = Actual Shear stress , τv

1014.25 kN 2

0.25  N/mm

= V/bd

Percentage of tensile of tensile steel = (100Ast)/(bd) =

 

0.19 %

Design of horizontal of  horizontal ties;

Due to horizontal earth pressure the vertical stem has a tendency of separating of separating out from the couterforts , Hence it should be tied to it by horizontal ties. At any depth(h) below the top the force causing the tendency of separation of  separation = γ.h.ka. Clear width of Couterfort of  Couterfort wall 8.56 h

Stem and Counterfort Wall Design for Abutment with Counterfort Wall

The Force at

1 m from bottom

168.26 kN/m 2

1201.83 mm

Area of steel required, Asv = Centre to centre spacing of the stirrup , Sv = (Ast*d)/Asv

376.418 mm

Provide 12 mm dia. 4 Leg stirrups 150 Spacing Design of Vertical ties; Due to horizontal earth pressure, the vertical stem has a tendency of separating out from couterforts, Hence it should be tied to it by horizontal ties. Max. net downward force at Point F =

514.674 kN 2

Area of steel required, Asv =

3676.24 mm

Centre to centre spacing of the stirrup , Sv = (Ast*d)/Asv

437.538 mm

Provide

16 mm dia.

8 Leg stirrups

150 Spacing

Max. net downward force at Point G =

28.95 kN 2

206.80 mm

Area of steel required, Asv = Centre to centre spacing of the stirrup , Sv = (Ast*d)/Asv Provide

10 mm dia.

759.557 mm

2 Leg stirrups

150 Spacing K

Design of Front Counterfort:

H

F Net downward pressure on heel slab at C = Net downward pressure on heel slab at B =

184.8 kN/m

2

34.89 kN/m

2

9.00

0.40 m 3.00 m 2.60 m

1.7

1.5 E

Width of Courterfortwall = Center to center spacing = Clear spacing provided =

G

80.47

5.00

C

B D

C' 401.29 551.20

The magnitude of reaction transferred to each counterfort at F and G Reaction at F = 554.40 kN/m Reaction at G = 90.70 kN/m Due to action of upward forces,the front counterfort will have a tendency to bend as a horizontal cantilever about section BH So, the face CH will develop compressive stresses and the face B'C' will develop tensile stresses. Total upward force = 1612.76 kN Distance of C.G. of the force from B' = 2.63 m BM at face of BB' = 4243.43 kN/m

M

Rb

Required depth of Stem slab =

Overall depth assuming diameter Overall depth, D =

=

32 mm bars in available overall depth provided

3805.43 mm Ok , Provided depth is greater

2 Layers and clear cover of  5000.000 mm

50 mm 3920

Available depth(GK) calculate by: Tanθ =FG/AB

 

Mθ =

Required tension steel, Ast = Required Number of reinforcement of  Providing the reinforcemt of

  

st

2



 j



d 

=

28 mm dia. @ layer each layer contained

1.000 45 Degree 3814.66 mm2 8 Nos. 4 bar

Stem and Counterfort Wall Design for Abutment with Counterfort Wall