THE COLUMN ANALOGY
I. INTRODUCTION
1. Purpose of the Monograph.The object of this bulletin is to
present some theorems dealing with the elastic analysis of continuous
frames. In ordinary cases these theorems are identical in form with
the theorems, with which every structural engineer is familiar, for
finding internal stresses in beams and struts.
The subject of structural mechanics is now experiencing demands
for greater precision by the very accurate analyses demanded in the
design of airplanes. In any case it is of the greatest importance to
isolate definitely those matters which are sources of uncertainty from
those which are certain and hence not proper fields of experiment.
Problems dealing with the analysis of restrained flexural mem
bersstraight beams, bents, archesoccupy a large space in struc
tural literature. The treatment often presented involves complicated
equations; in nearly all cases the method of solution is hard to
remember.
If the elastic properties of the different portions of the structure
are definitely known, the analysis of restrained members is essentially
a problem in geometry, because the member must bend in such a way
as to satisfy the conditions of restraint. The geometrical relations
involved are identical in algebraic form with the general formula for
determining fiber stress in a member which is bent.
Since the analysis of problems in flexure is a familiar procedure to
structural engineers, it is advantageous to state the relations involved
in the analysis of fixedend beams, bents, and arches in terms of the
beam formula. The advantages for structural engineers are similar
to those which result from using the theorems of areamoments in
finding slopes and deflections of beams; in some respects the concepts
involved are identical, and the use of the beam formula in the analysis
of restrained members may be thought of as an extension of the prin
ciples of areamoments. The general conception referred to in this
monograph as the "column analogy" includes the principles of area
moments and also the conception of the conjugate beam.*
In this bulletin it is shown that bending moments in arches,
haunched beams, and framed bents may be computed by a procedure
*See H. M. Westergaard, "Deflection of Beams by the Conjugate Beam Method," Journal of
Western Society of Engineers, December, 1921.
ILLINOIS ENGINEERING EXPERIMENT STATION
analogous to the computation of fiber stresses in short columns subject
to bending, and that slopes and deflections in these structures may
be computed as shears and bending moments, respectively, on
longitudinal sections through such columns.
The theorem makes available for the analysis of plane elastic
structures the literature of beam analysis, dealing with the kern, the
circle of inertia, the ellipse of inertia, graphical computations of
moments and products of inertia, and conjugate axes of inertia.
Certain terms are .defined in such a way that the method is ex
tended to include the effect of deformations due to longitudinal stress
and to shear in ribbed members, and to include trussed members.
The conceptions used in arch analysis by these methods make
possible a general statement of the relations of joint displacements to
joint forces, of which the familiar equation of slopedeflection* is a
special case, and hence make possible the convenient extension of
the method of slopedeflection, or of the theorem of three moments, to
include curved members and members of varying moment of inertia.
The method here presented has application in the fields both of
design and of research. In the field of design we use certain physical
properties of the materials, which are necessarily assumed. In re
search we may either resort to the laboratory and study by empirical
methods the properties of the structure as a whole, or we may study
only the physical properties of the materials themselves, and depend
on the geometrical relations to determine the properties of the struc
ture. It seems obvious that the geometrical relations are not them
selves a proper subject for experimental research.
The relations pointed out in this bulletin have at first been care
fully restricted to geometry, and the assumptions which are necessary
to apply this geometry to the design of structures are developed later
in the discussion.
2. Validity of Analyses by the Theory of Elasticity.The mathe
matical identity of the expressions for moment in an elastic ring and
for fiber stress in a column section has some value in considering in a
qualitative way the general validity of analyses based on the elastic
theory.
It appears at times that engineers are not altogether discriminat
ing in considering the value of elastic analyses, and seem to hold that
one must either accept as precise the results of such analyses or reject
entirely their conclusions.
*"Analysis of Statically Indeterminate Structures by the SlopeDeflection Method," Univ. of
Ill. Eng. Exp. Sta. Bul. 108, 1918.
THE COLUMN ANALOGY
Now no one but a novice accepts without discrimination the re
sults of the beam formula. It is open to many important objections,
such as lack of homogeneity of the material, effect of initial deforma
tions, and other defects; and yet it is difficult to conceive of modern
structural design existing without the beam formula, nor is anyone
seriously disturbed because lack of homogeneity modifies somewhat
the properties of the section, or by the fact that imperfect elasticity in
the material makes invalid the superposition of stresses determined by
the beam formula for different conditions of loading. Moreover the
beam formula becomes a most inaccurate guide to the maximum
stress in any section near the point of failure; and yet it is still true
that one can scarcely conceive of modern structural design without
the guidance of the beam formula.
Similarly we say that in an elastic structure the value of E may
vary from section to section, that imperfect elasticity makes super
position of stresses not quite correct, and that near failure the method
has only limited application. The normal process of structural design
is to determine moments and shears, and from these fiber stresses.
Whatever procedure is followed in the determination of the moments
and shears, the beam formula is used for final determination of stress.
There seem to be grounds for believing that the elastic analysis of an
arch or bent with truly fixed or truly hinged ends has greater validity
than does the method of analysis used later in design. The question
of foundation distortion and of its effect involve engineering judg
ment. Elements involving judgment should be clearly isolated so
that the limits of such judgment can be established.
3. Acknowledgment.The bulletin was written as a part of the
work of the Engineering Experiment Station of the University of
Illinois, of which DEAN M. S. KETCHUM is the director, and of the
Department of Civil Engineering, of which PROF. W. C. HUNTINGTON
is the head. The computations were made by M. F. LINDEMAN,
Research Graduate Assistant in Civil Engineering.
II. ANALYSES FOR FLEXURAL STRESS
4. General Equation of Flexural Stress.Equations for stress due
to flexure are usually based on the assumption that the variation of
stress over the crosssection may be represented by a linear equation.
This assumption is based on the assumption that the beam axis is
straight and also on the assumption, based chiefly on experimental
ILLINOIS ENGINEERING EXPERIMENT STATION
observations, of the conservation of plane right sections and of the
proportionality of stress to strain.
It will be shown later that none of these assumptions is necessary,
and that the same general form of equation may be used whatever
the facts as to variation of stress intensity over the section, provided
the facts as to the shape assumed by deformed sections and the
stressstrain relations are definitely known. For the present, however,
a linear equation of stress variation over the section will be assumed.
The stress will then have the general equation f =. (a + bx + cy), in
which the coefficients a, b, c, are to be determined from the statical
conditions which state that the sum of the fiber resistances must equal
the applied load and that the sum of the moments of these fiber
resistances about any axis in the plane of the section must equal the
moment of the applied loads about that axis.
Let x, y = co5rdinates of any point on the crosssection 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
Ix = fx2dA = moment of inertia about axis Y (along the
axis X)
I, = fy2dA = moment of inertia about axis X (along the
axis Y)
Ix, = fxydA = product of inertia about axes X, Y
P = normal component of external forces
Mx = moment of external forces about axis Y
My = moment of external forces about axis X
Also write
Ixy
M'x = MX  M y
IXV
I', = I,  I'
1 y  ly I z
THE COLUMN ANALOGY 11
All these terms are practically standard in the literature of flexure
except the "skew" terms designated by primes.
Write: f = a + bx + cy
Then, from statics, IV = 0
whence: P = ffdA = af dA + bf xdA + cf ydA
Since the axes are taken through the centroid, fxdA = 0 and
P
_fydA = 0, by definition of centroid. Then P = aA, a = 
A
The total moment about the axis of Y equals zero, whence
Mx = ffxdA = afxdA + bfx2dA + cfxydA
= bIl + clxy (1)
The total moment about the axis of X equals zero, whence
My = ffydA = afydA + bfxydA + cfy2dA
= blie + cly (2)
IX.
Multiplying (2) by , and subtracting from (1),
II,
M,  M my M'1,
b 
Y IV
Ixy
Multiplying (1) by , and subtracting from (2),
My  Mr I M'y
IV  I .
Hence
P M', M'..
f = + 7+<y
A i i
m *
12 ILLINOIS ENGINEERING EXPERIMENT STATION
At the neutral axis f = 0. The equation of the neutral axis,
then, is
P M'x M'y
A + I' x + y = 0
and the intercepts of the neutral axis are
P P
A A
xl = M' yl M'
I'x Itv
Equation 3 is a general equation of flexure for all cases if we as
sume linear variation of stress intensity. In order to apply it, it is
first necessary to know the area and centroid of the section and
Ix, I,, Ixy for two mutually perpendicular axes through the centroid.
The products of inertia should be computed first about the most con
venient axes and then transferred to parallel axes through the
centroid after this is located.
The foregoing brief statement of the formula contains all informa
tion necessary to its application. Like all mathematical relations,
it is subject to unlimited variation of form. The general formula
may be stated for principal axes, conjugate axes, neutral axis, kern
axis; the Spolygon may be computed; the moments and products of
inertia may be evaluated by summation, by integration, or graphi
cally by use of the properties of string polygons; these moments and
products of inertia may be combined to give the desired constants in
the function for (f) by use of the circle of inertia or the ellipse of in
ertia, using in the latter case the polar properties of the ellipse. These
are interesting lines of mathematical investigation; wherever they add
appreciable convenience to the solution of the problem, any selected
portions of the literature of flexure may be used.
It may sometimes be more convenient to concentrate all "prime"
expressions in one term of the formula. Write
P M'x M'y P M, M' FMX M'x1
fA' =x+ ,y A I x+ iy Y  I ,lz xI
the bracketed term reduces to
 Mz
My  M I . M', IxM
I,  ~I x
THE COLUMN ANALOGY
We then write
P M. M' ,
f A I x + i, y
Ix?
where y' = y  x "
Iv
This expression will, however, usually be found less convenient than
that given in Equation (3).
Some special cases of the general equation are more familiar.
If Ix = 0, as in a symmetrical section,
P M, My
f A + Ix±+ 1y (3a)
If P = 0, and Ixy = 0,
Ms M,
f= x + y (3b)
If one of the axes is normal to the plane of bending, Ix, = 0, and
P = 0.
Mx
f = I x, the usual beam formula. (3c)
If there is no bending,
P
f = A (3d)
The kern of a section may be defined as that portion of the section
within which a normal compressive force must act if all of the section
is to be in compression. It is sometimes useful where it can be con
veniently found.
If flexure is about one principal axis of a symmetrical section, the
stress at the outer fiber is
P Mx
A I
where x is the distance from the centroidal axis of Y to the outer
fiber.
ILLINOIS ENGINEERING EXPERIMENT STATION
The distance ek to the edge of the kern for this fiber can be found
from P Pek
e + x = 0
Ax1 x1
where px is the radius of gyration.
Let Mk be the moment about the edge of the kern,
Ix
Mk= Mx + P x
Ax,
P M. x / I\ Mk
Then f = +T x = KM +P ) = x (6)
In computing the stresses in frames it is often convenient to
compute the moments about the kern points and apply formula (6)
to obtain the stress directly.
5. Form of Computation Recommended.The arrangement of
computations in applying the general equation for flexural stress is
important where the computations are involved. It is convenient to
first compute the properties of the section and the bending moments
with reference to some convenient pair of axes. These properties are
later corrected for axes through the centroid of the section and then
corrected for lack of symmetry of the section.
A few illustrations are given to indicate the arrangement recom
mended. This arrangement is emphasized here because of its use
later in computations required in analyses of restrained beams.
Compressive fiber stresses are taken as positive and loads which
produce compression as positive. Co6rdinates will be taken as
positive when measured upward or to the right from the axes. Posi
tive moment produces compression on the positive side (top or right
hand side) of the section.
Problem 1.Unsymmetrical Section Unsymmetrically Loaded
It is desired to find the neutral axis and a measure of the fiber
stress in the section shown in Fig. 1. The section is unsymmetrical,
and the elements have widths so small that their moments of inertia
about their longitudinal axes may be neglected. The section is loaded
at point P, shown in the figure with a load of 10 000 lb.
THE COLUMN ANALOGY
Propert/es of ie Sectl'21
Coulc/ed
Area MVooment.s
2/87 6.748 0
a60 0o +2/.61
2/58 ÷/iS 790 3e97
4.945 +2.04 /.137
p p
A A
I lu
Proo&cts of /rert,~
07
0
1.8
54.0'"
Z.9
94.4
0.8
93.6
10.6
0
8.9
7.4
0
4.9
18.0
33.2
0.3
4.4
0
0
I7
ZO.3
Z0.3
Lc'ad7ý7
+/0000
~5 45
Co'vzputed
+60000
+4100
+45906'
+3Z60
45000
2300
42700
/3000
32760
Fiber Stress at a1n Point  7ix(Ver/Cisl Distance to ANeula/ Axis)
= X(1/er/c/ 91a'stace/0 to /Veutrl/ Ar4/s)
or $ '1'X(/'orizon/t/ /stance /o /Veaura/ Axls)
'7I X(lHorizo,a/l D/istnce /o Nettra/ Axis)
FIGa. 1. EXAMPLE OF AN UNSYMMETRICAL SECTION
Assume two convenient axes, in this case a horizontal axis XX
through the centroid of the portion of the section marked a and a
vertical axis YY through the centroid of the portion of the section
marked b.
Tabulate first the known properties of the section and of the load.
These are a description or key letter for each member, the length, the
width, the coordinates x and y to the centroid of each member, the
load, the co6rdinates of the load, or the moments about each of the
two axes.
x
4
+5
+41/
7//
a ZS9 030
i 6.00 0/O
c /079 0.6
Correct to
Ce4troil
Correct fo/r
Dissyffmetry
/11ercepos
for A/A
y
+15
15
Gi'ven
15ý1v'112
ILLINOIS ENGINEERING EXPERIMENT STATION
Now compute the statical moments of the areas of the members
about each of the two axes.
Also compute the products of inertia.* These are computed as
area times product of coordinates of the centroid plus the centroidal
product of inertia. The latter equals H42 area times product of pro
jections on the two axes. Signs are entirely automatic. It should be
remarked, however, that a product of inertia about the centroid is
dy
positive when dx for the member is positive, that is, when the member
slopes up to the right.
Add columns to get total area, total statical moments, total
products of inertia. If there are several loads or moments, add these
also for total load and total moments.
Correct to the centroid. Divide the statical moments by the area
to give the co6rdinates, x +0.41, y = 0.23, of the centroid.
Compute At2 = 0.8, Ay2 = 0.3, Axy = 0.5. Also compute
Px = +4100, Py = 2300.
The signs are still automatic. Subtract the corrections to give
products of inertia and moments about the centroidal axes.
Ixt
Correct for dissymmetry. Compute I,,  = +10.6, and
IX, IXV IxV
My i = +22 300, I, I  = +4.4, and M. I = 10 000.
The signs are automatic here also. Again subtract the corrections.
Now compute
P P
A A
x1= M  7.11, y M +2.13
These are the intercepts of the neutral axis on axes through the
centroid parallel to the original axes.
The fiber stress at any point may be found by multiplying the
M'1 32 700
vertical distance from the neutral axis by ,  or by
*A moment of inertia is merely a special case of product of inertia where the same axis is taken
twice.
THE COLUMN ANALOGY
Properh'es of t//e Sect ,/7
GYve,7
a 729 0.30 4
b 600 .aa/ 0
a .29 030 +4
Correct to
Cen7'ro/'i 0
/nferceA 73A
forA IA.
g
0
+3.5
0
t4Z
I *K
+dl
Co,'nouted
Stat/7a/
Area A/c,6,s/t
a ax cV
/187 8.748 0
a.600 0 +4l/
1/87 +S.748 0
497 0 +? O0
a7
a3
550
0.7
a?
0
73.Z
0
8.9
74
0
0
25.2
09
043
Load
Given
P
~/0aa&
+22
Cos'pQtea'
+2OO
0
+FOOo0
t 42O0
2420
E/Z'er Stress, atn'' PA'/,z = t X (Verf/ca/ o'/sftance to neutra/ aws)
S(Ver/li¢ oaV / ce /0to neu/t'/ a74is)
or  j X (AHorizonta/ d/isitare to nwtra/ arws)
= K('horkzonta/ d'/stance to neetral/ ax/is)
FIG. 2. EXAMPLE OF A SYMMETRICAL SECTION, UNSYMMETRICALLY LOADED
multiplying the horizontal distance of the point from the neutral axis
M' +23 600
by = 83.0
It will be observed that the sign of the stress is also automatic.
M'Since is negative, positive stress (compression) exists below the
Since  is negative, positive stress (compression) exists below the
it y
an"y^/.4
ILLINOIS ENGINEERING EXPERIMENT STATION
M'x
neutral axis. Similarly, since  is positive, compression exists to
I/x
the right of the neutral axis.
Problem 2.Symmetrical Section Unsymmetrically Loaded
If the section is symmetrical, the same order of procedure is fol
lowed, except that there is no product of inertia about the two axes
and no correction for dissymmetry.
The problem just solved has been somewhat modified in Fig. 2 for
a symmetrical section. The computations follow the order of the
previous problem.
Problem 3.Symmetrical Section Symmetrically Loaded
If, further, the section is loaded symmetrically about the axis of
symmetry, computations of the properties of the section about that
axis may be omitted. An illustration is shown in Fig. 3.
6. Transformed Section.In deriving Equation (3) for fiber stress
due to flexure it has been assumed that f is a linear function of x, y.
The same type of expression may also be derived if fn is a linear
function of x, y where n varies for different points on the section.
In this case the equations of statics are satisfied if in place of dA we
dA dA
write  so that fn  = fdA. The differential areas dA are then to
n n
be divided by the values of n.
fdA fdA
Let At = I I, =  y2
fdA fdA
I, =  x2 x"Ix, = xy
where x and y are measured from the centroid of the section having
dA
differential areas  .
n
The section thus defined, in which each area dA is replaced by an
dA
area  , may be called the "transformed section." For it we may
n
derive, by the algebraic process used in deriving Equation (3),
P M'A M'_
nf =  + Ir x + , y (4)
the "skew" terms indicated by primes being defined in their relations
THE COLUMN ANALOGY
Proer/'es of //hI Sect/i'7
Ar_ /logwets
a 71 ay
Z187 0
a.6(0 +./ 00
4.974 +a.lO0
p
krodeucs of /nerl'ly
ox +'~,
0
9.9
0
8.9
5.2
09
243
Load11
p
+10000
O z
20000
+ 4O0
24'00
r;
Fi Slless at, 7/V Po0n/t = 7 (xVer/cIa/ d61srnc' to n7etlra/ 7Arsl.)
= gd0(Vert/car/ /srnace A* :u.tro'/ w/is)
FIGa. 3. EXAMPLE OF A SYMMETRICAL SECTION, SYMMETRICALLY LOADED
to the other terms just as in Equation (3). From the values of nf
nf
we may get f= ! .
n
This method has been used in analyzing beams of reinforced
concrete on the usual assumptions as to the mechanics of such beams.
In this case it is assumed that plane sections remain plane and hence
that ds is a linear function of x, y. If the two sections between
E
which deformation occurs are parallel, as in a straight beam, ds is
constant across the section. Hence f is a linear function of x, y.
E
x
7.g9 030
b 600 0/0
a 7.29 030
Correfct to
Ce,/ro/o
/n/erco/As
of iNA.
y
0
*35
0
+049
'ZOO<
671VIII
ILLINOIS ENGINEERING EXPERIMENT STATION
The value of E here may be taken as relative. The modulus of con
crete in compression is taken as unity, that of concrete in tension as
E
zero (unlimited strain without stress), and that of the steel as E.
The section then is transformed by multiplying the area of the steel by
E,
E , and that of the concrete in tension by zero. Finally, the apparent
stress found in the concrete in tension is multiplied by zero, and that
E,
in the steel by E
This method has been traced out in detail in this elementary case
because it has broader usefulness. If the stressstrain diagram is
known for any material it is a simple matter to deduce stresses for
given loads on a section if we accept the assumption that plane sec
tions remain plane. In such a case we would first analyze on the as
sumption of Hooke's Law, then transform the section by multiplying
each differential area by the relative value of E corresponding to the
stress just deduced. This transformed section is then analyzed and
the deduced stresses multiplied by the relative value of E. Successive
revisions will furnish any desired precision.
Another important application occurs in the case of beams having
sharp curvature. Here it is usual to assume that plane sections
fds
remain plane and also that Hooke's Law holds. Hence  is a linear
E
function of x, y. The modulus E is constant, but ds varies across the
section, since the two sides of the differential element are not parallel.
The value of ds varies as the distance of each fiber from the center
of curvature of the beam (see Fig. 4). We may, then, transform the
section by dividing the width at any point by the radius of curvature
R at that point. The properties of this transformed section are then
determined. It will be shown later that these may all be deduced
from the known properties of the original section and the area of the
transformed section.
The values Rf are then found from Equation (4). These values of
Rf are then divided by the values of R to give the fiber stresses.
It should be noted here that the moment to be used in this case is
the moment about the centroid of the transformed section and not
about the centroid of the original section. The centroid of the orig
inal section has no special significance in this problem, and certain
awkward characteristics of the socalled WinklerBach formula arise
from neglect of this fact.
THE COLUMN ANALOGY
FIG. 4. SEGMENT OF A CURVED BEAM
In the usual treatment of curved beams the proportionality of
stress to strain has been assumed. This, however, is not necessary.
If Hooke's Law does not apply, but right sections plane before bend
fds
ing remain plane after bending, the value of  is a linear function of
x, y and the section may be transformed by dividing the differential
ds
areas by E . Hence sharply curved members of reinforced concrete,
such as sometimes occur, may be analyzed by use of the transformed
section. Brackets in rigid frames approximate this condition.
The transformed section could also be used either in combination
with or without Hooke's Law to deal with cases in which plane sec
tions do not remain plane in beams either straight or curved, provided
we know anything or wish to assume anything as to the shape as
fds
sumed by the plane after bending. In this case E is not a linear
fds E
function of x, y but E is such a function, where r = , defines the
curvature of the deformed section, as shown in Fig. 5. The section
'ec47?76, SeA1ed
FIG. 5. RATIO, r = ,
ILLINOIS ENGINEERING EXPERIMENT STATION
8
FIG. 6. CLOSED RING, CUT AND SUBJECTED TO AN ANGLE CHANGE
Er
can then be transformed by multiplying each differential area by  .
ds
A full understanding of the theory of the transformed section
seems desirable in computing stresses in any case in which the stress
distribution over the section is not linear. It has special advantage
in such cases in dealing with problems involving continuity or defor
mations because in such cases if the transformed section is used the
same methods are applicable in the analysis by the theory of elasticity
as would be used where the stress distribution is linear. This will be
explained in detail later.
III. THE COLUMN ANALOGY
7. Formulas Similar to Flexure Formula.The flexure formula,
then, is a solution not only of the problem to which it is usually ap
plied of determining stresses where the variation of stress over the
crosssection is linear and also, as shown in discussing the use of
transformed sections, where this variation is not linear, but is a gener
al type for the solution of certain algebraic problems. It will be shown
that the problem of analyzing arches, bents, and beams with fixed
ends in general is such a problem. The flexure formulas, at least for
the special case of symmetrical flexure, are familiar tools to struc
tural engineers, and therefore furnish a conveniently remembered
routine in treating certain problems in elastic analysis.
8. Geometry of'Continuity.Consider any closed ring as shown in
Fig. 6. Suppose this ring to be cut at A and that a certain rotation 4
takes place at B.
At A there is now produced
(a) A relative rotation of the two sides of the cut, = 4
0'r
THE COLUMN ANALOGY
(b) A relative vertical displacement of the two sides of the
cut, = ex
(c) A relative horizontal displacement of the two sides of the
cut, = 4y
If the ring is continuous there is actually no relative movement of
the two sides of the cut. Distinguish the rotations which occur
around the ring as those which would occur if the ring were cut and
those which result from the continuity at A. Call the first ,rota
tions due to the forces which are statically determinedand the
latter , rotations due to forces which are statically indeterminate.
Then, if continuity exists,
2S0,x = loix
Z0,y = TMy
The rotations 4i are due to equal and opposite forces on the two
sides of the cut section at A. Any moments in the ring due to these
forces will be a linear function of x, y. Call these moments mi
indeterminate moments. If for 0i we write mi mid ds, 
then f mi  ds = 4, (a)
m Oi xds = fj x (b)
fmi yds =f ,y (c)
where mi is a linear function of x, y. (d)
The values on the righthand side of the equations are assumed
to be known constants.
These relations are satisfied by an equation of which the equation
of flexure is a type.
Thus we may write mi = a + bx + cy where a, b, and c are
unknown coefficients to be determined from the three relations of
continuity just given. Designate the physical constants  the
mids
rotations per unit of moment per unit of lengthby the letter w (for
width of the elastic section as defined later).
Then fmiwds = afwds + bfwdsx + cf wdsy = Jf4
If the axes are taken through the centroid as defined by the
equations f wdsx = 0, and f wdsy = 0, a 
fwds
ILLINOIS ENGINEERING EXPERIMENT STATION
Also fm wdsx = af wdsx + bf wdsx2 + cf wdsxy
or = bf wdsx2 + cf wdsxy
and fmiwdsy = af wdsy + bf wdsxy + cf wdsy2
or = bf wdsxy + cf wdsy2
fwdsxy
Multiplying (2) by fwdsx and subtracting from (1)
fwdsy2
Multiplying (1)
Hence
f.x  fwdsxy
b fwdsxy
fwdsx2  fwdsxy fwdsy
fwdsxy
by fwdsx2 and subtracting from (2)
S fwdsxy
f0a fwdsy2
'C f wdsxy
fwdsy2  fwdsxy fw
m ds +
fwds
fwdsxy
fx  fy fwdsy2
fwdsxy
f wdsx2  f wdsxy fwdsy
Y:;wdsy2
f wdsxy
f ,y  f ,x f wdsx2
+ f wdsxy
fwdsy2  fwdsxy fwdsx2
If we conceive a narrow strip along the axis of the arch having a
variable width w =  then it is evident that wds corresponds to a
mids
differential area. If we treat this whole strip as an area A, we may
conveniently write
f wds = A
f wdsx2 = I=
f wdsy2 = I"
f wdsxy = I x
Ixy
Ix  1I  I'X
I,  I,  I't
= f.x (1)
= f Vy (2)
THE COLUMN ANALOGY
Also 4, corresponds to a load. If we call these known rotations
along the axis elastic loads, we may conveniently write
fre = P
f ,x = M:
f 0,y = MY
Ix2
and M.  My   M'=
My  M  MI'
P M'I M',
Hence m A ±= + + I, Y
9. The Analogy.If the effects of flexure only are taken into ac
count there exists, then, an exact algebraic parallel between the in
determinate moments in an elastic ring and the fiber stresses on plane
normal sections of a short column, as follows:
1. The stiffness Iof each short length of axis corresponds to a
m
differential area a.
2. The indeterminate moments m, correspond to stress inten
sities on the column section.
3. The known angle changes correspond to loads P on the
column section. These angle cbhanges may be due to external
forces acting on the structure, or to other causes, such as rotations
of abutments.
Consider any singlespan plane structure, with axis either straight
or curved, and with any variation in crosssection, subjected to known
loads. Draw any curve of moments for these loads consistent with
static equilibrium.
Picture a short length of column, a section of which has the same
shape as the side elevation of the beam axis and a very small width
varying along the axis as the elastic width defined above (w = ,ds
angle change for unit moment along a unit length).
Load this column with an intensity of load over these elastic areas
equal to the bending moment given by the curve of moments just
computed. The change in moment produced by restraint will now
equal the fiber stresses which would exist on crosssections of this
column. The total moment at any point equals the net intensity of
pressuredifference between load intensity at top and reaction
ILLINOIS ENGINEERING EXPERIMENT STATION
(a)eam w/IM Flcd Edds tHaiwnched
(c)ssuVCp.ort'ed
Pressure Corndu//
FIa. 7. TYPES OF ANALOGOUS COLUMN SECTIONS

"I/A
i
ý "T
I
(d)Clr1h viI' F//red zla's5,
Vegrf/Ml/ana7 Zt':Ir/w/ Oa,71/1g
THE COLUMN ANALOGY
intensity at bottomat the corresponding point on this column
section.
We have seen that a rotation in the beam corresponds to a load
on the analogous column. Equal and opposite rotations about two
centers produce linear movement normal to the line of centers.
Hence, in the analogy, a linear displacement in the beam corresponds
to a couple on the analogous column about the axis of displacement.
These rotations and displacements in the beam may be due to any
cause whatever, abutment displacement, temperature change, forced
distortion by jacking, or they may be imaginary displacements gen
erating influence lines according to the principle of influence lines as
stated by MfillerBreslau.
10. The Elastic Column and Its Load.The analogy just stated
furnishes a convenient mental picture of the relations of the moments
in restrained members. In Fig. 7 are shown some sketches corre
sponding to these conceptions. The shaded area is the section of the
column of any short length.
In Fig. 7a a straight beam of constant section and fixed at the ends
is loaded with a single concentrated load. The most convenient curve
of moments for such a load is obtained by treating the load as canti
levered from the nearer support. The analogous column has a width
_ 1
equalor proportionalto the value  or for the beam.
mds El
The intensity of load on the column equals the moment curve just
drawn.
In this case the average load intensity and the position of the
resultant are known by inspection, and hence the total load and its
moment are readily computed. Computations of the pressure inten
sity on the base of this short column will give the changes in bending
moment resulting from the fact that the beam is not cantilevered but
is fixed in position and direction at both ends.
In Fig. 7b the axis is assumed to be straight but the beam is not of
constant section. The width of the column is, then, not constant,
and the average load intensity and the position of the resultant load
are not evident by inspection. The problem is conveniently solved
by dividing the column section into a number of small lengths.
In Fig. 7a and 7b the elastic rings are closed by the earth, which
has zero elastic area. In Fig. 7c the elastic ring is completed by the
structure. The curve of moments due to the weight of water and to
the pressure head is drawn on the assumption that the top and sides are
simple beamsa stable condition. The elastic column and its load
ILLINOIS ENGINEERING EXPERIMENT STATION
are as shown. Pressure intensities on the base of the column are
changes in moment due to the fact that there are no hinges at the ends
of the top and sides, but that there is really complete continuity at
these points.
In Fig. 7d an arch is subjected to vertical loading on one side, and
to horizontal loading on the other side. Moment curves are drawn
independently for the two conditions of loading, the arch being as
sumed cantilevered from the left abutment for the vertical load, and
from the right abutment for the horizontal load. Note that it is not
necessary to assume the same statically determinate condition for the
beam for all parts of the load.
11. Signs in the Column Analogy.In flexural analyses it is con
venient to consider compressive force and stress as positive and to
measure co6rdinates as positive up and to the right from the centroid.
Positive loads on columns, then, are downward, and positive couples
are such as produce compression above and to the right of the centroid
of the column section.
In applying the column analogy bending moments in the beam
will be considered positive if they produce tension on the inside of the
elastic ring. Shears are positive if they accompany positive rate of
change of bendingmoment, increase up or to the right.
Positive rotations are such as would accompany positive bending
moments; positive displacements are such as would result from posi
tive shearing forces. A clockwise abutment rotation at the left end
of an arch, then, is positive; at the right end it is negative. An in
crease in arch span is a positive horizontal displacement. Settlement
of the right abutment is a positive vertical displacement.
Note, however, that rotations and displacements which actually
exist (abutment movements) are to be distinguished from those which
are resisted (shrinkage or distortions due to temperature changes).
In determining moments by the column analogy in the former case,
the sign of the displacement may be conveniently reversed.
12. Choice of Statically Determined Moments.The indeterminate
forces are equal and opposite on two sides of any section, and hence
satisfy the laws of statics whatever the determinate forces may be.
Also the indeterminate forces are computed to have values such that
continuity is preserved. Hence any set of determinate forces what
ever which will support the loads may be chosen in the first place,
since the conditions of both statics and continuity will be satisfied
by the solution.
THE COLUMN ANALOGY
It is, of course, important to choose the most convenient curve of
determinate moments. As a simple example, in the case of a fixed
beam with a moment load, (upper part of Fig. 8) the four sets of
determinate systems shown, among many, are available. In general
the first or second moment curves will be more convenient.
The moment at x may be written directly by finding the indeter
minate moment from the moment curve shown in (a) or in (b).
Similarly the end moments may be written directly. To write Mb
we use the moment curve in (a) and to write Ma we use the moment
curve in (b).
In the lower part of Fig. 8 are shown three of the possible elastic
loads on a rectangular bent carrying a single concentrated load. In
the first case the curve of statically determinate moments is drawn as
though the load were cantilevered from the left support, in the second
as though the girder were simply supported on the columns, and in the
third case as though the load were cantilevered from the right sup
port. The indeterminate forces and moments will be different in the
three cases, but the total moments will, of course, be the same. The
second moment curve will probably be found most convenient.
13. Components and Direction of Indeterminate Forces.The prob
lem of analysis of rigid frames and arches is essentially that of finding
the moments due to the indeterminate forces which are necessary to
preserve continuity. These moments equal zero along the line of
application of the indeterminate forces. This line of action, therefore,
corresponds to the line of zero stress in the analogous column. This is the
neutral axis for the given loading, and, as already found, it has inter
P/A P/A
cepts on the axes through the centroid of 1 =  M', andyi =  M
I' I'V
The stress in a beam at unit horizontal distance from the neutral
Sf M'x
axis is 3x  I' , and at unit vertical distance from the neutral axis is
bf M'y
8y. I'l "
The moment at unit vertical distance from the line of action of
the indeterminate forces equals the horizontal component of the
indeterminate force, and the moment at unit horizontal distance from
M'
the neutral axis equals the vertical component. Hence hi =
M',
I'= P
ILLINOIS ENGINEERING EXPERIMENT STATION
Bea'/7 wi/,/ ,ed Ends, Costa/, Sect/aon
e= l L1
Z7,
A .~  A
~0* t
* +
b
K
M ....... ......._... ....
c/~v~y/~7r 3e/2f
FIa. 8. POSSIBLE MOMENT LOADS FOR BEAM WITH FIXED ENDS
AND FOR RECTANGULAR BENT
T..~ 2'
....................
.................~
(a) a;5
(t)
(c)
THE COLUMN ANALOGY
A convenient basis for the computation of all indeterminate mo
ments is to locate first the neutral axis of the analogous column and
find one component along the axis.
The components of the indeterminate forces are changes in the
reactions from those existing under the static conditions assumed.
We may find the components of the reactions as V = v,  vi
H = h,  hi.
With the components of the reactions known it becomes a simple
matter to draw the pressure line.
The pressure line for the structure may sometimes be drawn
conveniently by superimposing the curve of statically determinate
moments on the neutral axis to a vertical scale of hi = 1 or to a hori
zontal scale of vi = 1 according to whether the moment is laid off
horizontally or vertically.
1/ 14. Application to Simple Cases.The principle of the column
analogy may be illustrated by application to a few simple cases.
Let it be required to compute the end moments on the beam with
fixed ends shown in Fig. 9a, assuming constant moment of inertia.
Consider the moment curve shown, produced by the load P on a beam
simply supported at its ends. Any static moment curve such as a
cantilever over length (a) or (b) might equally well have been used.
The centroid of a triangle using the notation of the figure may be
L+a
shown to lie at a distance  from one end.
3
In this case, then, the analogous column section is a narrow strip
1
of length L and width  . Both E and I, being assumed constant in
this case, may be given a relative value of unity. The column section
thus becomes simply L and the load mL. The outer fiber stresses in
2
the column, analogous to the end moments, may be found by the
usual column formula, or more conveniently in this case by taking
moments about the kern points. Then, f = Mky where Mk is the
In
moment about the opposite kern point, and Io is the moment of inertia
of the column section about its centroid.
ILLINOIS ENGINEERING EXPERIMENT STATION
7,Lee7m wl/h 7ea7d' End,
SV/ie get7a1
hyf/Toae'
Diey'rafly
"o/ogcm
beo/;:w
(b)Seam FA/ed o/ One EZd, /H/yea' a/ te Oter LEo½,
Concentzra''edz Loud
A
SmA/o/e Beam
And/ogous Co/lm
Secio/n/
wppor/)
(c)Ber? w/ih Oa,7 End F,/kea,
6//1 Rotat/on af"/?e 1 O/er /7,a'
U1C//7/1,
'?~/ od
FIG. 9. APPLICATION OF THE PRINCIPLE OF COLUMN ANALOGY TO SIMPLE CASES
THE COLUMN ANALOGY
Mk d
Mky 2 6Mk
Further, for rectangular columns, f   bd3  Ad
I o T1bdW Ad
mL b
6
2 3 b
Then, at the left end, f, = Ma  L . L m, 
a
And, at the right end, fb = Mb = M, L where m, is the simple beam
Pab
moment, m, = L
For a beam of uniform section fixed at A and hinged at B, subject
to a single concentrated load P, Fig. 9b, the hinge has an infinite
elastic area (unlimited rotation due to a moment) and hence both the
centroid and kern point of the infinite column section lie at the hinge.
mL . L + b
. L
Mky 2 3 L + b
Whence M = L = m L 2L
3
Suppose the moment corresponding to unit rotation of the free
end is required at each end of a supported beam which is fixed at
one end, Fig. 9c.
1
The analogous column is a strip of width E and length L loaded
with unit load at one end. Taking moments about kern points of
6Mk
A and B, write directly from the formula f = Ad '
2%L El L/3 El
Ma=6 L 4 Mb=6 L2
L L L L
L L
El ElL
Suppose a beam with fixed ends subjected to unit relative displace
ment of the supports. Displacement in a beam corresponds to mo
ment about the axis of displacement in the analogous column. For
unit relative displacement of the supports, then, we write directly
6M = 6 6EI
Ad L L2
E L
ILLINOIS ENGINEERING EXPERIMENT STATION
a b0
FIG. 10. ROTATION AND DISPLACEMENT OF THE ENDS OF A MEMBER
If the beam is subjected to loads and at the same time to rotations
and displacements of the ends, we may write
El El AEI
Ma = M'a + 404a  + 24b  6 L2
where M'a is the fixed end moment at A due to the loads, 4a and ,b
are rotations at A and B respectively, and A is the relative end
displacement.
A
In place of L write 4,, the angle of tipping due to end displacement.
I
Also write K = , and Ra instead of M'a.
Then Ma = 2EK(24a + 4b3'#) + Ra which is the equation
usually known in American literature as that of slopedeflection.
The signs used for 4 and , will depend on the convention of signs
adopted.
For some purposes this may be more conveniently written as
Ma = 2EK(20, + O,) + Ra
where 4, and # are measured from the axis of the beam before flexure,
and 0 from the chord after flexure. This expression can be derived
directly from the column analogy (see Fig. 10).
15. Simple Numerical Examples.The column analogy is pecul
iarly fitted to the direct solution of numerical examples.
(a) Beam Fixed at Ends
The curve of moments will be drawn for a single concentrated load
on a beam of constant section with fixed ends. Three curves of static
moments will be used. The beam and its load are shown in Fig. 11.
a
Of course for this case the formula M = m,  already derived is suffi
cient for finding the fixedend moments.
THE COLUMN ANALOGY
Fia. 11. BEAM WITH FIXED ENDS
From this formula
5 X 15 15
m,  20 X 10 = 37.5 Ma = 20 X 37.5 = 28.15
5
Mb = 20 X 37.5 = 9.38
In case I the load is assumed cantilevered from A, in case II the
load is assumed cantilevered from B, in case III the beam is treated
as simply supported at A and B.
In case I, the end moment determined by statics is 5 X 10 =
50
50, the average is    25 acting over an area 5 X 1. Hence
2f
m3= X/6' =37S5
A= 6 =00O
A 2
"="aZr" r 6.
/" +" T= 667 =&fr+.Oqe2'+8:
]:=X&?20'667
. 6^c2t8j=fe8./4^
Or ,"o,.l'7eg" o,6n:.s aiou," /,, Aern :
14 4 .= £ 7 .0  ff./4o
/l ' = 5  ( /(/ ) =   . 1
7 11 77 "
68 = ox+9.,) = 9,38
T"'*A^' 'y"'ns o'hout f/.'e ker>."
At **',^ = 6)_ 8.e
M, 30(106e7)  9397
t =  (+18. 14)=Ze.l,
, ,, (/l,e~xr.67x(+/o)_j,,.s#
A2 "';, o.= (^ 67 ^   ...¢
A, = /.yo(/4o.s") =g.,_a
Taki'/ngi :;"ori:n's crbol/t f/? ker,'.
Al "A s;'. = ^+^, r , +1z0./
667
A,= o(+Z6.,)==a./,
At C+375)x"167,(+/o)
Al "8, 6a67 =+9.38
A^s= o(9.8) =9.38
ILLINOIS ENGINEERING EXPERIMENT STATION
A
 N
k I +
4
N
tN
+
I
I
4
N
N
1
N
+
36 '
N
Vl
N
^
^
i§
t
%
THE COLUMN ANALOGY 37
E
n
cq
z
0
0
rfi
*
ILLINOIS ENGINEERING EXPERIMENT STATION
P = 25 X5 = 125. It acts at the centroid of the triangle of
20 1
moments. For the column section A = = 20, I = X 20 X 202
1 12
= 667. Applying these values
P Mxx 125 (125) X 8.33 X (±10)
A I  20 667
21.86 or +9.38
Plot this moment curve, and on it as a base plot the original curve of
moments.
The same procedure is shown for cases II and III.
(b) Simple Bents
Assume the rectangular bent shown in Fig. 12. Let the loads be a
vertical load of 10k on the top and a horizontal load of 6k uniformly
distributed along one side. It is desired to draw separate moment
curves for the two cases of loading.
Assume as convenient axes a vertical through the center of b and
a horizontal through the center of aa. Tabulate length L, moment
of inertia I, horizontal coordinate of centroid x and vertical coordin
ate of centroid y for each of the members a, b, and a.
Also record the elastic loads and their centroids. For the load of
10k, treating the girder as a simple beam, we have the moment curve
66.7
shown, average moment +  ,area loaded 3, and hence
P = +33.3 X 3 = +100. The centroid is as shown, and x =
1.7, y = +6. Whence M. = (1.7) X (+100) = 170.0 and
M, = (+6) X (+100) = +600.
For the horizontal load draw the curve of moments for the column
36
as a cantilever. Average moment  = 12; area loaded = 6.
Hence P = (12) X 6 = 72. Also x = 15, y = 3.
Mý = (15) X (72) = +1080 and M = (3) X (72) =+216.
Compute a, ax, a,, ax2 + ix, ay2 + i, for each member. The
1
centroidal moment of inertia (ix and iy) equals 2 a X (projection along
the axis)2. Find the totals.
18
Reducing to the centroid, x = 0, y = + 1  +1.2. Find Ax2,
Ay2, Px, Py, and subtract.
j
THE COLUMN ANALOGY
_ Properieos or S'e'cf'ontV
Ne Z'mr eng/1 I ar V 6y sy'/
a /ZO / /.26 0 0
144
S /1.5 6 3 +8 ÷+264 2
c 4.0 9 a +/1 + 6
d 4.1 5 . +/4. +// 8
/19. +47'6 E43
Correct to Cenroi' +2.82 /34
/£. SO59
A/7'.Csf/ /a ,Z,,
0
+128
+197
4
p
+430
+/15
+734
+734
+4?4
+/12
+14.5
FIG. 13. BENT WITH MONITOR
The intercepts of the neutral axis for the two conditions of loading
P/A P/A
are now found as x = M/and i = M,/I. For the vertical
M,
load compute hi , +2.1, and for the horizontal load compute
M=
vi = +0.37. The neutral axes are then plotted, and on them
the original curves of moments are drawn to the scale of distance.
+3565
+22395
+7725
2070
+5655
+ //./
ILLINOIS ENGINEERING EXPERIMENT STATION
/Rehywraede
N/o,"m/ni t4 ,'ber'r a at Pa/n' 0
44
24
18
6
+/2
÷Z4
5
180
/35S
24
+/fl0
+139
39
6
4X6
438
0
0530
49
Z530
/070
7
+/1Z
+/15
+75
9.0
7448
8
0
+80
+56
+75
625
9
0
140
3E09
140
73595
/0'
0
6
0
+90
/62Z
+ 68
337
0
360
162O
0
302.9
// /2 /3
+90 +/5
+2/0 + 5
/0
I f.
/4
+ 675
+840
+33/
/5'
+ 1350J
 5040
+ 2406
MSl«v
Ncz~'n/ a~" Po/4t 0:
0
___ + (+~7~)('2293) (+g~gs~l(44o) ft A'
355 /0700 2595
Fia. 14. UNSYMMETRICAL BENT
66.7
For the vertical load plot for static moments 2.1 = 31.7 t.; for the
36
horizontal loads, 0.369  97.3 ft.
In Fig. 13 is shown a reinforced concrete bent having a monitor.
The dead load on the roof is assumed uniform at 1000 lb. per hori
zontal foot.
Cooramares or
Po&/ 0 f241
3
75
7.5
40
/00
7f
/ z2
a /I 2
b /5 2
c /2 3
dO 30 3
e /S 2
/6
+ 8100
+ /020
+13860
0
+43600
+40 00
+14850
'saox
T&&?
+36.£SO
P01121'0 24a
faf
= +/1.7&W
*» '
THE COLUMN ANALOGY
In this case both load and structure are symmetrical, and there is
no need to compute moment of inertia about the axis of Y. The axes
are taken on the vertical center line and through the center of mem
bers aa.
The moment areas and their centroids for the different members
have been computed separately by breaking them up into trapezoids
and parabolas.
The same procedure is followed as in the preceding problem.
The neutral axis, however, is horizontal. hA, +11.1. Inter
P/A
cept of neutral axis yi = 3  3.92. The rise of the pres
200
sure line is 11.1 = 18.00 ft.
The signs of the intercepts of the neutral axis can usually be found
by inspection, since the neutral axis lies on the side of the centroid
opposite to the load.
In Fig. 14 is shown an unsymmetrical bent subjected to vertical
loads. The tabulation of elastic properties and of elastic loads follows
the procedure already explained. The elastic moments may also be
conveniently computed as previously explained. They have actually
been computed as the sum of the moment of the elastic load acting at
the centroid of the member plus the product of shear times elastic
centroidal moment of inertia. The method is explained later as an
extension of the analogy, but presents few advantages. The trial
axes are taken as the vertical through the center of member c and the
horizontal through the center of member a.
The correction to the centroid also follows the procedure already
explained.
The correction for dissymmetry is made as follows:
Ix, (2854\
for I, in column (6) write I,. * = (2854) 3209  2530,
I x, (2854\
for M.in column (15),write M,.  = (+25 230) 3209 22420
Ix, /2854\
for I, in column (9), write Ix,. = ( 2854) \13230 = 614
Ix /  2854\
for M, in column (16), write M,  = (+4745) =13230 1020.
Subtract these corrections to get I',, M'x, I',, and M',.
/
42 ILLINOIS ENGINEERING EXPERIMENT STATION
s .
VWZZi
4
z
THE COLUMN ANALOGY
NN~)
N N N
NNN
>1
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
N
NNNNN NN vN
N'0 N Ns ^ N NN N'0 1'
^^^^'3+4~«sr 4I
to N^ gi
K
K
K
K
NN
'0
N
N
N
N
N
'0
N
N
N
N
N
Ns
N
NN
'010
NN
N'0
'ON
t0
Ns
SN Mt; ts '
t + *
+++4+
N\N te
NNN i
NN9NNN
++
4+4
N~+.
++
NN
N
N
N
N
N
N
N N
NN
+ +
NN
NN
NN
Nt
N ~
NN
4+
NN
toN
N
N N
* +
N
N
N
N
N
to
+
to
N
'0
to
N
N
N
'4
N
N
N
N
to
N
~0
N
N
N
to
N
t~1
N
N
to
N
N
'0
N
N
N
'4
N
N
N
toN N
N N
+
to
N
N
to
to
N '0
N +
+ +
N
to
'0
N
N
I +
N
tt*
1+
k
'4
N
s^§
NN'0 <
N
N
N
N
It)
N
N
N
%<
+ I +
II I I]
n ?
I 7
^t4
'4I
KII
N
~t~N
K ~ K '~ ~N
K~t*~ KN
oil
~
N ~
N ~t t~
~ .~ ~
K
~N ~t~tb ~
~
S'o?
44 ILLINOIS ENGINEERING EXPERIMENT STATION
The bending moment at any point, such as joint o, may now be
found as M = m,  mi, as shown.
16. Arch Analysis.The column analogy affords an unusually
convenient means of analyzing reinforced concrete arches, because,
once understood, it furnishes a familiar order of procedure. The
actual computations are, of course, the same as presented by other
writers.
In the case of unsymmetrical arches the equation given in this
discussion seems to offer a much more convenient order of arranging
the computations than is found elsewhere. For this reason it has
seemed worth while to give in some detail the essential steps in the
analysis of an unsymmetrical arch.
(a) Unsymmetrical Arch
The arch analyzed is shown in Fig. 15. It has a span of axis of
60 feet divided into five panels of 12 feet each. The total rise is 20
feet and the difference in level between abutments is 15 feet.
The arch axis is first divided into fifteen segments of equal hori
zontal projections.
Use as convenient trial axes horizontal and vertical lines through
the highest point of the arch axis.
Tabulate first the known properties of the arch. These are the
length of each segment L along the arch axis, the distances x and y
to its centroid, the depth of the sections at their centers.
L L
Now compute the elastic areas a, each equal to  = 12  ; this is
for unit width of rib. From these compute the statical moments
ax and ay about the axes of x and y, the products of inertia about these
axes ax2, ay2, axy.
Now tabulate the m, values at centroids of sections for unit loads
at each of the panel points, A, B, C, D. The statically determinate
moments will be found for these loads cantilevered from the nearer
end of the arch. The statically determinate moment on any segment
between the load and the nearer abutment, then, equals the distance
from the load to the centroid of that segment. We then compute the
elastic load P = m,a, the moment of the elastic load about the axis
of Y, Px, and about the axis of X, Py. Note that for any segment
these three quantities may be written by multiplying m, by columns
(6), (7), (8), successively.
Sum the columns for elastic area, statical moments, products of
inertia, elastic loads, elastic moments.
THE COLUMN ANALOGY
Correct to the centroid. Compute = a and y = and the
x A A
corrections for the products of inertia V2A, y2A, xy A, and for the
elastic moments xa;P, yZP. Subtract the corrections.
Correct for dissymmetry. Write the value of ". IunderIand
I»
of  . My under M, and write the value of ". Ixy under Iy and of
I. M under My. Subtract the corrections.
Draw horizontal axis through the centroid. Now compute for a
load at each panel point the values of the components of the more
M't M'.
distant reaction vi = P= and h, =  and the intercept on the
P
X axis through the centroid, x\ =  A
Avi
Ribshortening has not been included in the computations. It
may be corrected for by computing the average intensity of com
pression for any given condition of loading and from this the change
of span which would take place if the arch were free to contract. This
change of span may then be treated as if it were due to change of
temperature, equivalent temperature change = fa.
Ee
The components and location of one reaction (at the more dis
tant abutment) now being known, it is easy to draw the pressure lines.
By scaling the ordinate from any pressure line to the kern point for
any particular crosssection and multiplying by the H value for that
pressure line, we can compute conveniently the moment about that
kern point. Thus we can draw influence lines if we wish, by plotting
kern point moments at various sections for different positions of the
unit load. From the kern moments the stresses can be computed
directly from the formula f  .
For temperature changes, the horizontal change in span if the arch
were free to expand multiplied by E is Ed X 60 ft. and the relative
vertical movement of the abutments multiplied by E is Edt X 15 ft.
Omit the constant multiplier Ed for the time being, and use it later
as a multiplier for the temperature stresses. It has been shown that
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46 ILLINOIS ENGINEERING EXPERIMENT STATION
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ILLINOIS ENGINEERING EXPERIMENT STATION
a linear displacement is analogous to a moment about the axis of
displacement. Hence we have on the analogous column elastic
moments Ms = ±15 and My = F60. That the signs of these mo
ments are opposite is determined by the fact that the change of span
is analogous to a moment about a line connecting the ends of the arch
axis. Such a moment on the analogous column produces compression
on the top and left or on the bottom and right of the section. After
the line of action of the temperature thrust has been determined, the
sign of the bending moment at any point is readily determined for a
rise or for a fall of temperature by observing that a thrust is required
to shorten the span, and a pull to lengthen it.
These moments M. = ± 15 and My = =F60 are now corrected
for dissymmetry. The H and V components of the temperature thrust
are then computed. The thrust, of course, passes through the elastic
centroid, just as the neutral axis of a beam passes through the centroid
for pure bending.
(b) Symmetrical Arch
If the arch is symmetrical, the procedure just given is shortened.
It is now necessary to consider only onehalf of the arch ring. By
inspection the product of inertia is zero and there is no correction for
dissymmetry.
In Fig. 16 is shown the analysis of an arch similar to the one just
shown except that the span is 5 panels of 16 ft. = 80 ft.
17. Haunched Beams.Because of the occasional importance of
haunched beams in continuous frames, examples of the analysis of
such beams have been included. The arrangement of the computa
tions is the same as in the analysis of arches except that there are no
y coordinates.
Attention is called to the computation of end moments resulting
from unit rotation of one end. These values are constants needed in
certain methods of analyzing continuous frames.* In order to com
pute these moments a unit load is applied at one end of the section of
the analogous column and the outer fiber stresses in the column are
determined.
A beam symmetrically haunched is shown in Fig. 17 and is anal
yzed for end moments due to a uniform load and also for end mo
ments due to a rotation at one end.
*"Continuity as a Factor in Reinforced Concrete Design," Hardy Cross, Proceedings, A. C. I.,
Vol. 25, 1929.
"Simplified Rigid Frame Design," Report of Committee, 301, Hardy Cross, AuthorChairman,
Journal, A. C. I., Dec., 1929.
"Analysis of Continuous Frames by Distributing FixedEnd Moments," Hardy Cross, Proceed
ings. A. S. C. E., May, 1930.
THE COLUMN ANALOGY
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FIG. 17. BEAM SYMMEmICALLY HAUNCHED
In Fig. 18 a beam unsymmetrically haunched is shown and this is
analyzed for end moments due to rotations at the ends.
18. Slopes and Deflections of Beams.
(a) Relation of the Column Analogy to Theorems of AreaMoments
The sum of the rotations, or change in slope, on the beam cor
responds to shear on a section through the analogous column, because
the product of moment by elastic area is rotation and the sum of load
ILLINOIS ENGINEERING EXPERIMENT STATION
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FIa. 18. BEAM UNSYMMETRICALLY HAUNCHED
intensity times area is shear. From the geometrical relations previ
ously explained, it follows that the statical moment of the rotations
about any section is the displacement at that section.
Slopes along the beam, then, correspond to shears on longitudinal
sections through the analogous column and deflections of the beam
correspond to bending moments on longitudinal sections through the
analogous column parallel to the line along which the deflection
is wanted.
THE COLUMN ANALOGY
The theorems of areamoments, then, are a part of the analogy.
The theorems dealing with the slopes and deflections of beams are
among the most useful and best known in the literature of structural
analysis. They deal with displacements relative either to a tangent
to the curved beam or to a chord of the curved beam.
They may be conveniently stated as follows:
(1) (a) Slope at any point measured with reference to a tangent
to the bent beam at another point may be found as area under the
M
El curve between the two points; (b) deflections at any point meas
ured with reference to a tangent to the bent beam at another point
may be found as the statical moment about the first point of the area
M
under the  curve between the two points.
El
(2) (a) Slope at any point measured with reference to any chord
of the bent beam may be found as shear at that point due to the area
M
under the  curve as a load on the chord acting as a beam simply
supported at its ends; (b) deflection at any point measured with
reference to any chord of the bent beam may be found as bending
M
moment at that point due to the area under the  curve treated as
El
a load on the chord acting as a beam simply supported at its ends.
All of these theorems are merely theorems of geometry stated in
terms convenient to the structural engineer. They neglect the effect
of distortions other than rotations, and are applicable to any curved
line where the angle changes are'very small if we substitute "angle
M
changes as loads" for "area under the  curve as a load" in the
I El
theorems.
In the column analogy the angle changes are treated as loads on
the analogous column. From the geometrical relations already
pointed out it is evident that the conception involved in the column
analogy is essentially that used in finding slopes and deflections.
If we use the column analogy in the extended form explained later
in which the angle changes are represented as forces on the analogous
column equal to moment times elastic area, and the linear distortions
are represented as couples on the analogous column equal to shear
times elastic centroidal moment of inertia, we may include at once
the effect of both angular and linear distortions. This is sometimes
a useful theorem. It is applied in the example of Fig. 14.
ILLINOIS ENGINEERING EXPERIMENT STATION
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Fic. 19. INFLUENCE LINES FOR SHEAR AND MOMENT IN AN ARCH
WITH FIXED ENDS
(b) Influence Lines by the Column Analogy
The column analogy may be combined with MillerBreslau's
principle to compute influence ordinates. According to this principle,
the influence ordinates equal the displacements of the load line which
would result from a unit distortion corresponding to the stress func
tion under investigation. For an influence line for moment, then, we
would apply a unit rotation, for shear a unit displacement, and so on.
The application for crown moment in an arch is shown in Fig. 19.
A unit rotation is applied at the crown and the displacements of
points on the arch axis are determined. Vertical displacements along
the arch axis are influence ordinates for crown moment due to vertical
loads, horizontal displacements are influence ordinates for horizontal
loads, rotations of the arch axis are influence ordinates for applied
couples.
All of these may be found as shears and bending moments on sec
tions through the analogous column when the column is loaded with
a unit load at the crown, as shown in the figure.
Similarly for unit crown shear, apply to the column at the crown a
unit couple about the vertical axis of the arch; for crown thrust, apply
THE COLUMN ANALOGY
to the column at the crown a unit couple about the horizontal axis
of the arch.
This procedure has great value in sketching influence lines. For
numerical computation it is probably as rapid and convenient to com
pute the influence ordinates by the elementary procedure of applying
unit loads at successive points on the arch axis.
19. Supports of the Analogous Column.
(a) Types of Supports
The analogous column is supported on an elastic medium. The
intensity of resistance offered by this medium is the indeterminate
moment, resisting distortion, or the moment resulting from the
restraint. The intensity times the area is the rotation produced by
this resistance.
If there is a hinge in the beam, the rotation would be infinite if
there were a constant moment at the hinge; the elastic area of the
hinge is infinite. But it is inconvenient to represent an infinite area;
instead we may choose a small area of infinite stiffnessa rigid point
support. A hinge in the beam, then, may be represented by a rigid
point support of the column.
A roller nest is equivalent to a rockertwo hinges on a line normal
to the roller bed. Hence a roller nest or rocker may be represented
by two point supports on a line normal to the roller bed.
A free end would be produced by three hinges not in line. Hence,
it may be represented in the analogous column by three point sup
ports not in line or by a fixed end.
It is interesting, though perhaps not very important, to note that
if the beam is unstable, the column is statically indeterminate; if the
beam is statically determinate, so also is the column; if the beam is
statically indeterminate, the column would be unstable if it were not
supported by the elastic medium.
In Fig. 20a is shown the analogous column and its load in the case
of a beam simply supported. The areamoment relation is familiar.
Figure 20b shows a threehinged arch and its analogous column.
Note that in statically determinate structures only one stable curve
of moments is possible and that there is no pressure on the elastic
baseno indeterminate moment.
Figure 20c shows an arch having a crown hinge. In analyzing this
case the rigid support is given an infinite area. The centroid, then,
lies at the hinge and the total elastic area is infinite.
Figure 20d shows a cantilever beam. Note that the free end of the
beam is rigidly fixed in the analogous column. The moment area
53 y_
ILLINOIS ENGINEERING EXPERIMENT STATION
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FIG. 20. TYPES OF SUPPORTS FOR ANALOGOUS COLUMNS
THE COLUMN ANALOGY
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w1117 Rolle Ale,5
FIG. 20 (CONCLUDED). TYPES OF SUPPORTS FOR ANALOGOUS COLUMNS
relation is familiar to most students, as is also the conjugate beam
relation.
In Fig. 20e is shown an arch having a roller nest equivalent to two
hinges on a line normal to the roller bed. The elastic area is infinite
and the elastic moment of inertia about the axis of the roller bed
the Y axisis also infinite.
(b) Reciprocality of Hinges and Supports in Beam and Analogous Column
Just as hinges in the girder or arch may be represented by rigid
supports in the analogous column, so rigid supports of the girder may
be represented by hinges or combinations of hinges in the analogous
column, since bending moment in the column corresponds to deflec
tion in the beam. If the restraint is in two directions, the column at
that point would contain two hinges at an angle with each other,
which is equivalent to a universal joint.
These relations do not, however, seem to be very useful, since it is
not convenient to analyze the stresses at the bases of such columns.
The analogy, however, is illustrated in Figs. 21a and 21b. In Fig. 21b
the analysis for moments in the beam is not affected by the hinge and
joint in the analogous column, while analysis for slopes and deflec
tions in the beam is facilitated by their use.
+ 56
ILLINOIS ENGINEERING EXPERIMENT STATION
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FiG. 21. TYPES OF SUPPORTS FOR ANALOGOUS COLUMNS
(c) The Conjugate Beam
The term "conjugate beam" is due to Professor II. M. Wester
gaard. In an article entitled "Deflection of Beams by the Conjugate
THE COLUMN ANALOGY
Beam Method"* he has shown a reciprocal relation between straight
loaded beams and certain imaginary beams which are subject to loads
equal in intensity to the curves of moments on the original beams.
The conjugate beam as presented by Professor Westergaard will
be seen to be the analogous column section. The conception, how
ever, can be extended to curved as well as to straight beams.
IV. EXTENSION OF THE ANALOGY
20. Introduction.The analogy between the computation of mo
ments in beams, bents, and arches and the computation of fiber
stresses on a section of an eccentrically loaded column as previously
stated is simple and convenient. The analysis thus far explained,
however, either neglects the effect of linear distortions within the
structure or makes separate allowance for the effect of such distor
tions. In this section of the bulletin it is explained that by an exten
sion of the analogy it is possible to include the effect of these linear
distortions directly in the analysis. In doing this, however, the sim
plicity of the picture is marred. The extension here presented is,
then, probably to be thought of as a very special tool to be used in
frequently, if at all. The completeness of analysis furnished by this
method of treatment justifies the inclusion of the material.
21. Internal Distortions.The beam formula may be modified as
follows: Moments may be separated into two parts, due to loads and
due to couples and the products of inertia may be separated into two
parts, due to the areas which make up the section and due to the
centroidal products of inertia of those areas. The terms in the beam
formula may then be redefined.
Write P = Spa
A = la
M, = 7pae, + SM,
My = Xpae, + ÷ZM,
I = 2;ax2 + 2i,
I, = 2ay2 + 2i,
Ix, = Zaxy + Zixy
Thus far in dealing with internal distortions only the angular
rotations which take place about the centroids of the small sections
into which the axis of the beam is divided have been considered.
*Loc. cit., page 7.
ILLINOIS ENGINEERING EXPERIMENT STATION
FIG. 22. TERMS USED FOR INTERNAL DISTORTIONS
Angular rotations have been treated as loads on the column section
and an elastic body which will suffer angular distortion from moment
has been treated as a part of the area of the column section. But a
linear movement will result from two equal and opposite angular
movements about different centers, and if these rotations are treated
as loads, they will be parallel loads of opposite sense and will consti
tute a couple. Hence any linear distortion within the body corre
sponds to a moment load on the analogous column section.
The bending moment on any differential portion of a member due
to unit forces through the elastic centroid of the member is x and the
relative displacement of the ends along any axis Y due to this moment
is axy (see Fig. 22). The total displacement then is faxy and is the
product of inertia of the elastic areas about the elastic centroid. But
it makes no difference whether this displacement is due to flexural
or to shearing distortions. Whatever its cause, we can still treat it
as if it were laxy and call it the centroidal product of inertia about
the axes X and Y.
Just as any portion of a section may be represented for purposes
of beam analysis by six quantities, namely, two coordinates of its
centroid, its area, and its products of inertia about two centroidal
axes, so the elastic properties of an elastic body may be completely
represented by corresponding elastic quantities. And just as these
six elementary properties of an area may be deduced from six other
propertiesnamely, the statical moments about any three axes in
the plane and the products of inertia about any three pairs of axes in
the plane, so the elementary elastic properties of the section may be
deduced if we have three displacements due to unit moment on the
body and three displacements due to equal and opposite forces at the
ends of the body.
This makes it possible to include automatically all types of stregs
distortion in beams directly in the analysis, if this seems desirable.
THE COLUMN ANALOGY
Moreover it is possible in members containing straight segments sub
ject to constant shear to evaluate the bending effects by separating
the effect of the total rotation and the relative displacement of the
ends due to bending without finding the centroid of the moment curve.
The elastic area of a body, as the term is used here, is the total
rotation produced in the body by opposite unit moments at its ends.
The elastic centroid is the center about which the rotation occurs.
The total displacement of opposite ends of a member along one
axis due to opposite unit terminal forces along any other axis is the
elastic product of inertia of the body about these axes. If the axes
are through the centroid, the displacement is an elastic centroidal
product of inertia. If the axes are coincident through the centroid,
the displacement is an elastic centroidal moment of inertia.
In computing the properties of a body as a whole, these centroidal
products of inertia may be treated just as they are treated in the com
putation of the properties of beam sections. The elastic products of
inertia of the analogous column, then, may be computed as the sum
of the elastic areas times the products of their distances from the axes
under consideration, plus the centroidal products of inertia of the
elastic areas.
These centroidal products of inertia are useful also in computing
the elastic moments on the analogous column, though the procedure
has no simple analogue in beam analysis. It has been explained that
known linear displacements may be treated as moment loads about
the axis of displacement. Such displacements are produced by thrusts
and shears on the portions of the elastic body. The displacement
along any axis due to a force along any line equals the product of the
force by the product of inertia of the body about the axis of displace
ment and the line of action of the force, as this product of inertia has
been defined.
The elastic moments on the analogous column may be taken for
each portion of the beam as equal to the moments at the elastic
centroid of that portion times the elastic area plus a couple equal to
the shear on that portion of the beam times the elastic centroidal
moment of inertia. In this way the elastic moments have been com
puted in Fig. 14.
V. PHYSICAL CONSTANTS OF DEFORMATION FOR 1 '
STRUCTURAL MEMBERS
22. Nature of Physical Constants.Thus far consideration of the
physical properties of the members has been restricted. No theory
ILLINOIS ENGINEERING EXPERIMENT STATION
has been propounded to predict the values of the elastic areas. If we
are dealing only with distortions produced by external loads, it does
not make any difference what are the absolute values of these elastic
areas, or moduli of distortion; it is the relative values only that are
wanted.
It seems somewhat unfortunate that the theory used in the anal
ysis of continuous frames has come to be known as the theory of
elasticity. In its simplest form it has nothing to do with elasticity in
the ordinary sense of the word. The theory of elasticity merely states
the geometrical conditions essential to continuity in terms of the
physical properties of the structure. If these physical properties are
known for the conditions of stress which actually exist, then the
theory may be applied. It is thus possible to apply the theory of
continuity in a perfectly definite way to plastic materials, such as
concrete, taking account of the variation of the plastic distortions
with both the intensity and the duration of the stress, provided the
properties of the material are accurately and definitely known.
Thus, suppose an exact analysis of a concrete arch is desired and
that the ratio of total stress to total deformation is known as a func
tion of both intensity and duration of stress. We first assume a
constant value of the ratio of stress to total deformation, which we
will call E, throughout the arch rib, and find all stresses. Since the
value of E is now a variable over each section, we transform the sec
tions as explained previously. Using the transformed sections, the
stresses are again found and the process is repeated to any desired
degree of precision.
We now know accurately the stress conditions at the time of load
ing. After an interval all values of E will be changed and we can re
peat the process just outlined. By successive repetitions we could
trace out the complete stress history of the structure.
Whether we could ever know the physical properties of the mater
ial with enough accuracy to take account of their variations is a ques
tion of fact to be considered separately. Whether the structure is
sensitive enough to such variations in properties to make the variation
in results secured from any such analysis appreciable is also another
matter. Of what importance such information would be in the safe
and economical design of the structure is still another matter.
Qualitative thinking along these lines will disperse certain illusions
which seem to be current as to mysterious results from plastic flow
and from time yield of concrete. The subject will not be pursued
here, since the monograph is restricted to geometrical relations. The
THE COLUMN ANALOGY
important point just now is to clearly distinguish those facts which are
purely geometrical from those which are necessarily a subject for
laboratory equipment.*
23. Method of Determining Physical Constants.Determination of
the elastic constants themselves involves a knowledge of the prop
erties of the material. In general they are to be determined as fol
lows: (a) apply a unit moment at each end of the body and determine
the magnitude of the rotation and the center of rotation (this gives
the elastic area and its centroid); (b) apply unit forces along an
axis Y at the ends of the member and through the elastic centroid and
determine the relative linear displacements of the ends along and
normal to axis Y. Similarly apply unit forces along an axis through
the elastic centroid along axis X, normal to axis Y, and determine
displacement along axis X (this gives Ix, Ixy, I,).
The elastic properties of a body as just definedthe elastic area,
elastic centroid, and elastic products of inertiaare true physical
properties for the stress condition in the body. If they are known for
each portion of the body, they can be computed for the whole. If
certain assumptions are made we can predict the properties of the
individual parts.
In the deductions which follow a constant value of E and the con
servation in bending of plane right sections is assumed. In most
cases these assumptions are very nearly correct. But these values
cannot be predicted with absolute precision because of chance
variations in the properties of the material.
The elastic properties of a body are defined for two ends at which
alone forces are supposed to be applied to the body. For another pair
of termini another set of elastic properties would be deduced.
The elastic properties for a given pair of termini may be deter
mined experimentally as follows:
Hold one of the ends rigidly and apply through a bracket attached
to the other end a vertical force and a horizontal force successively at
each of two points. In each case measure the vertical displacement
and the horizontal displacement of each of two points on the bracket
(see Fig. 23).
We now have sixteen quantities from which to deduce the six
quantities desired. If Hooke's Law holds for the material, only ten
of these quantities will be different and only six will be needed. Many
*For a treatment of these matters, see "Neglected Factors in the Analysis of Stresses in Concrete
Arches," by Lorenz G. Straub, presented as a thesis for the degree of Doctor of Philosophy at the
University of Illinois in 1927 and later published in part as "Plastic Flow in Concrete Arches," Proc., A.
S. C. E., Jan., 1930.
ILLINOIS ENGINEERING EXPERIMENT STATION
FIG. 23. EXPERIMENTAL DETERMINATION OF PHYSICAL CONSTANTS
different combinations of measurements may be used. The six
quantities may all be deflections due to unit forces, as follows:
Vertical at A due to vertical force at A = Ix + Ax2
Vertical at A due to vertical force at B = Ix + Ax (x + xi)
Vertical at A due to horizontal force at A = Ix, + Axy
Vertical at B due to vertical force at B = Ix + A (x + xl)2
Horizontal at A due to horizontal force at A = I, + Ay2
Horizontal at B due to horizontal force at B = I, + A (y + yi)
From these the values of A, x, y, Ix, Iy, and Ixy may be deduced by
simple algebra for known values of xi, yI.*
If Hooke's Law does not hold, different sets of measured displace
ments will not be consistent even though the measurements are made
with absolute accuracy, because the body does not have any one set of
elastic properties but has a different set for each different condition
of stress.
24. Computation of the Constants.Only a few of many possible
illustrations of the computation of elastic constants is given here.
(a) Hinges and Roller Nests
Since forces acting on solid foundations produce no deformation,
the elastic constants for the earth are taken as zero if we assume
immovable abutments.
Eccentric forces acting on a frictionless hinge will produce no
linear movement of the hinge but will produce unlimited rotation;
hence the elastic area of a hinge is infinite, and its elastic centroidal
products of inertia are zero.
Forces acting on a roller nest and inclined to its bed will produce
no rotation, no displacement normal to the bed, and unlimited dis
placement along the bed; hence a roller nest has zero elastic area, zero
*A procedure which is simpler algebraically is as follows: Apply unit moment and measure rotation
and horizontal and vertical displacement of one end. This gives the elastic area and locates the elastic
centroid. The elastic centroidal products of inertia may then be measured directly as displacements
along the centroidal axes due to unit loads through the centroid.
THE COLUMN ANALOGY
FIG. 24. FLEXURAL DISTORTION OF A BEAM
elastic centroidal moment of inertia about an axis normal to the bed,
infinite elastic centroidal moment of inertia about an axis lying in the
bed.
For a free end all elastic constants are infinite. The fact has no
particular significance except that the theorem which states the
column analogy is of perfectly general application to all plane beams,
whether statically determinate or not, both simply supported and
cantilevered.
(b) Flexural, Longitudinal, and Shearing Distortions in Straight Beams
The angle of flexure in a length of straight beam of constant sec
tion subjected to a given bending moment is readily computed by
geometry if sections plane before bending remain plane after bending.
In Fig. 24 let aa be given length L of the beam before bending and
bb after bending. The angle change 4 equals the change in length of
any fiber d divided by its distance from the neutral axis y.
Then 4 =_ d. From the definition of E, d = fL where f is the fiber
y E
f My
stress along any fiber. But   , where It is the moment of inertia
E I,
of the transformed section obtained by dividing the areas of the orig
inal section by their values of E. Then q = for straight beams.
It
Flexure of curved beams is discussed below.
PL
Longitudinal distortion is computed as  for a force through the
.At
centroid of the transformed section.
The shearing distortion per unit of length of beam equals the
shear divided by the continued product of area, the shearing modulus
of elasticity and a factor depending on the shape of the section.
ILLINOIS ENGINEERING EXPERIMENT STATION
(c) Straight Homogeneous Beam of Uniform Section
Consider a straight segment of a homogeneous beam of uniform
section. Let its properties be represented by length = L, area = A,
moment of inertia = I, modulus of elasticity = E, radius of gyra
tion = p.
Apply unit moment, unaccompanied by shear. Rotation is 
El
and centroid of rotations is at the midpoint.
Apply unit transverse shear, otherwise unaccompanied by mo
ment. Transverse displacement of end
(a) due to bending = + fx2da =  L2
12 El
L
(b) due to shearing distortions = + A , where n is the factor
referred to above and G is the shearing modulus of elasticity. Nor
mally, this displacement is unimportant. There is no longitudinal
displacement due to transverse shear.
Apply unit longitudinal force. Longitudinal displacement
L L
= + = + p2  . No transverse displacement.
AE El
The elastic properties, then, are:
Elastic area = 
El
Elastic centroidal moment of inertia about longitudinal axis
1 L L
12 EI L + AGn
Elastic centroidal moment of inertia along longitudinal axis =
AE
Elastic centroidal product of inertia about two axes = 0
Elastic centroid on axis at midpoint.
These six quantities completely describe, on the ordinary assump
tions of mechanics, the elastic properties of this body for terminal
forces.
(d) Bars of Trusses
The strain of a bar in a truss produces a rotation at the moment
A L
center for that bar . If a unit moment acts at that center, 0  .
r AEr
This determines the elastic area.
THE COLUMN ANALOGY
AA
A B
FIG. 25. DEFORMATION CONSTANTS FOR WEB MEMBERS IN A TRUSS
This expression is sometimes inconvenient for web members
where chords are not parallel, because the moment center does not lie
near the panel in which the section for stress is passed. If the chords
are parallel, it leads to indeterminate expressions.
In these cases it is convenient to replace the distant elastic weight
by two elastic weights and an elastic moment of inertia lying in the
L
panel. Thus the true elastic area at 0 is + L. If the areas at A
AEr2
and B are to have that at 0 as a resultant (see Fig. 25)
L bp L 1
a AEr2 p AErlr2 a
L ap L 1
Ab= AEr2 p AErlr2 b
r rl r r2
since  and 
bp p ap p
For product of inertia about axes normal to the lower chord to be
the same as that of Ao,
Io, = 0 = A.(ap)2 + Ab(bp)2 + I
L
AErr2 p2
ILLINOIS ENGINEERING EXPERIMENT STATION
a'dicks to a/0 f'be,
Raedis to c'frO')/
f fhe tri/7s/ormfed
FIG. 26. SEGMENT OF A CURVED BEAM
Evidently no correction for the product of inertia is necessary if
either axis is parallel to the lower chord.
The elastic areas Aa and Ab may be located on either chord and
the moment of inertia taken along that chord provided ap and bp are
computed along that chord. The positive elastic weight always lies
on the side of the panel next to the moment center.
If the chords are parallel, the elastic areas become zero, the mo
L
ment of inertia along the chord is still I = + AErr2 p2, but ri = r2.
This method of treating truss bars is sometimes advantageous for
computing stresses in indeterminate structures. In most cases
indeterminate trusses can be analyzed conveniently by other methods
than the column analogy.
(e). Beams Sharply Curved
Beams having a radius of curvature small compared with their
depth are common in machine parts; in structural design they occur
in thick arched dams and at the haunched junction of beams and
columns. The formula of Winkler, usually presented for the solution
of such beams, leads to certain complications when used to compute
deformations or for the analysis of such beams when they are statically
indeterminate. For this reason, a detailed explanation of the use of
the transformed section seems desirable here.
In Fig. 26 is shown a differential length of a beam having a center
of curvature at 0.
It has been shown that if the section is transformed by dividing
each differential area by its radius of curvature R we can write
P Me
fR = A + 7 y,
THE COLUMN ANALOGY
where f is the fiber stress at any distance yt from the centroid of the
transformed section
R is the radius of curvature of this fiber
P is the normal load
M, is the moment about the centroid of the transformed
section
At is the area of the transformed section
Is is the moment of inertia of the transformed section about
its centroid.
Also let
A = the area of the original section
Ro = the radius of curvature to the centroid of the original
section
Rt = the radius of curvature to the centroid of the trans
formed section.
By definition At = f
Taking moments about 0
I dA
Rf
S dA At
R
The moment of inertia about 0 =
R R2 = RdA = AR.
Reducing this to the centroid of the transformed section
A
It = ARo  AtRt2 = ARo  At  Rt = A(Ro  Rt)
If, then, the area and centroid of the original section are known,
it is necessary to find only the area of the transformed section in order
to compute directly all quantities needed.
The angle of rotation about the centroid of the transformed
section is
yt Eyt
S = f X (length of differential fiber under consideration)
But f Mtt
IR
ILLINOIS ENGINEERING EXPERIMENT STATION
R
Length of any fiber = R ds
Where ds is the length of the differential fiber along the axis of the
centroid of the transformed section.
ds
Hence  EI= R is the elastic area of the segment, or in general
A = angle of arc
EIt
If M, = 0, the section moves parallel to itself and
P P
A E(A ds = AE ds
E (AtR) AE
ds
Hence d is the elastic centroidal moment of inertia for ribshorten
EA
ing correction.
It will be seen, then, that all expressions for beams of sharp curva
ture take the same form as where the beams are straight if the axis of
the beam is taken as the axis defined by the centroids of the trans
formed sections instead of by the centroids of the original sections.
(f) Compound MembersBifurcated Members
The elastic properties of any elastic ring may be defined with
reference to the termini from the principles indicated. Thus, to
determine the elastic properties of the ring ABCDA, Fig. 27, with
reference to terminals A C, apply at C a unit moment, the ring being
cut at C and held at A. Now treat as a column section cut at A and
uniformly loaded along ABC and compute the shear at C and the
bending moment on vertical and horizontal sections through C. This
gives the elastic weight and its static moments about C, from which
the co6rdinates of the centroid may be computed.
Next apply at C a unit vertical force, the ring being cut at C and
held at A. Apply the column analogy and compute the bending
moments on the analogous columns on horizontal and vertical sec
tions at C, the section being cut at A. This gives moment of inertia
for vertical axis through C and product of inertia for vertical and
horizontal axes through C. These may then be reduced to the
centroid.
Finally apply at C a unit horizontal force, the ring being cut at C
and held at A. Compute the bending moment on a horizontal section
through C on the analogous column section cut at A. This is the
THE COLUMN ANALOGY
moment of inertia for a horizontal axis through C. This may now
be reduced to the centroid.
If loads occur within the ring, the elastic load on the ring may be
determined from the shear (elastic load) and moments on vertical
and horizontal axes through C (static moments of the elastic load
about C) for the elastic column section if cut at A as above.
The elastic load and its point of application and also the elastic
properties of the ring being known, the ring may be treated as is
any other member.
This procedure is relatively simple but involves a good deal of
computation. The problem, however, is not a very simple one.
Anyone who has occasion to use the method for numerical problems
will find that it leads to comparatively simple expressions.
(g) Successive Compounding
The compounding procedure just indicated may be extended in
definitely to include any number of branches. The most common and
important case is that of a series of continuous arches or bents. The
general procedure by this method in such a case is as follows:
(a) Apply a unit moment at the junction point of the outside
bent, find the elastic area and elastic centroid.
(b) Apply a unit vertical force at the junction and find the mo
ment of inertia about a vertical axis through this junction and the
product of inertia about horizontal and vertical axes through this
point. Reduce these to the centroid.
(c) Apply a unit horizontal force at the junction and find the
centroidal moment of inertia of the combination for a horizontal axis
through the junction. Reduce this to the centroid.
(d) Substitute these elastic properties for those of the pier in the
next bent; proceed as above and continue to include any desired
number of bents.
(e) In a similar way the elastic loads are to be combined for one
bent after another.
(f) The reactions having been determined for the last bent of the
series (which may be the center bent, combinations having been made
from both ends) these can be resolved back successively through the
series.
This method has enough value to justify reference to it, though,
even in the case of continuous arches, other methods are more
convenient.
ILLINOIS ENGINEERING EXPERIMENT STATION
A
FIG. 27. CLOSED ELASTIC RING
VI. APPLICATIONS OF THEOREM
25. Fields of Application of Theorem.The theorem here pre
sented has several fields of application:
(1) In the routine analysis of symmetrical and unsymmetrical
arches and bents for loads, either gravity or inclined, for temperature,
for shrinkage, or for abutment displacement.
(2) In determining moments at the ends of beams fixed at ends
and in determining other properties of such beams, such as the end
moments corresponding to unit rotation at end, for use in connection
with various methods of analysis of continuous girders or frames.*
(a) It is possible thus to determine constants for use in the
general equations of displacements. These equations state that
terminal forces and moments on a member are the sum of those
due to known loads on the member or distortions in the member
when the ends are fixed and to any displacements or rotations of
the ends, known or to be determined. A special case of these
equations, applicable where the members are straight and of uni
form section, is known in American literature as the equation of
slopedeflection.
Values of terminal forces in terms of the unknown terminal
displacements may be substituted in the equations of static
equilibrium of the joints. These equations may then be solved
for the terminal displacements.
When the joint displacements have been found the terminal
forces may be computed from the original equations of displace
ment.
(b) Constants may be determined for equations which state
the existence of continuity at the joints. Of these the theorems of
*Loo. cit., page 48.
THE COLUMN ANALOGY
three moments and of four moments are special examples of
limited application.
(c) Terminal forces may be determined for known loads or
internal distortions on the assumption that the joints are not dis
placed and then the unbalanced terminal forces may be distrib
uted among the connecting members in proportion to their re
sistance to end displacement. In a special case this is done by the
method of moment distribution. The constants needed may be
determined by the column analogy.
(d) In routine computations of slopes and deflections.
26. Methods of Analysis of Continuous Frames.This bulletin does
not deal primarily with methods of analysis of continuous frames, but
only with the analysis and elastic properties of the individual mem
bers of which the frame may be made up. Some comments on meth
ods of analysis of such frames is needed to explain applications of the
column analogy indicated.
Methods of analysis of continuous frames may be divided into
those involving internal work of the frame, such as the method of
least work, and those involving the geometry of continuity. That all
methods are really the same will at once be realized but their relations
to each other will not be discussed here.
Of the geometrical methods of analysis we may distinguish those
in which the displacements of the joints are treated as unknowns and
those in which the terminal forces acting on the members at the joints
are treated as unknowns. The former are represented in a special case
by the method of slopedeflection; the latter are represented by the
theorem of three moments and by the theorem of four moments.
The method in which the joint displacements are treated as un
knowns might be called the Method of Joint Displacements. In this
method the terminal shears and moments are written in terms of the
joint displacements. The equations of staticsthe forces balance at
each jointare then written for each joint. From these the terminal
displacement is computed. The terminal displacement being known,
the shears and moments may be found.
In the other method of analysis the terminal slopes and displace
ments are stated in terms of the terminal shears and moments. The
equations of continuity which state that the displacement of a joint
is common to the ends of all members meeting at the joint are then
written for each joint. From these equations the terminal shears and
moments are found directly. The equation of three moments is a
familiar illustration of an equation of joint continuity. A less familiar
ILLINOIS ENGINEERING EXPERIMENT STATION
illustration is the equation of four moments. This method might be
called the Method of Continuity.
It is well to distinguish the method of slopedeflection from the
equation of slopedeflection; the method, which seems to have been
restricted to straight members, consists in treating joint rotations and
displacements as unknowns in equations of static equilibrium for
the joints. The equation of slopedeflection is merely one of many
forms of the equation which relates the terminal forces to the loads
and terminal displacements of a straight beam. The method of slope
deflection may be used without using the particular form of equation
known as the equation of slopedeflection or the equation of slope
deflection may be used without using the method of slopedeflection.
If we write the end moments and forces in terms of loads and end
displacements, we can derive from these expressions for the end ro
tations and displacements in terms of loads and end moments and
forces. Displacements of the ends of all members at a joint are equal
and angular rotation at the ends of all members meeting at a joint are
equal. These displacements and rotations being in terms of end mo
ments and forces on the beams and of the physical properties of the
beams, the equations can be solved for the end moments and forces.
The Method of Terminal Force Distribution will not be discussed
here. It is more closely related to the Method of Continuity than to
the Method of Joint Displacements.
27. General Equation of Displacements and SlopeDeflection.The
general equations of end forces at any joint, A, of a structure may be
written as follows, provided, as is usually true, the principal axes of
the members are parallel and normal to each other:
da
Ma = OaNa  kbraNa + (Aa  Ab) y + M'a
F. A  A +b +Fad.a bdb + a,
Jo Io
Ma and Fa are total end moment and end force (thrust or
shear) in the member
(Note that in general there will be two equations of force at
any joint, one for horizontal and one for vertical forces.)
M'a = moment which would exist in the member if there were
no rotation or displacement of the joints.
F'a = force which would exist in the member if there were no
rotation or displacement of the joints.
4a, Aa are respectively rotation and displacement at the joint
considered.
THE COLUMN ANALOGY 73
kb, Ab are corresponding quantities at the other end, B, of each
member successively.
Na is the moment at joint A corresponding to a unit rotation of
this joint, the other end being fixed.
ra is the carryover factor at A (the ratio of the moment at B
due to a unit rotation at A to the moment at A due to such
a rotation.)
Io is the elastic moment of inertia about the centroid for any
member.
d is the distance of either end from the centroidal axis of a
member.
From these general forms and the equations of static equilibrium
we may write the equations for every joint in a complex structure
such as a continuous arch series or a Vierendeel truss. These equa
tions may then be solved simultaneously for displacements and from
these the moments, shears and thrusts may be determined.
The use of such equations, requiring a carefully selected conven
tion of signs with resulting possibility of error from this source and
involving simultaneous solution, is to be thought of as a research tool
and rarely as a tool of design.
If there are no joint displacements, or if it is convenient to make
separate allowance for such displacements, the process is much
simplified. We then have, from 2M = 0,
a M'= a + braNa
 Na zNa
the signs depending on the convention used.
This, perhaps, is more conveniently written,
M'/a raNa
0=  + Z 0
If connecting members are treated as prismatic,
I
N. = 4 r = V2
I
M'/a L
I I
2Z  2 
L L
74 ILLINOIS ENGINEERING EXPERIMENT STATION
If, further, the values M'a are due to known rotations of the bars,
6A I
M'a  L  6E  P = 6EK4I
L
IE
K K
'a = 23 24 K  20b 2ZK
which expression is convenient in finding secondary stresses in trusses.
Or this may be derived from the fundamental equation above
d
SM'0 ^4akrbNa 2(Aa  Ab)
'ka ~Na + ;Na ZKa
M'a = 0, N = 4 , r =  , Aa = 
d I 1
I L L
3ZK # SK #b
a = 22K 2ZK
These discussions include any combination of bars of any shape
or form provided the axes are parallel or normal to each other,, and it
is not very difficult to extend the method to include skewed axes.
The slopedeflection equation for prismatic beams,
2EI
M = (2a + kb  3P)
may be derived in a number of ways and follows from the first general
equation when
L2
Ao  Ab d2 4
Na = 4K, ra = Y, =  2d , I L L = 3K
12 1
The interest in the equations at present, however, is chiefly in the
utility of the column analogy in evaluating the constants M'a, F'a,
da, Io, Na, and V. for the members.
THE COLUMN ANALOGY 75
VII. CONCLUSION
28. Conclusion.The paper is restricted to geometrical relations
and does not discuss applications to design. Its thesis is simply that
moments, shears, slopes, and deflections of beams due to any cause
may be computed in just the same way and by the same formulas as
are used in computing reactions on a short column eccentrically
loaded or as would be used to compute shears and bending moments
on longitudinal sections through such a column.
The column analogy is a convenient tool of mechanics, a somewhat
mechanical device for a structural engineer, but one which seems to
give desired results with a minimum of thought as to method of
procedure or sign conventions. In the study of continuous frames it
becomes auxiliary to methods for the analysis of such frames.
The most obvious application is in the analysis of single span
structures; perhaps its most important applications occur in the study
of continuous bents, arches, and beams.