10. Anchor Block and Support Piers

13 February 2014

Anchor blocks and Support Piers Expansion joint

Support Piers -Rigidly hold the Penstock pipe. -Normally provided at every bend and at every 30m span of straight sections of Penstock line. - To be provided immediately upstream of PH to minimize the Penstock load on Turbine.

Support Piers supports the penstock vertically and prevent penstock from sagging -It allows axial movement of Penstock - Generally provided at the span of 3m interval in MHP. 1

A/B Design Procedure

Forces in (kN)

F1

Direction of Potential movement of A/B & S/P F1u (V )  F1u  sin 

α F1u  (Wp  Ww )  L1u  Cosα F1d  (Wp  Ww )  L1d  Cos 

F2 F2u   f  (W p  Ww )  L4u  cos

F1d (V )  F1d  sin 

F1u ( H )  F1u  cos

_

F1u

F1= Weight of Water and Pipe.

F1d ( H )  F1d  cos 

_ +

+

F1u and F1d are component of Combine Weight of Pipe and Water perpendicular to pipe.

β

F1d

Applies for both support piers and anchor block

Applies to both support piers and anchor block _ F (V )  (F  sin  ) 2u

F2=Frictional

1u

F2u ( H )  (F1u  cos )

F2

_ +

Remarks

+

forces due to the pipe sliding on the support.

α

Applies for both support piers and anchor block

The force acting on anchor block is sum of the forces acting on support piers between expansion join and the anchor blocks.

2

1

13 February 2014

Forces in (kN)

Direction of Potential movement of A/B & S/P

F3

 β α  F3  2  ρ  g  htotal  A  sin   2 

   F3 (V )  F3  cos  2     F3 ( H )  F3  sin 2 

_

_  αβ     2 

+

+

F3

β

F4u (V )  F4u  sin  F4u ( H )  F4u  cos

F4 d (V )  F4 d  sin 

_

F4 d ( H )  F4 d  cos 

α

F4u  Wp  L4u  sinα

+

F4d0=~

F4d is insignificant if expansion joint near downstream of A/B is used.

F5  7540  ΔT  (d  t)  t

_

F5 (V )  (F5  sin  )

F5  E *  T    (d  t )  t

F5

F5 ( H )  (F5  cos )

for steel

+ restrained by Anchor block, in

+

F5

the absent of expansion joint.

β

F5

Applies on Anchor block only if no expansion joint installed. _

F6

F6 (V )  (F6  (sin   sin  ))

α

F6  100  d

F6=Frictional force in the + expansion joint. (depend on

_

F6d

F6 ( H )  (F6  (cos  cos  ))

Applies on Anchor block that have (α & 𝛽) larger than 200.

F5= Thermally induced force

_

α

bend

F4=Component of Pipe weight + acting parallel to pipe

_

β

F4u

F4d  Wp  L4d  sin

F3=Hydrostatic Pressure on Applies on Anchor block that have horizontal & vertical bends.

α

 β α  F3  15.4  htotal  d 2  sin   2 

F4

     

Remarks

+

tightness of bolts etc.)

β

F6u

Applies to Anchor block only 3

Forces in (kN) F7  ρ  g  h total    (d  t )  t

F7

F7  31 h total  (d  t )  t

Direction of Potential movement of A/B & S/P F7u (V )  F7u  sin 

F7u ( H )  F7u  cos

 8  Q2 F8     d2 

   sin   - α    2  

 Q2 F8  2.5    d2 

   sin   - α    2  

F9 F9   water  htotal 

2  d 2  dsmall 4 big



F7d ( H )  F7d  cos

F7=Hydrostatic force on exposed + ends of pipe within expansion joint

_

F7d

F7d  31 h total  (d  t )  t

F8

_

F7d (V )  F7d  sin 

α

F7u  31 (h total  L4u sin  )  (d  t )  t

Remarks

+

Applies to Anchor block only (Usually insignificant)

β

F7u

_

    F8 (V )  F8  sin   2       F8 ( H )  F8  cos  2  

_

F8

F8=Dynamic force on bend due to

+

+

change in direction.

α

Applies to Anchor block only (Usually insignificant)

β

F9=Force

due to reduction in diameter of pipe.

F9

Applies to Anchor block only (Usually "0" "as same dia. is used in penstock")

F9

F10

F10 (V )  F10  sin 

F10   soil  h 2  cos(i)  K a Width

F10 ( H )  F10  cos

_ _ +

+

F10= Force due to soil pressure Applies to Anchorblock and support pier

Wb

Wb  Vblock   concrete Wb  Vblock  22

Ka 

cos(i ) 

cos2 (i )  cos2 ( )

cos(i ) 

cos2 (i )  cos2 ( )

Wb= Weight of bolck Applies to Anchorblock and support pier 4

2

13 February 2014

Checking for stability Resolve all forces on horizontal (FH) and vertical(FV) direction Find the Resultant of Forces.

 F    tan 1 v  F

R  FV2  FH2

 H 

  Fv FH

 FoS  1.5

  tan()

e

(from type of soil but not exceeding0.5 is recommended)

F r Resisting moment M R   V  FoS  2 Driving moment M a FH  a

FH a

Checking for Bearing failure M  FV

L  e  d   2 

Max. bearing pressure

R

FV

Checking for Overturning

d

FH

φ

Checking for sliding stability

e  eallowable

L 6

FH

r

FV

F  6e Pf  V 1  A  L 

L

M

d L/2

e

FV

f bearing capacityof soil  b  FoS  1.5 max.bearing pressure on soil Pf Unit Weight Friction angle Allowable Bearing Pressure (φ) (kN/m3) (kN/m2) Soft Clay and Silts 16 22 50 Firm clays 17 25 100 Stiff clays 20 30 200 Very stiff boulder clays 20 32 350

R

Thumb Rule (for small scale plant) -A/B for straight section at every 30m & at every bend - S/P at every 3m

Soil Type

 penstock dia.(mm)  Concrete volume(m3 )  1*    for straight section 300    penstock dia.(mm) Concrete volume  2 *   300 

   for ( -  ) less than 450  

5

Example γWater= 9.81 kN/m3 γconcretre= 22kN/m3 γsoil= 20kN/m3 (Stiff Clay) Wp = 0.44 kN/m Ww = 1.56 kN/m

L1u= 2m L1d= 2m L4u= 30 m L4d= ~ 0 d=450 mm hg=60 m hs= 48 α & β = 130 & 250 h1= 1.8 m h2= ~0 No. of support=8

Forces in (kN)

Block size (Assumed) Length, L= 3 m Height, H= 3 m Width, W=2m Height at bend center = 2,15m


F1

F1u  (Wp  Ww )  L1u  Cosα

α

β

F1u  (0.44  1.56)  2  Cos(13)  3.9

F1d  (0.44  1.56)  2  Cos (25)  3.63

F2 F2u  f  (W p  Ww )  L4u  cos

F2u  0.25  (2)  30  cos(13)  14.6

3.63

3.9

F1d  (Wp  Ww )  L1d  Cos 

Fh

Fv

F1u=-0.88

F1u=3.8

F1d=-1.53

F1d=3.29

± 14.2

± 3.28

Applies to both support piers and anchor block F1uper S/P 

F1u No. of S/P

± 14.6 α

Applies for both support piers and anchor block F2d = ~ 0, as expansion joint used just downstream of A/B

6

3

13 February 2014

Forces in (kN)


 β α  F3  2  ρ  g  htotal  A  sin   2 

F3

 β α  F3  15.4  htotal  d 2  sin   2   25  13  F3  15.4 108  0.452  sin    35.2 2  

F3= 11.46

F3=-33.28

F4=2.89

F4=0.67

F5=0

F5=0

F6=± 3.06

F6=±8.96

β Only applies to A/B

α

F4u  0.44  30  sin(13)  2.97

β

2.97

F4d  Wp  L4d  sin

F4d  0.44  ( 0)  sin (25)  0

~0

Only applies to A/B

F5

0

α

F5  E *  T    (d  t )  t

F5  7540  ΔT  (d  t)  t

FV

α

F4u  Wp  L4u  sinα

F4

35.2

 αβ     2 

Fh

β

0

for steel

As expansion joint are used, F5=0)

F6  100  0.45  45

F6

45 α

F6u  (100  d * (cos  cos  )) F6d  (100  d * (sin   sin  ))

β 45 Only applies to A/B

7

Forces in (kN)


F7  ρ  g  h total    (d  t )  t

F7

F7u  31 (108  30  sin 13)  (0.45  0.004)  0.004  5.7

F7d  31108  (0.45  0.004)  0.004  6.08

 0.452 F8  2.5    0.452 

0.26





2 F9   water  htotal  d 2  d small 4 big

F7u=5.55

F7u=1.28

F7d= -5.51

F7d= -2.57

F8=0.08

F8=-0.25

F9=0

F9=0

F10=22.76

F10=5.25

Wb=0

Wb=354.6

∑FH=52.1

∑FV= 345.5 8

α β

   sin  25 - 13   0.26   2  

F9

β

5.7

Fv

Only applies to A/B

 8  Q2    sin   - α  F8     d2   2     Q2    sin   - α  F8  2.5    d2   2   

F8

6.08

α

F7  31 h total  (d  t )  t

Fh

Only applies to A/B (Usually insignificant)



0 0 F9=0 (Usually, no change in diameter)

F10

F10 

F10  Ka 

 soil  h 2 2

20 1.82 2

cos(i ) 

cos2 (i )  cos2 ( )

cos(i ) 

cos2 (i )  cos2 ( )

Wblock

 cos(i )  K a  Width

i  cos(13)  0.37  2  23.36 

cos(13) 

cos2 (13)  cos2 (30)

cos(13) 

cos2 (13)  cos2 (30)

 0.37

F10=23.3676

Wb  Vblock   concrete Wb  Vblock  22

4

13 February 2014

Checking for stability

1.2 m 1.2 m

Resolve all forces on horizontal (FH) and vertical(FV) direction Find the Resultant of Forces. R  345.52  52.12  349.4

52.1

3m

Checking for sliding stability 0.57  345.5  3.77 52.1

  tan(30) 0.57

 FoS  1.5

Checking for Overturning

2  1.2   (3* 3) *1.5 -  *1.2  * 1.8  1.2 *  3  2   cg( x)   1.4   1.2  *1.2  3 * 3    2  

Checking for Bearing failure M  FV

e  eallowable

L  e   d    (1.72 - 1.5)  0.22 2 

 1.72

L 6

349.4

345.5 (neglectin g pipe)

Resisting moment M R (345.5 1.6  5.25 * 3)    4.52  FoS  2 Driving moment M a (52.1* 2.15  22.76 * 0.6)

d

x

e  0.24  0.5

52.1 22.76 5.25

a

0.6 1.6

1.4

345.5 L=3m

Max. bearing pressure

F  6  e  345.5  (6  0.22)  Pf  V 1   1   82.92 A  L  (3 * 2)  3 

f bearing capacityof soil 200  b   2.41  FoS  1.5 max.bearing pressure on soil Pf 82.92

2.15 M

52.1 d

L/2

e

345.5 349.4 9

5

10. Anchor Block and Support Piers

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