Auburn University Libraries
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3 1706 025 84833 9 '
File D 52.16/ 17 McCOOK FIELD REPORT SER IAL No. 1462
AIR SERVICE INFORMATION CIRCULAR
Vol. III
(AVIATION)
PUBLISHED BY THE CHIEF OF AIR SERVICE,.WASHINGTON, D. C.
April 30, 1921 No. 213
DEFLECTION OF BEAMS OF
NONUNIFORM SECTION
(AIRPLANE SECTION, S. & A. BRANCH REPORT)
Prepared by Engineering Division, Air Service
McCook Field, December 10, 1926
WASHINGTON
GOVERNMENT PRINTING OFFICE
1921
•
Ralph Brown Draughon
LIBRARY
MAR 28 2013
Non·Depoitory
Auburn University
J
DEFLECTION OF BEAMS OF NONUNIFORM SECTION.
The purpose of this report is to expl!i.in and illustrate
methods• of calculating the deflection of beams of nonuniform
section . The application of these methods to
airplane analyeis is given in article 17:{ of Structural
Deflign and Analysis of Airplanes. All methods of computing
the deflections of beams are based on the fundamental
differential equation of the. elastic curve of a be11m
under load :
(l•)
Where
:,; is measured alon!!' an axis tangent to the neutral axis
of the beam, ·
y is measured along an axis perpendicular to the axis
of x,
Mis the bending moment'acting on the beam,
Eis the modulus of elasticity of the beam,
I is the moment of inertia of the beam.
The methods considered in this report are based on t~e~
special theorem developed from equation (1), that the
deflecti.on of any point in a beam, x, from a line tapgent
to the elastic curve of the beam at any point x0, is equal
M
to the moment. of the area of the EI curve of the beam
between the two poinfu, about x. The proof of equation
(1) may be fo1md on pages 130134 of Boyd's Strength of
Materials. The proof of the special case is on pages 153
156' of the same work.
The graphical and analytical methods shown in this
report are logically identical, as both consist of computing
the moment of portions of the :r. curve about poinui
whose deflections are desired. It would be perfectly
poBBible to make part of the computations analytically
anl a part graphically, and in some cases that is very
desirable.
It, should he thoroughly understood that the deflections
obtained by theRe methods are deflections fyom a tangent
to the elastic curve of the beam. This is not generally
true of the deflection_ formula.fl found in the handbooks.
'!'he deflections found frem the formulas are measured
from a line, pB.BBing throug;h the supports, which .line, 1n
general, is not a tangent to the elastic curve. In such
cases the deflection from tc.e tan!!'ent must be modified
in a manner to be explained in order to obtain the deflection
from the line through the stlpporui. In the case of a
cantilever and of other be!l.ms so · fixed that the elastic
curve is horizontal over the supporui no modification is
necessary.
GRAPWCAL METHOD.
distance. of the force polygon, and a factor depending on
the scales of the force and funicular poly11:on (p. 343,
Spofford 's Tl).eory of Structure). Af.'. all the forces in this
work are vertical, the intercept in question "'ill be on a
· vertical line through the point.
In order to illustrate the application of this method an
example will be worked through. ABBU:D'le a beam 210
inches long, supporting a load varying uniforni.ly from ·7
pounds per inch to 21 pounds per inch, as shown by the
line ab in" figure 1. ABBume this beam to be supported
75 inches from the lPft end and at the riiht end. Assume
the moment at the righthand suppor~ to be zero. Asswne
th_e moment of inertia of the beam to vary as shown
in.figure 2. .
The first step is to ob:t.ain the moment curve from the
loading curve. To do this, divide the beam. int,o any
convenient number of sections. In Viis example the
beain is divided into 14 sections, each 15 inches long.
Any other number" of sections C'OUld be used, and if desired
they might be of unequal length. The reason for choosing
15 inches as the section length was· that it. gave a convenient
number of sections and each section has a constant
moment of inertia. Draw vertical lines through the
centers of the sections. These lines will represent closely
the lines of action of. the resultant loads on each section.
Draw the 'vertical line 014 in figure 3 and lay off on it.
01, 1 2, 23, etc., proportional to the loads on sections
1, 2, 3, etc., in figure L Select any conven.i.ent pole
distance, P, and draw the· "rays" P0, P1, P 2, etc.,
completing the force polygon. Construct t.he funicular
polygol!, figure 4. From any point, c, on the line of action
of force 1, draw "strinl/:8" parallel to P0 and P1 of
fi gure 3. From the intersection of the string Pl parallel
to· ray P1 and the line of action of force 2 draw a string
. P2 para.Ile! to ray P2. Continue until string Pl4 is
drawn from the line of action of force 14 to the line of
action of the rightha0:d reaction and parallel to ray P14.
Draw strin!!' PR from the intersection of PO and the line
of action of the lefthand reaction to the intersection of
PO a.i:id the .line ·of action of the leftha.rid reaction to the
intersection of Pl4 and t;he line of action of the righthii,nd
reaction. Draw the ray PR. ln this discUBBion a ·ray
is always a line in the force polygon, figure 3, and a string
iii a line in the funicular polygon, · figure 4. Every string
of the funicular polygon is parallel · to the ray of for,:e
polygon with thP. same name. · Tlte magnitude of the
lefthand reaction R1 represented in figure ~ by R0
and the righthand reaction by 14R. It is usually convenient
to compute the values of R1 .and R:i analytically
as a check on the graphical work at this stage. In this
case, taking momenui about R:,,
.The graphical method depends on th~ theorem generally R1Xl35=7X210X105+7X210X70=7X210Xl75.
u<ied f~r finding graphicaliy the moment of a number of . R,=1,905.5 pounds,
coplanar forces a.bout a point in that p~ne. The moment R2=14 X2101,905.5=1,034.5.
about any point of any number qf coplanar forces is,equal By scaling from figure 3
to the product of the iptercept on a line thro~h the point R]=l,912 pounds.
and para.Ile~ to the resultant of the forces, between the R2=1,028 pounds.
strings holding the resultant in equilibrium, the pole Error 6.5 pounds, or lj50 inch.
~21 (3)
s 6
3 4
(
/{./
60
so 5 ~
/'10 l/7L NT
~~
\~
\
~~
~
~I'\ ~"' \ '<:,..
... "N \ "~
\
\ I
V
4
IE'.
/0 · II
1 8 9
d
lO 110
Ft G./
75 60
40
OF IN UI TIii
Fl 6Z.
I
V
V
I )
p V
lY
I e ;
I ~ :y I !/I / ' f
' /,, .e.i.! /
7
I
F/_6.4
FIGS. 1, 2, 3, and 4.
13 /4 b
7S
V ~
I
/~
v
~
"
· ,4
FIG.!J
I/
This error is due in part at leaet to the fact that the can be shown to represent the bending moment ·at that
line of action of the load on each tiection was assumed to point on the beam. It is therefore evident that if the
be at the center of the section instead oi slightly to the divisions were made infinitesimal, the broken line Plleft
of the canter. This tends to make the value of R1 P2P3, etc., would become a smooth curve inscribed in
t.oo large. The funicular polygon of figure 4 is a bending that line.
moment diagram of the beam in figure 1 subjected to the Figure 5 is the M/EI curve of the beam. The ordinates
14 concentrated loads, one on ·each 15inch section. It of this curve are the ordinates from figure 4 divided by a
is evident that the bending moment curve for the uniform factor proportional to the moment of inertia .at the same
load is a .curve ins~ribed in the polygon shown. Thcl point, as shown in figure 2. The ordinates from which
proof of this is as follows: Take the division point between {he curve was plotted are those at the division points
any two sections, as point d_. between sections 9 and 10. b_etween the arbitrary sections into which the beam was
The forces to the right of dare 10, 11, 12, 18, '14, and R2, 'divided for purposes of computation. The sudden breaks
They are represented in figure 3 by 914 and 14R. This_ in the curve correspond to the sudden changes in the
result.ant is 9R, which is held in equilibrium by the rays .m.oment, 6f. inertia.
P9 and P~R. In the funicular polygon ef is the inter Figure 6 and figure 7 are developed from figure 5 in
cept on a vertical line tnrough d, between the strings P9 exactly the same manner that figure 3 and figure 4 were
and P7 R, and is therefore a measure of the bending moment developed from figure 1. A smooth curve inscribed in
at d. In a similar manner the ordinate of the funicular the broken line ghij represents the elastic curve of the
polygon of figure 4 at any division point between sections beam.

5
p
FIGS. 5, 6, and 7.
There are two modifications to this method which are The sc,cond modification is in the method used to allow
illustrated in figures 8, 9, and 10. One of these modifi for the . variation in I. In figure 5 the curve M/EI is
cations, shown in figure 9, is in the method of allowing p1otted, while in figure 8 the curv.e oi M is used. The
for a change in sign of M/EI. In figure 6 areas 1 to 7a ordinates. of the curve of figure S·are the same .as those in
are positive and are plotted from O down to 7a. Areas figure 4. The only difference between figures 4 and ·8 is
7b to 14 are negative and are plotted from 7a up to 14. that in the latter the base line is made horizontal for
All the rays are drawn from the same pole P. In figure 9 convenience. The areas of figure 8 are laid off as lengths
areas 1 to 7a are also plotted downward from o to 'fa and in figure 9 exactly as figure 3 was constructed from figure 1.
. rays are drawn from poles on the righthand side of the The · pole distance, however, is varied in proportion to I.
line 014. The ray p 2
_ 7a is prolonged to Pa,' making Thus the value of I for the first four sections of the beam
is 50, so the rays Pc0, P,1, P,2, P,3, and Pi4 are
P27a equal to P37a. The areas 7b to 14 are also plotted drawn Irom a pole 50 units.from the line 014. Between
downward, the rays being drawn from poles like Pa on sections 4 and 5, I increases to 60, so the ray P,4 is
the lefthand side of 014. It is easily seen that a given extended to P2, 60 units from 014 and the rays P.5 to
ray has the same direc;tion whichever way the force p
3
_g are drawn from poles 60 units from 014. The pole
poJygon is constructed. Suppose 7a7b in figure 9 had was changed from p
2
on the right to P
3
on the left after
been laid off from 7a up to 7b'. To prove that P27b' is drawing the ray ·P27aPa on account of the change in
parallel to P37b. In the triangle P27a7b' and P37a7b, sign of M. In a similar manner the poles P4 and P
5
are
Pa7a equals P2 7a, 7a7b equals 7a,7b' , and the angle located 75 and 40 units, resp~ctively, from 014 for the
P37a7b equals angle P27a7b'. Therefore P37b and sections oi the beam where the values of I are 75 and 40.
P27b' are parallel, the alternate interior angles being I The validity of the construction will be shown in the
equal, Q.KD. discussion of scales.
60684 0312
6
PROOF.
The proof that the curYe ghij of figure , represents the
elastic curYe of the beam is as follows : Assume the number
of sections in to which the area of the M/EI curve, figure 5,
is di,ided to be indefinitely increased. As the areas of
the indi,idual sections approach zero as a limit, the
funicular polygon becomes a curve inscribed in the broken
line ghij . Each tangent to this curve, as klm, will be the
string of the funicular polygon between the lines of action
oi the differential areas of thE. M/EI curve on each side
of t.he point of tangency and will be parallel to t.!:.e rays
of the force polygon holding in equilibrium the differential
area at the point of tangency. In figure 7 as <lrawn with
the beam di,ided into fourteen 15inch sections, klm is
t.he string between the lines of action of forces 9 and 10.
These are not t.rue forces as in figure 1, but rather areas
of the M/EI cune called forces by analogy to the construction
in figures 1 to 4. It is parallel to the ray P9
of figure 6, which is determined by the total area of the
M/EI C\IrYe up to the point d, figure 5, at the dividing
line between sections 9 and 10. No matter how many
sections the beam is di,ided into, the ray representing
the total area up to ·any given point, such as d, will remain
unchanged. The string klm, therefore, will continue to
be the string from the line of action of the area to the
left of d, to the area to the right of d, even when these
areas are oi diiierential size.
In figure 6 the area to the left of d is represented by
89 and is held in equilibrium by the rays P8 and P9.
As this area approaches zero, the two rays holding it in
equilibrium approach each other until they coincid e.
Thus the ray P 9 can be considered as two rays holding
in equilibrium a differential force at 9, figure 6 representing
a differential area at d, figure 5.
Consider any t.wo tangents to the curve ghij, figure 7,
as klm and ohp. These tangents are strings which hold
in equilibrium the forces representing the area of the
~I;EI curve between d and r, figure 5. The intercept on
an:, Yertical line between the lines klm and ohp ·will be
proportional to tJ1e moment of the area of the M/EI cune
between d and r, about the point where the vertical in
question intersects the beam. The intercept on a vertical
through one oi the points oi t.angenc~ ,\ill represent the
moment oi the area oi the sect.ion oi the )f/EI curve about
that end of the section . Whence, in general , t.he intercept
on any Yerti cal, between any tangent and the curYe, rep resents
the moment of the section oi t.he M/EI curve
between the point oi tangency of the tangent and the
Yertical , about the point on the beam located by the
vertical. Thereiore, from the original theorem, the curve
in figu re 7 represents the elastic cune oi the beam a.~ the
ordinate irom any point on it to any tangent to it is pro.
portional to the deflection oi the point from that tangent.
to the ordinates to the curve from a line, shj, drawn so as
to intersect the curve at the points representing the supports.
If the cantilever end were being considered by
itself, the deflections would probably be desired from the
line ohp, tangent to the elastic curve at the base of the
cantilever. It should be remembered in plotting a
reference line that a horizontal line will not in general
appear on the funicular polygon as horizontal. Therefore,
any reference line must be located by two known deflec "
tions or by the fact that it is tangent to the curve at a
known point or the deflection at one point is known and
also the fact that it is parallel to the tangent at some other
point. This last case occurs in finding the deflection
of an upper wing spar of an airplane like the Fokker D7,
where the deflections desired are those measured from a
line cutting the curve at the cabane strut point and
parallel to the tangent at the center line of the airplane.
Figures 8 and 10 are lettered in the same way as figures
5 and 7 and they may be referred to in the above proof
with identical results.
COMPUTATION OF SCALES.
One of the most important factors in the computati.on
of deflections by this method is the computation of the
correct sca,le. In figures 1 and 5, let 1 inch=q inches
for the linear scale_ In figure 1, let 1 inch=p pounds
per inch run.
Then l squareinch in figure l=pq pounds
For figure 3, let 1 inch=n square inch from figure 1.
Then in figure 3, 1 inch=n . p · q pounds.
If the pole distance in figure 3 is h inches, the scale of
ordinates in the bending moment polygon, figure 4, is
1 inch=npq2h inchpounds.
If the ordinates of the moment polygon are now divided
by the ordinates to the moment of inertia curve, and the
quotients plotted to a scale of 1 inch=m iuch of modified
bending moment ordinates, the scale for figure 5 is 1
inch=mnpq2h pounds per inches.3 Therefore, 1 square
inch of figure 5=mnp<f h pounds per inches. 2
For figure 6, let 1 inch=r square inch from figure 5=
rmnpq3h pounds per inches.2 If the pole distance in
figure 6 is hi inches, the scale of ordinates for figure 7 is ·
1 inch=rmnpq4hh1 pounds per in<::h. For deflections
this should be divided by E, so the scale of deflection
will be 1 inch=rmnpq4hhrfE inches. In the example
computation in this reportr=
0.5.
m=0.02 the reciprocal of the value of I for which
the ordinates of the curves of figures
4 and 5 are the samen
=0. 5.
p = 21 pounds per inch .
q =30 inches.
In order to find the deflection from the tangent to the
beam at any point it is necessary only to draw a tangent
to the elastic curve in figure 7 at that point and multiply 1
the ordinates from that tangent to the curve l>y the proper 1
scale. The deflections irom any other line can be simi1
larly obtained by plotting the line in the proper position I
and measuring the ordinates from it to the curve. Thus,
in a case similar to the one used for an example the deflection
would probably be desired from a line passing through
the two supports. The deflection will be proportional
h =2 inches.
h1 =2.5 inches.
Let E =1,600,000.
The scale of deflections therefore isl
inchnnnpq4hh1_0.02 X0.5X21X304X2.5 0_266 inch.
 E  1,600,000
The ordinates of figure 8 are the same as those of tigure 4.
Therefore, 1 inch=npq2h inchpounds and 1 square inch=
npq3h poundsinches.2
J
)
d1i, ~ .. _ .. ,.id
) ) )
,,/// \
\ .I ...........,,.,,.,, p~ 7 _.,,,./ ~  r I\ 16 8 9 d /0 II le. /3 14
I 2. 3 4 s 6 7 0\ d V • I\
[\ V " V ~ r. / ........ / _/
If I_G. B
~
/>4
....... .,,:::. .
i:, "'
L i :::::i
'~ < i, V  ..... r",.
I .. .,/  "
;:::::,,,  i,.,' .,.,....v ~ ~ ............. ,__ ~.... i\ ~ ~ ~
 _....... ~h  i\  .
v 
vv ~
..... r/G./0
'14
Fros. 8, 9, and 10.
8
Let 1 inch in figure 9 represent r square inch of figure 8.
Then 1 inch=rnpq3h pounds inches.2
The pole distance in figure 9 varies as the moment of
inertia and may be called kl.
The scale of deflection in figure 10 will then be
1 . h kirnpq4h krnpq4h . h
me = EI E me es.
Either the :r curve c:i,n be computed and further work
based on it, or the ¥ curve can be used and ,he momen~
of it divided by the value of E as the last operation before
getting the deflection. The latter method is usually
preferable, as it gives more convenient figures to work
with.
Moment,e_ are generally most easily obtained by the
In the present case the pole distance=0.05 I. There method of increment,e. This is based on the general
fore, k=0.05. The scale of deflection is formula
l . h=0.05X0.5X0.5X21X304X2 O 266 . h
me 1,600,000 . me .
The scale of figure 10 varies ·as the pole distance in
figure 9, and the deflections represented in figure 10 vary
inversely as the moment of inertia. Therefore, if the pole
distance varies directly as the moment of inertia, the ·
deflections will be repr€sented in figure 10 to a cons~nt
scale.
ANALYTICAL METHOD.
In the analytica~ method, the : curve is computed from
the load curve and is in turn treated aa a loading curve
and moment,e are figured from it. In this method it is
impossible to obtain the deflections from a line other than
a tangent to the elastic curve directly. The deflections
of the elastic curve and the desired base line must both
be computed from a tangent and the result,e l!,dded or
subtracted.
Where
Mb is the moment at point b,
Ma is the moment at p9int a,
Sa is the shear at point a,
Xis the distance from a to b,
Pis the sum of the loads between'a and b,
X0 is the distance ·from b to the resultant of the
loads P.
The beam is divided into sections and the moment found
at each division point between sections. The work can be
most eaaily done if the work is done in tabular form.
. M
Table I shows the computations for the moment and T
of the same example aa waa used to illustrate the graphical
method.
R1Xl35=7X210Xl05+7X210X70=7X210Xl 75.
l,470Xl75
R1 135 1,905.5 pounds.
~=1,034.5 pounds.
TABLE I.
Sta. w. Sum. AL. AS. .s. I ASXo. SAL. M. i I. MJI.

I
1 2 3 4 5 6 7 8 9 10 1:1
   
0 7. 00 0 .o I 50 0
15
15. 00 i5 112. 5 825.0 0
8.Q() 112.5 s25.o I 50 16. 5
17. 00 15 127.5 937.5 1667.5
30 9.00 240. 0 3450.0 50 69.0
19. 00
151
142.5 1050.0 3600.0
45 10. 00 382. 5 8100.0. 50 162.0
21.00 15 157. 5 1162. 5 5737.5
60 11. 00 540.C
8100.0 I 15000.0 { 50 300.0
15 I 60 250.0
23. 00 172.5 1275. 0 i
75 } 12. 00
15 1
1905.5 112.5 24375.0 60 406. 25
R1
187. 5 1
1193.0
25.00 11so5. o I
90 13.00 1387. 51
202. 5 1
 1005.5 7867.5 60 131.125
27.00 1~ I 1500. 0 15082. 5
105 14. 00  803.0 5715.0 60  95.25
29.00 15 217.5 1612.5 12045.0 '
120 15.00  585.5 16147.5 { 60 269. 125
31.00 15 232.5 1725. 0  !l782.5 75 215,3
135 16.00  353.0 23205.0 75 309.4
33.00 15 247.5 1837.5  5295.0
150 17.00  105.5 26662.5 "15 355.5
35. 00 15 262.5 1950.0  1582.5
165 18.00 157.0 26295.0 75 350.6
37.00 15 277.5 2062.5 2355.0
180 · 19.00 434.5 21877. 5 { 75 291.7
39.00 15 292.5 2175. 0 6517.5 40 546. 9375
195 20.00 727.0 13185.0 40 329. 625
41.00 15 307.5 2287.5 10905.0
210 '.?L 00 1034.51
7. 5 error. 40 0
21787. 5 21780. 0
Table I is arranged in the following form:
·Column l, is headed "Sta." and gives the distance from
the left end of the beam.
Colum.n 2 gives the loading in pounds per inch run at
the corresponding stations.
Column. 3 gives the sum of the adjacent figures in column
2.
Column 4 gives the distance between stationsin this
example 15 inches in every =e, but it is not at all necessary
to have the stations equidistant.
9
Column 5 is onehalf the product of the corresponding
values in columns 3 and 4. It is the load in pounds
carried by the section of the beam between adjacent stations.
:it is the quantity represented in the formula by P.
Column 6 gives the shear at each station and is the accumulated
sum of the values in column 5. It is represented
in the formula by s •.
Column 7 is the value of the section load from column
5 multiplied by the distance from the righthand end of
the section to the center of gravity of the load.. Usually
it is sufficiently accurate to assume the center of gravity
at the center of the se~tion, but in this case the true center
of gravity was used. "This is the quantity PX0 of the
formula.
Column 8 gives the value s. X, the product of the shear
at one station by the distance to the ne:xt station.
Column 9 gives the moment at the various stations.
It j.s the accumulated. sum of columns 7 and 8.
Column 10 gives the values of the moment of inertia
at the different,atations.
Column 11 is the quotient of the values of column ·9
divided by those in column 10.
This method of computing moments "by increments "
is extremely useful where the load curve is irregular.
It can usually be very easily checked. I.n this case it
is evident that the moment at station 210, the righthand
reac_tion, is zero. The method of increments gives ,it,
working from the left end, as +7.5 inchpounds corresponding
to an error of 7 .5/13.5=0.55 . pounds in the computation
of the reactions, which is negligible. If desired
the error can easily be distributed among the values for
the moment at the other stations.
The size of section that should be used varies with the
smoothness of the load cune, being relatively large for
smooth· curves. It ie advisable to have a division point
at every point at which the moment may be desired in
the further computation and every point where there is
a concentrated load or a sharp break in the loading curve,
or where the load curves passes through zero. There is
no necessity of having the sections equal, though it is
usually convenient in the computations,
TABLE II.
Sta. M[I. Sum. '1~. '1S. s. x,. <1SXo. S'1L. M. a, a,. a,.
   ~
75 406.25 0 0 0 0 0 0 0
537. 37 15 4030.3 8. 78
90 131.12 4030.3 35,386 0 35,386 0.02211 0.07000 0.09211
131.12 8. 5 557.3 5.61
98.5 0 4587.6 3,160 M,258 72,804 .04550 •. 10966 .15516
 95.25 6.5  309.5 2.16
105  95.25 4278. 1  668 29,819 101,955 .06372 .13999 .20371
364.37 15  2732.8 6.30
120 269.12 1545. 3  17,217 64,172 148,910 .09307 .20999 .30306
215.30 524. 70 15  3935.3 7.04
135 309. 40  2390.0  27,705 .· 23,180 144; 385 .09024 .27998 .37022
664. 90 15  4986.8 7.32
150 355.50 .  7376.8 ~ 36,503  35,850 72,032 .04502 .34998 .39500
706.10 15  .5295. 7 7.49
165 ~350.6() 12672.5  39,665 110,652  78,285  .04893 • 41997 .37104
642.30 15  4817.3 7. 73
180 291. 70 17489.8  37,238 190,088  305,611 .19101 .48997 •. 29800
5411.94 876.·56 15  6574.2 8.26 . 195 329.62 24064.0 : 54,302 262, 347  622,260 .38891 .55996 .17105
329.62 15  2472.1 10.00
210 0 26536.1  24,721 360,960 1007, 941 .62996 .62996 0
26636.1 199,473 808,468
TABLE III.
Sta. M/I. Sum. '1L. '1S. s. x,. '1SX,. S.1L.· M. a, a,. a,.

75 406. 25 0 0 0 O' 0 0 0
656.25 15 41l1H.9 8.10
60 .250. 00 4921.9 39,867 0 39,867 0.02492. 0.07000 0.()(508
I 300.00 462. 00 15 3466.0 8.25
45 162. 00 8386. 9 25,586 73,829 142,282 .08893  .13999  .05106
231.00 15 1732. 5 8.50 I 30 69.00 10119.4 H,726 125,804 282,812 .17676 .20999 ~ .03323
I
16. 50
85.50 15 641.2 9.85
I 15 10760.6 5,803 151,791 440,406 .27525  . 27998  .00473
16.50 15 123.8 10.00
I 0 0 10884.4 1,238 161,409 603,053 .31691  .34998 .02693
10
In Tables II a.nd III the column headed o1 gives the
deflection from a tangent to the elastic curve at the lefthand
reaction. The column headed o2 gives the vertical
distance from the tangent at the lefthand reaction to a
line through both supports. The cof umn headed o3 gives
the deflection irom the line through the two supports.
Tables II and III give the computations for the elastic
curve of the beam, Table II for the sections between the
supports and Table III for the cantilever. In both cases
the deflections are first computed for the tangent at the
lefthand support station 75. This is given in column 11.
Knowing the deflection of the righthand support from the
tangent it is a simple matter to compute the distance
between the tangent and a line through the supporls !l.t
any point. These values are given in column 12. Column
13 gives the deflection of the beam from the line joining
the supports which is the algebraic sum of the values in
columns 11 and 12. It should be noted that in Table II
the sections are not all.of the same length. The 15'inches
between stations 90 and 105 was divided into two sections
at station 98.5 because the values of M/L change sign at
that point.
Table IV gives 'the values of the deflections from a. line
through the supports· as obtained .by the three methods
used. The greatest divergence of the graphical method
froμr the analytical is a.bout 0.01 inch in a deflection of
0.40 inch or 2.5 per cent.
TABLE IV.Comparison of results deflections measured from a line through the s>upports.
.Analytical method. Graphical methods .
Sta. I Single ·pole. Multiple pole.
Dell.
Dell.
0.266 I Dell. I Ord. Error. Dell'. I Ord. Error.
I
I
0 0.02693 0.101 0.02392 0.09 0. 00301 0.02392 0.09 0.00301
15  .00473  .018  .00532  .02  .00059 ...: .00797  .03  .00324
30  .03323  .125  .03189  .12 .00134  .03455  .13  .00132
45  .05106  .192  .05050  .19 .00056  .'05316  .20  .00210
60  .04508  . 169  .04252  .16 .00256  .04518  .17  .00010
7;, 0 (j 0 0 0 0 0 0
90 .09211 .346 .08871
I
.33  .00340 • 00037 .34  .00174
105 .20371 . 764 .19934 . 75  .00437 .19934 • 75  .0043J
120 .30306 1.140 .29767 1.12  ·.00539 .29502 ' 1.11  .00804
135 .37022 1.392 .36412 I 1. 37  .00610 .36146 1.36  .00876
150 .39500 1.485
I
. 38804 1. 46  .00696 .38538 1.45  .00962
165 .37104 1. 395 .36412 1.37  .00692 . 36146 1.36  .00958
180 .29896 1.124 .29767 1.11  .00129 .29236 1. 10 .00660
195 .17105 .643 .17010 .64  .00095
I
.16744 . 63  .00361
210 0 0 I  0 0 0 0 0 0
The error is computed on the assumption that the snslytical ltlethod gave the correct values.
COMPARISON OF METHODS.
The accuracy and precision that ca.n be obtained by the
two methods a.re about equal. It is hard to say which
method should be used in any particular case. · That
depends ma.inly on the personal equation of the computer.
If he has only a 10inch slide rule, the graphical method
will often be desita.ble when he might use the analytical
method if he had a. 20inch rule or a calculating machine.
A cantilever is easier to compute analytically than is a simply
supported beam as the tangent a.t the support is usually
the axis from which deflections are ,to be measured.
In general, the analytical method is slower than the graphical
but the latter requires much closer attention to the
work as it is hard to construct the force polygon accurately
and to draw lines exactly parallel. It is, therefore, ha.rd
to keep from ma.king slips m the graphical method and
there are not 80 many opportunities for checking the accuracy
of the wotk as it proceeds.
Very often the computations for several beams can be
combined. This happens more often when the graphical
method is used than with the. analytical. In desigrting
internally braced wings for the Messenger airplane, the
spars were of uniform croBB section and the load oli the
front spar was a constant times the load on the rear, eo one
curve could be used for both spars. The scale of the loading
curve differed for the two spars and the other scales
varied in proportion. The lower wing spars had 11till
different loadings, but they held a constant proporti<)n
to the loadings of the upper wing spars. The cantilever
lengths differed for the two wings 80 one figure would
not do for both. wings, but the two figures could be drawn
together with most of the work common to the two figures.
By planning the work properly in such cases the labor
required to find the defleotions can often be very greatly
red~ced.
The following remarks are made regarding the two forms
in which the graphical method is given. When the
moment of inertia varies along a curve it is best to use the
M{I curve and a constant pole distance changing the pole
only when the values of M/I change sign. When the
moment of inertia changes at only a few points and is
· constant between these points it is easier to use the M
curve and pole distances proportional to the moment of
inertia. ':Chere is leBB chance for error when the pole is
changed from one side of the force polygon to the other to
indicate a change of sign in the curve from which the
force polygon is developed, but the construction requires
more room.
0
.:
J