Design of Anchor Block

Desing of Achor Blocks a Anchor Block No. 1 a1 a2 aH1

180 430

= = =

aH2

6.9

l1

=

6.9

l2

=

9.5

L2 D

= = = = =

4 0.88

 t

0

L1= 27.4

Upsteam length of pipe upto expansion joint Length of pipe upto 1st saddle 11 =

=

7.85 T/m 3mm 0.003m

Static Weight Per Meter  S1 = S2 = Dt   x 0.88 0.003 x 7.85  0.065 T / m

3

Area A 

D 2 4

 0.608m 2

H = 606.6-59262=13.98m Weight of water in the pipe per meter W1 = W2 = Q V

= = =

1.65m3/sec 1.85

Coefficient of friction between concrete and steel C = 0.45 Sin u2 = 0.682 Cos u2 = 0.731

3.042m/sec

Tan u2 = 1 1.1

Design of Anchor Block No. 1 Vertical component forces acting on the pipes are (i) Upper W

(i)

Lower W.

= =

=

0.933

1

1/2 (W. + S.) l. cos

0.608  0.065 x 6.9 x 0.951 2 1/2 (W. + S.) l. cos

a

=

2

0.608  0.065 x 9.5 x 0.731 2 W1

W2

1.2

=

Z-comp.

W {z

=

W {rSinu - cos 1

=

Y- comp

W {y

=

W {r Cos

=

x- comp

W {x

=

W {rSinu - sin

=

Z-comp.

W 2z

=

W 2 Sin Cos

=

1.594

Y-comp.

W 2y

=

W 2 Cos

=

1.709

X- comp.

W 2x

=

W 2 Sin Sin

=

0

Frictional force due to the upper portion of the penstoch pipe to the block is l   T1 = S1 L1 sin a 1  C (W1  S1 )  L1  1 Cos a 1 

= =

0

2

0.065x27.4 sin 18 + 0.45(0.608 + 0.065)(27.4 - 6.9) Cos 180 2 0.550 + 6.898 = 7.448 or = 6.348T

Considering the worst case we have T1 T2

=

=

l   S2 L2 sina2 C(21S21)L2  2 Cosa2 2  0.065 x 4.0 x sin 430 + 0.45 (0.608 + 0.065) (4.0 - 9.5/2) Cos 430 0.177 + 0.166 = 0.343 or -0.112

= = Considering the worst case we have T2

T1

T2

1.3

= T1x

=

Y - comp.

T1y

=

Z - comp.

T1z

=

X - comp.

T2x

=

 T1 .Cos a1 Sin a H  T 1 . Sin a 1  T1 .Cos a1 Cos a H  T2 .Cos a 2 Sina H

Y - comp.

T2y

=

 T 2 . Sin a

Z - comp.

T2z

=

P2

 g . D2

P1z

P2x

=

H

=

 P2 Cos a 1 Sin a

=

H

Cos = a1 . Cos a 2 , Cos a H  Sin a1 . Sin a 2

Considering the worst case we have Psz Z- comp.

0.179 T 0.026 T

=

0.026 T

= 0.179 x Cos 18 x Cos 0

Cos

Hydrostatic Pressure due to High Water Level Ps = a 2 HA r sin 1  2 Psx X-comp.

= =

0

P2z

Psy

=

0 0.179 x Sin 180

P2=Cos a1 Sin a H P =2 Cos a 2 Cos a H

Y- comp.

=

22 / 7 x 9.81 x 0.882

3

= = = 1.4

=

Where coeficient of frictioi f = 0.02

2 2 f Q21 L1 2x0.02 x 1.85 x 27.4

=

P2y

=

2

2x0.02 x 1.852 x 27.4 22 / 7 x 9.81 x 0.883

 =P1 Cos a 1 Sin a P1=Sin a1 P =1 Cos a1 cos a H

P2y

=

3 1

=

P1x

=

 T2 .Cos a 2 Sina H

Component force due to the friction of the inside water is calculated as P1 2 f Q21 L1 =

=

P2

0.343

X - comp.

 gD

P1

7.448

0 0,026 x Sin 430

=

-0.025 x Cos 430 x Cos 00

Ps (Cos a 1 2 Sin  / 2

=

Ps ( Cos a 1 Sin a 2 Sin  / 2

=

=

or

=

18 0 Sin 2 aH

=

Psy

=

=

Cos 180 . Cos 430. Cos 00 + Sin 180. sin 430 Cos (180 + 430) Cos 610 = 0 0 61 or 25

2 x13.98 x 0.608 x Sin

=

0

= 2

=

-0.977

Ps (Cos a1 Cosa H  Cosa 2 2 Sin  / 2

=

Ps (Cos a1 Cosa H  Cosa 2 2 Sin  / 2 or Psz

Considering the worst case we have

=

= 1.35

1.5 Contrifugal forces due to the motion of water acing on the curved portion. Pct 2 V2 A Sin = /2 2 x 3.0422 x 0.608 x Sin 610/2 = = g 9.81 Pctv Pef (Cos α1 Sin αH) X-comp = = 2 Sin = / 2 Pctv Pef (Sin α1 Sinα2) Y- comp = = 2 Sin = / 2 or = Pctv Considering the worst case we have = -0.126 Pctv

Z-comp

=

Pet(Cosα1 - Cos α2 CosαH) = 2 Sin = / 2 Psy = 0.295

F (X)

F(Y)

Considering the worst case we have

Forces Acting are

F(Z)

F(z) *2.4

W1

1/2 (W 1 + S1)1. Cosα1

= 20.201

0.00

2.100

0.682

W2

1/2 (W 2 + S2)2. Cosα2

= 2.337

0.00

1.709

1.594

T1

S1L1 Sin α1 + C (W 1+S1)(L1-l1/2)Cos α1

= 7.448

0.00

2.302

-7.083

T2

S2L2 Sin α2 + C (W 2+S2)(L2-l2/2)Cos α2

= 2.115

0.00

0.234

-0.251

P1

2

3

= 0.179

0.00

0.055

-0.17

2

3

= 0.026

0.00

0.018

-0.019

0.00

-0.977

1.35

0.00 0.00

-0.214 5.227

0.295 -3.602

2f Q L1 /g π D2

P2

2f Q L1 /g π D2

Ps

2 H A sin = /2 {2V2 Sin = / 2} / g

Pef

= 2.654 = 0.582

2 Weight of Anchor Block Penstock Diameter = Unit Weight of Concrete = 2.1 Figure ABCK V1

0.88 2.4

=

3.2 x 2.6 x 2 - 1.7 x 0.608

=

37.455 T

=

1.30 m

X1

=

2.6 / 2

=

1.60 m

Z1

=

3.2 / 2

=

1.70 x 1.30 x 2.6 - 2.10 x 0.608

=

W1

2.2 Figure EFIG V2

=

W2

=

4.4692 x 2.4

=

10.736 T

X2

=

2.6 / 2

=

1.3 m

Z2

=

0.9+ 1.7 / 2

1.75 m

=

0.9 x 0.6 x 2.6 -neg

=

2.3 Figure EJDK V3 W3

=

1.404 x 2.4

=

3.3696 T

15.606 m3

4.4692 m3

1.404 m3

0 -0.214 -0.501

0.125

Mx F(y) *1.3

-8.645 -1.85

6.795

X3

=

2.6 / 2

=

Z3

=

0.9 / 2

=

=

1/2 x 0.9 x 0.7x 2.6

2.4 Figure DJG V4 W4

0.45

0.619 m3

=

=

0.819 x 2.4

=

1.9656 T

X4

=

2.6/2

=

1.3 m

Z4

=

0.9 x 2/3

=

=

1/2 x 1.7 x 0.6x 2.6

=

1.326 m3

2.5 Figure HIG V5

0.6 m

W5

=

1.326 x 2.4

=

3.1824 T

X5

=

2.6/2

=

1.3 m

Z5

=

0.9 + 17 x 2/3

=

2.033 m

=

1/2 x 0.6 x 1.90 x 2.6

=

1.482 m3

2.6 Figure CFH V6 W6

=

1.482 x 2.4

=

3.5568 T

X6

=

2.6/2

=

1.3 m

Z6

=

26 + 1/3 x 0.6

=

S.N

Fy W1 37.4553 10.7261 3.3696 1.9656 3.1824 3.5568 60.2558

1 2 3 4 5 6 Total 3

1.3 m

X

2.8 m Z

1.3 1.3 1.3 1.3 1.3 1.3

Mx W 1 x X1 48.692 13.944 4.3805 2.5553 4.1371 4.6238 78.3327

1.6 1.75 0.45 0.6 2.033 2.8

MZ Remark W 1 x Z1 59.929 18.771 1.5163 1.1794 6.4698 9.959 97.8245

Total Forces and Moments ΣFx = 0.00 ΣFy

=

5.227 + 60.2558

ΣFz

=

-3.602 T

ΣMx

=

ΣMz

=

=

65.483 T

- 1.85 + 78.3326

=

76.483 T -m

6.795 + 97.8238

=

104.619 T - m

4 Sefty against Overturning ΣMz 4.1 X = ΣFy ex

=

Bx - X 2 >Bx 6

= = =

Hence the structure is safe against x-axis ΣMx = 4.2 Z = ΣFy ex

=

Bx - Z

=

104.619 =

1.598

65.483 2.6 -1.598 = 2 2.6 = 6

76.483

0.298 M 0.433

1.168

65.483 3.2 -1.168 =

0.432m

2 < 6

Bz

=

2 3.2 6

=

0.533

Hence the structure is safe against z-axis 5 Safe against Sliding = 5.1 fx

ΣFx

=

ΣFy Hence the structure is safe against sliding in x-axis = ΣFz = 5.2 fz ΣFy Hence the structure is safe against sliding in z-axis 6 Bearing Capacity of the Foundation

0

=

0

=

-0.055

<

0.6

65.2957

-3.602 65.483

<

0.6

2.208T

2.208T

0.682 2.1 0

9) Cos 180 = 6.348T

9.5/2) Cos 430 -0.112

0 2.302 -7.083 0 0.234 -0.251

= 0.02

0.055 -0.17

=

0.018

=

0.019

Cos 250

2.659 0 -0.977

0.576

1.35

0.582T

F(y) *1.3

Mz F(x) *2.4

6.795 0.00 6.795