Shell arch roof model under dead load and half side live load, September 1950

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Lehigh University
Lehigh Preserve
Fritz Laboratory Reports Civil and Environmental Engineering
1950
Shell arch roof model under dead load and half side
live load, September 1950
B. Thurlimann
B. G. Johnston
Follow this and additional works at:http://preserve.lehigh.edu/engr-civil-environmental-fritz-lab-
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Recommended Citation
Thurlimann, B. and Johnston, B. G., "Shell arch roof model under dead load and half side live load, September 1950" (1950).Fritz
Laboratory Reports.Paper 1462.
http://preserve.lehigh.edu/engr-civil-environmental-fritz-lab-reports/1462
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PROGRESS REPORT 213D
SHELL ARCH ROOF MODEr, UNDER SIMUI,ATED
DEAD LOAD AND HALF SIDE LIVE "LOAD
by
Bruno Thtir1imann"
Bruce G. Johnston
to
Roberts and Schaefer Engineering Company
Fritz Engineering Laboratory
Department of Civil Engineering and Mechanics
Lehigh University
Bethlehem, Pennae
September 19, 1950
TABLE OF CONTE!'JTS
Page No..
Sunnnary 1
Introduction 1
•
Part I: Theoretical Analysis 2
Problem 2
Ie Membrane Solution for a Cylindrical Shell 3
Continuous Over Two Spans
2. Loading of the Ribs 4
3.. Calculation of the Arches for Dead Load 6
(symmetrical load case)
a.} Arches Under Shear Forces S 6
b.) Arches Under String Force Y 9
4. Calculation of the Arches fer Anti- 11
s~~etrical Load
a.) Arches Under Shear Forces Sand Hori- 13
zontal Loads P
b.) Arches Under String Force Y 15
50 Deflection of the Arches 17
60 ApprOXimate Calculation of the Shell Forces 17
Part ~I: Numerical Analysis 18
Effective Cross Section of the Ribs 18
List of Principal Dimensions and Datas 20
1. Membrane Solution 21
2. Loading of the ribs due to the Boundary Forces
S and Y of the Shell 22
TABLE OF CON'TENTS (cont'd)
Page No.
30 Symmetrical Load Case (dead load) 23
4. Anti-symmetrical Load Case 25
5. Dead Load 25
a.) Ribs 25
=
b.) Shell Forces for w 0 27
6. Live Load over the Half Span 27
a Ribs 28
o )
b.) Deflection of the Ribs 29
Part III: Experimental Investigation 31
1. Description of Model and Test Set-up 31
2. Test Procedure 32
3. Test ResUlts 33
40 Comparison between Test Results and Analysis 34
5. Relation between the Model and an Actual 36
Structure
Conclusion 38
Tables I to VI 39-50
Figures 1 to 17
Notations
List of References
Progress Report 213D
Shell Arch Roof Model under Si~ulated Dead Load and
Ha1.f Side Live I,oad
by
Bruno ThUr1imann and Bruce Go Johnston
Summar::,:
The analysis of a shell arch roof model (Fig. 1) under
two cases of vertical loads simulating (1) dead load and
(2) uniformly distributed live load (e.g. snow) over the
half span (Fig. 2), is developed. The analytical results
are compared with experimental results obtained from tests
performed on an actual model.
Introduction.
Uniform load over the whole and over the half span are
the two ~ost L~portant load cases for the present type of a
structure. The half span load very nearly gives the maximum
extreme fi"ber stress due to bending.
"'
The actual loading of the model (Fig. 2) is not a un!'"
formly d:i.strlbuted one as assumed in the analysis. This does
not influence the stress distribution in the ribs to a great
extent. However, local differences between the experimental
and an.alytical shell forces, especially,the bending momer·:t.~"
must be expected.· The problem of' the ribs is therefore treated
in a more complete way. For the shell forces an approximate
procedure is used, similar to the one in Progress Report 213C.
The problem of the edge"'msmber distnrbance is not taken up
- 2 -
because a separation of the different influences for the
present type of a model is too involved.
Part I: Theoretical Analysis
Problem:
The structure, as shown in Fig. 1, is highly statioally
indeterminate. A first step to a solution is the analysis
~
of the shell as a continuous member over two spans,
supported by three diaphragms (Fig. 3). The boundary forces
(shear forces S) and the deformations along the boundaries
(strains c~in circumferential direction) give the loading
conditions for the ribs. The condition of continuity between
ribs and shell are fUlfilled, if the effective cross-section*
of the ribs is taken (See Ref. (1)+ p. 28 and Ref. (2)+ po 3).
The following problems have to be studied:
1. Membrane solution of a cylindrical shell, continuous
over two spans.
2. Ribs (effective section~) under the action of shear
force S.
3. Ribs (effective section under the action of a
string force Y, eliminating the relative difference
- - - - - - - - - - - - - - - - - - - - - - -
~ ~ ~ ~ ~ ~
* By "effective section" of the rib, the cross section consisting
of the rib and a flange of width equat to the effective width is
meant. On the following pages the term is used with this meaning.
+ (1), (2) See list of references at end of report.
.. 3 -
in strainc~betweenrib and shell due to the membrane
solution.
The dead load presents a symmetrical case of loading.
The ribs are twice statically indeterminate (Fig. 4).
Live load over the half span can be found by super
position of a symmetrical and an anti-symmetrical load case
(Fig. 4). Th~ advantage of this solution is the splitting
of the three redundant forces of the rib in two independent
groups of two and one redundant.
1.) Membrane solution for a cylindrical shell continuous
over two spans:
Dischinger, Ref. (4) p. 227, treats the problem of the
continuous stiffened cylindrical shell 1n a very complete
way. FIUgge, Ref. (3) p. 76, presents the case of a shell
continuous over two spans. In the following direct use of
the equations derived by FIUgge, Ref. (3), is made. The
notation is changed to conform with the notation used in
the previous progress reports.
The membrane forces in a continuous shell, Fig. 3, under
load of p lb/in2 of shell surface are: (FlHgge, Ref. (3),
-8
p. 79)
(1)
.. 4 ..
The shear force S becomes zero for
)R~
L 5t2 + 6(4+3v
Xo = 8 1,2 + 6(1+1:» R~ (4)
At t:JEl mlddle and outer ribs, the S-forc8s are:
I
Middle rib: S x=o = 2pxo sinw (5)
I
= ...
outer rib: S 2p(L-xo) sinCJ (6)
i=l.
The normal strain in circumferential direction is:
c (7)
OSc..:>
2.) Loading of the ribs:
The membrane solution resulted in boundary shear forces S,
E~.(5)and(6~ and in normal strain £win circumferontial direction,
Eq. (7) ~ The discontlnuity caused by the Cw between the rib
and the shell has to be eliminated.
The shear forces S give the following loading condition:
From Eq. (5) and (6)
S = S sinCJ (8)
rrl = - 2p(1,~o)
Where: Outer rib:
x=~
Middle rib: ~I = 2pX
x=o o
X as given by (4)
o
x = x =
Applying a string force Y at t and 0, the membrana
strains C may be eliminated in order to fUlfill the continuity
w
condition between shell and ribs The magnitude of Y 1s readily
determinad.* The strain twas function of T2 is:
- - - - - - - - - - - - - - - - - - - -
- - - - - - -
~ ~ ~ ~
* Ref. (5) gives a very complete treatment of the boundary
=
problem of ,::ylindrlca,l shells for the boundaries x const.
- 5 -
T at the point of application of the string force Y 1s
2
(Ref. (2), Eq. (3»:
=
Where: b effective width
SUbscript Y indicates that T 1s taken
2
at the point of application of Y
The strain cwdue to Y is:
To eliminate the membrane strain cG.J' Y has to be: (Eqo (7)
and (9»
pRS Y
- cr
COSGJ + '6Q = 0
Y = pRSb
oOS(.J
Putting pRsb = Y
=
y y coscJ (10)
In summary, the membrane solution of the shell has the
boundary forces S, Eq. (8) and Y, Eq. (10)0 Introduced in
the reversed direction they are the loads for the ribs.
As cross-section of the ribs the effective one is used to
take account of the interaction of rib and shell. (See Ref.
- 6 -
3.) Calculation of the arches ror dead load (sYmmetrical load case)
The arches in an actual structure are under the following
3 loads (compare Fig. 5).
Eq. (8)
a. ) Shear Porce: S =S- sine..>
Eq. (10)
=
b. ) String Force: Y ? cose..;
c.) Dead load of the rib ( cross section of the rib only).
The ribs are analysed for the first two loads. The dead
load of the rib itself is not treated because this load does
not enter into the problem under investigation and in a general
design problem it will not offer any difficUltyo·
a.) Arches under shear forces S:
In Fig. 6ao) the statically determinate base system is
shown. The normal force No and the bending moment M in the
o
base system may be obtained by integrating the contribution
of the distributed shear load S that is applied by the shell.
If ~ is the angle at which No and M are determined, let
o
represent the shear load applied over an incremental
shell distance at any angle ~ between 0 andw (Fig. 6b). Then
fc..>
d~
No = - SRS cos(w - Cot)
o
=
SUbstituting from Eq. (8), S S sin~ and performing the
integratton,
= - 1 -
No 2 SRS W sinGJ (11)
- - - - - - - - - - - - - - - - - - - - - - - - - -
~ ~ ~ ~ ~
.. In the test the "dead load" was simulated for the shell only
as a uniform.ly distributed "live load" acting over the entire
shell. See Fig. 2 and p. 31.