- Email: [email protected]
Comparison of modulus equations of frozen soil based on spherical template indenter
Cold Regions Science and Technology 170 (2020) 102911
Contents lists available at ScienceDirect
Cold Regions Science and Technology journal homepage: www.elsevier.com/locate/coldregions
Comparison of modulus equations of frozen soil based on spherical template indenter
T
Can-Jie Huanga,b, Ze Zhangb, , Hui-Jun Jina,b, Xin-Yu Lia, Xin Chenb,c, Wen-Jie Fengb ⁎
a
Civil Engineering College, Harbin Institute of Technology, Harbin 150090, China State Key Laboratory of Frozen Soil Engineering, Northwest Institute of Eco-environment and Resources, China Academy of Sciences, Lanzhou 730000, China c University of Chinese Academy of Sciences, Beijing 100049, China b
ARTICLE INFO
ABSTRACT
Keywords: Frozen soil Spherical template indenter test Compression modulus Strict solution Modified solution
In order to provide reliable modulus estimating equation, based on Spherical template indenter test and linear elastic theory, it obtained two different modulus formulations, called as strict solution and modified solution relatively. Following then, a general equation group was provided to illustrate the reason causing the difference of those two solutions. Using the testing data of frozen samples, the simulation was built to examine those two solutions with result that modified solution was better consistent with simulation and experimental data. Simulation results indicated that the frictionless hypothesis used for analysis was approximately correct within specific contacting area. By comparison with other experiment results, it found that the modified solution gave more safety value while the strict solution overestimated the compression modulus of frozen soil. Therefore, the modified solution is a feasible and reliable function applied for determining compression modulus of frozen soil.
1. Introduction The elastic modulus and long-term strength of frozen soil play important role both in theoretical analysis and engineering practice. With many tests acquiring those mechanical parameters, the spherical template indenter can approach accurate experimental results with high efficiency and convenient operations. The method proposed by Tsytovich (Tsytovich, 1947), later improved by other previous Soviet scholars, has been widely used in engineering studies of permafrost (Vyalov, 1945;Vyalov and Tsytovich, 1956; Zhemochkin, 1957; Bezukhov, 1961). Obtaining settlement curve directly through this test, it needs to determine the strength and modulus of frozen soil. This problem is so complex that it requires providing accurate description of contacting mechanics behavior and constitutive model first. (Kolesnikov Yu and Morozov, 1989). Both theoretical and experimental studies have been conducted to make it become the reliable and practical research method in frozen soil test (Chertolyas, 1977;Kolesnikov Yu and Morozov, 1989). For some experimental researches having been conducted, they concentrated on two aspects: 1) Comparing the experimental results from impression of spherical template and other traditional test methods to verify the efficiency of some theoretical solutions of modulus or long-term strength of frozen soil (Roman & Veretekhina, 2004); 2) Determining the relationship between the mechanical properties of ⁎
frozen soil and influence factors, such temperature, times of freezingthawing cycles, saline content, water content and so on (Roman, 1987; Roman and Zhang, 2010;Zhang et al., 2012; Zhou et al., 2014; Zhou, 2015; Fang et al., 2018). However, it's important to confirm the effectiveness of some solutions and find the optional method. This study tries to develop a suitable and simplified modulus equation for spherical template indenter test. Due to complexity of this subject, according to the experimental phenomenon, we had made four simplified hypothesis independently before analysis. Based on those hypothesis and simplified model, two equations of modulus were deduced according to elastic theory. Following then, it also illustrated the difference between those two modulus equations and the reason causing it theoretically in Section3. Because the hypothesis may be not feasible, it requires verifying them on the basis of experiment. With help of deformation sensors, the vertical deformations of spherical template indenter had been measured. By comparing the calculating result brought by two different modulus equations, some conclusions were given in Section 4. To better understand the proposed theory, the simulations were built for further discussion in Section 5. Besides, the modulus given by those two equations had been checked by other test results as to provide suitable modulus equation. Finally, some conclusions were drawn in Section 6.
Corresponding author. E-mail address: [email protected] (Z. Zhang).
https://doi.org/10.1016/j.coldregions.2019.102911 Received 22 February 2019; Received in revised form 6 September 2019; Accepted 1 October 2019 Available online 23 October 2019 0165-232X/ © 2019 Elsevier B.V. All rights reserved.
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
Fig. 2. Simplified model of geometrical size.
Fig. 1. Outward appearance and Structure of spherical template indenter (Zhang et al., 2012).
2. Brief introduction of spherical template indenter 2.1. Structure and operation The spherical template indenter is placed inside the thermo-container whose accuracy of temperature is 0.1 °C (Fig. 1(a)).The whole structure of spherical template indenter can be divided into two main parts: 1.the supporting system, which includes foundation plate (1), upright column (2); 2.the cantilever system, which consist of rack (6), knobs (5,10,13), the loading platform (8), slide bar (4), spherical template (9) and deformation transducer (12). The loading platform transmits the static load from weight (14) to spherical template with help of slide bar. Fixed by knob (13), the deformation transducer records the deformation of spherical template indenter in every time increment step. Before test, we can adjust the height between foundation plate and spherical template according to that of specimen and use the knob (10) to lock the slide bar first as to avoid sudden loading. According to national standard built within Russia (Gost 12248–96. Substrates, 1991), the sphere plate with diameter in 22 mm is applied both in our test and simulation.
Fig. 3. Curve of equivalent cohesive strength changing with time. Where: C0 represents the instant equivalent cohesive strength,which is calculated by Eq. (1) with time ranging from 0 to 25 s;C∞stands for long-term equivalent cohesive strength; Cf means the cohesive strength according to yield criterion of Mohr-Coulomb.
the frozen soil (Tsytovich, 1985), it is always considered as discrete, discontinuous and anisotropy material. However, for the perspective of statistic average, if the size of calculating units is much larger than the size of microscopic, the local difference between the results based on this hypothesis and the reality would have no significant effect on the whole. 2) Half-space hypothesis: It has shown that the deformation of spherical template was not > 2 mm, quite smaller than the height of specimen (20 mm). Besides, the diameter of compressing marker was no > 10 mm, quite smaller than 16 percentage of the diameter of specimen (61.8 mm).Therefore, it's reasonable to regard the specimen as hemi-spacious body. 3) Quasi-static hypothesis: From testing data, it revealed that the deformation of spherical template was no > 2 mm even after 24 hous or later. So, the loading rate is so slow that every moment of the loading state could be considered as static-equilibrium state. 4) Frictionless hypothesis: During the experiment, we use plastic wrap to seal the specimen for preventing water vapor escaping. And the loading surface of specimen was cut as smooth as possible. We assume the contacting surface is frictionless. In other words, there is only normal surface force distributing within the contacting surface.
2.2. Test principle of equivalent cohesive strength The test principle of approaching cohesive strength bases on the solution of ideal visco-plasticity theory without considering hardening or strengthening (Vyalov and Tsytovich, 1956).
Ceq = K
F DS
(1)
where: Ceq stands for equivalent cohesive strength; K is the reduced factor, usually equal to 0.18 if the specimen is frozen soil; D represents the diameter of sphere; S means the axis displacement of spherical template (Fig. 2). The relationship between equivalent cohesive strength Ceq and time tends to be similar as the curve shown in Fig. 3. In following part, we will examine whether the curve of the maximum shear stress-time calculated based on modulus formulation is consistent with the curve of equivalent cohesive strength-time.
3.2. Two modulus equations based on the model of double elastic spheres
3. Theoretical analysis
Let the radius of ball 1 and ball 2 as ρ1andρ2 relatively. Before compressing, the point sited at the surface of ball 1 (or ball 2), which has distance r with the symmetric axis is A (or B) (Fig. 4). The absolute coordinates of point A (or B) is (x, y, z1) (or (x, y, z2)) relative to Cartesian coordinates system. Because the absolute value of z1 (or z2)is
3.1. Assumption 1) Continuous medium hypothesis: Due to the complex components of 2
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
Fig. 4. The model of compressing process. Explanation: the position1 stands for the state when the gap between two ball eliminates at radius r.
much smaller than ρ1 (orρ2),ignoring high order value of z1 (or z2), we get:
z1 = r2/2 1; z2 = r 2/2
By comparing Eqs. (6) & (7), γ is modified as follow expression.
= 1/(D
(2)
2
Because the value of S is too smaller than that of D, γ is considered as constant. Due to the relatively smaller size of contacting surface, we can consider both of the balls as hemi-spacious bodies. According to half space linear elastic theory (Qian, 1956), regardless of the effect of friction,ω1andω2is obtained.
where: r = x 2 + y 2 . As shown in Fig. 4(a), if both of the balls have similar scale and belong to elastic material, we think that the contacting surface is very close to circular plane. The whole deformation process can be split into two stages: 1)firstly, the radius of contacting surface becomes r when gap between two balls eliminates with point C and D moving to point C1 and point D1 relatively; 2)secondly, two balls continue their deformation until radius of contacting surface approaches the maximum value r c. Letting the total deformation of Line CD as S, the deformation at second stage of ball 1 and ball 2 as ω1 and ω2 relatively, we obtain:
S = z1 + z2 +
1
+
2
=
1 (1/ 2
1
+ 1/ 2 ) r 2 +
1
+
2
1
S=
(3)
rc =
S
=
SD
(5) (6)
where: D is the diameter of spherical template. According to Fig. 4(b), the expression of rc is obtained.
rc =
S(D
S)
2
=
2 2
1
P dx dy R1
E2
(9)
2 1
+
2 2
1 E2
PdR1d = S
r2
(10)
where:R1stands for the distance between given point A (x,y,0)and calx )2 + (y y)2 . culating point B (x′,y′,0), R1 = (x It's so difficult to solve Eq. (10) directly that we should try another way. For the perspective of physical understanding, we know that the pressure concentrates on the small area of contacting surface and decreases with radius increasing. Therefore, the function of P is proposed as follows:
+ 1/ 2) . where: = We consider ball 2 expands to be half-spacious bodies. In other word, 1/ρ2 → 0.Thus, γ and rc should be rewritten as follows: = 1/D
P dx dy ; R1
E1
E1
(4)
1
2 1
1
1
1 (1/ 1 2
= 1/2
=
where: E1and E2represent compression modulus of ball 1 and ball 2 relatively;υ1and υ2stands for the Poisson ratio of ball 1 and ball 2 relatively; P is the vertical component of surface force distributing within the contacting surface; Ω symbolizes the distributing area of surface force. Substituting Eq. (9) into Eq. (3), and expressing it in polar coordinate system with respect to point A (Fig. 5), we obtain:
where:ω1 (or ω2) is the deformation of vertical direction of C1 (or D1). At the second stage, ω1 = ω2 = 0. Therefore, the expression of S is obtained.
r 2c
(8)
S)
P=
P0 2 rc rc
R2
(11)
where: P stands for vertical component of surface force at point C (Fig. 5); R symbolizes the radial distance between point C and central
(7) 3
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
Fig. 5. Diagram of integral area and polar coordinate system, Vertical cross section when?? = const.
point O;P0 is the vertical component of surface force at central point O; rc represents the radius of contacting surface. From Fig. 5(a), we know the radius of half-circular area O1AFD is equal to rc2 r 2sin2 .Therefore, it has: rcos + r 2c r 2 sin2
PdR1 =
r 2c r 2 sin2
rcos
P0 (r 2c rc 2
r 2 sin2 )
E2 =
P0 PdR1d = (2r 2c 4 rc
P0 =
(12)
2 1
+
E1
2 2
1
2
P0 (2r 2c 4 rc
E2
r 2) = S
r2
E2 = (14)
2 1
1
+
E1 2 1
1
S=
+
E1
2 2
1
2
E2
4 2 2
1
2
E2
2
P0 rc
2 P0 r 2c 3
(17)
Combining Eqs. (4), (16), (17),we get:
P0 =
rc =
S=
1/3
2
6F 5( 1 + 3 F(
1
+
2)
2)
1/3
(19)
4 2
9 F2 ( 16
+
1
2)
2
1/3
S0 = (
1
+
2 )|r = 0
2 2
1 0.5 2 2 S1.5 0 D 3F
r 2c
2 1
1 E1
R2
(D
E2 =
3F(1 4S1.5 0 (D
(25)
(26) (27) (28)
S0 )S0
P =
P0 rc
P0 =
3 F DS0
1
=S
(24)
8
rc =
E2 =
2 2
1/3
2)
3 F 2 (D S0 )S0
where: 1 + 2 = E + E ; = D S . 1 2 When r = 0, the elastic deformation of two balls at central point is obtained.
1
+
P0 =
rc =
2 1
1
P0 rc
(20)
1
3 F(
(23)
2 2)
S0 )0.5
(29)
According to Eqs. (11), (23), (24), (25),we obtain:
(18)
2
1/3
24F 2 2 1 + 2)
P= (16)
By analyzing the vertical direction of the equilibrium function of ball 1, it has:
F=
(22)
where: = Because the modulus of spherical template (it's usually made by steel) is much larger than the modulus of frozen soil, we let E1 → ∞. According to Eqs. (22)and (25), we will have safety value of E2 (or E2′). Therefore, based on Eqs. (11), (18), (19), (22), it has:
(15)
P0 rc
E1
1 . D
By comparing two sides of Eq. (14),we get:
=
5(
(13)
Substituting Eq. (13) into Eq. (10), we obtain:
1
2 1
1
However, if combining Eqs. (15)–(17), we can obtain:
rc =
r 2)
S 0 )0.5
3F
And then, the integral formulation is obtained. 2
2 2
1 4S1.5 0 (D
(r c )2
R2
DS0 /2 2 3 2 F(1 2) 1.5 0.5 4S0 D
(30) (31) (32) (33)
Based on the analysis above, Eqs. (26)–(29) are defined as modified solution while Eqs. (30)–(33) are defined as strict solution. The special forms of strict solution had been proposed by Bezukhov N. I. (Bezukhov, 1961). Roman.L.T. et al. tried to modify it and gave a
(21)
where: S0is the elastic deformation of two balls at central point. Substituting Eq. (8), Eq. (21) into Eq. (20), we get: 4
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
modified solution, whose analytical results are consistent with that of tri-axial test. (Roman &Veretekhina, 2004).However, Roman.L.T. et al. didn't gave reason why this special forms of solution is practical. By comparing Eqs. (26)–(29) with Eqs. (30)–(33), if ignoring value of S02,we get:
(42), letting r = rc, we have:
i Ai (2i + 1)!
N
F = 4 p0 r 2c
i=1
(45)
At contacting surface, ω1 + ω2 = 0, substitute it into Eq. (37). And we can us-e polynomial function to express S approximately as follows:
P0/P 0 = 1/2
(34)
rc/r c =
(35)
S
(36)
where: SNstands for vertical deformation of sphere plate when it's calculated by N order polynomial function;rc is the radius of contacting surface. Eqs. (43)–(46) build (N + 2) orders function groups between variable SN, rc,p0,A2~ANF,ρ1,ρ2, θ1, θ2 (it already exists A1 = 1). That means if the value of arbitrary five numbers of variables have been known, we can solve other (N + 2) number of variables theoretically. Usually, if F,ρ1,ρ2, θ1, θ2 are provided, the solution process of mechanical behaviors (S, rc,p0,A2~AN) is defined as forward analysis. In contrast, the solution process of mechanical property (θ1, θ2and so on) is defined as back analysis when F,ρ1,ρ2, S are given. It needs to provide the estimated value of polynomial truncation error of S calculated by function group above. Based on Taylor expansion, the expression of S can be written as follows:
2
E2 /E2 = 1/ 2
N n=1
According to Eqs. (34)–(36), it shows that the modified solution can make less pressure concentrating at small area but widen the distributing area of pressure by comparing with strict solution. Why it comes to different solutions based on same theory (Eqs. (26)–(33))? This difference must be illustrated clearly for the perspective of theory. 3.3. A basic solution for calculating arbitrary order of pressure and modulus For one reason proposed, both the actual size of the spherical template and specimen are not infinity. For another, the geometrical equation (Eq. (7)) is an approximate description of contacting surface at elastic state, which contradicts with Eq. (15). In fact, based on method of Taylor expansion, Eq. (3) should be written as follows: N 1
+
2
=S+ n=1
1 22n
(
1) n
1
(n
(2n 1)! [( 1)1 1) !n!
2n
+ ( 2 )1
2n]
n=1
2
=(
1
+
2)
PdR1d
R=
1 2 rc rc
r 2c
R1 = rcos + sin
RN <
f
1 2 rc rc
R2 = n=1
An (2n 1)!
cos rc
i=n+1
(i
Ai n) ! (i 1) !22i
1
=
r 2c
r 2sin2
n! 2p ( 0 1
+
2)
1
2n 1
+
rc
+
2 )rc
N i=1
i! (i
Ai 1) !22i
1
=S
S=
F=
2 2 r c p0 3
2
rc P0 (
1
+
rc
(2N + 1)
+
1
(2N + 1)
rc 1
2)
r 2n c
(46)
(47)
+ ( 2)
(2N + 1) ]
2(N + 1)
(2N + 1)
rc
(49)
(51) (52)
S1 <
3 8
3
rc
+
1
rc
3
1
rc
(53)
+ 2 ). where: = Eqs. (50)–(52) are as same as Eqs. (4), (16), (17) (modified solution). For this point of views, Eqs. (4), (16), (17) are the first order solution of general algorithm(Eqs. (43)–(46)). Eq. (53) means the modified solution has polynomial truncation error that makes the difference between strict solution and it. When N = 2, we obtain: 1 ( 1 2 1
2n 1
2
1) (43)
2p ( 0 1
2n]
r 2n c
(50) 2
S1
R1 = S
(42)
rc
(n = 1, 2, 3,…, N
(2N + 1)! + 1) !N!
22N + 1 (N
2n 1
where:A1 = 1;Anis the parameter without unit, n = 2,3, …,N; Substituting Eqs. (39), (42) into Eq. (38), by comparing with Eq. (37),we get: N
( 1) N (2N + 1)! [( 1) 22N + 1 (N + 1) !N!
SN =
S1 = r 2c
(41)
where: randrcare constant for given point A; θandβare variable for integral point B (Fig. 5). Based on Eqs. (37), it sets r2n as the single variable. By considering Eqs. (38)–(41), we propose: N
+ ( 2 )1
2n]
where: ∣ RN∣ means the absolute value of RN. Eq. (48) demonstrates solution of N order function group(Eqs. (43)–(46)) accompanies with N order of polynomial truncation error. Eq. (49) can be applied for estimating the upper limit of N order of polynomial truncation error based on function group (Eqs. (43)–(46)). For example, letting N = 1 and A1 = 1, we can obtain:
(40)
r 2sin2
+ ( 2 )1
(48)
(39)
r 2sin2 + (R1 rcos ) 2
2n
2n
where: ξ is the radius based on Taylor expansion method, 0 < ξ < rc. Therefore, it means:
(38)
R2
( 1) n + 1 (2n 1)! [( 1 )1 (n 1) !n!
22n 1
RN = S
Like similar reasons for providing Eq. (11),according to Fig. 5, we propose:
P = p0 f(R) = p0 f
( 1) n + 1 (2n 1)! [( 1)1 (n 1) !n!
Let N order Polynomial truncation error of S as RN, we can get:
r 2n
where: N stands for the order of Equation; (M)! represents factorial function of M, (M) ! = 1 × 2 × … × (M − 1) × M, M = n-1,n,2n-1. It also has:
+
22n 1
1
S=
(37)
1
1
SN =
S2 = r 2c
(44)
S2
By considering the vertical equilibrium function of ball 1 and Eq. 5
S=
3 4 rc ( 8 2p 0
(
1
1
1
3
+
+ 2)
2
3
)
rc
(54)
1 A + 2 2 16
(55)
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
A2 =
16 rc +
2p ( 0 1
F = p0 r 2c
|R2| = S
(56)
2)
2 A + 2 3 15 S2 <
5 16
(57)
rc
5
+
1
rc
5
rc
1
(58)
Combining Eqs. (54)–(56), we have contradict results because of Polynomial truncation error of vertical deformation. In conclusion, existence of Polynomial truncation error of vertical deformation results in difference between strict solution and modified solution. Besides, it may contribute to contradict solution result of N order function group. 4. Test condition and result 4.1. Sample preparation Fig. 6. Changing curve of vertical deformation versus time.
When the temperature is so low, the frozen soil would tend to be as similar as continuous and isotropic material, which meets the basic hypothesis of elastic theory. Thus, we select −20 °C as temperature environment. The main physical properties of Kaolin soil are listed in Table 1. It has definition as low liquid limit silts according to standard for engineering classification of soils made in China (GB/T 50145–2007, 2008). The remolded sample was kept in cylinder with diameter in 61.8 mm and height in 20 mm. Before experiment, the specimen had been saturated with sealed mold under vacuum to allow water equilibration. Kept sealed with mold, the specimens were put into the freezing cabinet and frozen at negative temperature at −20 °C for 24 h.
4%,which could be ignored. Letting ν2=0.35, based on Eqs. (1), (29), (33)) as well as the curve of vertical deformation tested (Fig. 6), the changing curves of compression modulus and equivalent cohesive strength with time have been obtained (Fig. 7). It indicates that the modulus decreases sharply at the beginning time from 5S to 160S while remains decreasing with time slowly after 160S (Fig. 7). In addition, both of the modulus and equivalent cohesive strength shows same changing regularity with time. 5. Discussion 5.1. Analyzing idea and the simulation model
4.2. Experimental process
Due to the complexity of this problem, we resort to simulation to check which modulus equation (Eq. (29) or Eq. (33)) is better. The structural diagram of research is shown in Fig. 8. Yield Criterion of DRUCKER-PRAGER (shortly written as D-P)has been proved practical in geotechnical engineering simulation. The linear kind of Yield Criterion of DRUCKER-PRAGER writes as follows:
Both the spherical template indenter and thermo-container were kept at −20 °C in order to insure steady temperature environment during testing. Because of the discrete problem, it requires testing at least six different points of one sample to determine average value curve of deformation. Following then, different loading weight were applied and compared to determine the suitable load which made deformation satisfy with requirement (Eq. (59)). By trials, we finally determined 15 kg weight as static load and continued the experiment for > 24 h. Thus, F = 15 × 9.81 = 147.15(N).
0.005D
S15 min
f=t
(60)
d=0
where: t stands for the parameter of deviator stress,t = q/g; q symbo1 2 2 2 lizes the octahedral stress, q = 2 ( 1 2) + ( 2 3) + ( 3 1) ; g is the shape function of f (Dong et al., 2013),g = 2K/ [1 + K + (1 − K) ∙ (r/q)3]; r has the meaning of the radius of yield
(59)
0.05D
ptan
( ) 27
1/3
surface within deviator stress plane, r = 2 J3 ; J3 is the third invariable of deviator stress; K represents the ratio of tensile yield strength with respect to the compressing strength according to tri-axial test, K = (3 − sinφ)/(3 + sinφ)，0.778 ≤ K ≤ 1 (Liu and Cheng, 2010)；P represents the hydrostatic pressure, P = (σ1 + σ2 + σ3)/3; d means the yield strength. According to Wang (2015), if determining the yield strength d through pure shearing test, d can be expressed as follows (Eq. (61)):
where:S15min stands for the deformation at fifteenth minute of the beginning; D represents the diameter of spherical template, D = 22mm. 4.3. Experimental result After the test lasting > 24 h, the average curve of vertical deformation was obtained as follows (Fig. 6). Though the Poisson ratio varies with temperature and water content as many studies having proved (Zhu and Zhang, 1982; Wang et al., 2008). According to Roman (2016), for frozen cohesive soils, the Poisson ratio is range from 0.35 to 0.4.So, if considering this, the relative error of modulus estimated by Eq. (29) or Eq. (33) is in range of
d=
3 2
1+
1 K
f
(61)
where:τf is the shear strength obtained from pure shearing test.
Table 1 Main physical properties of soil sample. Soil
Dry density (g/cm3)
Saturated density (g/cm3)
Saturated water content (%)
Liquid limit (%)
Plastic limit (%)
Plasticity index
Kaolin
1.39
1.9
36.29
43.22
32.43
10.79
6
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
Fig. 7. Changing in compression modulus and equivalent cohesive strength of frozen Kaolin soil versus time.
the law of Hooke, the stress tensor is expressed as follows: ij
=
+ 2µ
e ij
(63) E ; 2((1 + ))
E , 2 )(1 + )
µ= where: λ, μstands for Lame coefficient, = (1 ν andE symbolizes the Poisson ratio and modulus of specimen relatively;σijrepresents the component of stress tensor;εije means the component of elastic strain tensor; θ is the elastic volume strain, θ = εiie. Becauseτφr = τφz = 0,σθ is the principal stress. By considering of hypothesis of complete plastic principle, it regards σθ as second principal stress σ2.We get: 1/ 3
2
Fig. 8. The guideline of this study.
+ 2
z
r
±
2
z
2
+
2 rz
(64-1) (64-2)
= max
{
1
2
2
,
2
3
2
,
1
3
2
}
(64-3)
where:σr,σθ,σz stands for theradial, circumferential and vertical component of stress tensor relatively (Fig. 10a);τrzandτmaxrepresent the shear stress and it's maximum value relatively (Fig. 10c);σ1~σ3 are the principal stress,σ1 > σ2 > σ3. If σ1 > σθ > σ3, the hypothesis of complete plastic principle is correct. Then, components of surface force within the meridian contacting surface are obtained.
(62)
3 Ceq = 0
r
=
max
According to the test principle of spherical template test, letting τf = Ceq and φ = β = 0 (0 ≤ β ≤ φ), we obtain K = 1and d = 3 Ceq .Therefore, Eq. (60) is rewritten as follows:
f=q
=
As for elastic part of deformation, we choose linear elastic constitutive model. Because the beginning contacting point is close to the symmetrical center of specimen, we regard this subject as axial symmetrical contacting problem. The size and mechanical parameters of spherical template and soil sample are given in Table 2. Besides, the normal contacting property is considered as hard one while the tangential contacting property is regarded as frictionless. The deformation boundary of soil sample is bounded during the process of loading (Fig. 9.a). We adopt the deformation recorded at every moment as load. In order to have high calculating accuracy, there are high density of elements within the core cylinder with height in 10 mm and radius in 6 mm (Fig. 9.b). If having obtained the elastic strain tensor of yield points, based on
n
=
n
=(
2 r sin
z
+ r)/2
2 zcos
rz sin 2
sin 2 +
rz
(65-1) (65-2)
cos 2
Or the component of surface force within the meridian contacting surface can be expressed as:
Pr =
sin
r
+ cos
rz
(66-1)
Pz =
sin
rz
+ cos
z
(66-2)
where: the negative calculating value of Eqs. (65)–(66) indicates the direction of force is opposite to that having been made (Fig. 10.a~c);α characterizes the angle between the normal direction of contacting
Table 2 The size and physical parameter of the model. Object Sphere plate Specimen
Modulus (Mpa) 5
2 × 10 E2(orE2′)
Poisson ratio
Yield strength d (Mpa)
Plastic strain
Diameter (mm)
Height (mm)
0.3 0.35
335
0.00223 0
22 61.8
– 20
3 Ceq
7
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
Fig. 9. Constraint condition and element dividing in simulation.
surface and vertical direction;σn and τn represents the normal and tangential component of surface force distributing within the meridian contacting surface relatively (Fig. 10.b);PrandPz stand for the radial and vertical components of surface force within the meridian contacting
surface relatively. In following part, we select possible yield points (points labeled 1~5) along the contacting surface for studying (Fig. 10.d). Point2~4 represent some points, whose connecting line with point O divides the
Fig. 10. Components of stress or surface force within meridian interface. Explanation: Meridian interface is the plane cross the symmetrical axis. 8
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
Fig. 11. Principal stresses at possible yield points according to E2. Explanation: σ1~σ3are calculated according to Eq. (64).
total angle ϕ into one quarter, half and three quarter of it relatively.
maximum shear stress given by different kind of modulus are so close with each other for all points, which indicates the value of compression modulus has less influence on the value of maximum shear stress. According to Fig. 13a, b, the vertical component of surface force obtained from simulation remains more than half value of that given by strict solution for each moment. The distributing area given by simulation is much larger than the one given by strict solution. For this perspective, it concludes that the modulus given by strict solution (E2′) may overestimate the real modulus value of specimen as it originates from more pressure distributing within smaller area of contacting surface. However, if simulating according to the value of E2 (modified solution), we obtained the curve of vertical component of surface force, which's quiet similar to that given by modified solution not only for the value but also for the shape of curve (especially after 160S, Fig. 13c, d).Therefore, E2 may reflect the plastic property of specimen to a certain degree. According to Fig. 13a, c, it shows that vertical component of surface force obtains the maximum value at point 4 or nearby position. Besides,
5.2. Comparing the result brought by two modulus equations Firstly, it should testify the hypothesis of complete plastic principle. Due to similarity of simulating models (by E2 or E2′), it here only provides the calculating results according to E2 (modified solution, Fig. 11). It indicates that σθ is the second principal stress for all the possible yield points (Fig. 11).Thus, complete plastic principle is correct for plastic area. And we can make further analysis based on Eqs. (65)–(66). According to Eq. (64-3) and the result of simulation, we obtain the curve of maximum shear stress when inputting E2′(solved by strict solution) into simulation model (Fig. 12). From Fig. 12, we know that except point 1, all of the other points distributing along the contacting surface have already been in yield state. The maximum value of maximum shear stress is obtained no in point 5 but in point 4 (Fig. 12, Fig. 10d). Besides, it also reveals that the
Fig. 12. Curve of the maximum shear stress on possible yield points 9
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
Fig. 13. Comparison of vertical component of surface force.
it almost remains invariable with radial distance increasing before point 4, which's different from those of elastic solution. This result is consistent with solution of ideal visco-plasticity theory (Vyalov and Tsytovich, 1956), which proposed that the surface force remain same with radius increasing. According to Fig. 14, it indicates that the value of tangential component of surface force increases with radial distance increasing for each moment (especially after 5S). In addition, all of the curves reach the maximum value at point 5 while approach almost zero value at point 1, which meets the mechanic requirement (The shear stress approaches zero value at axial position in symmetric mechanic subject.). In the process of deducting the modulus equation, we make no consideration of shear effect along the surface. However, according to Fig. 14, the shear stress would have sharp mutation at contacting surface if considering spherical template as frictionless. Therefore, it should give further examination of frictionless hypothesis.
5.3. Examination of frictionless hypothesis ° n 2 2 Let f = n + n , tan = n andδ = α + β − 90 . Where: f stands for the joint force within meridian interface;β represents the angle betweenf and τn;δ symbolizes the angle between f and vertical positive direction, we let counterclockwise as positive rotation direction, which means the negative value of δ (or β) stands for clockwise rotation (Fig. 15). Following then, we obtain the curve of β and δ of different moment (Figs. 16, 17). According to Fig. 16, β decreases linearly with radial distance increasing before approaching three quarter of the radius of contacting surface whileβ has sharp reduction when reaching the edge. Besides, it also indicatesβ is range from 85°to 90°when radial distance increases to three quarter of radius of the contacting surface. So, it proves that the frictionless hypothesis is suitable only when the radial distance is less
10
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
Fig. 14. Comparison of tangential component of surface force.
than three quarter of the radius of contacting surface. According to both of the simulation based on E2 and E2′ (Fig. 17), δ increases with radial distance increasing to three quarter of radius of contacting surface while decreases enormously to negative value nearby the edge. Thus, it means that the direction of joint force within meridian interface can't remain at vertical direction. All of the research been done in this part tries to find better modulus equation by comparing simulation results with theoretical results. In fact, simulation is one kind of methods to examine the solutions. The important examination is some results coming from relevant tests directly. Therefore, we make reference from some similar experimental results to check the previous theory.
Fig. 15. The joint force f and it's direction within meridian interface
Fig. 16. Comparison of the angle between joint force and it's tangential component (β) 11
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
Fig. 17. Comparison of the angle between joint force and vertical positive direction (δ).
5.4. Examination from other experimental results
underestimates the value of modulus but the strict solution overestimates it. The main reason resulting in the difference of compression modulus given by spherical template indenter and other experiments is that the frozen soil is viscosity-elasticity-plastic materials, whose mechanic property can't be totally described by elastic theory. In fact, only knowing about vertical deformation of spherical template, we should find actual mechanics properties of specimens. This solution process has been defined as “back-analysis”. Problem included in “back-analysis” has different solutions depending on which basic theory or model selected (Bezukhov, 1961; Kolesnikov Yu and Morozov, 1989). However, based on spherical template indenter test and examination from simulating and testing, this paper provides a basic and efficient method to estimate the modulus of frozen soil with time changing. Besides, the solution method provided could be used in other engineering field like Brinell hardness test and so on.
Many studies have proved that the modulus of frozen soil increases almost linearly with temperature reducing (Zhu and Zhang, 1982). According to the previous testing results of modulus of frozen soil (Zhu and Zhang, 1982; Li et al., 2007; Yang et al., 2010), the curve of modulus changing with temperature is obtained (Fig. 18). Though they have not provided direct testing results of the modulus at temperature of −20 °C, we can use different experimental data to estimate it. The estimating value of modulus of frozen Kaolin soil at −20 °C, which has 36.29% saturated water content, is close to 335 Mpa. It usually adopts the deformation at time ranging from 5 to 25 s to calculate initial equivalent cohesive strength. Thus, we select the modulus of 5 s as the initial modulus. The strict solution of modulus equation provides estimating value of 420 Mpa while the modified solution approaches 297 Mpa, which shows that the modified solution
6. Conclusion According to hemi-spacious elastic theory and theoretical model of spherical template indenter, two group equations, noted as strict solution and modified solution, were derived. In order to find reasons causing different solutions, it proposed a general algorithm to calculate arbitrary order accuracy of compression modulus, surface force, radius of contacting surface relatively. Based on the tests, both the simulation built of compression process and experimental data were used to compare the calculated results of strict solution and modified solution. Therefore, the following conclusion could be acquired. (1) The modified solution is the first order approximate solution of the general algorithm. The strict solution overestimates the value of surface force but underestimate radius of contacting surface, which contributes to high estimating value of compression modulus as 2 times big as that given by modified solution. (2) The simulation indicates that not only the surface force, but also the maximum shear stress based on modified solution are closer agreement with the testing and simulation results. In contrast, the strict solution overestimates the value of compression modulus and surface force because of narrowing the pressure distributing area. (3) The simulation indicates that the frictionless hypothesis is
Fig. 18. The experimental data of modulus of frozen cohesive soils Source: Curve 1 is based on the experimental data obtained by Yang et al. (2010). Curve 2 Originates from the testing data got by Li et al. (2007).Curve 3 to 5 are according to the results provided by Zhu and Zhang (1982). 12
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
approximately correct when the radial distance is less than three quarter of radius of contacting surface. (4) The modified solution underestimates the compression modulus of frozen Kaolin soil while strict solution overestimates it according to other experimental data.
Li, S.Y., Lai, Y.M., Zhang, M.Y., et al., 2007. Study on distribution laws of compression modulus and strength of warm frozen soil. Chin. J. Rock Mech. Eng. 26 (S2), 4299–4305. Liu, S.T., Cheng, P.F., 2010. Based on ABAQUS for numerical analysis of soil constitutive model. Low Temp. Technol. 2, 90–92. Qian, W.C., 1956. Elastic Mechanics Theory. in Chinese. Chinese Science Press, Beijing, pp. 316–326. Roman, L.T., 1987. Frozen Peaty Soils As Beds for Structures. In Russian. Nauka, Novosibirsk. Roman, L.T., 2016. In: Zhang, C.Q., Zhang, Z. (Eds.), Frozen Soil Mechanics. Science Press, Beijing, China, pp. 150–159 trans. Roman, L.T., Veretekhina, É.G., 2004. Determination of deformation characteristics of permafrost from impression of spherical plate. Soil Mechanics and Foundation Engineering 41 (2), 60–64. Roman, L.T., Zhang, Z., 2010. Effect of cycles of freezing and thawing on the physical and mechanical properties of moraine loam. Soil Mech. Found. Eng. 47 (3), 96–101. Tsytovich, N.A., 1947. Determination of Cohesive Forces in Permafrost by the SphericalProbe Method. In Russian. FondyInst. Merzlotoved.Akad.Nauk SSSR. Tsytovich, N.A., 1985. In: Zhang, C.Q., Zhu, Y.L. (Eds.), The Mechanics of Frozen Ground. Chinese Science Press, Beijing Translated by. Vyalov, S.S., 1945. Analytical Formulation of Mechanics Property of Quasi-ViscosityElastic Material. in Russian. vol. 1(2). News of the USSR academy of sciences, The Technical Science, pp. 25–30. Vyalov, S.S., Tsytovich, N.A., 1956. Evaluation of the Bearing Capacity of Cohesive Soils by the Magnitude of Impression of a Spherical Plate. In Russian. vol. 3. Dokl.Akad.Nauk SSSR, pp. 1193–1196 6. Wang, P.F., 2015. The Study of Quasi-static Deformation and Failure of PBX Based on Drucker-Prager Criterion. Mianyang, Master Degree Paper of Institute of Systems Engineering. China Academy of Engineering Physics, pp. 15–16. Wang, Z.Z., Mu, S.Y., Nu, Y.H., et al., 2008. Predictions of elastic constants and strength of transverse isotropic frozen soil. Rock Soil Mech. 29 (S1), 475–480. Yang, Y.G., Lai, Y.M., Pu, Y.B., et al., 2010. Experimental analysis of damage of ice-rich frozen silt under uniaxial compression. Chin. J. Rock Soil Mech. 33 (10), 3063–3068. Zhang, Z., Ma, W., Zhang, Z.Q., et al., 2012. Application of spherical template indenter to long-term strength tests for frozen soil. Rock Soil Mech. 33 (11), 3516–3520. Zhemochkin, B.N., 1957. Theory of Elasticity. In Russian. Gosstroiizdat, Moscow. Zhou, H., 2015. The Change Rule of Frozen Loess Long-Term Strength Under Freeze-Thaw Cycle. Master Degree Paper of Lanzhou University, LanZhou City of China, pp. 29–38. Zhou, H., Zhang, Y.C., Zhang, Z., et al., 2014. Changing rule of long-term strength of frozen loess cohesion under impact of freeze-thaw cycle. Rock Soil Mech. 35 (8), 2241–2246 (2254). Zhu, Y.L., Zhang, J.Y., 1982. The elastic displacement and compressive displacement of frozen soil. J. Glaciol. Geocryol. 4 (3), 29–39.
Declaration of Competing Interest None. Acknowledgment This research was supported by Bureau of International Cooperation, Chinese Academy of Sciences (No.131B62KYSB20170012), National Natural Science Foundation of China (41771078, 41871061) and the Funding of the State key laboratory Frozen Soil Engineering (SKLFSE-ZT-19). References Bezukhov, N.I., 1961. Fundamentals of Elasticity, Plasticity, and Creep Theory. [In Russian]VysshayaShkola, Moscow. Chertolyas, N.F., 1977. Evaluation of the Strength and Deformation Properties of Cohesive Soils by the Impression of a Spherical Plate. ([In Russian]Author's abstract of dissertation for Candidate of Technical Sciences, Moscow). Dong, L.W., Zhang, M.Y., Xie, Y.Y., et al., 2013. Study on the parameters of soil models yield criterion in ABAQUS software. J. Qingdao Technol. Univ. 34 (1), 48–50. Fang, J.H., Chen, X., Xu, A.H., et al., 2018. The experimental study of the influence of freeze-thaw cycles on the physical and mechanic properties of Qinghai-Tibet red clay. J. Glaciol. Geocryol. 40 (1), 62–69. GB/T 50145-2007, 2008. Standard for Engineering Classification of Soil. In Chinese. Chinese Planning Press, Beijing, pp. 4–7. Gost 12248 - 96. Substrates, 1991. The Laboratory Method of Characterizing the Strength and Deformation Property of Frozen Soil. In Russian. The Micro Structure in Russia, Moscow. Kolesnikov Yu, V., Morozov, E.M., 1989. Contact Fracture Mechanics. in Russian. Nauka, Moscow.
13
Contents lists available at ScienceDirect
Cold Regions Science and Technology journal homepage: www.elsevier.com/locate/coldregions
Comparison of modulus equations of frozen soil based on spherical template indenter
T
Can-Jie Huanga,b, Ze Zhangb, , Hui-Jun Jina,b, Xin-Yu Lia, Xin Chenb,c, Wen-Jie Fengb ⁎
a
Civil Engineering College, Harbin Institute of Technology, Harbin 150090, China State Key Laboratory of Frozen Soil Engineering, Northwest Institute of Eco-environment and Resources, China Academy of Sciences, Lanzhou 730000, China c University of Chinese Academy of Sciences, Beijing 100049, China b
ARTICLE INFO
ABSTRACT
Keywords: Frozen soil Spherical template indenter test Compression modulus Strict solution Modified solution
In order to provide reliable modulus estimating equation, based on Spherical template indenter test and linear elastic theory, it obtained two different modulus formulations, called as strict solution and modified solution relatively. Following then, a general equation group was provided to illustrate the reason causing the difference of those two solutions. Using the testing data of frozen samples, the simulation was built to examine those two solutions with result that modified solution was better consistent with simulation and experimental data. Simulation results indicated that the frictionless hypothesis used for analysis was approximately correct within specific contacting area. By comparison with other experiment results, it found that the modified solution gave more safety value while the strict solution overestimated the compression modulus of frozen soil. Therefore, the modified solution is a feasible and reliable function applied for determining compression modulus of frozen soil.
1. Introduction The elastic modulus and long-term strength of frozen soil play important role both in theoretical analysis and engineering practice. With many tests acquiring those mechanical parameters, the spherical template indenter can approach accurate experimental results with high efficiency and convenient operations. The method proposed by Tsytovich (Tsytovich, 1947), later improved by other previous Soviet scholars, has been widely used in engineering studies of permafrost (Vyalov, 1945;Vyalov and Tsytovich, 1956; Zhemochkin, 1957; Bezukhov, 1961). Obtaining settlement curve directly through this test, it needs to determine the strength and modulus of frozen soil. This problem is so complex that it requires providing accurate description of contacting mechanics behavior and constitutive model first. (Kolesnikov Yu and Morozov, 1989). Both theoretical and experimental studies have been conducted to make it become the reliable and practical research method in frozen soil test (Chertolyas, 1977;Kolesnikov Yu and Morozov, 1989). For some experimental researches having been conducted, they concentrated on two aspects: 1) Comparing the experimental results from impression of spherical template and other traditional test methods to verify the efficiency of some theoretical solutions of modulus or long-term strength of frozen soil (Roman & Veretekhina, 2004); 2) Determining the relationship between the mechanical properties of ⁎
frozen soil and influence factors, such temperature, times of freezingthawing cycles, saline content, water content and so on (Roman, 1987; Roman and Zhang, 2010;Zhang et al., 2012; Zhou et al., 2014; Zhou, 2015; Fang et al., 2018). However, it's important to confirm the effectiveness of some solutions and find the optional method. This study tries to develop a suitable and simplified modulus equation for spherical template indenter test. Due to complexity of this subject, according to the experimental phenomenon, we had made four simplified hypothesis independently before analysis. Based on those hypothesis and simplified model, two equations of modulus were deduced according to elastic theory. Following then, it also illustrated the difference between those two modulus equations and the reason causing it theoretically in Section3. Because the hypothesis may be not feasible, it requires verifying them on the basis of experiment. With help of deformation sensors, the vertical deformations of spherical template indenter had been measured. By comparing the calculating result brought by two different modulus equations, some conclusions were given in Section 4. To better understand the proposed theory, the simulations were built for further discussion in Section 5. Besides, the modulus given by those two equations had been checked by other test results as to provide suitable modulus equation. Finally, some conclusions were drawn in Section 6.
Corresponding author. E-mail address: [email protected] (Z. Zhang).
https://doi.org/10.1016/j.coldregions.2019.102911 Received 22 February 2019; Received in revised form 6 September 2019; Accepted 1 October 2019 Available online 23 October 2019 0165-232X/ © 2019 Elsevier B.V. All rights reserved.
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
Fig. 2. Simplified model of geometrical size.
Fig. 1. Outward appearance and Structure of spherical template indenter (Zhang et al., 2012).
2. Brief introduction of spherical template indenter 2.1. Structure and operation The spherical template indenter is placed inside the thermo-container whose accuracy of temperature is 0.1 °C (Fig. 1(a)).The whole structure of spherical template indenter can be divided into two main parts: 1.the supporting system, which includes foundation plate (1), upright column (2); 2.the cantilever system, which consist of rack (6), knobs (5,10,13), the loading platform (8), slide bar (4), spherical template (9) and deformation transducer (12). The loading platform transmits the static load from weight (14) to spherical template with help of slide bar. Fixed by knob (13), the deformation transducer records the deformation of spherical template indenter in every time increment step. Before test, we can adjust the height between foundation plate and spherical template according to that of specimen and use the knob (10) to lock the slide bar first as to avoid sudden loading. According to national standard built within Russia (Gost 12248–96. Substrates, 1991), the sphere plate with diameter in 22 mm is applied both in our test and simulation.
Fig. 3. Curve of equivalent cohesive strength changing with time. Where: C0 represents the instant equivalent cohesive strength,which is calculated by Eq. (1) with time ranging from 0 to 25 s;C∞stands for long-term equivalent cohesive strength; Cf means the cohesive strength according to yield criterion of Mohr-Coulomb.
the frozen soil (Tsytovich, 1985), it is always considered as discrete, discontinuous and anisotropy material. However, for the perspective of statistic average, if the size of calculating units is much larger than the size of microscopic, the local difference between the results based on this hypothesis and the reality would have no significant effect on the whole. 2) Half-space hypothesis: It has shown that the deformation of spherical template was not > 2 mm, quite smaller than the height of specimen (20 mm). Besides, the diameter of compressing marker was no > 10 mm, quite smaller than 16 percentage of the diameter of specimen (61.8 mm).Therefore, it's reasonable to regard the specimen as hemi-spacious body. 3) Quasi-static hypothesis: From testing data, it revealed that the deformation of spherical template was no > 2 mm even after 24 hous or later. So, the loading rate is so slow that every moment of the loading state could be considered as static-equilibrium state. 4) Frictionless hypothesis: During the experiment, we use plastic wrap to seal the specimen for preventing water vapor escaping. And the loading surface of specimen was cut as smooth as possible. We assume the contacting surface is frictionless. In other words, there is only normal surface force distributing within the contacting surface.
2.2. Test principle of equivalent cohesive strength The test principle of approaching cohesive strength bases on the solution of ideal visco-plasticity theory without considering hardening or strengthening (Vyalov and Tsytovich, 1956).
Ceq = K
F DS
(1)
where: Ceq stands for equivalent cohesive strength; K is the reduced factor, usually equal to 0.18 if the specimen is frozen soil; D represents the diameter of sphere; S means the axis displacement of spherical template (Fig. 2). The relationship between equivalent cohesive strength Ceq and time tends to be similar as the curve shown in Fig. 3. In following part, we will examine whether the curve of the maximum shear stress-time calculated based on modulus formulation is consistent with the curve of equivalent cohesive strength-time.
3.2. Two modulus equations based on the model of double elastic spheres
3. Theoretical analysis
Let the radius of ball 1 and ball 2 as ρ1andρ2 relatively. Before compressing, the point sited at the surface of ball 1 (or ball 2), which has distance r with the symmetric axis is A (or B) (Fig. 4). The absolute coordinates of point A (or B) is (x, y, z1) (or (x, y, z2)) relative to Cartesian coordinates system. Because the absolute value of z1 (or z2)is
3.1. Assumption 1) Continuous medium hypothesis: Due to the complex components of 2
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
Fig. 4. The model of compressing process. Explanation: the position1 stands for the state when the gap between two ball eliminates at radius r.
much smaller than ρ1 (orρ2),ignoring high order value of z1 (or z2), we get:
z1 = r2/2 1; z2 = r 2/2
By comparing Eqs. (6) & (7), γ is modified as follow expression.
= 1/(D
(2)
2
Because the value of S is too smaller than that of D, γ is considered as constant. Due to the relatively smaller size of contacting surface, we can consider both of the balls as hemi-spacious bodies. According to half space linear elastic theory (Qian, 1956), regardless of the effect of friction,ω1andω2is obtained.
where: r = x 2 + y 2 . As shown in Fig. 4(a), if both of the balls have similar scale and belong to elastic material, we think that the contacting surface is very close to circular plane. The whole deformation process can be split into two stages: 1)firstly, the radius of contacting surface becomes r when gap between two balls eliminates with point C and D moving to point C1 and point D1 relatively; 2)secondly, two balls continue their deformation until radius of contacting surface approaches the maximum value r c. Letting the total deformation of Line CD as S, the deformation at second stage of ball 1 and ball 2 as ω1 and ω2 relatively, we obtain:
S = z1 + z2 +
1
+
2
=
1 (1/ 2
1
+ 1/ 2 ) r 2 +
1
+
2
1
S=
(3)
rc =
S
=
SD
(5) (6)
where: D is the diameter of spherical template. According to Fig. 4(b), the expression of rc is obtained.
rc =
S(D
S)
2
=
2 2
1
P dx dy R1
E2
(9)
2 1
+
2 2
1 E2
PdR1d = S
r2
(10)
where:R1stands for the distance between given point A (x,y,0)and calx )2 + (y y)2 . culating point B (x′,y′,0), R1 = (x It's so difficult to solve Eq. (10) directly that we should try another way. For the perspective of physical understanding, we know that the pressure concentrates on the small area of contacting surface and decreases with radius increasing. Therefore, the function of P is proposed as follows:
+ 1/ 2) . where: = We consider ball 2 expands to be half-spacious bodies. In other word, 1/ρ2 → 0.Thus, γ and rc should be rewritten as follows: = 1/D
P dx dy ; R1
E1
E1
(4)
1
2 1
1
1
1 (1/ 1 2
= 1/2
=
where: E1and E2represent compression modulus of ball 1 and ball 2 relatively;υ1and υ2stands for the Poisson ratio of ball 1 and ball 2 relatively; P is the vertical component of surface force distributing within the contacting surface; Ω symbolizes the distributing area of surface force. Substituting Eq. (9) into Eq. (3), and expressing it in polar coordinate system with respect to point A (Fig. 5), we obtain:
where:ω1 (or ω2) is the deformation of vertical direction of C1 (or D1). At the second stage, ω1 = ω2 = 0. Therefore, the expression of S is obtained.
r 2c
(8)
S)
P=
P0 2 rc rc
R2
(11)
where: P stands for vertical component of surface force at point C (Fig. 5); R symbolizes the radial distance between point C and central
(7) 3
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
Fig. 5. Diagram of integral area and polar coordinate system, Vertical cross section when?? = const.
point O;P0 is the vertical component of surface force at central point O; rc represents the radius of contacting surface. From Fig. 5(a), we know the radius of half-circular area O1AFD is equal to rc2 r 2sin2 .Therefore, it has: rcos + r 2c r 2 sin2
PdR1 =
r 2c r 2 sin2
rcos
P0 (r 2c rc 2
r 2 sin2 )
E2 =
P0 PdR1d = (2r 2c 4 rc
P0 =
(12)
2 1
+
E1
2 2
1
2
P0 (2r 2c 4 rc
E2
r 2) = S
r2
E2 = (14)
2 1
1
+
E1 2 1
1
S=
+
E1
2 2
1
2
E2
4 2 2
1
2
E2
2
P0 rc
2 P0 r 2c 3
(17)
Combining Eqs. (4), (16), (17),we get:
P0 =
rc =
S=
1/3
2
6F 5( 1 + 3 F(
1
+
2)
2)
1/3
(19)
4 2
9 F2 ( 16
+
1
2)
2
1/3
S0 = (
1
+
2 )|r = 0
2 2
1 0.5 2 2 S1.5 0 D 3F
r 2c
2 1
1 E1
R2
(D
E2 =
3F(1 4S1.5 0 (D
(25)
(26) (27) (28)
S0 )S0
P =
P0 rc
P0 =
3 F DS0
1
=S
(24)
8
rc =
E2 =
2 2
1/3
2)
3 F 2 (D S0 )S0
where: 1 + 2 = E + E ; = D S . 1 2 When r = 0, the elastic deformation of two balls at central point is obtained.
1
+
P0 =
rc =
2 1
1
P0 rc
(20)
1
3 F(
(23)
2 2)
S0 )0.5
(29)
According to Eqs. (11), (23), (24), (25),we obtain:
(18)
2
1/3
24F 2 2 1 + 2)
P= (16)
By analyzing the vertical direction of the equilibrium function of ball 1, it has:
F=
(22)
where: = Because the modulus of spherical template (it's usually made by steel) is much larger than the modulus of frozen soil, we let E1 → ∞. According to Eqs. (22)and (25), we will have safety value of E2 (or E2′). Therefore, based on Eqs. (11), (18), (19), (22), it has:
(15)
P0 rc
E1
1 . D
By comparing two sides of Eq. (14),we get:
=
5(
(13)
Substituting Eq. (13) into Eq. (10), we obtain:
1
2 1
1
However, if combining Eqs. (15)–(17), we can obtain:
rc =
r 2)
S 0 )0.5
3F
And then, the integral formulation is obtained. 2
2 2
1 4S1.5 0 (D
(r c )2
R2
DS0 /2 2 3 2 F(1 2) 1.5 0.5 4S0 D
(30) (31) (32) (33)
Based on the analysis above, Eqs. (26)–(29) are defined as modified solution while Eqs. (30)–(33) are defined as strict solution. The special forms of strict solution had been proposed by Bezukhov N. I. (Bezukhov, 1961). Roman.L.T. et al. tried to modify it and gave a
(21)
where: S0is the elastic deformation of two balls at central point. Substituting Eq. (8), Eq. (21) into Eq. (20), we get: 4
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
modified solution, whose analytical results are consistent with that of tri-axial test. (Roman &Veretekhina, 2004).However, Roman.L.T. et al. didn't gave reason why this special forms of solution is practical. By comparing Eqs. (26)–(29) with Eqs. (30)–(33), if ignoring value of S02,we get:
(42), letting r = rc, we have:
i Ai (2i + 1)!
N
F = 4 p0 r 2c
i=1
(45)
At contacting surface, ω1 + ω2 = 0, substitute it into Eq. (37). And we can us-e polynomial function to express S approximately as follows:
P0/P 0 = 1/2
(34)
rc/r c =
(35)
S
(36)
where: SNstands for vertical deformation of sphere plate when it's calculated by N order polynomial function;rc is the radius of contacting surface. Eqs. (43)–(46) build (N + 2) orders function groups between variable SN, rc,p0,A2~ANF,ρ1,ρ2, θ1, θ2 (it already exists A1 = 1). That means if the value of arbitrary five numbers of variables have been known, we can solve other (N + 2) number of variables theoretically. Usually, if F,ρ1,ρ2, θ1, θ2 are provided, the solution process of mechanical behaviors (S, rc,p0,A2~AN) is defined as forward analysis. In contrast, the solution process of mechanical property (θ1, θ2and so on) is defined as back analysis when F,ρ1,ρ2, S are given. It needs to provide the estimated value of polynomial truncation error of S calculated by function group above. Based on Taylor expansion, the expression of S can be written as follows:
2
E2 /E2 = 1/ 2
N n=1
According to Eqs. (34)–(36), it shows that the modified solution can make less pressure concentrating at small area but widen the distributing area of pressure by comparing with strict solution. Why it comes to different solutions based on same theory (Eqs. (26)–(33))? This difference must be illustrated clearly for the perspective of theory. 3.3. A basic solution for calculating arbitrary order of pressure and modulus For one reason proposed, both the actual size of the spherical template and specimen are not infinity. For another, the geometrical equation (Eq. (7)) is an approximate description of contacting surface at elastic state, which contradicts with Eq. (15). In fact, based on method of Taylor expansion, Eq. (3) should be written as follows: N 1
+
2
=S+ n=1
1 22n
(
1) n
1
(n
(2n 1)! [( 1)1 1) !n!
2n
+ ( 2 )1
2n]
n=1
2
=(
1
+
2)
PdR1d
R=
1 2 rc rc
r 2c
R1 = rcos + sin
RN <
f
1 2 rc rc
R2 = n=1
An (2n 1)!
cos rc
i=n+1
(i
Ai n) ! (i 1) !22i
1
=
r 2c
r 2sin2
n! 2p ( 0 1
+
2)
1
2n 1
+
rc
+
2 )rc
N i=1
i! (i
Ai 1) !22i
1
=S
S=
F=
2 2 r c p0 3
2
rc P0 (
1
+
rc
(2N + 1)
+
1
(2N + 1)
rc 1
2)
r 2n c
(46)
(47)
+ ( 2)
(2N + 1) ]
2(N + 1)
(2N + 1)
rc
(49)
(51) (52)
S1 <
3 8
3
rc
+
1
rc
3
1
rc
(53)
+ 2 ). where: = Eqs. (50)–(52) are as same as Eqs. (4), (16), (17) (modified solution). For this point of views, Eqs. (4), (16), (17) are the first order solution of general algorithm(Eqs. (43)–(46)). Eq. (53) means the modified solution has polynomial truncation error that makes the difference between strict solution and it. When N = 2, we obtain: 1 ( 1 2 1
2n 1
2
1) (43)
2p ( 0 1
2n]
r 2n c
(50) 2
S1
R1 = S
(42)
rc
(n = 1, 2, 3,…, N
(2N + 1)! + 1) !N!
22N + 1 (N
2n 1
where:A1 = 1;Anis the parameter without unit, n = 2,3, …,N; Substituting Eqs. (39), (42) into Eq. (38), by comparing with Eq. (37),we get: N
( 1) N (2N + 1)! [( 1) 22N + 1 (N + 1) !N!
SN =
S1 = r 2c
(41)
where: randrcare constant for given point A; θandβare variable for integral point B (Fig. 5). Based on Eqs. (37), it sets r2n as the single variable. By considering Eqs. (38)–(41), we propose: N
+ ( 2 )1
2n]
where: ∣ RN∣ means the absolute value of RN. Eq. (48) demonstrates solution of N order function group(Eqs. (43)–(46)) accompanies with N order of polynomial truncation error. Eq. (49) can be applied for estimating the upper limit of N order of polynomial truncation error based on function group (Eqs. (43)–(46)). For example, letting N = 1 and A1 = 1, we can obtain:
(40)
r 2sin2
+ ( 2 )1
(48)
(39)
r 2sin2 + (R1 rcos ) 2
2n
2n
where: ξ is the radius based on Taylor expansion method, 0 < ξ < rc. Therefore, it means:
(38)
R2
( 1) n + 1 (2n 1)! [( 1 )1 (n 1) !n!
22n 1
RN = S
Like similar reasons for providing Eq. (11),according to Fig. 5, we propose:
P = p0 f(R) = p0 f
( 1) n + 1 (2n 1)! [( 1)1 (n 1) !n!
Let N order Polynomial truncation error of S as RN, we can get:
r 2n
where: N stands for the order of Equation; (M)! represents factorial function of M, (M) ! = 1 × 2 × … × (M − 1) × M, M = n-1,n,2n-1. It also has:
+
22n 1
1
S=
(37)
1
1
SN =
S2 = r 2c
(44)
S2
By considering the vertical equilibrium function of ball 1 and Eq. 5
S=
3 4 rc ( 8 2p 0
(
1
1
1
3
+
+ 2)
2
3
)
rc
(54)
1 A + 2 2 16
(55)
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
A2 =
16 rc +
2p ( 0 1
F = p0 r 2c
|R2| = S
(56)
2)
2 A + 2 3 15 S2 <
5 16
(57)
rc
5
+
1
rc
5
rc
1
(58)
Combining Eqs. (54)–(56), we have contradict results because of Polynomial truncation error of vertical deformation. In conclusion, existence of Polynomial truncation error of vertical deformation results in difference between strict solution and modified solution. Besides, it may contribute to contradict solution result of N order function group. 4. Test condition and result 4.1. Sample preparation Fig. 6. Changing curve of vertical deformation versus time.
When the temperature is so low, the frozen soil would tend to be as similar as continuous and isotropic material, which meets the basic hypothesis of elastic theory. Thus, we select −20 °C as temperature environment. The main physical properties of Kaolin soil are listed in Table 1. It has definition as low liquid limit silts according to standard for engineering classification of soils made in China (GB/T 50145–2007, 2008). The remolded sample was kept in cylinder with diameter in 61.8 mm and height in 20 mm. Before experiment, the specimen had been saturated with sealed mold under vacuum to allow water equilibration. Kept sealed with mold, the specimens were put into the freezing cabinet and frozen at negative temperature at −20 °C for 24 h.
4%,which could be ignored. Letting ν2=0.35, based on Eqs. (1), (29), (33)) as well as the curve of vertical deformation tested (Fig. 6), the changing curves of compression modulus and equivalent cohesive strength with time have been obtained (Fig. 7). It indicates that the modulus decreases sharply at the beginning time from 5S to 160S while remains decreasing with time slowly after 160S (Fig. 7). In addition, both of the modulus and equivalent cohesive strength shows same changing regularity with time. 5. Discussion 5.1. Analyzing idea and the simulation model
4.2. Experimental process
Due to the complexity of this problem, we resort to simulation to check which modulus equation (Eq. (29) or Eq. (33)) is better. The structural diagram of research is shown in Fig. 8. Yield Criterion of DRUCKER-PRAGER (shortly written as D-P)has been proved practical in geotechnical engineering simulation. The linear kind of Yield Criterion of DRUCKER-PRAGER writes as follows:
Both the spherical template indenter and thermo-container were kept at −20 °C in order to insure steady temperature environment during testing. Because of the discrete problem, it requires testing at least six different points of one sample to determine average value curve of deformation. Following then, different loading weight were applied and compared to determine the suitable load which made deformation satisfy with requirement (Eq. (59)). By trials, we finally determined 15 kg weight as static load and continued the experiment for > 24 h. Thus, F = 15 × 9.81 = 147.15(N).
0.005D
S15 min
f=t
(60)
d=0
where: t stands for the parameter of deviator stress,t = q/g; q symbo1 2 2 2 lizes the octahedral stress, q = 2 ( 1 2) + ( 2 3) + ( 3 1) ; g is the shape function of f (Dong et al., 2013),g = 2K/ [1 + K + (1 − K) ∙ (r/q)3]; r has the meaning of the radius of yield
(59)
0.05D
ptan
( ) 27
1/3
surface within deviator stress plane, r = 2 J3 ; J3 is the third invariable of deviator stress; K represents the ratio of tensile yield strength with respect to the compressing strength according to tri-axial test, K = (3 − sinφ)/(3 + sinφ)，0.778 ≤ K ≤ 1 (Liu and Cheng, 2010)；P represents the hydrostatic pressure, P = (σ1 + σ2 + σ3)/3; d means the yield strength. According to Wang (2015), if determining the yield strength d through pure shearing test, d can be expressed as follows (Eq. (61)):
where:S15min stands for the deformation at fifteenth minute of the beginning; D represents the diameter of spherical template, D = 22mm. 4.3. Experimental result After the test lasting > 24 h, the average curve of vertical deformation was obtained as follows (Fig. 6). Though the Poisson ratio varies with temperature and water content as many studies having proved (Zhu and Zhang, 1982; Wang et al., 2008). According to Roman (2016), for frozen cohesive soils, the Poisson ratio is range from 0.35 to 0.4.So, if considering this, the relative error of modulus estimated by Eq. (29) or Eq. (33) is in range of
d=
3 2
1+
1 K
f
(61)
where:τf is the shear strength obtained from pure shearing test.
Table 1 Main physical properties of soil sample. Soil
Dry density (g/cm3)
Saturated density (g/cm3)
Saturated water content (%)
Liquid limit (%)
Plastic limit (%)
Plasticity index
Kaolin
1.39
1.9
36.29
43.22
32.43
10.79
6
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
Fig. 7. Changing in compression modulus and equivalent cohesive strength of frozen Kaolin soil versus time.
the law of Hooke, the stress tensor is expressed as follows: ij
=
+ 2µ
e ij
(63) E ; 2((1 + ))
E , 2 )(1 + )
µ= where: λ, μstands for Lame coefficient, = (1 ν andE symbolizes the Poisson ratio and modulus of specimen relatively;σijrepresents the component of stress tensor;εije means the component of elastic strain tensor; θ is the elastic volume strain, θ = εiie. Becauseτφr = τφz = 0,σθ is the principal stress. By considering of hypothesis of complete plastic principle, it regards σθ as second principal stress σ2.We get: 1/ 3
2
Fig. 8. The guideline of this study.
+ 2
z
r
±
2
z
2
+
2 rz
(64-1) (64-2)
= max
{
1
2
2
,
2
3
2
,
1
3
2
}
(64-3)
where:σr,σθ,σz stands for theradial, circumferential and vertical component of stress tensor relatively (Fig. 10a);τrzandτmaxrepresent the shear stress and it's maximum value relatively (Fig. 10c);σ1~σ3 are the principal stress,σ1 > σ2 > σ3. If σ1 > σθ > σ3, the hypothesis of complete plastic principle is correct. Then, components of surface force within the meridian contacting surface are obtained.
(62)
3 Ceq = 0
r
=
max
According to the test principle of spherical template test, letting τf = Ceq and φ = β = 0 (0 ≤ β ≤ φ), we obtain K = 1and d = 3 Ceq .Therefore, Eq. (60) is rewritten as follows:
f=q
=
As for elastic part of deformation, we choose linear elastic constitutive model. Because the beginning contacting point is close to the symmetrical center of specimen, we regard this subject as axial symmetrical contacting problem. The size and mechanical parameters of spherical template and soil sample are given in Table 2. Besides, the normal contacting property is considered as hard one while the tangential contacting property is regarded as frictionless. The deformation boundary of soil sample is bounded during the process of loading (Fig. 9.a). We adopt the deformation recorded at every moment as load. In order to have high calculating accuracy, there are high density of elements within the core cylinder with height in 10 mm and radius in 6 mm (Fig. 9.b). If having obtained the elastic strain tensor of yield points, based on
n
=
n
=(
2 r sin
z
+ r)/2
2 zcos
rz sin 2
sin 2 +
rz
(65-1) (65-2)
cos 2
Or the component of surface force within the meridian contacting surface can be expressed as:
Pr =
sin
r
+ cos
rz
(66-1)
Pz =
sin
rz
+ cos
z
(66-2)
where: the negative calculating value of Eqs. (65)–(66) indicates the direction of force is opposite to that having been made (Fig. 10.a~c);α characterizes the angle between the normal direction of contacting
Table 2 The size and physical parameter of the model. Object Sphere plate Specimen
Modulus (Mpa) 5
2 × 10 E2(orE2′)
Poisson ratio
Yield strength d (Mpa)
Plastic strain
Diameter (mm)
Height (mm)
0.3 0.35
335
0.00223 0
22 61.8
– 20
3 Ceq
7
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
Fig. 9. Constraint condition and element dividing in simulation.
surface and vertical direction;σn and τn represents the normal and tangential component of surface force distributing within the meridian contacting surface relatively (Fig. 10.b);PrandPz stand for the radial and vertical components of surface force within the meridian contacting
surface relatively. In following part, we select possible yield points (points labeled 1~5) along the contacting surface for studying (Fig. 10.d). Point2~4 represent some points, whose connecting line with point O divides the
Fig. 10. Components of stress or surface force within meridian interface. Explanation: Meridian interface is the plane cross the symmetrical axis. 8
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
Fig. 11. Principal stresses at possible yield points according to E2. Explanation: σ1~σ3are calculated according to Eq. (64).
total angle ϕ into one quarter, half and three quarter of it relatively.
maximum shear stress given by different kind of modulus are so close with each other for all points, which indicates the value of compression modulus has less influence on the value of maximum shear stress. According to Fig. 13a, b, the vertical component of surface force obtained from simulation remains more than half value of that given by strict solution for each moment. The distributing area given by simulation is much larger than the one given by strict solution. For this perspective, it concludes that the modulus given by strict solution (E2′) may overestimate the real modulus value of specimen as it originates from more pressure distributing within smaller area of contacting surface. However, if simulating according to the value of E2 (modified solution), we obtained the curve of vertical component of surface force, which's quiet similar to that given by modified solution not only for the value but also for the shape of curve (especially after 160S, Fig. 13c, d).Therefore, E2 may reflect the plastic property of specimen to a certain degree. According to Fig. 13a, c, it shows that vertical component of surface force obtains the maximum value at point 4 or nearby position. Besides,
5.2. Comparing the result brought by two modulus equations Firstly, it should testify the hypothesis of complete plastic principle. Due to similarity of simulating models (by E2 or E2′), it here only provides the calculating results according to E2 (modified solution, Fig. 11). It indicates that σθ is the second principal stress for all the possible yield points (Fig. 11).Thus, complete plastic principle is correct for plastic area. And we can make further analysis based on Eqs. (65)–(66). According to Eq. (64-3) and the result of simulation, we obtain the curve of maximum shear stress when inputting E2′(solved by strict solution) into simulation model (Fig. 12). From Fig. 12, we know that except point 1, all of the other points distributing along the contacting surface have already been in yield state. The maximum value of maximum shear stress is obtained no in point 5 but in point 4 (Fig. 12, Fig. 10d). Besides, it also reveals that the
Fig. 12. Curve of the maximum shear stress on possible yield points 9
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
Fig. 13. Comparison of vertical component of surface force.
it almost remains invariable with radial distance increasing before point 4, which's different from those of elastic solution. This result is consistent with solution of ideal visco-plasticity theory (Vyalov and Tsytovich, 1956), which proposed that the surface force remain same with radius increasing. According to Fig. 14, it indicates that the value of tangential component of surface force increases with radial distance increasing for each moment (especially after 5S). In addition, all of the curves reach the maximum value at point 5 while approach almost zero value at point 1, which meets the mechanic requirement (The shear stress approaches zero value at axial position in symmetric mechanic subject.). In the process of deducting the modulus equation, we make no consideration of shear effect along the surface. However, according to Fig. 14, the shear stress would have sharp mutation at contacting surface if considering spherical template as frictionless. Therefore, it should give further examination of frictionless hypothesis.
5.3. Examination of frictionless hypothesis ° n 2 2 Let f = n + n , tan = n andδ = α + β − 90 . Where: f stands for the joint force within meridian interface;β represents the angle betweenf and τn;δ symbolizes the angle between f and vertical positive direction, we let counterclockwise as positive rotation direction, which means the negative value of δ (or β) stands for clockwise rotation (Fig. 15). Following then, we obtain the curve of β and δ of different moment (Figs. 16, 17). According to Fig. 16, β decreases linearly with radial distance increasing before approaching three quarter of the radius of contacting surface whileβ has sharp reduction when reaching the edge. Besides, it also indicatesβ is range from 85°to 90°when radial distance increases to three quarter of radius of the contacting surface. So, it proves that the frictionless hypothesis is suitable only when the radial distance is less
10
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
Fig. 14. Comparison of tangential component of surface force.
than three quarter of the radius of contacting surface. According to both of the simulation based on E2 and E2′ (Fig. 17), δ increases with radial distance increasing to three quarter of radius of contacting surface while decreases enormously to negative value nearby the edge. Thus, it means that the direction of joint force within meridian interface can't remain at vertical direction. All of the research been done in this part tries to find better modulus equation by comparing simulation results with theoretical results. In fact, simulation is one kind of methods to examine the solutions. The important examination is some results coming from relevant tests directly. Therefore, we make reference from some similar experimental results to check the previous theory.
Fig. 15. The joint force f and it's direction within meridian interface
Fig. 16. Comparison of the angle between joint force and it's tangential component (β) 11
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
Fig. 17. Comparison of the angle between joint force and vertical positive direction (δ).
5.4. Examination from other experimental results
underestimates the value of modulus but the strict solution overestimates it. The main reason resulting in the difference of compression modulus given by spherical template indenter and other experiments is that the frozen soil is viscosity-elasticity-plastic materials, whose mechanic property can't be totally described by elastic theory. In fact, only knowing about vertical deformation of spherical template, we should find actual mechanics properties of specimens. This solution process has been defined as “back-analysis”. Problem included in “back-analysis” has different solutions depending on which basic theory or model selected (Bezukhov, 1961; Kolesnikov Yu and Morozov, 1989). However, based on spherical template indenter test and examination from simulating and testing, this paper provides a basic and efficient method to estimate the modulus of frozen soil with time changing. Besides, the solution method provided could be used in other engineering field like Brinell hardness test and so on.
Many studies have proved that the modulus of frozen soil increases almost linearly with temperature reducing (Zhu and Zhang, 1982). According to the previous testing results of modulus of frozen soil (Zhu and Zhang, 1982; Li et al., 2007; Yang et al., 2010), the curve of modulus changing with temperature is obtained (Fig. 18). Though they have not provided direct testing results of the modulus at temperature of −20 °C, we can use different experimental data to estimate it. The estimating value of modulus of frozen Kaolin soil at −20 °C, which has 36.29% saturated water content, is close to 335 Mpa. It usually adopts the deformation at time ranging from 5 to 25 s to calculate initial equivalent cohesive strength. Thus, we select the modulus of 5 s as the initial modulus. The strict solution of modulus equation provides estimating value of 420 Mpa while the modified solution approaches 297 Mpa, which shows that the modified solution
6. Conclusion According to hemi-spacious elastic theory and theoretical model of spherical template indenter, two group equations, noted as strict solution and modified solution, were derived. In order to find reasons causing different solutions, it proposed a general algorithm to calculate arbitrary order accuracy of compression modulus, surface force, radius of contacting surface relatively. Based on the tests, both the simulation built of compression process and experimental data were used to compare the calculated results of strict solution and modified solution. Therefore, the following conclusion could be acquired. (1) The modified solution is the first order approximate solution of the general algorithm. The strict solution overestimates the value of surface force but underestimate radius of contacting surface, which contributes to high estimating value of compression modulus as 2 times big as that given by modified solution. (2) The simulation indicates that not only the surface force, but also the maximum shear stress based on modified solution are closer agreement with the testing and simulation results. In contrast, the strict solution overestimates the value of compression modulus and surface force because of narrowing the pressure distributing area. (3) The simulation indicates that the frictionless hypothesis is
Fig. 18. The experimental data of modulus of frozen cohesive soils Source: Curve 1 is based on the experimental data obtained by Yang et al. (2010). Curve 2 Originates from the testing data got by Li et al. (2007).Curve 3 to 5 are according to the results provided by Zhu and Zhang (1982). 12
Cold Regions Science and Technology 170 (2020) 102911
C.-J. Huang, et al.
approximately correct when the radial distance is less than three quarter of radius of contacting surface. (4) The modified solution underestimates the compression modulus of frozen Kaolin soil while strict solution overestimates it according to other experimental data.
Li, S.Y., Lai, Y.M., Zhang, M.Y., et al., 2007. Study on distribution laws of compression modulus and strength of warm frozen soil. Chin. J. Rock Mech. Eng. 26 (S2), 4299–4305. Liu, S.T., Cheng, P.F., 2010. Based on ABAQUS for numerical analysis of soil constitutive model. Low Temp. Technol. 2, 90–92. Qian, W.C., 1956. Elastic Mechanics Theory. in Chinese. Chinese Science Press, Beijing, pp. 316–326. Roman, L.T., 1987. Frozen Peaty Soils As Beds for Structures. In Russian. Nauka, Novosibirsk. Roman, L.T., 2016. In: Zhang, C.Q., Zhang, Z. (Eds.), Frozen Soil Mechanics. Science Press, Beijing, China, pp. 150–159 trans. Roman, L.T., Veretekhina, É.G., 2004. Determination of deformation characteristics of permafrost from impression of spherical plate. Soil Mechanics and Foundation Engineering 41 (2), 60–64. Roman, L.T., Zhang, Z., 2010. Effect of cycles of freezing and thawing on the physical and mechanical properties of moraine loam. Soil Mech. Found. Eng. 47 (3), 96–101. Tsytovich, N.A., 1947. Determination of Cohesive Forces in Permafrost by the SphericalProbe Method. In Russian. FondyInst. Merzlotoved.Akad.Nauk SSSR. Tsytovich, N.A., 1985. In: Zhang, C.Q., Zhu, Y.L. (Eds.), The Mechanics of Frozen Ground. Chinese Science Press, Beijing Translated by. Vyalov, S.S., 1945. Analytical Formulation of Mechanics Property of Quasi-ViscosityElastic Material. in Russian. vol. 1(2). News of the USSR academy of sciences, The Technical Science, pp. 25–30. Vyalov, S.S., Tsytovich, N.A., 1956. Evaluation of the Bearing Capacity of Cohesive Soils by the Magnitude of Impression of a Spherical Plate. In Russian. vol. 3. Dokl.Akad.Nauk SSSR, pp. 1193–1196 6. Wang, P.F., 2015. The Study of Quasi-static Deformation and Failure of PBX Based on Drucker-Prager Criterion. Mianyang, Master Degree Paper of Institute of Systems Engineering. China Academy of Engineering Physics, pp. 15–16. Wang, Z.Z., Mu, S.Y., Nu, Y.H., et al., 2008. Predictions of elastic constants and strength of transverse isotropic frozen soil. Rock Soil Mech. 29 (S1), 475–480. Yang, Y.G., Lai, Y.M., Pu, Y.B., et al., 2010. Experimental analysis of damage of ice-rich frozen silt under uniaxial compression. Chin. J. Rock Soil Mech. 33 (10), 3063–3068. Zhang, Z., Ma, W., Zhang, Z.Q., et al., 2012. Application of spherical template indenter to long-term strength tests for frozen soil. Rock Soil Mech. 33 (11), 3516–3520. Zhemochkin, B.N., 1957. Theory of Elasticity. In Russian. Gosstroiizdat, Moscow. Zhou, H., 2015. The Change Rule of Frozen Loess Long-Term Strength Under Freeze-Thaw Cycle. Master Degree Paper of Lanzhou University, LanZhou City of China, pp. 29–38. Zhou, H., Zhang, Y.C., Zhang, Z., et al., 2014. Changing rule of long-term strength of frozen loess cohesion under impact of freeze-thaw cycle. Rock Soil Mech. 35 (8), 2241–2246 (2254). Zhu, Y.L., Zhang, J.Y., 1982. The elastic displacement and compressive displacement of frozen soil. J. Glaciol. Geocryol. 4 (3), 29–39.
Declaration of Competing Interest None. Acknowledgment This research was supported by Bureau of International Cooperation, Chinese Academy of Sciences (No.131B62KYSB20170012), National Natural Science Foundation of China (41771078, 41871061) and the Funding of the State key laboratory Frozen Soil Engineering (SKLFSE-ZT-19). References Bezukhov, N.I., 1961. Fundamentals of Elasticity, Plasticity, and Creep Theory. [In Russian]VysshayaShkola, Moscow. Chertolyas, N.F., 1977. Evaluation of the Strength and Deformation Properties of Cohesive Soils by the Impression of a Spherical Plate. ([In Russian]Author's abstract of dissertation for Candidate of Technical Sciences, Moscow). Dong, L.W., Zhang, M.Y., Xie, Y.Y., et al., 2013. Study on the parameters of soil models yield criterion in ABAQUS software. J. Qingdao Technol. Univ. 34 (1), 48–50. Fang, J.H., Chen, X., Xu, A.H., et al., 2018. The experimental study of the influence of freeze-thaw cycles on the physical and mechanic properties of Qinghai-Tibet red clay. J. Glaciol. Geocryol. 40 (1), 62–69. GB/T 50145-2007, 2008. Standard for Engineering Classification of Soil. In Chinese. Chinese Planning Press, Beijing, pp. 4–7. Gost 12248 - 96. Substrates, 1991. The Laboratory Method of Characterizing the Strength and Deformation Property of Frozen Soil. In Russian. The Micro Structure in Russia, Moscow. Kolesnikov Yu, V., Morozov, E.M., 1989. Contact Fracture Mechanics. in Russian. Nauka, Moscow.
13