0perational Method for Clapeyron's Theorem · 2018. 2. 11. · No.18 Operational Methocl for...

0perational Method for Clapeyron's Theorem

Norikazu YOSHIZAWA* and Bennosuke TANIMOTO**

(Received December 22, 1964)

SYNOPSIS

Various metkods for the analysis of coneinuous beams have been proposed

and investigated. They are compelled to treating simultaneous equations

involving many unkRowns, which have been recognized to be uRavoidably

necessary in the analysis of solid mechanics.

This paper takes a set of bending moments at any two consecutive

supports of a continuous beam as the matrix consisting 2-by-1 elements,

whick is referred to as the "eigenmatrix."i>

Then, one of these eigenmatrices can be shifted from one span to its

adjacent one by a certain shift operator matrix which is defiRed by the span

length and the rnoment of inertia of cross-section. By such a shift operation,

the bending moment at any support of the continuous beam can be obtained

readily and systematically, dispensing with simultaneous equations.2)

The infiueRce from external loads, settlement of suppots, and temperature

change is also expressible by a 2-by-1 shift operator matrix. For these

factors, it will be necessary to make a simple modification to the fuRdamental

operators.

This method will be of great advaRtage in the analysis of complicated

continuous beams.

INTRODUCTION

The generalized Clapeyron's theorem is written in the following form

for any consecutive two spans of the continuous beam illustrated in Fig.Ia.

M}-iCr,ii, + 2M}'(f'E', + ft1) + Mr+it?1

:k

**

Assistant of Civil

Nagano, Japan.Professor of Civil

Nagano, Japan.

Engineering, FacttIty of

Engineering, Faculty of

Engineering,

Engineering,

Shinshu

Shinshu

University,

University,

2 N. YosmzAwA and B. TANiMoTo No. 18

..

. 6(%r-i + l,-r) + 6E(.eq.zZtW,-r=i - Wr+1 -lr

Wr)

nv

F 3Ee(£r,-iATr-i + k'l.ATr)･<1)4), 7)

r-2. r-1

1l

r ･r-Al r+2･tt).L2

w?r-1

TWr

1Wr-i-1

wTr+2

l

'

l'r-2 lr-'J --I lr

l,r','f

(a)

Mr--1

r--1 lE

E

iVfr

r

-hr-1dTr--1

-ir-Jl

Afr-f-f

<b)

t

i

mArLS

lli1

i

tE

Qtr-1 Br-1

IiHA･ rlli

il･ ll ll li･

li il m

Qtr .8 ,･

(c)

Fig. 1.

In the above equation, the right side represents the infiuence by the external

load, the settlemeRt of support, and the difference in temperaturerespectively. ")

IntroduciRg the ratios

,ll, lr l'r l'r ==: I7 J.'lr' crr == ;',q' ler :lt,-,' (2)

Eq.1 yields

Mr wwi + 2Mi･ (1 + kr) + M} +ikr =r - 6 (%r mm ili, + Ur'/-. kr) ' ' '

- 6 ,.Lk, (a, wr-i - ao tv, ' wr {- wr÷i) + 3Es(Ihr.--',ATr-i + tll}AZk,)' (3)

No.18 Operational Method for Clapeyron's Theorem 3

and, for simplicity, using the symbols

6%r-1 6Ur lr"i' K}'= lr' l Ktr-1 ==

1,. (4) t s,=6iE.i.,e'fe,, z}. .. llL;, '") p= 32Es, )

the above equation becomes

Mr-i + 2M)' (1 + ler) ÷ Mr-t-ifer

= - (K'r-1 + K}-fer) - 6r(arWr-1 rm arWr -tVr -l- Wr+1)

+ B(ZV-iti Tr-i + Z3'ti T,'k･r)･ (5)

This is the fundamental equation adopted for use in the present paper.

CeNNECTION CONBITION

In order to derive the basic relation between bending moments at any

consecutive supports of a continuous beam, we shall consider the case in

which all theinfluences represented in the right side of Eq.5 equal to zero.

Taking out any set of three consecutive spans as shown in Fig.2, the

Clapeyron's theorem yields two equations

Mr-i + 2Ml' <1 + hr) + Mr+iler = O, l r (6) M･ + 2Mr+1(1 + kr") + Mr+2fer+1 = O･ 1

Mr-1 Mr Mr';-l Mr･+er-1 r r-f-1 r-2

X--ir-1rma'Y' ir 'mmmmS-l'i lr+1 -X'E

Lrs Nr･Fi Nr -l -l:'r

Fig. 2.

The above equations can be written in the matrix form

-1, 2(1+fer)nt -Mr-i- - ler, Oin -Mr+i + "O, 1 rm .M)･ " -2(1+ler+1), ler+1ww -Mr+2ww

which may also be written in the compact form

Cr{Nr-i, N?'+i}m-O,

= o, (7)

:v)

**)

The notations in this paper

It is assumed here that the

its netttral axis.

are indicated in Appendix III.

beam has a symmetrical cross-section with respect

<8)

to

4 N. YosHizAwA and B. TANiMoTo No.18in which

cr ==i Ioil 2(i t fe")I' I2 (i +k'" 'le,.i), fe ,O+ il l' (9)

and

-M)'-1- -Mr+i Nr-i == , Nr+i= ･ (10) ta -M.2mEq.7 or 8 is the connection equatioR for the two physical quantities Nr-i

and Nr+i. The physical matrix of the form of Eqs.10 will be referredto as

the "eigenmatrix" of the span considered. Contrarily, Eq.9 is one of the

operational matrices, with which definite operations are to be performed on

elgenmatrlces.

SMIFT OPERATOR

Eq.8 will yield the shift formulas

Nr+i == LrNr-i, Nr-i=L'rNr+i, (11)

in which the shift operators itqr and L'r represent

1 rm fer+i, O- -1, 2(1+kr)" Lr == - krkr +wwi mny 2(1+ kr+i), ler- -O, 1 -

1 n -kr+1, -2(1+kr)fer+1 - -rm k"ler"i .2 (1 + k?' +i), - kr +4(1 + le.)(1 + k..i) - ' (1 2)

and

Ml, '2(1+kr)- nv kr, Oh Ltr = m -O, 1 m -2(1+ kr+1>, fer+1nd

rmm fer + 4(1 + kr)(1 + kr+i), 2(1 + kr) fer+ie

-r , (13> - -2(1+fer+i), -kr+1 -respectively. It can be verified that

LrL'r =E <ee being the 2-by-2 unit matrix), (14)

which proves, from the practical standpoint, the correctness of the compu-

tatlons.

Lr is the rightward shift operator, or briefiy the right shiftor, since

Eq.11a suggests that the physical quantity Nr-i can at once be shifted

rightwards by premultiplication by the shiftor Lr. By similar reasoning,

No.18 Operational Methocl for Clapeyron's Theorem 5

L'r is the left shiftor.

Lr and L'r are, in the broad sense of words, the "ratios" between the

two physical quantities Nr-i and Nr+i･

VARIeUS FORMS OF FVNDAMENTAL SOLUTION

As a simple application of the preceeding reduction, a continuous beam

subjected to the external edge moments E!Jt and EI]t', which are given quanti-

ties, is taken as shown in Fig.3.') It will be found necessary to classify the

analysis of the continuous beam into several groups by the number of spans,

i.e, by the odd and even numbers of the constituent spans.

wtl' Wl

{/1 2 3 r-lr r-l-1 n-1 n n+I N) N i2 -r-dir-ilirU inrm'il irt /7 ' K ll

1. Continuous

When afour forms of

elgenmatnx

Solutien 1

In this

Ni ==[

in which DZ

Taking

N3 == L2Ni,

' -oddmnumber spans-Ilrmllllll..l

even-number spans-

Fig. 3.

Beam with Odd-Number Spans.

continuous beam consists of odd-number spans, there will be

solution, corresponding to the way of selection of the standard

Nr･

(Fig. 4).

case, the eigenmatrices are taken to be

tLII' N3"IIuaM,I･ Ns=:[MM:6,]･･Nn-i==[Mx'IJI (is)

and Ml' are the given external edge moments.

Ni as standard, Eq.11a yields the following shift formulas:

' Ns=L4N3 == L4L2Ni, '''''', Nn-ixLn-2Ln-4'''L4L2Ni･ (16>

*) To denote

used. The the letter n

support and

the support nttmber of a continuous beam, two letters r and n areletter r represents any support number and can be any integer, while

represents the support number at the extreme right or the next must be an even number: n = 2i (i = 1, 2, 3, ･･･).

6

I

Nl r2 N3 r, nt ny- tt:fTL Nn-1

t. Fig. 4.

The shiftors L.2, L4,･･･, g.n-2 are giveR by Eq.12. Then substituting

Eqs. 15 into Eq. 16c,

-anl-1rm -EMM = E..n-2 ecn-4-'' L4L2 ,

E)]t, -Mb-which can be transformed to

[Sz,I -t- I(lI M}i-i == Ln-2'''L2 egoZI -i- Ln-2'''L2 iOiI MIi,

from which it follows tkat

--1 IM#-i[ :== ILn-2inn-4 k2 I?], I-oil-

.M -DV 'O- × -Ln-2Ln-¢''L2 + ･ -O- -E}IIZt-

This is the desiyed final equation, and the present problem (Fig.4)

been solved dispensing with simultaneous equations.

Selutiosu 2 (Fig.5).

The eigenmatrices are taken to be

N2 == -ca- , tw4= -M"- , ..., N. ww2 .., -M]'-2- ,

.cam mMsm mMllve1rmand then the rightward shift operation yields

Nn-2 "" Emn-3 Ln-s'''Ls L3 N2-

E{.Ttt swt lv' ILd': ftflv 2Yflf lt(s ILIn-.2 2Lfn-1 -'qk

N2 r3 N4 r., --M[:n;I?II..3 Nn-2

Fig. 5.

N. YosmzAwA and B. TANiMoTo

r M-p M:, IL(; M,; IVfh-･2 IVf,t..i x.

No. 18

from

(17)

<18)

(!9)

has

(20)

(21)


To take boundary conditioRs into accoRRt, the Clapeyron's theorem for

the extreme right and the extreme Ieft censecutive two spans is writteR as

.,,-,M. ",),l]Il)[.(ii,?,:k,i]-:, il,kSi,'h,el .,, ,,l (22)

which can be written in the matrix forms

Li, 2P;Tl kL.3i(l]'Nfe;'l,k#J Ytilel-?'.. o. I (23)

' Eqs.23 are the necessary boundary equations which are expressed in

terms of the given external moments MZ and Mi'.

Substituting Eq.21 into Eq.23b, and referringto Eq.23a, the final solu-

tion is obtained as follows:

1vaMhl = - 1 Li, 22((1 -+i" ill'ii)Li)JE.nww3Lki-s'''L3I -ibtlllSn.i]' <24)

Selution 3 (Fig.6).

Using the foliowing method, the eigenmatrix at any odd numbered span

is determined directly. For example, taking Ns as standard, the right and

leftward shift operations from Ns yield

-mz r Ni= == Lr2Lt4Ns, -as- (25) :s) -ant-1" Nn-i= = Ln-2Ln-4'''LsL6Ns･ EEnt

w}' sz]}. f/"I'lr----ecl)-ll?-"-------III>---<ll!l---{IIII------------- N')

.- -- x

Writing

x- Ni -e-;T, N3 - Ns LL, L[`

Fig.

the above equations together,

[ee'i,,I M,t- -gl 2M-

･== +-Mn-1nv O, 1 - 0

-EMt .- wwM?Ii- -O, O-

L6

6.

ua･

M}l-1

- ----e- Nn..1 Ln--2

Lt2Lt4

Ln-2Ln-4'''L6

Ns, (26)

8 N. YosmzAwA and B. TANiMoTo

from which

ff M6- - o, o--i--wrZ- - Lt2L'4,

Mh 1,O O -- ca O,1 O -ecn-2Ln-4-''L6, HM.-i- - o, o- --mzt- This is the desired final solution. Eq.27 requires to compute

inverse, although this is of a simple form.

So}utieR 4 (Fig.7).

Taking N6 as standard, the right and leftward shift operations

Nn-2 = kn-3Ln-s'''ingL7N6, pa2 == L'3L'sN6･

tTLt"t2 M' itt' 7Mi M"Li2 .-bNI"-'i .'"XS(l t

No. 18

(27>

a 4-by-4

yield

(28>

N"- NL) - N4 'ntE-= N6 -m---N,n-2 s-- l-g L･r) L7･ Ln---3

Fig. 7.

Substituting Eqs.28 into Eqs.23 yields

Li, 2(! t2£,-ll) lr2il.-fe,2il.e,1(lel.'Ig5..ewSsl'1.,,-, -Ei'lii･fe.-,. 1 (2g)

Writing these equations together, the following final equation is obtained:

-M6- - L2 (1 + fe,. >, k,j ttmt3k., rs --iN EII)t -

=- ' (30) -M7m -Ll, 2(1 + kn-1>] gmn-3Ln-s'''L7 - rmEI])l'kn-1-

2. Continuous Beam with EveR-Number SpaRs

As described in the preceding article, there are also the following cases

of solution regarding the selection of standard eigenmatrix:

(a) Ni for the first span,

(b) N2 for the second span,

'Y') There are the following relations between the right

Ln-2Ln-4'''L4L2 = [Lt2Lt4''`Lln-4Lin-2]-1,

[Ln-2Lh-4'''L4L2]-i = Ll2Lt4'''Lrn-4Ltn-2.

Combining these relations with Eqs.25, we can get the

and left shiftors :

same result as soltttion 1.

No. 18

Solution 5

ES}--

ll 1 S2,.ll x N

X Ni r.2 N:i ri ---Lr.2 Nn-i --

Fig. 8.

Taking Ni as standayd, the rightward shift operatioR yielcls

Nn-i " Ln-2Ln-4'''L4L2Ni･

The right boundary equation is written to be

Ll, 2(1 ii- len>] Noe-1 '= - Tl'kn.

Substituting Eq.31 into the above equation,

mz- nto- Ll,2(1+kn)jLn-2Ln-4'''L2 -l- Mli =-℃}'len, Ol

from which the followiRg final equatioB is obtained:

rm --1 -orm ML) nd- - Ll, 2<1 + fen>JLn-2Ln-4'"L2 -1-

-mt- × Ll, 2(1 -Y len>J Ln-2Ln-4'''L2 +wr'kn ' uao-

Solntion 6 (Fig.9).

'

yr]lt m Mn--1 Mn. 's l.b M2

Operational Method for

(c) N2r-i for an arbitrary

(d) N2r for an arbitrary

<Fig. 8).

M2 M,3

23

Clapeyron's Theorem

odd numbered spaR,

even numbered span.

Mn-2 Mn"1 Mn n-2 n-1 n

and

' s n-l-1

K. N2 - L3

N.1･ - Ls

Fig.

" l;nl:?7-i Nn

9.

-1

9

(31)

(32)

(33)

(34>

10 N. YosmzAwA and B. TANiMoTo In this case, the standard eigenmatrix is taken as N2,

shift operation yields

Nn "= kn-ikn-3-''gsL3N2,

which is transformed to

Ol + M}i = Ln-iLn-3'''ksL3N2･ E)]'l' O

The left boundary equatien is written to be

sw't + L2(1 + k,), fe2] N2 = O･

Writing Eqs.36 and 37 together,

-sw'e- -O' -2(1+k,), fe, -

o+I A(L,+ N2=O, -Ln-iLn-3'''L3 EI]tt O

from which the final equation becomes

HN,un -Mhum -2(i+k2), k2, orm-i+Dt

-Mh ==- 1O - Ln-iLn-3'''g-3 it, IVC, om rms}ntSolutien 7 (Fig.10>.

wrt /--b' M2 M3 M･t Ms Mb' Mn..J (2----41)--<3 4 s 6' n,rmi

and the

.

fidrn

n

No. 18

rightward

･ffnptK xn+1 )

1

'

x S' NI T.2 N3 XT4 Ns r6 -n'rt,.2 Nn-1

Fig. 10.

For example, taking Ns as standard, the right and

operations from Ns yield

Nn-1 =Ln-2Ln-4'''LsL6Ns, Nl = L'2L'4N5' ,

The right boundary equation is 1 ' Ll, 2(1+kn)] Nn-i+EIiVt'kn=O. "

Substituting Eq.40a into the above equation,

-

leftward

(35)

(36)

(37)

(38)

(39)

shift

(40)

(41)

No.18 Operational Method for Clapeyron's Theorem

Ll, 2(1 + len>] Ln-2Ln-4'''a.sL6Ns + wrl'kn = e･

Eq. 40b is transformed to

-o lh -v Ml, - Lt2Lt4Ns -l- =: O.

1OWriting Eqs.42 and 43 together,

- L'2Lt,i

1M5.+ Ns+O -O- -Ll, 2(1 + fen>j Ermn-2Ln-4'''L.6- -MZ'kn-

from which the final equation becomes

-M6.- rmo, '-irm Wl- - Lt2Lt4

Mk ==- 1, O mM6rm -O, Ll, 2(1+fen)]Ln-2S.n-¢''L6- -MZ'knrm

SglutioR 8 (Fig. 11).

E)Jt -F-' M: ltf;s ?Vl.: M,-, Mn /1 "--.-llll>" t5 n

= o,

4 71+1 1rc -A K X"' NL' '"(Z,e Ni -cr,s -rmM Lnhi- Nn

Fig. 11.

Taking N4 as staRdard, the right and leftward shift

' NnmLn-ILn-3'''Lstw4, N2=L'3N4･The left bottndary eqttation is written to be

E!)Z + L2(1 -i- fe,), le,J N2 == O･

Using the same procedure as described in the preceding soltttion

and 47 may be written in the form

-L2(I+k,>, k,jLt3- MO- rmMZ-

A(C,-g- o =o, N,i + l - Ln-iLn-3'''g..s se}t o

g-}x,

x x 1 '/

11

(42>

(43)

(44)

<45>

N.u

operations yield

(46)

(47)

3, Eqs.46a

(48>

12 N. YosHizAwA and B. TANiMoTo No.18from which, the final equation becomes

-M,- -L2 (1 + k,>, k,j Lt3, o- -i -swl -

M, =- 1 O. (49) - L"-iLn-3'''Ls,

Mi, o EMt3. Continueus Beam with Clamped End.

The clamped end of a continuous beam (Fig.l2a,and 12b) may be

considered to be equivalent to the adjacent imaginary span which has an

infinitely great amount of momeRt of inertia as shown in Fig. 12c, and 12d. 5)

kt 23 :n--1nntJl' 't (a) I i (b) 1 Jo =- ,x) . In +i =-,oo , e I Ni 2 .3,i bt-1 nN" n'f'1 ' n.t.2 'i'L'"' lo r'L'rm' ti 'z,Ll"rm t2 r"31)L'j' l'`"Cilltn ic'Si''' zn "' llL lng'j

Fig. 12.

To derive the basic equation of the clamped end, assume that there are no

external Ioad, no settlement of support, and no difference in temperature in

this system.

C!amped at the Left End.

By Eq.1, the Clapeyron's theorem for the span li and the imaginary

span lo in Fig.12c yields

M6 tP- + 2M, (k' + -fl!,)+cai'- - o, (so)

from which it at once follows that

ca -=-2Mi, (51)and the eigenmatrix for the span li becomes

N,= llMthlll=I-II M,. (s2)

Two unknown elements are reduced to one, and the problem can besolved. ')

ttt t t tttt ttt

'it) Writing down the three moment equation for the spans li and l2, the support moment

Mk can be represented by the clamped end moment M!. Thus, in the case ofacon-

tinuous beam with clamped end, we can express the entire support moment as a function of the clamped end moment.

No.l8 Operational Method for Clapeyron's Theorem 13

Clamped at the Right ERd.

By the same procedure as in the preceding derivation, the eigenmatrix

Nn in Fig.12d is represented by

H-2- Nn= M}i+i･ (53) - 1-

EXTERNAL LATERAL LOAD

When a continuous beam subjected to external lateral loads, theClapeyron's theorern (Eq.5) takes the following form for any two consecutive

spans in a contiRuous beam:

M}e412rXitiiM+'(1,#1''M"r'+IXjk+1-:-1`5K']IJ+''KKr+1feEr'Li),l `5`'

in which

6 6X K'r-i =: l,rm, %r-1, K} =: 7.- Ur,

(55) *'*)

66 K'r= 7;J' %r, Kr-= l,+i Ur+i'

Eqs. 54 can be rearranged to the matrix form

1,i; 2`i ', k"'][t'M'I " I, ,, //'k,.,), kP.J[MMI::

-'K' r-1 + K}kr h x- , <56) -K'r -t- K}+lkr-ww

which can also be written in the compact form

Cr {Nr-i, Nr+i} =-Kr, (57)in which, Cr, Nr-i, and Nr+i have been defined by Eqs.9 and 10.

Kr is a 2-by-1 matrix which is determined by Ioading conditions as follows:

- K'r-i + Krkr rm

Kr= ･ (58) -K'r + K)'+lkr+1-

Eq.57 yields the shift formttlas

*,k) The valuescollected in

in Eqs.55

Appendix Iare for

defined

typicai

as the "Load Term'' inloading conditions.

this paper, and are

14 N. YosmzAwA and B. TANiMoTo No.18 gl:.Ikilew.i:ligei/tl,,l (sg)

in which the shiftors Lr and g'r have been given by Eqs.12 and 13.

Pr and P'r are the shiftors wkich represent the influence of the external

lateral load, and are defined by

1 - '(K'r-i+Kl'kr)kr+i rm P" == ferkr+i -2(K'.-i + Kl･fe,)(1 + kr+i) - (K'r + K}'+ikr+i) kr- '

(60>

-2(K'r + IL'+ikr+i) <1 + fer) - (K'r-i ÷ K}'fer)-

- -(K'r -l- Ki'-t-lkr+l) -

1. Contin"ous Beam with Oee-Number Spans Subjected tD Lateral Load.

When a continuous beam consists of odd-number spans, there are also

four cases of solution as described in the previous article.

Solution l <Fig.13).

Evzt x N '

Nl rr.,p2 N3 -E;/;ig7p,t --pm r[;,:IT-X;.,.pa.2. Nfi--1

Fig. 13.

Taking Ni as standard, the crightward shift operation yieids

Nn-i = Ln-2gdin-4Ln-6'''insL6g..4L2Ni + itoi-2Ln-4itn-6･''LsL6gmu4ge2

+ ±n-2g..n-4Em)t-6'''gansX6P4 +'''+ #ptn-2gen-4 + Pn-2

== kn-2#-n-4'''L2Ni + tcn-2 Ln-4{'''gig6(LP2 + P4)

+ P6'''1} ÷ Pn-4 + gen-2･ (61)

The above equation can be transformed to

bOt,I + Ici, l M}i-i = Ln-2Ln-4'''Lil-Moil + IOiIM{･i]

+Ln-2 Ln-4('''ge2'-')+gen-4 +Pn-2, (62)

from which it folSows that

!ve

"""ll 2 3 4

:[tt.

.I i tl-2n-1]

][II,iR" ..ss

nl

N


IAMa,Z[ -ww ILn-2Ln-4'''L2 IO, -- ･ I'gII.-ij-re Ln-2Ln-4･･-L2 IEi,iiZI

-}- SOz,1 rm ILn-2 ILn-4 ('''pl'') + pn-4I + pn-2I] (63)

This is the final equation for the present case. Comparing this with

Eq.19, it may be seen that these equations are quite similar, the only

difference being in the additional term due to loading condition postmultiplied

by Eq.63.

Selution 2 (Fig. 14).

t/ ll"" 2 3 t Yi "l n-,?S nnv2 nww1 l ; XN

K.: k N2 -m-'" --in - Nn-･) L3, P3 Ln tl, Pn-3 - Fig. 14.

The standard eigenmatrix is in `Lhis case taken to be N2, and

rightward shift operation yields

Nn-2 == Ln-3Ln-s'`'LsL3N2

-i- Ln-3Ln-s'''LsP3 +'''-i- Ln-3Pn-s + P"-3

"= Ln-3Ln-s'''LsL3N2

+ Ln-3 [Ln-s('''P3''') + Pn-s] + Pn-3･

The right and left boundary equatioRs are written to be

Li･ 2(i de-ew,,,nz2,i.; .E.MIk)tsi ne- :."i;i:2 k:i,l,iJifenrei)･ ]

Referring to Eq.64, and using the same procedure as described iR

vation of Eq.24, the final equation iR this case is obtained in the

-M,- nv 2(1+k,), k, --i N2 "" .m- , Mti - Ll, 2(1 + kn -1)] Ln-3Ln-s'''L3 m ww ev. × [?"i'lelli + (I'if52' tK",it2k'.-i)

then

the

form

the

(64)

(65)

deri-

16 N. YosmzAwA and B. TANIMOTO

Ln-3 [Ln-s('''P3'''> + ge n-s] + Pn-3 + Ll, 2(1 + kn-i)j

Comparing this result with Eq.24, the term due to loading

added in the postmultiplied matrix in the above equation.

Solution 3 (Fig. I5).

9.T} /-'1 2 iTg;IIR/'//,; ,-, 6 "-2 7i-f nvi (

No. 18

(66>

condition is

DLIti..ss

xnS

x

v Nl - L' L,, P,£,

L

N3 - LZ P,f

Taking any odd numbered

the final equation becomes

-M6- va

ca Mlz-1

Solution 4

S.}] L'A

! !1N

×

- #.p'2L'4,

- Ln-2'''L6,

-- Ml '

o + o

-EInt

(Fig. 16).

2 3 ,v 'IT

N- .- o Lci, Pa

Fig. 15.

elgenmatrlx,

o,

1,

o,

o,

Ln-2

6

o

o

1

o

-1

- -... Nh-1 Ln--L), Pn-2

for iRstance

L'2Pt4 + P'2

Ln-4('''P6''')+gen-4

7

- /1

, Ns as staRdard,

+ Pn-2

n-2 n-A tr Fll"i

, gAIJI

-- xn N

1

(67)

×

s

Taking

N6=

N2 t)/sp;'sNiitPtps-N6L;pi-i.M.,JtiMl;,..:,N'i'-fL'

J Fig. 16.

N6 as standard, the final equation becomes

-M,M - L2(1+k,), k,JLt3Lts --i-AL- -Ll, 2(1 + kn-i)I Ln-3Ln-s'''g-7 ww

/


× ISztfe.-S",Z" (k"'ll-; 522?'ikn")

+L2 (1+fe,), le,.1 -Lt3P's+Pt3M ww

rm ' - -･ (68) + Ll, 2<1 + hn-i)] Ln-3[k.n-s<'''P7'''> -l- Pn-s] wwi- Fn-3

2. CentiRueus Beam witk Even-Number Spans Subjected to Lateral Lead.

As in the preceding solutions, the final equation for this case will also

take the form postmultiplied by a certain matrix due to external load.

Assuming that, there acts an arbitrary series of external loads on tke

continuous beam as shown in Figs.8, 9, 10, and 11, solutions for each case

can be derived using the same procedures as described previously.

SolutioR 5.

TakingNi as standard, and shifting to the rightward direction as shown

in Fig.8, the final equation becomes

Mli m ut li, 2e+ kn)JLn-2Ln-4'''L2I?lrmlrm'

× ILI, 2(1 + len)jLn-2Ln-4'''L2 1swoZ" -+ EM'len + <K',t-i + Kltfen)

+ Li, 2(i + fen)j ILn-2[Ln-4 ('''p2'l) + pn-4] -t- pn-2I] (6g>

'Solutie" 6.

Taking N2 as standard in Fig.9, the final equation becomes

-M3.M M2 (1 + fe2), fe2, O- -1

M, =- 1 - Ln-iLn-3'''L3,

-WZ÷ (K,,+ Khk,) -

×-OM me -. (70> - Ln-1 [Ln-3 ('''P3''') + Pn-3] -i-Pn-1 EIIyzt

18 N. YosmzAwA and B. TANiMoTo No.18SolutioR 7.

Any odd numbered eigenmatrix is taken as standard in this case. Taking

Ns as stan(lard, the final equatioR is written to be

-ma"rm MO, --1 -g.'2g,.t4 ua == - 1,

th.ua- tO, Ll,2(1 + fen)] g-n-2･･･L6"

rm rmEIIrl- - (Lr2F'4 + Pt2) xo

-SM'kn + (K'n-1 + Knkn)

. <71) + Ll, 2(1 ml+ kn)J g.,n-2[E-n-4('''P6''') + Pn-4] + Pn-2

SDIution 8.

Any even numbered eigenmatrix is taken as standard in this case. Taking

N4 as standard, the final equation is written as follows, and may be com-

pared with Eq.49 which was derived for the unloaded case.

-M,M rfL2 <1 + k,), fe,]Lt 3, o- -1

- Ln-iE-n-3'''Ls,

- E"l +(K',+ Kl,k,)+ L2(1 + le,), fe,] P'3 -

xO, - -. (72) - Ln-i[Ln-3(''-Ps-'')+Pn-3]+ Pn-i

-E"Zll ' -M3. Clamped End of a Continugus Beam Subjected to Lateral Load.

Using the same procedure as described in the previous fundamental

solutioR, f.he clamped end of a continuous bearn can be analyzed for given

loading conditions.

Clamped at the Left End. tt In this case, the additional term due to loading condition

-6(S..''E2) (73)


is-to be added in the right-hand side of Eq.50, and the following relation

is readily obtained:

ca =-2M,-K,. (74)

In addition, writing down the three moment equation for the spans li

and l2,

Mi - (4Mi + 2Ki) (1 + k2) ÷ Mhk2 ww- ma (K'i+ Kafe2), (75>

the third support moment is obtained to be

i

1- -' Mh := k-. (3 + 4k,) M, + 2(1 + fe,) -K, - (Kt,+ K, le,)m . (76)

de -

In a similar manner, all support moments of a continuous beam caR be

represented by the clamped end morneRt Mi, and therefore the problem can

be solved if the right boundary condition is given.

Clamped at tke Rigkt End.

By the same procedure as descrlbed above, tke following relations are

obtained:

M}t =-2MLt+1-K'n (77>

L M}t-1 == (4 + 3len) Mi+1 +2(1 + fen) K'n- (K'n-1 + Kikn). (78>

In these equaeions, M)t+i represents the bending moment at the rigkt clamped

end.

SETTLEMENT OF SUPPORT AND DIFFERENCE IN TEMPERATVRE

Taking the infiuence due to settlement of support and difference in

temperature into consideration, the shift formulas can be represented by

N,'+i=Lr Nr-i+ ger+$r+Tr, (79)and

Nr-i == L'r Nr+i+ ge'r+$'r+ 'E"'r, (80)

tny WEhqilYh6oL;'esapnedct!.:I&yh.aVe been defined by Eqs･i2 and i3, and p, and pr.

20 N. YosmzAwA and B. TANiMoTo No.18 Sr, S'r, Tr, and T'r are 2-by-1 matrices which are defined by the

following formulas:

Shiftors for Settlement ef Support.

un - o"r (arWr-1-ctrWr - Wr -- Wr+1) kr+1 - lSr := ferkr+i ' o-r+i (evr+iwr - crr+iwr+i - wr+i + w7 +2) kr ' (81)

- -im 26r (arWr-1- arWr- Wr+ Wr+1) (1+ fer+1)-

-' O"r (arWr-1 - arWr - Wr + ZVr+1) + 20r+1 (ar+IWr ' ar+IWr+1 M

S,, .. -Wr+i+ZUr+2) (1+fer) . (s2) - -tir--1(crr+IWr-crr+IWr+1-Wr+1+Wr÷2) ,-

Shifters for Temperature Chaitge.

- (a-idT}-i+Z)'dZkr)kr-Fi - T" i= ferfePr+i m2(Z}-iATr-i mF Z)'ACZZxfer(1+kr+i) ' (83)

rm +(Z}ATr + Zi･+iA Traler+D ferm

-(ZL'-iATr-i + Z}ATrfer) ve 2<ZL･ATr + Z)'+idT}+ikr+i) (1 + kr)-

(Z5･ATr -- g･"riTr+ikr+i)

Using the same procedure as described in the case of external lateral

load, the infiuence due to settlement of support and difference in temperature

can also be taken into the analysis of coRtiRuous beams.

PARTIC(JLAR CASE

In the case of a continuous beam with equal span and equal cross-section,

the shiftors take the following simple form:

- 1, -4 Lr == , (85) - 4, 15-

15, 4 ec'r =: , (86) ww- 4, - 1-

"r == [4(K･;-IK"ki.)'-liillllt,+Ki.,)I' (87'


-4 (K'r + KU'+i) - (K'r-i + K}')-

- -<K'r+K}J+i) -

- Wr-1 + 2ZVr ' Wr+1

-4Wr-1 - 9Wr + 6Wr+1 um Wr+2-

- Wr-1 'i- 6W}' um 9Wr+1 + 4Wr+2

S'. =b , <90) - -Wr+2ZVr+1'Wr+2 -

1 Tr -- 2PZtiT , (91) -3

-3 T'r=: 2PZdT , (92) 1

GENERALIZED SOLUTION

As in the previous investigation of a continuous beam with simply

supported ends, the common form of final equation consists of

(1) n-by-n inverse to be determined by the beam construction,

(2) n-by-1 column matrix representing the edge moment, and

(3) n-by-1 column matrix representing the loading condition

(the number n representing 1, 2, 3, or 4).

Then the generalized solution can be written in the form

N=:= R[M+Q]. (93)

Here R represents an n-by-n inverse matrix and is designated as the

"Premultiplier." In the same manner, M is the "Edge Moment Matrix,"

and Q is the "Load Matrix."

When the both ends of a continuous beam are clamped, the final equation

takes the form

N== RQ. (94) In the case of a continuous beam with overhanging end (Fig.17a), the

load on the overhanging part can be reduced to the edge moment effect as

shown in Fig.17b. Therefore, the final equation is the same form as Eq.93.

22

li n-1

IIiix

lIi;

N. YosllIzAwA

P

tn

and B. TANIMOTO

n+1

17.

kin

Q

I1 n-1

No. 18

.Ev}i -- -Pl 7t

-,e･N

nN [ [-=; ln-i;1 lft t

(a) Fig.

The generalized formulas for all

Table I. The values of ee, gkafag, and

Ta'ole l. GeReralized Foxmulas

../ i lr,l -1 --1 (b)

ds of continuous beam are shown

are summarized ilt Appendix II.

for Continueus Beam

in

Type of Beam

Simplets-Simple

Free .vFree

SimpleNFree

(? e. '-

L " tt(ll "i

-) 1:/"',

tp=tt7 :/t it//t

Formulq

･1

N= R[M + Q]

Clamp -Simple

Clamp tvFree .-.4

Clamp -wClamp

ii]i N=R[{Nfi + Q]

N=RQ

Corresponding Table

Table III, IV

Table V, VI

'

Table VII, VIII

EXAMPLES

Practical applicatioRs of the present method will

ing examples.

Example 1.

DetermiBe the bending rnornents at the supports

with seven equal spans when the middle span 6)uniformly distributed load q.

1 2 3 . 4: "･ ll Mi ii ": ql 6

be given in thefollow-

of a continuous

alone is loaded

7' 8'

beam

by a

e･

ll･

a

b--

From Table II,

a

7@l

Fig.

is

m.-7l

18.

th x th

the load term obtained to be

No.18 Operational Method fer Clapeyron's Theorem ` 23

1 K4 =K'` == -2i-ql2. (95>

Substituting the above value into Eqs. 87 and 88,

p3 = Ir Z,1 - -l}qi2 I- ?I, pt,- Iww4il:1- {}qi2 1-il,

--K, - 1 --1- -4Kt,-K,- 1 - 3- P` == -4K, - K,,ww = mu4nqi2 - 3-, P'4 = - - K,, - =T z-gi2 -mi-, (96)

ps = I-4KKIil -- -l}qi2I-X, pts =: ImmoK'[ = -{}qi2Iegl,

and

P2 == P'2 == P6mP'6xO. (97)Then the shiftors Lr and L'r are given by Eqs. 85 and 86, and reduce to

--1, -4 ww -15, 4- .4, 15- --4, -1wwUsing the Solution 2 in Table III, the eigenmatrix is obtained as follows :

ww2(1+fe2), k2 --i -4 1 --i )R==- =- -- 1, -4 -2 Ll, 4] -Ll,2(1+k,)JL,L,- . - 4, 15-me

-- 780, 1- l 2911 ' me 209, -4-

(99)M -= o,

- (Kr,+ K,le,) -Q- (K', + K,fe,) + Ll, 2(1 + le,) ][LP3 + Ps]

oo = -1-qi2wwLi, 4Jll[-1:. 14s 1 lm Oil+I-ill- -- nv1-qi2 --4i-'

24 N. YosffizAwA and B. TANiMoTo No.18from which

N2 = IMAzl -- ew[ffwa +Q] = t/ilt:4 i'II･ (loo)

By the right shift operation, the unknown support moments are de-

termined as

-M4- ql21ma--1, -4"--1" - Oe- ql2--15rmN4 == mMtsm = LN2 + ge3 := 2'g'4" - 4, ls - - 4- + -m71 rm ": 2gtll4 rm- ls- '

rmMbrm ql2- 4'N6= -M÷-=LN4 -t- Ps= 2{tgZ4 nd -1rm' (lol)

The check calculation for the above obtained values may be carried out

by the left shift operation as follows:

rw7=nvillZ,- == gsl2, rm -g-, .... ･i

ql2 -15 Ns == E-' rw7 =: Egtli4 4, O･ K･

(102)

ql2 4 N3=: g-' ews+P'4 =: Egtz4 mls , O･K･

ql2 O Ni= E"' ee3 == twt l4 ml . O･ K･

Example 2.

The maiR pile of a cofferdam is strengthened by wales and struts as

shown in Fig. 19a. Calculate the bending moments at the strengthened points

of the rnaiR pile. The wales are arranged with ten equal spaces on the main

pile.

No.18 Operational Method for Clapeyron's Theorem 25 Support Moment (xql2or xe.Olwbff3)

11O IO -- e.O16 666 67

9 -e.O1443371

-8 -O.025 598 59

7 -O.e3317231

6- -O.04171228

O.6q-5 -O.04997857

O.7g-4 q -O.058 373 44

e.8g-3 .i -O.e6652766

e.9q-2 -O.07551591

1, .- O.081 408 71 --- g = wbff---.i main pile

(a) (b) Fig. 19.

The main pile is subjected to the hydrostatic pressure and can be con-

sidered as a continuous beam with the lower end clamped and the upper end

free as shown in Fig. 19b.

Using Table II, the load terrn is written down easily as follows :

l2 qla ql2 ' Ks= 6-d (8qa + 7qb) == '6d (8 + 6･ 3) = 'grt × 14.3 = 14. 3R,

K'i -= g-?o-- (7q. + sqb) =g6/-o-2 <7 + 7. 2> -- -/-i8- × i4.2 -- i4. 22, I

KIi == 12. 81, K'2= l2. n, Kl, =R.32, Kr, ,11. 22, i (103) "C)

K, = 9. 82, Kt, =: 9. 72, Ks == 8. 32, Kt, =:: 8. 22,

k:. :Igil :16,:- 21Iil kr. ZI]il ".17,i. :'.i'1.1 )

Substituting the above values into Eq.87, it follows that

- -(Kt,+K2,) - -27 -4 (Kt,+ Kl,) - (Kr, + Kb) rm m 84 - (104)

--21- --15 rm -9 ps, =2 , P, =2 , P, =A . -66 rm -48- L 30-

11wale

-"J

/ strut

10---'

9

N.,

8

@"

76

byn

i

54

ny"Y

I "a

N 32

.--"N l

.t 7' xX' st' t ･' ･･Z tt,' ./t

'

;') For simplicity, the notation 2= ql2/60 is adopted for use in this example.

N. YosmzAwA and B. TANiMoTo

Here tke shiftors Lr and L'･r are given by Eqs.85 and 86 and

stant value for the entire span.

The load on the overhanging part in Fig.19b may be reduced

the edge moment effect on the support 10, which is given by

1 2}]rl' = - 6-t ql2 = - 2.

Then, using the Solution 9 iR Table V, it follows that (n ;= 10 in

R =: I L,L,kL, I-i[ , 1-glli= [I-;: -l 4s 14 I-;I , I-g l[ 'i

O, -1 1 nv 70 226 , - 70 226, - 18 817m

ooEvg=: ==R , EDtt -1 -o -Q=: L4 - L3Pr L2P, - k..P6 - Ps rmK,-

M-2911, -10864' nt O lt -- 209, -780- --27-

- 10864, 40545m ]4.3- " 780, 2911-m 84rm

- I' lgl -,ggl I- i,il - I-il -,4,I I-i5,I - Im,,9I]

rm- 91 911. 2- =1 ' 343 021. 5

from which '

MMiM 2 -- 343 020. 5M N= -Mb- == R[M+Q] = 7o 2'i6 " 6os17.3 '

Using Eq. 74,

' -M,- - 1- -O- 2 -- 343020.5- Nl･== = Ml- = . m-2- -Kl- 70 226 rm-3!sigo.snd -Mlem

No. 18

have con-

to one of

(105)

this case)

(106)

(107)

(108)


Shifting to the upward direction, each support moment is determined sue-

cessively as follows :

-M3ww 2 -- 280 318. 3n N3-- -M4- ww- LNi + P2 = 7o 226 nt- 24s g6o.o- '

wwMsrm 2 -- 210 587. 7- N, -- M6 = LN3 + P4 == 70 226 - 17s 7s7. 2-'

un (109) N7 = Iilk]= LNs+ p6 = 'Jii7o&2'ww2'6"[: lg97 s767o31 s5",

Ng=` IMMb,,I = LN7+ ps = r7/ot'i'26 [I 6,O, ij],7I gl･ ?

Comparing Eq.109d with Eq.105 or 107, we can see the correctness of

computations. Thus the computation of tkis method can be carried out with

an automatic procedure. The values ef support moments are given in Fig.

19b.

Example 3.

' Find "he bending moments at supports of the continuous beam wlth

variable cross-section showR in Fig. 20.

l/1

21

q.

;st

/I

!f'y

f

Lttf

II,

ilii

tk

E l'

!･lf

z ''r

f.;l

2l-A-

3-- A4

.¥)..mutmugt-l-,7

.:/tt

l 21 3.l--

and

Frorn the given

therefore

From Table II, the

Fig.

configuration of

fer =:1 (r ==

load

20.

the beam,

2, 3, 4, 5,

- 1, -4

Lr- . - 4, 15-

term is obtained thus

it

6)

IS

'

evident that

(110>

(lll>


Ki :K',= {i- ql2 =ft,

Kh = Ki2 = Ks =: Kts=: K6 == Kt6 = pt,

K3 :Kt3 = 4pt,

K, == Kt4 ::= 9pt. i

Substituting the above values into Eq.87, it follows that

-2 ge2-nv pt, 3

-- 13rm I ge4 == Jt, 42

- 10 ge,=- pt. 7 38

UsiRg the Solution 21 in Table VIII,

R = [L,L,I-!I, I- 4i 3fe6I rm -i

..el-11 -,g--2llI, IvegI--i= -,3iol,,gl ,I

-O M -4Kt, - (Kt, -F K6)-

Q==L4k + -L4P2-Pa -Kl- - -Kt6

-- 1, -4rm2 rm o- - 2M M- 1, - 4- -- 2- 13-

== St -l- ,Lt- LS- H. 4, 15- rm1th -- 1- - 4, 15rm- 3m 42nv

- 31rr Ft, 129m tttttt tt t t t tt ttt tt ttttt ttt tttttt tt tt tttt tt t t ttttttttttttttttttttttttttttttttttt t tt tttt ttttt tt t t tt t

'i`) For simplicity, the notation pt = tql2 is adopted fgr use.

,

t'e

No. 18

(l12) ")

<113)

(114)


from which

N - [MM,i- =RQ =2･3a4o II1284g[. (11s)

The eigenmatrix for the first span becomes by Eq. 74 as follows :

Ni - ItX -IwwIIMi - },1 - ,st,, I-ve gg,il･ (ii6)

The unknown support moments are determined by the right shift operation

as follows:

-1207 No =m LN,+P2 == pt , . 2340 -6214 ' -"L (117>

. -4357 N, =LN3 + P4 -- rmi3'4o 242 '

The values above obtained may be checked by the following shift operation

N6 =- IillSI -- LN4+ps- ,g,, l' ll lgl IZ2ilil + 1- :gl pt

242 pt ' =Tl}'g'iifO -12gl' O･K･ (!ls)

CONCL{JSEONS

In conclusion, the follow}ng notes are g}ven :

1. The exact solution for all kinds of continuous beam can be obtained

by simple matrix algebra.

2. The bending moment at any support of a continuous beam can be

determined directly.

3. Compared with the moment distribution method, this method has a

kigher efflciency and perfect exactness. The eficiency will grow greater and

greater as the system become complicated.

4. The computatioR can be designed with the simultaneous checking

methed.

5. The calculated result is checked arbitrarily and effectively by the

right or left shiftors.

30 N. YosHizAwA and B. TANiMoTo No,18

ACKNOWLEDGEMENTS

The authors express their sincere thanks to Prof. K. SATo, Faculty of

Engineering, Shinshu University, for his lnspection of the manuscript from

wider point of view and for various useful suggestions. Their thanks are also

due to Prof. A. HAKATA, Faculty of Engineering, ShiRshu University.

REFERENCES

1) B.TANiMoTo, "Eigen-Matrix Method for Beams and Plates," Proceedings

of the A.S.C.E., Structural Division, Oct., 1963, pp. 173-215.

2) B.TANIMoTo, "Operational Method for Continuous Beams," Proceedings of

the A.S.C.E., Strttctural Division, Dec., 1964, pp. 213-242.

3) J.S.C.E., "Civil Engineering Hand Book" (in Japanese), GmouDoH Co.,

Ltd., 1964, pp. 65-69.

4) N.YAMAGucHi, "Applied Mechanics'' (in Japanese), ARusu Co., Ltd., 1952,

p. IIO.

5) I.KoNism, Y. YoKoo, and M.NARuoKA, "Theory of Structures" (in Japanese),

MARuzEN Co., Ltd., 1963, pp. 95-105.

6) S. TIMosHENKo, "Strength of Materials," Part I, 3rd] Edition, D.VAN

NosTRAND Co., Inc., 1955, pp. 204-209.

7) RoBERT B.B. MooRMAN, "Analysis of Statically Indeterminate Structures,"

JoHN WiLLEy & SoNs, Inc., l955, p. 153.

8) A.HiRAI, "Steel Bridge, III" (ln Japanese), GiHouDoH Co., Ltd., 1956,pp.

38-51.

No. 18

Table Il.

Operational Method for Clapeyron's Theorem

APPENDIXI. LOADTERM

LOAD TERM FOR OPERATIONAL METHOD

31

Loading Condition KE{iiill,:-za

i

Fl/l'.,'j'i,, ,l 1,' 1 ¢:e=tr"r=:i

'

4 .ww' t ""'

rr a --'b'1

-c-- t"

di. 1ntptllg

L--- t --'"

/il I･･ ,,

tg4rr --

ql,',.

ant

q--tl--"" 7t -ma'm

gg[III'i'l:,d,'/･i

as pi. ,irill' 1[ :l/,,

"U. e"L---- t -'

f,b (i+b)P

-II- gl2

oq4-tl--o."

(21-a)2

q (d2 - b2)(212 - b2 - d2)412

7

um ql2

60

g/" (i +3)(7 nv 3?,)

o6qnvoai, (3a2 - 15al + 2ol2)

qa26r6i2 (12a2 m 45al + 4el2)

l2

65(84a + 7qb)

l£t,(l3 - 2a21 + a3)

Kl

`flb, (i+a)p

-li- qi2

oq,-X (2i2 - .2)

1£/i2 (a2 - c2 )(212 - a[ - c2)

8- ql260

fli/Ilo (

a1+- l×,-3tA)

qa2

6el2(1012 - 3a2)

4qa2-6Cti}l2

(512 - 3a2)

l2

6-t(7qa + 8qb)

f/ (i3 - 2a2t + a3)

Note: The following relations are seen between the load

Method" and the "Slope Deflection Method":3)

K == - 2IL,B, K, = 2RZBA.

terms in the "Operational

APPENDIX II. VARIOUS FORMS OF SOLUTION FOR ALL KINDS OF

CONTINUOUS BEAM

The solution's' for all possible cases of continuous beams are collected and

classified ln the following Tables III, through VIII.

Using together with Table I, the analysis of all kinds of continuous bearns

can be carried readily and systematically.

32 N. YosmzAwA and B. 'IIANiMoTo

Table M SIISffPLE･-SIMPLE

No. 18

Shifting ProcedureIIlt

N R

ywt Abe

(-f2 Nl .....ip

Solution 1.

ii l'ig i?l, ll

-x･vl / ii -!

rmM,- mu

mlVCt-1 -

I:lliI

i[

LnT2Lnrm4'''L2

[ Ol --

'

-- 1-

- o-

-1

pe- t )tab-

("L2, N2･ ..- Solution 2.

Si,,li

-" x l' li/ liNY L i L

-ML-

-Mh-

'-Mk rm

ua

M2

-utt-1 -

111Ii"

1-Ii

I'

- 2(1+ks), fes --Ll, 2(1 + knrmi)jLn-3Ln-s'''L3nt

-1

(.Lr----<1 6 ) s" N- ig Ns- Solution 3.

t l

i

ll:

i

sR -qs-- N6 S Solutjon 4. s

t

-Mh-

-Mirm

- o, 6''- LtELt4,

1, O

O, 1- Ln-2'''L6,

- o, oT

-1

- t2(1 ÷ k,,), ksi Lt3Lr,., -

rmLl, 2(1 + fen･-i) Il-n-r)Ln-: '''L7-

-1

Tab}e IV SIMPLE--SIMPLEshifting pr5cedure t

it

i NI

wr

Nl ---v. - Solution 5.

yw･

i I[ua]

I

IEE)tm. ' ".c ' SJ}t

'N.- N2-- Solution 6:

i

A41,

ua

Tfuf}trm

R

Ll, 2(1+ fen)Jtn-2'''LE

nt o-

-1-

-1

L

IIl

Illl

(-ptim-- .t2> 4e-Ns -pt V Solution 7.

'

$.}t! 's)xt' Ap.- ---K(.il---GiL--g>-- .".m-L>

'iv ee4- Solution 8.

il

- 2(1 + fe,,), k2, O-

1 ww LnrmILn-3''"L3,

o

-1

-Mle-

ca

-M5-

-ua-

ca

mM)1rm

o, . LtsLk 1,

- O, Ll, 2(1 " fen)jLn-2'''L6-

-1

-L2(1 + le,･), k,jL'3,

- Ln-1'''Li,

o'1

o

-1

No. 18 Operational Method for Clapeyron's Theorem

(fer Odd-Number Spans)

33

M

- Ln-2Ln-4'''L2-mt-

"e-

ww o

+ EDtt

' EM-nvswttkn-i -

-- swt

o o- S!nt

' E[{} nv

-EI}ltfen-1 ww

Q

-I Ln-2[Ln-`("'P2"') + Pn-4]+ Pn-2 l

'(K,, + K,le,)

-(Ktn-2 ÷Kn-ilen-1)

+ Ll, 2(1 + kn-i)]IIILn-3[Ln-s("'P3''') +Pn-s] + Pnth3]-

Ln-21

Lt2Pt4 + Pt2

Ln-4 ("'P6'-') + Pn--4I + pn-2

-(Kt, + K,le2)

-(Ktn-2 + Kn-ilen-i)

"+- Ll,2(1 + fen-i)J ill

+ 2L(1 -f- k,), fe,][Lf,Pts -t- P'3]

Ln-3[Ln-s("'P7"') + Pn-s] -t- Pn-3

(fer Even-Nllmber Spans)

M

Ll,2(1 + fen)llLn-2"'L2

'EV}- × -l- EPIIlen pmo-

wwEVt -

o

s]zt

m owrt' kn

-EM -

o

mt

Q

(Kt.-1 + Knkn)

+ Ll, 2(1 + fen)1 l Ln-2[Ln-4 ('''P2'") + Pn-4] -l' Pn-2 l

-1

(Kt, + Khle2)

Ln-1[Ln-3("'P3''') + Pn-3] + Pn-t l

' -(Lt2Pt4+Pt2)

-(K'nmi -1- Knkn) ÷ Ll,2(1 + fen)A Ln-E("'P6''') -t- Pn-2 L

--1

(Ki, + K,le,) + L2(1 + le,), le,j Pi,

Lnml [Ln-3 ('''Ps'--) -t- Pn-3] 'tnv Pn-i 1


Table V. CLAMP-vSIMPLE

No. 18

Shiftin.cr Procedurei

11

N 1 R

(-

1 2

Ni -----i,.

Solution 9.

.eq ,,,I

n>ls. 7

rmM, -

-Mn-1-

E

:i

Ln-2Ln-4'''L2 IM

-2'

-1

o

I } lMi I i I

i

Solution

123 "n N2 '

Solution

10a.

10b.

i [ sJe' I

-dis- /nX l･il---/ I'

I

l･

[M,]

"M,rm

ca atgb-l

iil

'I

-2Ll, 2(1't-knnti)JLn-3'`'L3

3 +4k, - feo -

2, 1, O rm (3 + 4fe2)

-rm-rs. ' O, 1O, Ll, 2(1 -1- kn-i)jLn-3'''L3 -

-]

-1

sJeiWtg-'-e. s--"")

So}.ll. IioR. 11.

,l i

al 89' -.s

n

･-,-- Ns---e.-

Solution 12.

EUzt

x}

/

-M,

M!o

M,-M},rm1 rm

･ '-M, de

ca

mMb rm

l

t

=' 1

Lt2LkV6Lts,

Ln-2Ln--4'"Lie,

-1,

2,

o,

o,

o o-1 o

M =2,

3 -Y 4k,

le2

- O, Ll,

- Lt3LtsLt7

'

2(1 + fen.-i)ILn-3'''Lg

-1

Table VY. CLAISfP-vSIMPLE

Shiiting Procedure

Solution 13a.

-'"K Mt'

Nne7 Ni op

Solution 13b.

i

l N

If1:

1:I

[M,]

-M) un

Mhne!

rmM;1 m

R

i

H

X;lt

SV--ag:i> N2 ,ab s- solution 14' - m,

S,L-Gve-=ctPi) - Ng ---b.r s"- Solution 15.

i'

Ii

ii1

l11

-M2-

M,

Mi

-Mh.

-?va -

ca

.-Mieth

iii

lil,

Ll, 2(1 + kn)jLnm

rm 1' Ln-2'''L2

--2

O1,

--1 N IM2･-･L2

-2

M 1, O -1 'T O, 1 2(1 + k.n) .

1, O, 2, (3 + 4fe2)o, 1, o-, le2 .'- Ln-1"'L3 o,-l, Lt2L14Lt6Lts 2, O, Ll,2(1+,len)jLn-2'''Lio

o

o

1

e!

S"lti

-Nsi. -- I soiution i6. 1 11

uaMl

ma

ua

-Mb-1E

:1

:':' 2,'

3 + 4fe,

fe2

o,

o,

o,

o,

1,

o,

rm -1

-1

-1 -Lt3E.lsL17

-Ln-iLn-3'''Lg

No．18 Operational　Method　for　Clapeyron’s　Theorem

　　　　（for　Odd－Number　Spal裏s）

35

㎜0㎝

　　

@一號　　

@「

［顕～’々η＿1］

　　0　　－

　0＿購～～んη司＿

　　　ン

000

　　　一

一　〇　一

　9＿田～’々アz．．一1＿

Q

Lπ＿2L，～＿盗…L2

…〇一

　一1　一

一・司1・一（…・・…）＋P司・・。一・

L…（・＋・・一1）・ト・・一・L・・一・（…P3…）÷P司・ら一・

　　　　　　　　　　　　　　一κ1　　　　　　　｝　　　＋　秘一3Lπ一5■■．L3　　＿叢（1十々2）κ1　一　ゑ　（κ’1　＋　κ2々2）＿十κ’π＿2十・K麗＿1んη＿1

　　　　　　　　　　瓦　　　　　　　ゑ（κ’・＋嗣一十橘）κ・

（K1アz＿2十瓦～一．1んη＿1）＋L1・2（1＋婦1）」「しか・［L樋（…P・・一）＋P　・葺P囲L

一　工OKOO

一　　　　　｝

十

L’2［L’4（L’6P18÷P’6）＋P’4］＋P’2

Lπ＿2［Lπ＿4（…Plo…）十Pπ＿4］十Pπ．一2

一　　　　一ノ（1　　　　　　　－2　　　　1　　　　－L’・（L’・P’・＋P’・）一Pノ・病（1十々2）κ・「聾（κ11＋κ2々2）

（K1・一・＋K・・一・㈲・L…（・＋・・一1）ll…↓・一・（……）・町訊一・＝　　　　　　　　　　　　　　　　　　　　　　　　　　　　　［．

（for　Eve鷺・Nu紬er　Spa駐s）

i釦～1々・］

0

0

戴｝～’んπ

　　　ン

000勢

一

｝0

　0

分〕～ノ々π

㎜σ　一．．

　0

　0

1璽’＿

Q

　　　　　　　　　　　　　　　　　　　　　0κ’π＿ユ十Kπ々π÷L1，2（1十んη）」　L，置＿2L7z＿墨…L2

＿　　　　　　　　　　　　　　＿　　　　　＿一κ1＿

　　　　　　　　　　　＋L，、一2［L。一曇（…P。…）＋P，掃］＋P，囲

一L2z．＿含Lη＿4…L2

一　1．．6漏．

　　　一L7z＿2［L，z＿4（…P2…）十P7ト4］一Pπ＿2

＿一 P（L

（K’π＿1土《πん2ゆ

－」…

　　　　κ11（’1十」κ2々2　2（1十んの」K1

｝　　

　弓2　π

んP

　十

　う

　ら　P

　（

ゆの　

ヨ

々　㎜　

L｝　｝

　《

　L

　「

　　一

1一・…㌧

　　　　　　　　　　　　　　　　　　　｝0－　　　　L’2［L’墨（L’6P’8＋P’6）＋P’4］＋P’2＋

　　　　　　　　　　　　　　　　　　　＿κL・κ勉一・・魚璽デL…（・殉・｝｝｛一・（…ヒIG…・性・1豊・

徽　　　一十　2（1十秘）κ1　K／1一κ2島

　　　病一L’3（L’5P’7　十　P’5）一P13

　　　　　ん3　＿

一Lπ＿1［L7～＿3（…Pg…）十Pπ＿3］一Pπ＿1

一


Table VXI. CLAMP-CLAMP

No. 18

Shifting Procedure IN Il R

N1.

.Solution 17.

ll

'i I,

i i

I,

-M, rm

rmMh-

j

Iil1

!t

Lnrr2Ln-4'''L2 '1 -2

'

- 2-

-1

-1

N2. Solution 18.

I'

l･

I!

-Ml-i

hth-

1l

i

Ii

Ln-3Ln-s'''L3

-23+ 4k, le2

'

-' 4-3kn-i

2

-1

.- N7-Solution 19.

ilt

II･l･

ll

-Solution

Ns-20.

1Il

1

j

-Ml rm

ua

ta

rmMg-

-M,M

th

ca

.-

. .Mb.=.:,

l

iI

1, O,-2, O, e, - 2,

e, 1,

- Ll2Lt4Lte

-LnrELn-4'''L6

-1

- 2,

3+ 4k2im-･- t k2 o,

rmo,

e, - Lt3LtsLt7 o,

4 + 3len-i,

- Ln--3'''Lg- 2･

-1

Table Vgll. CLAMP--CLAMP

Shifting Procedure INl.

l1

R

Nl . Solution 21.

, mM,i

l rmMh.,

Np -ip- -i

Solution

l

ii

l･

i

ml

IiI

mI,

Lnm2Ln-4'''L2-1 -2

'

-- 4 - 3kn-

2

-!

22.

NM,

-Mh+1

"1 78 fl+1 ?-

--s- N7 --ep-

Solution 23.

MM,

Mn+i

ta

..Mh -l

M,

Mh+1

va

-Mb -

1

1

:!/

1!

i

Ln-itn-3"'L3

-23 + 4le,

fe2'

- 2M

-1

-1

.Hs-xg)-cuEFI -Ns-Solution 24.

! 1, 2, -2, O, e, (4 + 3kn),

- o, -21

-Lr2V4Lt6

- Lnm2Ln-4'''Ls

-1

I!

[III

Ii

-2, O, 3 + 4fe2 hn k2 , O,

e, - 2,

- e, 1,

-U3LtsW7

-tn-iLn-3'''Lg

-1

No. 18 Operational Method fer Clapeyren's Theorem

(for Odd-Number Spans)

37

Q

Ln-2Ln-4'''L2I2,]- -Ktn-1 nv

e- Ln-2IIILn-4("'P2''') + P"m"II m Pn-2

2(1 + kn-i)Ktnmt - (Kln-2 + Kn-!len-i)-

rm Kln-1 ' 2(1 + fe2) m Ln-3"'L3 le2

o-K,

Kfn-1

o

mK,

K, -Kt, + K2fe2

fe2- [Ln-3('''P3''') + P,t-3]

+

Lt2(Lt4Pt6 + Pi4) + Pi2

Ln-2[Ln-4('''Ps''') + Pn-4] + Pn-2

K, ww Kt, f.K2k2 - 2m(1fe+. k2) K,

Ktnrm2 + Kn-lkn-1 m 2(1 + lenml)Ktn-1

Ktn-1 ww

+

Lt3(LtsPt, -f- PIs) + Pt3

Ln-3[Ln-s('''Pg-'') +Pn-s] ÷ Pn-3

( for Even-Number Spans)

Q

Ln-2'''L2-o'

-Kl-+

-2(1 + fen)Kln - (Ktn-i + Knkn) -

- Ln-2[Ln-4('''P2"') + Pn-4] - Pn-2

Ln-iLn-3'''L3 Kl, + K2k2

le2

xi,

2(1 + fe2)

k2

-' -rm Kln-

+K,

o" Ln-i [Ln-3("'Ps'") + Pn-3 l - Pn-i

o KlKln-i ÷ Knlen ff 2(1 + fen)Ktn

+Ln-2l

Lt2(W,Pt6 + Pt4) -i- Pt2

IILn-4('''Ps''') + Pn-4] + Pn-2

- K,Kt, + K2le2

k2 Kln

o

2(1 + k,) Kl ke +

Ln-i [

L!3(LtsPt7 -I- Pls) + Pt3

Ln-3('''Pg"') + Pn-3] + Pn-i

38

r

n

lrM,･

L･

na･

EAr

Ur,

Wr

E

hr

A7';･

l'r

evr

fe7.

K･,

6r

p

NrGr

Lr,

Ex)i,

N. YosmzAwA and B. TANiMoTo No.18

A]PPENI])IXIII. NOTATION

The following symbols kave been adopted for use in this paper:

== support number countered from the extreme left support;

==r any even number representing the number of constituent span of

continuous beam ' ,== span length of r-th span countered from the extreme left span;

== bending momeRt at r-th support of continuous beam ;

= moment of inertia of cross-section of beam in r-th span ;

-- standard moment of inertia ' ' = section modulus of cross-section of beam in r-th span;

== modulus of elasticity ;

= area of bending moment diagram of r-th span calculated asa simple

beam for its loading;

E8r = reactioB at the left or right support of r-th span calculated as a

simple beam subjected to the moment diagram load A. ;

=: settlement of r-th support;

= coeMcient of thermal expansion;

= depth of cross-section of beam in r-th span ;

== difference iR temperature between the upper and lower sides of beam

in r-th span;

" ii･i. Ir,see Eq. 2a;

==

7l:L"---l,see Eq. 2b;

= 'l'/-,'/t',,See Eq. 2c;

K'r = load teym, see Eqs. 4a, 4b, and Table II;

6EL== '7r. kr, see Eq. 4c;

=: -23-Es, see Eq. 4d;

= "eigenrnatrix," see Eq. 10;

= connectioR matrix represeRting the connection conditions between

two eigenmatrices Nrffi and Nr+i, see Eq. 9; .ec'r = 2-by-2 matrices called the "shift operator" betweeR two physical

quantities Nr.i and nvr+i, see Eqs. 12, and 13;

= 2-by-2 unit matrix;

ept' == given external edge moments at the left and right extreme


supports of a continuous beam ;

Kr = 2-by-1 matrix representing external loading conditions, see Eq. 58;

Pr, P'r = shift operators consisting of 2-by-1 matrix, representing external

loading condition, see Eqs. 60 ;

Sr, S'pt == shift operators consisting ef 2-by-1 matrix, yepresentikg influence

of settlement of support, see Eqs. 8!, and 82;

Tr, T'r == shift operators consistin.ff of 2-by-1 matrix, representing influence

of temperature change, see Eqs. 83, and 84;

R = "premultiplier," see Eq. 93, Table I, and Appendix II;

M -- "edge moment matrix," see Eq. 93, Table I, and Appendix II;

Q = "ioad matrix," see Eq. 93, Table I, and Appendix I;

2 = tedl, see Example2;

" ,.. Elll;", see Example3;

L J = row vector; and

{ }= column vector.

PAGE

8

13

13

13

19

20

21

29

LINE

8

5

7

Eq. 54a

Eq. 77

Eq. 83

Eq. 92

5

ERRATA

FOR

Spans

beam subjected

a M>+1 + ･･･

K'n

4ATrkr(1 -F fer+i)

'

operatlon

READ

Spans.

beam is subjected

the

A4ml + ."

Ktm

ZrATrkr)(1 + fer+i)

operatlon:

0perational Method for Clapeyron's Theorem · 2018. 2. 11. · No.18 Operational Methocl for...

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