Ctn4Y
a· y
JUN 23 1923
U8RA¶3
THES I S.
AN ISNVESTIGATION AlND DEVELOPMEN-T OF A CERMTAIN
SLOPE DEFLECTION MITHOD AND ITS APPLIZ
CAT-ION TO RIGIDLY FRAMED STRUCI
TUMRES.
61
Prepared
BI
A
KALGC
1
B
by
SHIH.
To
T1: CLYka,
vng eeriYg%5
lc5ge.
for the
F1ulfilment of M.S. Degree in the
Department of Civil Engineering
of the
Massachusettes Institue of Teclhnology,
Mass.
JiTue, 1922.
F,
SJue
1, 1921.
Prof. A. L. Merrill,
Secretary of Faculty,
Massachusetts Institute of Technology.
Dear Sir;
In accordance with the requirement for the degree of Master
of Sience in Civil Engineering, I herewith submit my thesis entitled
" The investigation and deve.lopment of a certain slope-deflection
method for designing rioxidly framed structures."
Respectfully submitted,
Talean Shih.
.C)
-
Contents.
Page
Preface and acknowledgement.
...........................
1
Introduction.
1. A review of previous methods on rigidly framed
structure.
............................
2. General outline of the present method.
2
4
..............
3. Foundamental principals and assumptions of the
present slope deflection method.............77
4. Method of getting primary sections of the frame
for investigation.
..............
8
Part 1. Theories for getting the characteristics of a
given rigidly framed structure.
1. The change of slope of a beam.
2. The change of
'1ope of a column.
,..13
...........
..............
18
............
20
3. Ascertaining of moment of inertias.
momert factor
4. Cha;racteristic point of Contraflxureq,&
5. The procedure of computation.
20
..............
29
..... .,
..... .. 31
Appendix
Part ll.Moments and stresses due to vertical loads.
1., Several short-cut methods for finding moments
of rigid members
2. Separat•on
of moments and their sigrs.
.......
..... .32
t....37
.......
3. Deter~ination of shear and normal forces.
.......
40
4, Criteria for maximum combined stresses in
beams and columns,.
.............
41
Part III, Moments and stresses due to horizontal loads,
1. Derivation of expressions for moments due to
deflections.
.f.....r......,4
2. Single-story frames.
J.
K)K)-1
~3 ) V
5. Two-story frames.
.................. 51
4. Three- and more- story frames.
.................. 56
5. Secial cases.
.................. 60
6. Abstract on theory of determinants.
.................. 63
Part IV. Examples.
Example 1. Single bent subjected to vertical laod. .......... 68
Check by least work method.
Single bent subjected to horizontal load.
Check by least work method.
Example 2. Investigation of three story reinforced concrete
............. . ...........
building.
. . ..
71
Example 3. Method of application of theory of determinants...82
Conclusions.
.. **
4*
***
4***
***
***
*****
**....
04444*.
84
1
Preface.
This work is prepared for the fulfilment of M.S degree
in the Institute.
In doing it a time of about T~o
hours has been spent,
of which the most part was devoted in reading and researching.
Pains
have been taken to present the method in unique and simple manner and
to developý all the necessary expressions, so that the easiness of the
application of the method can be a~chieved. Examples are also given
at the end of the work, they serve the purposes of illustrating
the
method of application, checking up the correctness of this method,
and, at the same time, giving a brief comparison of this method with
those existing.
Oving to the limited time and the large amount of work
involved, some points might have been overlooked, although great care
has been taken by the writer.
The writer wishes to acknowledge his indebtedness
to
Professor Charles M. Spofford for the inspiring instruction received
from him as a student in his class on advanced structures.
He
par-
ticularly wishes to express his appreciation of the valuable instruction and criticism
on this work from Professor Hale Sutherland, under
whom the writer has taken the course on advanced structural design.
Kalean Shih
May, 1922.
t
1
2
INTRODUCTION
1. A review of previous methods on rigidly framed structure.
It is believed that a.wide knowledge of the analysis of rigidly framed structure is generally required
in gaining etconomies and in securing effective designs.
Notwithstanding the importance, the close investigation
of the stresses as they actually occur in this kind of structure is not usually attempted in practice. The reason is
simply due to the fact that the existing methods for investigation requires either considerable time to work out
formulas for statically indeterminate quantities to suit
a particular case in question or laborious work to apply
the different expressions already determined to get so
many unknown quantities.
In view of the ec4onomy of the
time element, several very approximate methods are in
general use, although they sometimes give results which
are serious in error. The methods that have been so far
developed can be classified as follow':
a) The approximate methods. They are originated by
various authorities ( Spofford, Fleming: Smith, Burt, Thayer, etc ). The methods consist. of certain combination.
of the following assumptions -.
:
1. Point of contraflexure of columns at mid-height.
2
I
-Point of contraflexure of beams at mid-length.
3-. Direct stresses in columns are propottional to
the distance of the column from neutral axis.
4. The shear on all interior columns is equal and
the shear on each exterior column is equal to one
half the shear on interior column.
5. Shear is distributed amongst columns in proportion
to their moments of inertia.
6. Shear is distributed amongst columns in proportion
to area of vertical rectangle of which the column
is
the axis,
As the methods are common in practice and can be found in
papers written by respective authors mentioned above, so they
will not be given here in further detail.
b) The more exact methods. These consistofthe methods
based on slope and deflection originated by Dr. CA.Melick
and Mr,. .F.Jonson. The original work is laborious and is
almost inpractical for buildings several stories high.
The further development of the slope and deflection
method by W. M. Wilson and several other men ( bulletins 80
and 108, Eng. Exp, Sta of Univ; QOf Ill.) rendered this method
generally applicable.
been derived.
Expressions for various cases have
It is considered simple, although in some
cases its application is restricted. ( See Conclution, Part 4).
4
c) The exact method.
This method is a development
by method of least work. The well-known papers regarding
to this method are by Prof. A. Smith (Journal of Western
Society of Engineers, Vo.l. XX ) and by Dr. Mikishi Abe ,
( Bulletin 107-, Eng. Exp, Sta. of Univ. of IIi..) .. The
method is exact and the direct stresses can be taken into
account.:
It is, of course:, very longj.
2. General outline of the present method.
The method herein developed and investigated
belongs to Class b, the more exact method, of the previous
discussion, or, in other words, the slope deflection method
treated in more unique manner.
The preference of this me-
thod will be discussed in the conclusion of Part IV of this
work.
The work is devided into 4 parts with an introduction at its beginning, in which, besides some general
discussion, a section for getting primary sections of a reinforced concrete frame is also given, Several formulas
derived from the writer"'s experience are presented, which,
when being applied, needs only a single slide rule operation
and are deemed simple and helpful.
The first part of the work, Part I, is the basis
of the wholed All t:he theories and derivations are included
in it. They serve in the main to get the characteristics
of a ggiven frame. By characteristics of a frameare here meant
the following':
I
V5
Taking each member of the frame separately and assu-ing one end being fixed, while the other end being sub~ad arzv
rc••cror..
jected to a unit moment, the expressions for the points of
contraflexures of-the memberacan be obtained. The position
of the points of contraflexure does not change for any mag-
nitude of the accually applied moment. ( Characteristic
I)
Taking each joint of t'he 4,ame. searately and assuming a unit moment being applied to the joint. Now since
it can be easily proven that in the strained position all the
members meeting at one point are subjected to the same change
of slope, the relation of transmission of the moment, or the
moment factors, can be derived.
The moment factor does not
change for any magn~itude of the actually applied moment.
( Characteristic IT )
At the end of Part I a summary of the procedure for
computation is also given.
IMost of the theories given in this part are taken
from Dr. Strassener's work appeared in the paper " Forscherarbeiten auf dem Gebiete des Eisenbetons, Hefte 26".
The expressions for momeat factors for cases where four
members meeti$3 at a joint are the writer's own developmen.t from the same principles' ; these complete the missing
link of Dr. Strassener's work.
The second part of this work, Part II,: treats the
method of application of the development given in Part I
to the case of a frame subjected to vertical loads. Section
2, the separation of moments and their signs, sho-ws the
unique manner of this method, in which Tt lies its superiority.
The two sections on several short-cut methods for finding
moments of rigid members and Criteria for maximum combined
Sstresses in beamse'and columns are collecteda
rom various
sources and seem to be helpful.
The third part of this work,
Part
lI:,1 treats
the method of application of the development gFiven in Part*I
4 loads.,
to the case of a frame subjected to xerwn4(
On ac-
count of the fact that the tops of all columns are subjected
to a deflection in this case, expressions for moments due to
unit deflection are derived. Now since the deflection and
the resulted moments are always ý:in direct proportion to
each other, advantage can be taken from this fact for deterLmining the moment of each member due to an ass~umed deflection
unity of each story of the frame.
From these moments the
horizontal shear subjected to each column can be found.
The
sum of the shears of all the columns in each story should
mat
eaily be equal to the external force applied at that
story., which causes the assumed unit deflection. by multiplying the mcment due to the unit deflection by the ratio
Acetually appli.ed force
External force causing unit defl.
the acctual moment of each member can be obtained.
Dr, Strassener gives some hint for attacking
such problem in his paper mentioned above, but the method
has not been treated Ay him in unique and simple manner.
The present method is a collection of the developments of
various engineers appearing in "Schweizerische Bauzeitung"
Sand
put up in simple and collective manner by the writer.
It should be noted that this method is. eore simple than
Wilson,'s method by reducing two equations from each story.
The fourth part of the work, Part IV, giveza
complete .illustration of the method and conclusions.
3. Foandamental principles and. assumptions of the present
slope deflection method.
The principles and the assumptions of the method in successive
sections given in this work are almost the same as are given
in
tWilson
's
paper ,. Bulletins80 and 108, Eng'g experimental
station of University of Illinois. The trzeatment and the
applications
of the principles and4assumpt-ons are entirely
different from those given in Wilson's work and the preference
of the present method will be discussed in the conclusions
vn P+
4t-- . The principles and assumptions adopted cmn
be
summarized as follows":
4
1)
The moment at an end of a member of a frame is a func-
tion of the changes in the slopes of the tangents to the elastic curve of the member at its ends and of the deflection of
one end of the member relative to the other end.
2): In the strained position, all the columns and girders
which intersect at one point have been subjected to the same
change in slope.
Upon these two principles the Part 1 of this work is based'..
3) The change in the length of a member due to direct
strees is equal to zero.
4) The .horizontal deflection of the tops of all columns
due to vertical load is equal to zero.
Upon these two assumptions the part 11 of this work is based.
5) The horizontal deflections of the tops of all columns
of a story due to horizontal load
.I
are equal..
Upon assumptions 3 and 5 the Part 111 of this work is based.
Other minor assumptions are,:
6)
The connections between the columns and girders are
perfectly rigid.
7)
The length of a girder is the distance between the
neutral axes of the columns which it
connects and the length
of a column is the distance between the neutral axes of the
ggirder which it
connects..
8) The deflec-tion of a member due to the internal shearing
stresses is equal to zero..
9) The wind load is resisted entirely by the rigid frame.
4:.
Method of getting primar-ry sections of the frame for in.vestigation.
In all the methods discussed in section 1, except
the approximate ones, it is required to know all the sections of the members before investigation.
This is,
how-
ever,,, the hardest task, which the designer has to encounter,
if
any exact computation is
to this fact, it
intended to make at all.
Owning
seems to the writer to be favorable to in-
clude this section in this work,
Now since the scope of s~trctIura1o.
•oditiOcns
is
so
wide that the degree of rigidity varies with all factors, like
supports, joints, materials,, etc. , it is impossible to present
.t
I
any method which can be applied to all cases.
lowing discussion it is,
In the fol-
therefore, confined to reinforced
concrete frames- which has absolute rigidity and the degree
of accuracy of the new slope deflection met-hod herein presented can be warranted.
The formulas given below are derived from the writer's
experience and are based upon 1 ': 2': 4 concrete ( .fc
fs
=
16000 and n
650 .
15):, which is the mixture used in almost
all casee of building frames..
formulas is
=
T hough the derivation of the
so simple and t heeir frms do not differ much
from ordinary ones, yet the merit of a single slide rule
operation can hereby obtained.
Formula I,, For rectangular beams.
( standard notations)
d= -. 4L
1080b
This formula is derived from ordinary chart for
rectangular beams.
For the specified stresses of 1 '
2 : 4
n1
= 108
1---- and whence
concrete
bd2
Formula 2, For T,-beams,
we have the formula.
neutral axis lying in stem.
d
35*b'
in which b" is the assumed flange width, and according to
the specification of Joint Committee
a.)
the beam.:,
It should not exceed-I- of the span length of
4
:j
b)
Its overhanging width on either side of the web
shall not exceed 6 times the thi'ckness of the slab.
It
e)
s-hall not exceed the distance c-c of the beams.
This formula is derived by examining a series of Tbeams together with the c-hartAgiven on page 364r-365 of
'Hool and Johnsont's Concrete Eng"'r
Handbook.
It is found
that the design of T-beams is always limited by the s.tress
of steel, while the stress of concrete is
below its limit.
generally far
The latter stress ranges from 300 to 400
for -competent design with cost ratio r
=
70 approximately.
Now let draw a horizontal line on diagram 8,
and J.'s
page 364H-i..
book, which passes the curve fc= 350 and so place
the line that half of the curve is above and half below it
it will be found that this: horizontal line corresponds
L - = 35 and whence we have the formula.
It has been tested
bd2
that this formula satisfifes most of the cases, especially
for getting primary s:ectionss
Formula 3. •.rt getting reinforcing steel of rectangular
and T.beams..
As
5
This formula is obtained by substituting j -7
general one A=
Formula 41.
---
jdfs:
For columns centrally loaded.
into the
P
Ac
Where Ac= the net area of cross section of the column, P= the
centrally applied load and C a constant given in the table
below, its value is different with different values of "p",,
the percentage of re.inforceme-nt.
f1
Ir
1
1"::2:4 (2000-.lb. concrete) n=15.
fc=450 for ordinary lateral ties.*
iO
.! ý o.ca
o.oz5
o.oC1o0oo
o
.o-o45i o5O
The values of C in the above table are computed from the
expression
C=fc[ 1+(n-l)p] , the formula is a general one
and requires no further discussion.
For beams reinforced with compressive steel and
columns subjected to eccentric loading, formulae given
in various text books are supposed to be used.
* Specification of Joint Committee.
Part 1.
THEORIES FOR GETTINRG THE CRARACTERIS'TICS OF A GIVEN RIGIDLY
FRAMED STRUCTURE4
T-he foundamental theories from which this part
of the work is based- are orighated by Prof. W. Ritter in
his book " Anweandng der graphiache Statik"' During the
last ten years research works have continuously been made
in Germany to study the effect of rigidnity of any kind of
framed strucature with the intention to obtain a method of
so.lution, which should be practical in actual use and, at the
same time, will give a degree of accuracy not far from the
exactness.
In year 1916 Dr. Strassener plblished a long
article treating this subject .inForsc-herarbeiten in Gebiete
des Elisenbetons" and since. then several articles regarding
this work have appeared during the-: years 1917-1921 in Sch1 s work
weizerische Bauze.itung' Tn both of Dr. Strassener'
and the articles in Schweizerische Bauzeitung the expressions given for moment, factors are confined to joints where
only three members meet. .I order to solve problems like
rigidly framed buildings; cases alwafs occur where four
f
laca
ADtrs
lougbpt, er, A,
ke in~Ine i r b ys of th
L)
u 3A din
Il
consequent# expressions for these, equations 10 to 14, are
I
developed by the writer.
!
1.
The change of slope of a beam.
Let- Ma= the moment appli'ed at the left end of a beam.
Mb=
"
""
"
"
" "
right "
"
ea= the slope angle ait the left end of a beam
b =1
due to Ma =
eb = the slope angle at the right end of a beam
due to Ma = Mb = 1
$= the slope angle at left end due to Mb= 1, or
the same at right end due to Ma= 1
and let x = the. distance of any section from left end
of the beam in question and x" the same from right
end, then the expressions for the slope angles
are.:
aa=·'4L *· J' x-'
E*dxx
I. xY b 1dx.
r
L2
2EI
.. (1)
EI.
For beams of constant cross section, ea
S
L
....
x'xadx
p·•1- *
and
=
=b
and
.
.......
(la
IET
The derivation of equation (1) can be obtained very easily
by following the discussions given in Morlefs Strength
Ma:terials
,
pages 170o-179,
of
og by Prof. Swain's theory given
in Transactions of A" S. C. T. 19184
SThe liit
*
of interati---------------n
n equaton------ is
ro
The limit of integration in equation (1) is f-r-or
--to
to0 .
r
~31
'I-i
kb
Er
A
L.
A
J#~)(a)
dV!
(3t
1de
(@}
(Amy
.1m
ZCL
K=
a
t, z
-[i
Wki.
;.
f
*I:
MP s
F e, 3
I
!
--
In case that the cross section of the beam is
not
constant and its moment of inertia varies, the integral
symbols of the above expressions can be rrplaced by the
symbol "E" and by looking the value dx as a part of the beam
having a length of x.
The solution of each expression can be
done graphically without difficulty,
In view of practical use,
two tables are hereby re-
produced from "Forscherarbeiten auf dem Gebiete des Eisenbetons, heft xxvi".
In order to facilitate the use of the
tables, the following equation is given':
e-,2
=
," a
El
Equation (l
b )
-. -b
.(1b
6F1E
is only applicable to symmetrical
beams, which type is used in almost all cases in building
construction'. The factors ,
and
L
are the slope
2EI,
6E I,
angles 6 and 3 respectively for beams of constant moment
of inertia I,,
and the factors Ta and Pb are constants found
either from table 1 or table 11 corresponding to the type
of haunches that the beam has:.
These constants depend also
upon the following items of the beak:
KL = length of haunch, and
CtI
C,=
--
,
d
d-a
ds
a
-
or
r
.
.....
. .. ....
.
. .....
.*
. 1 for beams having constant width
l
(2
)
..
~1118~08~18~2~B~-·d··r;-*i~~--.rr··
.
, • rli :.· · ;·.-~ur;~--r
Li-
-
-----
.
~-__1__1_-
--
------
-
.
Table I. For beams with straight haunches.
'Pa= upper vmlue, 'b= lower valuel
I ..% - - ---
~----·
-
-----·
----------- ·"·
-------------------
-
C2
0
0.4
0.8
1.0
-- `~-----p~
.432
.518
61.2
1/3
1 "0
1.0
.742
.843
.621
.758
--':-----UY*-r·i
1/4
.
.0
.0
.806
.907
.716
.856
1.4
IT.2
.,,• .
"• .
.375
.331
.295
.. 460
.41 4
.376
"" -C-~UII~------------.·.·~--11111~-,
-.
r~uC~--~k~--Llur
.583
.729
· 1?·1-····-1-~---
.688
.839
554
705
·-- r··l··
3-----
.665
.825
1.0
1.0
.845
939
.773
.905
.750
.894
.732
*884
.266
.344
~l~·r~------YrrC1-··r
.511
i
8.48
.813
.803
.706
.869
~~--~-~~-
j717T7J
U7LZIF2
ý---4_
.633
-.---
j7TIZ77
3.0
.222 .156
.. 294 . .21 5
-4 .2I·lr··~-~.u·5·_
4-9
.481
6"69
i-~
.718
.876
3.0
$~O
.530
.'86
4 -- -5
1__-·
ly$
1.6
1.6·-W··-II·~
4-------
L.
.438
.643
.64
.601
I
.611
.. 578
.787
.761
.689
L.859
.663
S842
-.l·l...----~-·l-.sp~__~.II
U7CLJ27J
jc~
01
Ob2
L . _ _.....
---- ·-
. ...
.
,. ...
.4
1¾
1.0
1.0
1/3
1.0
0.8
-- n
1.0
1.2
PA4
.545
.670
.507
.635
.475
.605
1.0
1.0
1C
1
.73 C
.820
.592
I-~
.7tl-O
.819
.913
1-·4111-·-----·1-~-·
.728
.859
-~---~-·U··
.,696
8339
.865
.949
..796
r772
.917
.905
--
.671
.821'
I-- -
.892
.. 967
.. 837
..946
-w~
YCz3
~--:6-
.0-4
1 2 0
I
-
3.0
.753
.895
.8T'8 .803
1 3P'9 .. 931
S.832
.793
1,
.738
.886
r··
L.,604
.771
.L---·.L
-
I
.724
.703
..878
.865
--------·- I~
.790
.779
.925
.920
~7IZEL7
.336
.459
~--.-·---~-L-.I
,650
.806:
i------~--)ll*
.406
.535
.449
.5 79
i.~,__~-___~-
1.0
1.0
.
-
f·~-
.........---...
-U~Llii-~~C~
__
_1
..
!
t
1/4
_~I_,,
...
"~'`~'
""i"~~;"----~"~`
"--'~"i^'~`~-~"'
~-"--~~~~~-~~--~~'~'~-^~~'~~~I-"-"~~I
..... ~~~--...~~--...--.I.--~..
-.
.-------~--~-------- ~--~ ~-·------------~--·---------
t
1/2
... .
Table II. For beams with parabolic haunches,
•a= upper value, 'b = lower value.
·--------- · ·-- ·--C
IC
~. !i
..
.7862
.911
.557
.73-2
·~--·I-~···
I
.668
.842
--~---~------L--~I.734
.896
jy2IrIyj
2.
The change of the slope of a column.
Let , M = the moment applied at the top of a column,
Md =
"
"
foot
"
"
%c = the change of slope at the top of a column
due to Me = Md
1
ed = the change of slope at the foot of a colunr
1
due to
Pc
=
= Md = 1
the change of slope at top due to Md = 1, or
the change of slope at foot due to Mc = 1.
The slope angles are expressed as (fig. 3 )':
l
SJ. xedx
xd
1
=-
and
J
dx'x dx
-.--
h2
1
h
X"`dx
El
(3
El
It is generally that the columns are built as
shown in fig. 4 and in that case
h'
d
(2h-h')
*
----
2hEFI
C
H
(3a)
2hEIl
Sc= .
(.-2.
6h 2 EI
)
c
wher h' is the clear height of the column and h is the distance between the foot of the column and the neutral axis
of the adjourning beams
With constant I = Ic throughout the column, then
h = h', and
e = 8c
e-=--h-2O
&JI
dp hence
and
-=
h
6EI3
e••
,,
•,
The strengthening of the top part of the column
due to beam haunches does not exert an influence so serious
as the haunches to the beam itself;, however, the effect depends largely u.pon their,relative stiffness. With strong
column and weak beams, i.,e. beams easily subjected to deformation:,, the effect is very small and h' will be almost
equal to h. In the contrary, the whole strengthening effect
should be taken into consideration to warrant the correctness of the design.
* The derivation of equations (31, (3 ) and (3b ) can be
in
made without difficulty from the references given
Section I. The limits of integration of equation (3a )
are from 0 to h'.
3
E.c..an
of mome,nt o
Q_.
!As
ne tia m
I'n reinforced concrete buildings the moment of
inertia for beams and girders is generally not the .same
throughout the .e.vb:er,-it is therefore necessary to compute
and use the least moment of inertia for a given member.. A
chart for determini:ng that for T-sections. is reproduced here
from the same book refered above, which can be used for Tcolumns and sometimes for beams and girders. The moment of
inertia is expressed as '
•
bI
Whr "i~is
.
a c
'on:s
1 .* (4)
."
ob.t.
"-U
Where u is a constant obtained
from the chart.
By characteristic point of contraflexure is meant
the zero moment point in a rigid member resulted by applying
a unith.moment at one end.
By moment factor is here understood as a facter by
which the moment applied at a rigid joint is to be multiplied
so as to obtain the- actual portion of the moment resisted by
each elastic member meeting at the joint,;
Now let the charater-istic point of contra.flexure
and its distance from either end of a beam or a column be denoted in the way as shown in fig'. 7a and let
a
=
the c-hange of. slope at the left end of a beam
due to Ma = If
the change of sl.ope at the rfight end of a beam
due to Mb
,
cc c
the change of slope at the top of a column due
1, and
to Me
ad = the change of slope at the bottom of a column
due to Md 1,
6
then from Fid. b, we have
ab
F.
...
=
XPRIWI
qI
I I
f.
d
U. =.1!5
.
e
.4'
.40
.435
.55
.50
• o0 i
-4 0
.5
I5
dlo
t
•
uLAxS
•Zo
*zS
Crk fcfincns
Norn~nV
.o
of
,45
lncr
o'
T-5econ
.55
.50
r.
2b~
cS5-
22
The moment M=1 applied at B will produce a moment Ma at A.
The magnitude may be obtained by simple proportion, whence,
a
L-a
Now by definition ebis the slope angle produced by
moments M=1 both at ends A and B (fig
7a_c_d_e
86c), then from figures
it is evident that
Ma
ab =b
a
= eb- (1 + L-a
SL-a
With the same conception and draw similar figures we can
derive expressions for ma, ac, and rd.
a =
and
•aa L-b
ab =b,-
- L- -
c C C
h
h-d
L-a
ad = 9 d- -h
hac
Thus we have
8, at left end of a beam
at right end of a beam
* f,
.
(C5)
at top of a column
c
,
at foot of a column
The values of 8 and P are given by previous equations, and
mostly, for beams either of constant cross section or of
syfmmetrical in form, aa
=
ab =
a
. The values of a and b will
be given later.
To ascertain the moment factors the transmission of the moment should be carefully considered.
In order
to avoid confusion the following set s of notations are
used:
na = a fraction of the moment applied at a continuous
column, which is taken by the
._..
nb = a ftaction of the moment applied at a continuous
column, which is taken by the•_•l•UbJi._e
n, =a fraction of the moment applied at a continuous
beam, which is taken by the _he•...te_~
ig
nI = a fraction of the moment applied at a continuous
beam, which is taken by the_Bkpa•,_Jadt_,,
nul= a fraction of the moment taken by a column, when
the moment is transmitted to it from any member
adjourning its ".pjz
end.
nlu = a fraction of the moment taken by a column, vhen
the moment is transmitted to it from any member
adjourning its L.oj_ end
nrl= a fraction of moment taking by a beam,, when the
moment is transmitted to it from any member adjourning its ri•ght end.
n Ir= a fraction of moment taking bya-beam-, when the
moment is transmitted to it from any member ad-ýjourning its lfi end.
For clear understanding of thezse notations figure 16 given
in the section on separation of moments will be refered.
Let us denote the resulting slope angle common to two
rigidly connected bars due to M = 1 by "y" with suffixes
.
-IC.
5-
tc
ii
'
kM
-(
c
I/
a'l
'-(I
COL)
(c)
F;9. T1
cIL
or fc
S~ry
M Lr
-rcol
or
C..
('r~
(a)
\I
'rr
T
7~r7
~4 .i
r6
3, '
f'Z-'r:
r
or
l;
I
\1\
~--*--- (eb)
/
,
-r
,
i/
b,, cý, and be representing the angles common to beams, columns,
beam and column respectively.
Now let the four members given in figjures 7 b andc
be arranged
as shown in figure 8, first consider the two
vertical members,
or columns, be connected rigidly
and
are continuous and a moment M = 1 is incurred at the joint,
fig. 8 a, then the slope at the top of the lower column
= nb
= ( 1-na)
a,
and of the bottom of the upper column
-na
d = ( 1-nb)
d
By assumption II these two slopes should be equal
to the common slope angle Yc at the joint,
therefore
have,
ac
na
+
ac
0d
d.
nb =
and
-
**....
a c ad
ac
+
d
(6)
we
Similarily from fig. 8b we can derive
Lb
nl=-
ab:+ &,
Yb -
U
a•
aa+b
(7)
I·
-aB
_
s~a+ag~
Again consider the case shown in fig.8c. we have, f-or a moment
coming from column to beam,
Ybc= nrlab= (1,nrl) "c
and for amomemt coming from beam to column,
Ybc = nulac = (lnnul) ab
therefore.,
rl
n b+i
Lc+-b
Ybc )= me
71
(b+ac
4.0·
-
,**
*I.-O.~li
(Q\
In the simila:r way we can derive expressions fornnlr, nitetc
ey
examin'e.the
equations (6), (7),and (8),-the following rule
holds true':
-If a moment M=1 is applied at the joint of two bars,
x and y, then the portion of the moment which will come to bar x,
at bar
=
, ,, , , .,. ...
; .. (9)
i at bar x* a at bar y
and the portion of the moment which will come to bar y
n -a
at bar x
.'-T•• bar x+% at bar y
By a it is understood the angle caused by a moment M=1 applied
only to the bar in question.
n
*I Of at *.! at 01 *1 0./0
0
i'.
(M~)
(L\•
I,
~il
77~T6
27
I;
-i1S
///
d
1
_-c
C
---a--
A:Z7-......-7§S
-
-
A- -i
....................
- _,.-.5__F
• "; i•
IQI-----•<:_.
....
-----
"'
,/-;
5 s•
'•;•<:•:
..........
-5; (• ,_•
1'
So far we only have considered a combination of two *usbers,
for cases as shown in fig. 8c and 8d, equation (9) can still
be applied. By looking the two continuous members as a single
member and according to assumption (2) we have,
r
n-r=-
Yc+"b
nul=
-
%b +Yc
nla
n
Yb+
ec+Yb
In the same way we can obtain the value of n for other comb ination of members..
To find the values of a, b, and c, d, distances of point of
contraflexure from ends of beams and columns respectively,
as shown in figures 9 and 10 ýgeneral expression is here first
derived:
Let p = change of slope of the support due to M=l1 and k
= change of slope at ends of the member due to subjected load.
Using other notations as used before,then refer to fig.(11)
it is evident
-paa= ka
a+ Ga- ) + Mbb
-b Mp=kl, + M (Ob-e
) + MaP
If the member is only subjected to a moment, then
oa + M (ea-)
-PaM
-P b
=
+ MbB
- + MbGOb- ) + Map
(ilL
::
Now since M=- a
•••
b
6
), by s•ubstitution
can
-ae
find
we can find
(11)
a+pa
Similar1'-ily. for expressions of b, c, and d.
From equation (11) and kee in mind that p is the angle made
by the support due to M=1 the following expressions can be
derived':
FProm fig. (9)
a=
b"= b'8L _n _a
_L_.L_
ea+nrlb,
eb' +nlr a
'
cA= h'Bc
d=
-
he
c,+nluad,
d+nul
d,.-,-*****
(12)
c
Where nrl, nul, etc. are those given in equation (8)
tif
h"
c ?,
C.'
a
'cI+ Yb
L
.
.
- * -* ***
(13)
a+ Ye
From Fig. (10)
Same expressions as equation (12)
except
****(14)
arl,nuletc. are those given inequation (10)
For beams and columns of constant moment of inertia, the expressions in equations (5), (12) and (14) can be very much ,
simplified and in that case::,.
.=L
6E
nd
aI,
8=--
=
29I
Equation (5)
c
3P,
=
h
6EIc
and
ec
=h__
3e
2E:Ic
is simplified to
'(a = p[ 3
Mc = ic[3-
L 4]
L-b
h
a b=p
]
3_
'L ]
L-a
d = p-L[ 3- ..
*** *******(15)
.]
h.-c
h-d
and all values of 8"s in equations(12), (13) and (14) can be
substituted by 3-times their correspondingp 's.
5. The procedure of computation
For investigating an existing building or a,building,
of whose members the sections have already been approximately
determined by the method given in the introductory, the distances of the points of contraflexure, a,b,c, and d from each
end of the meymber have first to be found. In order to do this
the following steps have to be taken:
(1) Determine the moment of inertia of each member by the
aid of the given chart, if necessary.
(2) Determine the values of E and P of each member by
the aid of equation (1) and (3) and by the given tables, if
nacessary.
(3) Beginning from the columns of the lowest storT,
which are generally fixed or rigid. The distance d will be':
Taking care of the strengthening effect of the column
head,
S3h-.2h'
2h-A'
.
h'
. (16)
3
d=- h.- .**....*
and when the same is neglected,
(1T)
3
In the first case th:eangle at the top of the column,
=
h13
----*
4EIc(3h[h-h']+h'2)
and in the second,
dc =--
4EIc
or C . -
*
.(.
3
2
** (1-)
(19
8c
(For derivation of equations 16, 11, 18, andl9 see
appendix.)
In order to proceed the computation of angle'%y'
column, it
is nacessary first to determine
of each
' (equation 6 )
"s
in determining the latter , the distsnce"c"o the next upper
column has to be first assumed (eqauation 5 ). c=•-.h is in
4
most of the cases close enough.
With this assumption and the values of Yec thus found
we can determine-, from left to right the distance a , and
from right to left the distance b by eqinuation. (13) for the
beam elose.to the wall columnP and by equation (14) for beams
in interior spans.
(4) After knowing the location of points of contraflexure
so proceed to determine the:,'distance'd: of the columns in the
story next above. In doing that use equation (7) to compute
Yb, equation ?(10) nlu, and finally equation (14) gives the required istance d.
0(5) The location of the points of contraflexure for beams
above the second story will then be calculated. The angle m for
the column below can be found either by equation (5) or (15).
In order to do this it is again nacessary to assume the distance
c =1 h of the column in question. In the same way as before we
4
a4b,for the beams.
compute Yc and the dista:nces a,a•g
(6)
Repeat the process mentioned in (4) and (5), we can
obtain all the distances a, b, c, and d for all stories of the
building. Beginning from top we can finally determine the
last unknowjn : i.e. the distancec, which was assumed before
for each story of the building.
In case that the final value c obtained in (6) varies
a great dea-l compared with the assumed value, so the process
may be repeated by assuming c = iits•- final value obtianed in
the first computationL IN general, however, the first computation gives result sufficiently accurate, while in the
second computation the result is almost exact. The degree of
exactness to be attained depends:, of course, upon the character of the building and the opinion of the designer.
N•t•'l
For c =
I
h and for constant moment of inertia from
4
equation (5),1 YCv
cd
I
=
h
V•T
6•
.,L.
or
or
,
d
5
3
31
Appendix.
The
Derivation of equations 16, 17, 18, and 19.
From equation (11)
·¶ !·*~··t
i· :·
9 d+
(A)
Pd
From equation (3a)
Od ~-d
" '2h- ~
-2h'
-- AA
3hr.2j
.=
2
6c ih1~*
Bc-~~h:2
2hETT
8h"ET
also from equation (5)
cCd "
C
h"
**
h
*..;***
*.- (B)
Substituting the values of edand Scin (A),we have
d3.(h-Zh'
-..
.
2
3h" (2hi-h') + hBIcpd
For continuous and rigidlj fixed columns Pd =. 0
, then
h,2h'
d=- h'-
3* (2h-h')
If h = h'. then
d =
........ ....
h.
(17)
3
Similarlj
assume pd = 0 and substituting the values
of OC,
pc and d in (B), we obtain
.h'.
-
3h2 3h-2h' (h-h"_
c hE6h
"I
)-6hh'Ic(3h-2h")
OEc3(2hh'
(18)
* * *·~
i"] +h'')
4ETI c (3h[ h-h
If h
=
hr', then
4EI
*... *
or, since P
3
h
Cg
2 Pc.
.
, * * (19)
-PART 2
MOME1MTS AND STRESSES DUE TO VERT'ICAL LOADS.
In computing moments and stresses due to vertical
loads a general assumption has here been made that the deflection due to vertical load at the. top of the columns is equal
ýto zero. The theories followed in this section are mostly
given in Prof. W. Ritter's book " Anwendungen der Graphische
Statik." It is endeavored, however,in this section to put
thdse theories into practical adoption: for the particular
use to solve rigidly framed buildings. The use of moment
factors developed in part 1 is a special feature, upon which
the separation of the moment at each joint is based. This
principle can be adopted for any kind of rigidly framed structures. In view of practical use several short cut methods
for finding moments of a rigid member are presented here
also.
ae•k~rs.
From equations (11)a,
(11)b, and (11) if
are not =0 we 'have,: from eq-ua~tion
Pa,
a-.a
,
:((11,
'
Pb = -eb-
b
Substitute these values, in (11)a and simplifying
ing expressions result':
+
+
a L-a
Mb
L 'b
MaL =
the follow-
*.a-- -a
_a
b
ka and kb
..
b . kb
L
Refer to fig. (12) and by geometrical relation, we can see at
once:,
kaa
a
b
S =m
.. kb
-kb
i
For members of constant moment of inertia ka and kb =EI
-ALLx)
L
)=. L(,rk 0
6ERI
substitute these expressions into former equation-,
we
obtain,
Sa a - 6 L-.,
a
2
L
L
L
( Morley's Strength of Materials, page 1 4•6.) and
b
Sb
-.
L2
L
-b
L
A
'
*-
Where A is the bending moment diagram due to the applied load
when the member is simply supported.
T is the distance of the centroid of the area A from the
revre
origin A, and (L-r) is its distance from BforSc-, s
q and q' are the second factor of the right hand expression of each equation.
If we substitute c amd d for a and b respectively into equation
(20) , the expressions for Sc and Sd for a vertical member result. All the short-cut methods given below are based on equa-tion (2.0).
For beams of varying cross section separate discussions
are made in Case A and Case B.
--
Ca
A. Uniform distributed load . Fig. 13.--
In this case the moment diagram for a simply supported
beam is a pa:rabola:, Calling t-he moment at center =M, then,
A=2/3 ML,
L-'Z=--L
hence q=q'=2M.
2 '
From simple geometrical relation of similar triangles it is
evident that the graphical solution of Saand Sb shown in fig.
13 holds. The intersectilon of the two crossing lines coinsideswith the apex of the parabola.
For bea:ms of varying cross, section , we have for equation _,20)9 ,
ka=kb
8
=
fL dx,
I-x'x dx
L2B
tw
(Morley's ,page 183)
(Part I, )
EIx
By putting these two angles in the form.of a ratio-the factor
34
i1
S-_dl both in the Inumera:tor and in the denominator caacels
I
0
x
out, therefore the variation of moment of inertia of the
member does not effect the values of- Saand :Sbin this case.
-Case B.. Single concentrated load-. Fig. 14-A§ shown in the figure, the moment diagram for a
simply supported beam under this case is a tri.angle, whose
A=1/2 LM, Yfrom A = 1/3 (L+x) and from B
1/3 (L+•i), hence
from equation (20)
qkL..
q,=
)
L
and
L
To find the values of qand q' without calculation, lay off
the length L either side from the load, join the ends with
the apex of the moment,diagram,and extend the lines as shown,
then q and 4'qare obtained, sior. by similar triangles the
relation
q ":.
M = L+x : 'L
and q' :M = L+x":L
exists. Again dra&- crossing Jines in the way shown, then the
values of Sa and Sbcan be found. A s:imilar relation holds, f.e.
S a: a
and
=
q:L
Sb ":b = q"': L
Now in case that the cross section of the beam varies, the
s olution of the problem is somewhat difficult here, for the
fc+k--or :
-
J- dx
( compare Case A.)
become
Ix
o
and
dx
iLdx
•o IX
+ x
xx
in the expression for angle k,
in the expression for angle B ,
and these factor-s will not cancel out when expressed inithe f
formof a ratio, therefore the variation of the cross section
of the beam does have effect on the values of Sa.and Sb
To take care of this effect the work is complicated.
In order to facilitate speedy solution, a table is reproduced
tere. from "StrassenerJs Forscherarbeiten". The values of S a
and Sb can be taken directly from th±p ta2ble. For single con•....
i-P
,b
I
I·~
Ir
-
IIr~t
1
b
~C-------t------------I
I
I
Il
191
_ ,
iC----
_ _
-Fc----.
2.
i
r
;-I
_ i
Lk_/
Fi7
I?
z.
/:
Sr~a
6
1~
\
r'
bb
Me
c* C")
r'
-
-t-
- >.---+
L
I
o
-
r
'"
N
t -h---9----·r
I
/:
-I
-\ -
7
N
/.
-
W47
I
Ii
ti
N.K
I
/ -
~"
,
-
i
:!,
-M~d
Fig .15
.9
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L
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.4-.
--
7
-
I
o.89o
a 00~
0.5sS
05,52.9
~-:-~
---
·
-
·
-
37
centrated load multiply the table value by the load and for
severa:l concentrated loads or for a combination of concentrated load and uniform load take each load separately and
sum up the moments.
Case A and Case B hold also for vertical members.
Columns subjected to crane load. Fig. 15.
-- qCas_.
The applied crane load P causes a moment P'e:, which
is equavalent to the moment caused by a horizontal couple
H as shown. Lay off CC' eýqual to the moment P'e. Draw line
CK parallel to J'D and join LK as shown, then the resulting
area bound by the broken lines CK, KL, LD, and CD will be
the moment area of the column due to the crane load P with
no regard to t'he continuity of the column. The value of A and
x can be easily computed. Compute q and q' from equation (20),
and lay off the same equal to CC" and DD" in the figure,.then
join C"D and CD" , the values of Seand Sd are found. The
proof is the same as the first two cases.
The moments due to vertical load will, in general,
be investigated by considering each single span separately
as shown in fig. 1&, in which it brings out the way of the
separation of moments onto the columns and beams surrounding
the loa.ded central span.
If the moments Ma and Mb for the loaded span be determined according to the method given in the previous section,
then at the column to the left of the load, the beam will
take a moment = nri'Ma and the total moment which falls to
NAMa
Ma - nrlMa. From this
the adjacent columns at left =
moment the lower column takes an amount = nb 'AM
A and the
Similasxrily at the column
rest falls onto the upper column.
to the right of the load, the beam takes a moment = nr' Mb,
the total moment which falls onto the adjacent columns at
right = AMb Mb - nlrMb, from which the lower column takes
a portion = nbAMb while the upper takes the rest.
The tra'sition of moment lines for unloaded spans
_
__
b
it,
A
I
---I
~
F
"KY.
rn(,Py
flri
, rl
A
lb
1/-~L
P
jIMA
I
P
-I-t
I
J~TJ-
'R9. la
I
I
-
,
r39
is as followsý:
For all spans to the right of t-he loadedi t--heamoment line
passes through .the point B,
for all spans to the left of the loaded the moment line
passes t~hrough the point A,
for all the columns abcve the moment line passes through
the point C, and
for all the columns below t-he moment. line passes thru.
the point D.
The pointsA, B, C, and D are the points of contraflexure
as marked in fig:ure 6, part 1.
For the.j.oints in stories next above or below the moment
transmitted over will be sepatated again. Let the moment to.6
be reseparated be called McandMd., then at the joint next above
the reduced column moment
nlu' Mc, the total moment to be
resisted by the beams = AM= Mc - nluMc, from which the left
beam takes a portion
beam.;
=
nlAM c and the rest falls to the right
Similarlly at th.e joint next below t-he reduced column
moment = nul'Md, the beams have to resist a moment = AMd=
Md- nul Md:, from which the beam at left takes the portion
nl'AMd';', while that at right takes the rest.
The moment tra.nsmitted and to be reseparated becomes
smaller and smaller , the further is the span from the loaded one.
At the end columns the moment carried over by the beam
should be separated in such a way that it will wholly be
taken by the columns, so in that case AMO-Maand AMb= Mb.
In the above discussion we ha:ve only considered the
absolute magnitude of the moments. Hoever,in ca:se that
several spans are lo-aded, t-he moment due to the load in
one span may in most c:,ses cancel that due to the load in
the other. In order to get the algebric sum of the moments
due to the loadings in various spans and to avoid confusion
the following conventional signs will be found helpful=
.f
" The moment is considered positive, if the elastic curvature eof the beam axis at the portion considered
slopes downward, and nagativer, if it slopes upward..
The moment is considered positive, if the elastic curvature of the column axis at the portion considered
slopes to the right, and negative, if it slopes to the left."
In figure 15 the positive moment is drawn below
and to the right of the beam and column axes respectively and
the negative moment, above and to the left.
Keeping the above signs in mind it is very easy
to determine the character of the moments by considering the
elastic deflection of the member..
3. De-.e..ai
Leq~t..
.
ra
For all members of a building , if their moments
at joints are known, we can find the shear and normal forces
very easily. For rectangular construction l:dvantage can be
taken from the fact that, just as the shear is perpendicular
to the normal force, the column axis is perpendicular to the
beam axis, therefore we can find the normal force of a column
from the shear of a beam and conversely the normal force of a
beam from the shear of a column.
Let Q be the normal force or the shear required,
Qobe the same corresponding to a simply supported member, and,
as before , Ma and Mbthe moment at the. left and the right end
of a fixed beam, and Mcand Md the moment at the top and the
bottom of a fixed column respectively, then, taking positive
shear as that directed to the top
at the left section of a
beam, and. that directed to the left at the lower section of
a column, we have :
For the shear of the beam or normal force of the column in
= Q0-
question
* * (21)
-
For the shear of the column or normal force of the beam in
question
a = Qo +
.(22)
c-d
h
For columns , since they are generally free from external
forces:, then Qo= 0 , and
-.--.
S.
A
..
**
(23)
h
The derivation of the above expressions follows immediately
by taking EM = 0 of the member considered.
Both the maximum stresses in beams and columns occur
from the source of maximum bending moment and direct
stress
resulting from normal force. For columns, especially those
of very slender section, the deflection, which we so far assume
to be ziero, may cause an additional stres4, but the stress
due to maximum moment is , in almost all cases , greater than
that due to maximum deflection and these tiro will not occur
simultaneously.
If we load alternate bays of the structure
so as to give maximum bending moment to the column in question, (seF d-0iscussions given later) the deflection ca:used
by the load on either side is of compensating character and
the stress resulted therefrom, if any, is generally small
and negligible. From figiure 16 it can be easily seen the
following criteria for maximum moments and hence the maximum stresses of the beams and columnst
For beams maximum positive moments are caused by
loading all the odd number of bays in both directions from
the bay in question, counting the latter zero . The negative
moments are caused by loading all the even bays in the direction and all the odd bays away from the direction of the beam
end in question, counting in the same manner the loaded bay
zero.
For interior columns maximum momemts are caused by
loading the bay adjacent to the column in question for the
full hight of the structure, and, where possible
y loading
alternate bays in both directions from the bay.
For exterior columns loadings causing maximum
moments are similar to interior columns.
For corner columns the moment may be considered as
being introduced into the column by the two girders meeting
at right angles to each other, and the moments may then be
combined t-b give a diagonal resulting moment.
The maximum stresses in all members will be found
by combining the stress caused by the maximum momemt with
the normal force found by methods given in section 3.
PART 3.
MOMTENTS AND STRESSES DUE TO HORIZONTAL L ADS.
In high building design the stresses caused by the
wind load applied in horizontal direction play such an important r6le that a close investigation always seems to be
justified. For buildings, of which the width is small compared
with its height, the wind stress- is always a great factor.
In this part of the work the general theories given in Prof. W.
Ritter's and Dr. Strassener's books will be followed. It is
endeavored, however, that the theories are extended and put
in unique manner for the practical solution of high buildings
due to horizontal loads.
In dealing with horizontal loads several approximate methods given by various authorities have been known,
and they are generally deemed as simple and practical. Nevertheless , none of them seems to be consistant with the actual
condition. The exact method given in bulletin 80 of engineering experimental stat;;on of the University of Illinois
and sevetal other known methods derived from the theory of
least work are all too laborious and they are, therefore,
restricted in almost all cases in its practical adaptibility.
In the preceeding parts we have made one assumption, that is, the horizontal deflection of a rigid joint
due to vertical loads is equal to zero. This assumption, though
it does not conform to the theoretical correctness,
is rea-
sonable for the solution of the moments and stresses due to
vertical loads, since the effect is
small and negligible.
In dealing with heor'~Khal loads this assumption will no more
hold, we have here two kinds of moment to consider:
a. Moments caused by loads considering the rigid
joints not being subjected to any deflection.
b. Additional moments caused by resulting deflection.
The present method will be described under various sections
given below,, from one-story--frame
to the general case of
many-story frames'"
1. Derivation of expressions for moments due to deflections..
--,~-~-~~~~-~------------------ -~-i
I
UI
/ >I
'I
/M
Sif
CS 8
/·"
~
i
i'
i
-~n~,··:--···-? P
--
7i'/~~i
1/i
j
j
M4
"FL.f]
]:s• .\' .
Considering the case as shown in figure (1i) the column has
an elastic s-ufport at its lower end, the upper end is
to move and subjected to a horizontal load
free
H, then the
resulting deflection is
Hh2 d
A-= H'+
and the angle, through which the top of the column is turned,
is
,=
where X' = h2(x
' + Hhjd
d -
Pc)
and a'= had
For expressions of ad and Pc refer to Part 1.
By substitution, we have
A= Hh2 (Oc+Od- 1c )
and
,r= Rh(xd+ Pc)
Deviding the former expression by the latter, we have
45
=h-
ad Od
Part 1
But from equation (-),
S-----
ad + ?d
-d
..
therefore
hd
..
or,
.
h--ý
The moment at the foot, of the column due to the deflection
A caused by the horizontal load H is then
Md6= -
-
h=
A
ad+
0
d
(From equation X)
-d
or, since ad.+kd.
d
Md
5
* * ** *
(24)
(h-d)hpci
Equation (24) is the general expression for moment due to
deflection, when the deflecting end is free to move.
Again refer to figure 18 the column has elastic
supports at both ends and theck-clu•;i
distances c and d,ai
of the member, .i.e.
already fou.nd according to the method
given in Part 1. First assume the top end of the co.lumn is
free to move and is subjected to the shear Q, then we have
the moment.
h 2 pc
Similarly by assuming the lower end of the column being free
to move and subjected to the horizontal shear Q, we have for
MA--^
.
M
,
=
h
cA
hBc
Now in the actual condition of the column the ends does not
move freely.
As soon as the deflection A occurs at one end,
there produces moments Mc and Mdat their respective ends; although
these end moments will offset a part of the angular turning
of the column ends, they,. however, do not have any effect on
the value of the moments Sc and Sd, since by hypothesis of
Part 1, point C is the zero-moment point of the column due to
a moment acting at the bottom end of the column, and point D
is the corresponding point of the same due to a moment at the
top end.
From above discussion we can have the following ex-6
pression to find the distance zc, distance of zero-moment
point from the top end.
S
-Sd
zc - c
h-zc-d
z = d + h-d-c
Sd
81
S
By substituting the values of Scand Sd in the above expression
we have
ch
c+d
which gives the location of the zero-moment point.
Now at point C the moment Sc also equal to Q(z- c), from
which relation we obtain
= A.
h2pc
c
c-
C
From the figure it is evident that
Mc= Q'ze
and
Yd=-
Q( h -
zc )
By proper substitution the equation for moments is expressed
in the following general form:
Mc= m'C
Md= m'd
Where a is a constant
...........
= -
(25)
.
Pch(h
"-d)
In computing the moments of a building frame the value of A
in equation (25) can be assumed as unity.
After Mcand Md
due to unit deflection are found, we compute the shear Q
of each column from the following equation:
h
for moments of opposite signs, and
S. ....
....... (26)
( Md - Mc )
h
for moments of the same sign.
The sum of the internal shearshould be equal to the external
load 1.
The actual moments are found by multiplying the values
of Mc and Md due to unit deflection by the factor
-
The separation of the moments will be done by thea
aid of moment factors as described in Part 1 and 11.
In order to make the method clear the application
of the method to various cases
tions following.
are given in different
sec-
2,
.S9igie-1toryfraue4
V-5.;
V--~
7
7/77
Let us assume the case given in figure 19, a single
story, two span frame acted upon b y a horizontal force H at the
top girder.
The columns may be either hinged, elastic', or
fixed.
The horizontal force H causes a deflection of the
columns at the top end in the direct-ion of the applied horizontal load.
The deflections at all columns are the: same,
because the top ends of the columns are connected by a con;
tinuous girder and the effect of normal stress can be neg.e.
lected.. The magnitude of the deflection due to the applied
horizontal load is unknown, we can, however, first assume
a unit deflection and by the aid of equation (25) the moments
due to that deflection can he solved.
From the shear calcu-
lated therefrom we can determine the force
the assumed deflection.
coefficient
H_
,
By multi'plying the moments by the
the required moments due to t-he horizon-
tal load H can be found.
foll ows:
i.which produces
The procedure of computation is as
f'
777
'147-
(a)
(b)
F9 2
r I
A
*11
A
i
'I
~iITL·_
/
A
IL'
i
U0)
/
'9~zz (Al)
A~--
'i
F
Si!
FT
ti
1
7-7
(Ca)
77747
774-;
77
(6)
1. Compute the characteristics of the frame, i.e. distances
a, b, c, and d, and the moment factors of the corresponding
members by the method given in Part 1.
2 . Treat each column separately and assume the column in
question either as fixed or elastic, as the case may be, and
the rest as free to move. ( Figures 21 a, 22 a, and 23 a.)
Compute the moment of each member in each case by equation (25)
in mhich the value of A is taken as unityl. The reason for
taking each column separately is for the convenience of separating the moments by moment factors to each member.
The
moment diagrams in each case are shown in figures 21 b, 22 b,
and 23 b. In actual computation these diagrams need not be
drawn and the work can be done mentally-.
3 .Taking
the algebraic sum of the moments, it gives the
combined diagram figure 24 b. The resulting moment in each
member is that due to the assumed unit deflection.
4,
Qa,
Apply equation (26) and find the values of Q1,
Q2,
and
fig. 25, then the sum of the shears must be equal to the
load P, the load which causes the unit deflection and produces the moments, or
P = Q,
+
Q,
+
Q,
5. Find the coefficient of Ji with which the moments
in figure 24 b are to be multiplies to give the required moments of the members of the frame due to the horizontal load
H.
Fig. 26 shows a general:case for the srparation of moments
due to deflection b. moment factors.
3. Twow-storv frames.
For two story frames the moment of each member
are due to the combined effect of horizontal loads H. and
H2 , fig. 27, Through t-e influence of theseloads the frame
changes its shape as shown in fi;. 28, i.e. all the joints
deflect horizontally in the, direction of the load.
Now
since the magnitude of the horizontal deflection is not known
so we have to get at it in some indirect way
-as before,
by the aid of an assumed deflection and the moments
therefrom.
resulted
5:t
C
__
---1
i
~4_---
rr
.,,,-
'I
'
/T
'
--- -1 - I
-;i-
2;
I1
T4
rr
I
I
7
C)t
C4
(C',)
TP'
!4
)
1
Ii7f
(b)
FTg9o
%M
F ,
I
, -Sdr
's
r,,,
~(
g
1
--
---.
-f.r _
-,--+
F~
N
.
tg3
l
is
?_
----------
PI
F=.35
Fi~
9.~
-
-
-----
----
-
------
---
~--------------
For two story frame we have to consider two conditions of deformation. For the first condition assume the
top girder of the frame is subjected to a deflection A=unity,
while the lower girder remains in its original position ( fig.
29 a.).
The moment of each member can be computed in the same
way as for one story frame, in this we again consider each
of the three columns separately and compute moments by equation (25). The moment for each case is separated and the
algebric sums for the three: cases are found.
The external
forces acting on the frame, which produces the deformation
of condition 1 are (fig. 29 d)
On upper girder
P,
=
Q.
+
Q&
+
R,
=6
Q
+
2q
+ Q8 +
Qa
On lower girder
al
+ Q2, +
•
and on the footings
S, = Qz + a,
+
Qs
The suffix given to P, R, and S
denote the number of the con-
dition of deflection considered.
For the second condition a deflection of A = unity
at lower girder will be considered. The ujper part of the
frame is free to move, so as soon as a deflection is given to
the lower girder,
the upper girder will have a deflection
greater than A. The latter will be assumed as being restricted from producing a deflection greater than A, so that
the relative position of the upper part of the girder to the
lower part does not change. ( The deflection of the upper
girder cannot be assumed as being restricted to zero, as
in that case another moment due to deflection in opposite
direction will be resulted, which violets the condition of
moment separation ).
After the moments in three cases are
combined and shears found, the external forces will be
( fig. 30c ) "
On uppe"girder,
R,2 =Q, + Q1 + Q*
on l.ower girder,
P =
, + ~Q
+ Q, + 1 + Q2 + QS
and on the footings,
S,
= Q1
Now we have two loacL::
+
Q2
+
Q 3
diagrams on hand, figures 29 c and 3o c
The first produces a deflection unity at the upper girder and
the second produces the same at the lower.
diagrams is also shown the elastic behaviia
In both of these
of the frame
against the acting-loads. From the moments resulted from
these assumed deflections we can determine the maentsdue to
loa4
any horizontalby some simple modification.
Let t-he moments caused by the loads in figi 29 c
be called am
and those caused by the loads in fig .30 c be
called a,:, then by addition we have a combined diagram with
forces
PL- R,
and
acting on upper girder,
" loker
F,R,
"
-
" on
o, the supports.!
Fig
31
The combined moments are then = al + im2 .
It should be kept in mind that this combined load.
diagram gives a unit deflection both at upper and lower girders.
Now in order to conform the actual deflection for a specific
case let the load5t.. in diagrams29 c and 30 c be multiplied
by the coefficients x and y respectively, the combined loads,fig(32),will now be :
6PI - xR,
on the upper girder,
xP, . 7R,
on the lower g.irder.
and -yS,
2 +
x S2 on the supports.
Simila.r'ily tne moments = xm i
+
ya,
It is evident that for the applied loads Hi and H2
=R,Hz
P,-
and
xP,
gR~ =
tH,
From these two expressions the coefficients x and y
can be found and the moments computed from
M = xm, + y2
In practice in order to find the maximum moments it is
aenerally required to treat the horizontal loads as moving
load and in that case the values of influence moments have to
be computed, i. e.
t-he moments due to a horizontal load unity
at each panel point separately,, thus the conditional equations
are as follows':
For a load unity acting at panel point 1,( fig.- 34)
P
R, = 0
xP,
and
M-
"owergirder.
=
R,
-
1
xm 1 + ylam
2
For a load unity acting at panel point 2, (fig.35) at upperl girder,
-Pt - xz Rt = 1
Xo
and
t
P1
-I
TR
= 0
i = x 2 m1 + y7
MR
2
For the load condition shown in fig. 33,
M=
+
l
, 2H
frames
4, Three-and- more-stor~
The method of computation given above can be extended to the solut.ion of three and more story frames. There
are as many conditions of deflection as are number of stories.
From the load diagrams the elastic behavior of the frame
against the influence of horizontal loads will be given. For
any specific caze each condition of deflection gives a load
diagram and a corresponding sets of coefficients..
There
are always as many conditional equations as are unknown
coefficients'
To make the method more evident it
-.
•.,
.,,,
Lor a toerL
.,,
Ll
suory fibE
Iram
,
Cf,
r,,,
,~,,
is given below
f
LL~,·
h1nUa.flectZio[n conUUiLUS W.imn I'hlr
corresponding moment, and load-diagrams (figures 32 a-c) and
the conditional equations for the determination of coefficients for each case of loading:
For the specific case s'hown in figure 40 the con,
ditional equations are?:
xP, - yR, + ~R
NP,
wR" - zR," -,
zP + xRI
and the moments
M
=
-H,
-
H',
xmi + yMa, + za 3
the
If the horizontal loads are taken as moving load, then~conditional equations and expressions for influence moments for
a unit load applied.at each panel point separately are as
follows':
For a load unity acting at first floor girder,Fig..37, the
conditional equations are":
7-
_
_
_
__
42
.
e-o
H..I
-4-·-
mi~
-- ·
7
r
7T
T~
.38
9
=a)
Ii,
W-1i
r'1:~
0o
-b
(bA
£'1
t4
7
17-4-
V;4~Ac
9L."3
L
w
I5w
'C
rc\4'
4i
.2
77J
-P--;·1 -
p
\i
ii
! <I.
·
3,
77,r,-
,,,,-----
3 -5+o rj-YFra me
__
Locz4 5.tMrent-
4
K- *--
_~i_
~T
F
-,
7p.
/
n
.(P)
-J1
r
I
I
xlP, r. yiR,
+ agR s =
= 0
aR,"
-
4P, -•R"j
=31
R ' - y,
Z- P, *·
From which the coefficients xyy, and z, can be solved,
The
moment expression is
M = x ai
+ y.mi.. + Z.as
For a load unity acting at second floor girder:,, fig.- 38, the
conditional e.quations are-:x 2 P,,
P2
.+ zR
-
=
7-Z 2 R
x22 R
0
"=1
z P, + 1R1 ' -, RS2"=0
and the moment express:ion is
M2 = x 2l.,
+ y,2,m2 + z a0
s
For a load unity acting at the top girder; fig, 39, the conditional equations are:
x,P, - y•,R' + z-,R"'= 1
- xqRa023s =
as2`P
z
,
+PxsRI'"~ 3
+
R2 '
= 0
and also the moment expression is
M, = x.,m,1 + y m. +zm3,
s
For the specific case; fig. 4po:,. the expression for total moment
due to the combined effect HI, 'H
M' HzM
+'H2
and:H, is given as.
M- +÷'HMs
In all these cases the signs of the moments should be taken
into due consideration.
1
In computing the moments it is eecommended to use
t~e 4Follow;vy
+fabLac
f 0 r-v,;
p-uzW
rim
TaTobux
Lorrm
aQ
PoIn
0±ember
a
cnvc
for Conm4#prfng
--L
n+
I
Slome~i
M
4
or
H il~+ 9
kMOac~i~th
of
Tcrce
-
ry
-Frame
-
I
I
I
I
I;
I
I
.
Y.
3m
+v
+
.F..r
lone~r
+ol9ýr 9
c-V
Momcaorh
a:
---
1
`
r
9 Coy
fO;Ld
icy
t
e 3pe;
2d
-- ·--------- ~---I-,.-..
-
---
~
M
---
5.
eial
caseThe. usefulness of the methods is n.ot only limited
to cases discussed abbve, but it can be adopted in almost all
kinds of framed structure with some aodification.
of this method it
By means
is especially simple and exact for the s.o-
lution of truss-formed frames and truss-bents,,
Let take the case s-hown in fig. 41, a truss4f ored
frame., -It is evident that this truss can be considered as a
frame of the same type formerly considered except it
is layed
in a horizontal position and, instead of one side, both sides
are fixed-, i.e. fixed at both supports-.:
The deflection con-
dition can be so considered that:, while one vertical member
is subjected to a deflection A due to any vertlcal load, t-he
ends of other members joining this vertical are all subjected
to the same deflection, whi:le the rest verticals are kept in
their original position by some imaginary external forces.
HRere again the moments of all members due to the deflection
of each indevidual member for each deflection condition in
question are to be determiined and these moments combined to
give Ia moment diagram for each particular deflection condition.
Figureha 43 gives the three cases of deflection condi-
tions with their respective moments and load diagrams;
For the case the truss being subjected to the loads
W1 , W•,
and Wa (fig. 4z) the conditional equations for solving
the coefficients x:, Y, and z are as follows':
I
-
I
V--
I-_j71-L
I---
I·
"`f
.~...I
KA
Fe
.4
SI
Ii-
)--~I-
-
T f;7
...
i. ..•
-t"-':
/
I,
;
_
'\
-1
\
AI
i,
ii
-
--.
,I
i
AF.. ....5I--.._...T
..
-1
g..
÷
i..
-4i,...
...
.....
..~L.!
.
...,
•.
.lz.
m
I
......
':
..
:..
F1.
I
1 - /9
-
_____
IL
43
VT
1
_
_
--I--
IT
A~
}F
~__
~_I_
--
"7
1ý"P5
j1Pz1B
1~
-
A
I-
Iti
n7T·
4
-
* -TI.-
~_
__
-
--
----
--
--
-----
--
---------------
xP 1
yR-
+ zR'
zR," =ZTh
-xR;"+yP,
xR 1
I
w yR' + zP, = W
and the moment expression is
M = xm
i
+ Ym2 + z•aI
If the loads are to be treated as moving load;, a unit load can
be plased as moving from one panel point to the other and the
three sets of coefficients can be solved from similar equations given above.
To find the reactions A and B at the supports the
f,ollowing equations -hold':
xA1.'- yA2 + zAs = A
xB, - yB, + zB , = B
and
A + B = ~I
+
W2 + Ta
If the truss and the loads are quite unsymmetrical
then there will be an imaginary horizontal force H acting at
the upper chords for each load diagram, fig.4:4. This force
is equal to the algebraic sum of the shears in the verticals,
or the unbalanced force in each case. We have now an additional deflection condition to consider, fig. 45. In figi 45
the value P4 is evidently equal to the algebrkic sum of F's
in fig. 44, and the values of Ri', R4 " and R" are equal to the
direct stress of the corresponding verticals, or the end
shear of the corresponding upper chords. The four coeffi-..
cients can be sdlved from the following four conditional
equations':
yR2 ' + zR '"+ vR 4 ' = WI
xP2
1
xR`"+ jP,- .zR
xR1M -yR"
-x1
+
vR" = VW
'I
+ P + vR
+ yR - 2+ zH3 + VP
=0
4
and the moment expression is
M= X2
1
+ 'i2
+z 3
s
+ v1i,
In case that there is accually a horizontally applied load H
acting on the upper chordi, then in solvingthe coefficients
t:he left hand expression should be put equal to H, the applied
load, instead of 0.
If the truss is connected with two posts in the form
of a bent as shown in fig. 46, then another deflection condi-.
tion,fig. 47 , has to be taken care of and the conditional
equations will be raised to five in number.
of determinants.
6. Abstract on theory
------------------From the above discussion it
is seen that for each
For
story of the frame there is an unknown coefficient.
buildings more than fo•r- or five stories high, the solution
It seems
of these coefficients may be felt handicapped.
to the writer that the use of the method of determinants
The following abstract on this method is
is preferable.
taken from "Hawkes Advanced Algebra." and will be found
helpful in most cases.
a) Definations':
Let us solve the equations
ax + bly =C
and
a 2 x + b2 Y
=
c
2
64 .
Multiply the first equation by b2 and the second by b1 , we
obtain
x + b b,2
ab
-ubtracting, we obtain
or if
=
azb 2 x + bzb 2 y
aib,
, a.b L
x
(a,b
2
b,c2
= bze
=
- a,.b.)x
b 2,c
-bc,.
OI
b 2 cz
azb 2
and
•
1
y=
ab.
ac,
c
a ,-c1
aazb,
a 2 bb
We note that the denominators of the expressions for
x and y are the same.
This denominator we will denote syn,
bolically by the following notatiod:
a z b 2 -a
2b
b
= a
a2
b2
The symbol in the right hand member is called a "Determinant".
Since there are two rows and two columns, this
determinant is
said to be of the "second order/ The left
hand member of the equation is
of the determinant..
called the " Development"
The symbols a ,ba,a
2
,.b
-
are called
" Elements "of the determinant, while the elements az and b2
are said to comprise its Principal Diagonal".
For a determinant of the third order, we may combine the terms
as fo.llows'-.
a1 bz c1
a2
b2
c2
= a1 (b cz-bd
•c-bc)
c) - a,(bcc-, bec)
+ a (b
a3 b9 cs
b2ý c
Sb, 3,
3
bb
a
'bs3
e2
c
bi ci
Cb
.
b
a bt c 2
65
We observe that the coefficient of a1 is the determinant ~that we obtain by erasing the row and column in which a1
lies;.
A similar fact holds fEor ýthe coefficients of a, and a3 .
The determinant obtained byýsing the row and column in which
a given element lies is called the ".'Minor" of that element.
0- c
Thus
.
is the minor of a,,
1
b3 c8
We notice that in the above
development by minors the sign of a given term is + or-
ac-
cording as the sum of the number of the row and the number of
th'e columnof the element in that term is.even or old.
Rules.:
1.
The development of a determinant of the nth
order is equal to the algebraic sum of the terms consisting of
letters following each other in the same order in which they
found in the principal diagonal but in which the subscripts
take on all possible permutations. A term has the positive
or the negative sign according as there is an even or an odd
number of inversions in the subscripts.
2. For the development of a determinant by minors
write in succession the elements of any row or column, each
muItiplied by its minor. Give each term a + or a, sign according as the sum of the number of the row and the number of the
If in a series of positive integers a greater integer precedes a less, there is said to be an inversion, Thus in the
series 1 43 2 there are three inversions.
006
column of the element in that term is even or odd.
Develope
the determinant in each term by a similar process until the
value of the development can be determined directly by nultiplication.
3.
The value of one of the variables in the
solution of n linear equations in n variables consists of
1.o1
r
ction
+ whose den minator is the determinant of +he s s
%
CL.
.A'
LIJ'
JIJLJ
Ud
.
I
LA
U-L
U.LL
A
U
b
tem and whose numerator is the same determinant, except that
the column which contains the coefficients of the given
variable is replaced by a column consisting of the constant
terms.
Principles':
1.
If every element of a row or a column is
multiplied by a number m, the determinant is multiplied by m.
2. The value of a determinant is not changed,
if the columns and rows are interchanged.
3.
If two columns or two rows are interchanged,
the sign of the determinant is changed-.
4. If a determinant has two rows (or two columns)
or any row (or column) being m times any other row (or column)
its value is zero.
5.
If each of the elements of any row or any
column consists of the sum of two numbers, the determinant may
be written as the sum of two determinantsi
6. The value of the determinant is unchanged if
the elements of any row (or column) are replaced by the ele-
I
ments of that row (or column) increased or diminis-hed by a
multiple of the elements of another row or column.
A numerical.example for illustrating the use of determinants-
for the solution of the coefficients
is given at the end of Part IV.
S..
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CON0CL
~ ITNS.
T
As the least work method has the special merit for its
exactness, the method herein described possesses the preference
for its simplicity.
The good features of this method can be sum-
marized as follows;
1. In applying this method it is not necessary to remember
any formula, as soon as the general principles involved are understood.
2. The computation can be done with single slide rule operations. There is no long expressianito be computed, this lessens
the
liability of making numerical errors.
3. As soon as the so-called characteristics have been comouted for a given framed structure, the moments can be found for any
kind of loading.
ease.
Influence table can also be prapared with graat
With the same amount of work it does not seem to be possible
to do that in applpin~ both of the least work and the ordinary slope
deflecti on method.
4. The moments of all members and at all points of a member
can be comnuted at one time.
This is illustrated in the example of
the three story-building.
5. Irregular structures and structures subjected to eccentric
load,-,as crane load, can be investigated without the least difficulty,
as the comnutation of characteristics and the seParation of moments
can be handled in usual manner.
6.- The amount of labor to be put in for investipgation
can
vary according to the degree of accuracy required, as the separation
of moments is quIte elastic.
It can be carried from a fairly ac-
curate result to one, which is almost exact.
7. The variation of cross section of a member can be
taken care of, as it only involves the additional labor to multiply the slope ang-le by a constant.
The writer wants to add a few words on the necessity of
a close investigation for buildings by this method. It is
a well-
known fact that for hiýgh buildings. on account of wind load and the
econony of the structure, a close investigation is always necessary.
For ordinary buildings up to four to five story high such investigation has never been attempted in ordinary office practice, because
the engineer a do not deem it
to do it
lustified to snend the labor and time
by means of exact methods Iknown to them.
this method appears,
it
Now as simnle as
seems to the writer that it does pay the
small amount of labor involved in this method,
for it
eliminates the
"'uesswork which is p'enerally madei in desiýninF. wall columns or
columns subjected to crane load for ordinary office endm_ mill buildin-5s.
BITI 0•
•R•EY.
Tulletins #lo7, #108 and #80, Experimental Station of University of
Illinois.
Morley, Strength of Materials.
V. RL-Utter, Die Anwendung der Graphische Statik.
Strassener, Forscher-Arbeittmg auf dem Gebiete des Eisenbetons.
Strassener, Die durchlaufende Bogentraeger.
Schweizerische Bauzeitung, 1917-1920.
Hool & Johnson, Concrete Engineers' Handbook.
Huette, Des Ingenieurs Taschenbuch.