GERMAN QUANTIFIERS: DETERMINERS OR ADJECTIVES?

Description

GERMAN QUANTIFIERS: DETERMINERS OR ADJECTIVES? Stefanie Dipper Institute of Linguistics University of Potsdam Proceedings of the LFG05 Conference University of Bergen Miriam Butt and Tracy Holloway King (Editors) 2005 CSLI Publications

• structure analysis of quantifiers
• ordinary adjective—functions as a specifier
• determiners
• determiner
• further details on the implementation
• dp
• adjectives
• definite article
• quantifier
• declension

Sujets

• adjectifs

Informations

• Publié par: udwi
• Nombre de lectures: 17
• Langue: English
• Poids de l'ouvrage: 1 Mo

Extrait

Free Vibrations of Nonuniform Timoshenko Beams II
C.H. von Kerczek
1. Introduction:
I present here some vibratory characteristics (eigenvalues and eigenfunctions) of
Timoshenko beams (T-beams) with variable cross section shape and/or a variable elastic
property along the length of the beam. In this study I have developed a 2nd order finite
difference method for solving these problems. In the case of uniform prismatic beams the
eigenvalue problem can be solved exactly. I use these exact solutions to validate the finite
difference solution of the differential equations.
2. Formulation of the T-beam theory:
Timoshenko beam (T-beam) theory is a well known extension of the Classical Euler-
Bernoulli (E-B) beam theory. EB- theory is taught in elementary Mechanics of Materials
courses in most engineering curricula. T-beam theory is taught in advanced mechanics of
Materials courses. Therefore we will not derive the T-beam theory here, but only state the
relevant equations, scaling and notation. A derivation of the T-beam theory can be found in
Reference [1].
The the beam lies along the x axis in the x-y-z coordinate system with its left end at
x=0 and its right end at x=L. The forces and the motion of the beam are only in the x-y plane
with y(x,t) denoting the displacement of the neutral axis of the beam. The beam is
characterized by a cross section area equation S(x,y,z) which is symmetric across the x-y
plane; S(x,y,z)=S(x,y,-z). The beam material is characterized by an elastic modulus E(x), a
poison ratio ν and a mass density ρ. The differential equations governing the deflection y(x,t)
and the shear angle f(x,t) of the beam are
2∂ y ∂ ∂ y
(1a)  A =  AG −f px ,t2  [ ]∂ x ∂ x∂ t
2∂ f ∂ ∂ f ∂ yI = EI  A G −f(1b) 2 [   ∂ x ∂ x ∂ x∂ t
E
G=where t denotes time, , κ is a cross section shape factor, p(x,t) is the force
21
distribution exciting the beam, A(x) is the cross section area and I(x) is the cross section area
2nd moment about the z axis passing through the centroid of the area.
2
Ax= dydz and Ix= y dydz∬ ∬ .
Sx, y ,z Sx, y , z
1The beam's elastic modulus E, cross section area A and 2nd moment I can vary along
E ,A and Ix. Say E(x), A(x) and I(x) have the reference values respectively at some 0 0 0
location on the beam. Then we write
E=E E'x, A=A A 'x and I=I I'x(2a,b,c) 0 0 0
so that E', A' and I' are dimensionless shape functions of x. If E', A' and I' are constant in x,
then they are each equal to 1. Now apply the following scalings:
2(3) p=F p'/L, x=Lx ', y=Ly', t=t' L /E0 0
and substitute (2) and (3) into equations (1) and dispense with the primes. There results
2∂ y ∂ ∂ y q
(4a) = EA −f 
2  [ ]A ∂x ∂ x A∂t
2∂ f 1 ∂ ∂ f A ∂ y
(4b) = EI RE −f2    I ∂x ∂ x I ∂ x∂ t
F 0 Lq=p' =p =where (dimensional p) is a rescaled force distribution,
E A E A 21
0 0 0 0
2L A0and R= . R is the natural slenderness ratio for this problem (the inverse of the square
I0
of the dimensionless radius of gyration of the beam reference cross section). .
Reconsider the dimensionless time variable t. Rescale it as and substitute t=R
this into equations (4) to obtain
21 ∂ y ∂ ∂ y q
(5a) = EA −f 
2  [ ]R A ∂ x ∂ x A∂
21 ∂ f 1 ∂ ∂ f A ∂ y
(5b) = EI  RE −f .2    R I ∂ x ∂ x I ∂ x∂
Multiply equation (5a) by R and rearrange it to obtain
2∂ y 1 ∂ ∂ y qR
= R E A −f (6a) .2 [  ]A ∂x ∂ x A∂
Then multiply equation (5b) by I and rearrange it to obtain
2∂ y I ∂ f ∂ ∂ f
 RE A −f = − EI(6b) .2   ∂ x R ∂ x ∂ x∂
2∂ y
fIn equation (6b) as R→∞ , so that substituting (6b) into the right side of (6a) and
∂ x
taking the limit R→∞ one obtains
2 2 2∂ y −∂ ∂ y
Ax = EI qx,R(7) .2 2 2 ∂ ∂x ∂ x
The force term qR in equation (7) remains finite as R→∞. In the above scaling we have
2 2 3F L A F L L0 0 0qR=p' =p' =p =w . The last equality is the normal scaling for the E-B
E A I E I E I0 0 0 0 0 0 0
beam equations. So by replacing qR by w in equation (7) we have recovered the E-B beam
theory. Furthermore, we have the new scaling of dimensional time t=Tτ, where
4L A0 . T=
I E 0 0
Equations (5) with q=0 are our basic working equations for vibration of T-beams. These
equations are supplemented by an appropriate combination of boundary conditions (BCs)
a yxl=0, b fxl=0, c f 'xl=0, d y'xl−fxl=0(8)
where xl denotes the location of the boundary condition, usually xl=0 and 1.
In T-beam theory a cantilever end does not impose condition y'(xl)=0 as in E-B theory.
Instead the shear angle f is made zero there (condition b). Likewise, in T-beam theory the
condition of no bending moment and no shear force at a free end requires conditions (c) and
(d) respectively instead of the conditions y''(xl)=0 and y'''(xl)=0 as in E-B theory.
3. The Calculation of Vibration Characteristics of T-beams:
To compute the vibration characteristics of T-beams we set q=0 and assume the time
variation in equations (6) to be of the form
Yxy = expi(9) .[ ] [ ]f Fx
2Then equations (6) become our eigenvalues (evs), ω /R, and eigenfunctions (efs), (Y and F,)
problem
2 1 d dY
− Y=  E A −F(10a) [  ]R A dx dx
and
31 1 d dF  RE A dY2
(10b) −  F= EI  −F .   R Ix d x dx Ix dx
There are many numerical ways to calculate the evs and efs of equations (10). We will do this
by a second order finite difference (FD) matrix method which is very easy to implement.
3.1: FD matrix method:
First equations (10) are expanded as
2 EA' EA'd Y dY dF 4 E  − E − F=− Y(11a) 2 A dx dx Adx
2 EI'd F dF E A dY E A 4E  R −R F=− F(11b) 2 I dx I dx Idx
where (…)' denotes the x-derivative of the quantity in the parenthesis.
My implementation partitions the interval [0,1] into a uniform N point grid as
x ∈[0=x ,x ,...,x =1] h=x−x =constantwhere for j=2,3,...,N. Furthermore I assign j 1 2 N j j−1
Y=Yx and F=Fx the grid point values of Y and F to be for j=1,2,...,N. Equations (11) j j j j
are evaluted at each interior grid point j=2,3,..,N-1 with the derivatives replaced by finite
difference formulas, for example for Y,
2Y −Y Y −2YYdY j−1 j1 d Y j−1 j j1
= and =(12) .2 2   dx 2h dx hj j
Now I construct matrix derivative operators that obtain the derivatives at all the interior grid
points simultaneously. Define the N component row differencing operators
d1=[0,0,...,−1,0,1,...,0,0](13a) j
d2=[0,0,...,1,−2,1,...0,0](13b) j
for j=2,3,...,N-1, where the first non-zero element in these row operators are at the position j-
1. Define the vector of discrete values of Y and F as
 TY =[Y , Y ,...,Y ,F ,F ,...,F ](14) ,1 2 N 1 2 N[]F
Where the superscript T denotes transpose the row vector into a column vector.
The first and second discrete derivatives of Y and F at the interior points j=2,3,...,N-1
4can now be computed by the simple vector operations
2 d  1 d  1d1 Y d2 YY Yj j= and =(15) for j=2,3,...,N-1.2 2dx[] 2h [] [ ] dx h [ ]F d1 F F d2 Fj jj j
By ,By ,Bf ,BfFor the time being let each be N component row vectors that 1 N 1 N
x =0 and x =1implement the some type of condition on Y and F at respectively. For 1 N
Y '0=Y 'example say we want to evaluate the derivative . The central differencing formula 1
Y(13a) cannot be used at the end point j=1 because there is no value , but it is possible to 0
2use a quadratic (to maintain O(h ) accuracy) forward differencing formula given as the N
By =[−3,4,1,0,...,0] By ,By ,Bf ,Bfelement row vector . Presently we will leave the 1 1 N 1 N
open to later implement the BCs. But an example of just taking the first derivative of a discrete
g=sin x x ∈[0,0.1,0.2, ...,0.9,1]function, say for (N=11) we construct the matrix j j j
derivative operator D1 as
By1
d12
d12
1 . D1g= g(16) 2h .
.[d1 ]N−1
ByN
By =[0,0,...,0,1,−4,3]where and D1 is a NxN matrix. Formula (16) is the first derivative N
operator that produces the derivative values at all the grid points at once. The Figure 1 below
g'shows a comparison of the exact g'(x) with the discrete derivative computed by D1 at j
xthe discrete points .j
5
Figure 1 A comparison of the numerical vs the exact derivative of
sin(πx). o are the exact values and * are the numerical values obtained
using formula (16).
It can be seen in Figure 1 that formula (16) performs quite well except at the end points which
are not quite as accurate. This is of no consequence since this can be made much more
accurate by making N larger, which will be almost always the case. Typically, we will not need
to compute the derivatives at the end points since BCs specify exactly what is to be imposed
By ,By ,Bf ,Bfat the ends. These conditions will be specified by the end vectors . Before 1 N 1 N
dealing with these let us construct the matrix difference equations for the interior points of the
interval.
Define the matrices
 By Bf 00 1 1
1d1 d2 d2 22 2 2
d1 d2 d2 13 3 3 3
1 11 . . . .D21= D22=D1= I1=(17) , , and 2 22h h h. . . .
. .. .[ ]d1 [d2 ] [d2 ] [ ]1N−1 N−1 N−1 N−1
 By Bf 0 N N 0
6where denotes a row of N zeros, 1 denotes a row of zeros except at po