anchor block type 2

7/27/2019 Anchor Block Type 2

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Manual for Anchor Block, Saddle Support & Machine Foundation Design Aids © 2004, Prepared By Pabitra Gurung 1

Design of an Anchor Block – Type 2

Model of Type 2 Anchor Block:

Example: Anchor block of Chandra

Jyoti MHP , Kavre

H 1 = 1.3 m L = 1.0 m w = 0.9 m Z = 0.3 m

h1 = 0.6 m h2 = 0.3 m

X = 0.6 m i = 10°

l c = 0.3 m hc = 0.2 m

α = 10° β = 35°

Pipe Diameter, d = 130 mm

Pipe Thickness, t = 4 mm

Discharge, Q = 20 lps

Gross Head, h gross = 42 m

Surge Head, h surge = 95 m

Distance to u/s support pier, L2u = 4m

L1u = 2 m

Distance to d/s support pier, L2d = 4m

L1d = 2 m

Distance to u/s expansion joint = 18m

L4u = 18 m

Number of piers at upstream = 4

Relation Used for Area Calculation:

A1 H 1 N 1.08 m2

A2 H 2(L – N) 0.16 m2

A3 21 N(H 2 + M – H 1 ) 0.03 m2

A4

2

1 M(L – N) 0.04 m2

Relation Used for Centre of Gravity Calculation:

X1

2

1 N 0.42 m

X2

2

1(L + N) 0.92 m

X3

3

2

N 0.55 m

X4

3

1(L + 2 N) 0.89 m

Relation Used for Calculating Deduction Volume Occupied by Penstock:

V p

4

πl u (d u + 2t)2 +

4

πl d (d d + 2t)2 0.02 m3

WhereY = H 1 – Z –

αcos2

2t d + – X tanα

H 2 = Y – (L – X)tan β +

βcos2

2t d ++ Z

M = Y – H 2+[ 3

5 Z +

)90(cos2

2

θα −−+ t d

]sinθ

Y = 0.82 m

H 2 = 0.93 m

M = 0.44 m





N = X +[ 3

5 Z +

)90(cos2

2

θα −−+ t d

]cosθ

θ =2

1( 180 – α – β )

l u =αcos

X

l d =βcos

X L −

N = 0.83 m

θ = 67.5°

l u = 0.61 m

l d = 0.49 m

Fur ther Design Procedure for the Block,

Total head, htotal = h gross + h surge = 42 + 95

= 137 m

Block volume excluding pipe volume, (V) = w ( A1 + A2 + A3 + A4 – l c hc ) – V p

= 0.9 (1.08 + 0.16 + 0.03 + 0.04 – 0.3 × 0.2) – 0.02 m

3

= 1.10 m3

Unit weight of concrete, ( concreteγ ) = 22 kN/m3

Weight of block, WB = V concreteγ

= 1.10 × 22

= 24.20 kN

Unit weight of concrete, ( steel γ ) = 77 kN/m3

Weight of pipe, WP = steel t t d γπ )( +

= π × 0.134 × 0.004 × 77= 0.13 kN/m

Unit weight of concrete, ( water γ ) = 9.81 kN/m3

Weight of water, WW = water

d γ

π×

4

2

=4

)13.0( 2π × 9.81

= 0.13 kN/m

WP + WW = 0.26 kN/m

Calculation for relevant forces:

1. F1u = αcos)( 1uW P LW W +

= 0.26 × 2 × cos 10° = 0.51 kN

2. F1d = βcos)( 1d W P LW W +

= 0.26 × 2 × cos 35° = 0.43 kN

3. Frictional force per support pier:

= αcos)( 2uW P LW W f +±

Where f = 0.6 for steel on concrete

= ± 0.6 × 0.26 × 4 × cos 10° = ± 0.61 kN per support pier





Since there are 4 support piers then,

F2u = ± 0.61 × 4 = ± 2.46 kN

Note that F2d is zero since an expansion joint is located immediately downstream of the

anchor block.

4. F3 =

−

2sin

42

2 αβπγ

d htotal water

=

−×

×××2

1035sin

4

13.013781.92

2π

= 7.72 kN

5. F4u = αsin4u P LW

= 0.13 × 18 × sin 10° = 0.41 kN

But F4u is insignificant since a is less than 20° and could have been ignored. ∴ F4u ˜ 0F4d is negligible since an expansion joint is placed immediately downstream of the anchor

block, i.e., L4d ˜ 0 and therefore F4d ˜ 0

6. F5 = t t d T Ea )( +∆ π

Since the expansion joint is installed between the blocks,

F5 = 0

7. F6 = d 100±

= ± 100 × 0.13 = ± 13 kN

8. F7 = t t d htotal water )( +πγ

F7u = 9.81 × (137 – 18 sinα ) × π × 0.134 × 0.004

= 2.21 kN

F7d = 9.81 × 137 × π ×0.134 × 0.004

= 2.26 kN

Note that as discussed earlier the resultant of these forces is insignificant.

9. F8 =

−

2sin

82

2 αβ

πd

Q

=

−

××

2

1035

sin13.0

02.082

2

π

= 0.01 kN

Note that as discussed earlier, this force is insignificant.

10. F9 = )(4

22

d utotal water d d h −π

γ

Since the pipe diameter does not change,

F9 = 0

11. Soil force, F10 Unit weight of concrete, (?soil) = 20 kN/m3

For stiff clay and stiff sandy clay (Ø) = 30°





K a =φ

φ

22

22

coscoscos

coscoscos

−+

−−

ii

ii= 0.355

F10 = i

h

w K soil

a cos2

2

1γ

But in this example,

h1 – h2 = 0.6 – 0.3 = 0.3 < 1, So F10 is insignificant, therefore F10 = 0

Resolu tion of forces:

α = 10°, β = 35°

Forces (kN) X – component (kN) → + Y – component (kN) ↓ +

F1u = 0.51 – F1u sinα = – 0.09 + F1u cosα = + 0.50

F1d = 0.43 – F1u sin β = – 0.25 + F1u cosβ = + 0.35

F2u = ± 2.46

± F2u cos α = ± 2.42

Negative during 1expansion

Positive during 2contraction

± F2u sin α = ± 0.43

Negative during 1expansion

Positive during 2contraction

F3 = 7.72 + F3 sin

+

2

αβ= + 2.95 – F3 cos

+

2

αβ= – 7.13

F6 = ± 13

± F6 (cosα – cos β ) = ± 2.15

Positive during 1expansion

Negative during 2contraction

± F6 ( sin β – sinα ) = ± 5.20

Positive during 1expansion

Negative during 2contraction

F7u = 2.21 + F7u cosα = + 2.18 + F7u sinα = + 0.38

F7d = 2.26 – F7d cos β = – 1.85 – F7d sin β = – 1.30

F8 = 0.01 + F8 sin

+

2αβ ˜ 0 – F8 cos

+

2αβ = – 0.01

WB = 24.20 0 + 24.20

SUM H ∑ = + 2.67 1Expansion

H ∑ = + 3.21 2Contraction

V ∑ = + 21.76 1Expansion

V ∑ = + 12.22 2Contraction

Note that forces are positive in X-direction is towards the right and Y-direction downwards.

Sum of horizontal forces that act at the bend, = x F H 10−∑ 1Expansion case = + 2.67 – 0 = + 2.67 kN → 2Contraction case = + 3.21 – 0 = + 3.21 kN →

Sum of vertical forces that act at the bend, = B y W F V −−∑ 10 1Expansion case = + 21.76 – 0 – 24.20 = – 2.44 kN ↑ 2Contraction case = + 12.22 – 0 – 24.20 = – 11.98 kN ↑

Calculate for the centre of gravity of the block from the upstream face of the block taking the

moment of mass. The effect of the pipe passing through the block is considered negligible, so need

not be calculated.

X =concrete

concrete

i

ii

w

w

A

X A

ρ

ρ×

∑∑

=2.03.004.003.016.008.1

15.02.03.089.004.055.003.092.016.042.008.1×−+++ ××−×+×+×+× = 0.51 m

∴The weight of the block WB acts 0.51 m from point O.





A force diagram on the block is as shown in Figure;

Checking for Safety of the Block:

1. Checking safety against overturning

1

Expansion Case

Taking sum of moment about point O with

clockwise moments as positive:

∑ M at O,

= 24.20×0.51+2.67×0.82 – 2.44×0.60

= 13.06 kN-m

V

M

∑∑

=76.21

06.13= 0.60 m

V

M Le

∑

∑−=

2

= ¦ 0.50 – 0.60¦ = 0.1

6

baseallowable

Le = =

6

1= 0.17

allowableee <∴ OK

2Contraction Case

Taking sum of moment about point O with clockwise moments as positive:

∑ M at O,

= 24.20×0.51+3.21×0.82 – 11.96×0.60

= 7.80 kN-m

V

M

∑∑

=22.12

80.7= 0.64 m

V

M Le

∑∑

−=2

= ¦ 0.50 – 0.64¦ = 0.14

6

baseallowable

Le = =

6

1= 0.17

allowableee <∴ OK

Since eallowablee < for both cases, the structure is safe against overturning.

2. Checking safety on bearing capacity

For stiff clay allowable bearing pressure is 200 kN/m2

1 Expansion Case

base P =

+

∑

basebase L

e

A

V 61 =

×+

× 1

1.061

9.01

76.21= 38.68 kN/m2

allowablebase P P < OK





2Contraction Case

base P =

+

∑

basebase L

e

A

V 61 =

×+

× 1

14.061

9.01

22.12= 24.98 kN/m2

allowablebase P P < OK

In both cases allowablebase P P < = 200 kN/m2, ∴the structure is safe against sinking.

3. Checking safety against sliding

1 Expansion Case

V H ∑<∑ µ

µ = 0.5 for concrete/masonry on soil

2.67 kN < 0.5×21.76 kN

2.67 kN < 10.88 kN OK

2Contraction Case

V H ∑<∑ µ

µ = 0.5 for concrete/masonry on soil

3.21 kN < 0.5×12.22 kN

3.21 kN < 6.11 kN OK

Since, V H ∑<∑ µ in both cases the structure is safe against sliding.

∴The anchor block is stable.

anchor block type 2

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