Abutment Design

Design of Abutment Section Design constants Concrete Grade M25  Allowable stress in bending compression compression =  Allowable stress in direct compression compression =  Allowable stress in bending tension tension = Modulus of elasticity = Steel Percentage 0.25 2 Shear stress,kg/cm 2.3 Steel Grade Tension in flexure, shear or combined bending Direct compression Modulur ratio = n = Es/Ec Computed Q = 1.1

0.5 3.1

8.33 6.25 0.61 25000 0.75 3.6

2

N/mm 2 N/mm 2 N/mm 2 N/mm 1 4

1.25 4.4

Fe415 2

200 N/mm 2 170 N/mm 10

= = =

Design will be done first for vertical load and longitudinal moments as per charts of Turner & Mourier . Then stresses on effective section due to transverse moment are superimposed to get final stresses in the member. Bottom section of abutment & reinforcement in it will be as shown below : Length at bottom section of abutment =

6600 mm =

6.60 m

Width of abutment bottom section=

1200 mm =

1.2 m

Equivalent Equivalent length length of abutment abutment =

6600 mm

62.5

6475 mm

62.5

1200

6600 6600 mm

Use 25 mm dia bar @ Area of steel in each long face = Ratio of steel in long face = np =

(21669.27/(6600x1200)) =

(10x0.00274)=

0.0274

Distance from outer face to cg of steel   d' d' = 62.5 mm d'/h =

(62.5/1200)

Used chart for =

150 mm C/C 2 21669.27 mm

0.05 0.05

0.003

Check for Stresses : Case (I)

( Maximum longiudinal case Non-Seismic )

P = ML = MT =

4568 KN 2212 KN-m 0 KN-m

e =

(2212/4568) =

h/e =

(1.2/0.4842) =

Now from chart of h/e =

2.478

We get C =

9.5

2.478

and np=

& k =

Stress in extreme fiber of concrete =

0.4842 m

0.0274 0.43

(9.5x2212x10^6/(6600x1200x1200))  

2

2.21 N/mm

=

Stress in tensile steel =

(10x2.2111x((1-0.05)/0.43)-1)  

2

26.74 N/mm

=

Case (II) (Maximum longitudinal moment case Seismic ) Length at bottom section of abutment without semicircular portion = Width of abutment bottom section=

6600 mm

1200 mm =

owa e stress n concrete =

1.2 m

2

mm

.

2

300  N/mm

 Allowable stress in steel = . Equivalent length of pier =

6600 mm

62.5

6475 mm

62.5

1200

6600 mm 2

21669.27 mm

 Area of steel in each long face = Ratio of steel in long face = np =

(10x0.00274) =

d' = d'/h =

(21669.27/(6600x1200)) = 0.0274

62.5 mm (62.5/1200)

Used chart for =

0.05 0.05

0.00274

Check for Stresses :

P= ML = MT =

4247 KN 6435 KN-m 0 KN-m

e =

(6435/4247) =

1.5152

h/e =

(1.2/1.5152) =

0.792

Now from chart of h/e = We get C =

0.792 10.5

k =

Stress in extreme , fiber of concrete =

and np=

0.0274 0.23

(10.5x6435x10^6/(6600x1200x1200))  

=

Stress in tensile steel =

2

7.109 N/mm

(10x7.109x((1-0.05)/0.23-1))  

=

2

221.90 N/mm

Distribution Reinforcement : Provide minimum reinforcement 0.15% on each face. Distribution reinforcement is 0.25 % of gross area. Providing main reinforcement 25 Tor @ 150mm c/c on Both-Face Providing side reinforcement 25 Tor @ 150mm c/c on Both-Face Providing distribution reinforcement 16 Tor @ 150mm c/c on Both-Face Provide,10mm Tor open links @ 150mm C/C verically and 150mm C/C horizontally in staggered manner  ec or ear :

For max longitudinal force From load combinations = 347.29 KN/m width Shear stress = V / (b x d) = 0.31 Mpa 2  Area of steel in longitudinal direction = 21669 mm Steel ercenta e = 0.23 . Permissible shear stress = 0.20 Mpa (from IRC:21 , Table-12B) Shear to be resisted by shear reinforcement = 347.29x10^3-0.2x1000x1137.5 = 119.79 KN Providing 1L, 10mm Tor stirrups, spacing required = 223.63 mm However provide,10mm Tor open links(or S-loops) with hook @ 150mm C/C verically and 150mm C/C horizontally in staggered manner