A constitutive model of frozen saline sandy soil based on energy dissipation theory
A constitutive model of frozen saline sandy soil based on
energy dissipation theory
Yuanming Lai a ,b ,*,Mengke Liao a ,c ,Kai Hu a
a State Key Laboratory Frozen Soil Engineering,Cold and Arid Region Environmental and Engineering Institute,Chinese Academy of
Sciences,Lanzhou,Gansu 730000,China
b School of Civil Engineering,Lanzhou Jiaotong University,Lanzhou,Gansu 730070,China
c University of Chinese Academy of Sciences,Beijing,100049,China
a r t i c l e i n f o
Article history:
Received 3February 2015
Received in revised form 10October 2015
Available online 2November 2015
Keywords:
B:Anisotropy
B:Mechanical properties
B:Plastic deformation
B:Thermomechanical
Frozen saline soil a b s t r a c t A series of triaxial compression tests are carried out for frozen saline sandy soil with Na 2SO 4contents 0.0,0.5,1.5,and 2.5%under con ?ning pressures from 0MPa to 16MPa at à6 C,respectively.The test results indicate that,the Critical State Line (CSL)of frozen saline sandy soil is curve and is not through the origin in (p ,q )plane,and the soil particles have the properties of initial anisotropic rotational angle and loaded anisotropy in process of loading.In order to describe the deformation properties of frozen saline sandy soil,a new double yield surface constitutive model is proposed based on the triaxial compression tests in this paper.The proposed model contains the in ?uence of salt contents on me-chanical characteristics,so it is suitable for describing the stress e strain relation of frozen saline soil.The proposed model has the following properties:(1)By de ?ning a modi ?ed
effective stress p *,a critical state strength envelope function is established according to the
Modi ?ed Cam Clay model.The envelope approximates a straight line under low con ?ning
pressures,but it is curve downward under high con ?ning pressures due to pressure
melting.(2)The effect of the initial anisotropic rotational angle and loaded anisotropy,
during process of loading under plastic volumetric compression mechanism,on yield
surface of rotational hardening is taken into account.(3)A paraboloid yield surface
function,including the rotational hardening law induced by loading,is proposed under
plastic shear mechanism.The universality of the proposed model is veri ?ed by the test
results of frozen saline sandy soil under different stress paths.Finally,in order to further
study the applicability of the proposed model in this paper,the stress e strain relation of
the cemented clay is simulated by it.And the in ?uences of the pressure melting phe-
nomenon and the rotational hardening rule on the calculated results of the proposed
model are investigated.The research results indicate that the proposed model can simulate
not only the mechanics properties of materials whose CSL is straight but also those of
materials whose CSL is curved,other than predict the deformation regularity of frozen
saline sandy soil well.
?2015The Authors.Published by Elsevier Ltd.This is an open access article under the CC
BY-NC-ND license (https://www.sodocs.net/doc/2a16202907.html,/licenses/by-nc-nd/4.0/).
*Corresponding author.State Key Laboratory Frozen Soil Engineering,Cold and Arid Region Environmental and Engineering Institute,Chinese Academy of Sciences,Lanzhou,Gansu 730000,China.Tel.:t869314967288.
E-mail address:ymlai@https://www.sodocs.net/doc/2a16202907.html, (Y.
Lai).
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International Journal of Plasticity
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0749-6419/?2015The Authors.Published by Elsevier Ltd.This is an open access article under the CC BY-NC-ND license (https://www.sodocs.net/doc/2a16202907.html,/licenses/by-nc-nd/4.0/).
International Journal of Plasticity 78(2016)84e 113
https://www.sodocs.net/doc/2a16202907.html,i et al./International Journal of Plasticity78(2016)84e11385 1.Introduction
Frozen soil consists of mineral particles,ice inclusions,liquid water and gaseous inclusions(Tsytovich,1985).And the frozen saline soil is a type of frozen soil which contains solution of soluble salt and salt crystal.Test results show that the precipitation of salt crystals at negative temperature is strongly sensitive to temperature(Wan and Lai,2013).Compared with general frozen soil,the structure of the frozen saline sandy soil is changed with the precipitation of ice crystal and salt crystal in the freezing process,so the theoretical modeling of the constitutive relation of frozen saline frozen soil is more complex and dif?cult than that of other soils.With the development of society and economics,many engineering projects,such as highways,railways,houses,airports and oil pipelines,will be constructed in cold saline soil regions.At present,the study on mechanical property for frozen saline soil is rarely performed,so it is signi?cant and essential to investigate the mechanical behaviors of frozen saline soil for engineering design.
Based on the theory of critical state soil mechanics,Roscoe et al.(1963)proposed Cambridge Clay model for soft soil. Afterward Roscoe and Burland(1968)presented Modi?ed Cam Clay https://www.sodocs.net/doc/2a16202907.html,de and Duncan(1978)formulated a dilatancy model for sandy soil,which is called Lade-Duncan model.Since then,the constitutive model research of geotechnical ma-terials becomes a hot issue.In the case of Rock Mechanics,Khan and co-workers have developed constitutive models to capture elasto-plastic behavior of Berea sandstone under a large range of con?ning pressures(Khan et al.,1991,1992).In damage constitutive model study on the brittle material,Shojaei et al.(2013)built a damage constitutive theory based on the dynamic dissipation work method for polycrystalline materials under dynamic loading,and developed a continuum damage mechanics(CDM)constitutive model to describe elastic,plastic and damage behavior of porous rocks(Shojaei et al.,2014).For unfrozen geomaterials,some researchers proposed a lot of strength criteria and constitutive models to solve different en-gineering problems(Li and Dafalias,2000;Altenbach et al.,2001;Hashiguchi and Tsutsumi,2003,2007;Hashiguchi,2005; Vorobiev,2008;Steinhauser et al.,2009;Shen et al.,2012;Xie and Shao,2012;Yao and Wang,2014;Mortara,2015).The deformation behaviors of soils are usually identi?ed by two basic plastic?ow mechanisms:one is related to plastic shear,and the other to plastic volumetric compression(Xie and Shao,2006).In order to describe the two plastic?ow mechanisms well,a double yield surface model has been established by combining a cap surface for plastic compression and a cone surface for plastic shear(Peric and Ayari,2002;Lai et al.,2010).Because of the structural complexity of frozen saline sandy soil,it is dif?cult to describe the mechanical properties by a single yield surface,so the mechanical characteristics of plastic volumetric compression mechanism and plastic shear mechanism are used to investigate the constitutive relation of frozen saline sandy soil in this paper.
Many researchers have made a lot of researches to describe the mechanical characteristics of general frozen soil (Chamberlain et al.,1972;Bragg and Andersland,1981;Fish,1991;Hashiguchi,2005;Zhang et al.,2007;Lai et al.,2009;Yang et al.,2010;De and Pereira,2013;Painter and Karra,2014).By the test and numerical methods and selecting different test materials and theory,the various affecting factors on mechanical behaviors for frozen soil have been studied and different constitutive models have been established up to the present.The more detailed frozen soil test methods and research pro-cesses were introduced in the authors’previous works(Lai et al.,2009,2014).However,it should be pointed out that stress e strain relationship of frozen saline sandy soil is more complex than that of unfrozen soil or general frozen soil.During the freezing process,ice crystals,salt crystals,and soil particles are cemented together to make the frozen saline sandy soil have tensile capacity.The characteristics of frozen soil are similar to that of cemented clay.In the study on mechanical behavior of cemented clay,Suebsuk et al.(2010)proposed a constitutive model,based on the Structure Cam Clay(SCC)model proposed by Liu and Carter(2002),to simulate the mechanical behaviors of cemented clay better.Gao and Zhao(2012)and Nguyen et al.(2014)proposed a modi?ed SCC model and its simulated results agree well with the test results of cemented clay with different cement contents.However,with the increase of con?ning pressure,the frozen saline sandy soil has a char-acteristic of pressure melting but the cemented clay does not have it.The pressure melting of frozen soils results in that with the increase of con?ning pressure,the critical state line(CSL)of frozen saline sandy soil will gradually bend downward,while the critical state line of cemented clay is linear.The critical state line of the frozen saline sandy soil will degenerate as that of cemented clay if the in?uence of pressure melting is not considered.
During the process of shearing loading or isotropic compression,the unfrozen geotechnical granular materials appear different degree of anisotropic properties(Anandarajah,2008;Yin and Chang,2010;Rowshanzamir and Askari,2010;Liu et al.,2013;Fonseca et al.,2013).Based on theory and tests,many researchers have made a lot of studies on anisotropic properties for granular material and soft soil.Voyiadjis et al.(1995)proposed a granular material constitutive relation based on anisotropic rotating yield surface,and studied the deformation of anisotropic elasto-plastic damage for concrete(Voyiadjis et al.,2008).Cleja-Tigoiu(2000)established an anisotropic?nite elasto-plastic constitutive equation by combining the ki-nematic hardening law of material.Based on test data of soft soils,Wheeler et al.(2003)proposed an anisotropic elasto-plastic constitutive model called as S-CLAY1,in which a rotational yield surface and a rotational hardening law were used, and the sensitivity of rotational hardening parameters was analyzed.In the previous study,it is found that the plastic?ow directions does not coincide with the hydrostatic axis(Lai et al.,2009)when the isotropic consolidation was completed and the shearing began,and the initial anisotropy is produced in the process of consolidation.From the triaxial tests of frozen saline sandy soil,it is found that the soil particles generate initial anisotropic angle before shearing loading due to the consolidation pressure.With the increasing of shearing loading,the anisotropic rotational angle of frozen saline sandy soil decreases from initial rotation angle below horizontal axis to zero,and then increases to the maximum rotational angle above horizontal axis.
In the geotechnical engineering theoretical analysis and study,the thermodynamics theory is introduced to solve the problem of related theory.Based on thermodynamic framework,Collins and Hilder (2002)built a theory foundation of anisotropic elasto-plastic rotational hardening model by triaxial tests,and studied the forms of Helmholtz free energy rate and the parameter's choice of dissipation rate.Voyiadjis et al.(2011)proposed a thermodynamic damage and healing process constitutive model for viscoplastic materials,which contains two new yield surfaces for the damage and healing processes that take into account the isotropic hardening effect.Xiao (2014)proposed a rate-independent ?nite elastoplastic equation with thermo-coupled effects to bypass the yield condition and loading-unloading conditions.The purpose of this paper is to propose a constitutive model which can describe the mechanical behaviors and the particle rotational rule of frozen saline sandy soil.Based on the thermodynamic theory,we proposed a new rotational hardening double yield surface constitutive model of frozen saline sandy soil considering initial anisotropy and loaded anisotropy.By employing double yield surface functions and non-associated plastic potential functions,the proposed model contains plastic compression mechanism and plastic shear mechanism.Additionally,the in ?uences of rotational hardening rule on the calculated results are investigated to validate the advantages of the proposed model of frozen saline sandy soil.It is found that the calculated results of the proposed model are agreement well with the test results.
2.Test conditions and results
2.1.Description of the test
The tested Na 2SO 4saline sandy soil was collected from permafrost regions in Qinghai-Tibet plateau and its grain size composition is given in Table 1.Firstly,in order to desalinate the soil,salt in saline sandy soil was removed ten times with distilled water.The desalinated soil sample was dried at 105 C for 24h,and then pulverized.Secondly,the dry preparation soil samples are selected to make up the saline sandy soil whose salt contents are 0,0.5,1.5,and 2.5%(The salt is anhydrous Na 2SO 4?ne particles),respectively.The freezing temperatures of saline sandy soil with salt contents 0.0,0.5,1.5,and 2.5%are à0.19 C,à1.47 C,à1.54 C and à1.58 C,respectively.To ensure the uniformity of the samples,the saline sandy soil was mixed with water at moisture content of 13%,and then kept for 12h without evaporation to allow the water to be uniformly distributed in the soil.
When the water was uniformly distributed in the soil,the prepared saline sandy soil was ?lled in a cylindrical mold to make cylindrical soil specimens with target dry density 1.89g/cm 3under certain compression rate.The specimens were prepared as cylinders with 6.18cm in diameter and 12.5cm high.Then,the specimens in the molds were saturated under a vacuum.In order to avoid large frost heaving and prevent moisture migration,the soil specimens were placed in a refrig-eration unit and frozen quickly at a temperature below à30 C.After 48h of freezing,the molds were removed and the specimens were mounted with epoxy resin plates on both ends and covered with a rubber sleeve to avoid moisture evap-oration.Finally,the specimens were kept in an incubator for over 24h at the test target temperature of à6 C,such that the specimens reached a uniform temperature.
The con ?ning pressures were from 0to 16MPa,and the strain rate of axial loading _ε
a ?1:67?10à4s à1was employed in the conventional triaxial test according to the GB/T 50123-1999(People's Republic of China National Standard,1999).Each sample was loaded until failure and the test duration was about 20min,so the conventional triaxial test is fast shear test,and the deformation of frozen soil is mainly elastic e plastic deformation,and the in ?uence of loading rate on mechanical properties can be neglected.Therefore,during so short test time,the visco-plastic deformation is not taken into account in the analyses of test results.Based on the conventional triaxial fast shear test results,the rate-independent mechanical properties of frozen soil are investigated in this paper.
2.2.Test results and analyses
The present study used a cryogenic triaxial apparatus improved from the MTS-810material test machine which was illustrated by Lai et al.(2014).After the specimens mentioned above were prepared,they were placed into the pressure cell of the MTS-810material test machine and a series of triaxial compression tests were performed.Firstly,the temperature in the pressure chamber was set at à6 C with a precision of ±0.1 C;then,the prepared specimen was consolidated under a preset pressure for 2h prior to the axial loading.In the axisymmetric case of specimen,the principal stress s 2?s 3,and the effective principal stress of axial load is s 1.We suppose that the stress and strain be positive under compression condition like the assumption in geotechnical engineering.
The stress and strain for conventional triaxial test are expressed as:
Table 1
Physical parameters of saline sandy soil (unit:%).
Composition of particle diameters
>0.50mm
0.50e 0.25mm 0.25e 0.075mm 0.075e 0.005mm <0.005mm 3.72710.57344.36035.920 5.420
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p ?13es 1
t2s 3T;q ?s 1às 3(1a)εv ?ε1t2ε3;εg ?23eε1àε3T(1b)
where p ,q ,εv ,and εg are hydrostatic pressure,deviator stress,volumetric strain,and shear strain,respectively.The normal components of the stress and strain tensors are assumed to be positive under compression condition like the general assumption in geotechnical engineering.A series of triaxial compression tests were carried out at à6 C at the salt contents of 0.0,0.5,1.5,and 2.5%,respectively.The test results of frozen saline sandy soil with salt content 2.5%are shown in Fig.1.
3.Thermodynamics basic framework
The theory of thermodynamics has better been developed and applied to research of geotechnical constitutive relation (Collins and Kelly,2002).Based on the general thermodynamic method,many new critical state constitutive models can be developed (Collins and Kelly,2002;Collins and Hilder,2002).
3.1.The ?rst and second laws of thermodynamics
The ?rst law of thermodynamics states that there exists a state function called the internal energy (Ziegler,1977;Collins and Houlsby,1997),such
that:
Fig.1.Test result curves of frozen saline sandy soil with salt content 2.5%under various con ?ning pressures.
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_U ?s ij _εij àq k ;k (2)where U ,q i ,s ij ,and εij are internal energy in unit volume,heat ?ow vector,stress tensor,and strain tensor,respectively.When the deformation of material is irreversible,the internal dissipation rate _f
satis ?es the Clausius e Duham inequality of the second law of thermodynamics,which is expressed as:
_f ?T _s i ?T e_s à_s r T?T _s à àq k ;k tq k T ;k T !0(3)
where _s
i is internal entropy rate when the material deformation is irreversible;_s r is external entropy rate and _s r ?àgrad q k T ;_s
is system entropy generation rate;T is absolute temperature.Substituting Eq.(2)into Eq.(3),we obtain:
_f ?s ij _εij à _U àT _s àq k T ;k T !0(4)
Supposing that j ?U àTs is the (Helmholtz)free energy of unit mass,then,we have _j
?_U àT _s às _T .Based on Eq.(4),the following formula can be obtained:
_f ?s ij _εij à _j ts _T àq k T ;k !0(5)
3.2.Thermodynamics basic equations
If the continuum element undergoes isothermal condition and irreversible deformation process,the in ?uence of tem-perature on the dissipation function can be omitted,i.e.,_T
?0and T ,k ?0in Eq.(5).Based on the conventional triaxial fast shear test,the deformation of frozen soil is mainly elastic e plastic deformation,and the in ?uence of loading rate on me-chanical properties cannot be taken into account.Therefore,the visco-plastic characteristics are not considered in the deformation of the material in this study.For the elasto-plastic problem of rate-independent materials in isothermal con-dition,the elasto-plastic constitutive relation can be expressed by incremental forms.The incremental constitutive laws can be established based on the thermodynamics theory (Collins and Houlsby,1997;Collins and Kelly,2002).The incremental work done by the applied stresses is the sum of the increment in the free energy function,j ,and the increment in the dissipation function,f .Both functions are de ?ned in per unit volume.Here,based on the above conditions and simplifying Eq.
(5),the incremental work-energy function can be expressed as (Collins and Hidder,2002;Collins and Kelly,2002):
s ij d εij ?d j tdf ;where df !0(6)where s ij d εij ,d j and df are incremental work,differential of the free energy and dissipation potential increment in unit volume,respectively.The total strain tensor εij can be expressed as εij ?εe ij tεp ij ,so we have d εij ?d εe ij td εp ij .For the non-coupled elastic material,the instantaneous elastic modulus is independent on the internal variables.Based on the energy decomposition principle,the free energy in unit volume can be expressed as (Collins and Houlsby,1997,Collins and Kelly,2002;Ulm and Coussy,2003):
j εe ij ;εp ij ?j e εe ij tj p εp ij (7)
where j e eεe ij Tand j p eεp ij Tare elastic free energy and plastic free energy in unit volume,respectively.The differential of the free
energy is given by:
d j ?v j
e v εe ij d εe ij tv j p v εp ij d εp ij (8)
From Eqs.(6)and (8),the following formula can be obtained:
s ij àv j e v εe ij !d εe ij ts ij d εp ij àv j p v εp ij d εp ij àdf ?0(9)The dissipation incremental function df is not a state function,because it only depends on the plastic strain increment.It cannot depend on the total strain increment since a purely elastic deformation would not produce dissipation.For energy dissipation of elasto-plastic deformation process in the rate independent case,the increment of dissipation function df is
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?rst-order homogeneous function of generalized increment d εp ij (Collins and Kelly,2002).According to the Euler's theorem of homogeneous function,the following formula can be obtained:
df ?v edf Tv d εp
ij d εp ij (10)
It is supposed that the internal variables εp ij
are “generalized displacements ”in thermodynamics,and the “generalized force ”,conjugated work of “generalized displacements ”,is v edf Tv ed εp ij
T,which can be called dissipative stress (Ziegler,1983;Ziegler and Wehrli,1987).In dissipation stress space,the dissipation yield condition can be obtained by eliminating the component d εp ij .
Substituting Eq.(10)into Eq.(9),we have:
s ij àv j e
v εe ij !d εe ij t24s ij àv j p
v εp ij àv edf Tv d εp ij 35d εp ij ?0(11)
In any process of loading under isothermal condition,for any d εe ij and d εp ij ,whilst Eq.(11)is satis ?ed if:
s ij ?v j e v εe ij
(12a)
and s ij ?v j p v εp ij tv edf Tv d εp
ij ;where df !0(12b)
For isothermal condition,the relationships between the stress components and the free energy as well as dissipation potential,Eqs.(12a)and (12b),are deduced from the ?rst law and the second law of thermodynamics.If the soil does not yield,only the elastic deformation appears,then df ?0,the relationship between stress components and free energy can be expressed by Eq.(12a).In contrast,if the frozen soil generates plastic (or irreversible)deformation,then df >0,Eqs.(12a)and (12b)must be satis ?ed simultaneously during the process of loading.Eq.(12a)denotes the elastic part of constitutive law.And
in Eq.(12b),v j p
v εp ij is called as shift stress,and v edf Tv ed εp ij
Tdissipation stress.Eq.(12b)demonstrates that the true stress is equal to shift stress plus dissipation stress (Collins and Hilder,2002),and it illustrates that the relationship between stress components and the plastic free energy as well as dissipation potential under plastic state.
According to the work-energy theorem,the external work increment of geomaterial under the isothermal condition is expressed as d U ?s ij d εij .Thus,whilst the plastic work increment is the product of the true stress with the plastic strain increment,the plastic dissipation increment is the product of the dissipative stress with the plastic strain increment.And the total plastic work increment contains the plastic dissipation increment and plastic free energy increment.Thus,the external work increment is also expressed as:
d U ?s ij d εij ?v j
e v εe ij d εe ij tv j p v εp ij d εp ij tv ed
f Tv d εp ij d εp ij (13)
The second law of thermodynamics requires the dissipation increment df !0,but it does not restrict the increment of shift work v j p
v εp ij d εp
ij ,which can be positive or negative.During the process of loading,if the stress state is only elastic,or the current stress state is in elastic unloading,the deformation is elastic and there is no energy dissipation.Then,the free energy is elastic free energy,and its relation with the stress components is described by Eq.(12a).If an irreversible deformation occurs in the current stress state,the relationships between the stress components and free energy as well as dissipation potential should be described by Eqs.(12a)and (12b)simultaneously,and then the dissipation increment df is always larger than or equal to zero.
For multi-yield-surface elasto-plastic problem,the total strain and total strain increment are given by:
εij ?εe ij tX
n k ?1εp k ij ;d εij ?d εe ij tX n k ?1d εp k
ij (14)where εe ij and d εe ij are elastic strain and elastic strain increment,respectively;εp k ij and d εp k ij (k ?1,2,…,n representing the k -th
yield surface)are inelastic strain and inelastic strain increment,respectively.
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4.Elasto-plastic constitutive model of frozen saline sandy soil
For the elasto-plastic constitutive relation of frozen saline soil,according to the continuum mechanics theory,the total strain of frozen saline soil can be decomposed into elastic strain and plastic strain,such as Eq.(14).For the deformation of geomaterial,the volumetric and shear strain components are considered.The thermodynamics results in the principal stress space in Section 3can be deduced to those in the p e q plane.In Section 4.1,the determination methods of elastic deformation law and elastic parameters are investigated by isotropic compression loading-unloading-reloading tests and triaxial shear loading-unloading-reloading tests for frozen saline sandy soil.According to the concept of critical state line,the characteristic of critical state of frozen saline sandy soil is studied in the Section 4.2.
The deformation behaviors of soils are usually identi ?ed by two basic plastic ?ow mechanisms:one is related to plastic shear,and the other to plastic volumetric compression (Xie and Shao,2006).In the Section 4.3,the double yield surface constitutive model is used to study the plastic volumetric compression and shear mechanisms.In the following section,the stress state and deformation law of the frozen soil will be described by the variables in the p e q plane,i.e.,the generalized stresses p and q and the corresponding generalized strains εv ;εg ;εe v ;εe g ;εp v and εp g .
4.1.Elastic deformation
In the sample preparation,water and salt were uniformly distributed in the soils,so the samples of salt soil are uniform.The soil specimens were placed in a refrigeration unit and frozen quickly at a temperature below à30 C,to ensure the uniform distribution of ice crystals and salt in the frozen soils.Under the condition of small elastic deformation,in this paper,in order to reduce the model parameters,the isotropic elastic stress e strain relationship is used in the proposed model by referencing the general practice of the anisotropic elastic e plastic constitutive model for rotational yield surface in the present studies (Wheeler et al.,2003;Yin and Chang,2010;Liu et al.,2013).Based on the test results of isotropic compression loading-unloading-reloading tests,triaxial shear loading-unloading-reloading tests and conventional triaxial tests,we studied elasto-plastic deformation mechanism of the frozen saline sandy soil under the loading condition of constant axial strain rate.In the part of elastic deformation,there is only elastic free energy j e eεe Tproduced within materials,so that the dissipation function does not exist in the elastic deformation,which has been described in the Section 3.By using the related test results and combining with elastic free energy function,the elastic strain increment and the elastic strain can be calculated.For the elasto-plastic decoupling material,a purely elastic deformation would not produce dissipation.
The elastic part of constitutive law Eq.(12a)only relates to elastic free energy,and the following differential expression can be given:
d s ij ?v 2j
e v εe ij v εe kl d εe kl (15a)
Under the hypothesis that the elastic behavior is isotropic and independent on the third invariant,in p e q plane of general stress space,Eq.(15a)can be rewritten as:
dp ?v 2j e
v àεe v ád εe v and dq ?v 2j e v εe g d εe g (15b)
As long as the formulation of elastic free energy in unit volume expression j e (εe )is obtained,based on Eq.(15b),the stress e strain relationship of material can be calculated from elastic free energy.In other words,the free energy also char-acterizes the elastic parameters.In p e q plane,the two main elastic parameters,bulk modulus K and shear modulus G ,can be obtained by test methods.In general,the bulk modulus K can be determined by the isotropic loading-unloading-reloading compression test,and the shear modulus G by the loading-unloading -reloading triaxial shear test.
(1)The determination of bulk modulus K
In the study of nonlinear elastic model,according to the Domaschuk e Villiappan method of K e G model (Li,2004),the bulk modulus can be obtained by the isotropic compression test (q ?0),the bulk modulus is obtained by K ?dp /d εv ,then K ?p a d (p /p a )/d εv ?pd [ln (p /p a )]/d εv .p a is atmospheric pressure.In the isotropic compression tests with loading e unloading e reloading cycles,?ve different stress levels were used to investigate the in ?uence of con ?ning pressures (or stress levels)on bulk modulus of frozen sandy soil.Based on the research by Lai et al.(2014),in isotropic compression tests the loading and unloading rate of 0.1MPa/min was used.When the con ?ning pressure reaches the preset value,it begins to unloading to 0.5MPa,and reloading at the next set con ?ning pressure value at the same loading rate after unloading.Based on the results of loading-unloading-reloading isotropic compressive tests under different pressures,the ln (p /p a )àεv curve can be described in Fig.2.From Fig.2,the closed-hysteresis loops can be seen clearly.Supposing k v ?d [ln (p /p a )]/d εv ,its physical meaning is that k v is chosen as the slope of the unloading curve or the connecting line of the hysteretic loop ends,that is to say,it is the ratio K /
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p ,as shown in Fig.2.The deformation corresponding to unloading and reloading process in closed-hysteresis loops can be seen as elastic deformation when the volumetric strain is smaller than 0.5%,i.e.,the total strain is taken as the elastic strain,D εv ?D εe v .From each k v and the unloading compression pressure,the bulk modulus K can be obtained by:
K ?p D ?ln ep =p a T =D εe v ?p ?k v (16)By averaging the results of several tests,the bulk modulus K is determined by the test results.During each conventional triaxial test,the con ?ning pressure s 3is expressed by s c and treated as a constant.According to the results of isotropic compression test (at this situation,q ?0,the hydrostatic pressure p is equal to s c ),the relationship between the bulk modulus K and con ?ning pressure s c ,shown in Fig.3,can be expressed by
K ?K p p a exp n s c p a !(17)
where p a is atmospheric pressure,p a ?0.10133MPa,and K p and n are material parameters relating to salt contents,K p ?1441.7exp (0.5161S );n ?à0.008S 2t0.006S t0.0301,in which S %is the mass fraction of anhydrous Na 2SO 4.
In the process of conventional triaxial loading test,for a speci ?c compression pressure s c and in interval D p ,the bulk modulus K given by Eq.(17)is approximately taken as constant.When Eq.(17)is substituted into K ?dp =d εe v ,the elastic
volumetric strain increment d εe v can be given by:
d ε
e v ?dp K p p a exp n s c p a !(18)
From Eqs.(15b)and (18),we can obtain:
Volumetric strain ε / %
l n (p /)Fig.2.Isotropic compression loading e unloading-reloading test curve when salt content
2.5%.
Fig.3.Bulk modulus changes of frozen saline sandy soil with different salt contents versus the con ?ning pressures.
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v 2j e v àεe v á2?K p p a exp n s c p a !(19)
(2)The determination of shear modulus G
In order to obtain the shear modulus,the loading-unloading-reloading triaxial shear tests under different con ?ning pressures were carried out.In the process of test,loading,unloading and reloading rates were equal to the loading rate of
conventional triaxial tests _ε
a ?1:67?10à4s à1which is mentioned above.When the shear stress reached 70%shear strength of the samples in the process of loading,the unloading was applied until the shear stress was equal to 0.3MPa.After that time,another reloading was carried out.And the q àεg curve can be obtained by the loading-unloading-reloading triaxial shear tests,shown in Fig.4.Because the unloading deformation is elastic,the shear modulus G can be de ?ned by G ?dq =e3d εe g T.
Fifteen loading-unloading-reloading triaxial shear tests under ?ve group con ?ning pressures were carried out.From each hysteretic loop,the shear modulus value can be got.Finally,the shear modulus G under the different con ?ning pressures can be obtained by calculating the average values under each con ?ning pressure.The relationships between shear modulus G and con ?ning pressure s c are shown in Fig.5and the relational function of G às c is expressed by:
G ?p a a g ln s c a tb g !(20)
By ?tting the G às c curve,the material parameters a g and b g can be obtained as:a g ?à260.78S 2t1151.5S à311.08;b g ?675.63S 2à2377.27S à5010.03,where S %is the mass percent of anhydrous Na 2SO 4.
From the de ?nition expression of G and Eq.(20),the elastic shear strain increment d εe g can be obtained by:
d ε
e g ?dq 3p a a g ln s c p a tb g !(21)
From Eqs.(15b)and (21),we can obtain:
v 2j e v εe g ?3p a a g ln s c a tb g !(22)
By integrating Eqs.(19)and (22)together with the condition j e j t ?0?0and j e j εe ?0?0(the reason is that at initial time the elastic strain is equal to zero),under the hypothesis that s c is seen to be a constant for every con ?ning pressure,the elastic free energy in unit volume j e can be calculated as follows:
j e ?12K p p a exp n s c p a !àεe v á2t32p a a g ln s c p a tb g ! εe g 2(23)
where K p ,n ,a g and b g are material parameters related to salt content S .
D e v i a t o r s t r e s s q / M P a Shear strain ε / %Fig.4.Triaxial shear loading-unloading-reloading test curve.
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4.2.Critical state line
For the elastio-plastic model,the response of a material to loads is eventually tended towards an ultimate condition in which plastic shearing could continue inde ?nitely without changes in volume or effective stresses.This condition of perfect plasticity has become known as a critical state (David,1990).For the soils and other granular materials,if continuously distorted until they ?ow as a frictional ?uid,the CSL is determined by q f ?Mp (Roscoe and Burland,1968)in critical state,where M and q f are critical stress ratio and critical generalized shear stress,respectively.During the process of loading,the stress ratio h does not exceed the critical stress ratio M ,i.e.h ?q /p M (Scho ?eld and Wroth,1968).Because there are tensile properties of frozen saline soil,the CSL of frozen saline soil is different from that of other nature soils.Since frozen saline sandy soil contains water and salt,the non-adhesive characteristics of the sandy soil will be changed by the bonding of ice,salt crystal and soil particles at low temperature.Nguyen et al.(2014)proposed that the CSL of the cemented clay eventually reduced to an asymptote coinciding with the CSL of natural clay.
Fig.6illustrates that the CSLs of frozen saline sandy soil,obtained by conventional triaxial compression tests,are nonlinear curves and approximately straight lines under low con ?ning pressures,and the slope of the straight line is called as initial critical stress ratio M 0.Since there pressure melting exists,the adhesive force of frozen saline sandy soil and the stress ratio decrease gradually,and the CSL is gradually bended downward with the increase of hydrostatic pressure.Therefore,if the pressure melting is not considered,the critical state line of frozen saline soil is the same as that of cemented clay.
The CSLs are approximately linear under low con ?ning pressure,but the slopes of the CSLs are similar for frozen saline sandy soil with different salt contents.If the con ?ning pressures exceed a certain value,the effect of salinity on its CSL is signi ?cant.Based on the Critical State Soil Mechanics and Modi ?ed Cam Clay model (MCC),the CSL variation law of frozen saline sandy soil is studied by mean effective stress degradation equations.For conventional triaxial test,the CSLs are shown in Fig.6,and the mean effective stress is de ?ned by:
p *?p tp t
(24)
Fig.5.Shear modulus changes versus the con ?ning pressures for frozen saline sandy soil with different salt
contents.
Fig.6.Test values and ?tting curves of CSL for conventional triaxial test by different salt contents.
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p t ?C l 0 1àp l exp p l (25)
where p t is a function of hydrostatic pressure,describing the increment of mean effective stress;C l is a intercept of CSL in p e q plane.It stands for the cohesion between soil particles due to ice and salt crystal precipitation during freezing process (Nguyen et al.,2014).M 0is the initial critical stress ratio,describing the critical stress ratio of initial stage loading in the conventional triaxial compression test.Besides,the parameter M 0is a function of temperatures and salt contents for frozen saline soil.The parameter B l is related to pressure melting characteristics,which re ?ects q increase slowly,invariability or decrease with the increase of hydrostatic pressure p in CSL,respectively.
Based on the Modi ?ed Cam Clay model (MCC)(Roscoe and Burland,1968),the CSL envelope function of frozen saline sandy soil can be expressed by q f ?M 0p *?M 0(p tp t ),and the following formula can be obtained:
q f ?M 0p tC l 1àp B l exp p B l (26)
In p e q plane,the CSL is strength envelope line.If C l ?0,it means that there is no cohesion between the soil particles,the strength envelope line is changed into critical state line of Modi ?ed Cam Clay model (MCC).The CSLs,shown in Fig.6,are obtained from the results of conventional triaxial compression test,and their ?tting parameters are given in Table 2.
4.3.The plastic mechanism for frozen saline sandy soil
Generally,the plastic deformation behavior of geomaterial can be identi ?ed by two basic plastic ?ow mechanisms:plastic volumetric compression mechanism and plastic shearing mechanism (Peric and Ayari,2002;Xie and Shao,2006).In this study,we use k ?1and k ?2to express the plastic volumetric compression mechanism and plastic shearing mechanism for frozen saline sandy soil,respectively.Under plastic volumetric compression mechanism,the plastic volumetric deformation is calculated by the hydrostatic pressure in the principal stress space or/and by the mean normal stress in p e q plane,respectively.However,under plastic shear mechanism,the plastic shear deformation is re ?ected by the deviator stress in principal stress space or/and general shear stress in p e q plane,respectively.Therefore,the compression and shear yield mechanisms are used in the plastic analyses.
(1)The plastic deformation of volumetric compressive mechanism
The plastic energy dissipation in unit volume is related to plastic strain,based on the thermodynamics theory,referring to the dissipation increment function of anisotropy model presented by Collins and Kelly (2002),it for frozen saline sandy soil can be improved as:
df 1? A d εp 1v ta d εp 1g 2tB d εp 1g 2!1=2(27)
where A and B are functions of hydrostatic pressure variable (i.e.p and p 0).The parameter p 0is the hydrostatic pressure of the intersection between yield surface rotational axis and yield surface in p e q plane shown in Fig.7(b).The parameter a is rotational angle of yield surface and anticlockwise rotation is de ?ned as positive.
The proposed model,in Eq.(26),is changed into Modi ?ed Cam Clay model (MCC)when C l is equal to zero and the cor-responding yield surface schematic diagram is shown in Fig.7(a).The parameter p x ,shown in Fig.7(a),is hydrostatic pressure value of intersection between the initial yield surface and MCC.Then the expression r ?p 0/p x ?2can be got,where the parameter r is called as spacing ratio (Yu,1998).Generally speaking,different materials have different values of the parameter r .From the test results of frozen saline sandy soil,it can be found that the value of parameter r ranges from 1.0to 2.0.
When the loading condition is isotropic compression for geomaterials,Collins and Hilder (2002)proposed that the dissipation incremental function of volumetric plastic deformation,based on Modi ?ed Cam Clay model (MCC),can be expressed as df ??e1à2=r Tp tp x d εp v .In this paper,considering the characteristic of pressure melting and initial anisot-ropy,taking modi ?ed mean effective stress p *as hydrostatic pressure,and assuming A ??e1à2=r Tp *tp 0x 2and B ?eM 0p 0x T2,Table 2
The relationship between CSL parameters and salt contents.
Salt (S%)
0.0%0.5% 1.5% 2.5%M 0
1.0934 1.2032 1.2268 1.2777B l
18.11215.41917.86919.570C l 5.6140 4.8022 5.5427 5.7303
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where p 0x ?p *0=r ,the schematic diagram of initial anisotropy yield surface is shown in Fig.7(b).Then,df 1!0and the modi ?ed
consolidation pressure is expressed by:
p *0?p 0tp t
(28)
From Eq.(27),the dissipation stresses can be given by the following formulae:p 0?v edf 1Tv àd εp 1v
á?A àd εp 1v ta d εp 1g ádf 1(29a)t 0?v edf 1Tv àd εp 1g á?A àd εp 1v ta d εp 1g áa tBd εp 1g df 1(29b)
Under the plastic volumetric compression mechanism,the plastic free energy increment in unit volume can be expressed as:
d j p 1?p 0x
d εp 1v ta d εp 1g (30)
From Eqs.(29)and (30),in the true stress space we have:p *?v j p 1p 1v tv edf 1Tv àd εp 1v á?p 0x tA àd εp 1v ta d εp 1g á1(31a)
q ?v j p 1v εp 1g tv edf 1Tv àd εp 1g
á?a p 0x tA a àd εp 1v ta d εp 1g átBd εp 1g 1(31b)
From Eq.(31a),we can obtain:p 0?p *àp 0x ?A àd εp 1v ta d εp 1g ádf 1(32)
From Eqs.(29)and (31),we have:
t 0àp 0a ?q àa p *?Bd εp 1
g df 1(33)
Substituting Eq.(27)into Eqs.(32)and (33),according to the determination method of yield condition proposed by Collins and Hilder (2002),the yield condition in dissipative stress space can be given
by:
Fig.7.Yield surface schematic diagrams:(a)Modi ?ed Cam Clay model,(b)The proposed model of this paper for initial anisotropy.
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et 0àp 0a T2B tep 0T2
A ?1(34a)
Eq.(34a)is yield surface function in dissipative stress space.If a ?0,the yield surface function is expressed by a standard elliptical yield surface in dissipative stress space (p 0àt 0),otherwise a sloping ellipse yield surface whose rotational angle is a in dissipative stress space.Substituting Eqs.(32)and (33),A ??e1à2=r Tp *tp 0x 2and B ?eM 0p 0x T2(where p 0x ?p *0=r )into Eq.(34a),the yield function of anisotropic rotational hardening in true stress space can be obtained by: q àa p *
00x 2t p *àp 0x e1à2=r Tp *tp 0x !2?1(34b)
Using the method of determining the CSL envelope,the yield surface of frozen saline sandy soil can be obtained by letting r be equal to 2in Eq.(34b).Thus,the initial yield surface function is de ?ned by:
f 1initial ?àq àa p *á2àM 20p *àp *0àp *á?0(35)
Supposing that the hardening parameter is p *m ,in the initial yield surface p *m ?p *0,thus,the loading yield surface function
can be obtained as follows:
f 1?àq àa p *á2àM 20p *àp *m àp *á?0(36)
If p t ?0in Eq.(24)and a ?0,then Eq.(36)becomes loading yield surface function of Modi ?ed Cam Clay model (MCC).Referencing the methods proposed by Huang et al.(1981),the plastic ?ow direction vectors can be calculated by subtracting the elastic stain from the total strain based on the conventional triaxial test results,and they are drawn in Fig.8.From Fig.8,it can be found that the plastic ?ow directions of frozen saline sandy soil are not vertical to the yield surface when Eq.(36)is chosen as the yield function (i.e.the curve of r ?2in Fig.8).Therefore,the non-associated ?ow rule is employed to solve the deformation problem of frozen saline sandy soil (for example,the curve of r ?1.5in Fig.8).
By taking plastic potential function similar to yield surface function and substituting p 0x ?p *m =r into Eq.(34b),the
different plastic potential surfaces can be obtained from different spacing ratio values,and the plastic potential function is given by:
g 1?
q àa p *
00x 2t p *àp 0x e1à2=r Tp *tp 0x !2à1?0(37)
Based on a series of test results,the plastic potential function can better describe the plastic ?ow when the spacing ratio r is equal to 1.5in Eq.(37),which is applicable to the non-associated ?ow rule.
Since unloading swelling of isotropic compression tends to a line in double logarithm space,mobile hardening law associating to plastic volumetric strain under plastic compressive mechanism can be obtain
by:
Fig.8.Plastic ?ow directions and potential surfaces of compressive mechanism when salt content 2.5%and s 3?16MPa.
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p *m àεp 1v á?ep 0tp t Texp εp 1v p !(38)
where c p is material parameter.The initial yield surface is determined when completing isotropic consolidation of the sample,
here p *m ?p *0.Because there is not dilatancy in the loading hardening process of plastic volumetric compressive mechanism,
the differential of mobile hardening is expressed as:
dp *m ?p *m p
where <>are Macaulay brackets,which indicates that
Under the plastic volumetric compression mechanism,we suppose that only plastic volumetric strain in ?uences the
rotational angle of yield surface.In other words,the rotational angle a is only a function of εp 1v .During the process of loading,
the rotational angle will gradually increase to a maximum value,but it no longer increases until the specimen failure.Referring to the rotational hardening law proposed by Wheeler et al.(2003),if we consider only the effect of plastic volu-metric strain,the rotational hardening formula can be presented by:
d a ?eu àa Tb
where u and b are maximum yield surface rotational angle and material parameter of rotating speed under the action of loading,respectively.The parameters u ,a ,and a 0are shown in Fig.9and the parameter a 0is initial rotational angle of yield surface,which could be positive or negative.Its value is different for different soils (Wheeler et al.,2003).
In this model,parameters p *m and a are hardening variables under compressive mechanism,which re ?ect the changes of modi ?ed hydrostatic pressure in the processes of loading and rotation of the yield surface,respectively.When a ?u ,the rotational angle of yield surface gets to maximum and the rotational angle increment d a ?0.
(2)The plastic deformation of shear mechanism
For plastic deformation under plastic shear mechanism,when the stress level reaches the initial shear yield surface,materials appear shear plastic ?ow during the process of loading.The plastic dissipation increment function under shear mechanism is assumed as:
df 2? A 0àd εp 2v á2tB 0 d εp 2g 2!1=2(41)
In this study,we suppose the parameter A 0?r (p àr )and B 0
?(M 0r )2,where r is shift stress.Under the plastic shear mechanism,the plastic shear strain increment is much larger than the plastic volumetric strain increment,so df 2!0can be ensured.Thus,the plastic free energy increment in unit volume under plastic shear mechanism is de ?ned
by:Fig.9.Conceptual diagram of yield surface and rotational hardening law.
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d j p 2?r d εp 2g (42)
Like the deduction of Eqs.(29)e (34),the shear yield surface equation in p e q plane can be obtained by:
p àr r t q 0r
2
?1(43a)
From Eq.(43a),the initial shear yield surface can be given by:f 2initial ?q 2tM 20r ep à2r T?0
(43b)If the rotational hardening parameter q only relates to plastic shear strain,we suppose r ?à12C l M 0and substitute it into Eq.(43b),and then,the parabolic yield surface equation is obtained as:
f 2?q 2à12q M 0C l p tC l M 0 ?0(44)
The test results and the ?tted shear rotational yield surface are shown in Fig.10,respectively.The rotational hardening parameter q is a function of plastic shear strain,q ?q eεp 2g T.The rotational hardening increment is given by:
d q ?d q àεp 2g
ád εp 2g d εp 2g (45)
According to the conventional triaxial shear test results of frozen saline sandy soil,the plastic potential function is expressed as:
g 2?q 2à1k s q M 0C l p tC l 0 ?0(46)
where k s is material parameter.
4.4.The constitutive relations of frozen saline sandy soil
The plastic deformation mechanism of frozen saline sandy soil is decomposed into that of the compression mechanism and the shear mechanism.The total strain increment is the sum of the elastic strain increment and the plastic strain increment.From the theory analyses in Sections 4.3,when r ?2and k s ?1,at this situation we have f 1?g 1and f 2?g 2,i.e.,the ?ow law is associated.However,generally speaking,the non-associated ?ow rule is more suitable for geotechnical materials.According to the test result analyses of the frozen saline soil,the stress e strain relationship of frozen saline sandy soil is still not described well by a double yield surface model with the associated ?ow rule,because its deformation forms are dilatancy
p / MPa
Fig.10.Shear rotational yield surface when salt content 2.5%.
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on low con ?ning pressures and strain hardening on high con ?ning pressures.Thus,the ?ow rule of the frozen saline sandy soil should be non-associated,and the spacing ratio r is equal to 1.5.
The ?ow rule for each loading mechanism can be expressed as:
d εp 1ij ?d l 1v g 1v s ij and d εp 2ij ?d l 2v g 2
v s ij (47)
where d l 1and d l 2are plastic multipliers of plastic volumetric compression mechanism and plastic shear mechanism,respectively.
From Eqs.(36)and (44),the consistency conditions of plastic volumetric compression mechanism and plastic shear mechanism can be given by:
df 1?v f 1v s ij d s ij tv f 1v p *m v p *m p 1ij d εp 1ij tv f 1v a v a p 1ij
d εp 1ij ?0(48)df 2?v f 2v s ij d s ij tv f 2v q v q v εp 2ij d εp 2ij ?0(49)
Substituting Eqs.(48)and (49)into Eq.(47),in p e q plane we have:
d l 1?1A 1 v f 1v p dp tv f 1v q dq ;d l 2?1A 2 v f 2v p dp tv f 2v q
dq (50)
where:A 1?à(
v f 1v p *m v p *m v εp 1ij tv f 1v a v a v εp
1ij )T (v g 1v s ij )(51)
A 2?à(v f 2v q v q v εp 2ij )T (v g 2v s ij )(52)
During a general loading history,the plastic volumetric compression mechanism and plastic shear mechanism can be activated either separately or simultaneously.Four distinct constitutive domains can be identi ?ed as follows:
1)If f 1<0,f 2<0,the stress state of the frozen soil sample is fully inside the elastic domain during the process of loading and
only elastic deformation occurs d εij ?d εe ij ,or leads to an elastic unloading due to external conditions.At this situation,no plastic ?ow occurs and therefore d l 1?d l 2?0in Eq.(50),namely d εp ij ?0.
2)If f 1?0,df 1?0but f 2<0or f 2?0,df 2<0.The plastic volumetric compression mechanism is activated while the plastic
shear mechanism does not appear.Then,the plastic multiplier d l 1>0is determined by Eq.(50a)and d l 2?0.At this
situation,the plastic strain increment can be expressed as d εp ij ?d εp 1ij .3)On the contrary,if f 2?0,df 2?0but f 1<0or f 1?0,df 1<0.The plastic shear mechanism is activated while the plastic
volumetric compression mechanism is not generated.Then,the plastic multiplier d l 2>0is determined by Eq.(50b)and
d l 1?0.At this situation,th
e plastic strain increment can be expressed as d εp ij ?d εp 2ij .4)For a general plastic stress state,i
f f 1?0,df 1?0and f 2?0,df 2?0,both of the plastic mechanisms are activated.The plastic
multipliers d l 1>0and d l 2>0can be determined by Eq.(50)(a and b),respectively.At this situation,the plastic strain increment can be expressed as d εp ij ?d εp 1ij td εp 2ij .
Based on the four distinct constitutive domains mentioned above,the detail strain increment expression of Eq.(14)can be deduced by the following methods.
Under the plastic volumetric compression mechanism,from Eq.(36),the plastic parameters in p e q plane can be given by:
v f 1*
?à2a àq àa p *áàM 20p *m t2M 20p *(53)dp *dp ?1àC l M 0p B 2l exp p B l (54)
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v f 1*m
?àM 20p *(55)dp *m dp ?àC l M 0p B 2l
exp p B l (56)v f 1v p ?v f 1v p *dp *dp tv f 1v p *m dp *m dp ?"1àC l
M 0p B 2l
exp p B l #h à2a àq àa p *áàM 20p *m tM 20p *i tM 20p *(57)v f 1?2àq àa p *á(58)v f 1?à2p *àq àa p *á(59)
From plastic potential function Eq.(37),we obtain:v g 1v p *?2àp *àp 0x áe2à2=r Tp 0x ?p 0x te1à2=r Tp *?3à2a àq àa p *áàM 0p 0x á2(60)
v g 1v p 0x ?à2àp *àp 0x áe2à2=r Tp *?p 0x te1à2=r Tp *?3à2àq àa p *á2M 20àp 0x á3(61)
dp 0x ?1dp *m ?à1C l 0p B l
exp p l (62)v g 1v p ?v g 1v p *dp *tv g 1v p 0x dp 0x (63)v g 1v q ?2àq àa p *áàM 0p 0x á2(64)
Substituting Eqs.(39),(40),(55),(59),(63)and (64)into Eq.(51),we can obtain:
A 1?à
M 20p *p *m c p t2eu àa Tb p *àq àa p *á!
The hardening modulus A 1has two hardening parameters εp 1v and a .For plastic volumetric compression mechanism,the
in ?uence of deviator stress is not taken into account,and when the volumetric strain increases gradually,the deformation is strain hardening.Thus,the ?rst two parts of Eq.(65)is no-vanishing.The plastic potential surface corresponding to initial yield surface in p -q plane is shown in Fig.8.If the A 1is vanishing,which means frozen soil under volumetric compression critical state,the plastic volumetric strain increment d εp 1v /0under plastic volumetric compression mechanism.
Substituting Eqs.(53)e (65)into Eqs.(47),(50)and (51),the sub-plastic strain increment under plastic volumetric compression mechanism can be given by:
&d εp 1v d εp 1g '?1A 126664 v g 1v p *dp *dp tv g 1v p 0x dp 0x dp v f 1v p v g 1v p *dp *dp tv g 1v p 0x dp 0x dp v f 1v q v f 1v p v g 1v q v f 1v q v g 1v q
37775&dp dq '(66)
Under plastic shear mechanism,from Eqs.(44)and (46),we obtain:
v f 2v p ?à12q M 0C l (67)
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v f2
?2q(68)
v f2 v q ?à1
2
M0C l
pt
C l
M0
(69)
v g2 v p ?à1
2
k s q M0C l(70)
v g2
v q
?2q(71) Substituting Eqs.(67)e(71)into Eq.(52),we have:
A2?1
M0C l
pt
C l
d q
dεp2g
2q(72)
The expression A2is hardening modulus under plastic shear mechanism.When the deformation characteristic of frozen
soil is transformed from shrinkage to dilatancy,d q
dεp2g
is equal to zero and it is an unstable https://www.sodocs.net/doc/2a16202907.html,ually,the corresponding to point can be avoided in the computer's program calculation by the increment method of the theoretical model.The sample
may be of shrinkage or dilatancy when the value of d q
dεp2g
is larger than zero or smaller than zero,respectively.Substituting Eqs.
(67)e(72)into Eqs.(47),(50)and(52),the plastic strain increment under plastic shear mechanism is given by:
&
dεp2v dεp2g '
?1
A2
2
64
1
k s q2M20C2
l
àqk s q M0C l
àq q M0C l4q2
3
75
&
dp
dq
'
(73)
Finally,substituting Eqs.(18),(21),(66)and(73)into Eq.(14),the following incremental constitutive relation of frozen saline sandy soil in p e q plane can be derived:
&
dεv dεg '
?C11C12
C21C22
!&
dp
dq
'
(74)
where:
C11?1
K
t1
A1
v g1
v p*
dp*
dp
t
v g1
v p0x
dp0x
dp
v f1
v p
t1
A2
1
4
k s q2M20C2
l
(75a)
C12?1
1
v g1
v p*
dp*
t
v g1
v p0x
dp0x
v f1
v q
à1
2
qk s q M0C l(75b)
C21?1
1
v f1
v p
v g1
v q
à1
2
q q M0C l(75c)
C22?1
t1
1
v f1
v q
v g1
v q
t1
2
4q2(75d)
The elastic deformation increment can be calculated by Eq.(18)and Eq.(21).The plastic deformations are dominated by the plastic shear mechanism and plastic volumetric compression mechanism.In this simulation,employing non-associated rules,the plastic volumetric compression mechanism are simulated by the elliptical yield criterion Eq.(36)and the plastic potential function Eq.(37),as well as the hardening law Eqs.(38)and(40),while the plastic shear mechanism are described by the parabolic yield criterion Eq.(44)and the plastic potential function Eq.(46),as well as the hardening law Eq.(45).The total volumetric and shear strain increments of frozen saline sandy soil can be simulated by Eq.(74).
5.Experimental results used for parameter determination and model validation
In the proposed model,there are4parameters in bulk modulus and shear modulus,which are K p,n,a g and b g,relating to salt content S.And there are3material parameters M0,B l and C l in CSL function,which relate to salt content S.Additionally, there are8material parameters in the analyses of hardening laws(i.e.,hardening parameters a0,u,b,c p under plastic volumetric compression mechanism,and hardening parameters w,a s,b s and c s under plastic shear mechanism).From a series of conventional triaxial test results,the relationship between the model parameters and the con?ning pressure can be
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determined.Finally,substituting all the determined parameters into the proposed model,and comparing the test results with the calculated values,the validity of proposed model in this paper can be demonstrated.
5.1.Parameter determination and analyses of proposed model
Among all of the model parameters,the elastic parameters (K p ,n ,a g and b g )and the critical state parameters (M 0,B l and C l )have been determined in the mentioned above,using the determining method of Lai et al.(2010).In this section,there are 8parameters,under the process of plastic compression mechanism and plastic shear mechanism in the proposed model,to be determined by the conventional triaxial test results.
(1)Determining parameters a 0,u ,b and c p under plastic volumetric compression mechanism
If sample is in volumetric strain hardening stage,the parameter u is de ?ned by:
u ?M f àa 0(76)where M f is the stress ratio of intersection between stress path and CSL,called as critical stress ratio.a 0is initial rotational angle of anisotropy,or the slope angle of normal consolidation line (Wheeler et al.,2003;Nakano et al.,2005).Along the loading path,we can obtain M f by the critical state point (p ,q f ).The stress ratio h is expressed as h ?q p tC l =M 0 u ?à3:2?10à5 s 3 p a 2t0:0045 s 3p a t0:4204àa 0;R 2?0:973(77) In this study,the test results of frozen saline sandy soil with salt content 2.5%are taken as an example to analyze the characteristics of plastic ?ow.According to Modi ?ed Cam Clay model shown in Fig.7(a),the plastic ?ow direction goes along p axis and the yield surface inclination is zero when generalized shear stress q is equal to zero.From the theory of Collins and Hilder (2002),the initial inclination of the yield surface is equal to the inclination angle of the normal consolidation line shown in Fig.9.Therefore,the yield initial rotational angel a 0can be determined by the intersection angle between the long axis of initial yield ellipse and p axis.From the plastic ?ow direction shown in Fig.8,the initial rotational angle of yield surface a 0is identi ?ed when generalized shear stress q is equal to zero,and its result is shown in Fig.11. From Fig.11,the relationship between initial rotational angle a 0and con ?ning pressure s 3is obtained as: a 0?à1:231?10à5 s 3 a 2t2:816?10à3s 3a à3:3345;R 2?0:858(78) Material parameter b in Eq.(40)is a physical quantity of rotating speed.Its rotation rule is shown in Fig.12.When the parameter u has been determined,referencing the methods presented by Wheeler et al.(2003),integrating the rotational hardening rule,given by Eq.(40),can obtain the following formula: I n i t i a l r o t a t i o n a l a n g l e α / r a d Confining pressure σ / MPa Fig.11.Initial rotational angles under different con ?ning pressures when salt content 2.5%. https://www.sodocs.net/doc/2a16202907.html,i et al./International Journal of Plasticity 78(2016)84e 113 102 a ?eu àa 0T"1àexp àà b εp 1v á During the process of loading,when a increases to approach maximum a max ?u ,then,we have d a /0(Usually choosing a smaller datum as d a ,we take 10%as the maximum amount of rotation in this study).Substituting the corresponding εp 1v into Eq.(79),the parameter b can be determined.From the triaxial test results,the parameter b under different con ?ning pres-sures when salt content 2.5%is shown in Fig.13,and the b às 3relationship is expressed as: b ?0:0331 s 3 p a 2à7:0627s 3p a t456:2;R 2?0:937(80) Based on the test results,the parameter c p in Eqs.(38)and (39)can be determined.Under plastic volumetric compression mechanism,from Eqs.(24)and (36),the following formula can be derived: p *m ?p tp t t?q àa ep tp t T 2 M 0t (81)The relationship between hardening parameter p *m and plastic volumetric strain εp 1v under different con ?ning pressures is shown in Fig.14when salt content 2.5%.From Fig.14,it is found that the hardening parameter p *m increases exponentially versus the plastic volumetric strain εp 1v . From the hardening law described by Eq.(39)and Fig.14,we have: Volumetric strain εp / % R o t a t i o n a n g l e α / r a d Fig.12.Relationship between rotational angle and volumetric strain with different values of parameter b when salt content 2.5%. β Confining pressure ratio ( σ / p ) Fig.13.Parameter b under different con ?ning pressures with salt content 2.5%https://www.sodocs.net/doc/2a16202907.html,i et al./International Journal of Plasticity 78(2016)84e 113 103 8字模型与飞镖模型模型1：角的8字模型 如图所示，AC 、BD 相交于点O ，连接AD 、BC ． 结论：∠A ＋∠D ＝∠B ＋∠C ． O D C B A 模型分析 证法一： ∵∠AOB 是△AOD 的外角，∴∠A ＋∠D ＝∠AOB ．∵∠AOB 是△BOC 的外角， ∴∠B ＋∠C ＝∠AOB ．∴∠A ＋∠D ＝∠B ＋∠C ． 证法二： ∵∠A ＋∠D ＋∠AOD ＝180°，∴∠A ＋∠D ＝180°－∠AOD ．∵∠B ＋∠C ＋∠BOC ＝180°， ∴∠B ＋∠C ＝180°－∠BOC ．又∵∠AOD ＝∠BOC ，∴∠A ＋∠D ＝∠B ＋∠C ． （1）因为这个图形像数字8，所以我们往往把这个模型称为8字模型． （2）8字模型往往在几何综合题目中推导角度时用到． 模型实例 观察下列图形，计算角度： （1）如图①，∠A ＋∠B ＋∠C ＋∠D ＋∠E ＝________； 图图① F D C B A E E B C D A 图③ 2 1O A B 图④ G F 12 A B E 解法一：利用角的8字模型．如图③，连接CD ．∵∠BOC 是△BOE 的外角， ∴∠B ＋∠E ＝∠BOC ．∵∠BOC 是△COD 的外角，∴∠1＋∠2＝∠BOC ． ∴∠B ＋∠E ＝∠1＋∠2．（角的8字模型），∴∠A ＋∠B ＋∠ACE ＋∠ADB ＋∠E ＝∠A ＋∠ACE ＋∠ADB ＋∠1＋∠2＝∠A ＋∠ACD ＋∠ADC ＝180°． 解法二：如图④，利用三角形外角和定理．∵∠1是△FCE 的外角，∴∠1＝∠C ＋∠E ． ∵∠2是△GBD 的外角，∴∠2＝∠B ＋∠D ． ∴∠A ＋∠B ＋∠C ＋∠D ＋∠E ＝∠A ＋∠1＋∠2＝180°． （2）如图②，∠A ＋∠B ＋∠C ＋∠D ＋∠E ＋∠F ＝________． 图② F D C B A E 312图⑤ P O Q A B F C D 图⑥ 2 1 E D C F O B A （2）解法一： 如图⑤，利用角的8字模型．∵∠AOP 是△AOB 的外角，∴∠A ＋∠B ＝∠AOP ． ∵∠AOP 是△OPQ 的外角，∴∠1＋∠3＝∠AOP ．∴∠A ＋∠B ＝∠1＋∠3．①（角的8字模型），同理可证：∠C ＋∠D ＝∠1＋∠2．② ，∠E ＋∠F ＝∠2＋∠3．③ 由①＋②＋③得：∠A ＋∠B ＋∠C ＋∠D ＋∠E ＋∠F ＝2（∠1＋∠2＋∠3）＝360°． 解法二：利用角的8字模型．如图⑥，连接DE ．∵∠AOE 是△AOB 的外角， ∴∠A ＋∠B ＝∠AOE ．∵∠AOE 是△OED 的外角，∴∠1＋∠2＝∠AOE ． ∴∠A ＋∠B ＝∠1＋∠2．（角的8字模型） ∴∠A ＋∠B ＋∠C ＋∠ADC ＋∠FEB ＋∠F ＝∠1＋∠2＋∠C ＋∠ADC ＋∠FEB ＋∠F ＝360°．（四边形内角和为360°） 练习： 1．（1）如图①，求：∠CAD ＋∠B ＋∠C ＋∠D ＋∠E ＝ ； 图 图① O O E E D D C C B B A A 解：如图，∵∠1=∠B+∠D ，∠2=∠C+∠CAD ， ∴∠CAD+∠B+∠C+∠D+∠E=∠1+∠2+∠E=180°． 故答案为：180° 解法二： 美国常青藤名校的由来 以哈佛、耶鲁为代表的“常青藤联盟”是美国大学中的佼佼者，在美国的3000多所大学中，“常青藤联盟”尽管只是其中的极少数，仍是许多美国学生梦想进入的高等学府。 常青藤盟校(lvy League)是由美国的8所大学和一所学院组成的一个大学联合会。它们是：马萨诸塞州的哈佛大学，康涅狄克州的耶鲁大学，纽约州的哥伦比亚大学，新泽西州的普林斯顿大学，罗德岛的布朗大学，纽约州的康奈尔大学，新罕布什尔州的达特茅斯学院和宾夕法尼亚州的宾夕法尼亚大学。这8所大学都是美国首屈一指的大学，历史悠久，治学严谨，许多著名的科学家、政界要人、商贾巨子都毕业于此。在美国，常青藤学院被作为顶尖名校的代名词。 常青藤盟校的说法来源于上世纪的50年代。上述学校早在19世纪末期就有社会及运动方面的竞赛，盟校的构想酝酿于1956年，各校订立运动竞赛规则时进而订立了常青藤盟校的规章，选出盟校校长、体育主任和一些行政主管，定期聚会讨论各校间共同的有关入学、财务、援助及行政方面的问题。早期的常青藤学院只有哈佛、耶鲁、哥伦比亚和普林斯顿4所大学。4的罗马数字为“IV”，加上一个词尾Y，就成了“IVY”，英文的意思就是常青藤，所以又称为常青藤盟校，后来这4所大学的联合会又扩展到8所，成为现在享有盛誉的常青藤盟校。 这些名校都有严格的入学标准，能够入校就读的学生，自然是品学兼优的好学生。学校很早就去各个高中挑选合适的人选，许多得到全国优秀学生奖并有各种特长的学生都是他们网罗的对象。不过学习成绩并不是学校录取的惟一因素，学生是否具有独立精神并且能否快速适应紧张而有压力的大一新生生活也是他们考虑的重要因素。学生的能力和特长是衡量学生综合素质的重要一关，高中老师的推荐信和评语对于学生的入学也起到重要的作用。学校财力雄厚，招生办公室可以完全根据考生本人的情况录取，而不必顾虑这个学生家庭支付学费的能力，许多家境贫困的优秀子弟因而受益。有钱人家的子女，即使家财万贯，也不能因此被录取。这也许就是常青藤学院历经数百年而保持“常青”的原因。 布朗大学(Brown University) 1754年由浸信会教友所创，现在是私立非教会大学，是全美第七个最古老大学。现有学生7000多人，其中研究生近1500人。 该校治学严谨、学风纯正，各科系的教学和科研素质都极好。学校有很多科研单位，如生物医学中心，计算机中心、地理科学中心、化学研究中心、材料研究实验室、Woods Hole 海洋地理研究所海洋生物实验室、Rhode 1s1and反应堆中心等等。设立研究生课程较多的系有应用数学系、生物和医学系、工程系等，其中数学系海外研究生占研究生名额一半以上。 布朗大学的古书及1800年之前的美国文物收藏十分有名。 哥伦比亚大学(Columbia University) 私立综合性大学，位于纽约市。该校前身是创于1754年的King’s College，独立战争期间一度关闭，1784年改名力哥伦比亚学院，1912年改用现名。 第四章景观模型制作 第一节主要工具的使用方法 —、主要切割材料工具的使用方法 （—）美术刀 美术刀是常用的切割工具，一般的模型材料（纸板，航模板等易切割的材料）都可使用它来进行切割，它能胜任模型制作过程中，从粗糙的加工到惊喜的刻划等工作，是一种简便，结实，有多种用途的刀具。美术刀的道具可以伸缩自如，随时更换刀片;在细部制作时，在塑料板上进行划线，也可切割纸板，聚苯乙烯板等。具体使用时，因根据实际要剪裁的材料来选择刀具，例如，在切割木材时，木材越薄越软，刀具的刀刃也应该越薄。厚的刀刃会使木材变形。 使用方法：先在材料商画好线，用直尺护住要留下的部分，左手按住尺子，要适当用力（保证裁切时尺子不会歪斜），右手捂住美术刀的把柄，先沿划线处用刀尖从划线起点用力划向终点，反复几次，直到要切割的材料被切开。 （二）勾刀 勾刀是切割切割厚度小于10mm的有机玻璃板，ABS工程塑料版及其他塑料板材料的主要工具，也可以在塑料板上做出条纹状机理效果，也是一种美工工具。 使用方法：首先在要裁切的材料上划线，左手用按住尺子，护住要留下的部分，右手握住勾刀把柄，用刀尖沿线轻轻划一下，然后再用力度适中地沿着刚才的划痕反复划几下，直至切割到材料厚度的三分之二左右，再用手轻轻一掰，将其折断，每次勾的深度为0.3mm 左右。 （三）剪刀 模型制作中最常用的有两种刀：一种是直刃剪刀，适于剪裁大中型的纸材，在制作粗模型和剪裁大面积圆形时尤为有用；另外一种是弧形剪刀，适于剪裁薄片状物品和各种带圆形的细部。 （四）钢锯 主要用来切割金属、木质材料和塑料板材。 使用方法：锯材时要注意，起锯的好坏直接影响锯口的质量。为了锯口的凭证和整齐，握住锯柄的手指，应当挤住锯条的侧面，使锯条始终保持在正确的位置上，然后起锯。施力时要轻，往返的过程要短。起锯角度稍小于15°，然后逐渐将锯弓改至水平方向，快钜断时，用力要轻，以免伤到手臂。 （五）线锯 主要用来加工线性不规则的零部件。线锯有金属和竹工架两种，它可以在各种板材上任意锯割弧形。竹工架的制作是选用厚度适中的竹板，在竹板两端钉上小钉，然后将小钉弯折成小勾，再在另一端装上松紧旋钮，将锯丝两头的眼挂在竹板两端即可使用。 使用方法：使用时，先将要割锯的材料上所画的弧线内侧用钻头钻出洞，再将锯丝的一头穿过洞挂在另一段的小钉上，按照所画弧线内侧1左右进行锯割，锯割方向是斜向上下。 二、辅助工具及其使用方法 （一）钻床 是用来给模型打孔的设备。无论是在景观模型、景观模型还是在展示模型中，都会有很多的零部件需要镂空效果时，必须先要打孔。钻孔时，主要是依靠钻头与工件之间的相对运动来完成这个过程的。在具体的钻孔过程中，只有钻头在旋转，而被钻物体是静止不动的。 钻床分台式和立式两种。台式钻床是一种可以放在工台上操作的小型钻床，小巧、灵活，使 O D C B A 图12图E A B C D E F D C B A O O 图12图E A B C D E D C B A H G E F D C B A 第一章 8字模型与飞镖模型 模型1 角的“8”字模型 如图所示，AB 、CD 相交于点O ， 连接AD 、BC 。 结论：∠A+∠D=∠B+∠C 。 模型分析 8字模型往往在几何综合 题目中推导角度时用到。 模型实例 观察下列图形，计算角度： （1）如图①，∠A+∠B+∠C+∠D+∠E= ； （2）如图②，∠A+∠B+∠C+∠D+∠E+∠F= 。 热搜精练 1．（1）如图①，求∠CAD+∠B+∠C+∠D+∠E= ； （2）如图②，求∠CAD+∠B+∠ACE+∠D+∠E= 。 2．如图，求∠A+∠B+∠C+∠D+∠E+∠F+∠G+∠H= 。 D C B A M D C B A O 135E F D C B A 105O O 120 D C B A 模型2 角的飞镖模型 如图所示，有结论： ∠D=∠A+∠B+∠C 。 模型分析 飞镖模型往往在几何综合 题目中推导角度时用到。 模型实例 如图，在四边形ABCD 中，AM 、CM 分别平分∠DAB 和∠DCB ，AM 与CM 交于M 。探究∠AMC 与∠B 、∠D 间的数量关系。 热搜精练 1．如图，求∠A+∠B+∠C+∠D+∠E+∠F= ； 2．如图，求∠A+∠B+∠C+∠D = 。 O D C B A O D C B A O C B A 模型3 边的“8”字模型 如图所示，AC 、BD 相交于点O ，连接AD 、BC 。 结论：AC+BD>AD+BC 。 模型实例 如图，四边形ABCD 的对角线AC 、BD 相交于点O 。 求证：（1）AB+BC+CD+AD>AC+BD ； （2）AB+BC+CD+AD<2AC+2BD. 模型4 边的飞镖模型 如图所示有结论： AB+AC>BD+CD 。 描写常青藤优美句段写常青藤作文散文句子 描写常青藤优美句段写常青藤作文散文句子第1段： 1.睁开朦胧的泪眼，我猛然发觉那株濒临枯萎的常春藤已然绿意青葱，虽然仍旧瘦小，却顽强挣扎，嫩绿的枝条攀附着窗格向着阳光奋力伸展。 2.常春藤是一种常见的植物，我家也种了两盆。可能它对于很多人来说都不足为奇，但是却给我留下了美好的印象。常春藤属于五加科常绿藤本灌木，翠绿的叶子就像火红的枫叶一样，是可爱的小金鱼的尾巴。常春藤的叶子的长约5厘米，小的则约有2厘米，但都是小巧玲珑的，十分可爱。叶子外圈是白色的，中间是翠绿的，好像有人在叶子上涂了一层白色的颜料。从叶子反面看，可以清清楚楚地看见那凸出来的，一根根淡绿色的茎。 3.渴望到森林里探险，清晨，薄薄的轻雾笼罩在树林里，抬头一看，依然是参天古木，绕着树干一直落到地上的常春藤，高高低低的灌木丛在小径旁张牙舞爪。 4.我们就像马蹄莲，永不分开，如青春的常春藤，紧紧缠绕。 5.我喜欢那里的情调，常春藤爬满了整个屋顶，门把手是旧的，但带着旧上海的味道，槐树花和梧桐树那样美到凋谢，这是我的上海，这是爱情的上海。 6.当我离别的时候，却没有你的身影；想轻轻地说声再见，已是人去楼空。顿时，失落和惆怅涌上心头，泪水也不觉悄悄滑落我伫立很久很久，凝望每一条小路，细数每一串脚印，寻找你 的微笑，倾听你的歌声――一阵风吹过，身旁的小树发出窸窸窣窣的声音，像在倾诉，似在安慰。小树长高了，还有它旁边的那棵常春藤，叶子依然翠绿翠绿，一如昨天。我心头不觉一动，哦，这棵常春藤陪伴我几个春秋，今天才惊讶于它的可爱，它的难舍，好似那便是我的生命。我蹲下身去。轻轻地挖起它的一个小芽，带着它回到了故乡，种在了我的窗前。 7.常春藤属于五加科常绿藤本灌木，翠绿的叶子就像火红的枫叶一样，是可爱的小金鱼的尾巴。常春藤的叶子的长约5厘米，小的则约有2厘米，但都是小巧玲珑的，十分可爱。叶子外圈是白色的，中间是翠绿的，好像有人在叶子上涂了一层白色的颜料。从叶子反面看，可以清清楚楚地看见那凸出来的，一根根淡绿色的茎。 8.常春藤是多么朴素，多么不引人注目，但是它的品质是多么的高尚，不畏寒冷。春天，它萌发出嫩绿的新叶；夏天，它郁郁葱葱；秋天，它在瑟瑟的秋风中跳起了欢快的舞蹈；冬天，它毫不畏惧呼呼作响的北风，和雪松做伴常春藤，我心中的绿色精灵。 9.可是对我而言，回头看到的只是雾茫茫的一片，就宛如窗前那株瘦弱的即将枯死的常春藤，毫无生机，早已失去希望。之所以叫常春藤，可能是因为它一年四季都像春天一样碧绿，充满了活力吧。也许，正是因为如此，我才喜欢上了这常春藤。而且，常春藤还有许多作用呢！知道吗？一盆常春藤能消灭8至10平 一、常青藤大学 目录 联盟概述 联盟成员 名称来历 常春藤联盟（The Ivy League）是指美国东北部八所院校组成的体育赛事联盟。这八所院校包括：布朗大学、哥伦比亚大学、康奈尔大学、达特茅斯学院、哈佛大学、宾夕法尼亚大学、普林斯顿大学及耶鲁大学。美国著名的体育联盟还有太平洋十二校联盟（Pacific 12 Conference）和大十联盟（Big Ten Conference）。常春藤联盟的体育水平在美国大学联合会中居中等偏下水平，远不如太平洋十校联盟和大十联盟。 联盟概述 常春藤盟校（Ivy League）指的是由美国东北部地区的八所大学组成的体育赛事联盟（参见NCAA词条）。它们全部是美国一流名校、也是美国产生最多罗德奖学金得主的大学联盟。此外，建校时间长，八所学校中的七所是在英国殖民时期建立的。 美国八所常春藤盟校都是私立大学，和公立大学一样，它们同时接受联邦政府资助和私人捐赠，用于学术研究。由于美国公立大学享有联邦政府的巨额拨款，私立大学的财政支出和研究经费要低于公立大学。 常青藤盟校的说法来源于上世纪的50年代。上述学校早在19世纪末期就有社会及运动方面的竞赛，盟校的构想酝酿于1956年，各校订立运动竞赛规则时进而订立了常青藤盟校的规章，选出盟校校长、体育主任和一些行政主管，定期聚会讨论各校间共同的有关入学、财务、援助及行政方面的问题。早期的常青藤学院只有哈佛、耶鲁、哥伦比亚和普林斯顿4所大学。4的罗马数字为"IV"，加上一个词尾Y，就成了"IVY"，英文的意思就是常青藤，所以又称为常青藤盟校，后来这4所大学的联合会又扩展到8所，成为如今享有盛誉的常青藤盟校。 这些名校都有严格的入学标准，能够入校就读的学生，必须是品学兼优的好学生。学校很早就去各个高中挑选合适的人选，许多得到全国优秀学生奖并有各种特长的学生都是他们网罗的对象。不过学习成绩并不是学校录取的惟一因素，学生是否具有独立精神并且能否快速适应紧张而有压力的大一新生生活也是他们考虑的重要因素。学生的能力和特长是衡量学生综合素质的重要一关，高中老师的推荐信和评语对于学生的入学也起到重要的作用。学校财力雄厚，招生办公室可以完全根据考生本人的情况录取，而不必顾虑这个学生家庭支付学费的能力，许多家境贫困的优秀子弟因而受益。有钱人家的子女，即使家财万贯，也不能因 幻想之旅角色模型制作流程 1.拿到原画后仔细分析角色设定细节，对不清楚的结构、材质细节及角色身高等问题与 原画作者沟通，确定对原画理解准确无误。 2.根据设定，收集材质纹理参考资料。 3.开始进行低模制作。 4.制作过程中注意根据要求严格控制面数（以MAX为例，使用Polygon Counter工具查 看模型面数）。 5.注意关节处的合理布线，充分考虑将来动画时的问题。如有疑问与动作组同事讨论咨 询。 6.由于使用法线贴图技术不能使用对称复制模型，可以直接复制模型，然后根据具体情 况进行移动、放缩、旋转来达到所需效果。 7.完成后，开始分UV。 分UV时应尽量充分利用空间，注意角色不同部位的主次，优先考虑主要部位的贴图（例如脸，前胸以及引人注意的特殊设计），为其安排充分的贴图面积。使用Relax Tool 工具确保UV的合理性避免出现贴图的严重拉伸及反向。 8.低模完成后进入法线贴图制作阶段。 现在我们制作法线贴图的方法基本上有三种分别是： a.在三维软件中直接制作高模，完成后将低模与高模对齐，然后使用软件工具生成法 线贴图。 b.将分好UV的低模Export成OBJ格式文件，导入ZBrush软件。在ZB中添加细节 制作成高模，然后使用Zmapper插件生成法线贴图。 c.在Photoshop中绘制纹理或图案灰度图，然后使用PS的法线贴图插件将灰度图生成 法线贴图。 （具体制作方法参见后面的制作实例） 建议在制作过程中根据实际情况的不同，三种方法结合使用提高工作效率。 9.法线贴图完成后，将其赋予模型，查看法线贴图的效果及一些细小的错误。 10.进入Photoshop,打开之前生成的法线贴图，根据其贴在模型上的效果对法线贴图进行 修整。（例如边缘的一些破损可以使用手指工具进行修补，或者在绿色通道中进行适当的绘制。如需加强某部分法线贴图的凹凸效果可复制该部分进行叠加可以起到加强 https://www.sodocs.net/doc/2a16202907.html, 有意向申请美国大学的学生，大部分听过一个名字，常青藤大学联盟。那么美国常青藤大学盟校到底是怎么一回事，又是由哪些大大学组成的呢？下面为大家介绍一下美国常青藤大学联盟。 立思辰留学360介绍，常青藤盟校(lvy League)是由美国的七所大学和一所学院组成的一个大学联合会。它们是：马萨诸塞州的哈佛大学，康涅狄克州的耶鲁大学，纽约州的哥伦比亚大学，新泽西州的普林斯顿大学，罗德岛的布朗大学，纽约州的康奈尔大学，新罕布什尔州的达特茅斯学院和宾夕法尼亚州的宾夕法尼亚大学。这8所大学都是美国首屈一指的大学，历史悠久，治学严谨，许多著名的科学家、政界要人、商贾巨子都毕业于此。在美国，常青藤学院被作为顶尖名校的代名词。 常青藤由来 立思辰留学介绍，常青藤盟校的说法来源于上世纪的50年代。上述学校早在19世纪末期就有社会及运动方面的竞赛，盟校的构想酝酿于1956年，各校订立运动竞赛规则时进而订立了常青藤盟校的规章，选出盟校校长、体育主任和一些行政主管，定期聚会讨论各校间共同的有关入学、财务、援助及行政方面的问题。早期的常青藤学院只有哈佛、耶鲁、哥伦比亚和普林斯顿4所大学。4的罗马数字为“IV”，加上一个词尾Y，就成了“IVY”，英文的意思就是常青藤，所以又称为常青藤盟校，后来这4所大学的联合会又扩展到8所，成为现在享有盛誉的常青藤盟校。 这些名校都有严格的入学标准，能够入校就读的学生，自然是品学兼优的好学生。学校很早就去各个高中挑选合适的人选，许多得到全国优秀学生奖并有各种特长的学生都是他们网罗的对象。不过学习成绩并不是学校录取的惟一因素，学生是否具有独立精神并且能否快速适应紧张而有压力的大一新生生活也是他们考虑的重要因素。学生的能力和特长是衡量学生综合素质的重要一关，高中老师的推荐信和评语对于学生的入学也起到重要的作用。学校财力雄厚，招生办公室可以完全根据考生本人的情况录取，而不必顾虑这个学生家庭支付学费的能力，许多家境贫困的优秀子弟因而受益。有钱人家的子女，即使家财万贯，也不能因此被录取。这也许就是常青藤学院历经数百年而保持“常青”的原因。 动画精度模型制作与探究 Animation precision model manufacture and inquisition 前言 写作目的：三维动画的制作，首要是制作模型，模型的制作会直接影响到整个动画的最终效果。可以看出精度模型与动画的现状是随着电脑技术的不断发展而不断提高。动画模型走精度化只是时间问题，故精度模型需要研究和探索。 现实意义：动画需要精度模型，它会让动画画面更唯美和华丽。游戏需要精度模型，它会让角色更富个性和激情。广告需要精度模型，它会让物体更真实和吸引。场景需要精度模型，它会让空间更加开阔和雄伟。 研究问题的认识：做好精度模型并不是草草的用基础的初等模型进行加工和细化，对肌肉骨骼，纹理肌理，头发毛发，道具机械等的制作更是需要研究。在制作中对于层、蒙版和空间等概念的理解和深化，及模型拓扑知识与解剖学的链接。模型做的精，做的细，做的和理，还要做的艺术化。所以精度模型的制作与研究是很必要的。 论文的中心论点：对三维动画中精度模型的制作流程，操作方法，实践技巧，概念认知等方向进行论述。 本论 序言：本设计主要应用软件为Zbrsuh4.0。其中人物设计和故事背景都是以全面的讲述日本卡通人设的矩阵组合概念。从模型的基础模型包括整体无分隔方体建模法，Z球浮球及传统Z球建模法（对称模型制作。非对称模型制作），分肢体组合建模法（奇美拉，合成兽），shadow box 建模和机械建模探索。道具模型制作，纹理贴图制作，多次用到ZBURSH的插件，层概念，及笔刷运用技巧。目录： 1 角色构想与场景创作 一初步设计：角色特色，形态，衣装，个性矩阵取样及构想角色的背景 二角色愿望与欲望。材料采集。部件及相关资料收集 三整体构图和各种种类基本创作 2 基本模型拓扑探究和大体模型建制 3 精度模型大致建模方法 一整体无分隔方体建模法 二Z球浮球及传统Z球建模法（对称模型制作。非对称模型制作） 三分肢体组合建模法（奇美拉，合成兽） 四shadow box 建模探索和机械建模 4 制作过程体会与经验：精度细节表现和笔刷研究 5 解剖学，雕塑在数码建模的应用和体现（质量感。重量感。风感。飘逸感） 2019年美国常春藤八所名校排名享有盛名的常春藤盟校现在是什么情况呢？接下来就来为您介绍一下！以下常春藤盟校排名是根据2019年美国最佳大学进行的。接下来我们就来看看各个学校的状态以及真实生活。 完整的常春藤盟校名单包括耶鲁大学、哈佛大学、宾夕法尼亚大学、布朗大学、普林斯顿大学、哥伦比亚大学、达特茅斯学院和康奈尔大学。 同时我们也看看常春藤盟校是怎么样的？也许不是你所想的那样。 2019年Niche排名 3 录取率5% 美国高考分数范围1430-1600 财政援助：“学校选择美国最优秀的学生，想要他们来学校读书。如果你被录取，哈佛会确保你能读得起。如果你选择不去入学的话，那一定不是因为经济方面的原因。”---哈佛大三学生2019年Niche排名 4 录取率6% 美国高考分数范围1420-1600 学生宿舍：“不可思议！忘记那些其他学校的学生宿舍吧。在耶鲁，你可以住在一个豪华套房，它更像是一个公寓。一个公寓有许多人一起住，包括一个公共休息室、洗手间和多个卧室。我再不能要求任何更好的条件了。这个套房很大，很干净，还时常翻修。因为学校的宿舍深受大家喜爱，现在有90%的学生都住在学校！”---耶鲁大二学生 2019年Niche排名 5 录取率7% 美国高考分数范围1400-1590 综合体验：“跟任何其他学校一样，普林斯顿大学有利有弊。这个学校最大的好处也是我选择这个学校的主要原因之一就是它的财政援助体系，任何学生想要完成的计划，它都会提供相应的财政支持。”---普林斯顿大二学生 2019年Niche排名 6 录取率9% 美国高考分数范围1380-1570 自我关心：“如果你喜欢城市的话，宾夕法尼亚大学是个不错的选择。这里对于独立的人来说也是一个好地方，因为在这里你必须学会自己发展。要确保进行一些心理健康的训练，因为这里的人通常会过量工作。如果你努力工作并且玩得很嗨，二者都会使你精疲力尽，所以给自己留出点儿时间休息。”---宾夕法尼亚大一学生 2019年Niche排名7 录取率7% 美国高考分数范围1410-1590 综合体验：“学校的每个人都很关心学生，包括我们的身体状况和学业成绩。在这里，你可以遇到来自世界各地的多种多样的学生。他们在学校进行的安全防范教育让我感觉受到保护。宿舍生活非常精彩，你会感觉跟室友们就像家人一样。总之，能成为学校的一员我觉得很棒，也倍感荣幸！”---哥伦比亚大二学生2019年Niche排名9 录取率9% 美国高考分数范围1370-1570 学术点评：“新的课程培养学术探索能力，在过去的两年中我 我国很多学子都想前往美国的常青藤大学就读于研究生，所以美国常青藤大学研究生申请条件都有哪些? 美国常青藤大学研究生申请条件： 1、高中或本科平均成绩(GPA)高于3.8分，通常最高分是4分，平均分越高越好; 2、学术能力评估测试I(SAT I，阅读+数学)高于1400分，学术能力评估测试II(SAT II，阅读+数学+写作)高于2000分; 3、托福考试成绩100分以上，雅思考试成绩不低于7分; 4、美内国研究生入学考试(GRE)成绩1400分以上，经企管理研究生入学考试(GMAT)成绩700分以上。 大学先修课程(AP)考试成绩并非申请美国大学所必需，但由于大学先修课程考试对于高中生来说有一定的挑战性及难度，美国大学也比较欢迎申请者提交大学先修课程考试的成绩，作为入学参考标准。 有艺术、体育、数学、社区服务等特长者优先考容虑。获得国际竞赛、辩论和科学奖等奖项者优先考虑，有过巴拿马国际发明大赛的得主被破例录取的例子。中国中学生在奥林匹克数、理、化、生物比赛中获奖也有很大帮助。 常春藤八所院校包括：哈佛大学、宾夕法尼亚大学、耶鲁大学、普林斯顿大学、哥伦比亚大学、达特茅斯学院、布朗大学及康奈尔大学。 新常春藤包括：加州大学洛杉矶分校、北卡罗来纳大学、埃默里大学、圣母大学、华盛顿大学圣路易斯分校、波士顿学院、塔夫茨大学、伦斯勒理工学院、卡内基梅隆大学、范德比尔特大学、弗吉尼亚大学、密歇根大学、肯阳学院、罗彻斯特大学、莱斯大学。 纽约大学、戴维森学院、科尔盖特大学、科尔比学院、瑞德大学、鲍登学院、富兰克林欧林工程学院、斯基德莫尔学院、玛卡莱斯特学院、克莱蒙特·麦肯纳学院联盟。 小常春藤包括：威廉姆斯学院、艾姆赫斯特学院、卫斯理大学、斯沃斯莫尔学院、明德学院、鲍登学院、科尔比学院、贝茨学院、汉密尔顿学院、哈弗福德学院等。 1 O D C B A 图1 2图E A B C D E F D C B A O O 图12图E A B C D E D C B A 第一章 8字模型与飞镖模型 模型1 角的“8”字模型 如图所示，AB 、CD 相交于点O ， 连接AD 、BC 。 结论：∠A+∠D=∠B+∠C 。 模型分析 8字模型往往在几何综合 题目中推导角度时用到。 模型实例 观察下列图形，计算角度： （1）如图①，∠A+∠B+∠C+∠D+∠E= ； （2）如图②，∠A+∠B+∠C+∠D+∠E+∠F= 。 热搜精练 1．（1）如图①，求∠CAD+∠B+∠C+∠D+∠E= ； （2）如图②，求∠CAD+∠B+∠ACE+∠D+∠E= 。 2 H G E F D C B A D C B A M D C B A O 135 E F D C B A 2．如图，求∠A+∠B+∠C+∠D+∠E+∠F+∠G+∠H= 。 模型2 角的飞镖模型 如图所示，有结论： ∠D=∠A+∠B+∠C 。 模型分析 飞镖模型往往在几何综合 题目中推导角度时用到。 模型实例 如图，在四边形ABCD 中，AM 、CM 分别平分∠DAB 和 ∠DCB ，AM 与CM 交于M 。探究∠AMC 与∠B 、∠D 间的数量关系。 3 105O O 120 D C B A O D C B A 热搜精练 1．如图，求∠A+∠B+∠C+∠D+∠E+∠F= ； 2．如图，求∠A+∠B+∠C+∠D = 。 模型3 边的“8”字模型 如图所示，AC 、BD 相交于点O ，连接AD 、BC 。 结论：AC+BD>AD+BC 。 2021中考数学易错题飞镖模型8字模型探究试题模型一：角的飞镖模型基础 结论：C + ∠ ∠ = ∠ B + A BDC∠ 解答： ①方法一：延长BD交AC于点E得证 ②方法二：延长CD交AB于点F得证 ③方法三：延长AD到在其延长方向上任取一点为点G得证 总结： ①利用三角形外角的性质证明 模型二：角的8字模型基础结论：D ∠ ∠ = + + C B A∠ ∠ 解答： ①方法一：三角形内角和得证 ②方法二：三角形外角【BOD 】的性质得证总结： ①利用三角形内角和等于 180证明 推出 ②利用三角形外角的性质证明 角的飞镖模型和8字模型进阶 【例1】如图，则= ∠E D B A + C + + ∠ ∠ ∠ + ∠ 解答： ①方法一：飞镖ACD得证 ∠E + D C A B ∠ ∠ = 180 ∠ + + ∠ + ②方法二：8字BECD得证 + ∠ ∠E B A + C D ∠ = + 180 + ∠ ∠ 【例2】如图，则= E ∠F + D C A B ∠ ∠ ∠ + + ∠ ∠ + + 解答：飞镖ABF+飞镖DEC得证 ∠F + ∠ E D B + A C ∠ = ∠ + 210 ∠ ∠ + + 【例3】如图，求= E D ∠F B A + C ∠ + ∠ + ∠ ∠ + ∠ + 解答：8字模型得证 ∠F + ∠ E D A B C + 360 + = ∠ ∠ ∠ + ∠ + 【例4】如图，求= ∠D C A + B ∠ + ∠ + ∠ 解答：连接BD得飞镖BAD+飞镖DBC得证 + ∠D A ∠ C B = + ∠ 220 + ∠ 【例5】如图，求= ∠H G ∠ F + D A C + E B + ∠ + ∠ ∠ + + ∠ + ∠ ∠ 解答：飞镖EHB+飞镖FAC得证 ∠H ∠ + + ∠ G F A B C D E ∠ + + = 360 ∠ ∠ ∠ + + ∠ + 模型三：边的飞镖模型基础 结论：CD + > AC BD AB+ □所需要的设备有：电脑，设计软件AutoCAD，雕刻机，工作台，油漆喷枪等。 □所需要的原材料有：各种厚度的有机玻璃板，各种厚度的PVC板，普通海绵，大孔海绵，背胶纸，各色绒线末，粗鱼线，铜丝电线，0.5mm漆包线，涂料，各色油漆，绒面墙纸，三氯甲烷，干花，发胶，小彩灯等。 □所需要的工具有：美工刀、锯条刀、木工工具、电工工具等。 一、沙盘台子 首先，要将顾客交付持房地产平面布置图和施工图纸研究透，组装部根据平面布置图及沙盘的比例来制作沙盘的台子。台子一般做成台球桌状，如果是大型的沙盘，要做成几个小台子，拼到一起。 二、PVC板喷漆 喷漆部根据楼房图纸的设色调出相应颜色的油漆来，喷在相应的PVC板上，送到设计部进行雕刻。 三、雕刻楼房部件 设计部根据施工图按比例设计出楼房的结构，并在电脑上分解成不同的板块，按施工的要求设计出墙面的花纹、房顶的瓦棱、窗子等，然后发送到雕刻机在PVC板上雕刻出楼房的板块，送到制作部制作。 四、组合楼房 制作部根据设计部送来的楼房板块，根据说明和粘合方式，用三氯甲烷将PVC板块粘合成楼房的大致形状。窗子的形状是直接雕刻在PVC板上的，用薄而透明的有机玻璃板粘在内部窗子的位置作为窗子的玻璃。 五、置景 置景部根据组装部所作的台子和平面布置图，在台子上划分出平面布局，用绿色绒面墙纸作为草地粘在绿化区，大孔海绵浸上绿色油漆晾干，裁成长条作为绿化带粘在小灌木区。如果布局中有水和湖泊，可以用波纹面的有机玻璃板，背面喷湖蓝色漆，裁成河流或湖泊的形状放在相应的位置。若是有高地，可将有机玻璃板或PVC板层层堆积并修整成形，再抹上涂料填充缝隙，晾干后覆上草地。用灰色的背胶纸粘成公路，用白色背胶纸刻成公路线标粘在上面。 六、制作配件 制作部将铜丝电线剥皮，将铜丝拧成树干的形状，喷上漆。普通海绵浸漆，晾干后粉碎，将树干的枝丫浸胶，粘上碎海绵，做成树。若是绿树，海绵可浸绿漆，若是秋天的树，可浸橙色漆。柳树可用0.2mm的漆包线拧成树干与树枝，然后在树枝上粘上绿色绒线末。松树是将粗鱼线剪成细段，用夹子夹住，再将两根0.5mm的漆包线夹住绞动，松开夹子，就成了松树的形状，修剪一下，粘上绿色绒线末即可。其它的花草可以用干花剪下来染色来制作。用医用棉签或牙签做成路灯。泡沫塑料可以用刀片雕刻成假山石的形状，喷上漆。 七、整体组合 置景部将制作部送来的花草树木及楼房按布置粘在相应的地方。组装部根据每栋楼房所在的位置，打孔并装上小彩灯，使楼房模型内部能发光，如同开灯的效果，并接好线路。 留学美国常春藤八大院校 美国常春藤声誉： 几乎所有的常春藤盟校都以苛刻的入学标准著称，近年来尤其如此：在过去的10多年里常春藤盟校的录取率正在下降。很多学校还在 特别的领域内拥有极大的学术声誉，例如： 哥伦比亚大学的法学院、商学院、医学院和新闻学院； 康乃尔大学的酒店管理学院和工程学院； 达特茅斯学院的塔克商学院（TuckSchoolofBusiness）； 哈佛大学的商学院、法学院、医学院、教育学院和肯尼迪政府学院； 宾夕法尼亚大学的沃顿商学院、医学院、护理学院、法学院和教 育学院； 普林斯顿大学的伍德鲁·威尔逊公共与国际事务学院； 耶鲁大学的法学院、艺术学院、音乐学院和医学院； 美国常春藤八大名校【哈佛大学】 哈佛大学（HarvardUniversity）是一所位于美国马萨诸塞州波 士顿剑桥城的私立大学，常春藤盟校成员之一，1636年由马萨诸塞州 殖民地立法机关立案成立。 该机构在1639年3月13日以一名毕业于英格兰剑桥大学的牧师 约翰·哈佛之名，命名为哈佛学院，1780年哈佛学院更名为哈佛大学。直到19世纪，创建了一个半世纪的哈佛学院仍然以英国的牛津大学、 剑桥大学两所大学为模式，以培养牧师、律师和官员为目标，注重人 文学科，学生不能自由选择课程。19世纪初，高等教育课程改革的号 角在哈佛吹响了，崇尚“学术自由”和“讲学自由”。“固定的学年”和“固定的课”的老框框受到冲击，自由选修课程的制度逐渐兴起。 哈佛大学是一所在世界上享有顶尖大学声誉、财富和影响力的学校， 被誉为美国政府的思想库，其商学院案例教学也盛名远播。作为全美 的大学之一，在世界各研究机构的排行榜中，也经常名列世界大学第 一位。 美国常春藤八大名校【耶鲁大学】 耶鲁大学（YaleUniversity）是一所坐落于美国康涅狄格州纽黑 文（纽黑文市CityofNewHaven）的私立大学，创于1701年，初名“大学学院”（CollegiateSchool）。 耶鲁起初是一所教会学校，1718年，英国东印度公司高层官员伊莱休·耶鲁先生向这所教会学校捐赠了9捆总价值562英镑12先令的 货物、417本书以及英王乔治一世的肖像和纹章，在当时对襁褓之中的耶鲁简直是雪中送炭。为了感谢耶鲁先生的捐赠，学校正式更名为 “耶鲁学院”，它就是今日耶鲁大学的前身。18世纪30年代至80年代，耶鲁在伯克利主教、斯泰尔斯牧师、波特校长等的不懈努力下， 逐渐由学院发展为大学。至20世纪初，随着美国教育的迅猛发展，耶 鲁大学已经发展到了惊人的规模，在世界的影响力也达到了新的高度。耶鲁大学是美国历建立的第三所大学（第一所是哈佛大学，第二所是 威廉玛丽学院），该校教授阵容、学术创新、课程设置和场馆设施等 方面堪称一流，与哈佛大学、普林斯顿大学齐名，历年来共同角逐美 国大学和研究生院前三的位置。哈佛大学注重闻名于研究生教育，威 廉玛丽学院闻名于本科生教育，耶鲁则是双脚走路，都非常，在世界 大学排名中名列前茅。 美国常春藤八大名校【宾夕法尼亚】 宾夕法尼亚大学(UniversityofPennsylvania)是一所私立大学， 是在美国开国元勋本杰明·富兰克林的倡导下于1740年建立起来的。 它是美国东北部常春藤大学之一，坐落于合众国的摇篮——费城，独 建筑制作项目流程 1、制作前期策划 根据甲方提供的平面图、立面图、效果图及模型要求，制定模型制作风格。 2、模型报价预算 预算员根据[1]、模型比例大小、材料工艺及图纸深度确定模型收费、签订制作服务订单。 3、制作组织会审 技术人员将核对分析图纸，确定模型材质、处理工艺、制作工期及效果要求。 (1)建筑制作进程： 建筑制作师根据甲方提供的图纸施工制作，效果以真实、美观为原则。所有建筑均采用AutoCAD绘图，电脑雕刻机切割细部、建筑技师手工粘接的流水线作业法，既保证了各部件的质量又保证了工期。 (2)环境景观设计制作进程： 总体环境将由专业景观设计师进行把控。专业制作人员结合图纸进行设计制作。原则是根据甲方的设计图纸再现设计师的设计意图。切不可胡乱操作，自由发挥。同时使用仿真树木、小品、雕塑等进行点缀，使得整个景观部分美观精致。 (3)建筑环境灯光组装： 灯光系统根据甲方要求进行设计制作，体现沙盘的夜景效果。 4、制作完工检验 质检部经理及项目负责人对照图纸，进行细部检查和调整。 5、模型安装调试 模型服务人员在模型展示地现场调试安装清洁，达到甲方满意后离开。 编辑本段 建筑模型分类 黏土模型 黏土材料来源广泛取材方便价格低廉经过“洗泥”工序和“炼熟过程其质地更加细腻。黏土具有一定的粘合性可塑性极强在塑造过程中可以反复修改任意调整修刮填，补比较方便。还可以重复使用是一种比较理想的造型材料,但是如果黏土中的水分失去过多则容易使黏土模型出现收缩龟裂甚至产生断裂现象不利于长期保存。另外，在黏土模型表面上进行效果处理的方法也不是很多,黏土制作模型时一定要选用含沙量少，在使用前要反复加工，把泥和熟，使用起来才方便。一般作为雕塑、翻模用泥使用。 油泥模型 油泥是一种人造材料。凝固后极软，较软，坚硬。油泥可塑性强，黏性、韧性比黄泥（黏土模型）强。它在塑造时使用方便，成型过程中可随意雕塑、修整，成型后不易干裂，可反复使用。油泥价格较高，易于携带，制作一些小巧、异型和曲面较多的造型更为合适。一般像车类、船类造型用油泥极为方便。所以选用褐油泥作为油泥的最外层是很明智的选择。油泥的材料主要成分有滑石粉62%、凡士林30%、工业用蜡8%。 石膏模型 石膏价格经济，方便使用加工，用于陶瓷、塑料、模型制作等方面。石膏质地细腻，成型后易于表面装饰加工的修补，易于长期保存，适用于制作各种要求的模型，便于陈列展示。 塑料模型 塑料是一种常用制作模型的新材料。塑料品种很多，主要品种有五十多种，制作模型应 美国常青藤名校的由来 美国常青藤名校的由来 以哈佛、耶鲁为代表的“常青藤联盟”是美国大学中的佼佼者，在美国的3000多所大学中，“常青藤联盟”尽管只是其中的极少数，仍是许多美国学生梦想进入的高等学府。 常青藤盟校(lvy League)是由美国的8 所大学和一所学院组成的一个大学联合会。它们是：马萨诸塞州的哈佛大学，康涅狄克州的耶鲁大学，纽约州的哥伦比亚大学，新泽西州的普林斯顿大学，罗德岛的布朗大学，纽约州的康奈尔大学，新罕布什尔州的达特茅斯学院和宾夕法尼亚州的宾夕法尼亚大学。这8所大学都是美国首屈一指的大学，历史悠久，治学严谨，许多著名的科学家、政界要人、商贾巨子都毕业于此。在美国，常青藤学院被作为顶尖名校的代名词。 常青藤盟校的说法来源于上世纪的50年代。上述学校早在19世纪末期就有社会及运动方面的竞赛，盟校的构想酝酿于1956年，各校订立运动竞赛规则时进而订立了常青藤盟校 的规章，选出盟校校长、体育主任和一些行政主管，定期聚会讨论各校间共同的有关入学、财务、援助及行政方面的问题。早期的常青藤学院只有 哈佛、耶鲁、哥伦比亚和普林斯顿4所大学。4的罗马数字为“IV”，加上一个词尾Y，就成了“IVY”，英文的意思就是常青藤，所以又称为常青藤盟校，后来这4所大学的联合会又扩展到8所，成为现在享有盛誉的常青藤盟校。 这些名校都有严格的入学标准，能够入校就读的学生，自然是品学兼优的好学生。学校很早就去各个高中挑选合适的人选，许多得到全国优秀学生奖并有各种特长的学生都是他们网罗的对象。不过学习成绩并不是学校录取的惟一因素，学生是否具有独立精神并且能否快速适应紧张而有压力的大一新生生活也是他们考虑的重要因素。学生的能力和特长是衡量学生综合素质的重要一关，高中老师的推荐信和评语对于学生的入学也起到重要的作用。学校财力雄厚，招生办公室可以完全根据考生本人的情况录取，而不必顾虑这个学生家庭支付学费的能力，许多家境贫困的优秀子弟因而受益。有钱人家的子女，即使家财万贯，也不能因此被录取。这也许就是常青藤学院历经数百年而保持“常青”的原因。 布朗大学(Brown University) 课题： 5-4建筑模型制作步骤 教学目标：（结合岗位知识、能力、素质目标确定） 1.通过讲授建筑模型的具体制作步骤，让学生了解建筑模型制作过程中各环节所需要具备的能力，培养学生的模型制作技能。 教学重难点分析： 1.教学重点： （1）建筑模型制作步骤 （2）建筑模型制作技能点 2.教学难点： 通过课程讲解，加深对建筑模型成型特点、以及技术介绍，并让学生熟悉建筑模型分类及其作用，培养学生对建筑模型类别和作用的掌握。 教学过程： 1.导课： 由多媒体PPT展示建筑模型制作过程的图片，从建筑模型图片引入课程，全方位观察建筑模型的制作全过程，导入课堂新课程——建筑模型制作步骤。 2.教学内容： （1）建筑模型制作步骤： 1.绘制建筑模型的工艺图： 首先确定建筑模型的比例尺寸，然后按比例绘制出制作建筑模型所需要的平面图和立面图。 2.排料画线： 将制作模型的图纸码放在已经选好的板材上，仵图纸和板材之间夹一张复印纸，然后用双面胶条固定好图纸与板材的四角，用转印笔描出各个而板材的切割线。需要注意的是图纸在板材上的排列位置要计算好，这样可以节省板料。 3.加工镂空的部件 制作建筑模型时，有许多部位，比如门窗等是需要进行镂空工艺处理的。可先在相应的部件上用钻头钻好若干小孔，然后穿入钢丝，锯出所需要的形状。锯割时需要留出修整加工的余量。 4.精细加工部件 将切割好的材料部件，夹放在台钳上，根据大小和形状选择相宜的锉刀进行修整。外形相同的部件，或者是镂空花纹相同的部件，可以把若干块夹在一起，同时进行精细的修整加工，这样可以很容易地保证花纹的整齐。 5.部件的装饰 在各个立面黏结前，先将仿镜面幕墙及窗格子处理好，再进行黏结。 6.组合成型 将所有的立面修整完毕后，对照图纸精心地黏结。中考必会几何模型：8字模型与飞镖模型
美国常青藤名校的由来
第四章 景观模型制作
1第一章 8字模型与飞镖模型(1)
新整理描写常青藤优美句段 写常青藤作文散文句子
关于美国常青藤
角色模型制作流程
什么是美国常青藤大学
模型制作方法
2019年美国常春藤八所名校排名
美国常青藤大学研究生申请条件都有哪些
初中数学优质专题：8字模型与飞镖模型
2021中考数学易错题飞镖模型8字模型探究试题
模型的制作工艺及流程
留学美国常春藤八大院校
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美国常青藤名校的由来
最新模型制作教案5-4建筑模型制作步骤教案(精)
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