PALAMANERI HEP (480 MW)

STRUCTURAL DESIGN OF CONCRETE LINING IN ROCK

Concrete Lining without Reinforcement

Finished diameter of Tunnel = 7.2 mAxial Force = 250 kN (Tension)Moment = 1 kNmThickness of Lining = 300 mm

= 300 X 1000

= < 3.2 N/mm2(clause: B - 2.2.1, IS 456 2000)

=Ast = 230

Provide 16 Ф @ 175 c/c in the middle of section =And dist. Steel as 16 Ф @ 200 c/c as dist. Reinforcement

Concrete Lining without Reinforcement

Finished diameter of Tunnel = 7.2 mAxial Force = 500 kN (Tension)Moment = 2 kNmThickness of Lining = 300 mm

= 300 X 1000

= < 2.8 N/mm2(clause: B - 2.2.1, IS 456 2000)

=Ast = 230Provide20 Ф @ 135 c/c in the middle of section =

And dist. Steel as 16 Ф @ 200 c/c as dist. Reinforcement =

ROCK CLASS A (VERY GOOD ROCK)

Grade of Concrete = M-20

Tensile Stress developed in Concrete 250 X 10 3

0.883 N/mm2

250 X 10 3 1087 mm2

1149 mm2

ROCK CLASS B (FAIR ROCK)

Grade of Concrete = M-20

Tensile Stress developed in Concrete 500 X 10 3

1.67 N/mm2

500 X 10 3 2174 mm2

2319 mm2

10.05 mm2

Concrete Lining without Reinforcement

Finished diameter of Tunnel = 7.2 mAxial Force = 890 kN (Tension)Moment = 7 kNmThickness of Lining = 300 mm

= 350 X 1000

= < 2.8 N/mm2(clause: B - 2.2.1, IS 456 2000)

Provide 25 Ф @ 115 c/c in the middle of section And dist. Steel as 16 Ф @ 200 c/c as dist. Reinforcement

Concrete Lining without Reinforcement

Finished diameter of Tunnel = 7.2 mAxial Force = 1300 kN (Tension)Moment = 10 kNmThickness of Lining = 450 mm

=450 x 1000

≈ 2.89 < 3.2 N/mm2(clause: B - 2.2.1, IS 456 2000)

Concrete Lining without Reinforcement

Finished diameter of Tunnel = 7.2 mAxial Force = 1470 kN (Tension)Moment = 13 kNmThickness of Lining = 500 mmGrade of Concrete = M-25

ROCK CLASS C (POOR ROCK)

Grade of Concrete = M-20

Tensile Stress developed in Concrete 890 X 10 3

2.54 N/mm2

ROCK CLASS D (VERY POOR ROCK)

Grade of Concrete = M-25

Tensile Stress developed in Concrete 1300 X 10 3

ROCK CLASS E (VERY POOR ROCK)

=500 x 1000

= 2.94 < 3.2 N/mm2(clause: B - 2.2.1, IS 456 2000)

Concrete Lining with Reinforcement

Finished diameter of Tunnel = 7.2 mThickness of Lining = 500 mmAxial force 'T' = 1810 kNMoment 'M' = 15 kNm

=500 x 1000

= 3.62 > 3.2 N/mm2(clause: B - 2.2.1, IS 456 2000)

Hence, Reinforcement is needed to resist the tensile forces

Here, Tensile Force is very large , as compared to the Moment , so eccentricity will be small

Eccentricity 'e' = M/T = = 8.29 mm

1810 X 103

In this case, the line of action of the forces lies within the section between thetwo layers of steel as shown in the figure below ;

↓ b ↓

D

At1 d'

Tensile Stress developed in Concrete 1470 X 10 3

ROCK CLASS F (EXTREMELY POOR ROCK)

Tensile Stress developed in Concrete 1810 X 10 3

15 X 10 6

D↓

Axis through C.G. of steel→ T

At2 d'

↑ ↑

The whole section is in tension and the concrete area is ineffective.

Taking moment of tensile forces about the bottom steel,

=

= 1810 X 103 (170 - 8.3)=

0.87 X 415 (170 + 170)

T

=T - σst1 X At1

=1810X103 - 0.87X415X2500

=0.87 X 415

Provide 20 Ф @ 125 c/c on each face

= T Ac + m(Ast)

=494974 + 11 X 5026

= 3.29 > 3.2 N/mm2(clause: B - 2.2.1, IS 456 2000)

Just greater than permissible , so it is O. K.

d1

e−−−−−−−−−−−−− d2↑

The total tensile force acts at a distance 'e' from the c. g. of steel areas, if σst1 and σst2 are stresses in

steel areas , At1 and At2 , respectively.

σst1 X At1 X (d1 + d2) = T (d2-e)

or, At1 T (d2-e)

σst1 X (d1 + d2)

2384.2 mm2

And σst1 X At1 + σst2 X At2 =

or, At2 σst2

2513.2 mm2

Tensile Stress developed in Concrete

1810 X 103

(VERY POOR ROCK) Concrete Lining with Reinforcement (for internal pressure)

Finished diameter of Tunnel = 10 mThickness of Lining = 600 Mm (= D)Axial force 'T' = kNMoment 'M' = kNmGrade of Conc. = M-20Cover To R/F = 50 mmPermissible Stress in Concrete = 2.8Permissible Stress in Steel in = 230direct tension and flexural tension (Clause: B - 2.2, IS 456 : 2000)

Here, Tensile Force is very large , as compared to the Moment , so eccentricity will be small

As per Carranza -Torres article – Mechanical analysis of circular lines with particular reference to composite supports (Page 10-12)-Tunnel underground space technology(2009)r = 5b = 0.5 ms = 0.5 mn=b/s= = 1

For steel

E1 = 200000 Mpa

A1 = 0.00475 m2

v1 = 0.25

I1 = 0.0000513 m4

D1 = 1013333333.33

= 1013.33 MN

K1 = 10944000

= 10.944 MN/m2

For shotcretet2 = 0.4 M

E2 = 20000 Mpa

v2 = 0.15

A2 = (1/n) x bt2 – A1 = 0.19525 m2

I2 = (1/n) x bt2^3 /12 – I1 = 0.0026153667 m4

= ###

= 3994.88 MN

= 53511338.45

N/mm2

N/mm2 (grade fe 415)

D1 = E1A1/(1-v12)

K1 = E1 I1/(1-v12)

D2 = E2 A2/(1-v22)

K2 = E2 I2/(1-v22)

= 53.51 MN/m2

= 0.393 m

= 25488 Mpa

= 10892 Mpa

K1/n(K1+K2) = 0.169791987187491K2/n(K1+K2) = 0.830208012812509

Beam L/C Node Fx kN Fy kN Mz kNm3 1 LOAD CASE 1 3 -356.113 78.508 -83.448 -14.17

4 362.927 47.144 100.278 17.034 1 LOAD CASE 1 4 -362.695 45.52 -100.278 -17.03

5 363.575 92.709 72.226 12.265 1 LOAD CASE 1 5 -370.201 32.465 -72.226 -12.26

6 361.883 106.933 27.96 4.756 1 LOAD CASE 1 6 -372.789 14.567 -27.96 -4.75

7 384.847 119.215 -37.296 -6.337 1 LOAD CASE 1 7 -400.706 6.648 37.296 6.33

8 429.65 104.391 -95.398 -16.208 1 LOAD CASE 1 8 -440.48 33.535 95.398 16.20

9 450.455 71.348 -117.876 -20.0120 1 LOAD CASE 1 20 -423.403 29.977 -51.386 -8.72

21 423.754 126.511 -0.418 -0.0721 1 LOAD CASE 1 21 -421.853 6.412 0.418 0.07

22 422.898 149.983 -77.467 -13.1522 1 LOAD CASE 1 22 -411.476 -12.024 77.467 13.15

1 413.196 168.226 -174.202 -29.5823 1 LOAD CASE 1 2 -351.648 147.109 6.359 1.08

3 362.039 -20.235 83.448 14.1724 1 LOAD CASE 1 1 -353.646 220.668 174.202 29.58

2 367.41 -92.08 -6.359 -1.08-29.57810375

y 125 mmA 4750 mm2I 51300000 mm4fy 230 N/mm2M 29580000 knmF 89900 N

teq = sqrt (12(K1+k2)/(D1+D2))

Eeq=n(D1+D2)/(bteq)

G= Eeq/2(1+v)

M1(kNm) = MK1/n(K1+K2)

σt/0.6fy + σbt/0.66fy IS 800:1984, Cl 7.1.2

0.13714721587 + 0.474809113253049 = 0.611956329 <1 OK

Page -8

FS = 0.67A = 392987.2

= 2.5 MpaN = 1466370.3 - 1.534401E-06

N Q Q-1000000.0 551310.2 -551310.2-500000.0 793601.3 -793601.3

0.0 977580.2 977580.2500000.0 1132043.8 -1132043.8

1000000.0 1267826.2 -1267826.21500000.0 1390411.2 -1390411.22000000.0 1503031.4 -1503031.42500000.0 1607782.1 -1607782.13000000.0 1706113.5 -1706113.53500000.0 1799078.5 -1799078.54000000.0 1887470.1 -1887470.14500000.0 1971903.6 -1971903.65000000.0 2052867.2 -2052867.25500000.0 2130756.6 -2130756.66000000.0 2205897.5 -2205897.56500000.0 2278561.8 -2278561.87000000.0 2348979.3 -2348979.37500000.0 2417346.5 -2417346.58000000.0 2483832.5 -2483832.58500000.0 2548584.7 -2548584.79000000.0 2611732.0 -2611732.09500000.0 2673388.2 -2673388.2

10000000.0 2733654.0 -2733654.0

N = σcA/FS – 9Q2FS/4σcA

mm2

σt

Q2

0.0 1000000.0 2000000.0 3000000.0

-2000000.0

0.0

2000000.0

4000000.0

6000000.0

8000000.0

10000000.0

12000000.0

Column A

Q( in N)

N (

in N

)

As per Carranza -Torres article – Mechanical analysis of circular lines with particular reference to composite supports (Page 10-12)

-69.28 -71.51 -284.6083.25 72.78 290.15-83.25 -72.73 -289.9659.96 73.09 290.48-59.96 -74.43 -295.7723.21 73.04 288.84-23.21 -75.25 -297.54-30.96 78.11 306.7430.96 -81.32 -319.39-79.20 87.55 342.1079.20 -89.75 -350.73-97.86 91.91 358.55-42.66 -85.33 -338.07-0.35 85.74 338.010.35 -85.36 -336.50

-64.31 86.07 336.8364.31 -83.76 -327.72

-144.62 84.74 328.465.28 -71.19 -280.46

69.28 72.71 289.33144.62 -72.69 -280.96-5.28 74.38 293.03

-89.74501508078

M2(kNm) = MK2/n(K1+K2)

Fx1 = FxD1/(n(D1+D2)) +M(D2K1 – D1K2)/(nR(D1+D2)(K1+K2))

Fx2 = FxD2/(n(D1+D2)) -M(D2K1 – D1K2)/(nR(D1+D2)(K1+K2))

0.0 1000000.0 2000000.0 3000000.0

-2000000.0

0.0

2000000.0

4000000.0

6000000.0

8000000.0

10000000.0

12000000.0

Column A

Q( in N)

N (

in N

)