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粘土和砂土统一的应力路径无关硬化参数本构模型

粘土和砂土统一的应力路径无关硬化参数本构模型
粘土和砂土统一的应力路径无关硬化参数本构模型

A uni?ed constitutive model for both clay and sand with

hardening parameter independent on stress path

Y.P.Yao

a,*

,D.A.Sun b ,H.Matsuoka

c

a Department of Civil Engineering,Beihang University,Beijing 100083,China b

Department of Civil Engineering,Shanghai University,Shanghai 200072,China

c

Department of Civil Engineering,Nagoya Institute of Technology,Nagoya 466-8555,Japan

Received 31August 2006;received in revised form 30March 2007;accepted 30March 2007

Available online 21May 2007

Abstract

A uni?ed constitutive model for both clay and sand under three-dimensional stress conditions is derived from the modi?ed Cam-clay model,by taking the following two points into consideration.First,a transformed stress tensor based on the SMP (spatially mobilized plane)criterion is applied to the Cam-clay model.The proposed model consistently describes shear yielding and shear failure and com-bines critical state theory with the SMP criterion for clay.Secondly,a new hardening parameter,which is independent of the stress path,is derived in order to develop a uni?ed constitutive model for both clay and sand.It not only describes the dilatancy for lightly to heavily dilatant sand,but also reduces to the plastic volumetric strain for clay.The validity of the hardening parameter is con?rmed by the test results of triaxial compression and extension tests on sand under various stress paths.Only ?ve conventional soil parameters are needed in the proposed model.

ó2007Elsevier Ltd.All rights reserved.

Keywords:Clay;Dilatancy;Elastoplastic model;Sand;Stress path;Three-dimensional stress

1.Introduction

The modi?ed Cam-clay model,called the Cam-clay model for short in this paper,which is suitable for clay (in this paper,the term clay is to be interpreted as normally consolidated clay),was proposed by Roscoe and Burland [18].The Cam-clay model and many other models have been generalized by assuming a section of the yield surface to be circular in the p -plane.The mean stress p e?r ii =3Tand deviator stress q e?????????3=2p ?????????

s ij s ij p Tare used as the stress parameters in the models,where s ij e?r ij àp d ij Tis a devia-toric stress tensor and d ij is Kronecker’s delta.That is to say,the criterion of the Extended Mises type (g =q /p =const.)is adopted for the shear yielding and the shear failure of clay in the Cam-clay model.Shear yielding is caused by an increase in stress ratio g =q /p ,while com-pressive yielding is caused by an increase in the mean stress p.However,as experimental evidence shows,the Extended Mises criterion grossly overestimates the strength in triaxial extension,and also results in incorrect intermediate stress ratios in plane strain [22].In contrast to the Extended Mises failure,the SMP failure criterion [8],which is consid-ered to be a three-dimensional extension of the Mohr–Cou-lomb criterion,is a failure criterion that explains the high quality test results for soils.A transformed stress tensor has been proposed by Matsuoka et al.[9]which makes the SMP criterion become circular in the transformed p -plane.A revised transformed stress tensor ~r

ij for the SMP criterion is developed in this paper based on the transformed stress method [9]and the work of Yao and Sun [24,25]in order to take the plastic strain increment direction into account additionally.The new transformed

stress tensor ~r

ij is applied to the Cam-clay model.The pro-posed model consistently describes the shear yielding and shear failure of soils under three-dimensional stresses,both

0266-352X/$-see front matter ó2007Elsevier Ltd.All rights reserved.doi:10.1016/https://www.doczj.com/doc/366953942.html,pgeo.2007.04.003

*

Corresponding author.Tel.:+861082316111.E-mail address:y-pyao@https://www.doczj.com/doc/366953942.html, (Y.P.Yao).

https://www.doczj.com/doc/366953942.html,/locate/compgeo

Available online at https://www.doczj.com/doc/366953942.html,

Computers and Geotechnics 35(2008)

210–222

of which obey the SMP criterion,and combines critical state theory with the SMP criterion for clay.

Traditionally,the plastic volumetric strain e p

v is taken as

the hardening parameter in the Cam-clay model,which is not appropriate for dilatant sand.To date,a lot of harden-ing parameters which can describe the dilatancy of soils have been assumed(e.g.,[5,15,2,11,1,6]),and several dilat-ant plasticity models have been proposed(e.g.,[4,16, 17,27,3,26]).In this paper,a new hardening parameter,that is independent of stress path,is derived by considering uni?ed yield and plastic potential functions which are the same those for Cam-clay.As will be described later,the physical meaning of this hardening parameter is clear. The proposed hardening parameter not only describes the dilatancy from lightly to heavily dilatant sand,but also

reduces to the plastic volumetric strain e p

v for clay.The

validity of the hardening parameter H is con?rmed by tri-axial compression and extension test results on sand under various stress paths.

The ability of the proposed model to predict the drained behavior of normally consolidated clay and saturated sand is examined along various stress paths under triaxial com-pression and extension conditions.The results predicted by the proposed model agree well with the test results.Only ?ve soil parameters are needed in the proposed model. The values of these parameters can be determined by a loading and unloading isotropic compression test and a conventional triaxial compression test.In this paper,the term stress is to be interpreted as e?ective stress.The imple-mentation of the model into the?nite element method fol-lows the methods described in Sheng et al.[20]and Sloan et al.[21].

2.A transformed stress tensor based on the SMP criterion

A transformed stress tensor,by which the SMP criterion is made circular in the transformed p-plane,has been pro-posed by Matsuoka et al.[9].A revised transformed stress tensor~r ij is developed in this paper based on the trans-formed stress method[9]and the work of Yao and Sun [24,25]in order to take the plastic strain increment direc-tion into account additionally.The outline of the new transformed stress tensor~r ij is introduced as follows: The SMP criterion can be written as

I1I2=I3?conste1Twhere I1,I2and I3are the?rst,second and third stress invariants,respectively.The solid curve in Fig.1is the shape of the SMP criterion in the p-plane.The transformed stress is deduced from what makes the SMP curve in the p-plane become a circle,the dot line of Fig.1,with the center being the origin in the transformed p-plane(~p-plane).

When p=const,we assume that~r is equal to r0,where~r is the stress radius in the p-plane of the transformed stress space and r0is the stress radius in the p-plane of the ordin-ary stress space when h=0.From Eq.(1),r0can be expressed by Matsuoka et al.[9]~r?r0?

???

2

3

r

q??

???

2

3

r

2I1

3

???????????????????????????????????????????????

eI1I2àI3T=eI1I2à9I3T

p

à1

e2T

The generalized deviatoric stress in the transformed stress space~qe?

????????

3=2

p

~rTcan be written as

~q?q??

2I1

3

???????????????????????????????????????????????

eI1I2àI3T=eI1I2à9I3T

p

à1

e3T

To realize the transformation from r ij to~r ij,the follow-ing equations should be made:

~p?p

~q?q?

~h?f

1

ehT

8

><

>:e4T

where~p is the mean transformed stress;~h and h are Lode’s stress angles in the transformed stress space and the ordin-ary stress space respectively,which are one-to-one corre-spondence with intermediate principal stress parameters~b and b(=(r2àr3)/(r1àr3))in the transformed stress space and the ordinary stress space respectively;f1(h)is a func-tion of h which also can be expressed by b.So,Eq.(4) can be rewritten as

~p?p

~q?q?

~b?f

2

ebT

8

><

>:e5Twhere

~b?~r2à~r3

~r1à~r3

~p?1

3

e~r1t~r2t~r3T?1

3

~r ii

~q?1??2p

?????????????????????????????????????????????????????????????????????????

e~r1à~r2T2te~r2à~r3T2te~r3à~r1T2

q

?

??????????????????????????????????????????????

3

2

e~r ijà~p d ijTe~r ijà~p d ijT

q

8

>>>

>>>

<

>>>

>>>

:

e6T

f1(h)or f2(b)is a function to transform the?ow direction of the plastic strain increment in the ordinary stress space into the?ow direction in the transformed stress space.By

Y.P.Yao et al./Computers and Geotechnics35(2008)210–222211

f 1(h )or f 2(b ),the deviation of the plastic strain increment direction from the stress direction in the ordinary p -plane can be taken into consideration,although the directions of stress and plastic strain increments in transformed p -plane are coincident.

A reasonable plastic potential surface for frictional materials should be between the curved-triangle shape (I 1I 2/I 3=const)and the circular shape (q /p =const)in the p -plane.So,the ?ow direction of the plastic strain increment is also between the normal to the triangular shape of SMP criterion and that to the circular shape,as shown Fig.1.The following f 2(b )[24]has the above characteristics

f 2eb T??????r 1p t?????r 3p ?????r 2p t?????r 3p b ??????r 2p à?????r 3

p ?????r 1p à?????r 3p e7T

So,the transformed stress tensors under the general stress

state can be obtained as ~r

ij ?p d ij tq ?q s er s

ij àp s d ij Te8T

and

p s ?r s ii

=3q s ??????????????????????????????????????????????????????????e3=2Ter s ij àp s d ij Ter s ij àp s d ij Tq r s

ij ?eI s 1r ik tI s 3d ik Ter kj tI s 3d kj T

à1I s 1??????r 1p t?????r 2p t?????r 3p I s 2??????????r 1r 2p t?????????r 2r 3p t?????????r 3r 1p I s

3???????????????r 1r 2r 3

p 9

>>>>>>>>>>=>>>>>>>>>>;e9T

When the stress tensor r ij is given,the transformed stress

tensor ~r

ij can be calculated from Eq.(8).From the above derivation,it can be shown that the shape of the SMP cri-terion becomes a cone with the axis being the space diago-nal ~r

1?~r 2?~r 3in the transformed principal stress space.Its cross-section is a circle with the center being the origin

O in the ~p

-plane (see Fig.1).Noting the similarity in the shapes of the Extended Mises criterion in the principal stress space and the SMP criterion in the transformed prin-cipal stress space,we can revise existing elastoplastic mod-els such as the Cam-clay model by using the transformed

stress tensor ~r

ij based on the SMP criterion.3.Basic method for constructing a uni?ed hardening

parameter independent of stress path

Hardening of materials is due to the occurrence of plastic strain.Therefore,a parameter representing the degree of hardening must depend on the plastic strain.Hardening parameters are internal variables that are used to indicate the rate of plastic deformation,and should be a function of plastic volumetric strain and plastic shear strain.For a single yield surface model,the hardening parameter should have the following properties:(1)its increments should be the same from one point on a yield surface to di?erent points on another yield surface,and (2)its increment should

also be same from one point to another point along di?erent stress paths.Here,the hardening parameter is di?erent with the hardening modulus which obviously changes from point to point.However,the plastic strain is usually dependent on the stress path.For example,the plastic volumetric strain and plastic shear strain for sand are all dependent on stress path (the test results in the following section can con?rm this characteristic).So,the plastic strain increments instead of the total plastic strains should be used to construct a uni-?ed hardening parameter for various kinds of soils.That is to say,the hardening parameter should be a function of the plastic volumetric strain increment d e p v and the plastic shear strain increment d e p

d .It is assumed that th

e plastic volumet-ric strain increment and plastic shear strain increment can be related through the stress-dilatancy equation

d e p v =d e p

d ?f 1eg T.For example,th

e stress-dilatancy equation in the original Cam-clay model is d e p v =d e p d ?M àg (stress ratio g =q /p ;M is the value o

f

g at the critical state),so that f 1(g )=M àg .Therefore,we can choose either the plastic volumetric strain increment d e p v or the plastic shear strain increment d e p

d as a basic parameter to construct th

e harden-ing parameter.However,plastic shear strain does not occur during isotropic compression,which means that it is impos-sible to describe hardening under isotropic compression by use o

f the plastic shear strain increment.Therefore,we have to choose the plastic volumetric strain increment d e p v to con-struct the hardenin

g parameter.Realizing that the incre-ment of the hardening parameter should be independent of the stress path,we assume there exists a stress function R (g )suc

h that the integral R d e p v

R eg Tis independent of stress path.Thus,this integral can be used as a general hardening parameter

H ?Z

d e p v

R eg Te10T

In the Cam-clay model,R (g )=1.A suitable R (g )for both clay and sand will be introduced in the next section.4.A uni?ed hardening parameter for both clay and sand A basic method for constructing a uni?ed hardening parameter independent of stress path has been introduced in the preceding section.We will now derive a new harden-ing parameter for both clay and sand in this section.The modi?ed Cam-clay model is considered to be one of the best basic elastoplastic models for clay,which is better than the original one [19].In the modi?ed Cam-clay model,the yield and plastic potential functions are assumed to be of the same form as follows:

f ?

g ?ln p p 0tln 1tq 2M p

2 àZ d e p v

c p ?0e11T

where p 0is the initial mean stress,d e p v e?d e p

ii Tis the plastic volumetric strain increment and c p is written as

c p ?

k àj 1te 0

e12T

212Y.P.Yao et al./Computers and Geotechnics 35(2008)210–222

with k being the compression index,j being the swelling in-dex,and e 0the initial void ratio at p =p 0.It is reasonable that Eq.(11)is also chosen as the plastic potential function for dilatant sand,because this equation predicts that the plastic volumetric strain increment is positive before the characteristic state (g =M )[7]and negative after the char-acteristic state (see Fig.2).To adopt an associated ?ow rule in new model,a yield function similar to Eq.(11)is also as-sumed for sand.However,the plastic volumetric strain cannot be used as the hardening parameter for sand be-cause it is dependent on the stress path and does not in-crease monotonically with loading.In this paper,a new hardening parameter H is derived to describe the hardening behavior of clay and sand.The yield and plastic potential functions for sand are written as

f ?

g ?ln p p 0tln 1tq 2

M 2p 2 àH ?0e13T

Nova and Wood [14,15]have chosen a combination of

the plastic volumetric strain and plastic deviator strain as a hardening parameter.In the present model,the harden-ing parameter is considered to be a combination of the stress tensor r ij and plastic strain increment tensor d e p ij ,i.e.,a plastic work type of hardening parameter.Thus,the following hardening parameter is assumed by using the plastic strain invariants

H ?Z d H ?Z

c 1er ij T

d

e p v tc 2er ij Td e p

d ??e14Twher

e c 1(r ij )and c 2(r ij )are the functions o

f the stress tensor,respectively,and d e p d

?????????

2=3p ?????????????????????????????????????????????????????????????????

ed e p ij àd e p v d ij =3Ted e p ij àd e p

v d ij =3Tq Tis the plastic deviator strain increment.In the modi?ed Cam-clay model,the fol-lowing stress-dilatancy equation is adopted:d e p v d e p d

?M 2àg 22g e15T

Substituting Eq.(15)into (14)gives

H ?Z c 1er ij Td e p v tc 2er ij T2g M àg 2

d e p v

!?Z c er ij Td e p

v

e16T

where c (r ij )is a function of the stress tensor.Note that c (r ij )in Eq.(16)is also equal to 1/R (g )in Eq.(10).Fig.3

shows the constant mean stress path (path AB)and the iso-tropic compression stress path (path AC)along which the hardening parameter changes from H 0to H .The process for ?nding the hardening parameter H is (1)to derive a general equation of H along path AB and (2)to determine an explicit equation of H along path AC.4.1.Along the constant mean stress path AB

After substituting Eq.(16)into Eq.(13),the total di?er-ential form of the yield function is expressed as d f ?

o f o p d p to f o q d q to f o H d H ?o f o p d p to f o q d q àc er ij TK o f o p

?0e17T

So the proportionality constant K can be written as K ?

1c er ij T

o f

o p

d p to f

o q

d q o f

o p

e18T

Based on Eq.(13),the following two di?erential equa-tions can be obtained:o f o p ?

1p M 2àg 2

M tg 2e19To f o q ?1p 2g M tg 2

e20T

By substituting Eqs.(19)and (20)into Eq.(18),the plas-tic deviator strain increment along a constant mean stress path is as follows:d e p d

?K o f o q ?

1c er ij T1p 4g 2

M 4àg 4

d q e21T

Fig.4shows the results of triaxial compression tests on clay and sand (data from [12]).It can be seen from Fig.4a that the shapes of the curves g $e d for clay and sand are alike.The stress ratios (q /p )at failure for clay and sand are M and M f respectively.In fact,M f is not constant dur-ing shearing.If the variation of M f is considered during shearing,more complex behavior (e.g.,the softening)of soils can be described [23].In this paper,M f is assumed to be constant for the sake of simplicity.The parameters M and M f are similar to those of Nova and Wood [15].

Y.P.Yao et al./Computers and Geotechnics 35(2008)210–222213

Therefore,compared with the equation of the plastic devi-ator strain increment for clay in the Cam-clay model when the mean stress is constant(Eq.(22)),the equation of the plastic deviator strain increment for sand is assumed to be Eq.(23)when the mean stress is constant

d e p

d ?c p

1

p

4g2

Màg4

d qefor clayTe22T

d e p

d ?q

1

p

4g2

M4

f

àg4

d qefor sandTe23T

where q is a constant.Eqs.(22)and(23)can also be written in the linear forms

M4àg4 4g2?c p

d g

d e p

d

efor clayTe24T

M4

f

àg4 4g2?q

d g

d e p

d

efor sandTe25T

The validity of Eqs.(24)and(25)is con?rmed by the tri-axial compression results in Fig.4b.So,the assumed form of Eq.(23)is appropriate.It is worth noting that M f=M for clay and the elastic deviator strain is very small under a constant mean stress in Fig.4b.By combining Eqs.(21) and(23),c(r ij)in Eq.(21)is expressed as

cer ijT?1

q

M4

f

àg4

Màg4

e26T

Substituting Eq.(26)into Eq.(16)gives

H?

Z

d H?

Z1

q

M4

f

àg4

Màg4

d e p

v

e27T

4.2.Along the isotropic compression stress path AC

When g=0(path AC),Eq.(27)becomes

H?

Z

d H?

Z1M4

f

M

d e p

v

e28T

Moreover,under the isotropic compression condition

(g=q/p=0)and e p

v

?c p lnep=p0T.In addition,Eq.(13)

becomes H=ln(p/p0)when g=0.So,the following equa-

tion can be obtained from the above two equations:

H?

Z

d H?

Z d e p

v

c p

e29T

Letting Eq.(28)equal Eq.(29)gives

q?c p

M4

f

M

e30T

Finally,we can obtain the following equation of the new

hardening parameter for sand by substituting Eq.(30)into

Eq.(27):

H?

Z

d H?

Z1

c p

M4

M

f

M4

f

àg4

Màg4

d e p

v

e31T

Comparing Eq.(10)with Eq.(31)gives

RegT?c p

M4

f

M4

M4àg4

M4

f

àg4

e32T

If M=M f,Eq.(31)becomes H?

R

d H?

R

d e p

v

=c p,which

is the same as the hardening parameter for clay in the

Cam-clay model.H in Eq.(31)is a uni?ed one for both

clay and sand.

5.A uni?ed elastoplastic model for both clay and sand

In the proposed model,the equations of the yield locus

and plastic potential remain the same as in Cam-clay,but

the transformed stress tensor~r ij(based on the SMP crite-

rion)and the new hardening parameter are adopted to

model the mechanical behavior of clay and sand under

three-dimensional stresses.

The total strain increment is given by the summation of

the elastic component and the plastic component as usual:

d e ij?d e e

ij

td e p ije33T

Here,the elastic component is given by the following

equation

d e e

ij

?

1tm

E

d r ijà

m

E

d r mm d ije34T

where m is Poisson’s ratio,and the elastic modulus E is ex-

pressed as

214Y.P.Yao et al./Computers and Geotechnics35(2008)210–222

E ?

3e1à2m Te1te 0T

j

p

e35T

The plastic component is given by assuming the ?ow rule

not in r ij space but in ~r ij space.d e p ij ?K

o g

o ~r

ij e36T

where the plastic potential function g (or the yield function

f ),the hardenin

g parameter H ,the proportionality con-stant K and the di?erential equation o g =o ~r

ij are given,respectively,as follows:

f ?

g ?ln ~p ~p 0tln 1t~q 2

M ~p 2 àH ?0e37T

H ?Z d H ?Z

1c p M 4M 4f M 4f à~g 4M 4à~g 4d e p v e38TK ?c p M 4f M 4M 4à~g 4

M 4

f à~

g 4d ~p t2~p ~q

M 2~p 2à~

q 2d ~q e39To g o r ij ?1M ~p 2t~q 2M 2~p 2à~

q 23p d ij t3e~r ij à~p d ij T !

e40T

In the above equations,the deviator stress ~q and the trans-formed stress ratio ~g in ~r ij space,M e~g at critical state)and M f e~g at peak)are written respectively as follows:~g ?~q =~p

e41T~q ?????????????32

~s ij ~s ij r e42TM ?

6sin /c

3àsin /c e43TM f ?

6sin /

3àsin /

e44T

where ~p 0e?p 0Tis the initial mean stress,and (/c ,/)are the

internal friction angles at the characteristic state and shear failure,respectively.The main features of the proposed model for both clay and sand are presented as follows.5.1.For sand

The stress-dilatancy equation of the proposed model is expressed as follows:d e p v

d e p d

?M 2~p 2à~q 22p q e45T

Eq.(45)can be drawn as Fig.5a in the ~q =~p $àd e p v =d e p

d plan

e and as Fig.5b in the q =p $àd e p v =d e p

d plan

e under triaxial compression and extension conditions.It is seen from these two ?gures that the unique relationship between

~q =~p and àd e p v =d e p

d can explain th

e di?erence in the q =p $àd e p v =d e p d relations between triaxial compression and extension.

Under triaxial compression and extension stress condi-tions,the yield function in Eq.(37)is plotted in Fig.6,where CL and FL show the characteristic state line and

the shear failure line,respectively.~r

a and ~r r are the trans-formed stresses corresponding to r a and r r respectively,in which r a and r r are the axial and radial stresses in triaxial stress conditions.It is seen from Fig.6that although the yield curves in triaxial compression and extension are sym-metrical with respect to the ~p -axis in the ~p $e~r

a à~r r Tplane (see Fig.6a),the yield curves are not symmetrical with respect to p -axis,and the value of q in triaxial extension is smaller than that in triaxial compression in the p $(r a àr r )plane at the same p (see Fig.6b).This trend is similar to the test results from various kinds of soils (e.g.,[10]).

How the proposed model describes the dilatancy of soil is explained as follows.From Eq.(38),we obtain d e p v

?c p M 4f M 4M 4

à~g 4M 4f à~g 4

d H e46T

Since d H is always larger than or equal to 0,the follow-ing conclusions can be obtained from Eq.(46):(1)~g ?0(isotropic compression condition):d e p v ?c p d H .(2)06~g 0.

(3)~g ?M (characteristic state condition):d e p

v ?0.(4)

M <~g 6M f (positive dilatancy condition):d e p v <0.

As indicated above,the dilatancy of clay and sand are

reasonably described by the new hardening parameter H .The validity of H and the other quantities used usually as hardening parameters will be checked in the various stress paths as follows.

Fig.7shows the stress paths in triaxial tests on Toyoura sand (data from [11])in terms of the relation between mean stress p and deviator stress q .The values of the principal stress ratios are the same (r 1/r 3=4)at points F and F 0in Fig.7.We will check the stress path dependency of four quantities,the plastic volumetric strain e p v ,the plastic devi-ator strain e p d ,the plastic work W p

and the proposed hard-ening parameter H ,in four kinds of triaxial compression tests (paths:ADEF,ABCF,AGF and ABEF)and three kinds of triaxial extension tests (paths:AD 0F 0,ACF 0and AF 0).

Figs.8–11show the variations of those quantities along the seven kinds of stress paths under triaxial compression and extension conditions.It is obvious from Figs.8and 9that the plastic volumetric strain e p v and the plastic devi-ator strain e p

d ar

e unsuitable for the hardening parameter for sand because these quantities depend on the stress paths at the same stress state.From Fig.10the plastic work W p is almost independent o

f the stress paths only in triaxial com-pression or triaxial extension,but its value at the stress state F is di?erent from that at F 0although the stress invar-iants (~p and ~q )of the stress states F and F 0are the same.The other problem induced by the plastic work hardenin

g parameter is that sand will always harden even thoug

h it is at a peak or failure state.Hence,the plastic work is also not good for the hardening parameter in general stress states.However,Fig.11shows that the values of the

Y.P.Yao et al./Computers and Geotechnics 35(2008)210–222215

proposed hardening parameter H are uniquely determined at the same stress state,regardless of the stress path in tri-axial compression and extension and the previous stress history.So,the proposed hardening parameter H is a state quantity,and we employ it as a new hardening parameter for sand.

5.2.For clay

As mentioned before,if M=M f,the new hardening parameter H becomes the plastic volumetric strain,which is the same as the hardening parameter for normally con-solidated clay in the Cam-clay model.Therefore,in this case the di?erence between the proposed model and the Cam-clay model is only the stress tensor used.Let us dis-cuss the critical state conditions in three-dimensional stres-ses in detail.The critical state conditions of the proposed model in three-dimensional stresses can be expressed as follows:

216Y.P.Yao et al./Computers and Geotechnics35(2008)210–222

~g cs?~q cs=~p cs?M

d e p

v =d e p

d

?0

e p

v

?c p?lne~p cs=~p0Ttln2

e p d !1

9

>>>

=

>>>

;

e47T

where the su?x cs means the critical state.When Eq.(47)is satis?ed,soil will be continuously distorted.Eq.(47)is the same as the critical state conditions of the Cam-clay model in triaxial compression,so we might say Eq.(47)is the extension form for the critical state conditions of the Cam-clay model under three-dimensional stresses.

6.Predictions versus experimental results

A series of triaxial compression and triaxial extension tests on normally consolidated Fujinomori clay and saturated Toyoura sand have been completed by Nakai and Matsuoka[12]and Nakai[11].Fig.12shows the initial yield surfaces from Eq.(37)and the test stress paths con-ducted in triaxial compression and extension for clay and sand.The test data are used to examine the capability of the proposed model in predicting drained behavior of clay and sand.The values of soil parameters used in the model are M=M f=1.45,k/(1+e0)=0.0508,j/(1+e0)= 0.0112and m=0.3for Fujinomori clay,and M=0.95, M f=1.66,k/(1+e0)=0.00403,j/(1+e0)=0.00251and m=0.3for Toyoura sand,respectively.The above soil parameters are determined from isotropic compression tests and conventional triaxial compression tests.

6.1.Path p=const

Fig.13compares the predicted and observed results for the drained behavior of Fujinomori clay and Toyoura sand under triaxial compression and extension conditions when

Y.P.Yao et al./Computers and Geotechnics35(2008)210–222217

218Y.P.Yao et al./Computers and Geotechnics35(2008)210–222

p=196kPa.It can be seen from this?gure that the predic-Array tions(solid lines)of the proposed model agree well with the

observed test results(marked s)for clay and sand at con-

stant mean stress under triaxial compression and extension

conditions.

6.2.Path r3=const

Fig.14compares the predicted and observed test results

for the drained behavior of Fujinomori clay and Toyoura

sand under triaxial compression and extension conditions

when r3=196kPa.It can be seen from this?gure that

the predictions(solid lines)of the proposed model agree

well with the observed test results(marked s)for clay

and sand with increasing mean stress under triaxial com-pression and extension conditions.6.3.Path r 1=const

Fig.15compares the predicted and observed test results for the drained behavior of Fujinomori clay and Toyoura

sand under triaxial compression and extension conditions when r 1=196kPa.Since the most of the stress path in tri-axial compression for clay is within the initial yield surface (Fig.12a),the predicted strain is a little smaller than the

Y.P.Yao et al./Computers and Geotechnics 35(2008)210–222

219

test results(Fig.15a).But,it can be seen from Fig.15a and b that the predicted results(solid lines)agree well with the test results(marked )for clay under triaxial extension con-dition and for sand under triaxial compression and exten-sion conditions.

6.4.Stress paths including constant and decrease in stress ratio in triaxial compression for sand

Fig.16shows two special stress paths,which contain a constant and a decrease in stress ratio under triaxial compression,for tests on a sand.The values of the sand parameters used in the model are M=1.08,M f=1.68,k/ (1+e0)=0.00462,j/(1+e0)=0.00259and m=0.3, respectively.The above soil parameters were determined by an isotropic compression test and a conventional triax-ial compression test.

Fig.17shows the predicted and observed test results for the drained stress–strain behavior along the stress paths ABC and DEF,respectively.It can be seen from this?gure that the results(solid lines)predicted by the proposed model agree well with the test results(marked s)in those special stress paths.

6.5.Stress paths in plane strain for clay

In order to validate the present model under plane strain condition,the model is used to predict the results of the plane strain tests on Fujinomori clay[13].

Fig.18shows the stress paths tested in the plane strain tests.The tests are conducted along four kinds of stress paths(AB,AC,AB0and AC0)under plane strain condition from K0-consolidation state(point A:r y=196kPa, r x=r z=98kPa).The model parameters used in the pre-dictions are the same as the above for Fujinomori clay.

Figs.19and20show the comparison between predicted and measured stress–strain behavior and intermediate prin-cipal stress along paths AB and path AC,respectively, without change in principal stress directions.

Figs.21and22show the comparison between predicted and measured stress–strain behavior and intermediate prin-

220Y.P.Yao et al./Computers and Geotechnics35(2008)210–222

cipal stress along paths AB0and path AC0,respectively, with change in principal stress directions.

It is seen from Figs.19–22that the present model gives relatively good predictions of the measured stress–strain behavior and the measured intermediate principal stress values.

Therefore,it can be seen from the above comparisons that the proposed model can reasonably describe the stress–strain characteristics of clay and sand under three-dimensional stresses,as well as the dilatancy of sand along various stress paths.

7.Conclusions

(1)A new hardening parameter is derived by considering

uni?ed yield and plastic potential functions for both clay and sand.It not only describes the dilatancy for lightly to heavily dilatant sand,but also reduces to the plastic volumetric strain for clay.The validity of the hardening parameter is con?rmed by triaxial compression and extension tests on sand along vari-ous stress paths.

(2)An elastoplastic model is proposed by applying the

transformed stress tensor~r ij(based on the SMP crite-rion)and the new hardening parameter H to the Cam-clay model.The proposed model can reason-ably describe the stress–strain behavior of clay and sand under three-dimensional stress states.

(3)The?ve soil parameters(k,j,M,M f and m)in the

proposed model can be determined by a loading and unloading isotropic compression test and a con-ventional triaxial compression test.Acknowledgements

This study was supported by the National Natural Sci-ence Foundation of China,NSFC(No.10672010and No.50479001).The authors thank Prof.S.W.Sloan and Dr.D.C.Sheng at the University of Newcastle,Australia, for their helps in improving the quality of the paper.The authors are also appreciative of the contributions made by D.C.Lu and W.Hou at Beihang University,China, who made part of the predictions in this paper. References

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室内传播和路径损耗计算及实例(完整版)

室内传播与路径损耗计算及实例 RFWaves公司 Adi Shamir 摘要:通过对传播路径损耗得估算来预测无线通信系统在其工作环境下得性能;解释了自由空间传播损耗得计算;电磁波在介质中得发射与反射系数得理论计算就是预测反射与发射系数得工具。下面得一些实例与模型就是在2、4GHz工作频率时给出得。 ------------------------------------------------------------------------------------------- 1、简介 大多数无线应用设计人员最关心得问题就是系统能否正常工作在无线信道得最大距离。最简单得方法就是计算与预测:a)系统得动态范围;b)电磁波得传播损耗。 动态范围对设计者而言就是一个重要得系统指标。它决定了传输信道上(收发信机之间)允许得最大功率损耗。决定动态范围得主要指标就是发射功率与接收灵敏度。例如:某系统有80dB得动态范围就是指接收机可以检测到比发射功率低80dB得信号电平。传播损耗就是指传输路径上损失得能量,传播路径就是电磁波传输得路径(从发射机到接收机)。例:如果某路径得传播损耗就是50dB,发射机得功率就是10dB,那末接收机得接收信号电平就是-40dB。 2.自由空间中电磁波得传播 如上所述,当电磁波在自由空间传播时,其路径可认为就是连接收发信机得一条射线,可用Ferris公式计算自由空间得电波传播损耗: Pr/Pt= Gt、Gr、 (λ/4πR)2 (2、1) 式中Pr就是接收功率,Pt就是发射功率,Gt与Gr分别就是发射与接收天线得增益,R就是收发信机之间得距离,功率损耗与收发信机之间得距离R得平方成反比。公式2、1可以对数表示为: PL=-Gr-Gt+20log(4πR/λ)=Gr+Gt+22+20log(R/λ) (2、2) 式中Gr与Gt分别代表接收天线与发射天线增益(dB),R就是收发信机之间得距离,λ就是波长。 当λ=12、3cm时(f=2、44GHz)可得出: PL2、44=-Gr-Gt+40、2+20log(R) (2、3) R得单位为米。 图2-1表示了信号频率2、44GHz,天线得增益为0dBi时得自由空间得损耗曲线。 注意:在此公式中收发天线得极化要一致(匹配),天线得极化不同会产生另一损耗系数。一般情况下对于理想得线极化天线,极化损耗同两个天线得极化方向得夹角得余弦得平方成正比。例如:两个偶极天线得方向夹角为45°时,极化损耗系数为-3dB左右。

简化的路径损耗模型

简化的路径损耗模型 信号传播的复杂性使得用一个单一的模型准确描述信号穿越一系列不同的环境的路径损耗的特征非常困难。准确的路径损耗模型可以通过复杂的射线追踪模型或者经验测量获得,其中必须满足严格的系统规范,或者基站和接入点的布局必须在最佳的位置。然而,出于对不同系统设计的通用权衡分析,有时候最好的方式是用一个简单的模型抓住信号传播的本质特征,而不是求助于复杂的路径损耗模型,后者也仅仅是真实的信道的近似。这样,下面这个路径损耗(以距离为自变量的函数)的简单模型成为系统设计的常用方法。 (2.20) 如果用dB衰减的形式表达,则为: (2.21) 在这个近似公式中,K是无单位常数,取值取决于传播、天线参数和阻塞引起的平均衰减,d0是天线远场的参考距离,γ是路径损耗指数。由于在天线近场存在散射现象,模型(2.20)通常只适用于传播距离d>d0,其中室内环境下假设d0的范围是1-10米,室外环境下假设d0的范围是10-100米。K的值小于1,而且通常被设定为在距离d0处的自由空间路径损耗(这个设定已经被经验测试数据证实): (2.22) 或者K也可以由在d0处的测量数据决定,并且进行进一步的优化,以便模型或者经验数据之间的均方误差(MSE)能够最小化。γ的值取决于传播环境:对于近似遵循自由空间模型或者双路径模型的传播来说,γ值相应地取为2—4。在更复杂的环境中,γ值可以通过拟合经验测试数据的最小均方误差(MMSE,Mimimum Mean Square Error)来取得(如下面的例子所示)。或者γ值也可以由考虑了载频和天线高度的经验模型(如Hata模型、Okumura模型等)来取得。表格2.1概括了900MHz下不同的室内环境和室外环境下的γ值。如果载频更高,则路径损耗指数γ也会更高。主要指出的是,室内环境下γ的取值范围变化比较大,这是由地板、隔墙和物体引起的信号衰减导致的。

船舶强度与结构设计_授课教案_第四章应力集中模块

第四章应力集中模块 一、应力集中及应力集中系数 在船体结构中,构件的间断往往是不可避免的。间断构件在其剖面形状与尺寸突变处的应力,在局部范围内会产生急剧增大的现象,这种现象称为应力集中。 由于船体在波浪上的总纵弯曲具有交弯的特性,应力集中又具有三向应力特性,严重的应力集中更易于引起局部裂纹和促进裂纹的逐渐扩展。第二次世界大战中和大战后,由于结构开口引起应力集中从而产生裂缝导致船体折断的事故占整个船体结构海损事故总数中的极大部分。因此,在第二次世界大战后,关于船体结构的应力集中问题,曾引起了造船界的普遍重视,开展了大量的研究工作。现在,对这个问题已经有了比较清楚地了解。 由于应力集中是导致结构损坏的一个重要原因,结构设计工作者在设计中必须始终注意这个问题。再进一步对船体结构中比较突出的几个应力集中问题及该区域的结构设计作一些介绍。 通常,用应力集中系数来表示应力集中的程度。应力集中区的最大应力m ax σ或m ax τ分别与所选基准应务0σ或0τ之比值,即 0max 0max ττσσ==k k 或 (1)

称为应力集中系数。基准应力不同,应力集中系数也不同。所以,给定应力集中系数时,应指明基准应力的取法。 间断构件的应力变化规律以及应力集中系数的大小很大程度上决定于这些构件的形状。目前,已经能够确定各种形状的间断构件的应力集中系数。 二、开口的应力集中及降低角隅处应力集中的措施 在大型船舶上,强力甲板上的货舱口、机舱口等大开口,都严重地破坏了船体结构的连续性。当船舶总纵弯曲时,在甲板开口角隅外的应力梯度急剧升高,引起严重的应力集中,造成船体结构的薄弱环节。关于舱口角隅处应力集中的确定,导致去除方角而采用圆弧形角隅,并在角隅处采用加复板或厚板进行加强,同时要采用IV 级或V 级的材料。 1.开口的应力集中 关于孔边的应力集中,可用具有小椭圆开孔的无限宽板受位抻的情况来说明(见下图)。应用弹性理论可求得A 、B 两点的应力分别为: ?????-=+=σσσσB A p a )21( (2) 式中σ为无限远处的拉伸应力; a b /2=ρ为椭圆孔在A 点的曲率半径;

电缆损耗计算公式

电缆损耗计算公式 如果从材料上计算,那需要的数据比较多,那不好算,而且理论与实际差别较大。嗯,是比较正常的。常规电缆是5-8%的损耗。一般常用计算损耗的方法,就是通过几个电表的示数加减计算的。因为理论与实际的误差是比较大的,线路老化,会造成线路电阻变大,损耗增大。7%的损耗,是正常的。还需要你再给出一些数据…如电阻率等… 185的铜线,长度200米,电 缆损耗是多少。 电缆线路损耗计算一条500米长的240铜电缆线路损耗怎么计。 首先要知道电阻: 截面1平方毫米长度1米的铜芯线在20摄氏度时电阻为0.018 欧,R=P*L/S(P电阻系数.L长度米.S截面平方毫米) 240平方毫米铜线、长度500米、电阻:0.0375欧姆假定电流100安培,导线两端的电压:稀有金属3.75伏。耗功率:37.5瓦。 急求电缆线电损耗的计算公式? 线路电能损耗计算方法A1 线路电能损耗计算的基本方法是均方根电流法,其代表日的损耗 电量计算为:ΔA=3 Rt×10-3 (kW·h) (Al-1)Ijf = (A) (Al-2)式中ΔA——代表日损耗电量,kW·h;t——运行时间(对于代表日t=24),h;Ijf——均方根电流,A;R——线路电 阻,n;It——各正点时通过元件的负荷电流,A。当负荷曲线以三相有功功率、无功功率表示时:Ijf= = (A) (Al-3)式中Pt ——t时刻通过元件的三相有功功率,kW;Qt——t时刻通过 元件的三相无功功率,kvar;Ut——t时刻同端电压,kV。A2 当具备平均电流的资料时,可以利用均方根电流与平均电流的等效关系进行电能损耗计算,令均方根电流Ijf与平均电流 Ipj(代表日负荷电流平均值)的等效关系。 3*150+1*70电缆300米线路损耗如何计算 300*0.01=3米也就是说300米的主材消耗量是3米.如果工作量是300米的工程,那么造价时的主材应申请303米.但如果是300米的距离敷设电缆时,需考虑波形弯度,弛度和交叉的附加长度,那么就应该是(水平长度+垂直长度)*1.025+预留长度,算完得数后再乘以1.01就是主材的最后消耗量。 一般电缆的损耗怎样计算 理论上只能取个适当的系数,如金属1.01~1.02,非金属1.04~1.05。要确切的得称重收集数据并总结归纳可得。 电缆线用电损耗如何计算?如现用YJV22-3*150+1*70 电缆线。 电缆电阻的计算: 1、铜导线的电阻率为:0.0175hexun1 Ω·m, 根据公式:R=P*L/S(P电阻系数.L长度米.S截面平方毫米),电缆的电阻为:R=0.0175*260/70=0.065Ω; 2、根据用公式P=I2R计算功率损耗。

路径损耗模型和参数-ITU

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10KV电缆的线路损耗及电阻计算公式 线损理论计算是降损节能,加强线损管理的一项重要的技术管理手段。通过理论计算可发现电能损失在电网中分布规律,通过计算分析能够暴露出管理和技术上的问题,对降损工作提供理论和技术依据,能够使降损工作抓住重点,提高节能降损的效益,使线损管理更加科学。所以在电网的建设改造过程以及正常管理中要经常进行线损理论计算。 线损理论计算是项繁琐复杂的工作,特别是配电线路和低压线路由于分支线多、负荷量大、数据多、情况复杂,这项工作难度更大。线损理论计算的方法很多,各有特点,精度也不同。这里介绍计算比较简单、精度比较高的方法。 理论线损计算的概念 1.输电线路损耗 当负荷电流通过线路时,在线路电阻上会产生功率损耗。 (1)单一线路有功功率损失计算公式为 △P=I2R 式中△P--损失功率,W; I--负荷电流,A; R--导线电阻,Ω (2)三相电力线路 线路有功损失为 △P=△PA十△PB十△PC=3I2R (3)温度对导线电阻的影响: 导线电阻R不是恒定的,在电源频率一定的情况下,其阻值 随导线温度的变化而变化。 铜铝导线电阻温度系数为a=0.004。 在有关的技术手册中给出的是20℃时的导线单位长度电阻值。但实际运行的电力线路周围的环境温度是变化的;另外;负载电流通过导线电阻时发热又使导线温度升高,所以导线中的实际电阻值,随环境、温度和负荷电流的变化而变化。为了减化计算,通常把导线电阴分为三个分量考虑:1)基本电阻20℃时的导线电阻值R20为 R20=RL 式中R--电线电阻率,Ω/km,; L--导线长度,km。 2)温度附加电阻Rt为 Rt=a(tP-20)R20 式中a--导线温度系数,铜、铝导线a=0.004; tP--平均环境温度,℃。 3)负载电流附加电阻Rl为 Rl= R20 4)线路实际电阻为 R=R20+Rt+Rl (4)线路电压降△U为 △U=U1-U2=LZ 2.配电变压器损耗(简称变损)功率△PB 配电变压器分为铁损(空载损耗)和铜损(负载损耗)两部分。铁损对某一型号变压器来说是固定的,与负载电流无关。铜损与变压器负载率的平方成正比。 配电网电能损失理论计算方法 配电网的电能损失,包括配电线路和配电变压器损失。由于配电网点多面广,结构复杂,客户用电性质不

3路径损耗模型-ITU

ITU-R P.1238-5建议书 用于规划频率范围在900 MHz到100 GHz内的室内无线电 通信系统和无线局域网的传播数据和预测方法 (ITU-R第211/3号课题) (1997-1999-2001-2003-2005-2007年) 范围 本建议书介绍了在900 MHz 至100 GHz频率范围内的室内传播的指导原则,主要内容如下: –路径损耗模型; –时延扩展模型; –极化和天线辐射图的效应; –发射机和接收机选址的效应; –建材装修和家具的效应; –室内物体移动的效应。 考虑到 a)正在开发将在室内工作的许多短距离(工作范围短于1 km)的个人通信应用; b)正如许多现有产品和热门的研究活动所表明的那样,无线局域网(RLAN)和无线专用交换机(WPBX)需求很旺盛; c)希望设立无线局域网标准,可与无线和有线通信都兼容; d)采用非常低功率的短距离系统在移动和个人环境下提供业务有许多优点; e)在建筑物内的传播特性和在同一区域内许多用户引起的干扰这两方面的知识,对系统的有效设计是非常重要的; f)用于系统初步规划和干扰估算的通用(即与位置无关)模型和用于某些细致评估的定型(或具体地点)模型都是需要的; 注意到 a)ITU-R P.1411建议书为频率范围在300 MHz到100 GHz的室外短距离电波传播提供了指导,并且该建议也应该作为同时存在室内和室外传播条件的那些情况下的参考文件。 建议 1 对工作于900 MHz到100 GHz之间的室内无线电系统的传播特性进行评估时,采用附件1中的资料和方法。

附件 1 1 引言 室内无线电系统的传播预测在某些方面是与室外系统有区别的。跟室外系统中一样,根本目的是保证在所要求的区域内有效覆盖(或在点对点系统情况下保证有可靠的传播路径)和避免干扰,包括系统内的干扰以及其他系统的干扰。然而,在室内情况下,覆盖的范围是由建筑物的几何形状明确地限定的,而且建筑物本身的各边界将对传播有影响。除了一建筑物的同一层上的频率要重复使用外,经常还希望在同一建筑物的各层之间要频率共用。这样就增添了三维干扰问题。最后,距离很短,特别是使用毫米波频率的场合,意味着无线电路径附近环境的微小变化可能会对传播特性有重大的影响。 由于这些因素的复杂性,若要着手室内无线电系统的具体规划,就需要知道特定位置的详细情况,如几何形状、材料、家具、预期的使用模型等。但是,为了进行系统初步规划,必须估计出覆盖该区域内所分布的移动站所需要的基站数目以及要估计与其他业务的可能干扰或系统之间的潜在干扰。对这些系统规划的情况而言,通常必须要有代表该环境中的传播特性的模型。同时,为了完成计算,该模型不应该要求使用者提供许多输入信息。 本附件主要说明了在室内无线电环境中遇到的传输损伤的通用的、与位置无关的模型和定性的建议。如有可能,也给出与位置有关的专用模型。在许多情况下,基本模型可用的数据受限于频率或试验环境。当可以取得更多的数据时,希望将附件中的建议加以扩充。同样,要根据使用这些模型过程中取得的经验来改善这些模型的精度。但是,本附件代表了目前可以使用的最佳建议。 2 室内无线电系统中的传播损伤和质量的度量标准 室内无线电信道的传播损伤主要由下列因素所造成: —来自房间内的物体(包括墙和地板)的反射和物体附近的衍射; —穿过墙、地板和其他障碍物的传输损耗; —高频情况下能量的通道效应,特别时走廊中这个效应更明显; —房间中人和物体的运动,包括在无线电链路的一端或两端可能的运动,而引起的传播损伤如下: —路径损耗——不仅有自由空间损耗,还有由于障碍物以及穿过建筑物材料传输引起的附加损耗,并且由于通道效应,自由空间损耗可能会减小; —路径损耗随时间和空间的变化; —从波的反射分量和衍射分量而引起的多径效应; —由于移动终端的随机位置变化而引起的极化失配。 室内无线通信业务可以由如下特性来表征: —高/中/低数据速率;

室内传播和路径损耗计算及实例(完整版)

室内传播和路径损耗计算及实例 RFWaves公司 Adi Shamir 摘要:通过对传播路径损耗的估算来预测无线通信系统在其工作环境下的性能;解释了自由空间传播损耗的计算;电磁波在介质中的发射和反射系数的理论计算是预测反射和发射系数的工具。下面的一些实例和模型是在工作频率时给出的。 ------------------------------------------------------------------------------------------- 1.简介 大多数无线应用设计人员最关心的问题是系统能否正常工作在无线信道的最大距离。最简单的方法是计算和预测:a)系统的动态范围;b)电磁波的传播损耗。 动态范围对设计者而言是一个重要的系统指标。它决定了传输信道上(收发信机之间)允许的最大功率损耗。决定动态范围的主要指标是发射功率和接收灵敏度。例如:某系统有80dB的动态范围是指接收机可以检测到比发射功率低80dB的信号电平。传播损耗是指传输路径上损失的能量,传播路径是电磁波传输的路径(从发射机到接收机)。例:如果某路径的传播损耗是50dB,发射机的功率是10dB,那末接收机的接收信号电平是-40dB。 2.自由空间中电磁波的传播 如上所述,当电磁波在自由空间传播时,其路径可认为是连接收发信机的一条射线,可用Ferris公式计算自由空间的电波传播损耗: Pr/Pt= . (λ/4πR)2 式中Pr是接收功率,Pt是发射功率,Gt和Gr分别是发射和接收天线的增益,R是收发信机之间的距离,功率损耗与收发信机之间的距离R的平方成反比。公式可以对数表示为: PL=-Gr-Gt+20log(4πR/λ)=Gr+Gt+22+20log(R/λ) () 式中Gr和Gt分别代表接收天线和发射天线增益(dB),R是收发信机之间的距离,λ是波长。 当λ=时(f=可得出: =-Gr-Gt++20log(R) () R的单位为米。 图2-1表示了信号频率,天线的增益为0dBi时的自由空间的损耗曲线。

低压线路损失计算方法

1.输电线路损耗 当负荷电流通过线路时,在线路电阻上会产生功率损耗。 (1)单一线路有功功率损失计算公式为 △P=I2R 式中△P--损失功率,W; I--负荷电流,A; R--导线电阻,Ω (2)三相电力线路 线路有功损失为 △P=△PA十△PB十△PC=3I2R (3)温度对导线电阻的影响: 导线电阻R不是恒定的,在电源频率一定的情况下,其阻值 随导线温度的变化而变化。 铜铝导线电阻温度系数为a=0.004。 在有关的技术手册中给出的是20℃时的导线单位长度电阻值。但实际运行的电力线路周围的环境温度是变化的;另外;负载电流通过导线电阻时发热又使导线温度升高,所以导线中的实际电阻值,随环境、温度和负荷电流的变化而变化。为了减化计算,通常把导线电阴分为三个分量考虑:1)基本电阻20℃时的导线电阻值R20为 R20=RL 式中R--电线电阻率,Ω/km,; L--导线长度,km。 2)温度附加电阻Rt为

Rt=a(tP-20)R20 式中a--导线温度系数,铜、铝导线a=0.004; tP--平均环境温度,℃。 3)负载电流附加电阻Rl为 Rl= R20 4)线路实际电阻为 R=R20+Rt+Rl (4)线路电压降△U为 △U=U1-U2=LZ 2.配电变压器损耗(简称变损)功率△PB 配电变压器分为铁损(空载损耗)和铜损(负载损耗)两部分。铁损对某一型号变压器来说是固定的,与负载电流无关。铜损与变压器负载率的平方成正比。 配电网电能损失理论计算方法 配电网的电能损失,包括配电线路和配电变压器损失。由于配电网点多面广,结构复杂,客户用电性质不同,负载变化波动大,要起模拟真实情况,计算出某一各线路在某一时刻或某一段时间内的电能损失是很困难的。因为不仅要有详细的电网资料,还在有大量的运行资料。有些运行资料是很难取得的。另外,某一段时间的损失情况,不能真实反映长时间的损失变化,因为每个负载点的负载随时间、随季节发生变化。而且这样计算的结果只能用于事后的管理,而不能用于事前预测,所以在进行理论计算时,都要对计算方法和步骤进行简化。为简化计算,一般假设: (1)线路总电流按每个负载点配电变压器的容量占该线路配电变压器总容量的比例,分配到各个负载点上。 (2)每个负载点的功率因数cos 相同。 这样,就能把复杂的配电线路利用线路参数计算并简化成一个等值损耗电阻。这种方法叫等值电阻法。

塑料产品结构设计准则

产品结构设计准则--壁厚篇 基本设计守则 壁厚的大小取决於产品需要承受的外力、是否作为其他零件的支撑、承接柱位的数量、伸出部份的多少以及选用的塑胶材料而定。一般的热塑性塑料壁厚设计应以4mm为限。从经济角度来看,过厚的产品不但增加物料成本,延长生产周期”冷却时间〔,增加生产成本。从产品设计角度来看,过厚的产品增加引致产生空穴”气孔〔的可能性,大大削弱产品的刚性及强度。 最理想的壁厚分布无疑是切面在任何一个地方都是均一的厚度,但为满足功能上的需求以致壁厚有所改变总是无可避免的。在此情形,由厚胶料的地方过渡到薄胶料的地方应尽可能顺滑。太突然的壁厚过渡转变会导致因冷却速度不同和产生乱流而造成尺寸不稳定和表面问题。 对一般热塑性塑料来说,当收缩率”Shrinkage Factor〔低於0.01mm/mm时,产品可容许厚度的改变达;但当收缩率高於0.01mm/mm时,产品壁厚的改变则不应超过。对一般热固性塑料来说,太薄的产品厚度往往引致操作时产品过热,形成废件。此外,纤维填充的热固性塑料於过薄的位置往往形成不够填充物的情况发生。不过,一些容易流动的热固性塑料如环氧树脂”Epoxies〔等,如厚薄均匀,最低的厚度可达0.25mm。 此外,采用固化成型的生产方法时,流道、浇口和部件的设计应使塑料由厚胶料的地方流向薄胶料的地方。这样使模腔内有适当的压力以减少在厚胶料的地方出现缩水及避免模腔不能完全充填的现象。若塑料的流动方向是从薄胶料的地方流向厚胶料的地方,则应采用结构性发泡的生产方法来减低模腔压力。 平面准则 在大部份热融过程操作,包括挤压和固化成型,均一的壁厚是非常的重要的。厚胶的地方比旁边薄胶的地方冷却得比较慢,并且在相接的地方表面在浇口凝固後出现收缩痕。更甚者引致产生缩水印、热内应力、挠曲部份歪曲、颜色不同或不同透明度。若厚胶的地方渐变成薄胶的是无可避免的话,应尽量设计成渐次的改变,并且在不超过壁厚3:1的比例下。下图可供叁考。

室内传播和路径损耗计算与实例(完整版)

室传播和路径损耗计算及实例 RFWaves公司 Adi Shamir 摘要:通过对传播路径损耗的估算来预测无线通信系统在其工作环境下的性能;解释了自由空间传播损耗的计算;电磁波在介质中的发射和反射系数的理论计算是预测反射和发射系数的工具。下面的一些实例和模型是在2.4GHz工作频率时给出的。 ------------------------------------------------------------------------------------------- 1.简介 大多数无线应用设计人员最关心的问题是系统能否正常工作在无线信道的最大距离。最简单的方法是计算和预测:a)系统的动态围;b)电磁波的传播损耗。 动态围对设计者而言是一个重要的系统指标。它决定了传输信道上(收发信机之间)允许的最大功率损耗。决定动态围的主要指标是发射功率和接收灵敏度。例如:某系统有80dB的动态围是指接收机可以检测到比发射功率低80dB的信号电平。传播损耗是指传输路径上损失的能量,传播路径是电磁波传输的路径(从发射机到接收机)。例:如果某路径的传播损耗是50dB,发射机的功率是10dB,那末接收机的接收信号电平是-40dB。 2.自由空间中电磁波的传播 如上所述,当电磁波在自由空间传播时,其路径可认为是连接收发信机的一条射线,可用Ferris公式计算自由空间的电波传播损耗: Pr/Pt= Gt.Gr. (λ/4πR)2 (2.1) 式中Pr是接收功率,Pt是发射功率,Gt和Gr分别是发射和接收天线的增益,R是收发信机之间的距离,功率损耗与收发信机之间的距离R的平方成反比。公式2.1可以对数表示为: PL=-Gr-Gt+20log(4πR/λ)=Gr+Gt+22+20log(R/λ) (2.2) 式中Gr和Gt分别代表接收天线和发射天线增益(dB),R是收发信机之间的距离,λ是波长。 当λ=12.3cm时(f=2.44GHz)可得出: PL2.44=-Gr-Gt+40.2+20log(R) (2.3) R的单位为米。 图2-1表示了信号频率2.44GHz,天线的增益为0dBi时的自由空间的损耗曲线。 注意:在此公式中收发天线的极化要一致(匹配),天线的极化不同会产生另一损耗系数。一般情况下对于理想的线极化天线,极化损耗同两个天线的极化方向的夹角的余弦的平方成正比。例如:两个偶极天线的方向夹角为45°时,极化损耗系数为-3dB左右。

电缆电路功率损耗计算

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(3)根据电流与损耗功率决定电缆电阻P=I×I×R (5) 根据电阻率与长度决定电缆截面积 ρ=RS/L 电阻率请询问电缆厂家 几种金属导体在20℃时的电阻率

已知电缆长度,功率,电压,需要多粗电缆 电压380V,电压降7%,则每相电压降=380×2= 功率30kw,电流约60A,线路每相电阻R=60=Ω 长度1000M,电阻 铝的电阻率是,则电缆截面S=1000×=131㎜2 铜的电阻率是,则电缆截面S=1000×=77㎜2 由于电机启动电流会很大,应选用150㎜2以上的铝缆或95㎜2以上的铜缆 电压降7%意味着线路损耗7%这个损耗实际上是很大的。如果每天使用8小时一月就会耗电500度, (农电规程中电一年就是6000度。 压380V的供电半径不得超过500米) 电缆选型表

基本含义:H—电话通信电缆 Y—实心聚氯乙烯或聚乙烯绝缘 YF—泡沫聚烯轻绝缘 YP—泡沫/实心皮聚烯轻绝缘 V—聚乙烯 A—涂塑铝带粘接屏蔽聚乙烯护套 C—自承式 T—石油膏填充 23—双层防腐钢带线包铠装聚乙烯外被层 33—单层细钢丝铠装聚乙烯外被层 43—单层粗钢丝铠装聚乙烯外被层 53—单层钢丝带皱纹纵包铠装聚乙烯外被层 553—双层钢带皱纹纵包铠装聚乙烯外被层

线路电能损耗计算方法

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K2=[α+1/3(1-α)2]/ [1/2(1+α)]2 (A2-2) 当f <,且f >α时,按二阶梯持续负荷曲线计算K2: K2=[f(1+α)-α]/f2 (A2-3) 式中f——代表日平均负荷率,f=I pj/ I max,I max为最大负荷电流值,I pj为平均负荷电流值; α——代表日最小负荷率,α=I min/ I max,I min为最小负荷电流值。 A3 当只具有最大电流的资料时,可采用均方根电流与最大电流的等效关系进行能耗计算,令均方根电流平方与最大电流的平方的比值为F(亦称损失因数),F=/,则代表日的损耗电量为: ΔA=3FRt×10-3(kW·h) (A3-1) 式中F——损失因数; I ——代表日最大负荷电流,A。 max F的取值根据负荷曲线、平均负荷率f和最小负荷率α确定。 当f >时,按直线变化的持续负荷曲线计算F: F=α+1/3(1-α)2 (A3-2) 当f <,且f >α时,按二阶梯持续负荷曲线计算: F=f (1+α)-α (A3-3) 式中α——代表日最小负荷率; f——代表日平均负荷率。 A4 在计算过程中应考虑负荷电流引起的温升及环境温度对导线电阻的影响,具体按下式计算: (1+β1+β2) (Ω) (A4—1) R=R 20 β =(I pj / I20)2 (A4—2) 1

10kV线路损耗计算

10kV线路损耗计算 1、线路资料 线路长度:7km,导线型号:JKLYJ-150, 配变容量:2800kV A 2、线路参数计算: 20℃时铝绞线交流电阻率:31.5Ω·mm2/km,则R=L·ρ/S=7×31.5/150=1.47Ω。 3、损耗计算 ⑴、按用户功率因数达0.9来计,只考虑有功电量。 P=3UIcosφI= P/(3Ucosφ) (U=10.5kV cosφ取0.9 R= 1.47Ω) ΔP=3*I2R=0.01646P2(w)=1.646×10-5 P2 (kw) ,即线路有功功率损耗与有功负荷的平方成正比。 P总=P+ΔP 同时乘以等效时间τ,即电量W总=W+ΔW。 ΔW=ΔPτ=1.646×10-5 P2τ=1.646×10-5 PW=1.646×10-5 W2/τ 按一班制,等效时间τ取240小时,则 ΔW=6.86×10-8 W2(kw·h) (W单位为kw·h) 即线路有功电量损耗与用户有功电量的平方成正比。 ⑵不考虑功率因数达标,同时考虑有功电量和无功电量。 ΔP=R*(P2+Q2)/ 1000U2(除1000是将R折算为kΩ) ΔW=ΔPτ= R*(W2+V2)/1000U2τ(U=10.5kV R= 1.47Ω) 按一班制,等效时间τ取240小时,则 ΔW=ΔPτ=1.47*(W2+V2)/26460000=5.56×10-8(W2+V2) (kw·h) (W单位为kw·h,V单位为kvar·h)

两种方式计算比较: 由此可见,采用同时考虑有功电量和无功电量计算方式较为客观,在功率因数为0.9时,两种方式线损一致。在功率因数低时,线损增加。 若要采用固定线损率方式,根据配变容量2800kV A,每月电量估计在30~40万度,固定线损率取2.4%较为合理。

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