Shakedown analysis of lined rock cavern for compressed air ... · PDF fileBackground - CAES Compressed air energy storage (CAES) systems Lined or unlined rock caverns Gas pressure:

Juan Wang

The University of Nottingham, Ningbo, China

Pin-Qiang Mo

State Key Laboratory for GeoMechanics and Deep Underground Engineering, China University

of Mining and Technology, Xuzhou, China

Hai-Sui Yu

School of Civil Engineering, University of Leeds, Leeds, UK

Shakedown analysis of lined rock cavern for compressed air energy storage

International Geotechnics Symposium cum International Meeting of CSRME 14th Biennial National Congress December 14-17, 2016

Contents

Background

An analytical approach

Results

Conclusions

Background - CAES

Compressed air energy storage (CAES) systems

Lined or unlined rock caverns

Gas pressure: 10 to 30 MPa

Repeated compression-decompression cycles

Background – carven behavior

Stress state ≥ Yield

P0 < Stress state < Yield

∆

=2C (Tresca,

cylindrical case)

>∆ <∆

P0 P1 P2

P1 ≤ Pe

P1 P0

P1 P0 P3

P0

Pe

P2

P3

Pre

P0

Pe …

Elastic shakedown Alternative plasticity (reverse yielding)

Creeping effect is not considered

deformation deformation

The cavern may exhibit plastic strain during each loading cycle as yielding occurs in loading

and unloading processes.

Reverse yielding can lead to low-cycle fatigue of elastic-plastic materials. E.g. failure will

occur in steel specimen after 1000 cycles for a strain range of around 0.02 (Stephens, 1980).

Johansson (2003) suggested to check low-cycle fatigue of steel lining of lined rock cavern.

Low-cycle fatigue needs to be considered in the survivability analysis of rock openings

subjected to multiple attack (Balachandra, 1978).

Shakedown analysis can provide the maximum capacity of a structure against reverse

yielding under cyclic loads based on elastic-plastic theory.

Melan’s shakedown theorem:

For the CAES carven problem, if the stress state after unloading is smaller than the yield

criterion, shakedown will occur; otherwise, reverse yielding will happen.

Background

𝑓(𝜎𝑒 + 𝜎𝑟) ≤ 𝑌

Two layers of isotropic homogenous linear elastic cohesive-frictional

materials (Mohr-Coulomb criterion)

Combine both cylindrical scenario (k=1) and spherical scenario (k=2)

Reverse yielding occurs or not in both materials when fully

unloaded?

An analytical approach – problem definition

a

(a) Loading

Elastic-B

Elastic-A

Plastic-B

Plastic-A

′′

′′

(b) Unloading

b

r σr

σθ

r′′

σ′′r

σ′′θ

a′′= 0

b′′

cA

cB

Material A

Material B

Material A

Material B

Yield or not?

When loaded: (Mo et al., 2014)

When unloaded:

in which:

An analytical approach

𝛼𝐴𝜎𝑟,𝐴′′ − 𝜎𝜃,𝐴

′′ = 𝑌𝐴

𝛼𝐵𝜎𝑟,𝐵′′ − 𝜎𝜃,𝐵

′′ = 𝑌𝐵

𝛼𝑛 =1 + 𝑠𝑖𝑛𝜙𝑛

1 − 𝑠𝑖𝑛𝜙𝑛

𝑌𝑛 =2𝐶𝑛 𝑐𝑜𝑠𝜙𝑛

1 − 𝑠𝑖𝑛𝜙𝑛

𝛼𝜎𝑟 − 𝜎𝜃 = 𝑌

𝑌 =2𝐶 𝑐𝑜𝑠𝜙

1 − 𝑠𝑖𝑛𝜙

𝛼 =1 + 𝑠𝑖𝑛𝜙

1 − 𝑠𝑖𝑛𝜙

𝛼𝐴𝜎𝑟,𝐴 − 𝜎𝜃,𝐴 = 𝑌𝐴

𝛼𝐵𝜎𝑟,𝐵 − 𝜎𝜃,𝐵 = 𝑌𝐵

~𝑓 𝑎 , 𝑟

~𝑓 𝑏, 𝑟

where n = A or B

?

𝜎𝑟,𝐴′′ = 𝜎𝑟,𝐴 + ∆𝜎𝑟,𝐴

𝜎𝜃,𝐴′′ = 𝜎𝜃,𝐴 + ∆𝜎𝜃,𝐴

𝜎𝑟,𝐵′′ = 𝜎𝑟,𝐵 + ∆𝜎𝑟,𝐵

𝜎𝜃,𝐵′′ = 𝜎𝜃,𝐵 + ∆𝜎𝜃,𝐵

? 𝑓 𝑎 , 𝑟 𝑓 𝑏, 𝑟

For material A:

An analytical approach

∆𝜎𝑟,𝐴 𝑟=𝑎

= − 𝑎′′ − − 𝑎 = 𝑎 − 0

∆𝜎𝑟,𝐴 𝑟=𝑏

= − 𝑏′′ − − 𝑏 = 𝑏 − 𝑏

′′

∆𝜀𝜃,𝐴 𝑟=𝑏

= ∆𝜀𝜃,𝐵 𝑟=𝑏

∆𝜀𝜃,𝐴 𝑟=𝑏

=1

𝑀𝐴−

𝜈𝐴

1 − 𝜈𝐴 2 − 𝑘∆𝜎𝑟,𝐴

𝑟=𝑏+ 1 − 𝜈𝐴 𝑘 − 1 ∆𝜎𝜃,𝐴

𝑟=𝑏

∆𝜀𝜃,𝐵 𝑟=𝑏

=1

𝑀𝐵−

𝜈𝐵

1 − 𝜈𝐵 2 − 𝑘∆𝜎𝑟,𝐵

𝑟=𝑏+ 1 − 𝜈𝐵 𝑘 − 1 ∆𝜎𝜃,𝐵

𝑟=𝑏

∆𝜎𝑟,𝐴 = 𝐴 +𝐵

𝑟𝑘+1− 𝑝0

∆𝜎𝜃,𝐴 = 𝐴 −𝐵

𝑘𝑟𝑘+1− 𝑝0

c

c

For cavity elastic expansion/ contraction from p0:

𝑓 𝑎 , 𝑟

𝑀𝑛 =𝐸𝑛

1 − 𝜈𝑛2 2 − 𝑘

∆𝜎𝑟,𝐴 =

∆𝜎𝜃,𝐴 = 𝑓 𝑎 , 𝑟

Yield or not?

When loaded: (Mo et al., 2014)

When unloaded:

in which:

An analytical approach

𝛼𝐴𝜎𝑟,𝐴′′ − 𝜎𝜃,𝐴

′′ = 𝑌𝐴

𝛼𝐵𝜎𝑟,𝐵′′ − 𝜎𝜃,𝐵

′′ = 𝑌𝐵

𝛼𝜎𝑟 − 𝜎𝜃 = 𝑌

𝛼𝐴𝜎𝑟,𝐴 − 𝜎𝜃,𝐴 = 𝑌𝐴

𝛼𝐵𝜎𝑟,𝐵 − 𝜎𝜃,𝐵 = 𝑌𝐵

~𝑓 𝑎 , 𝑟

~𝑓 𝑏, 𝑟

?

𝜎𝑟,𝐴′′ = 𝜎𝑟,𝐴 + ∆𝜎𝑟,𝐴

𝜎𝜃,𝐴′′ = 𝜎𝜃,𝐴 + ∆𝜎𝜃,𝐴

𝜎𝑟,𝐵′′ = 𝜎𝑟,𝐵 + ∆𝜎𝑟,𝐵

𝜎𝜃,𝐵′′ = 𝜎𝜃,𝐵 + ∆𝜎𝜃,𝐵

? 𝑓 𝑎 , 𝑟 𝑓 𝑏, 𝑟

Yield or not?

When loaded: (Mo et al., 2014)

When unloaded:

in which:

An analytical approach

𝛼𝐴𝜎𝑟,𝐴′′ − 𝜎𝜃,𝐴

′′ = 𝑌𝐴

𝛼𝐵𝜎𝑟,𝐵′′ − 𝜎𝜃,𝐵

′′ = 𝑌𝐵

𝛼𝜎𝑟 − 𝜎𝜃 = 𝑌

𝛼𝐴𝜎𝑟,𝐴 − 𝜎𝜃,𝐴 = 𝑌𝐴

𝛼𝐵𝜎𝑟,𝐵 − 𝜎𝜃,𝐵 = 𝑌𝐵

~𝑓 𝑎 , 𝑟

~𝑓 𝑏, 𝑟

𝜎𝑟,𝐴′′ = 𝜎𝑟,𝐴 + ∆𝜎𝑟,𝐴

𝜎𝜃,𝐴′′ = 𝜎𝜃,𝐴 + ∆𝜎𝜃,𝐴

𝜎𝑟,𝐵′′ = 𝜎𝑟,𝐵 + ∆𝜎𝑟,𝐵

𝜎𝜃,𝐵′′ = 𝜎𝜃,𝐵 + ∆𝜎𝜃,𝐵

? 𝑓 𝑎 , 𝑟 𝑓 𝑎, 𝑟 𝑓 𝑎, 𝑟

𝑎,𝑠𝐴 r = a

𝑓 𝑏, 𝑟

For material B:

An analytical approach

∆𝜎𝑟,𝐵 𝑟=∞

= − 0 − − 0 = 0

∆𝜀𝜃,𝐴 𝑟=𝑏

= ∆𝜀𝜃,𝐵 𝑟=𝑏

∆𝜀𝜃,𝐴 𝑟=𝑏

=1

𝑀𝐴−

𝜈𝐴

1 − 𝜈𝐴 2 − 𝑘∆𝜎𝑟,𝐴

𝑟=𝑏+ 1 − 𝜈𝐴 𝑘 − 1 ∆𝜎𝜃,𝐴

𝑟=𝑏

∆𝜀𝜃,𝐵 𝑟=𝑏

=1

𝑀𝐵−

𝜈𝐵

1 − 𝜈𝐵 2 − 𝑘∆𝜎𝑟,𝐵

𝑟=𝑏+ 1 − 𝜈𝐵 𝑘 − 1 ∆𝜎𝜃,𝐵

𝑟=𝑏

∆𝜎𝑟,𝐵 = 𝐴 +𝐵

𝑟𝑘+1− 𝑝0

∆𝜎𝜃,𝐵 = 𝐴 −𝐵

𝑘𝑟𝑘+1− 𝑝0

c

c

For cavity elastic expansion/ contraction from p0:

𝑓 𝑎 , 𝑟

𝑀𝑛 =𝐸𝑛

1 − 𝜈𝑛2 2 − 𝑘

∆𝜎𝑟,𝐵 =

∆𝜎𝜃,𝐵 = 𝑓 𝑎 , 𝑟

∆𝜎𝑟,𝐵 𝑟=𝑏

= − 𝑏′′ − − 𝑏 = 𝑏 − 𝑏

′′

Yield or not?

When loaded: (Mo et al., 2014)

When unloaded:

in which:

An analytical approach

𝛼𝐴𝜎𝑟,𝐴′′ − 𝜎𝜃,𝐴

′′ = 𝑌𝐴

𝛼𝐵𝜎𝑟,𝐵′′ − 𝜎𝜃,𝐵

′′ = 𝑌𝐵

𝛼𝜎𝑟 − 𝜎𝜃 = 𝑌

𝛼𝐴𝜎𝑟,𝐴 − 𝜎𝜃,𝐴 = 𝑌𝐴

𝛼𝐵𝜎𝑟,𝐵 − 𝜎𝜃,𝐵 = 𝑌𝐵

~𝑓 𝑎 , 𝑟

~𝑓 𝑏, 𝑟

𝜎𝑟,𝐴′′ = 𝜎𝑟,𝐴 + ∆𝜎𝑟,𝐴

𝜎𝜃,𝐴′′ = 𝜎𝜃,𝐴 + ∆𝜎𝜃,𝐴

𝜎𝑟,𝐵′′ = 𝜎𝑟,𝐵 + ∆𝜎𝑟,𝐵

𝜎𝜃,𝐵′′ = 𝜎𝜃,𝐵 + ∆𝜎𝜃,𝐵

? 𝑓 𝑎 , 𝑟 𝑓 𝑎 𝑓 𝑎, 𝑟

𝑎,𝑠𝐴 r = a

𝑓 𝑏, 𝑟

Yield or not?

When loaded: (Mo et al., 2014)

When unloaded:

in which:

An analytical approach

𝛼𝐴𝜎𝑟,𝐴′′ − 𝜎𝜃,𝐴

′′ = 𝑌𝐴

𝛼𝐵𝜎𝑟,𝐵′′ − 𝜎𝜃,𝐵

′′ = 𝑌𝐵

𝛼𝜎𝑟 − 𝜎𝜃 = 𝑌

𝛼𝐴𝜎𝑟,𝐴 − 𝜎𝜃,𝐴 = 𝑌𝐴

𝛼𝐵𝜎𝑟,𝐵 − 𝜎𝜃,𝐵 = 𝑌𝐵

~𝑓 𝑎 , 𝑟

~𝑓 𝑏, 𝑟

𝜎𝑟,𝐴′′ = 𝜎𝑟,𝐴 + ∆𝜎𝑟,𝐴

𝜎𝜃,𝐴′′ = 𝜎𝜃,𝐴 + ∆𝜎𝜃,𝐴

𝜎𝑟,𝐵′′ = 𝜎𝑟,𝐵 + ∆𝜎𝑟,𝐵

𝜎𝜃,𝐵′′ = 𝜎𝜃,𝐵 + ∆𝜎𝜃,𝐵

𝑓 𝑎 , 𝑟 𝑓 𝑎 𝑓 𝑎, 𝑟

𝑎,𝑠𝐴 r = a

𝑓 𝑏, 𝑟 𝑓 𝑎, 𝑟 𝑓 𝑎 , 𝑏, 𝑟

𝑎,𝑠𝐵 r = b

Assume 𝑏′′= 0

𝑎,𝑠 = 𝑚𝑖𝑛 𝑎,𝑠𝐴, 𝑎,𝑠𝐵

Results – numerical validation (cylindrical case)

(a) Cylindrical case: r=a

(b) Cylindrical case: r=b

0 -5 -10 -15 -20 -250 -20 -40 -60 -80 -100 -12015

10

5

0

-5

-10

-15

150

100

50

0

-50

-100

-150

P (kPa)

q/2

(kP

a)

P (kPa)

q/2

(kP

a)

Initial state

Final state

Elastic loading

Plastic loading

Ela

stic

un

load

ing1

1

sinφ2

sinφ1

1sinφ2

1sinφ1

End of loading

(a) Cylindrical case:

r=a

0 -50 -100 -150

150

100

50

0

-50

-100

-150

P (kPa)

q/2

(kP

a)

End of loading

End of unloading

Cyclic load

/ = 4 𝐶A/𝐶B = 50 𝜙A = 𝜙B = 30° 𝐸A = 𝐸B = 100 𝑀

𝜈A = 𝜈B = 0.2 0 = 10𝑘

𝑷𝒂,𝒔/𝑷𝟎 = 𝟐𝟎. 𝟑𝟐

𝑎 = 20.32 0

𝑎 = 20.32 0 × 1.1

𝑝 =𝜎𝑟+ 𝜎𝜃

2; 𝑞 = 𝜎𝑟 − 𝜎𝜃

Results – numerical validation (spherical case)

P (kPa)

0 -100 -200 -300 -400 -500 -600

q (

kPa

)

1000

500

0

-500

-1000

-1500

0 -5 -10 -15 -20 -25

P (kPa)

q (

kPa

)

30

20

10

0

-10

-20

-30

-40

(a) Spherical case: r=a (b) Spherical case: r=b

1

1

2

2

sin3

sin6

2

2

sin3

sin6

Failure line for compression

Failure line for extension

1

1

1

1

sin3

sin6

1

1

sin3

sin6

Unloading

(b) Spherical case:

r=a

P (kPa)

0 -100 -200 -300 -400 -500 -600

q (

kPa

)

1000

500

0

-500

-1000

-1500

End of loading

End of unloading

Cyclic load

/ = 4 𝐶A/𝐶B = 200 𝜙A = 𝜙B = 30° 𝐸A = 𝐸B = 100 𝑀

𝜈A = 𝜈B = 0.2 0 = 10𝑘

𝑷𝒂,𝒔/𝑷𝟎 = 𝟏𝟏𝟓. 𝟎𝟓

𝑎 = 115.05 0

𝑎 = 115.05 0 × 1.1

𝑝 =𝜎𝑟+ 2𝜎𝜃

3; 𝑞 = 𝜎𝑟 − 𝜎𝜃

Results – Lined rock carven

Shakedown limit for a spherical rock cavern with steel lining

(a) (b)

φrock = 37° Erock = 10GPa

νrock = 0.3

a = 2 m

σc,rock = 10 MPa

σy,steel = 355 MPa

Esteel = 198 Gpa

νsteel = 0.3

Shakedown of rock

Shakedown of steel

Rock strength σc (MPa)

Pa

,s /

P0

5

10

15

20

25

30

35

40

0 10 20 30 40 50

Thickness of steel lining (m)

0 0.1 0.2 0.3 0.4 0.5

160

0

20

40

60

80

100

120

140

Pa

,s /

P0

• P0 = 2.5MPa (around 100m depth) • Rock friction angle 𝜙 = 37⁰ • Rock unconfined compression

strength 𝜎𝑐 = 10MPa • Rock cohesion = 𝜎𝑐 1 − 𝑠𝑖𝑛𝜙 /

2𝑐𝑜𝑠𝜙 = 𝜎𝑐/4 • Rock tensile strength = 𝜎𝑐/3 • Rock Young’s Modulus = 10GPa • Rock Poisson’s ratio = 0.3 • Steel yield strength = 355MPa • Steel Young’s Modulus = 198GPA • Steel Poisson’s ratio = 0.32 • Cavern radius = 2m

Analytical solutions for shakedown conditions of a cavity in two layers of cohesive-

frictional materials are developed.

It has been proved by numerical simulations that the proposed analytical solutions

provide admissible cyclic loading conditions for the cavity to avoid reverse yielding.

The shakedown limit pressure of underground lined rock caverns for the CAES systems,

could then be determined by using the developed solutions, which can avoid low-cycle

fatigue.

Further work will be conducted considering concrete lining and better estimation of 𝑏′′.

Concluding remarks

谢 谢

Thank You