A dynamic model of a double-helical planetary gear set

A dynamic model of a double-helical planetary gear set
A dynamic model of a double-helical planetary gear set

A dynamic model of a double-helical planetary gear set

Prashant Sondkar ?,Ahmet Kahraman

Department of Mechanical and Aerospace Engineering,The Ohio State University,201W 19th Ave.,Columbus,OH 43210,USA

a r t i c l e i n f o a

b s t r a

c t

Article history:

Received 26July 2012

Received in revised form 11July 2013Accepted 14July 2013

Available online 8August 2013A linear,time-invariant model of a double-helical planetary gear set is proposed in this paper.The model formulation allows the analysis of a gear set with any number of planets,any planet phasing and spacing configurations and any support conditions.Torsional,transverse,axial and rotational (rocking)motions of gears and the planet carrier are included in this three-dimensional model.Planets are allowed to be positioned equally or unequally spaced around the sun gear.The natural modes are computed by solving the corresponding Eigen value problem.The modal summation technique is used to find the forced response due to periodic gear mesh excitations under constant or proportional modal damping conditions.An example gear set consisting of four equally-spaced planets is considered at the end to demonstrate the influence of key design parameters including right-to-left stagger angle,and support conditions on the dynamic response of the double-helical planetary gear system.Numerous modes with dominant tilting motions are predicted to underline the essence of using a 3D formulation for double helical gears.Staggering of gear teeth is shown to impact the dynamic response of the gear system substantially.Different values of stagger are shown excite different types of modes,resulting in different dynamic response curves.It is also found that left and right sides of double helical gears can carry equal or different dynamic load amplitudes based on the right-to-left stagger values.Likewise the influence of the sun support conditions is shown to be more pronounced for the case of a non-zero stagger.

?2013Elsevier Ltd.All rights reserved.

Keywords:

Gear dynamics

Double-helical planetary gear sets Planet mesh phasing

Double-helical gear stagger

1.Introduction

Planetary gear sets are used commonly in many automotive,industrial,aerospace and wind turbine gearbox applications.The typical applications include jet propulsion systems,rotorcraft transmissions and passenger vehicle automatic transmission and transfer cases.A simple planetary gear set consists of a central sun gear (s ),three or more identical planet gears (p )in mesh with the sun gear as well as an internal ring gear (r ).The planets are held by a rigid carrier (c )through rolling element bearings in between.One of the main advantages of planetary gear sets is that the input torque is split into number of parallel planet paths,providing increased power density (power-to-weight ratio).Planetary gear sets consist of either spur,(single)helical or double-helical gears.Spur planetary gear sets can be commonly found in heavy machinery and off-highway gearboxes and transmissions,the helical planetary gear sets are the norm for all automotive applications,while double-helical planetary gear sets are often found in high-speed aerospace and rotorcraft applications.Their high load carrying capacity and ability to cancel out the axial thrust forces make double helical planetary gear sets suitable for such applications.

Dynamic analysis of a double-helical planetary gear set,as any other gear system,is required for two main reasons.One reason is associated with the resultant noise outcome.High-frequency dynamic forces created at the gear meshes are transmitted to gearbox housing and the surrounding support structures to generate structure-borne noise.Therefore,reduction of vibrations necessitates a

Mechanism and Machine Theory 70(2013)157–174

?Corresponding author.Tel.:+16143124253;fax:+16142923163.E-mail address:sondkar.1@https://www.360docs.net/doc/4f2205910.html, (P.

Sondkar).0094-114X/$–see front matter ?2013Elsevier Ltd.All rights reserved.

Contents lists available at ScienceDirect

Mechanism and Machine Theory

j ou r n a l h o m e p a g e :w w w.e l s e v i e r.c o m /l o c a t e /m e c h mt

158P.Sondkar,A.Kahraman/Mechanism and Machine Theory70(2013)157–174 Nomenclature

C Damping matrix

D Diameter

e Transmission error

b e Amplitude of l-th harmonics of transmission error

F Force acting on planet i due to coupling between carrier and planet i

F spi Dynamic mesh force for sun–planet i gear mesh

F rpi Dynamic mesh force for ring–planet i gear mesh

F Force vector

FW Face width

I Diametral mass moment of inertia

J Polar mass moment of inertia

k Planet bearing stiffness

k sp Mean value of sun–planet i gear mesh stiffness

k rp Mean value of ring–planet i gear mesh stiffness

k x?Support bearing stiffness in x direction(?=s,r,c)

k y?Support bearing stiffness in y direction(?=s,r,c)

k z?Support bearing stiffness in z direction(?=s,r,c)

kθx?Support bearing stiffness inθx direction(?=s,r,c)

kθy?Support bearing stiffness inθy direction(?=s,r,c)

m Mass

M Moment acting on planet i due to coupling between carrier and planet i

M Mass matrix

N Number of planets

p Relative mesh displacement

q Forced vibration response vector

Q Mode shape

r?Base radii of gear?(?=s,r,pi)

r c Planet pin hole radius

T Torque

x Forced vibration response in x direction

y Forced vibration response in y direction

z Forced vibration response in z direction

Z Number of teeth on gear

αpi Position angle for planet i

βHelix angle

γspi phase angle between the s-pi mesh on the left side and the reference s-p1mesh

γrpi phase difference between the r-pi mesh and the r-p1mesh on the left side

γstg phase difference between the left and right side due to intentional nominal stagger of the teeth γrs phase difference between the reference s-p1mesh and the r-p1mesh on left side

εStiffness multiplier for proportional damping

ζDamping factor

ηMass multiplier for proportional damping

θx Forced vibration response inθx direction

θy Forced vibration response inθy direction

θz Forced vibration response inθz direction

ΘDynamic compliance

σPhase angle of l-th harmonics of transmission error excitation

κPhase difference

?Transverse pressure angle

ψAngle made by plane of action with vertical y axis

ωNatural frequency

ωm Mesh frequency

Subscripts

b Bore

c Carrier

better understanding of the dynamic behavior is crucial.The other main reason stems from the durability and reliability of the planetary gear set.Dynamic gear mesh and bearing forces that alternate about the static forces transmitted,cause dynamic stresses and dynamic factors to potentially impact the fatigue lives of the gears and bearings adversely.Hence,dynamic analysis of a double helical planetary gear set becomes a critical step in the design process which aims at developing quiet and reliable transmissions.

Numerous analytical models on planetary gear dynamics consisting of spur or helical gears can be found in the literature while research on double-helical gears is quiet limited.Most of these models are focused on analytical work comprising of discrete-parameter models with gears bodies assumed to be rigid and tooth flexibilities are modeled as a springs.In their review paper,Yang and Dai [1]summarize these torsional,two-dimensional (2D)and three-dimensional (3D)models employing linear or nonlinear formulations as well as constant or periodic gear mesh stiffnesses.Types of analyses performed in these published studies also vary as some focused solely on prediction of natural modes [2–4],some others emphasized the excitation mechanism and cancelation/minimization of these excitations by properly phasing the planets [5–7],while a number of studies predicted forced vibration response and dynamic tooth load [8–12].More recently,Parker et al.[13]and Kahraman et al.[14]used a deformable-body contact model developed by Vijayakar [15]to study the dynamic behavior of planetary gear sets with flexible rims.

Literature on double-helical gears is quiet sparse and limited to single gear pair.Thomas [16]developed an analytical model for investigating load distribution and transmission error of a double helical gear pair under quasi-static conditions.The model was later used by Clapper and Houser [17]to investigate the root stresses of double helical gears and comparisons were made to experiments performed.Zhang et al.[18]carried out noise optimization of double helical parallel shaft gearbox by developing a 3D finite elements model.Jauregui and Gonzalez [19]developed single degree-of-freedom model to study axial vibrations of double helical gear pair due to manufacturing errors.Quasi-static and dynamic analyses of a double-helical gear pair was carried out by Ajmi and Velex [20]by combing a 12degree-of-freedom (DOF)model of the left side of gear pair with another 12-DOF model of the companion right side pair using Euler beam elements.The model is then analyzed to study the effect of floating pinion and staggering of teeth on the quasi-static and dynamic behavior of a gear pair.Anderson et al.[21]conducted experiments on double-helical planetary gear set to measure efficiency,vibration amplitudes and stress levels.

To the best knowledge of the authors,no published study is available on the dynamic modeling of double-helical planetary gear sets.Accordingly,overall aim of this study is to investigate the dynamic behavior of double-helical planetary gear system through the development of general analytical model.The proposed model will be used to investigate the free and forced vibration characteristics of double-helical planetary gear sets within ranges of several key design parameters.

2.Modeling methodology

A three-dimensional discrete model of an N -planet double-helical planetary gear set is proposed.A deformable-body dynamic i Index for number of planets l Harmonics number m Mean value mesh Gear mesh

n Number of beam element r Ring gear

pb Planet bearing pi Planet i gear s Sun gear

sup Support bearing x x direction y y direction z z direction θx θx direction θy θy direction θz θz direction

λNumber of degrees of freedom

Superscript L Left side of double helical gear R Right side of double helical gear

159

P.Sondkar,A.Kahraman /Mechanism and Machine Theory 70(2013)157–174

6(N +3)degree-of-freedom (DOF)model of a helical planetary gear set according to Kahraman [11]using the method proposed by Ajmi and Velex [20].The following assumptions are made in the model formulation:

(i)The bodies representing the sun,planet and ring gears and the carrier are assumed to be rigid.

(ii)Each gear mesh flexibility is represented by a linear spring acting on the plane of action normal to gear tooth surfaces

(inclined by helix angle β).

(iii)Time-varying component of mesh stiffness due to fluctuation of number of tooth pairs in contact is neglected as this was

shown to be a good assumption for a helical gears [5,22].

(iv)Planets are assumed to be identical to each other such that each planet-sun and planet-ring mesh can be assumed to have

the same properties.However,if needed the stiffness and displacement excitation at each mesh can be defined individually through minor revision to the proposed model.

(v)Frictional forces arising from tooth sliding are considered to be negligible in accordance with the off-line-of-action gear

vibration measurements of Kang and Kahraman [23].

(vi)Left and right sides of double-helical gears are assumed identical in geometry expect the hand of the teeth is opposite,but

staggered by a certain angle with respect to each other.

(vii)Damping of the system is represented by either a constant (uniform)modal damping or proportional damping.

Fig.1shows the overall dynamic model of an entire double-helical planetary gear set with only one planet shown for simplicity purposes.Under the assumptions listed above,the model formulations will be done first for three basic sub-systems:(i)a sun –planet i pair (left or right side),(ii)a ring –planet i pair (left or right side),and (iii)a carrier –planet i pair (left or right side).A beam formulation will then be introduced to combine left and right sides of double helical gear set.A general assembly process will then be employed to obtain overall mass and stiffness matrices including support bearing conditions.2.1.Formulation of sub-systems

2.1.1.A sun –planet i pair

Fig.2illustrate an dynamic model of an external gear pair,which represents the one side (either left or right side)of the sun gear (subscript s )meshing with planet-i (subscript pi )located at arbitrary position angle αpi .As shown in Fig.2,the plane of action of the gear pair makes an angle Ψspi with the vertical y axis.This angle can be defined in terms of the transverse pressure angle ?sp of the sun –planet pair and αpi as:

ψspi ?

?sp ?αpi ;??sp ?αpi ;

T s :Counterclockwise ;T s :Clockwise ;

(e1T

where T s is external torque acting on the sun gear.The undamped equations of motion for the s-pi pair are derived using the helical gear pair formulations of Kahraman [11].These equations for sun gear motions are

m s y

€s t eTtk sp cos βcos ψspi p spi t eT?0;m s x €s t eTtk sp cos βsin ψspi p spi t eT?0;m s z €

s t eT?k sp sin βp spi t eT?0;I s θ€

ys

t eTtk sp r s

sin βcos ψspi p spi

t eT?0;I s θ€

xs t eTtk sp r s sin βsin ψspi p spi t eT?0;J s θ

€zs t eTtk sp r s cos βp spi t eT?T s =2N eT:e2a –f T

The corresponding equations of motion for the degrees of freedom of planet i are

m p y

€pi t eT?k sp cos βcos ψspi p spi t eT?0;m p x €pi t eT?k sp cos βsin ψspi p spi t eT?0;m p z €

pi t eTtk sp sin βp spi t eT?0;I p θ

€ypi t eTtk sp r p sin βcos ψspi

p spi t eT?0;I p θ€

xpi t eTtk sp r p sin βsin ψspi p spi t eT?0;e3a –f T

160P.Sondkar,A.Kahraman /Mechanism and Machine Theory 70(2013)157–174

Fig.1.

Dynamic model of a double-helical planetary gear set.

161

P.Sondkar,A.Kahraman /Mechanism and Machine Theory 70(2013)157–174

In these equations,m ?,I ?and J ?are mass,the diametral mass moment of inertia and the polar mass moment of inertia of one side (left or right),and r ?is base circle radius of gear ?(?=s,pi ).k sp is the average gear mesh stiffness for sun planet pair.P spi (t )represents the relative displacement of the s-pi mesh in the direction normal to tooth surface such that

p spi t eT?

y s ?y pi cos ψspi tx s ?x pi sin ψspi tr s θzs tr p θzpi h i

cos βtr s θys tr p θypi cos ψspi tr s θxs tr p θxpi sin ψspi t?z s tz pi h i

sin β?e spi t eT:

e4T

where e spi (t )is loaded static transmission error excitation at the s-pi mesh and βis the helix angle.

Eqs.(2)and (3)are written in matrix form as

M s

00

M p

!q €s t eTq €pi t eT()tk sp K 11spi K 12spi sym :K 22spi

2435q s t eTq pi t eT()?f sm tf si t eT

f spi t eT():e5T

2.1.2.A ring –planet i pair

Next consider the meshing of the same planet i with the ring gear (subscript r )on one side of the double-helical system as shown in Fig.3.With pi being at the same angular position αpi ,the plane of action of this internal mesh makes an angle Ψrpi with the vertical y axis,which is defined in terms of the transverse pressure angle ?rp of the ring –planet pair and αpi as

ψrpi ?

?rp tαpi ;??rp ?αpi ;

T s :Counterclockwise ;T s :Clockwise

:(e6T

pi

y pi

z 162P.Sondkar,A.Kahraman /Mechanism and Machine Theory 70(2013)157–174

The undamped equations of motion for the r-pi pair shown in Fig.3are derived for the motions of the ring and planet i ,respectively,as

m r y

€r t eTtk rp cos βcos ψrpi p rpi t eT?0;m r x €

r t eT?k rp cos βsin ψrpi p rpi t eT?0;m r z €

r t eTtk rp sin βp rpi t eT?0I r θ

€yr

t eT?k rp r r

sin βcos ψrpi p rpi

t eT?0;I r θ€xr t eTtk rp r r sin βsin ψrpi p rpi t eT?0;J r θ€zr t eTtk rp r r cos βp rpi t eT?T r =2N eT;e7a –f T

m p y

€pi t eT?k rp cos βcos ψrpi p rpi t eT?0;m p x €

pi t eTtk rp cos βsin ψrpi p rpi t eT?0m p z €

pi t eT?k rp sin βp rpi t eT?0I p θ€

ypi t eTtk rp r p sin βcos ψrpi

p rpi t eT?0;I p θ€xpi t eT?k rp r p sin βsin ψrpi p rpi t eT?0;J p θ€zpi t eT?k rp r p cos βp rpi t eT?0:

e8a –f T

In Eq.(7a –f),m r ,I r and J r are mass,the diametral mass moment of inertia and the polar mass moment of inertia of one side of

the ring gear,respectively,and r r is base circle radius of the ring gear,respectively,and k rp is the average gear mesh stiffness of the ring –planet pair.Similar to the s-pi formulation,p rpi (t )represents the relative displacement of the r-pi mesh along the plane of action normal to the tooth surfaces

p rpi t eT?

y r ?y pi cos ψrpi t?x r tx pi sin ψrpi tr r θzr ?r p θzpi h i

cos βt?r r θyr tr p θypi cos ψrpi tr r θxr ?r p θxpi sin ψrpi tz r ?z pi h i

sin βte rpi t eT

e9T

zc

163

P.Sondkar,A.Kahraman /Mechanism and Machine Theory 70(2013)157–174

with e rpi (t )being the loaded static transmission error excitation at the r-pi mesh.Eqs.(7)and (8)can be put into the matrix form as

M r 00

M p

!q €r t eTq €pi t eT

&'tk rp

K 11

rpi

K 12

rpi

sym :K 22

rpi

"#q r t eT

q pi t eT

&

'?

f rm tf ri t eT

f rpi t eT

&

'

:

e10T

2.1.

3.A carrier –planet i pair

Fig.4shows a one side of a carrier –planet i pair,with planet i positioned at an angle αpi and attached to the carrier via a bearing.Rotation center of the planet i represented by the z pi axis is at a distance r c from the rotational axis z c of the carrier.The planet bearing

is modeled as a diagonal stiffness matrix K bp ?Diag k y k x k z k θy k θx 0??

that couples planet pi to the carrier c along a circle of radius r c .Bearing forces and moments due to any arbitrary motions of pi and c of this particular side are defined as

F y t eT?k y

y c ?y pi tr c θzc cos αpi h i ;F x t eT?k x x c ?x pi ?r c θzc sin αpi h i

;

F z t eT?k z z c ?z pi ?r c θyc cos αpi tr c θxc sin αpi h i

;

M y t eT?k θy θyc ?θypi

;

M x t eT?k θx θxc ?θxpi

:e11a –e T

Here,pi can rotate in the θz direction with no resistance such that M z (t )=0.With these bearing forces and moments,the Eqs.of motion of the c-pi pair are written as

m c y €c t eTtF y t eT?0;m c x €c t eTtF x t eT?0;m c z €c t eTtF z t eT?0;

I c θ€yc t eTtM y t eT?r c cos αpi F z t eT?0;I c θ€xc

t eTtM x

t eTtr c

sin αpi

F z

t eT?0;

J c θ€

zc t eT?r c sin αpi F x t eTtr c cos αpi F y t eT?0;e12a –f T

m p y €pi t eT?F y t eT?0;m p x €pi t eT?F x t eT?0;m p z €

pi t eT?F z t eT?0;I p θ€

ypi t eT?M y t eT?0;I p θ€

xpi t eT?M x t eT?0:e13a –e T

Eqs.(12)and (13)are written in matrix form as

M c 00

M p

!q €c t eTq

€pi t eT&'tK 11

cpi K 12

cpi sym :

K 22

cpi

"#q c t eT

q pi t eT

&

'

?00

&':

e14T

2.2.Connection of the left and right sides

The sub-system models shown in Figs.2to 4consist of only one side of the double helical gears.In an actual double-helical system,left and right sides of a gears and the carrier are either one-piece (for planet and sun gears)or connected rigidly (for the ring gear and the carrier).While they might be perceived to be rigid,these connecting structures have certain flexibility as measured in comparison to gear mesh stiffnesses.As done earlier by Ajmi and Velex [20],left and right sides of gear (s ,r ,pi )are connected by finite beam elements to capture such compliances accurately.Also for practical reasons,use beam elements between left and right sides creates additional finite element nodes to be used for connecting this model to a given drivetrain in a more accurate way.

In Fig.5,a double-helical gear is divided into three pieces:left side at inner diameter D b ,the right side at the same diameter,and connecting structure of outside diameter D g .The connecting structure spans from gear face mid-point of the right side to the gear face mid-point of the left side.Here,partitioning is done such that the total mass of the gear equals the sum of individual masses of the left 164P.Sondkar,A.Kahraman /Mechanism and Machine Theory 70(2013)157–174

are also handled the same way by using a two-element finite element model,thus representing each gear by three nodes.While not obvious here,the additional node of the beam elements created in middle is useful in interfacing this model with larger driveline model where gear set is employed.With this,the sub-matrices for the connecting structures are given as

K?e?

K11?e1K12?e10

K22?e1tK11?e2K12?e2

sym:K22?e2

2

64

3

75;M

?e

?

M11?e1M12?e10

M22?e1tM11?e2M12?e2

sym:M22?e2

2

64

3

75:e15a;bT

where?=s,r,pi,c and subscript e represents beam elements for each component.

These sub-matrices corresponds to the displacement sub-vector

q?e?

q?

àá

L

q?

àá

M

q?

àá

R

8

<

:

9

=

;:e16T

where subscripts L,R and M indicate left side,right side and middle nodes of each components respectively.

2.3.The overall system equations

These sub system matrices defined by Eqs.(5),(10)and(14)are assembled systematically along with right-to-left coupling matrices defined by Eq.(15)to obtain the overall equations of motion of a double-helical planetary gear set consisting of N planets (a total of18(N+3)degrees of freedom)as

Mq€teTtC q_teTtK mtK b

eTq teT?F teTe17aT

where the overall displacement vector q(t),the mass matrix M,gear stiffness matrix K m and the force array F(t).Arrays q(t)and F(t)are defined below while M and K m are given in the Appendix A.

q teT?

q se teT

q re teT

q ce teT

q p1e teT

?

8

>>>

>>>

<

>>>

>>>

:

9

>>>

>>>

=

>>>

>>>

;

;F teT?

F s teT

F r teT

F p1teT

?

8

>>>

>>>

<

>>>

>>>

:

9

>>>

>>>

=

>>>

>>>

;

:e17b;cT

11

element 1

beam

element 2

right side

gear

segment

Fig.5.Beam elements devised to connect left and right sides of a gear.

165 P.Sondkar,A.Kahraman/Mechanism and Machine Theory70(2013)157–174

The sub-arrays included in F (t )are defined as

F s t eT?

f sm tX N i ?1f si !L 0f sm

tX N i ?1

f si !R

8>>>>>>><>>>>>>>:9

>>>>>>>=

>>>>>>>;;F r t eT?f rm tX N i ?1f ri !L 0f rm

tX N i ?1

f ri !R

8>>>>>>><>>>>>>>:9

>>>>>>>=

>>>>>>>;;

F pi t eT?f spi tf rpi

L

0f spi tf rpi

R

8><>:9>=>;

:e17d –f T

A support stiffness matrix K b is defined to include the matrices for the sun (K bs ),ring (K br ),and carrier (K bc )supports.

K b ?Diag 0

K bs

K br

K bc

0…0? :

e17g T

Here,K bs ,K br and K bc are 6×6matrices that are applied to the middle nodes of the respective connecting structures which

can be defined as

K b ζ?Diag k y ζ

k x ζ

k z ζ

k θy ζ

k θx ζ

k θz ζ?

?:

e17h T

where ?=s ,r ,c ..C is the proportional damping matrix that is given as C =εK +ηM where εand ηare constants.In Eq.(17g,h)the sub-stiffness matrix K b ζrepresenting fixed member has very high k θz ζvalue while other two central members would have k θz ζ=0.

In Eq.(17c),individual forcing vectors forming F (t )include the transmission error excitations that are given as part of the relative gear mesh displacements in Eqs.(4)and (9).These periodic excitations (a total of 4N of them for an N -planet gear set)at individual gear meshes have the same fundamental frequency that is equal to the gear mesh frequency ωm .The transmission error excitation at a given s-pi or r-pi mesh can be computed by using a gear load distribution model [24]as well as the average gear mesh stiffnesses k sp and k rp .While each of these periodic excitations e spi (t )and e rpi (t )along the meshes of the right and left sides of the gear set have the same waveforms at the same harmonic amplitudes,they possess a unique phasing relationship defined by N ,αpi (αp 1=0and i [1,N ]),and the numbers of teeth on the sun,ring and planet gears.

The s -p 1mesh on the left side of the gear set is chosen here as the reference mesh.The transmission error excitation at this reference mesh is given in Fourier series form as

e L eT

sp 1t eT?

X L l ?1

b e spl cos l ωm t tσspl ;

e18a T

where b e spl and σspl are the amplitude and phase angle of the l -th harmonic of this excitation as predicted by the gear load

distribution analysis [24].The transmission error functions on other s-pi meshes on the left side can be defined relative to the reference s -p 1mesh as

e L eT

spi t eT

?

X L l ?1

b e spl cos l ωm t tσspl tl γspi

;

i ∈2;N ? :e18b T

Here γspi is the phase angle between the s -pi mesh on the left side and the reference s -p 1mesh [11,25]

γspi ?

Z s αpi ;?Z s αpi ;

for CW planet rotation ;for CCW planet rotation :

(e18c T

where Z s is the number of teeth on the sun gear.With γstg defined as phase difference between the left and right side due to

intentional nominal stagger of the teeth (e.g.γstg =πfor 50%stagger and γstg =0for no stagger),the excitation on the s -pi on the right side of the gear set is defined as

e R eT

t eT

?

X L b e cos l ωt tσ

166P.Sondkar,A.Kahraman /Mechanism and Machine Theory 70(2013)157–174

Meanwhile,the transmission error excitation at the r -p 1mesh on the left side is given as

e L eT

rp 1t eT?

X L l ?1

b e rpl cos l ωm t tσrpl tl γrs ;

e19a T

where γrs is phase difference between the reference s -p 1mesh and the r -p 1mesh,both on the left side as defined in Ref.[25].b

e rpl and σrpl are the amplitude and phase angle o

f the l -th harmonic of the excitation.Similarly,excitations on any other r -pi meshes on the left side are given as

e L eT

rpi t eT

?

X L l ?1

b e rpl cos l ωm t tσrpl tl γrpi tl γrs

;

i ∈2;N ? :e19b T

Here γrpi is the phase difference between the r -pi mesh and the r -p 1mesh on the left side:

γrpi ?

?Z r αpi ;Z r αpi ;

for CW planet rotation ;for CCW planet rotation

(e19c T

where Z r denotes number of teeth on ring gear.With the same stagger phase angle γstg defined between the left and right sides,

excitations of the ring –planet meshes of the right side of the gear set are defined as

e R eTrpi t eT

?

X L l ?1

b e rpl cos l ωm t tσrpl tl γrpi tl γrs tl γstg ;

i ∈1;N ? :e19d T

It is noted here that the phase angles γstg ,γspi ,γrpi and γrs define the relative phasing amongst all 4N gear meshes of planetary

gear set in terms of number of teeth of the gears and planet position angle.2.4.Solution methodology

The mass and stiffness matrices,M and K =K m +K b ,in Eq.(17a)is considered for free vibration analysis of a given system.The corresponding Eigen value problem for a undamped system yields natural frequencies ωλand normalized mode shapes Q λ(λ∈[1,N dof ]).The response of the given system to transmission error excitations can be obtained using Modal Summation Technique.Based on phasing information the forcing vector can be expressed as sum of 4N vectors (each representing the excitation at one gear mesh)as

F t eT?

X 4N k ?1

F k t eT:

e20T

Response to each individual forcing vector F k (t )is obtained by modal summation as

q k t eT?e

F k k ?p X L l ?1X N dof λ?1

Θλl j ωm eTb e ?pl cos l ωm t tσ?pl tκ

;

e21a T

where ?=s ,r ,j ?????????

?1p and κrepresents appropriate phasing terms defined by Eq.(18)or Eq.(19).e F k is the amplitude of F k (t ).Θλl (j ωm )is dynamic compliance matrix given by

Θλl j ωm eT?Q λQ Τλω2

λ?l 2ω2m à

átj 2l ζλωλωm eT

;e21b T

where Q λis the λ-th normalized mode shape.Modal or proportional damping can be employed in this formulation.For modal damping ζλ=ζand for the proportional damping ratio ζλof the λ-th mode is

ηtεω2λ

167

P.Sondkar,A.Kahraman /Mechanism and Machine Theory 70(2013)157–174

where ηand εare the proportional damping constants.The overall displacement vector is given as the of steady-state responses to each of these excitations

q t eT?

X 4N k ?1

q k t eT:

e22T

With q (t )known,dynamic mesh forces at each of the 4N gear meshes can be obtained as (F ?pi )L /R (t )=k ?p p ?pi (t ),?=s ,r .

Likewise the components of the support bearing forces are computed by using user defined support stiffness matrices and the corresponding gear (or carrier)displacements.With the linear system and periodic excitation,displacements and corresponding dynamics mesh forces will be periodic and amplitudes of these periodic forces (F ?pi )L /R ,?=s ,r can be calculated at different mesh frequencies to study the dynamic behavior of the system under transmission error excitations.3.Example results

A double helical planetary gear set consisting of four equally spaced planets (N =4)with a stationary (non-rotating)carrier is used here as an example system.Table 1lists its basic gear design parameters.Total number of degree of freedom for the model is N dof =18(N +3)=126.Individual contact analyses of one sun –planet and one ring –planet mesh were carried out by using a gear pair load distributions model [24]under specified loading condition to determine the average mesh stiffnesses values (k sp and k rp ),and the harmonic components of the transmission error excitations (amplitudes b e spl and b e rpl ,and phase angles σspl and σrpl in Eqs.(18)–(19)where l ∈[1,3](Table 2)).Connecting structures between left and right sides have the dimensions listed in Table 1according to Fig.5.Table 1also lists the support stiffness values for the sun and ring gears and the carrier as well as the planet bearing stiffness values.

Eigen value solution is carried out for the system defined in Table 1to predict the undamped natural frequencies ωλas listed in Table 3.The corresponding modes are classified as in-phase,sequentially phased and counter-phased adapting the planetary gear set mode classification of Refs.[7,11]to describe the 2D (planar)modes from 2D models.Such classifications are not fully descriptive here,however,since the most modes exhibit 3D motions.The modal summation procedure described in Section 2was employed to predict the forced response vector q (t )to excitations given in Eqs.(18)and (19).Fig.6plots the maximum dynamic mesh force amplitudes (F spi )L and (F rpi )L for sun –planet and ring –planet meshes of the left side,respectively,as a function of gear mesh frequency (ωm =Z s Ωs =Z r Ωr for the case of fixed carrier where Ωs and Ωr are the rotational speeds of the sun and ring gears in Hz,

Table 2

Transmission error excitations.Harmonics b e spl (μm)b e rpl (μm)σspl (rad)σrpl (rad)10.4050.483?0.769?0.70920.0880.135?1.299?1.543Table 1

Basic design parameters of the example gear system.

Sun

Planet Ring Carrier Number of teeth

4739125

–Normal module (mm) 1.81 1.81–Helix angle (°)

21.521.5–Normal pressure angle (°)22.522.5–Base radius (mm)41.834.7111.2–Mass (kg)

2.40.73 6.814.4Mesh stiffness (N/μm)564531–k y ζ,k x ζ(N/μm)

1001001000–k θy ζ,k θx ζ(1e6Nm/rad)5510–D g (mm)7525064–D b /D o (mm)30

263.8

42

ε 1.35e ?6η

50

168P.Sondkar,A.Kahraman /Mechanism and Machine Theory 70(2013)157–174

ωm [Hz]

600

1200

1800

2400

3000

0200040006000800010000

(b)

9451

2994

(F rpi )L

(F spi )L

Table 3

Predicted natural frequencies and mode types.Sr.no.Mode no.Frequency (kHz)Mode type

110.987In phase

22(2) 1.262Sequentially phased 33 1.902In phase

44(2) 1.965Sequentially phased 55(2) 2.076Sequentially phased 66 2.506In phase 77 2.518In phased

88(2) 2.522Sequentially phased 99(2) 2.994Counter phased 1010 3.166Planet mode ???

?

21228.398Planet mode ??

?

?

2325(2)8.852Sequentially phased 24269.037In phase

2527(2)9.451Sequentially phased ???

?

282911.047Planet mode ?

?

?

?

169

P.Sondkar,A.Kahraman /Mechanism and Machine Theory 70(2013)157–174

respectively).Fig.7shows the corresponding dynamic gear mesh forces (F spi )R and (F rpi )R for the right side meshes.In both figures,results for three right-to-left gear tooth stagger phase angles of γstg ?0;π2and πare compared.Several observations can be made from Figs.6and 7:

?Maximum dynamic sun –planet mesh force amplitudes at all planet meshes at a given (left or right side)are equal (i.e.F sp 1=F sp 2=F sp 3=F sp 4).The same is true for the ring –planet mesh forces as well.

?Each of the resonance peaks is associated with a particular mode excited by a certain harmonic amplitude l of the excitations.The frequencies of these modes as well as the harmonic exciting the mode are specified as labels for each resonance peak.Shapes of some of the excited modes are illustrated graphically in Fig.8.

?Maximum dynamic mesh forces vary considerably with γstg considered.Not only amplitudes of response but also resonance frequencies change with γstg .Different values of stagger excite different types of modes.For γstg =0,excited modes exhibit motion with both left and right sides of double helical gears move together as one piece as in the case of modes at natural frequencies of ωλ=1262and 8852Hz,both of which are classified as sequentially phased modes in Table 3.If the modal displacements for these modes are applied to relative gear mesh displacement expressions of Eqs.(4)and (9)with transmission error terms discarded,one would arrive at templates for these modes in the form

e23T

ωm [Hz]

(F rpi )R

(F spi )R

600

1200

1800

2400

3000

0200040006000800010000

9451

2994

(b)

Fig.7.Maximum dynamic mesh force amplitudes at the right side (a)s-pi and (b)r-pi meshes.

170P.Sondkar,A.Kahraman /Mechanism and Machine Theory 70(2013)157–174

It is noted in these modes that the right and left side gear meshes are in unison,and hence get excited for γstg =0.On the other hand the sequentially phased modes at ωλ=2076and 3757Hz exhibit equal but opposite motions on the right and left sides such that

e24T

As a result,they are excited by the excitations for γstg =π.Also it is evident from Figs.6and 7that second harmonic of excitation for γstg =πexcites the modes with left and right sides moving in unison exhibiting relative mesh displacements similar to Eq.(23)(ωλ=2ωm =8398Hz).Any stager angle other than 0and πexcites both types of modes but with less amplitudes,as illustrated in curves for γstg =π/2.It is noted here that connecting left and right sides of double helical gears rigidly instead of by Euler beam elements would not be able to capture the dynamics associated with these models.

?For the cases when γstg =0or π,maximum gear mesh forces at the left and right side meshes are equal:(F spi )L =(F spi )R and (F rpi )L =(F rpi )R while (F spi )L ≠(F spi )R and (F rpi )L ≠(F rpi )R for any other γstg .

In Fig.9,the influence of floating the sun gear radially on the forced response curves is demonstrated for both γstg =0and π.In the model,radial floating condition for sun gear is achieved by reducing the radial sun support stiffnesses significantly.Modes

Fig.8.Representative mode shapes at (a)ω22=8398Hz and (b)ω27=9451Hz.

171

P.Sondkar,A.Kahraman /Mechanism and Machine Theory 70(2013)157–174

of the gear set with the stagger condition of γstg =πexhibit changes at lower frequency ranges (say ωm b 5000Hz)when the sun gear is switched from a radially fixed (piloted)condition to a radially floating condition as shown in Fig.9(b).Very little influence of sun gear support is evident in Fig.9(a)for γstg =0.4.Conclusions

In this study,a linear,time-invariant model of a double-helical planetary gear set was proposed.The model allows the analysis of a double-helical planetary gear set with any number of planets,any planet phasing and spacing configurations and any support conditions.The model included all rigid body degrees of freedom of gears and the carrier in a three-dimensional manner as well as the coupling amongst them.The model allows planets to be positioned equally or unequally spaced around the sun gear with the assumption that any unbalance issues associated with unequal planet spacing are addressed independently.The model captures the phasing relationships between the planet meshes as well as the right-to-left phase differences associated with the staggering the teeth of each double helical gear.Free and forced vibration analyses of the model were performed by solving the governing Eigen value problem to the corresponding undamped system and applying the modal summation technique with proportional damping.

An example set of analyses were performed to point various unique features of double-helical planetary gear sets.Numerous modes with dominant tilting motions were demonstrated to underline the essence of using a 3D formulation for double helical gears.Results also indicated that staggering of gear teeth influences the dynamic behavior of the gear system substantially.Different values of stagger excite different types of modes,resulting in different dynamic response curves.It is also noted that left and right sides of double helical gears can carry equal or different dynamic load amplitudes based on the right-to-left stagger

ωm [Hz]

(F rpi )L (F spi )L Fig.9.Effect of radially floating sun gear on dynamic mesh force amplitudes for (a)γstg =0and (b)γstg =π.

172P.Sondkar,A.Kahraman /Mechanism and Machine Theory 70(2013)157–174

Acknowledgments

Authors thank Pratt &Whitney for supporting of this research activity.

Appendix

A

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174P.Sondkar,A.Kahraman/Mechanism and Machine Theory70(2013)157–174

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中考必会几何模型:8字模型与飞镖模型

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

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。

1第一章 8字模型与飞镖模型(1)

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 。

新整理描写常青藤优美句段 写常青藤作文散文句子

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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+

留学美国常春藤八大院校

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

美国常青藤名校的由来

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)

初中数学常见模型之8字模型与飞镖模型

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 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 。

美国常青藤大学

一、常春藤大学的综合考察 1 (一)秉承传统,注重办学特色 2 (二)讲求质量,保持领先地位 4 (三)避免封闭,与盟校切磋交流 7 二、常春藤大学的分别考察 (一)哈佛大学 9 (二)耶鲁大学 13 (三)哥伦比亚大学 18 (四)普林斯顿大学 23 (五)康乃尔大学 25 (六)宾西法尼亚大学 27 (七)布朗大学 29 (八)达特茅斯学院 34 三、常春藤大学面向未来开展教育创新 36 (一)调整课程体系 37 (二)倡导学术诚信 37 (三)加强科研攻关 37 (四)推动数字建设 38 (五)服务经济发展 38 (六)加强国际交流 39 (七)营造优美校园 39 (八)提高管理效率 40

美国常春藤大学 美国著名的八所常春藤大学均位于纽约领区,在一定程度上体现了我领区的优秀教育资源。本资料基于教育组领事对全部常春藤大学的实地考察编写而成。 一、常春藤大学的综合考察 美国常春藤大学历史悠久,崇尚学术,注重特色,塑造校风,规范办学,教授水平高,学生质量好,在美国乃至全世界高教领域都有较大影响,“常春藤”(Ivy League)几乎成了一流大学的代名词,成了这些名牌老校的无形资产。在21世纪的头几年,常春藤大学通过委任资深校长,聘用杰出教师,招收优秀学生,致力于发现知识、传授知识、培养人才、贡献社会,并制定出面向未来的规划,意欲在激烈的竞争中,始终处于不败之地。 (一)秉承传统,注重办学特色 源于竞技,代表名校。20世纪三、四十年代,美国各大学之间体育比赛如火如荼,并将体育比赛成绩与学校水平相提并论。各校为在比赛中独占鳌头,纷纷降低录取条件,提供高额奖学金,争招体育运动尖子,客观上造成为比赛而比赛,为名次而降分的局面,引起一些名校的不满。1945年,哈佛大学(建于1636)、耶鲁大学(1701)、哥伦比亚大学(1754)、和普林斯顿大学(1746)、四校达成共识,声明取消体育项目奖学金,并结成橄榄球竞赛联盟,在四校间进行比赛。“四”的罗马数字为I V,音同I V Y,意为长春藤,于是有人称这些大学为长春藤大学。 1954年,康乃尔大学(建于1868)、宾夕法尼亚大学(1740)、达特茅斯学院(1769)、布郎大学(1764)入盟。八校共同签署协议,决定各种体育比赛项目均在八校间开展。这八所名校皆为私立,除康乃尔建于19世纪外,其余都有数百年历史,比美国建国还早。这些大学均有古色古香的建筑物,墙壁上爬满了常春藤,从而强化了对常春藤的称谓。由于常春藤大学的历史久、水平高、名气大,尽管学费高昂,人们仍然心向神往。一些人上名校不只为读书,也为广泛交友,八面得风,预埋一张无形的社会关系网。 挟其声威,广揽财源。借助于在学术上的显赫气焰,常春藤大学一般很会造势。每年召开很多会议,邀请知名校友和著名财团、政要成为校董,以使财脉不段。校友乐于回馈母校,报答社会,向母校捐赠成为风尚。 学术自由,独立办学。常春藤大学享有充分的自主权。开什么专业,由学校决定;开什么课,由院系决定;如何上课,由教师决定。这些大学认为,学术自由是宪法修正案中明确赋予大学的权利,惟有如此,方能不媚权贵,追求真理,培育英才。当然,自由是相对的,有自由就有责任。 发展强项,注重特色。常春藤大学根据实际情况规划学校发展,有所为有所不为。注重特色,讲究比较优势,扬长避短。哈佛的政府学院历来第一,哥大的国际关系学院首屈一指,宾大的沃顿商学院傲视同行,耶鲁的法学院赫赫有名。达特茅斯学院曾有发展成研究型大学的机会,但它不改校名高扬本科教育的旗帜;布郎一贯实施课程的完全选修制,让学生自主学习;普林斯顿尽管财力雄厚,但不设医学院,集中力量办好原有系科;康乃尔农牧见长,农业推广项目享誉全美。 优美环境,优良校风。常春藤大学校园美丽,清静整洁,建筑典雅,错落有致,春夏秋冬各具风光,是理想的学习研究之地。这些大学提倡严谨诚信,以人为本,培养品德高尚、独立思考、身体健康、具有领导才能和创新精神的杰出人才。 依法治校,规范管理。常春藤大学均设有法律事务处,依据学校规模,内设3----18名专职法律顾问,总顾问常由一名副校长兼任。他们对外协调法律纠纷,努力维护学校权益,对内理顺部门关系,协助校长监管学校运作和教师操行。同时,各校均有一整套详尽的管理制度,实行规范管理。学校总部负责重大事项,院系则安排具体的教育教学活动,各司其职,各负其责。校董会(Corporation)是学校最高决策机构,由上院(Board of Fellows)和下院(Board of Trustees)组成。校董会的主要职责是:保证办学诚信,任命校长,确保教学管理有序,通过预算,筹款并管理大学的基金,却保足够的物质设施,总揽长远规划,充当大学于社区的桥梁,保持大学自治,仲裁校内纠纷。

常青藤代表着什么教育理念

来源:南方人物周刊 作者:南方人物周刊特约撰稿薛涌最后更新:2012-08-29 08:27:02 “常青藤”在中国已经甚为耳熟。人们对于耳熟的东西,也往往误会很多。10年前,中国掀起了“建设世界一流大学”的运动,常青藤作为“研究性大学”被当成模仿的标本。如今,中国的留学潮从研究院蔓延到了本科,常青藤的博雅教育、“完人”(well-rounded person)培养的理念,也被广为讨论,成为批判中国“应试教育”的有力武器。另外,在制度上,常青藤基本属于私立,和官办的中国高等教育体制形成了鲜明的对照。这似乎也成为市场效率的明证,值得中国的大学借鉴。 在我看来,对常青藤的这些认识,都有简单化之嫌。我个人作为美国教育制度的介绍者,对此也并非没有责任。现在的常青藤固然多为研究性大学,但却来自于和研究性大学非常不同的传统。教育的“完人”理想固然可贵,但这种贵族性的理念长期以来被用于维护上流社会的既得利益,有着相当丑陋的历史。相比之下,常青藤录取中的“应试化”,倒是一种进步的趋势。作为私立的常青藤,充分利用了市场模式,比起欧洲的官办大学来显示出巨大的优势。但是,常青藤在战后的进步,往往和联邦资金的大量介入、政府的一系列法规有密切关系。总之,常青藤有着悠久丰富的传统,但很难代表一个一成不变的教育理念。理解常青藤,就必须分析其复杂的历史基因。 耶鲁大学校园,布什和克里都曾在这里加入“骷髅会”(一个秘密精英社团,成员包括许多美国政界、商界、教育界的重要人物) “牛桥传统” 美国的高等教育是几大传统的汇流。追根寻源,哈佛、耶鲁、普林斯顿等常青藤盟校属于盎格鲁-萨克逊传统,或称“牛桥传统”(Oxbridge),即牛津-剑桥所代表的高等教育。这派的要旨,是培养“完人”,强调通才教育而轻视专业研究。牛津、剑桥到20世纪初尚不授予博士学位。人,而非专业,是教育的核心。 平民社会急剧扩张,几所贵族气十足的常青藤自然无法满足高等教育的需求。南北战争后,联邦政府通过了“颁地法”,即1862年的Merrill Land Grant Act和1887年的Hatch Act,拨给各州大量的联邦土地,让其用卖地所得的款项建立和发展州立大学,其宗旨是传授实用的生产技艺,特别是农业生产的技艺。密歇根、威斯康星以及加利福尼亚的州立大学体系,就是在这个时期成型的。直到今天,州立大学大多比较强调实用技能,也是和这种下里巴人的起源有关。 到19世纪末,德国的研究型大学崛起,其基本理念是把大学建成学术工厂,以学术带头人为中心组成专门的科系,研究人员在严格的分工下推进知识的新边疆。这种专业化体制,使德国在科学研究上突飞猛进,很快就成为诺贝尔奖得主最多的国家。爱因斯坦等一代科学巨子,实际上都是德国体制的产品。美国高等教育界人士显然在第一时间注意到了这一体制的优势,仿照德国模式建立了一系列私立的研究型大学。芝加哥大学、约翰·霍普金斯大学等等,都是在这一潮流中问世的。到20世纪,这种研究型大学和顶尖的州立大学汇流,强调专业训练,强调课程,和标举“完人”教育理想的常青藤形成了鲜明的对照。 在南北战争后,美国进入了急剧的工业扩张,一跃而成为世界第一大经济体。在这样一个激荡的年代,常青藤依然还是盎格鲁-萨克逊白人清教徒(WASP)精英阶层的私人俱乐部。其学生来源,主要是纽约、新

中考数学必会几何模型:8字模型与飞镖模型

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 =________.

中考必会几何模型:8字模型与飞镖模型

8字模型与飞镖模型模型1:角的8字模型 ∠C . O D C B A 模型分析 ∵∠AOB 是△AOD 的外角,∴∠A +∠D =∠AOB .∵∠AOB 是△BOC 的外角, ∴∠B +∠C =∠AOB .∴∠A +∠D =∠B +∠C . ∵∠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: (2)如图②,求:∠CAD +∠B +∠ACE +∠D +∠E = .

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