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do - Tunnel Radius - 3.5 m

- Angle of Internal friction - 30 Degree

c - Cohesive Strength - 0.1 Mpa

GSI - - 60

- Unit weight of Rock mass - 26 kN / cu m

H - Height of over burden - 100 m

do - Tunnel radius - 5 m

Po - In-situ Stress

- H

- 2.6 Mpa

Pi - Internal Support Pressure

- 1.2 Mpa (Figure-1)

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ci - Lab Uniaxial compressive strength

-

- 1.1017 Mpa

cm - Rock mass Strength/In situ Uniaxial compressive strength

- 2c cos /(1-sin )

- 0.35 Mpa

dp - Plastic zone radius

i - Tunnel Sidewall deformation/Tunnel closure

k - or

cm/Po - Rock mass Strength/In situ Stress

- 0.13

Pi/Po - 0.46

dp/do - 1.20

i/do - 0.005

- i /do

dp - 4.19 m

i - 0.02 m

0.019 e 0.05 GSI/ cm

(1+sin )/(1-sin 1/ 3

Strain of the tunnel

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Definitions of Dimensions:

b - 1500 mm

t - 150 mm

b 1 - 780 mm

b 2 - 330 mm

t 2 - 95 mmd 1 - 16 mm

d 2 - 20 mm

- 30 Degree

Cover - 30 mm

h g - 195 mm

bg - 180 mm

Lo - 2000 mm

d 3 - 6 mm @ 150 mm c/c

LoRock Bolts

ELEV

b

b 1

d 1

t

t 2d 3

d 2

b 2

PLAN

195

180 Lattice Girder

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h - t + t 2

- 225 mm

d g - h g - 0.5(d 1 + d 2)

- 167 mm

Definitions of Material Properties:

f cd - 20 Mpafyd - 435 Mpa

- 0.85

A st 1 - 402 Sq mm

A st 2 - 314 Sq mm

A st 3 - 28 Sq mm

A c - b t + 0.5 t 2 b 2 - A st 1 - A st 2

- 221110 Sq mm

b eff - 0.2 (b-b 2) + 0.24 L o+b 2

- 1044 mm

A s11 -

- 599 Sq mm (only in tension)

h cm -

- 84.71 mm (Calculation of center of mass)

Eight Strain conditions to derive M-N envelope:

1 Pure axial force

2 Zero strain in tensioned bars (positive Moment)

3 Equilibrium (Max strain in shotcrete and yielding in rebars at the same time - P

4 Pure moment (Positive)

5 Pure tension

6 Pure Moment (Negative)

7 Equilibrium (Max strain in shotcrete and yielding in rebars at the same time - N

8 Zero strain in tensioned bars (Negative Moment)

1 Nd - f cd A c + f yd (A st1 + A st2)

- 4070 kN

5 N tog - f yd (A s11 +A st2)

- 397 kN

3 cu - 0

syd - 0

d - hg - 0.5 d 2

- 185 mm

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X bal - cu d

cu + syd

- 116 mm

0.8 X bal < t

0.8 X bal- 93 mm

< 150 mm

f c - f cd 0.8 X bal b eff - ### N

f s1 - f yd A st1 - 174835 N

f s2 - f yd A st2 - 136590 N

N ud - + f s1 - f s2 - 1680 kN

M ud - f c (h cm -0.4 X bal) + f s1 (h cm - 0.5 d1) + f s2 (d - h cm)

- 62.84 kN m

2 h c3 - 0.8 d - 148 mm

A c3 - t b eff + (0.5 b 2

- 162006 sq mm

Center mass of compression zone

90

- (h - h c3)2 / tan )

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4 x - f yd A st2

f cd 0.8 b eff

- 9.62 mm

Mu - f yd A st2 (d -0.4 x)

- 24.74 kN m

5

6 d 1 - h -0.5 d 1

- 217 mm

h c -

- 94.04 mm

M u2 - f yd A s 11 (d 1 - 0.67 hc)

- 40.10 kN m7 X bal2 - cu d

cu + syd

- 136 mm

h c2 - 0.8 X bal2

- 109 mm

f c2 - - 346632 N

f s11 - yd A s 11 - 260395 N

f s2 - f yd A st2 - 136590 N

N ud2 - + f s2 - f s11 - 223 kN

M ud - f c2 (h-h cm -0.67 h c2) + f s1 (d-h cm ) + f s2 (h cm - 0.5d1)

- 17 kN m

8

Sqrt ((A s11 f yd tan )/ f cd)

f cd (h c2) 2/tan

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0 0

M ud3

M ud

M u

Moment = 0 = 0

-M u2-M ud3

-M ud4

0 0

N d

N ud3

N ud

0 0

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Nforce = -N tog =

0 0

N ud2

N ud4

N d

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sitive Moment)

egative Moment)

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Beam L/C Node Fx kN Fy kN Mz kNm

3 SELF WEIGHT+ROCK LO 2 941.45 103.39 0

39 -937.7 -52.77 48.8

34 SELF WEIGHT+ROCK LO 15 199.74 15.72 -262.21

25 -207.5 82.83 242.46

35 SELF WEIGHT+ROCK LO 25 213.39 -30.56 -242.46

26 -236.46 126.68 196.236 SELF WEIGHT+ROCK LO 26 246.88 -72.47 -196.2

27 -284.72 163.8 126.68


28 -351.43 190.03 39.66


29 -434.81 200.59 -56.26

39 SELF WEIGHT+ROCK LO 29 456.35 -126.28 56.26

30 -531.52 190.48 -149.45


31 -636.55 155.09 -225.5241 SELF WEIGHT+ROCK LO 31 651.99 -53.2 225.52

32 -743.32 91.03 -267.95

42 SELF WEIGHT+ROCK LO 32 748.18 26.43 267.95

33 -844.31 -3.35 -259.2

43 SELF WEIGHT+ROCK LO 33 833.28 135.39 259.2

4 -931.83 -127.64 -181.81


40 -933.95 -2.14 65.96


41 -930.2 48.48 51.48


42 -926.45 99.11 5.36


43 -922.7 149.73 -72.4


4 -918.95 200.36 -181.81

Beam L/C Node Fx kN Fy kN Mz kNm

Max Fx 3 IGHT+R 2 941.45 103.39 0

Min Fx 34 IGHT+R 15 199.74 15.72 -262.21

Max Fy 43 IGHT+R 33 833.28 135.39 259.2Min Fy 38 IGHT+R 29 434.81 -200.59 56.26

Max Fz 43 IGHT+R 33 833.28 135.39 259.2

Min Fz 53 IGHT+R 43 922.7 -149.73 72.4

Max Mx 40 IGHT+R 30 552.26 -103.44 149.45

Min Mx 43 IGHT+R 33 833.28 135.39 259.2

Max My 42 IGHT+R 32 748.18 26.43 267.95

Min My 43 IGHT+R 33 833.28 135.39 259.2

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Max Mz 41 IGHT+R 32 743.32 -91.03 267.95

Min Mz 34 IGHT+R 15 199.74 15.72 -262.21

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1.00 m

0.030.15 m

0.10 m0.03

0.15 mCompression Load @ Moderate eccentricity

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from book

Page 19

b=D=m=dc=dt=

d=Bending moment = M

Tensile force = T

let us assume, x = position of N.A from top of compr face=therefore, k = x/d

No. of bars = n_c=dia of bolts = dia_c = dc =

No. of bars = n_t=

dia of bolts = dia_t = dt=

Eccentricity (e) = M/P =

Asc = n_c * pie * (dia_c) * (dia_c)/4 =Ast = n_t * pie * (dia_t) * (dia_t)/4 =

taking the moments of internal and external forces about the centre of tensile

i.e.,

53100000

thus,

again, equating the sum of internal forces to the external forces, we have

i.e.,299975.08

thus,

hence, from stress diagram, x=1*d/(1+t/(m.c')) =

Now assume an intermediate value between (1) and (2) for x, say x=

substituting in eqn (A), we geti.e., bxc'(d-x/3)/2 + (mc-1)Asc.c'(x-dc)(d-dc)/x = P(e+D/2 -dt)

51854933.33

bxc'(d-x/3)/2 + (mc-1)Asc.c'(x-dc)(d-dc)/x = P(e+D/2 -dt) ---

bxc'/2 + (mc-1)Asc.c'/x(x-dc) Ast.t =P ----------------> (

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from book

Page 20

i.e., c' =

substituting in eqn.(B), we geti.e., bxc'/2 + (mc-1)Asc.c'/x(x-dc) Ast.t =P

294975.33

thus,

hence, from stress diagram, x=1*d/(1+t/(m.c')) =

thus,

and,

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from book

Page 21

400 mm800 mm1960 mm60 mm

740 mm200 N-m per metre width400 kN

450 mm -------> (1)0.61

625 mm

6

25 mm

500 mm

2943.75 mm22943.75 mm2

steel, we have

*c' + 47708375 *c' = ###

c' = 3.33 N/sq.mm

+ 233845.15 + -2943.75 *t = 400000

t= 45.46 N/sq.mm

430.77mm ------> (2)

436mm (say)

*c' + ### *c' = ###

---> (A)

B)

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from book

Page 22

3.38 N/sq.mm

+ 236158.83 + -2943.75 *t = 400000

t= 44.55 N/sq.mm

437.07mm

c' = 3.38N/sq.mm (compressive)

t = 44.55N/sq.mm (tensile)

tc = 83.14N/sq.mm (compressive)

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Sheet6

Page 23

b= 1000 mmD= 225 mmm= 19dc= 30 mmdt= 30 mm

d= 195 mmBending moment = M 200 per metre width

Tensile force = T 400 kN

let us assume, x = position of N.A from top of comp 135 mm -------> (therefore, k = x/d 0.69 med to be between 0.

No. of bars = n_c= 2dia of bolts = dia_c = 20 mm

No. of bars = n_t= 2

dia of bolts = dia_t = 32 mm

Eccentricity (e) = M/P = 500 mm

Asc = n_c * pie * (dia_c) * (dia_c)/4 = 628 mm2Ast = n_t * pie * (dia_t) * (dia_t)/4 = 1607.68 mm2

taking the moments of internal and external forces about the centre of tensile steel, w

i.e.,

10125000 *c' + ###

thus, c' = 18.88 N/sq.mm

again, equating the sum of internal forces to the external forces, we have

i.e.,1274377.8 + 253595.94 +

thus, t= 701.62 N/sq.mm

hence, from stress diagram, x=1*d/(1+t/(m.c')) = 65.97 mm ------> (2

Now assume an intermediate value between (1) an 90 mm (say)

substituting in eqn (A), we geti.e., bxc'(d-x/3)/2 + (mc-1)Asc.c'(x-dc)(d-dc)/x = P(e+D/2 -dt)

7425000.00 *c' + ###

bxc'(d-x/3)/2 + (mc-1)Asc.c'(x-dc)(d-dc)/x = P(e+D/2 -dt) -------> (

bxc'/2 + (mc-1)Asc.c'/x(x-dc) Ast.t =P ----------------> (B)

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Sheet6

Page 24

i.e., c' = 24.99 N/sq.mm

substituting in eqn.(B), we geti.e., bxc'/2 + (mc-1)Asc.c'/x(x-dc) Ast.t =P

1124432.96 + 287688.25 +

thus, t= 629.55 N/sq.mm

hence, from stress diagram, x=1*d/(1+t/(m.c')) = 83.83 mm

thus, c' = 24.99N/sq.mm

t = 629.55N/sq.mm

and, tc = 474.76N/sq.mm

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Sheet6

Page 25

)to 0.7)

e have

*c' = ###

-1607.68 *t = 400000

)

*c' = ###

)

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Sheet6

Page 26

-1607.68 *t = 400000

(compressive)

(tensile)

(compressive)

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