HEAD RACE TUNNEL - HORSE SHOE SHAPE: DESIGN OF CONCRETE LINING: Head Race Tunnel is a part of water conveyance system connecting intake shaft and surge shaft. The tunnel proposed has a length of 705 m and a finished diameter of 3.0m (Horse shoe). Concrete lining of 20cm thick is proposed for the tunnel. The slope of the tunnel up to surge shaft is 1 in 100. The sill level of tunnel at intake is 736.50m. . GEOLOGY OF HEAD RACE TUNNEL: The proposed 3m dia HRT runs for a length of about 705m below the hill ranges of the right bank of river. The maximum cover above the HRT is 85m and minimum cover is 40m. The detailed surface geological mapping reveals that the HRT is crossing 2 valleys. The alignment is covered by slope wash materials and over burden ranging from 3m to 15m and rock is exposed in the valleys. The rock type is Biotite Geniss, which is striking NNE-SSW dipping steeply towards east. Three sets of joints are observed. The project authorities have already taken up 6 bore holes along the HRT. These bore holes have indicated sound rock at a depth of 3m to 18m below the ground. The exploration carried out has brought out that the sufficient rock cover over the tunnel crown is available and the tunneling is expected to be largely smooth and the rock mass media is anticipated to be self supporting, expect between Ch 390m and Ch 405m where fractured zone is anticipated. In general, the RQD values range from 70% to 85% except in bore hole no. BH-5, which is drilled 15m in overburden and 5m in rock where RQD is 38%. However, rock bolting is required in jointed rock reaches. References: CBIP-manual on planning and design of hydraulic tunnel. IS-4880 (part IV)-1971 IS-5878 (part V) – 1971 IS-4880 (part VI) – 1971 IS-456 - 2000

7 5

15

0

1 0 °

9 0 °

2 2 .5 °

R=

30

00 R = 3 0 0 0R = 3 0 0 0

R = 1 5 0 0

P C C L IN IN G

2 0 0 T H IC K

D E S IG N F IN IS H L IN E

M IN IM U M E X C A V A T IO N L IN E (A )

P A Y L IN E (B )

3 5 D ia G R O U T H O L E S

D A T A

Grade of concrete for lininig = M-20

Finished Internal dia of the tunnel = 3.00 m

Assumed thickness of concrete lining (t) = 0.20 m

Internal radius of the tunnel ( r ) = 1.50 m

Mean radius of the tunnel ( R ) = 1.60 m

Young's modulus of lining material (M:20 concrete)

= 1.6154 T/sqm

Moment of Inertia of concrete lining ( I ) b = 1, d = 0.2 m

= 0.00067 m4

Unit weight of water w = 1.0 t/cum

Unit weight of concrete Wc = 2.40 t/cum

Total rock load on mean diameter (P)

Calculation of Rock load: Ref: CBIP - Dec 1984, publication No. 178, Manual on planning and design of hydraulic tunnel, page 18 table 4.2

2) Rock load for massive moderately jointed rock condition =0 to 0.5B where B = width of the opening (tunnel)

Hydrostatic pressure between RL 757.72 & RL 732.5 considered= 25.22m (see note)

=

25% hydrostatic pressure is considered.

Net hydrostatic pressure= 25.22/4 = 6.305 t/sqm

Vertical load with hydrostatic pressure = ( 0 to 0.5B)x + 25% Hydrostatic head (0.25x3.55)x(2.45-1)+6.305

= 7.59 say 7.6

t/sqm t/sqm

Note: FRL=757.72, spring level of tunnel at surge shaft = 732.50

sub

V V

7 .6 .t/s q m

RADIAL SHEAR

Due to uniform vertical load (t)

Due to conduit weight (t)

Due to contained water (t)

Total (t)

0 0 0 0 0

45o -0.25P = -.25x7.6 = - 1.90

-0.8976 WctR = -.8976x2.4x.2x1.6 = -0.6893

-0.4488wr2

- 0.4488x1x1.52

-1.0098

-3.59

90o

0 +0.1667wctR = 0.1667x2.4x.2x1.6 = 0.128

0.0833wr2

0.833x1x1.52

0.1874

0.3154

135o 0.25P = 0.25x7.6 = 1.90

0.6732wctR 0.6732x2.4x0.2x1.6 =0.5170

0.3366 wr2

0.3366x1x1.52

0.7573

3.1743

180o 0 0 0 0

Max Radial Shear = 3.59t Shear stress = 3.59/(100x20) = 1.79 Kg/sqcm < 1.83kg/sqcm (safe)

BENDING MOMENT

Due to uniform vertical load (t-m)

Due to conduit weight (t-m)

Due to contained water (t-m)

Total (t-m)

0 0.125PR 0.125x7.6x1.6 =1.52

0.4406WctR2

0.4406x2.4x0.2x1.62

=0.5414

0.2203 wr2R 0.2203x1x1.52x1.6 =0.79308

2.855

45o 0 -0.03344 WctR2

= -0.03344x2.4x0.2x1.62 = -0.04109

-0.0167wr2R - 0.0167x1x1.52x1.6 = - 0.0601

-0.1021

90o

-0.125PR -0.125x7.6x1.6 = -1.52

-0.3927WctR2 = -0.3927x2.4x0.2x1.62 = -0.4825

-0.1963wr2R

-.1963x1x1.52x1.6 -0.70668

-2.709

135o 0 0.03344 WctR2

= -0.03344x2.4x0.2x1.62 = -0.04109

0.0167wr2R 0.0167x1x1.52x1.6 = 0.0601

0.10112

180o 0.125PR 0.125x7.6x1.6 = 1.52

0.3448Wct R2

0.0448x2.4x0.2x1.62

0.4237

0.1724wr2R 0.1724x1x1.52x1.6 0.62064

2.56

M = Qbd2

d = M/Qb = 2.855x1000x100

13 x 100 = 14.82 cm Actual thickness provided is 20 cm. As per IS-4880 (Part V) 1972, clause 7.6 min. thickness of 18 cm is recommended. However provide 20 cm thick concrete lining of grade M20

NORMAL THRUST

Due to uniform vertical load (t)

Due to conduit weight (t)

Due to contained water (t)

Total (t)

0 0 +0.1667wctR = 0.1667x2.4x.2x1.6 = 0.128

-1.4166wr2

- 1.4166x1x1.52

-3.187

-3.059

45o 0.25P =0.25x7.6 = 1.90

1.1332 WctR = 1.1332x2.4x.2x1.6 = 0.870

-0.7869wr2

- 0.769x1x1.52

-1.770

1.0

90o

0.5P =0.5x7.6 = 3.80

+1.5708wctR = 1.5708x2.4x.2x1.6 = 1.206

-0.2146wr2

-0.2146x1x1.52

= -0.4828

4.5232

135o 0.25P = 0.25x7.60 = 1.90

0.4376wctR 0.4376x2.4x0.2x1.6 =0.3361

-0.4277 wr2

-0.4277x1x1.52

= - 0.9623

1.274

180o 0 -0.1667WctR = - 0.1667x2.4x.2x1.6 = - 0.128

-0.5834 wr2

-0.5834x1x1.52

= - 1.3126

- 1.446

Maximum negative thrust = - 3.059 t (this indicates tension at the crown) Tensile stress in concrete at crown = (3.059 x 1000)/(100x20) = 1.529 Kg/sqcm < 5.85 Kg/sqcm. Hence safe.

METHOD OF CONCRETE LINING

As per IS 5878 (Part V) 1971 clause 5.1(b) page 5 the concrete shall be placed at Kerbs first followed by side walls and arch and finally invert. As per IS 5878 (Part V) 1971 clause 9.1- page-7, the concrete may be placed either by a concrete pump or a pneumatic placer through steel pipes. In the invert, the concrete may be placed by direct placement also. As per IS 5878 (part V) 1971 clause 2.1.1, page 4: All the specifications for course aggregate cement and water required for normal concrete in accordance with IS 456-2000 shall hold good in this case also. However the slump shall not be less than 10cm at the point of placing and the sand content shall be more than what is specified in IS-456-2000. Where concrete is placed directly as in case of inverts, the slump should be reduced to 3.0cm. The maximum size of aggregate shall not exceed 38mm. As per IS 5878 (part V) 1971 clause 2.1.12 page 4: For easy flow of concrete it is advisable to use air entraining agents to entrain up to 3 to 4% of air. As per IS 5878 (part V) 1971 clause 2.1.1, page 8: GROUTING: PACK GROUTING: As per IS 5878 (Part V) 1971 clause 14.1- page-8, concrete lining of underground work shall be grouted to pack the spaces between rock and concrete lining. Pack grouting shall be done not earlier than 21 days after the concrete lining has been placed. Grouting shall be done with low pressure not exceeding 5 kg/sqcm or as required depending on the circumstances. The grouting shall start from the bottom holes first followed by the holes in the sides and finally through the holes in the crown. PRESSURE GROUTING: As per IS 5878 (Part V) 1971 clause 15.1- page-8, when it is found that the rock strata is highly fissured or poor or there is seepage of water, it is necessary to do high pressure grouting deep holes either before the lining is placed or after the lining is placed according to circumstances met with.

DESIGN OF ROCK BOLTS

Ref: IS 11527 – 1985 – Criteria for structural design of Energy dissipators For spillways IS - 4880 – (part VI) - 1971 DATA: Fb1 = Permissible bond stress between steel and grout = 6 Kg/sqcm Fb2 = Permissible bond stress between grout and rock = 4 Kg/sqcm L = length of Anchors =( AstxFt)/(Wr-W) Ast = Area of steel Ft = Tensile yield stress of steel with FS 1.2 Wr = Density of Rock = 2.6 t/cum W = Density of water = 1 t/cum Fu = Net downward force / unit area Fu/Ft = Anchorage steel / unit area n = no. of Anchors/sqm 1/n = spacing of Anchors B = width of opening of tunnel = 3.55 m Vertical load without hydrostatic pressure on the crown of the tunnel: (concrete lining is provided ) = (0 to 0.5B)Wr = 0.25x3.55x2.6 = 2.307t/sqm Fu =2.31t/sqm

Ast = (Fu/Ft) = (2.31/4.167) = 0.55 sqcm/sqm

No. of Anchor rods/Sqm = n = Ast/a Provide 25mm dia rods for Anchor a = area of the rod = 4.906 sqcm n = 0.55/4.906 = 0.1121

Spacing of Anchors = 1/n = 1/0.1121 = 2.98 m c/c How ever provide a spacing of 2.0 m c/c CALCULATION OF LENGTH OF ANCHORS IN ROCK: As per IS 11527 a) L= (Ast x Ft )/(WR) = 4.906 x 0.55 / 2.6 = 1.04 m b) By considering Bond strength between steel and grout Fb1 = 6 kg/sqcm Increase this by 40% = 8.4 kg/sqcm

Bond length = st

xdxlx

= 3.14x2.5xlx8.4 -----1 Load carried by each rod = (2)2 x 2.31 = 9.24 t ----2 Equate 1 & 2 3.14x2.5xlx8.4 = 9.24 l = (9.24x1000)/(3.14x2.5x8.4) = 140.12 cm = 1.40 m c) By considering Bond between grout and rock Permissible bond strength = 4 Kg/sqcm

Bond strength = gr

xdxlx

3.14x3.8xlx4 = 9.24 l = (9.24x1000)/(3.14x3.8x4) = 193.59cm = 1.94m

d) As per IS 4880 (part 6) 1971 page 8 : l i.e length of the Anchor rod shall be > b/4 to b/3 i.e 3.55/4 = 0.8875m, 3.55/3 = 1.18m max of above = 1.94 m Provide Anchor rod of 25mm dia of 2.0 length at 2.0 m c/c However, the length of the rod also depends upon the depth of the maximum loosening of the rock during execution. In such a case the length of the Anchor rod shall be greater than the thickness of the loosening rock.