advanced study of column analogy
DESCRIPTION
columnTRANSCRIPT
bO
ADVANCED STUDY OF THE COLUMN ANALOGY METHOD
by
DAVID K. C. CHENG
5. S., Taiwan Provincial Taipei Institute of
Technology, 1957
A MASTER'S REPORT
submitted in partial fulfillment of the
requirements for the degree
MASTER OF SCIENCE
Department of Civil Engineering
KANSAS STATE UNIVERSITYManhattan, Kansas
1964
11
sir'^V TABLE OF CONTENTS
SYNOPSIS ... iii
INTRODUCT ION 1
DERIVATION 2
Formulas for a Symmetrical Cross Section 2
Formulas for an Unsymmetrical Cross Section 8
SIGN CONVENTION 12
GENERAL PROCEDURE 13
APPLICATION OF THE METHOD 14
Loaded Gable Frame 14
Loaded Double Gable Frame 17
CONCLUSIONS 53
ACKNOWLEDGMENT 54
NOTATION'
55
REFERENCES 57
Ill
SYNOPSIS
The purpose of this report is to show a general derivation and
application of the method of column analogy. This method can be used
in finding the moments in third degree statically indeterminate struc-
tures. It is also very effective in calculating the stiffness and
carry-over factors for nonprismatic members for moment distribution.
The application section of this report shows how this method
can be applied to actual problems of finding moments and stiffness
and carry-over factors. A comparison of the results of the column
analogy method and the moment distribution method is presented, which
demonstrates the fact that these results are essentially identical
for the problem solved.
INTRODUCTION
The method of column analogy was first discovered in 1930, by
Professor Hardy Cross (l, 6). This method can be used in computing
the moments in rigid frames and the stiffness and carry-over factors
of the moment distribution method. Professor Cross stated that, "The
column analogy is a mathematical identity between the moments pro-
duced by continuity in a beam, bent or arch and the fiber stresses
in a short column eccentrically loaded." (l)
The method of column analogy is one of the most important methods
used in solving rigid frames or any structures which can be considered
to be closed rings. These include pipes, culverts, tanks, arches,
bents, and single span fixed ended beams.
In general, the method of column analogy can be applied to the
three types of structures shown in Fig. 1. (6)
V L_i— :
(a) Single SpanFixed EndedBeam
(b) Bent
W» '-, .:.
(c) Culvert
Fig. 1
^Numbers in parentheses refer to corresponding items in the
References.
The earth support is considered as a portion of the ring with
EI of infinite magnitude. These three types of structures are sug-
gestive of the many structures to which the method is applicable.
The column analogy method is used to compute the fiber stresses
in an analogous column loaded with the statically determinate moment
diagram from the reduced ring. The axis of the analogous column is
the same as that of the ring; the thickness of the analogous column
is equal to 1/EI of the ring. E and I can be either constant or vari-
able. The fiber stresses in the analogous column are numerically equal
to the moments required to restore continuity to the reduced ring struc-
ture. Thus by adding the determinate moment at a section to the fiber
stress in the analogous column at that section one obtains the total
moment in the actual structure for the section investigated.
DERIVATION
Formulas for a Symmetrical Cross Section
A ring structure as shown in Fig. 2(a) is being considered. The
ring is loaded with a set of forces P. Under the action of P, there
will exist shear, thrust (or tension) and moment at section A. F is
used to indicate the resultant of these shears, thrusts (or tension),
and moments on each side of section A.
After the ring is cut at A, the combined effect of the external
loads P, and the forces F, would keep it continuous at this section.
Assume that these two sets of forces, P and F, are acting separately.
(1) The loads ?1
, Y>2
, P„ , Pn _ 15
Pnwill cause
bending of the structure and produce statically determinate moment, Md ,
(a) Geometry of closedring
A—~7\
// EI
V.
(b) Portion of the analogouscolumn of the ring
Fig. 2. A portion of theanalogous column
(c) A portion of the analogouscolumn loaded with M^
diagram and reactionforce F
at any section throughout the ring. Owing to this moment there will
be relative rotation and relative displacement of both ends at the
section where it is cut. The rotation in a small length ds is
Md
* ds
dQ = Q»d EI
u;
In Fig. 2, since both A and A are very small, AA* and ABx y
are assumed to be mutually perpendicular, then
A ABC c-» AAA'D
^x d = y
( 2 )
AA« T§ KZ)
or
and,
Avd = X
AA" AB(3)
Solving equations (2) and (3) for Ad and A , yields
A , _ AA' (x) , .
and,
AB
Since 9 is very small, it can be assumed equal to its own
tangent, then
M. ' ds0= AA-ML then0d = f_J
AB J EIAB
Substitute into equations (4) and (5), then
^yd=/Md(ds)x ( 6 )J EI
and,
A = I u{ ds )y•
( 7 )
xd J dEI
(2) As a result of the F forces, which are acting on both sides
of the section at A, after the ring is cut, there is a statically
indeterminate moment Mi
at every section throughout the whole ring.
By the same reasoning as found in section (l), the rotation and de-
formation caused by the indeterminate moment WL, are
Q. = I M. ( ds )J EI
(8)
and,
yi = !m. (ds ) ..O)
J XEI
xi = (m. ( ds )(10)
J EI
Since the forces P and F are assumed to keep the ring continuous,
the total rotation and total displacement due to both determinate
moment, NL, and indeterminate moment, M._, must be equal to zero. Hence,
G , + 0. =0 (n)d l
A . +A • = (12)yd *-*yi
A . +A . = (I 3 )" xd xi
or substituting equations (l), (6), (7), (8), (9), and (10) into
equations (ll), (12), and (13) respectively.
(m (ds) + (m. (ds) =0 (14)* EI J EI
JM, (ds)x + [m. (ds)x = (15)J EI J
EI
[m (ds)y + (m. (ds)y = (16)J EI J EI
The above equations are called the equations of moments in a
closed ring.
(3) Take a small element ds at some distances x and y along
the X and Y axes respectively from the centroid of the analogous column.
M, is the load acting on the top of the analogous column and F is the
reaction force acting at the bottom of the column, as shown in Fig. 2(c).
ds/EI is the area of the portion of the column. Considering the con-
ditions of equilibrium, the following equations are obtained.
IV =
or,
and,
or.
M ( ds ) + F( ds ) = (17)EI EI
Zm = oXX
M ( ds )y + F( ds)y =0 (18)d
EI EI
and,
or.
IM =* yy
M, ( ds )x + F( ds )x = (19)EI EI
Integrating to include the entire column, equations (17),
(18), and (19) can be written respectively as
/m ( ds ) + (f( ds ) = (20)J d
EI J EI
Jm ( ds )y + ( F( ds )y =J d EI ^ EI (21)
fM ( ds )x +JF( ds )x = (22)
J dEI 'EI
In computing the fiber stresses F of a column, either con-
centrically loaded or eccentrically loaded, the following formula is
used.
M • y M * x
F =£ + -^ + -*A I I
* y
In comparing equations (14), (15), and (16), and equations (20),
(21), and (22), Mj_ is equal to F. The final moment in the ring is
equal to M^ - M. , or M - F. This is the basic principle of the column
analogy method. (6)
Formulas for an Unsymmetrical Cross Section
The formulas for an unsymmetrical cross section have been de-
rived by Chu-Kia Wang as follows (8). Assume a column as shown in
Fig. 3a, which is subjected to a set of downward loads P^ P^ P3
>
acting on the points (Xp Y^, (X2 , Y
2) , and (Xg, Yg), etc.
These loads can be transferred to the centroid of the cross section of
the column as indicated in Fig. 3b, as a downward concentric load and
one bending moment about each axis.
(x2_iVol
(x3 ,y3 )
l(x1,y
1)x
(a) A column loaded with
eccentric loads
(b) The transformation
of load
Fig. 3. Eccentrically loaded column
and its transformation.
According to the principles of statics, after the transfer, the
new load P, and the new moments, Mx
and M , may be written as
(24a)P = P
x+ P
2+ P
3
\ =Vl + P
2y2+ P
3Y3 < 24b >
\ = P1X1+ P
2X2+ P
3X3
(24C >
The pressure F at any point can be expressed as
F = a + bx + cy (25)
where the constants a, b, and c can be determined from the three
equations of statics, as follows:
P =] FdA (26a)'
=j
F(dA)(y) (26b)J
=J
F(dA)(x) (26c)y
Substituting equation (25) into equation (26a), then
A
P =
'o
=j F(dA)J
rA
=J
(a + bx + cy)dA
/A /A /A=
j dA + (x)dA +1 (y)dA
Since
fJ dA = A
10
('x)dA =J
fA
J(y)dA =
Hence.
-J*dA
= aA (27)
Substituting equation (25) into equation (26b), then
M =JF(dA)y
'0
r
r
=1 (a + bx + cy)(dA)y
_(A2
-J (ay + bxy + cy )dA
Ik ik(k
2= a y(dA) + bj (xy)dA + c
J (y )dA
= + b(l ) + c(l )xy x
b(l ) + c(l ) (28)xy x
Substituting equation (25) into equation (26c), then
=JF(dA)x
y -o
rk- l (a + bx + cy)(dA)xJ
11
-\M = \ (ax + bx + cxy)dAy
(A /A
9/A
= aJ
(x)dA + bj (x )dA + cj (xy)dA
= + b(l ) + c(l )v
y xy
- b(l ) + c(l )
Y xy(29)
Solving equations (27), (28), and (29) for the constants a, b, and c.
_ Pa "A
M.. -xy
b = Y X Ix.
r-2-xy
Iv(1
- T~rr '
x y
(Mx
- M^
xy
2
xy
x y
Let
xyM' = M - M (
y)
x x y jy
i
M« = M _ M (
xy)
y y x i
I = I (1x x^ I ' I
X y
12
Txv2
I =1 (1 --^ )
y y i • i
then
A
My
I
y
Mx
c =,
I
Substitute the values of a, b, and c into equation (25),
theni
'
FP
+ ^_ (x)+ -i— M (30)
SIGN CONVENTION
The sign convention of this report is that used by Chu-Kia
Wang. (7) This convention is known by most engineers as the regular
beam sign convention.
(1) Loading on the top of column is downward if M, (statical
moment, or moment due to the applied loading on the simple beam AB as
determined by the law of statics) is positive, which means that it
causes compression on the outside. The definition of the terms inside
and outside is as indicated in Fig. 4.
13
outside
| 1
^ inside p
outside
Fig. 4. Simple Ring Type Structure
(2) Upward pressure on the bottom of column, M. (indeterminate
moment, or moment to be determined to satisfy the conditions of
geometry), is positive.
(3) The moment at any point in the structure is equal to
M = M . - M. , which is positive if it causes compression on the outside.
(4) The sign of the moment-arms, x and y, is taken from the
coordinate system. That is, positive up and to the right from the axis
of the coordinates.
GENERAL PROCEDURE
There are five steps in finding moments by the method of the
column analogy.
(1) Make the ring (or any structure which can be considered
as a ring) statically determinate.
(2) Compute the statically determinate moment Md
, which is
produced by the external loads P, according to the laws of statics.
(3) Apply the M, diagram which is obtained in the second step,
as a load, on the top of the analogous column.
14
(4) Compute the fiber stresses in the analogous column by the
laws of statics.
(5) The final moment in the ring (or any structure which can
be considered as a ring) is
M = M , - M.d 1
or,
M = Md
- F
In a manner similar to that for finding moments, the stiffness
and carry-over factors for members with either constant or variable
cross section can also be found by using the method of column analogy.
The details of finding these factors are illustrated in the section,
Applications of the Method.
APPLICATION OF THE METHOD
Loaded Gable Frame
Assume a gable frame with one axis of symmetry and loaded with
a 15 Kip external concentrated load as indicated in Fig. 5a. Assume
that ends A and E of the frame are fixed. The dimensions of each member
of the frame are given in Fig. 5a.
•The properties of the analogous column of Fig. 5b are
(assume EI = l):
A = 2(l/2 x 36)*+ 2(30) = 96
Y = 36(10) + 60(35) = 2^^,96
15
I =2 -J2 (1)(30)3+ 30(1)(9.4)
2+
1 / / w v2,20 N ,o
2-75 (l/2)(36) (35) + 18(15. 6T = 19.78
I = 2(30)(30)2+ 2 -i5(l/2)(36) 2
(g) + 18(15)'
= 54,000 + 2(1350 + 4050)
= 64,800.00
(a) Loaded Gable Frame (b)- Analogous-columnsection(considering EI = 1)
Fig. 5
Cut the right support, E, of the frame in order to make the
whole frame a determinate structure as shown in Fig. 6a. The de-
terminate moment diagram, which is produced by the external load, is
treated as a load acting on the top of the analogous column as shown
in Fig. 6b.
The total load on the top of the analogous column, and the
moments about the axes of the cross section of the analogous column
are calculated to be:
16
15
225
15
(300)
(3001
(a) Basic determinatestructure
(b) Analogous column withM. diagram as a load
Fig. 6
17
P * 1/2(18)(150) + 30(300)
= 10350.00
M = 30(300)(9.4) - 1/2(18)(150)(8.9)
= 84,600 - 12,000
= 72,600.00 (clockwise)
My = 30(300)(30) + 1/2(18)(150)(25)
= 270,000 + 33,750
= 303,750.00 (clockwise)
The calculations and the results are shown in Table 1.
Loaded Double Gable Frame
The following double bay gable frame, shown in Fig. 7, is
analyzed by both the column analogy method and the moment distribution
method
.
30.0'
20.0*
1.0 K/foot
45.0' 20.0'_ 45.0'
Fig. 7. Double bay gable frame
<xu,
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Solution by the Column Analogy Method
I. Properties of the analogous column section . Take a portion
ABCD of the frame as a free body with its two ends fixed as shown in
Fig. 8(a). Solve this part of the frame as an independent structure by
the method of column analogy. The length of the analogous column
section is the same as that of the actual frame. The width of the
analogous column section is a constant l/EI, as shown in Fig. 8(b).
It will be convenient to let EI = 1, so that the width at every point
of the analogous column is one.
Area of column section = (30)(l.O) + (25)(l.O) + (47.5)(l.O)
= 102.5
- _ (30)(15) + ' 25)(37.5) + (47.5)(37.5)
102-5
= 3168 - 5 = 31 feet102.5
x = (25) (10) + (47.5)(42.5)
102.5
= 2268.7 = 22.1 feet (from left)102.5
3
Ix= -J^ 30^ (30)(16)
2+ -^(25)
3(§)
2+ (25}(6.5)
+ -j^ 47 ' 5 )
3(~^)
2+ (47.5)(6.5)
2
= (2250 + 7680) + (469 + 1056) + (890 + 2006)
= 14,351.0
20
1 3 9I = 12(30)(1.0) + (305(22.33)
+ T2(25) Hlr> + (25)(12.33)2
+ -^(47.5)3
( ~~) 2+ (47.5)(20.4/
= 46,932.6
Ixy
= (30)(-22.l)(-16.0) + (25)(-12.l)[6.5)
+ (47.5)(20.4)(6.5)
= 10,608.0 - 1966.3 + 6298.5
= 14,940.2
I2
i' =i (i - *y)
I Ix y
= 14351.0(1 -
( 14351. 0)(46932.6)
= 14351.0(1 - 0.333)
= 14351.0(0.667)
= 9572.1
I2
I .= 1 1 - *Y)
y y I • IJ
x y
= 46932.6(0.667)
= 31304.0
21
c
^^"^\ D
fB
' ~ " ^31.0
22.1s
1 A
(a ) Free-body (b) Analogous columnsection
Fig. 8
(a) Load on theanalogous column
1.0 (radian)
(b) Load after transfer tothe centroid of theanalogous column section
Fig. 9
22
II. Stiffness and carry-over factors . A load of 1 radian is
applied at the right end, D, of the analogous column section and trans-
ferred to the centroid of the section as a concentrated load and two
moments as indicated in Fig. 9(a) and (b), from which the following
relationships are obtained.
M = (l.0)(-1.0) = -1.0
My= (1.0)(42.9) = 42.9
M ° = M - M ( —22
—
)x x y I
;
y
' - -U.O) - (42.9X |||^ ,
= -16.9
M'
= M - M (XY
)
Y y x ix
. 42.9 - U.0)(^g)
= 42.9 + 1.C4
= 44.0
The joint moments are tabulated in Table 2.
The stiffness factor at D and the different carry-over factors
are obtained as follows:
The stiffness factor at D is SD 0.072.
The carry-over factors are:
(1) From D to C.
0.017°DC ~ 0.072 °* 236
5
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24
(2) From D to B.
r - 0-020 _ n 97oCDB " 0.072 °* 278
(3) From D to A.
0.038 „ ___°DA 0.072
III. Moments at each joint due to the applied loading . Cut the
left end A of the free-body ABCD, as a redundant and make the whole
free-body a determinate structure, as indicated in Fig. 10(a). The
determinate -moment diagram of the free-body is plotted on the com-
pression side as shown in Fig. 10(b). Use this diagram as a load
acting on the top of the analogous column.
P = —(25.0) (200.0) + (47.5)(200.0) - f- (45.0)(252.13)
+ _i(47.5)(2112.5) - 200.0)2
= 49051.0
M = (1666.67)(7.1) - (9500.0 - 7563.75)(20.4)
- (45448. 0)(45 - 15 - 2.1)
= -154262.5 (counterclockwise)
Mx= -(16666. 67)(10.25) - (9500.0 - 7563.75)(6.5)
- (45448. 0)(4.0)
= -211461.0 (counterclockwise)
25
M • = M . M ( -£L-
)
x x y Iy
,~ -w 14940.0 n
= -211461.0 - (-1542o2.5)(-46932#6'-)
M ' = -162094.0
M - M (
y x I
xy
14940.0 .
-154262.5 - (-211461. 0)t 14351 . )
= 65657.0
1.0 K/foot
I(l)(20-+45)'2
= 2112.5
(a) Loaded determinate frame (b) M. diagram, plotted on the
compression side of the frame
Fig. 10
26
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27
the
frame. as shown in Pig. ll(a).
ture by the column
:n section is the
same i i- ~ s c ° :umn
section is a constant It will be con-
ic let EI = ] rY section of "ens analo-
column section is 1
ion = (3( + ..'; .5] .1.0]
>.5
_LiL_.~ 23 :om rig/.
102.5
, _ :. 1)
: -:• (25)(37.5) * (47.5)(37-5;.
102.5
= 3 (from the
_ ] ;3o.o)(i<
s)3
( t~—
;
2- -s)
2
1/ H I . S>
+ _i_(25.0)3(4|-)
2+ (25.0}(6.5)
2
12 25
= 143.
28
Y
23.84'
p
^̂"~*""-^
X
31.0'
G
H
(a) Free-body (b) Analogous columnsection
Fig. 11
1.0 (radian)
//TV^:V,
(a) Load on the analogouscolumn (b) Load after transfer to
the centroid of the anal-
ogous column section
Fig. 12
29
I =-Y2(30.0)(1.0)
3+ (30.0)(23.84)
2
+ -j^47.5)3
{ ^775")2+ (47.5)(1.34)
2
+ ^-(25.0)3(-||)
2+ (25.0)(31.16)
2
50274.4
I = (25)(1.0)(-31.16)(6.5) + (47.5)(1.0)(1.34)(6.5)xy
+ (30)(1.0)(23.84)(-16.0)
= -16140.6»
2
I •= I (1 - ^X X j
jx y
)
= 14351.0 1 -
•
(16140.6)2
J.*"TO %J J. • \J JL
(14351.0)(50274.4)
= 9170.3
I*=I (l - ^)
y. y iix y
= 50274.4 1 -(16140. 6)
2
(14351.0) (50274.4)
= 32125.3
By the same technique and procedure which have been used previously,
a load of 1 radian is applied at the left end, D, of the analogous column
section and transferred to the centroid of the section as a concentrated
30
load and two moments as denoted in Fig. 12(a) and (b). The moments
at each point and stiffness factor at D, and carry-over factors from
D to different points are obtained as follow:
Mv = (1.0)(-1.0) = -1.0
M = (1.0X-41.17) = -41.17
IxvM = M - M (
—ii
—
)x x y i
'
= -(1.0) - (41.17) - (I6140 ' 6 >
50274.4
= -14.20
I
m = m - m (xy
)
y y x iX
-<- l7> - <-'-<»«$
= -42.30
The stiffness factor at D, Sq = -0.065
The carry-over factors are:
(1) From D to F.
(2) From D to G.
(3) From D to H.
rr= 0-0202 = 0.311
DG 0.0650
= 0-0263 =CDH 0.0650 °* 405
gt-H
HO<Q<O-1
<i—
1
oHai
QXaa.Q
&-
O<->
Xo<
§
X
ai
CQ<
X
Sin-.
31
CM COin in O vOvO r-i CM CM
O O O• • • •
i
O1
O O1
^-> ^—s *-^ OO •
• • • r-i
1 r-l
r—
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I
CO1
- - • *^_^ • r- •v_^ O ^-^ O*** O \0 y***» t-H ^"*s • <""* •
r~ <-< r~ CM O f—1 O O COCNJ .-1 CM •—
1
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T • V O T O • <tf .
r—t O •—
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cm ^r O —
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CM O CO O • CO • • CM •
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1
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r-vO
in t- r- r»vO v£> \£>
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• • •
11 O O
O O O
a UU O X
32
IV. Moments at each joint of the free-body DFGH, due to the
applied load . Cut the right end H of the free-body DFGH, and make
the whole free-body a determinate structure, as indicated in Fig. 13(a).
The determinate moment diagram of the free-body is plotted on the com-
pression side as shown in Fig. 13(b). Use this diagram as a load
acting on the top of the analogous column.
P =-|-(45.0)( 1012.5) + (1012.5)(25) - ~(20)(50)
+ -~(25.0)(2112.5 - 1012.5)
= 53583.33 (upward)
Mx= -(15187.5)(11.5) - (25312.5 - 666.67)(6.5) - (13750)(4.0)
= -389845.0 (counterclockwise)
M = (15187. 5)(9. 92) + (25312.5 - 666.67)(31.16)
+ (13750.0)(34.5)
= 1,257,404.0 (clockwise)
M ' = M - M ( —^1—
)
x x y IJ
y
»x . , -16140.6 .
Mx
= -(389854.0) - ( 1257404. 0)( 50274.4)
- -27185.6
33
M '• = M - M (xy
)y y x i
X
= 1257404.0 - (-389854. 0)(-I 6140 - 6
)
14351.0'
- 818818.2
1.0 K/foot
3
H
|(1.0)(45)'
i(l.0)(65)
= 2112.5
(a) Loaded frame
Fig. 13
(b) M, diagram, plotted on
the compression side
The calculations and the resulting moments at each joint of the
free-body due to the applied load are shown in Table 5.
CO
SQLUI—
l
-1aa.<ai•x(-
OHWQIPQit,
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o>-)
Xo<ai
H
ai
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I
I
in
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-1-
OCM
COCOin
vO
if)
CO
t-c\
I
o
ID
CM
CMI
nOON
CM
II
if) ,
CM oi—
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inCM oCO «—
t
COCO•
r-coo
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vD
aII -_4
C»J
If) COcm CO^<f\ ONCO CO
ini
vO vOr- r-
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11
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34
o•
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35
V. Moments at each joint due to sidesway at point D .
Fig. 14. Sidesway at D
vJxy^f- + H/y2(-ff-)
+ M Jy
ds _ ,.
ds
EI= AH
,, i ds , .. ( ds , ,, / dsvlx— + H
Jy— + mJ—
Since Jxda = 0, jyda = 0, then
vjx 2
-ff-+ Hjxy
=
ds
EI=
vJxy ^f
. + Hjy2 -|!
_ .Ah
36
or.
V(V + H(lxy
}= °
V(I ) + H(I ) =AHxy' x'
Solve for V and H.
-AH I
v = -rr-( -f*->X
AHH = 7
\
M = M . - M.d i
= Md
- (Vx + Hy) (G)
The calculations and the resulting moments at each joint of
the free-body due to sidesway at D, are indicated in Table 6.
By the same technique which has been used above, when sidesway
exists at the joint D of the right side free-body DFGH, the following
relationships and the moments at each joint of the free-body due to
this sidesway can be obtained. (7)
VI + HI =0y xy
VI + HI = -(AH)xy x
Solve for V and H from these two equations above, then
S
>
X X< 1—
1
II
^.Q >b~ X<><»COai • • •• •Qi-<
CO
oH04 <—
-
3 XQ ^-^
<—-.
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< I—
1
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1 1—
1
h-
1
11o•-J
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u ><ai
H<
£ai2gosUJ •ua: asH .
1
•
vO
LU.-1
pa<H
•pc•Hot->
37
CMin IT)
COin
I
o
I
o of—
i
i
CM 00r» cmin •-io i
in
CM
o•sr •
X o o<
1
COr—
(
CO
1—
1
1
11
CM1
>—
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•
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r-i-«*^•sTo o
X CO •
i
f—
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CMCM
I
CQ
COI
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II
CMr-na-
CMCM00
I
—' II
CMCM
I
00 oo
38
H , xyV T" < i )
y x
h =-AHI
'
x
M = M ,- M.
d i
Fig. 15. Sidesway at D
= M - (Vx + H'y)
The calculations and the resulting moments at each joint of the
free-body due to sidesway at D, are obtained as shown in Table 7.
The fixed-end moment of bar DE, due to sidesway at D, is ob-
tained as follows:
Fig. 16. Fixed-end moment of bar DE, due
to sidesway at joint D
39
PL!
UJX
I
U!-I
COCO
r-CM
CMCO
I
iT)
CO
COCOCO
I
CO
•
srCOCO
I
Q
5CO
'x
o CO• 00
C J •
r* CO(H ino CM
•^ 11
oTrH
^.—•*"- \Do •
X • CM
<;o in
i >-\ +o11
"N COo 00
•
o COr- lOt-t CMo +
COI
o
OCM
^ 11
00
oHUJ
QXoIX,
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O*->
g<UJ
X
X<
>
1
if)
COT.—
1
vO1-4
lD •
CM o«—
(
CMCM iT)
CO 1
CMi
\D o
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*Q co
CMt-
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COCM
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<#—
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c
oO
40
MDE= M
ED" 6EI (^H)
302
= 207.27
Distributions of fixed-end moment at joint D, due to the applied
load and sidesway at D are indicated in Tables 8 and 9. The ratio of
shear stresses, due to the applied load and the sidesway at D, is
—(1113.14 - 531.48 + 547.17 + 273.59 + 1216.9 - 1330.08)k = - _3pJ
-|q(338.57 - 149.27 + 197.37 + 50.74 + 336.45 - 396.40)
= - 3.51•v
M = Moment, due to the applied load + (-3.51) (Moment, due to side-
sway at D)
M^ 1113.14 - (3.51)(338.57) = -7.50
MBA= -MBC
= -531.48 - (3.51)(-149.28) = -7.47
Mqb = -MqD = -507.06 - (3.51)(-107.1) = -131.08
MDC= -2401.74 - (3.51)(-649.61) = -121.56
MDE = 547.17 - (3.51)(107.37) = 170.00
MdF = 1854.57 - (3.51)(542.27) = -48.8
MFD
= MFG
= - 1102 - 33 " (3.51)(-382.12) = -238.16
%F = %H = - 1050 ' 51 " (3.5l)(-289.82) = -33.20
MHG= -1330.08 - (3.5l)(-396.4) = -61.32
MED= 273.59 - (3.51)(50.74) = 95.49
41
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42
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43
Solution by the Moment Distribution Method
The distribution of fixed-end moment due to the applied load is
indicated in Table 13. The fixed-end moment at each joint due to side-
sway at B, D, and G respectively are obtained as follows:
1.155( A B)
Fig. 18. Sidesway at B
TABLE 10.- FIXED-END MOMENT DUE TO SIDESWAY AT B
MAB - MBA
MbC " MCB
MCD ~ MDC
.*>;'
6EI( A B)
30^
6EI( 1.1555 AB)
252
6EI(0.973 Ab,2
66. 67
-110.88
26.0
(47.5)'
44
0.973( A D) 0.973( A D)
Fig. 19. Sidesway at D
TABLE 11.- FIXED-END MOMENT DUE TO SIDESWAY AT D
Ar) _ i*0*i
"bc*' CB6EI(1.155 AD)
252
-110.88
MCD
= MDC6EI(0.973 Ap)
(47. 5)2
26.0
^E= MED
6EI( A D)
30
-66.67
MDF= MFD
6EI(1.155 Ap)
252
110.88
Mnn =FG
,V1
GF6EI(0.973 Ap)
(47.5)2
-26.0
45
0.973(-4 G)
3?W
Fig. 20. Sidesway at G
TABLE 12.- FIXED-END MOMENT DUE TO SIDESWAY AT G
MDF - MFD- 6EI( 1.155 &G)
252
*h*<"110.88
FG*" ^GF
6EI(0.973 AG)
(47.5)2
26.0
MGH " M
HG
6EI(AG)
30z
-66.67
The distributions of the fixed-end moments due to the sidesway
at point B, D, and G respectively are indicated in Tables 14, 15 and 16.
46
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II
50
The shear condition, caused by the applied load.
9.70
^"~~N0.478 84 * 24 /£T^\4.21 1 26 . 63 /tf~~N 6 . 33c— c— c-
4.65 \ _y 41.92
The shear condition caused by sidesway at B.
74.22 4.82 -15.34 0.767
0.767
70.38 V / -7.68
The shear condition caused by sidesway at D.
40.48
/B 2.02
-86.26
^~
20.08
/D ^ 5.43
-76.51
The shear condition caused by sidesway at G.
1.71 33.00
/p \l.65
0.87 16.56
63.22
-0.92 /.—^ 0.046
-0.45
24.28
0.046
-48.00
/G ^1.55
1.57
51
0.478 + 4.82k1
- 0.767k2
- 0.046k3=
4.210 + 2.02k! - 5.430k2 + 1.210kg =
6.330 + 0.086k! + 1.65k2- 1.550kg =
Solve for k, , k , and k_, from these three equations.
kx= -0.11
k2= -0.30
kg = 1.80
MAR = 4.64 - (0.113(70.38) - (0.30)(20.08) + (l.8)(0.87) = -7.54"AB
MgA= -MgC
= 9.70 - (0.11)(74.42) - (0.30)(40.48) + (l.80)(l.7l)
= -7.51
MCB
= "MCD= - 137 ' 34 " (0.1l)(-50.48) - (0.30)(-42.86)
+ (l.80)(-6.76) = -131.13
Mj^ = -145.02 - (0.11)(30.9) - (0.30)(25.22) + (l.80)( 18.56)
= -121.59
M^ = 84.24 - (0.11)(-15.34) - (0.30)( -86. 26) + (l.80)(33.00)
= 170.21
Mnr,= 41.92 - (0.1l)(-7.68) - (0.30)(76.5l) + (l.80)(l6.56) = 95.49
ED-
MDF= 60.68 - (0.11)(-15.52) - (0.30)(61.02) + (l.80)(-51.56)
= 48.8
52
o
"ocHR) »\
+-> TDja C •
-^ CJ
+> —^ o •He e *i
(0
h c C(Jl o cIS .1-1 •H.— -J t'j
-o 3 01
XI o+J •»-i Mc h aQ) +J ee U3 oc •H o£ TJ
aen +J x:c C *J•H CJ
+J £ ciH o o3 1O -aO o CD
N x: -P-t-> +J
o ox: >.-HH Xi a
Dl
53
MFD
= MFG
= 46,35 " (°- n )(-52 * 08 ) ~ (0.30)(46.35)
(l.80)(-52.04) = -238.9
%F ="%H
= - 126 ' 72 ~ (0.11)(0.92) - (0.30)(-24.28)
(1.80)(48.00) = -33.25
MHG= 63.22 - (0.1l)(-0.45) - (0.30)(l2.15)
(l.80)(0.87) = 61.19
CONCLUSIONS
It appears that the method of the column analogy can be applied
in finding moments and stiffness and carry-over factors for statically
indeterminate structures with either a symmetrical or unsymmetrical
cross section, in a direct manner.
The comparison vof the results of the column analogy method and
the moment distribution method reveals that the results, which are
obtained by these two methods, are essentially identical, as shown in
Fig. 17 and Fig. 21. The column analogy method took much less time
than the moment -distribution method, when they were both used to
solve the double bay gable frame in this report.
54
ACKNOWLEDGMENT
The author wishes to express his deep appreciation to his
major professor, Dr. Robert R. Snell, for his advice, guidance,
suggestions, and encouragement during the preparation of this re-
port.
55
NOTATION
x, y = Coordinates of any point on the cross-section along any two
mutually perpendicular axes X and Y through the centroid of
the section,
f = Intensity of normal stress at point x, y.
A = Area of section.
I =J y dA i— Moment of inertia about X axis (along the Y axis).
( 2Iy -
J x dA Moment of inertia about Y axis.
I = IxydA Product of inertia about axes X, and Y
P = Normal component of external forces.
*
M = Moment of external forces about X axis,x
M - Moment of external forces about Y axis.
K \ - V-if"'
X
,11=1-1 (xy
)
X
I'M - I L-2L.)y y xf i
J
y
= A relative rotation of the two ends of the cut.
A = Relative horizontal displacement of the two ends of the cut.x
y = Relative vertical displacement of the two ends of the cut.
56
M - Determinate moment throughout the entire ring due to the loads
P.., P , .... after the ring is cut.
M - Indeterminate moment due to the forces F acting on the ends
of the cut.
57
REFERENCES
1. Cross, HardyThe Column Analogy. University of Illinois EngineeringExperiment Station Bulletin 215, 1930.
2. Cross, HardyContinuous Frames of Reinforced Concrete. New York:Wiley, 1932.
3. Fif, Walter Maxwell and Wilbur, John BensonTheory of Statically Indeterminate Structures. New York:Edward Arnold Ltd., 1937.
4. Micholas, James P.
Theory of Structural Analysis and Design. New York:The Ronald Press Co., 1958.
5. Shedd, Thomas Clark and Vawter, JamesonTheory of Simple Structures. New York: John Wiley and Sons.
1949.
6. Sutherland, Hale and Bowman, Harry Lake
Structural Theory. New York: Wiley, 1950.
7. Wang, Chu-KiaStatically Indeterminate Structures. New York:
Toronto, and London: McGraw-Hill Book Co., 1953.
8. Williams, Clifford D.
Analysis of Statically Indeterminate Structures. 4th Ed.
Scranton, Pennsylvania: International Textbook Co., 1959.
ADVANCED STUDY OF THE COLUMN ANALOGY METHOD
by
DAVID K. C. CHENG
B. S. , Taiwan Provincial Taipei Institute ofTechnology, 1957
AN ABSTRACT OF A MASTER'S REPORT
submitted in partial fulfillment of the
requirements for the degree
MASTER OF SCIENCE
Department of Civil Engineering
KANSAS STATE UNIVERSITYManhattan, Kansas
1964
The purpose of this report is to show the general derivation
and application of the method of column analogy. This method is very
useful in finding the moments in third degree statically indeterminate
structures.
The formulas of this method are derived from two sets of equations.
The first set is found by setting the rotation and displacements caused
by the determinate moments equal and opposite to those caused by the in-
determinate moments. In other words, by enforcing the continuity re-
quirements of the structure.
6d= -e
i.
A j = -A .
yd yi
A = -A .
xd xi
or, it can be written as:
jMd _!_( x) +
jMi _|_(x) = o
JMd -ff-(y)J «i -ff-(y)
=
The second set of equations is obtained from the equations of
static equilibrium for an imaginary column loaded with the Mj diagram
of the ring. The analogous column has the same axis and the same length
as the ring and has a thickness of l/EI.
SV =
(u ds ( ds _ .
JMd.-Er
+jMi-Er -°
ZMy
= 0.
jMd-|f-(x) + JM.-|f-(x) =
ZM =x
J^ffrM /vi-M - °
In comparing these two sets of equations we see that if M. = F,
they are identical. Therefore, we can determine the indeterminate
moments in the ring by calculating the fiber stresses in the analogous
column. The formula for computing the fiber stresses in a column,
either concentrically loaded or eccentrically loaded, is:
p Mx (y)
M (x)
A Xx
xy
In the application section of this report is shown the general
use of the column analogy method in finding moments in a frame, fixed-end
moments and stiffness and carry-over factors for members of a frame to be
used in the moment distribution method.
The comparison of the results of the column analogy method and
the moment distribution method reveals that the results, which are ob-
tained by these two methods, are essentially identical. The column
analogy method took much less time than the moment distribution method
when they were both used to solve the double bay gable frame in this
report.