Analysis of continuos folded plate roofs.pdf

7/28/2019 Analysis of continuos folded plate roofs.pdf

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~ ~

\ I

ANALYSIS OF CONTINUOUS

FOLDED PLATE ROOFS

by

YUNG-PING WANG ) lq

A

THESIS

submitted to the f a c u l t y of the

UNIVERSITY OF MISSOURI AT ROLLA

in p a r t i a l fu l f i l lmen t of the r e q u 1 r ~ m e n t sf o r the

Degree of

MASTER OF SCIENCE IN CIVIL ENGINEERING

Rol la , Missour i

1965

Approved by

M 1\ ~ w( a d v i s o r ) .c ? a d ~ ;y J5 @ a n


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ABSTRACT

This work presen ts an:. a n a ly s i s of cont inuous folded

p l a t e roofs consider ing the e f f e c t s o f r e l a t i v e j o i n t

displacements . For t h i s ana lys is the normal modes of

the l a t e r a l beam v ibra t ion were used a s the form o f the

def lec t ion cu rve and the load ing was s inuso ida l . By us ing

symmetry and anti-symmetry, a poss ib le method o f analyz ing

pr i smat i c fo lded p l a t e roof s comprising one bay t ransverse ly but cont inuous over tw o o r more spans long i tud ina l ly

i s sugges ted .

11


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ACKNOWLEDGEMENTS

The w r i t e r i s deeply g r a t e f u l to Professor Je r ry

H. Bayless , h is adv iso r, fo r a s s i s tanc e and i n s t ruc t io n

i n th e development of t h i s ,study. He would a s lo l ike

to express h i s s incere thanks to Dr. William L. Andrews,.

Professo r o f C i v i l Engineer ing , fo r h is valuable sugges-

t ions andc r i t i c i s m s .

i i i


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TABLE OF CONTENTS

ABSTRACT. . . . . . . . . .ACKNOWLEDGEMENT . . . . . .LIST OF ILLUSTRATIONS . . . . . LIST OF TABLES . . . . .

... . . . .

. . . . .

. . . . .

. . . .

. . . . .

Page

i i

. i i i

v i

LIST OF SYMBOLS . . . . . . . . . . . . v i i .I .

I I .

INTRODUCTION . REVIEW OF LITERATURE

. . . . . . . . . . . . 1. . . . . . . . . 4I I I . PROPOSED METHOD OF ANALYSIS OF CONTINUOUS

FOLDED PLATE ROOFS ... . .IV. CONCLUSIONS . . . . . . . . . . . . . . .

8

. 58

APPENDIX . . . . . . . ' . .BIBLIOGRAPHY. . . .VITA. . . . .

. . . . 60. . . . . . . . . . . . .

64

. 66. .


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Figure

1

2

3

4

5

6

LIST OF ILLUSTRATIONS

Page

Normal Curves . . . . . . 12

Dimensions of Example 1 . 19

Resolu t ion of Ridge Loads . . 20

LongitudinaT St resses a t a Jo in t of Two

Adjacent P la t e s . . . . 22Equilibrium o f Horizonta l Forces

(a) Basic Loading o f Example 1

(b) Longi tud ina l St r e s s e s a t 0.4L

. 24 28.

. ' . . . 33

(c) Longi tud ina l St resses a t In te rmed ia te

Support . . . . . . . . 337 Wil l i o t Diagram fo r Rela t ive J o i n t Displacement 39

8 F i n a l Longi tud ina l S t ~ e s s e sof Example 1 . 43

9 F i n a l Transverse Moments of Example 1 . . 43

10 F i n a l Ve r t i c a l J o i n t Set t lements o f Example 1 43

11 Rela t ionsh ip between Moments and Shearing

12

13

Forces fo r a Uniformiy Loaded Folded Pla te

with Three Continuous Spans . . . . . . ....Dimensions of Example i . . . . . . . . . . .

Moment D is t r ibu t ion Due to an Arbi t r a ry

45

46

Rota t ion o f Example 2 . . . . . . . . . 51


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Table

I .

I I

I I I

LIST OF TABLES

G e n e ra l D a t a o f Example 1 . . . . . . .Page

29

.30l a b Moments Due t o E x t e r n a l Loads . . . R e s o l u t i o n o f Ridge Loads . . . . . 30

IV Free Edge S t r e s s e s R e s u l t i n g from th e

Elemen ta ry A n a l y s i s . . . . . 31S t r e s s D i s t r i b u t i o n . . . . . . . . . .. . . . .3 2

VI Slab Ac t ion and P l a t e Loads Due t o an A r b i t r a r y

.Ro ta t i on . .. . . . . . . . . . . . . . . . . .3 6V II R e s o l u t i o n o f J o i n t Re a c t ions f o r t h e Cor r e c -

t i o n A n a l y s i s . . . . . . . . . . . . . . . .37V II I F re e Edge S t r e s s e s f o r an A r b i t r a r y R o t a t i o n . 37

IX S t r e s s D i s t r i b u t i o n R e s u l t i n g from an A r b i t r a r y

X

XI

XII

X I I I

XIV

R o t a t i o n . . . . . . . ) 8F i n a l L o n g i t u d i n a l S t r e s s e s o f Example 1 . . 1

. .4 2i n a l Tr a nsve r se Moments o f Example 1 .

F i n a l D e f l e c t i o n s o f Example 1 . .

Genera l Data o f Example 2. . . . .Slab Moments Due t o E x t e r n a l Loads

..

.

. 42

47

47

XVF re e Edge S t r e s s e s from th e Elemen ta ry A n a l y s i s .4 8

XVI S t r e s s D i s t r i b u t i o n R e s u l t i n g from t h e

Elemen ta ry A n a l y s i s . . . . . 50

XVII S t r e s s D i s t r i b u t i o n R e s u l t i n g from an

A r b i t r a r y R o t a t i o n 52

XVIII F i n a l L o n g i t u d i n a l S t r e s s e s o f Example 2 56


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Table Page

XIX Fina l T ~ n s v e r s eMoments and De f l e c t i o n s of

Example 2. " . . . .. . )7

v i i


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LIST OF SYMBOLS

~ c ross sec t iona l area of pla te n

= ~ r 1 z o n t a 1length of p la te n ( s l s b seot1on) ca r ry over f a c t o r from j o in t n to j o i n t n+l

= s t r e s s d i s t r i bu t ion fac to r a t j o i n t n of

p la te n+l

f t , f b , f n = long i tud ina l f i be r s t r e s s e s 1 n the p l a t e s a t

E

I

h

L

M

top , a t bottom and a t j o i n tn

= modulus of e la s t1o1 ty

= moment of i n e r t i a

= p l a t e height ( s lab sec t ion)

= span l eng th

= bending moment

= normal load= r e s u l t a n t shear ing force a t j o i n t n

= p la t e loads on p la t e n due to s lab reac t ion

a t j o in t n-1

R = s lab reac t ion

kn = fac to r fo r the ac tua l j o in t displacement

r e s u l t i n g from an a r b i t r a r y ro t a t i on o f pla te n

S = sec t ion modulus. Tn = l o n g i t u d i n a l shear ing force a t j o i n t n

t = p la t e t h i ckness

Vn = v e r t i c a l jo1nt se t t l ement o f j o i n t n

v = shearing s t ressesI ' 1 ' .


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= d e f l e c t i o n of plate n in 1ts own plane

~ the angle between n and n+l a t j o in t n= r e l a t ive j o in t displacements of p l a t e n

= the angle between p la te n and the horizonta l


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2

Since one o f the assumptions made in folded plate

roof deAign i s t h a t the suppor t ing members {diaphragms,beamA, frames, e t c . ) are : infi .nltely s t i f f in t h e i r own

planes and comple te ly f l ex ib l e normal to t h e i r own planes ,

folded p la t e s t ruc tu re s continuous over two spans l ong i -

tud ina l ly might be cons idered as two separa te spans with

one end simply supported and the o the r b u i l t - i n . I f there

a re more than two spans, the s t r u c tu r e could be analyzed

by assuming t h a t the middle spans have both ends b u i l t - i n

and the e x t e r i o r span has on e simply suppor ted end and one

b u i l t - i n end.

I t i s necessary to s e l e c t a s inuso ida l load so as to

d e f l e c t the s lab to conform with the deformed p la t e s . Th e

d i s t r i b u t i o n of these s inuso ida l loads a long the s t ruc tu re

i s according to the normal funct ion of f ree v ib r a t i o n ,

which wil l make the pla te def lec t ion propor t iona l to the

load d i s t r i b u t i o n . The use of the normal funct ion r e s u l t s

in a considerab ly s impl i f i ed procedure fo r f ind ing the

s t r e s s e s and d e f l e c t i o n s in cont inuous s t r u c t u r e s , r egard -

l e s s of the type of e x t e r n a l load a c t i n g on the s t ruc tu re .

In ana lyz ing cont inuous fo lded p la t e s t r u c t u r e s , thef o l l ~ w i n gbasic assumptions w i l l be fol lowed which are

recommended by the ASCE Task Committee:

( l) Th m b ~ 1 $ lt ~ h a m o K ~ n ~ a u and l l n ~ ~ r l r- 1 - ~ t l a .

(2) The ac tua l de f l ec t ions are minor- r e l a t i v e t o the


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overa l l conf igura t ion of the s t ruc ture . Conse-

quent ly, equilibrium conditions fo r the loadeds t r u ~ t 1 1 r emay be developed using the c o n f i g u r a ~

t ion of the undeflected s t ruc ture .

( J ) The pr inc ip le of super-posi t ion holds; th i s

assumption i s ac tua l ly derivable from the pre

vious two assumptions.

(4) Longi tudinal jo in t s a re fu l ly monoli thic with

the s lab ac t ing continuously through the jo in t s .

(5) Each suppor t ing end diaphram i s i n f i n i t e l y s t i f f

p a r a l l e l to i t s own plane but 1s per fec t ly

f lex ib le normal to 1 ts plane.


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4

I I . REVIEW OF LITERATURE

The p r inc ip l e of fo lded p la t e oons t ruo t lon was f i r s t

d ~ v e l o p e dby Mr. o. Ehlers and Mr. Creamer(l , 2 ) 1n Germany

in 19JO. They considered the var ious p l a t e elements as

beams suppor ted a t the j o i n t s and end diaphragms. Alrmg

the long1 tud 1nal edges , the p l a t e s were assumed to be con-

nected by hinged j o i n t s . They proposed a fo lded pla te

theory based on a l i n e a r v a r i a t ion of l o n g i t u d in a l s t r e s sin each pla te bu t neg lec ted the e f f e c t s of the r e l a t ive

displacements of the j o i n t s . Since t h a t t ime , there have

been numerous papers wri t t en on the sub jec t . Messers.

Winter and P e i ( J ) publ i shed a paper in 1947 in which they

t ransformed the a lgeb ra ic so lu t ion in to a s t r e s s d i s t r i -

bution method, which has the advantage o f numer ica l sim-

p l i o i t y over the a lgeb ra ic procedure.

The f i r s t method to take in to account the e f f e c t of

r e l a t i v e j o i n t displacement was proposed by Messers.

Gruber an d Gruenlng( 4 ,5) . For determinat ion of the r idge

moments an d disp lacements , Mr. Gruber developed hi s so lu

t i o n in the form o f s imul taneous d i f f e r e n t i a l equat ions

of the four th order. Consequently, he concluded t h a t the

in f luence of the r i g i d connect ions ought not to be

neg lec ted .

aeoently, M ~ .I . Oaarar< 6 > e n ~M ~ .o. 1 t ~ h a k 1 ( ? . S )

have introduced methods which cons ider sepa ra te ly the


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6

ha s been used and has given sa t i s fac to ry r e s u l t s 1s topropor t ion the long i tud ina l s t r e s se s over the support and

a t midspan on the bas is of the moments crea ted 1n a con-

t inuous beam whose spans a te equal to those of the folded

p la te . In t h i s approximation, the t ransverse d i s t r i bu t ion

i s based on an e ffec t iv e span length equal to the dis tance

between the po in t s of i n f l ec t io n of the oontinuouR beam.

Mr. Ashdown(l 2 ) presented a complete ca lcu la t ion fo r

a th ree span continuous pr ismat ic roof but neglec ted the

e f f e c t of the r e l a t iv e jo in t displacement . He assumed

t h a t a p l a t e which i s continuous over suppor t ing s t i f f e n -

e r s can be considered as an ordinary continuous beam fo r

the de te rmina t ion o f the long i tud ina l bending momP.nts a t

the ends of any span.

As fo r the continuous folded p la t e s t ruc tu re consid-

ering the e f f e c t of r e l a t iv e jo in t displacement , Mr. D.

Yitzhaki(? ,S) o r i g i n a t e d the p a r t i c u l a r loading and s lope -

def lec t ion method for analyz ing continuous two span folded

p l a t e s t ru c tu res .

An an a ly t i ca l s o l u t i o n f o r the i n t e r i o r panel of amult ip le span, mult ip le bay, r i b l e s s pr ismat ic s h e l l was

presented by Lee, Pulmano and Li n ( l J ) in February, 1965.

The general approach i s s imi l a r to the t rea tment of con-

t inuous r i b l e s s cy l ind r i ca l s h e l l s , but the study i s

l imi ted to the 1 n v e s t 1 ~ a t 1 o no f the i n t e r i n ~n A n ~ ln ~


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7

loads uniformly d i s t r i bu t ed in the longitudinal d irect ion.

I tis a lso necessary to solve 8 r simultaneous l inear

equat ions , where r i s the number of p l a t e s , fol"' each

harmonic of ttlhe t r igonometr ic se r i e s .

The method developed in t h i s thes i s i s a synthesis

of many methods out l ined above. I t can be appl ied to

mult i - span continuous folded p l a t e roofs under symmetrical

loadings which include d i s t r i bu t ed loads, concentrated

loads and inc l ined loads . In order to make a comparison,

the author of t h i s paper used the same assumptions o f

loading and o t h e r condi t ions o f Mr. Yitzhaki and

Mr. Ashdown and extended Mr. Gaafar ' s method to two an d

th ree-span continuous folded p l a t e roofs .


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I I I . PROPOSED METI10D OF ANALYSIS OF CONTINUOUS

FOLDED PLATE ROOFS

The nomplexity of the ana lys i s of continuous folded

p l ~ t e si s du e pr imar i ly to the f a c t t h a t end r e s t r a i n t of

continuous fo lded p l a t e s c rea te s l ong i tud ina l s t r esses a t

the in te rmedia te suppor ts which are i n f i n i t e l y s t i f f in

the plane of loads and are assumed as clamped.

8

In dea l ing with continuous folded p la t e s with twoequal spans, s ince the loading i s symmetrical about the

in te rmedia te support , only on e span need be inves t iga ted .

The s t a t i c a l behavior of every span i s t h a t of a single

s h e l l , b u i l t - i n a t the middle t r aver se and f r ee ly sup

ported a t the ou te r t r ave rse . The s t r e s s e s an d e l a s t i c

curves a re s imi l a r to t h a t of a beam with one en d b u i l t - i n

and the other f r ee ly supported. For th ree-span and mul t i

span continUOllS fo lded p l a t e s , the same assumption wi l l be

made in e x t e r i o r spans , and the support condi t ion of

in te rmedia te spans w i l l be cons idered as b u i l t - i n a t both

ends.

The a n a ly s i s i s div ided in to three pa r t s in the same

manner as the method o f ana lys i s fo r simply supported

s h e l l s , and in a d d i t i o n , the e f f e c t of con t inu i ty over

the suppor ts i s considered .


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A. Elementary Analys is

The f i r s t s tep in the a n a ly s i s i s the computation

of the fo rces and of the t r ansver se and l ong i tud ina l

s t r e s se s ac t ing a t the edges of each p l a t e element ,

neg lec t ing the e f f e c t of the r e l a t i v e displacement of the

j o i n t s .

The roo f i n the t r ansver se d i r ec t io n i s considered

9

to be a cont inuous one way s l ab supported on r j g i d sup

por t s a t the j o i n t s . All loads ca r r i ed t r ansver se ly to

the j o i n t s a re considered t o be t r an s fe r r ed longi tudinal ly

to the end support ing members by the p l a t e s ac t ing as

inc l ined simple beams. The reac t ions a t the jo in t s are

reso lved in to p l a t e loads i n the p lanes of the p la te s .

Longi tudinal s t r e s s e s w i l l be determined from these pla te

loads , and cor rec ted i n a manner s im i l a r to the moment-d i s t r i b u t i o n method. From the equal ized edge s t r e s se s ,

the p la t e de f l ec t ion a t 0.41 of the e x t e r i o r span and a t

mid-span of the middle span w i l l be obta ined .

B. Correc t ion Analys is

The second s t ep i n the ana lys i s i s to provide fo r the

e f f e c t t h a t the r e l a t i v e t r ansver se disp lacement of the

j o in t s has on the t r ansver se and l o n g i t u d in a l s t r e s s e s .

This opera t ion i s most e a s i l y accomplished by apply

ing a r b i t r a r y r e l a t i v e j o i n t displacements success ively to

each p la t e . and computing the resu l t ing p l a t e def lec t ions .


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A number of simultaneous equations equal tc the

number of re s t ra ined p l a t e s canbe s e t

up from thegeometrical r e l a t i on and aolved fo r the aotual rel&tlve

jo in t displacements.

C, Superposit ion

10

The r esu l t s of the elementary ana lys is are added

a lgebra ica l ly to the corresponding values jn the correc

t ion ana lys is to give the f i na l forces , moments, s t re ssesan d displacements.

D. Normal Curves

The pr inc ipa l problem assoc ia ted with the analys is

of folded p la tes i s t h a t of making the displacements

computed from the long i tud ina l behavior compatible with

the displacements obtained from the t ransverse behavior.

In a s t r i c t sense no t only must t h i s equa l i ty o f

displacements be s a t i s f i e d at. a few points along a s t r i p ,

but the requirement should be sa t i s f i ed a t a l l poin ts on

the surface . To secure t h i s , i t i s necessary to express

the ex te rna l loads as a sinusoidal load.

In the case of s ingle-span roofs symmetrically loadedwith respect to the middle of the span, the re la t ive

def lec t ions can be represented by h a l f of a s1ne curve,

instead of assuming them to vary as the e l a s t i c l ine of

the corresponding loaded beam.


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11

In the case of mult ispan roofs , or of roofs on wh1oh

the loads are f a r from being symmetrical about the middle

o f the span, t h i s sine curve t reatment cannot be used with

accuracy, and a spec i f i c form of e l a s t i c curves, known as

the normal modes of l a t e r a l beam vibrat ions have to be

a d o ~ted .

The form of the def lec t ion curve of a folded plate i s

the same as t h a t of a beam, which depends mainly on i t s

support condit ions, regardless of the longitudinal var ia

t ion of the load. The use of normal curves would grea t ly

simplify the ana ly t i ca l t reatment in continuous folded

plate . design for- the two most 1mpol"tant cases : (1) the

beam b u i l t - i n a t one end and f ree ly supported a t the o t h e ~

and. (2) the beam with both ends b u i l t - i n .

"Normal Mode" of vibra t ion of the beam i s a def in i teshape 1n which the beam wi l l def lec t while vibrat ing

harmonical ly. The mathematical expressions which define

the normal modes are ca l l ed charao t e r i s t i c funct ions.

For each type of beam with specif ied end condi t ions

there i s an i n f i n i t e s e t of these functions.

The funct ion of the normal modes wil l be derived

from the condit ion of iden t i ty in form of the load an d

the corresponding e l a s t i c curves, expressed in the form:

N ( X ) = ky ( X ) ( 1 )


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Figure 1 . Normal Curves

Subs t i tu t ing t h i s r e l a t io n in to the d i ffe ren t i a l

equation of the e l a s t i c curve

12

4N(x) = EI d y ( x )

dx 4(2 )

The load and def lec t ion curves wil l be expressed 1n

the form N(x) = N0 f (x) , y = y0 f (x ) , where N 0 , y0 , areth e maximum ordinates of the load an d deflect ion curves,

f(x) i s a funct ion of the coordinate x defining the shape

of the normal mode of v ib ra t ion under consideration, whichis re fe r red to as the normal funct ion . Equation ( l )

becomesIV

EI f (x ) = kf(x) ( 3)

from which the normal functions for any p ar t i cu la r case

can be obta ined , and the genera l solu t ion of th i s equa-

t ion wil l have the following form:f(x) = c1 (cos nx +co sh nx) + c2 (cos nx +co sh nx)

where

+ C)(sin nx + s inh nx) + c4 (s1n nx + sinh nx) (4)

~n l } JET


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In Eq. (4) c1 , c2 , c3 , c4 are constants which should bedetermined i n each p ar t i cu la r case from the condit ions a t

the ends of the beam.

1 . Beam with one end b u i l t - i n , one end simPlY

supported.

1 )

Assuming t h a t the l e f t end ( x ~ o )i s simply supported,

the fol lowing end condit ions are obtained:

(a ) f ( x ) =0, x=O,)

(b) f(x):O, x=L,

( c ) f ' ( x ) = O , x=L, (d ) f"(x)=O, x=O.

The condi t ions of (a) and (d) y ie ld c1 =C 2=0 1n the.general so lu t ion of Eq. (4). The remaining two cond1t1ons

give the fol lowing equat ions :

s1 nh nL) = 0 ( 5 )

c3 (cos nL + cosh nL) + c4 (cos nL - cosh nL) = 0 (6)

A so lu t ion fo r the cons tan ts c3 and c4 , d i ff e r en t fromzero, can be obtained only when the determinant of Eqs.

(5 ) and (6 ) i s equal to zero. Therefore,

tanh nL = tan nL (?)The consecutive roo ts of t h i s equation a re :

3.9266023 7.06858275 10.21017613 1 ) . 35176878

For purposes of design only the f i r s t term needs to

be ~ s e l o i , .The e ff e o t o f th I\.\OceeC!.1:nc term w i l l 'bo

important only in the v ic in i ty of the suppor ts , and wil l


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not p ~ o d u o eany s ig n i f i can t s t resses a t the sect ion of

maximum def lec t ion and maximum moment 1n the span.

Subs t i tu t ing the n1L value into Eq. (5) And Eq. (6).

the r a t i o c3;c 4 fo r the f i r s t mode of Vibration can be

ca lcu la ted and the shape of the dAfleot ion curve wil l

then be obtained.

From Eq. (8 ) , i t was found t ha t the m a x ~ m u mdef lec

t ion would occur a t approxtmately x = 0.419L, an d the

maximum moment a t x = 0. )8)L. I t would n ot make a l a ~ g edi ffe rence i f 0.4L i s s l ec ted fo r maximum moment and

maximum def lec t ion . This approximation, while acceptable

fo r determining the c r i t i c a l s t resses and moments, tends

to obscure the exact d i s t ~ i b u t i o nof s t r esses .

When f(x)x=0. 4L = 1.0641)76, the fol lowing equations

are obtained:

def lec t ion curve -

fyN= 1 . 06 t137 6(s in ) .9266f + 0.02787494 s1nh).9266f>moment curve -

-1.0641376 L2 II -L 2 I IfMN = () .9266)2 X 0.93586229 fyN =13.56 fyN

= 0 9 J S ~ 62 2 9( - s i n ) . 9 2 6 6 f +0.02787494 sinhJ.9266f)I

shear curve -r = r ~v-:ol 626jl.379t? ...

'" . f r ~ z .3~ r ) r i i iSN ) ,92 ) x 0.9?21251 y ~ 5 . )1 yN t r : 9 7 ~ 1 2 5 ' f( .. ece , . 9166f + 0. 02111494 oosh,, 9166f)


24/74

load curve -

= t 4 fiV L4 IVfN J N = --=--- f( ) .9 2

6 6 ) ~y 237.72 yN

l X X= l.o6 4137 6(sinJ.9266E + 0.02787494 sinhJ.9266 1 )

maximum de f l ec t ion -

N L 4 M L 20 0

Yo = 2J7.72EI = l - ) ~ . 5 - 6 ~ E - I

maximum moment -

minimum moment -

maximum shear -

y EI0so = 55.30 )L

-N L0

N0= 4 . ) 0

a t x = 0.4L

a t X = 0 .. 4L

2

a t x = L

2. Beam with both ends b u i l t - i n .

In the case of a beam with both ends f ixed the

boundary cond i t ions a re

(a) f(x}=O, x=O, (b ) f ' (x)=O, x=O,

( C ) f ( X ) =0 , X =L , (d) f ' (x )=O, x=L,

In order to s a t i s f y the cond i t ions ( a ) and (b) the

cons t an t s o1 a\14 g' s:hul4 l.;)e eq\Ulll to aero U'l Jtq, ( 4.)

15


25/74

l6

and from condi t ions (o) and (d ) we obta in

C2 (cos nL - cosh nL) + c4 (s1n nL - s inh nL) 0 (q )

c 2 (s1n nL + s i n h nL) + c4 ( .. cos nL+oosh nL) = 0 (10)

in which the frequency equation wil l be

cos nL cosh nL = 1

The f i r s t four consecutive roo ts of t h t s equat ion

are as fol lows:

4.7)00408 7.8532046 10.9956078 14.1371655

(11)

Subs t i tu t ing the n 1L value in to Eqs. (9) and (10),

the shape of def lec t ion curve wil l be obta ined , when

f ( x ) 1.58815 a t X = 0 .5L

def lec t ion curve -

moment curve -t._ I ,JJ81S ~ f ..rz L l .r--

M N - ( 4 : 7 J ) Z !/II.IZ.IS"6.S' !JI'I 1'1.13 yN

shear curve -

l /. S331,f" L l 1IL L 3 / . .1/l'J"N- (417J)lJCI,,fW{) -411 - ~ a ~ J J!Jit/


26/74

17

load O U l" Ve -..r L+ .:. 1..4 [ . j lrv l +. 7S ) !I N * ' .roo t..r !IN

- 1 r : ~ I S '( r ~ s h 4 7 . 1 f '- c o $ -I.I.J;f-J- q p . l i ' ( S I N A ~ 1 J - f - -SJit/4?4)1

maximum def lec t ion -

Yo = .500.,5,5EI =

maximum moment -

minimum moment -

maximum shear -

N Ls = 0 ~

0 J.7.5

l7 . lEI a t X = O.,SL

a t X = 0.,5L

a t x = 0

In comparing the def lec t ion curves caused by the

normal curve load and uniform load fo r d i f f e r e n t support

condit ions.(Appendix) i t i s observed t ha t the discrepancy

in the ord ina te s of the normal curve corresponding to the

ordina tes of the def lec t ion curve caused by uniform load

i s ev iden t ly qui te small . Hence, the e r r o r introduced

in to the a n a l y s i s by replacing a de f l ec t i on curve w1th a

normal curve can be neglec ted .


27/74

18

The e l a s t i c curve of a beam with on e f ixed en d an d

one simply s ~ p p o r t e den d tha t ca r r i e s a uniform load,

having the maximum def lec t ion y 0 a t 0.4L, i s expressed by

3 4fyw = ).86(-Lx - ~ + 2x )

L3 ~

The e l a s t i c curve produced by a uniform load fo r a

beam with both ends f ixed , having the maximum def lec t ion

y 0 a t mid-span, i s

E. Continuous Folded Pla tes with Two Equal Spans.

A continuous prismat ic folded pla te of the shape

shown in Figure 2, with two equal spans, and continuous

over the middle t raverse w i l l be analyzed. Since the

loading i s symmetrical about the center l ine suppor t ,

only one span need be considered.

1. Resolut ion of r idge loads.

Consider a prismat ic folded pla te loaded along a l l

j o i n t s . Since in the ac tua l s t ruc ture there are no

supports a t the var ious j o in t s , forces of equal but

opposite magnitude to the reac t ions are appl ied to the

p la te s t r u c t ~ e .These ridge loads are assumed to be

res i s ted by the p la t e s act ing long1tud1nal+Y as deep

beama. Per th1a ~ u ~ o e ethe ~ e a o t i o n sa re resolved i n to

components p a r a l l e l to the p la t e s as shown in Figure J.


28/74

19

J o i n t s

Traver ses

P l a t e s

-Suppor t ing member

(a )

Figure 2. Dimensions of Example 1


29/74

s,,n+,

(b )

F igu re J .

Sm-t,n

( o)

Resolut ion or Ridge L0ads


30/74

From Figure

8n+l ,nRn+l

J ( o ) y using the sine

=sln(90 -

~ n +2

)

s in etn+l,n+2

cos tSn+ 2= R

n+l sin oln+l ,n+ 2

law

By the same reasoning Rn i s resolved in to i t s com

ponents sn ,n+l ' and sn,n-1

. cos ~ nS = Rn,n+l n s in ~ +ln,n

I t 1s seen t h a t the t o t a l load act ing in the plane

o t p la t e n+l 1s

pn+l = 5n+l ,n - 5n,n+lcos ~ n +2 cos ~ n

= Rn+l - Rs in dn+l,n+ 2 n s in ~ n , n + l

The general form of pla te load Pn wil l be

.21

cos ~ n + l

s in ~ , n + lcos n- l

- R 1 - - : - - ~ ~ - = -n- s in ~ - l , n

(12)

2 . Stress d is t r ibu t ion method.

These p la te loads are appl ied to the p la te s as loads

acting along the e n t i ~ elength as shown in Figure 4. In

6omput1ng the s t r e s se s the p la t es are assumed a t f i r s t to

act independently of each other. Moreover 1 t i s assumed

t h l ~ the pla tes ar$ hQmo&eneoua ana. thet>4ttol"G t h ~ s t r e sd

is equal to the moment divided b y ~ t h esect ion modulus.

Because o f di fference in loading and depth the


31/74

common Junction. Since junction n i s common to both

plates nand

n-1;the

s t r a in sand hence the

s t r essthere

must be the same fo r both pla tes . But th is can g e n e ~ a l l y

only be possible 1 f a longitudinal shearing force Tn i s

acting along t h i s j o in t which tends to equal ize the

s t resses 1n both p la tes meeting a t the common Junct ion.

7 4 ~ Z -th1. /fl,- - - j ~ :6 =T 6:!- fh ! +__:r_h, ,4 h - A

- ~ 6.Z]:f = 74T. " , . z l,q

(a )

,.,..,

/_! 11 ~ ~ ~ : ~.,

1 + ~ 14~ -i,.,_,

Tn-t ~ ~ z . . , f . ~ " .,c,;_,..(b )

P!gure 4. Long1 tud1nal Stresses a t e Jo in t ot Two

Ad i a c e n t rna tes


32/74


33/74

24

For pla te n+l a t junc t ion n the d i s t r i b u t i o n fac to r

i s :

A0n , n + l = n (l6b)An+ An+lNow i t i s seen from Figure 4 t h a t the snearing force

T causes a t j unc t ion n-1 of pla te n the s t r e s s -2Tn/Annan d a t junc t ion n+l o f pla te n+l the s t re ss 2Tn/An+l

Comparing these s t r e s s e s with those caused by Tn a t

junction n, i t w i l l be found t h a t they are minus one-ha l f

of t h e i r magnitude. This denotes t h a t the car ry-over

fac to r i s -1 /2 .

) . Shearing s t r e s s e s .

For a. complete design , i t i s necessary to check the

shear ing s t r e s s e s . The shearing s t re sses v a t an y poin t

in the folded p l a t e s are induced by the shear ing forces T,

which can be ca lcu la t ed from the equil ibrium of the hor i

zontal forces (F igure .5).

Figure .5. Equilibrium o f Horizonta l Forces


34/74

T = J dAThe r esu l t an t shearing forces N can be obtained by

N = j TdxThus beg1n1ng a t the l e f t edge, the resu l tan t forces

a t the r idges wi l l be:

N1 = -1 /2 ( f 0 + f 1 )A11 /2 ( f 1 + r 2 )A 2

1 /2 ( f 2 + f))AJ

The genera l form can be wri t ten a s :

The l o n g i t u d i n a l shearing force N1 a t any poin t

between j o in t s i s

NY = N2 1/2 (f2 + fy) ty

or N = N - 1/2 t y ( f n - l + f h-Y)y n-1 n-1 h-1/2 ty f l.n h

The r e s u l t a n t shear ing force a t the middle of the

plates can be wr i t t e n :

Ny

(17)

(18)

(19)

(20)

Since the va r i a t i on of the . longi tud1nal shear ing

force NY i s s imi l a r to the moment Mrt due to the load Pn'~ ' v a r ~ par.bol1oal1t.


35/74

26

For a simply supported s t ruc ture , subjected to a

uniformly d i s t r ib u ted load , M=

wL 2/8 , and the momentmax

a t an y distance x from the support 1s

M = wx(L-x} = M 4 ; ~ J L - x lX 2 max L2N = N 4x{L-xly max L2

Because'Ny i s propor t ional to Mx' then

NY = (Nma:x/Mmax> /Mx' anddN Nv = 1 _ _ z = 4 max (L-2x}

t dx tL2

N dM!. max xvmax = t M dxmax

= Nmax wL/2 =t wL2/8

N V= max xtMmax

N4 maxtL

and i f loaded by a s ine curve load the shearing s t r essbecomes

M = M s i n ~max L7 t X

N = Nmax s in -r-N 1 t

max ' c o s ~tL L=

(21)

(22)

( 2 ) )

(24)

(25)

(26)

( 27 )

Therefore, combining Eqs. (23) , and (27) , the t o t a l

shearing s t r e s s can be obtained. For prac t i ca l design

the shearing s t r e s s obtained by the sine curve load or

normal curve load i s quite small compared with the value

obtained by the e l e m e n t a ~analys1st hence, the s&oond.tet'lll, Eq, ( .27), ean ba negleotad..


36/74

.

27

Theore t ica l ly, fo r a beam fixed a t one end, supported

a t the other, subjec ted to an uniform d i s t r i b u t e d load,

th e shearing s t r e s se s can be obtained from the following

der ivat ions. From Table I of the Appendix,

Mmax = wL2/14.28

)wL wx 2MX = 8 X - ~ = M (5.36x _max L

2N = N ( .2..:..J.2.x_ 7, 14x )

y max L L22v = 1 N (..2..:..J.Q _ 14,28x )

t max L L2

27 .14x )

L2

For a beam f ixed a t both ends, subjected to a

uniform dis t r ibu ted l oad , the s h e a ~ l n gs t r ess wi l l be

expressed as fo l lows:

from Table I I of the Appendix,

= M ( 2x 2max L

N = N (2Lx - 2y max

v _ 1 N ( 2 _ 24x)- t max L L2

Prac t i ca l l y, as the shear ing s t r e s se s .are small

t h ~ o u g h o u tthe e n t i ~ es t ~ u e t u ~ e ,the v a l u ~ w111 b

{28)

(29 )

(30)

(Jl)

(J2)

(JJ)

obtained by consider ing the p la te as a simply supported

beam fo r convenience and s impl ic i ty.


37/74

EXAMPLE 1.

The fo lded p la te roof with two equal spans shown inFigure 2 wi l l be analyzed fo r i t s own weight only.

The loading was computed as fo l lows:

Weight of p la t e = 1/4 x 150 = 37.5 p sfWeight o f edge beam = 150 x 7/12 x 4 = 350 l b / f t .

Table I provides the general data of the cross

sect ion.

a . Elementary Analysis:

28

i . Transverse s lab ana lys is . A uni t s t r i p taken

.from the fo lded p la tes 1s assumed to a c t as a continuous

one way slab on unyielding supports . The transverse s lab

moments are determined in Table I I , an d the reac t ions a t

each j o i n t a re computed.

The moment d i s t r ib u t io n fac tors a t j o in t 2 are- 3/4 1D21 - l + 314 = 0.428 n 23 = 1 ~ + - - J ~ / ~ = 0.572

Figure 6 ( a ) . Basic Loading of Example 1

The f ixed end moment wi l l be

MF 2 l = 1/8 X 7. 794 X 9 X .37. 5 = )28. 8 f t - l b / f t .

MF 2 '3 = MF 32 = 1/12 X 8. 86) X 9 X 37 5 = 249.18 ft-ll:/ft


38/74

29

TABLE I

General Data

(a) Pla tes '

Plate h, in t , i n A, in s , i n p5 sins& cosNo. f e e t in . sq. f t . cu. f t .

1 4 .07 2.333 1.556 90 1.00

02 9 .0 3 2. 250 3.375 30 0,50 0.8663 9 .0 3 2.250 3.375 10 0.174 0.985

(b) Jo in t s

Jo in t ~ s i n ~ co t oe.

0 0 0 01 60 0.866 0.5762 20 0,342 2.750

3 20 0.342 2.750

(c) Moment d i s t r i b u t i o n constants

Jo in t P la t e Rela t ive S t i f f n e s s Dis t r ibu t ion

.o 1 --- ----1 l KlO = 0 0

2 Kl2 = 4 1

2 2 K21 = 3/4(4) = 3 0,4283 K ~3 = 4 0.572

3 3 K32 = 4 0,.500

4 K34 = 4 0.500". , . . -


39/74

10

, -. - - ---

Jo in t

1.?

) 0

TABLE I I

Moments, Shears and Jo in t Reactions in Transverse

One Way Slab a t 0.4L from the Outer Support

1 2 3 Jo in t

12 21 23 32 Member

0.428 0.572 0 Dis t . fac tor- - ~ - - - -- ~ - - - - - - - H ~

328.8 .-249. 2 249.2 F. E. Moment-34.1 -45.5 Dis t r ibu t ion

-22.8 Carry over

I 9 4 ~ 7-294.7 226.4 Fina l moment-37.8 I 37.8 7.7 -7 .7 M/hcos 9S168.8 168.8 . 168.8 I 168.8 wh/2

480.9. )83.08 )22.1 J o i n t reac t ion

i1 . Longi tudinal pla te analy.!....

(1) Pla te Loads. The v er t i ca l jo in t reac t ions

are resolved in to components p a r a l l e l to the con t i

guous p la tes by using Eq. (12). The pla te loads

act ing on each pla te are tabulated in Table I I I .

TABLE I I I

Resolution of Ridge Loads

(1) (2) (3 ) (4) (5) (6)Reaction cos .n+l =(l)x(2) cos n - l R Pla te

l b . / f ~ n-1 Loadss in cJ. s in ~ n - l x ( 4 )n l b . / f t .480.9 480.90~ R ~ n ? . 8 7 7 . 110l.q5 0 0 1101.9.5


40/74

(2) Free edge s t r e s se s . I t i s assumed tempor

a r i l yt h a t

eachpla te

bendsindependently

dueto

p la te l oads . The maximum s t r e s s an d def lec t ion

occur approximately a t 0.4L from the ou te r support .

The moment due to a uniform load w i l l be ( r e f e r to

Table I of th e Appendix)

31

M0.4LPL 2

(34)14.28

M' PL 2 (35)= s = 14.28 s

The f ree edge s t r e s s e s are t abu la ted in Table IV

(3) Free edge s t r e s s d i s t r i b u t i o n . The f ree

edge s t r e s se s a re d i s t r i b u t e d in order to determine

the a c t u a l edge s t r e s s e s , which must be equal a t the

j o i n t .

TABLE IV

Free Edge Stresses

Plate Pla te s 2 2 fb = - f t

12

3

Load L 65l b / f t . cu. f t . 14,28

=14.28 k ip / sq . f t .

480.9 1.556 295.87 91.451101.9

3.375295.87 96.60

-44 .36 3.375 295.87 -3 .89

The f ree edge s t r e s s d i s t r i bu t ion i s shown in

Table V; and i s p l o t t e d in Figure 6. The s t r e s s

d i s t r i b u t i o n f ac to r s , by us ing Eq. (16) , a re


41/74


42/74

f- .........~ - ~ - - : - ; 3'--...,"""


43/74

J4.

(4} Pla te def lec t ions . From the equal ized edge

s t re sses , the p la t e def lec t ions a t o.4L can be com

puted. For a uniform load , the def lec t ion l s (from

Table I of the Appendix)

ML 2Yo.4L = 12.99 EI ( ) 6 )

in which the moment a t x = o.4L l s

f b - f

M0.4L = 2 t S ( J? )

subs t i tu t ing Eq . ( )7) in to Eq . (J6)

1 fb - f t SL 2Yo.4L = 12.99( 2 } ~

For a rec tangular pla te ~ = ~

1 fb - f t L2Yo.4L = 12.99( h )E ( 38)

Assuming E i s 105 k i p / s q . f t . , the p la t e def lec

t ions in terms of the free edge s t r e s se s a t X = 0.4Lare found as fo l lows:

2(-27.84 - 15.86)x65 = _ 0 0162 f t12.99x9xE

2(-4.99 + 27.84)x65 = 0 00826 . f t12.99x9xE

(48.38 +4 . 4 9 ~ x 6 5 2

= 0.0433 f t .12.99x xEb. Correct ion Analysl s:

1. TQlll syer me .slab anBlrv:s ih The anal;ys t s i s made

fo r an a r b i t r a r y ro ta t io n of the p la t e a t the sec t ion

o.4L f ~ o mthe ou te r suppor t . The f ixed end moment a t


44/74


45/74

(a)

ptJ_o

(b )

0

10

TABLE VI

Slab Action and Plate Loads Due to an Arbi t rary

Rotat ion of Pla te 2

For an a r b i t r a r y r o t a t i o n o f Pla te 2

1 2 3 J o i n t12 21 2 ) 32 Member

1 .000 0.428 0 .572 0.5000 Dis t . f a c t o r

-3 .0 0 0 F. E. moment1 . 286 1 .7 1 4 Dis t r ibu t ion

0 .857 Carry over

-1 .7 1 4 1 .7 1 4 0 .857 Fina l moment0.220 -0 .2 2 0 -0.290 0 .290 m(hcos )

0 .220 - 0 . 5 1 0 0.580 Jo in t reac t ion

For an a r b i t r a r y ro ta t ion of Pla te J

1 2 3 J o i n t

12 21 Member3 32r- - - - - -1 .000 0.428 0 .572 0.500 Dis t . f ac to r

J6

- - ------

______.. __. . _ , . ~ ~ --6 . 000 - 6 . 000 F. E. moment

2.568 3 .432 Dis t r ibu t ion1 .716 Carry over

- ~...... ~ - ~ - ~ - - -

~ . 5 6 8-2 . 568 -4 .2 8 4 Fina l moment-0 .329 0 ~ ) 2 9 0 .773 -0 .7 7 3 m/(hcos )

- 0 . ) 2 9 1 . 1 0 2 -1 .5 4 6 Jo in t reac t ion


46/74

37

TABLE VII

Resolut ion of Jo in t Reactions

(a) For an a rb i t r a ry ro ta t ion of Pla te 2

(1) (2) ( J) (4) (5 )

cosn+l cos,Sn-i R Pla tePlate Reaction =(1 )) X n-1 Loadssine( sinot i x{4>n (2 n- k / f t .

1 0.22 0.222

-0.51 2.-877 -1.467 0 0 -1.467J 0.58 2.;877 1.669 2 . ~ 3 5 -1 .29 2.961

(b ) For a? a rb i t r a ry ro ta t ion of Pla te 3

1 -0.329, -0.3292 1.102 2.877 J . l 7 0 0 3.170 ,JJ -1.546 2.877 -4.448 2.5)5 2.79 -7.242

.

TABLE VIII

Free Edge Stresses fo r an Arbitrary Rotation

Plate s L2 65 2 fb ::: - f tWlate Load

k / f t . cu . f t . 17.53 = 17.53 kip/sq . f t ..(a} For an a rb i t r a ry ro ta t ion of Plate 2

1 0.22 1.556 241.02 45.87

2 -1.467 3. 375 24'1. 02 -104.763 2.961 3.375 241.02 211.45

, ,

' - ' '{b) For an a rb i t r a ry r o t a ~ i o nof Plate 3 .

l ... o.,29 1.556 241.02, -68.592 3.170 3.375 241.02 ' 226.383 -7.242 3.375 241.02 -51?.17


47/74

TABLE IX

St ress Dis t r ibu t ion Resul t ing from an Arbi t ra ry Rotation

(a} For an a r b i t r a r y ro ta t ion of Pla te 2

(b) For an a r b i t r a r y ro ta t ion of Pla te 3

~ 1 2 9 . 0 1189.44 1 8 9 . 4 4 - 3 4 0 . 4 5 ~ 3 4 0 . 4 5428.83 Tota l


48/74

Jl.,

Figure 7. Wil l io t Diagram for Relative Jo i n t Displacement

\. V

1.:0'


49/74

40

For (b) an a rb i t r a ry ro ta t ion o f Plate J

" C-129.01 - 189 444)2

y l = X 65 = -0,12)8 f t .1).56 X X E

". 2

Y2 = (182 1 44 + ~ 4 01 4 5 l X 6 j = 0 18)4 f t .1) .56 X 9 X E "

" ~ - J 4 01 4 ~ - 428 1 8)} X 6 ~2 = -O 266 3y3 = f t .13.56 X 9 X E The general expression of the geometrical r e l a t io n -

ship between def lec t ions and ro ta t io n s , as shown in

Figure 7, is

Yn-1A = - ~ = - . : ; ; ; ; . . , _ -+ y (co t 1 + co t ol ) -n sinotn-l n n - nYn-+1

s in ot n(43)

The f in a l def lec t ions must be expressed in terms of

numerical r esu l t s obtained from the elementary ana lys i s ,

Y10 , Y20 , YJO' plus those for the various ro ta t ion so lut ions , each mult ip l ied by an unknown fac to r kn

The arb i t ra ry ro ta t io n was

EI 6 f t - k= 1 x knh2n

(44)

hence2 1 9_2 X ] ) X 12~ 2 h2 X= k2 o.622k 2I

2k2 =

(1./4)) X 105=

~ 3 = o.622k 3

by geometrical re la t ionsh ips , ~ s i n gEq. (4J)

. 6 2 = o .622k 2 = - l . l .5y 1 + 3.32y 2 - 2.92y 3. " f "

= - l . l . 5 (ylO + Y1 k2 + Y1 k3) + ) .. )Z{y 20 + Y2k2 + Yt))


50/74

41

~ 2 = 0 .0238- o. 7442k 2 + 1.6724k 3 = o.622k 2 !45)

s imi la r ly,

A J = -0 .1571 + 1.1887k 2 - 2. 7790kJ = O. 622kJ ( 46)

Solving Eqs. (45) and (46)

k2 = -0 .0726c. Superposi t ion:

k = -0.07013

The f in a l value of the long i tud ina l s t r e s se s , t rans-

verse moments and def lec t ions by combining the elementaryana lys is and the cor rec t ion analysis are. summarized in

Tatiles X, XI , and XII , and are plo t ted in Figures 8 , 9

and 10. The stresses a t the middle suppor t can be obta in

ed using Eqs. (35a) and (42) .

. TABLE X

Longitudinal St resses

(a) Longitudinal s t r esses a t 0.4L

~ o i n t sElementary Correction Analysis Total FinalAnalysis Rotation Rotation Correction Va.J.uef:

o f Plate 2 of Pla te 3 ' k/sqft

0 48.23 -4.96 9.04 4.09 52.321 -4 .99 6.59 -13.28 -6.69 ~ 1 1 . 6 8

2 -27 .84 -10.64 23.87 13.23 ~ 1 4 . 6 1J 15.86 12 . 99 -30.07 -17.08 -1 .22

(b) Longitudinal s t resses a t . the middle support

0 -89.2.3 7.49 -13.67 -6.18 95.401 8.91 -9.95 20.07 10.11 19.032 49.70 16.07 -J6 .07 -19 .99 . 29.71- ' " ' " ' 0 41"J, _ ,Q l..h 4'5.4'5 25.81 - 2 . 5 0


51/74

42

TABLE XI

Transverse Bending Moments a t 0,4L

J o i n t Elementary Correct ion A n a l ~ s i s Total FinalAnalysis ------ - -- Rotat ion . Correct ionota t ion Values

o f Pla te 2 of Pla te 3 f t - l b .

1 0

2 -294.72 -124.44 180.02 .55.58 ~ 2 3 9 . 1 4

3 ~ 2 2 6 . 4 0 62.22 -300.31 -238.09 r-464.49

TABLE XI I

Defect ions a t 0.4L

Jo in t Elementary Correc t ion Analysis Tota l Fina lAnalysis Rota t ion Rotat ion Correct ion Values

of Pla te 2 of Pla te '3 f t1 0.0433 -0 .0090 0.0087 -0 .0003 0.0432 0.0083 0.0060 -0 .0129 -0 .0069 0.0014

3 -0 .0162 ' -0 .0082 0.0187 0.0105 -0 .0057

In the above ana lys i s the intermediate suppor t ing

s t i f f e n e r i s assumed to be a r i b . A t i e between po in t 2

an d 6 would e f f e c t a sav ing , bu t could be omit ted because

of headroom and appearance. . As the shearing s t r e s s e s .are

smal l , only a nominal amount of . re inforcement i s provided


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\

\

& s: ' o o

- - -\----


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44

F. Folded Pla tes Continuous Over Three Spans.

Since the load ing and span are symmetrical about the

cen te r l i n e , only the cen te r and l e f t ex te r io r spans w l ~ l

be inves t iga ted . These w i l l be considered ind iv idual ly.

The ex te r io r spans h ~ v ethe same behavior as t h a t analyzed

in the previous example of two equal spans , an d the center

port ion has both ends f ixed.

There e x i s t s a considerable difference in the d e t e r-

mlnation of the long i tud ina l moment over each of the two

inner t raverses in comparison w i t ~ the moment over the

middle t raverse of two equal spans. That i s , the end

moment, wL 2/8 , o f a s ing le beam subjected to a uniformly

d i s t r i b u t e d load with one simply supported en d and one

f ixed en d i s exact ly equa l to the moment a t the middle

support of a cont inuous beam with two equal spans. Therefo re , th e s t re sses a t the middle t r ave rse are propor t ional

to the maximum s t r e s se s a t 0.4L of the span. But the

longi tudina l moment over each of the two inner t raverses

in a continuous fo lded p l a t e with three spans i s n o t the

same as the en d moment o f a beam with on e end simply

supported and one end f ixed .

I t i s known t h a t the e ff ec t of cont inu i ty over the

supports on s t r e s se s in she l l s i s s i mi l a r to the e ff e c t

o f cont inui ty on s t r e s s e s 1rt.ord1nary beams. Thus, fo r

the ~ u r p o a eo t eva lua t ing the s t re sses on each of the

inner t r ave r se s , the bending moment w i l l have to be


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c a l c u l a t e d from the th ree moment equat ion o r one of the o ~ r

acceptable methods incommon

use.The shear ing s t r e s s in mult i -span continuous folded

p la t e s wi l l ' be fu r th e r i nves t iga ted and w i l l be emphasized

in Example 2.

Figure 11 shows a moment diagram fo r a uniformly

loaded fo lded p la t e with t h ree continuous spans.

Figure 1 1 . Rela t ionsh ip between Moments and Shearing

Forces fo r a Uniformly Loaded Pla te with

Three Continuous Spans

As expla ined before , the shear ing force N can be

ca lcu la ted s ince i t i s propo r t i ona l to the bending moment.

Nrnax rep re sen t s the shea r ing force a t mid-span. The bend

ing moment on the p la t e a t a d i s t ance x from the f i r s t

i n t e r i o r support , consider ing continuous beam a c t i o n only,

i s

!!!. ' " t XM = (L - x) + 'M + (M - M ) L (47)X 24 x { L - x ~ ' " t ~N = N + N + (N - N ) (48)X max L2 L


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47

TABLE X I I I

General Data

!Plate h , in t , in A, in S, in sin cos f t . in . sq. in. cu. in .

l 2.667 5 160 85 3 90 1.000 02 9.000 3 324 5832 45 0.707 0.7073 9.000 3 324 5832 00 0 1.000

TABLE XIV

Slab Moments Due to Externa l Loads'

l 2 ?_. Jo in t- ., - - ~ - ' " ' - - -...............12 21 23 ~ ~ Member

------- - . ----

0.428 0.572 0.572 Dist . fac to r--- -- --. - .. --------

7.16w -6 . 75w . 6.75w F. E moment-0 .17 -0 .23 0 .23 Dis t r ibu t ion

f - - - - - - - - - - - ~ -. .-- . - - - ~ - - -+ .. -- ... . . . . . . . ~ - - - -0.12 -0 .12 Carry over

I -0 .05 -0 .07 0 .07 Dis t r ibu t ion------- r - - - - - - - - - . - - - - - - - - - - -I 0.03 -0 .03 Carry over!: -0 .01 -0 .02 - 0 .02 Dis t r ibu t ionr - - - - - - ~ - - -.. -... . - ~ - - - - -' 6.9lw -6 .9 lw 6.91w Fina l moment

- - - - - ~ - - - ~ - - - - - ~ -- l .09w i l .09w M/hcos

------I 4. 5ow r ~ i ~ / 2.50w 4.50w 4.50w

Assuming w = 51.54 p s f , wt. of edge beam = 254 l b / f t42 0 .520 .520 J o i n t r eac t ion


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49

St ress Dis t r ibu t ion Fac to r s :

32 4 - 60 11 = 160 + 32 4 - O. 7D12 = 1 - 0 .67 = 0 .33

D 2 ~ = D23 = 0 .5 0

The s t r e s s d i s t r 1 but ions are performed in Table XVI.

In determining th e de f l ec t io ns , E i s assumed to be

2 x 10 6 p s i .

For a uniform load , the def lec t ions a t 0.4L i n the

e x t e r i o r span a re :

- ( -63 .4 + 6. 4 ) X 30 2 X 12 = 0 00263 iy 20 - 12 .99 X 9 X E - n.

and a t

Y1o= ( 2 1 8 , 0 + 6 l . 4 ) 302

12.99 X 2. 7 X E . X X 12

mid-span of the middle span are as

~ - 6 6 ~ z + 6 ~ 2 l2 12Y2o = 16 X 9 X E X 40 X =

Ylo = C23o . s+66 .z ) 40 2 x 12 =16 X 2 . 67 X E X

b. Correct ion Analys i s :

= 0.04380 in .

fol lows:

-0.003986 in .

0.06678 in .

In determining the e f f e c t of the r e l a t ive d isp lace

ments of the j o i n t s , a u n i t t r ansver se s t r i p i s consid-

ered , and the f ixed end moment a t edge 3 i s

3EIA = 3 X 2 X 103 X 144 X 1/12 X (3/12)3 X 1/12MF = h ~ 92

= 1.1574 f t . kip per f t .

The f ixed end moments are d i s t r i b u t e d , and the mom-

ents and the sh e a r fo rces are shown i n Figure 13 , and the


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51

'I z

Z./!22. M E M 8 E R -

c.4ZS o.S72 D .P-trooi ~ -E:f!t .. - - -

ap61 6 6 4 D I S / .f - - - - - - +-- __,_ ... -..- ..... - - ~ - - - ~ - - - ~ - -- 664'664- , : INFJL M .

I +IO.JJ ~ - - - - - - - -

-1oS" j SHEn,e

Figure l J . Moment Dis t r ibu t ion Due to an Arbitrary

Rotation o f Example 2

The free edge s t r e s s e s due to the ro ta t ion a t pla te

2 can be obtained.

For the exte r ior span:' ' -148 X 30 2 X 12

Pla t e 2 fb = - f t = 17 53 x 5832 = -15.63 ps i

P la te 1 ' ' = 105 . 0 X 102 X 12fb = - f t 17.53 X 85J 75.40 ps i

For the middle span:

Pla te 2 f ' = - f ' = -148 x 402 x 12 16 .7 ib t 29.2 x 5832 = - ps

Pla te 1 ' 1 = 105.0 X 40 2 X 12 = 8 psifb = - f t - ~ 9 ~ 2x.853 5

These free edge s t re sses aga i r i ~ h o wincompat ib i l i t ies

which must be removed by s t r e s s d i s t r ib u t io n (Table XVII).


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52

TABLE XVII

S t re s s Dis t r ibu t ion Resu l t ing from an Arbi t rary Rotat ion

The ca lcu la ted d e f l e c t i o n s due to the ro t a t ion o f

P la te 2 are as fo l lows:

For e x t e r i o r span:

y2' = ( - 2 9 . 4 - ~ , o )x 3 0 2 x l 2 = -0 .001697 in .13,5 x 9 x E

' = ( 5 2 . 4 + 2 9 . 4 l x 3 0 2 x l 2 =Y1 1 ) . 56 X .2, 67 X E o.o1220 in .


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::r-r

correc t ion so lu t ions mul t ip l i ed by i t s respectiveAn The

f i n a l resu l t s are showp i n Tables XVIII an d XIX.

The value of the def lec t ion which i s para l le l to the

pla te element, shown in Table XIX, i s a re la t ive value

because an a rb i t r a ry modulus of e l a s t i c i t y was used.

The v er t i ca l def lec t ion o f any j o i n t can be ca lcu la t

ed fro,m the p la te def lec t ions . The re la t ionsh ips between

these def lec t ions are shown i n Figure 7 and a re expressed

as follows:

The shearing fo rces N along the j o in t s may be ca lcu

la ted from Eq. (17)

N1 = -1/2 ( -4)6 .8 + 126.7) x 160 = 24,800 lb .

N2 = 24800 - 1 /2 (126.7 + lJ .O) X )24 = 2200 l b .The shearing s t r e s se s are computed from Eq. (49) and

Eq. (50) as fol lows.

Pla te 1 . - At the suppor ts o f the ex te r io r spans, the302pos i t ive simply suppor ted bending moment i s P2 8 =

112.5P 2 f t -1b .

then,Nmax = 24800 x ~ = 22100 lb

from Eq. {50),

= 4 X 22100 24800 - 5 4 15 ' - " l 4 lb / f tv l 4. 5 x 360 ~ 4. 5 x 360- 7 ... 3 - . . ~ 9 sq

V' = 0 '


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55

At th e inner su p p o r t x = L1v

1 = -54 . 7 - 1 5 . 3 = -7 0 .0 l b / s q f tAt each suppor t of the c e n t e r spant the simple span

4o 2moment i s P - a - = 20 0 P 2 f t - l b2 9

Nmax

200= 24800 X 12 6.4 = 39,400from Eq. ( 49)

v = 39400 x 4 = 73 1 b / s q f t .1 4 .5 X 480

Pla t e 2 . - At th e suppor t of th e e x t e r i o r spans a t

J o in t 1

Nmax = 22100 l b

v 1 = j ~ g ~0 0 - 3 x4 ~ ~ g= 8 2 - 2 3 =59 1 b / sq f t .and a t J o i n t 2

Nmax = 22oo x i ~ g : a= 1960 1b

V = 4 X 1 ~ 6 0 2200 =2 3 X 3 0 - 3 X )6 0 7. 2 8 - 2. 04 = 5 .24 l b / s q f t .

At x = L123 = -105 l b / s q f t .

v 2 = -7 .28 - 2 .04 = 9 . 3 2 l b / s q f t .At each suppor t of t he c e n t e r span , and a t J o i n t 1 ,

200Nmax = 24800 x 12 6. 4 = 39200 1b

v l

and a t J o i n t

= 4 x 34200 = 109 l b / s q f t .3 x so200 .

2, Nmax = 2200 ~ 12 6.4 = )480 l b

'.

v 2 = 3 j 8 ~ f 8~ = 9.65 l b / s q f t .


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56

TABLE XVIII

Longi tud ina l St resses

(a ) Longi tudinal s t r e s s e s a t the intermediate support

~ o i n tI Elementary Correc t ion Tota l Finall-j Analysis Analys i s Correc t ion Values

i psi'

0 -436.8 -436.81 126.7 126.72

1).013.0

(b ) Longitudinal s t r e s s e s f o r the e x t e r i o r span a t o.4L 1

0 218.0 52.4 - ) .44 214.61 -6) .4 -29.4 1.93 -61.42 -6.4 9.0 -0.59 7. 0

(c) Longi tudinal s t r e s s e s f o r the c e n t e r span a t mid-span

0 230.5 .56.0 -5.60 224.91 -66.7 -31.5 3.13 -63.62 -6.9 9.6 -0.9.5 -7.9


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TABLE XIX

Transverse Moments and Deflections

( I ) Transverse Moments

(a) Transverse moments fo r the ex te r io r span a t 0.4L 1~ o i n t Elementary Correc t ion Tota l Fina 1 values

Analysis Analysis Correction f t ~ l ' b / f t .

2 -355.0 664.0 -4 ) . 6 -398.6

(b ) Transverse moments fo r the c e n t e r span a t mid-span

2 -355.0 664.0 -66 .0 -421.0

(II} Deflect ions

(a ) Deflections fo r the exte r ior span a t 0.4L 1

1 0.04380 0.01220 -0.0008 0.0430 in .2 -0.00263 -0.001697 0.0001 -0.0025 in .

{b) Deflections f o r the cen te r span a t mid-span

1 0.06678 0.01840 -0 .0018 o 0649 in .2 -0.00399 -0.00256 0.000,3 -O.OOJ7 in .


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58

IV. CONCLUSIONS

The proposed method o f ana lys i s of folded p la te s

developed in t h i s paper y i e l d s s a t i s f a c t o r y r e s u l t s fo r

the ana lys i s of cont inuous fo lded p la t e roofs in compari

son to the values obta ined by Yitzhak i ' s s lope-def l ec t ion

method.

Although the s tudy presen ted here in suggests a

p rac t i c a l method to des ign continuous folded pla te roofswith symmetrical l oad ing , i t can a lso be applied to

symmetrical fo lded p l a t e r o o f s , unsymmetrically loaded,

by div id ing the unsymmetr ical load i n t o symmetrical and

ant i -symmetr ical loa,ds. The f i n a l s t r e s s e s and def lec

t ions wil l be obtained by superimposing the resu l t s of the

two cases .

The determinat ion o f th e spacing of the in termediate

supports would be based on an economic study. The th ick

ness , depths , the magni tude of the angles between the

ind iv idual p l a t e s , and the r i g i d i t y of the t ransverse

s t i f f e n e r a re a l l impor tan t fac to r s which wil l a f f e c t the

spacing of the in te rmed ia te support s .

The s t i f f e n e r s must be designed to car ry t h e i r own

dead load plus the r e a c t i o n s imparted to them by the

shearing forces from th e ad jo in ing p l a t e s . The s t r e s s e s

and the design of th e i n t e rmedia te s t i f f e n e r need to be

fu r the r inves t iga ted .


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TABLE I I . Deflections, Moments and Shears Produced by Normal Load and Uniform

Load fo r A Beam with Both Ends Fixed

Type of Load Normal Load Uniform LoadN w

Deflect ion ~ ~ ~ -I . S " ~ I Sr c ~ h 4 . 7 . ; s f - 0 S { { . 7 J f : J - 4 f J 3 L r { J I N A 4 l " f - J / N 4 7 . J - f J ]. ~ z . XJ .!S-4Curve '.jW*I'{L.z

.lLJ_,. .t_4)

Moment ~ Lz .1L ~ . L'l_z L ( ~Curve

J7.JJ '!JN-1 . 2 ,'.J.-~ h 4 7 . J f +cas4.7..Jf ~ 7 3 ' . z . J { S . N M 7 . 3 f - o ~ - s w ~ 7J.t-;J'Mw=a! 1 , . . . - 2 ! - r ~

Shear Curve ~ ~ t . , ~ ~ J t - = - / . ; - : r o { { . J I N I J 4 7 . J j... S I N 4 7 . J f r 4 . 1 J . f ( C o S } , . ~ 7 1 f - ~ > e t > S - / . 7 j21-L'

. ~J : ; ; . r r ; - , . s f s i ) r ~ / , ~7J { - -e o s 4 7 J ~ 9 - ~~ 2 2 . j { . 5 1 N A 4 - 7 J f - . J I N 1 l J } J }fM L-1- .11.Load Curve w = 33 4 ~ w- I

Maximum ciL"" M.Lz wL4 ~ z .

Deflect ior Y.,:a .J"cq.r.rO l7 1 E L ~ - 38'9Ez-= ''&.~ - -. riMaximum

z M ... wLz. d ~ E r 9 .ll. 1'1' .t. IZ I E.I YoMoment ' Z)tZ = z ,

~ t . ~X&J.fl i.

Minimum . N1,.,;,- - N . L ~ wLz.Moment /7.7.[""111-;,--;-r

Maximum ~ -N.L. S.: wJ .

Shear :J.7'3". - 2

x - ~- . -

0' \N


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BIBLIOGRAPHY

1 . Ehlers , G., Ein neues Konstruktionsprinzip, Bauingen

i e u r i Vol. 9, 1930.

2. Craemer, H., Theorie d er Falwerke, Beton und Eisen,

Vol. 29, 1930.

3. Winter, G. and Pei , M . , Hipped Pla te Const ruct ion.

Journai ACI, Vol. 4J, 1947. pp. 505-532.

4. Gruber, E .,B ~ r e c h n u n g

Pr i smat i sche r Scheibenwerke,In te rn a t l . Assoc. of Bridge and S t r u c t u r a l Engr.

Memoirs, Vol. 1, 1932.5. Gruening, G., Di e NebenspannUngen i n Pr1smat1schen

Faltwerken, Ingen ieur-Arch iv. , Vol. J, N o . 4 ,

1932.

6. Gaafar, I . , Hipped Pla t e A n a l y ~ i sConsidering Jo in t

Displacement, Transac t ions , ASCE, Vol. 119, 1954.

7. Y1 tzhaki , D. , P rl sma t i c and Cy l i n d r i c a l She l l Roofs;

Haifa Science Publ i shers , Haifa , I s r a e l , 1958.

B. Yi tzha.ki, D. and Reiss M., Analysis o f Folded Pla t e s .

Proc. ASCE, Vol. 88, 1962, p . lOq.

9. Phase 1 Report on Folded P la te Construct ion. Proc.

ASCE,Vol. 89, 196J, p. 36510. Gruber, E ., Die Durchlaufender Prismatischen F a l t w e r k ~

In te rn a t l . Assoc. of Bridge and Struc tura l Engr.

Mem.o1rs, Vol. 12, 1957.


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1 1 . D i r e c t S o l u t i o n o f Folded Plate Concrete Roofs.

Advanced Engineer ing Bul le t in No. 3, PCA, 1960.

1 2 . Ashdown, A . , Th e Design of Pr ismatic Structures .

Concre te Pub l i ca t i ons Ltd . , London, 1951.

1 ) . Lee , S . a n d Pulmano, V ., Ribless Folded Plates , Pro9.

ASCE, Vol. 91 , 1965, p. 253.

14. Simpson, H . , Des ign o f Folded Pla te Roofs. Proc. ASCE

Vol . 84 , 1958 .

15. Traum, E. , Des ign of Folded Pla t e s . Proc. ASCE, Vol.

85, 1959, P 103.

16. Ta b l e s of' C h a r a c t e r i s t i c Functions Representing No_rmal

Modes o f Vi b r a t i o n of A Beam. The Universi ty of

Tex as P u b l i c a t i o n . No. 4913, 1949.

1?. Timoshenko , s . , Vib ra t io n Problems in Engineering.

D . Van Nostrand Co. , I n c . , New York; 1954.

18. Dunham, c. , Advanced Reinforced Concrete, McGraw-Hill

C o . , New York, 1964.


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VITA

Yung-Ping Wang, son of Mr. and Mrs. Rex K. z. Wangwas born on September 9, 1937 a t Shanghai, China.

v v

He graduated from Taiwan P r o v i n c i a l C h ~ e nKuo Middle

School , Taipe i , in 1956, and received the degree of

Bachelor of Science in C iv i l Engineer ing from Taiwan

C hr i s t i an College i n June, 1960. He served ROTC as a

Second Lieutenant in the Chinese. Army fo r on e yeara f t e r '

being graduated from the co l l ege in Taiwan.

In November, 1961, he accepted.employment as a

s t r u c t u r a l designer in Taipei-Keelung Highway No. 2 Con-

s t r u c t i o n Office , Taiwan Highway Bureau, f o r about two

year s .

He enro l l ed a t the Universi ty of Missour i a t Rolla ,

i n January, 1964, fo r graduate study i n th e f i e l d of

C1v i l -Engineering

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