elastic a
TRANSCRIPT
1
Exact and Numerical Solutions for Large Deflection of Elastic
Non-Prismatic Beams
by
Farid A. Chouery, PE, SE1
©2006, 2007, 2008 Farid A. Chouery All Rights Reserved
Abstract:
The solution for large deflections of beams that has not been solved in general in 260 years is
now presented in this paper for point loads and moments in any directions along the beam with
various end conditions. Also, the solution can be used for non-prismatic beams with various end
conditions and numerical solution is presented to obtain exact solutions. Curvilinear beams and
extensibility along the beam are also addressed.
Introduction:
The large deflection of beams has been investigated by Bisshopp and Drucker [1] for a point load
on a cantilever beam. Timoshenko and Gere [2] developed the solution for axial load. Virginia
Rohde [3] developed the solution for uniform load on cantilever beam. John H. Law [4] solved it
for a point load at the tip of the beam and a uniform load combined. In this paper the general
solution developed for a prismatic beam and in some cases for non-prismatic. However,
numerical integration maybe needed along with solving compatibilities equations for the
constants of integrations. A more general and preferable numerical solutions for a non-prismatic
beam is also given using only many point loads acting with an angle on the beam with a moment
on the node representing the approximate load. This point load can take any direction on the
1 Structural, Electrical and Foundation Engineer, FAC Systems Inc. 6738 19
th Ave. NW, Seattle, WA 98117
2
beam bending in the x-x direction or y-y direction of the moment of inertia. Thus the load is to be
resolved to x-x direction and y-y direction of the moment of inertia in each orthogonal deflection
given two non-linear differential equations. By solving each non-linear differential equation the
orthogonal deflection components can be obtained.
An approximation attempt has been investigated by Scott and Caver [5] for all problems in
which the moment can be expressed as a function of the independent variable. However, the
solution presented here is not an approximation and is of a closed form. As a consequence one
can take a load function and divide it into line segments for a non-prismatic beam with no axial
loads and by using the proposed solution the large deflection can be calculated with reasonable
accuracy. Thus, by taking smaller and smaller increments the accuracy is improved.
The solution is based on solving the non-linear differential equation of Bernoulli-Euler beam
theory;
( ) )(
)(
)(13
2 xEI
xM
y
y=
+ &
&&…………………………………………………………………. (1)
Where M(x) is the moment in the direction that corresponds to the moment of inertia I(x), E is
the modulus of elasticity, y is one of the orthogonal deflection say for Ix-x. Thus, if the other
deflection is desired, then another equation is needed where M(x) is the moment in the direction
that corresponds to the moment of inertia Iy-y(x). It is assumed the modulus of elasticity is
constant and the bending does not alter the length of the beam. Three different closed form
solutions are investigated for three different cases of the non-linear differential equation. In many
ways M(x) is not known until the final deflection is known so will assume M(x) is known in the
equations.
3
Case I:
In this case assume )()(
)(xf
xEI
xM= , a function of x only. Thus,
( ))(
)(13
2
xf
y
y=
+ &
&& ………………………………………………….…………… (2)
Let θ
θθ
θ2
2
cos and
cos
1)(1tan
&
&&&& ==+⇒= yyy
Where θ is a new variable, then substitute θ in Εq. 2 yields,
∫∫ ∫ +=+=⇒= 11 )(sinor )( cos )(cos CdxxfCdxxfdxf θθθθθ&
( ) 2
12
1
2 )(1sin1cos
+−±=−±= ∫ Cdxxfθθ
Thus,
[ ]21
1
)(1
)(
cos
sin
∫
∫+−
+±==
Cdxxf
Cdxxfy
θθ
& ………………………………………………...……. (3)
[ ] 22
1
1
)(1
)()( Cdx
Cdxxf
Cdxxfxy +
+−
+±= ∫
∫
∫ ………………………………………………….. (4)
Where, C1 and C2 are constants of integration. This off course the solution Scott and Carver
approximated as an infinite series not realizing it can be expressed in a closed form. Note: Eq. 4
gives an integral solution where if the moment is approximated by a curve it can give a better
approximation than small deflection equations approximations. It is seen that if the denominator
of Eq. 4 is approximated as a unity it would give the standard solution for small deflection
approximations. In the case of large deflection it is better to use point loads and moments for
approximating the general loading because it will be shown that it can be presented as elliptical
4
integrals of the first and second kind. Elliptical solutions enable us to have a closed form solution
as it has been successfully done through recursion [7, 8] instead of integration.
Case II:
In this case assume )()(
ygEI
yM−= , a function of y only. This happens in buckling problems
and the moment of inertia is considered constant. Thus,
( ))(
)(13
2
yg
y
y−=
+ &
&& ……………………………………………….……………… (5)
Let 322
11
1
x
x
dx
dy
dy
xd
xdx
xd
xy
xy
&
&&&
&
&
&&&
&& −=−=−=⇒=
By replacing yields,
( ))(
)(13
2
yg
x
x=
+ &
&& …………………………………………………..……………… (6)
By following case I analysis with interchanging x by y the solution becomes:
[ ]21
1
)(1
)(
∫
∫+−
+±=
Cdyyg
Cdyygx& ……………….…………………………………….……...……. (7)
[ ] 22
1
1
)(1
)()( Cdy
Cdyyg
Cdyygyx +
+−
+±= ∫
∫
∫ ………………………………………...………….. (8)
5
Case III:
In this case assume )(),(
byaxhEI
yxM+= , a function of x and y. This happens in combine
bending and buckling problems with point loads and the moment of inertia is considered
constant. Thus,
( ))(
)(13
2
byaxh
y
y+=
+ &
&& ……………………………………………………..……………… (9)
Consider the rotation of axis and let:
yba
bx
ba
av
yba
bx
ba
au
2222
2222
+−
+=
++
+=
……………………………………………………..………….. (10)
Where u and v are the new variables then ( )ubahbyaxh 22)( +=+
b
baBvuBvu
b
bay
a
baAvuAvu
a
bax
2 where)()(
2
2 where)()(
2
2222
2222
+=−=−
+=
+=+=+
+=
……………………………..………. (11)
And,
)1(
)1(
vBdu
dy
vAdu
dx
&
&
−=
+=
…………………………………………….………………………….………… (12)
6
( )32
2
1
2
)1(
1
1
1
1
1
1
1
1
1
v
v
Ab
a
vAv
v
du
d
b
a
dx
du
v
v
du
d
b
a
dx
yd
v
v
b
a
v
v
A
B
dx
dy
&
&&
&&
&
&
&
&
&
&
&
+
−=
+
+−
=
+−
=
+−
=
+−
=
………….……………......……. (13)
Or
( ) ( ) 2
32
22
222
22
22
222
22
2
3
2
2
2332
1
4
1
11
1
)1(
2
)(1
+−
−+
+−
++
−=
+−
++
−=
+
ab
ab
ab
abv
v
ba
ba
v
v
b
av
v
Ab
a
y
y
&
&&
&
&&
&&
&
&&
…………………….……………. (14)
Let
θθ
θ
2
2
22
22
2
22
22
22
22
cos1
tan1
&
&&
&
+−
−=
+−
−=+−
+
ba
abv
ba
ab
ba
abv
…………………………………………………..…….. (15)
Where θ is a new variable, then substitute θ in Εq. 14 yields;
( ) ( )
( )[ ]21
22
1
2222
1cos
sin cos
∫
∫
++−−±=
++−=⇒+−=
Cduubah
Cduubahubah
θ
θθθ&
………………....………….. (16)
7
( )( )[ ] 22
22
2
1
22
1
222
22
22
22
222
22
22
1
1
tan1
ba
ba
Cduubah
Cduubah
ba
ba
ba
ab
ba
abv
+−
+++−
++
+−
−=
+−
−
+−
−=
∫
∫m
& θ
………………………………..……………. (17)
( )( )[ ] 222
22
2
1
22
1
222
22
22
1
1 Cu
ba
badu
Cduubah
Cduubah
ba
bav +
+−
+++−
++
+−
−= ∫∫
∫m ………………..... (18)
Thus, after integrating with respect to u, substitute x and y from Eq. 10 and an explicit equation
of deflection in x and y is obtained. Another way of calculating x and y is pick u find v from Eq.
18 then find x and y from Eq. 11.
Hence, the solution for the nonlinear differential equation has been obtained for three cases and
applications will follow.
Application for Case I - Numerical Solution for Any Load Function Non-Prismatic Beam:
This example is to demonstrate the solution for a cantilever beam. Other boundary conditions for
beams are similar. First divide the beam into segmental beams of each length Li and on each
node of the segment insert the equivalent load Pi and moment Qi to approximate the real load,
see FIG.1. The moment on the segment beam at xi is:
8
nn
n
j
jjjn
ii
i
j
jjji
xxxQxxPM
xxxQxxPM
xxxQxxPQxxPM
xxxQxxPM
<≤+−=
⋅
⋅
<≤+−=
⋅
⋅
<≤+−++−=
<≤+−=
−
−
=−
+=
∑
∑
1
1
0
1
1
0
211110001
100000
for )(
for )(
for )( )(
for )(
………….........……………… (19)
Where, all xi are unknown.
Now the moment of inertia:
∑=
==i
j
iTiTii LLLII0
e wher)( ………………………………………………………….. (20)
Where, the moment of inertia is approximated at each beam segment to be constant2 and the
moment of inertia function is assumed continuous. Thus:
+−== ∑
=
i
j
jjj
ii
ii QxxP
EIEI
Mxf
0
)(1
)( ……………………………….....…………………. (21)
Substitute Eq. 21 in Eq. 3 and find the slope on the segmental beam i yield:
12
0
2
0
2
for
1)()(5.01
1
1)()(5.01
)( +
=
=≤≤
+
−+−−
+
−+−
=
∑
∑ii
i
i
j
jjjj
i
i
i
j
jjjj
i
i xxx
CxxQxxPEI
CxxQxxPEI
xy&
………………………..…....…………… (22)
Apply compatibilities yield:
2 Note: this approximation does not mean there are stress singularities due to sharp corners at the discontinuities
where each segment meets. All it means is the actual deflection of that segment can be approximated with the
deflection of a beam with constant moment of inertia.
9
ii
xxxyxy iiiii
Bm. Seg. to1 Bm. Seg.
at )()( 1
−
==− && ……………………......……………… (23)
At x = L, where L is the length of the beam at x = xn = L,
0)()( == Lyxy nnn&& ……………………………………………………………....…………. (24)
Where n is the number of beam segments and n+1 is the total numbers of beam segments; apply
Eq. 24 in Eq. 22 yields:
0 since )()(5.01
11
or
0 1)()(5.01
0)(
1
0
2
1
1
1
1
0
2
1
==
−+−−==
=+
−+−⇒=
∑
∑
−
=−−
−
−
=−
nn
n
j
jjjj
n
n
n
n
j
jjjj
n
n
QPxLQxLPEI
CC
CxLQxLPEI
Ly&
…....….……. (25)
When applying Eq. 23 for all i yield:
11111 1210 CCCCC n ==⋅⋅⋅⋅⋅⋅=== − and
12
0
2
0
2
for
1)()(5.01
1
1)()(5.01
)( +
=
=≤≤
+
−+−−
+
−+−
=
∑
∑ii
i
j
jjjj
i
i
j
jjjj
i
i xxx
CxxQxxPEI
CxxQxxPEI
xy&
…………………………….…………… (26)
assuming ii II =−1 at the joints (See Appendix B). Now impose the length of the beam segment
to be un-extendible, yields,
( ) )(11
2dxxyL
i
i
x
xii ∫
+
+= & or
∫∑
+
+
−+−−
=
=
1
2
0
2 1)()(5.01
1
i
i
x
xi
j
jjjj
i
i
CxxQxxPEI
dxL ……………….....…….………. (27)
10
Eq. 27 does not lend itself to a simple solution (see appendix D for setting up Elliptic functions)
and numerically complex. To simplify the equation assume the increments are small enough such
that the slope throughout the interval of xi ≤ x ≤ xi+1 is the same (see Appendix A for the general
solution), so:
)()( 1+≅ iiii xyxy && ………………………………………………………...………………….. (28)
Thus,
( )222 )(1 iiiiii xyxyxL &+∆=∆+∆= ……………………………......……………………… (29)
Or
2
0
2
1
1)()(5.01
1
+
−+−−
−=
∑=
+
CxxQxxPEI
xxL
i
j
jijjij
i
ii
i …………………......………….. (30)
Thus, at i = 0 yields:
[ ] [ ]
111)()(5.0
11
2
01
2
000
2
000
0
01
0
C
xx
CxxQxxPEI
xxL
−
−=
+−+−−
−=
……………...………. (31)
At i = 1 yields
[ ] [ ]
1)()(5.01
)()(5.01
1
2
111
2
111
1
100
2
100
1
121
+−+−+−+−−
−=
CxxQxxPEI
xxQxxPEI
xxL
[ ]2
101
2
101
1
12
1)()(5.01
1
+−+−−
−=
CxxQxxPEI
xx ……………………......……………. (32)
Substitute x1 – x0 from Eq. 31 in Eq. 32, for a given C1 and x2 – x1 is found.
11
At i = 2 yields
[ ] [ ]
1)()(5.01
)()(5.01
1
2
122
2
122
2
010
2
020
2
232
+−+−+−+−−
−=
CxxQxxPEI
xxQxxPEI
xxL
………………......…………………………… (33)
Substitute x1 – x0 from Eq. 31 and x2 – x1 from Eq. 32 in Eq. 33, for a given C1 and x3 – x2 is
found, where x2 – x0 = (x2 – x1) + (x1 – x0).
Thus, if guessing C1 then find xi+1 – xi , can be found since the denominator of Eq. 30 is always
known from previous equations. And since 00
0
1 )( xLxxxx n
n
i
ii −=−=−∑=
+ , x0 can be found.
Therefore for a given C1 x0 , x1 , x2 , ……. , xn-1 can be solved, then proceed by checking the
end slope of Eq. 24 or Eq. 25. If it is not satisfied update C1 with numerical analysis until all the
variables are found. For the deflection from Eq. 4 yields:
12
0
2
0
2
for 2
1)()(5.01
1
1)()(5.01
)( +
=
=≤≤+
+
−+−−
+
−+−
= ∫∑
∑iii
i
j
jjjj
i
i
j
jjjj
i
i xxxCdx
CxxQxxPEI
CxxQxxPEI
xy
………………………....…………………. (34)
To find C2i assume compatibility and enforce:
.....
2 find and )()(
2 find and 0)()(
2111
1
etc
Cxyxy
CLyxy
nnnnn
nnnn
−−−−
−
=
==
And the solution is found numerically.
12
Application for Case II - Numerical Solution for Any Load Function Non-Prismatic Beam:
This solution is very similar to the application of Case I or the previous application but instead of
using x and xi substitute for y and yi and can find C1, y0 , y1 , …… , yn-1 and C2i for the
deflection of Eq. 8 see FIG.2.
Application for Case III - Numerical Solution for Any Load Function Non-Prismatic Beam:
Again implement the solution using similar procedures as in the application of Case I, see FIG.3.
If assume at xi the resultant moment in the x and y direction is Mxi and Myi and if assume at xi the
resultant force in the x and y direction is Rxi and Ryi , then the moment for Eq. 9 becomes:
yi
yi
ii
xi
xi
ii
i
yi
i
i
xi
i
R
Myy
R
Mxx
EI
Ryy
EI
Rxxbxaxh
+=+=
−−−−=+
and
here w)()()(
………………………….…………. (34.1)
Thus, if translating temporarily the axis to a local axis with ii yx and Eq. 9 becomes:
( ) i
yi
i
i
xiiiiii
EI
Rb
EI
Raybxaybxah
y
y−=−=+=+=
+ and where, )(
)(13
2&
&& …………. (34.2)
Note: for segment i-1 and segment i at x = xi the resultants are the same so
11 and −− == iiii bbaa ………………………………………………………….……. (34.3)
Thus if impose
ii
xxxyxy iiiii
Bm. Seg. to1 Bm. Seg.
at )()( 1
−
==− &&
Which means from Eq. 13 and 34.3 yields:
iiiii uuuvuv ==− at )()( 1&& ……………………..…………………………. (35)
13
This implies 11111 1210 CCCCC n ==⋅⋅⋅⋅⋅⋅=== − , assuming ii II =−1 at the joints (Se Appendix
B). u0 , u1 , ……. , un can be found by using the approximation:
)()(or )()( 11 ++ ≅≅ iiiiiiii uvuvxyxy &&&& ……..……………………......………………….. (36)
Thus,
222
222
2
2
2
222 )1(
2)1(
2
−
++
+
+∆=
∆
∆+
∆
∆∆=∆+∆= i
i
ii
i
i
ii
i
i
i
i
iiiii v
b
bav
a
bau
u
y
u
xuyxL &&
………………………..………………………… (37)
Or
2222
1
)(1)(1
2)(
−+
++−≅ +
i
ii
i
iiii
iiib
uv
a
uvbauuL
&&……..…………………..………….. (38)
Which is again for a given C1 ui‘s can be found from Eq. 38. When using Eq.10:
02
1
2
1
1
0
1
0
1 )( uLba
auuuu
nn
n
n
n
i
ii −+
=−=−−−
−−
=+∑ . Thus, all ui‘s can be obtained by back substitutions
with checking if C1 satisfies:
0
1
1
)(1
)(1
or
0)()(
2
1
2
1
1
2
1
2
1
1
1
1
1
1 =
++
+−
=
+
−
==
−−
−
−−
−
−
−
−
−
Lba
av
Lba
av
b
a
uv
uv
b
a
Lyxy
nn
nn
nn
nn
n
n
ni
ni
n
n
nnn
&
&
&
&
&&
…………………….………………… (39)
Now, find C2n-1 from the last segment to satisfy
14
Lba
aL
ba
avuv
nn
n
nn
nnnn
2
1
2
1
1
2
1
2
1
1)(
−−
−
−−
−
+=
+= …………………………...………………….. (40)
and find all of C2i from the compatibility equation of deflection yi and using Eq. 11:
.....etc
2 find and )()(
2 find and )()(
3212222
211111
−−−−−−−
−−−−−−
−=−
−=−
nnnnnnn
nnnnnnn
Cuvuuvu
Cuvuuvu
When done find all of v0 , v1 , ……… , vn-1 then substitute in Eq. 11 to find xi and yi .
Application Examples of Case III Numerical Solution:
(a) Fishing Pole: an application for Case III is shown on FIG. 4, where the beam has an angle
α with the horizontal and a load P0 and P1 hanging from the beam. One common
application is a fishing pole that has a fish that has the load P0 from the vertical and P1 =
0. In this case when varying the angle α with the horizontal the moment changes and the
deflection curve change giving a smaller or bigger moment with various elastic curves.
This affects the ability to pull the fishing with the real and having various controls on
catching the fish by puling or letting go the line. An experienced fisherman does this
procedure naturally and gets the credit for not loosing the fish. A good fishing rod would
be designed to have a moderate elastic curve configuration when varying P0 and the angle
α. The equations for a non prismatic fishing pole can be:
:becomes Eq.9in function theThus sin and cos 11 iyx hPPPP αα ==
15
[ ]
[ ]
αα
αα
αα
sin and cos
where,
)()(sin)(cos)(),(
or
sin)(cos)(),(
00
0
0
0
0
0
00
0
0
000
∑∑∑
∑
∑
∑
∑
==
=
=
=
=
=
−====
−+−=−−−=
−−−=
i
j i
j
i
i
j i
j
ii
j i
j
i
j i
jj
ii
j i
j
i
j i
jj
i
iiii
i
j
jj
i
j
iii
EI
Pb
EI
Pa
EI
P
EI
yP
y
EI
P
EI
xP
x
yybxxayyxxEI
Pybxah
yyxxEI
Pybxah
In this situation each beam segment can be translated in a new local axis by ii yx and to
have Eq. 10 ready for rotation of axis to satisfy Eq. 9 and then translate to the global axis.
For example translate the local axis by ii yx and , and substitute in Eq. 17 and 18 the new
coordinates )( and )( ii vvuu −− for u and v in the solution of Eq. 17 and Eq. 18, where
ii vu and are obtained from substituting ii yx and in equation 10 with the replacement of
ii bbaa by and by .
(b) Curved Beam: Another application example where the beam is a non-prismatic curved
beam. Thus, if subdivide the curved beam to a smaller segments cantilever straight
beams3 with constant moment of inertia as in using FIG. 3. If αi is the angle the
segmental beams make with the horizontal, then the problem can be solved by taking
each deflection curve derived from the local axis of the segmental beam and rotated by
the angle αi then translate by ii yx and to the global axis and the problem can be solved
numerically.
3 Note: this approximation does not mean there are stress singularities due to sharp corners at the discontinuities
where each segment meets. All it means is the actual deflection of the slightly curved segment can be approximated
with the deflection of a beam with straight segment with a constant moment of inertia.
16
(c) Bow and Arrow: The ancient structural problem in archeries or shooting bow and arrow
can finally be solved. Amazingly, can design a curved non-prismatic beam to give the
proper deflection curve for a human precise measurement giving the best comfortable
result for a more accurate bull’s-eye. Possibly, a unique design for each athlete, so
putting tension by changing the string size can be less desirable. To simplify the
equations and show the example, a non-prismatic curved beam will not be used but use a
non-prismatic straight beam, see FIG. 5. The function hi in Eq. 9 becomes:
−−
−−−
−=
−−
−=
2)(
)()(
2)(
1
2cot)(
2)(
1),(
2
0
2
0
00
0
00
0
000
P
xLL
xLyy
Pxx
EI
Pyy
Pxx
EIybxah β
2
0
2
0
00
2
0
2
000
00
)(
)(
2 and
2
where
)()(
2)(
)()(
2)(
1
2cot)(
2)(
1),(
xLL
xL
EI
Pb
EI
Pa
yybxxa
P
xLL
xLyy
Pxx
EI
Pyy
Pxx
EIybxah
i
i
i
i
ii
i
i
iii
−−
−−==
−+−=
−−
−−−
−=
−−
−= β
In this situation each beam segment can be translated by 00 y and x in its local axis to
satisfy Eq. 9 and then to the global axis. For example translate by x0 and y0, and substitute
in Eq. 17 and 18 by the new coordinates (u-ui) and (v-vi) for u and v of Eq. 17 and Eq. 18,
17
where ii vu and are obtained from substituting 00 y and x in equation 10 with the
replacement of ii bbaa by and by .
Column with Load through Fixed Point: The problem of column with load through fixed
point was presented by Timoshenko and Gere [2]. Jong-Dar Yau [6] presented a solution
for Closed-Form Solution of Large Deflection for a Guyed Cantilever Column Pulled by
an Inclination Cable. A more general problem is to allow the fixed point D to have a
coordinate point (xd, yd) instead of the coordinate point (xd, 0) as in FIG. 6. As if the tip of
the column is attached by a cable with a shackle to point D and the shackle is being
tightened. The column is assumed a non-prismatic. This situation can also happen in a
vertical fishing pole, where the fish pulls with an angle β. Thus, moment becomes:
2
0
2
0
0
2
0
2
0
0
00
2
0
2
0
00
2
0
2
0
00
00
00
)()(
)( and
)()(
)(
where,
)()(
)()(
)()(
)()(
)()( ),(
cos)(sin)(
)()(
xxyy
xx
EI
Pb
xxyy
yy
EI
Pa
yybxxa
xxyy
xxyy
xxyy
yyxx
EI
Pybxah
PyyPxx
PyyPxxM
dd
d
i
i
dd
d
i
i
ii
dd
d
dd
d
i
ii
xy
−+−
−−=
−+−
−=
−+−=
−+−
−−−
−+−
−−=
−−−−=
−−−−=
ββ
FIG.6 - Column with Load through Fixed Point (d)
In this situation each column segment can be translated by 00 y and x in its local axis to
satisfy Eq. 9 and then to the global axis. For example translate by x0 and y0, and substitute
in Eq. 17 and 18 by the new coordinates (u-ui) and (v-vi) for u and v of Eq. 17 and Eq. 18,
18
where ii vu and are obtained from substituting 00 y and x in equation 10 with the
replacement of ii bbaa by and by .
The ± sign in Eq. 3, 7 and 17:
The ± sign in the solution of Eq. 3, 7 and 17 can be used interchangeably when the slope of the
deflection curve goes to infinity at a point and the slope change sign. Another word the
deflection curve is not a function anymore and becomes circular. In Eq. 17 ∞⇒v& when
bay /⇒& . At that point the correct sign of Eq. 17 must be used.
Other End Condition: See Appendix E
Curvilinear Beams:
For a curvilinear beam with a function y = R(x) the new radius of curvature must satisfy the
following equation:
old
new
rEI
xMr
1)(
1
+= …………………………………………………………………. (41)
So that if M(x) = 0 the radius of curvature does not change and remain of the function y = R(x).
Thus Eq. 1 becomes:
( ) [ ])(
)(1
)(
)(
)(
)(13
23
2
xt
xR
xR
xEI
xM
y
y=
+
+=+ &
&&
&
&&………………………………………… (42)
And solution is as Eq. 3 and Eq. 4 by replacing f(x) by t(x).
19
Extensibility:
In order to account for extensibility of the beam will analyze a beam segment Li. Let αi be the
directional angle of the load Pi and θi the angle at Pi representing the slope of the beam at that
point as in Fig. 7.
For extensibility of a small arc length ds in Li expressing the change in length (shortening) εi due
to the axial load in Li as:
dsEA
Pi
i
x
xi
ii
i ∫+ −
−=1
)sin( αθε ……………………………………………….……………… (43)
Expressed as (ΣPL/AE) where Ai is the area of the segment Li and the load Pi is the resultant at
every ds. In a shortening condition, the shape of the deflection curve did not change only the
curve has shrunk. The resultant moment is affect by extensibility due to the change of
iii uyx or , for a new iii uyx or , . Then for a given deflection curve derived without including
extensibility that gives iii uyx or , yields:
20
( )
( )( )
( )
( )( )
⇐+
−−=
−−=+=
⇐+
−−=
+
−−=
−−=
−−=
−−=
−−=+=
∑∫
∑∫∑
∑∫
∑∫
∑∫
∑∫
∑∫
∑∫∑
=
==
=
=
=
=
=
==
+
+
+
+
+
+
+
+
1
sincos
sin)sin(
ˆˆ
1
1sincos
/1
1sincos
sincos
sincos
sincoscossin
cos)sin(
ˆ
0
2
0
0
0
2
0
2
0
0
0
0
0
1
1
1
1
1
1
1
1
j
j
x
xj
jjj
i
i
j
x
xj
jj
i
i
j
jii
j
j
x
xj
jjj
i
j
j
x
xj
jjj
i
j
j
x
xj
jjj
i
j
j
x
xj
jjj
i
j
j
x
xj
jjj
i
i
j
x
xj
jj
i
i
j
jii
dxy
y
EA
yPy
dsEA
Pyjyy
dxyEA
yPx
dxdxdyEA
dx
dyP
x
dxds
dx
EA
dx
dyP
x
dxEA
ds
dx
ds
dyP
x
dsds
dx
EA
Px
dsEA
Pxixx
j
j
j
j
j
j
j
j
j
j
j
j
j
j
j
j
&
&&
&
&
αα
θαθ
ε
αα
αα
αα
αα
αθαθ
θαθ
ε
…………………. (44)
Alternatively, if would like to include the effect extensibility on the moments then Eq. 27
becomes4:
4 Note the extensibility Eq. 45 is slightly conservative since ∫ ≠ iLds when integrate from 1 to +ii xx .
21
[ ]
[ ]
[ ] ( ) [ ]
[ ] [ ]
[ ][ ] [ ]
[ ]
[ ] [ ]
[ ] ( ) ( )EA
Pxx
EA
Pyydxxy
dxEA
xyPdxxy
dxxyEA
xyxy
xyP
dxxy
dxxyEA
dy
dx
ds
dyP
dxxy
dxxyEA
Pdxxy
dsEA
PdxxyL
or
dxxyL
i
iiii
i
iiiix
xi
x
xi
iiiix
xi
i
x
xi
i
i
i
i
i
i
x
xi
i
x
xi
iiix
xi
i
x
xi
iiix
xi
x
xi
iix
xii
x
xiii
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
αα
αα
αα
αα
αθαθ
αθ
ε
sincos)(1
sincos)()(1
)(1
sin)(1
1cos
)(1
)(
)(1
)(1
sincos
)(1
)(1sincoscossin
)(1
)sin()(1
)(1
11
2
2
2
22
2
2
2
2
2
2
2
1
11
11
11
11
11
1
−+
−++=
−−+=
+
+−
+−+=
+
−
−+=
+−
−+=
−−+=
+=−
++∫
∫∫
∫∫
∫∫
∫∫
∫∫
∫
+
++
++
++
++
++
+
&
&&
&&&
&
&
&&
&&
&
&
………………………………….. (45)
This equation can be adjusted with the rotation of the axis for αi of Case III and implemented per
Eq. 13 and instead of updating xi in Eq. 27 update iu and Eq. 17 when substituting Eq. 17 in Eq.
13 to obtain )(xyi& of Eq. 45 to find xi. This assumes the total deflection curve is shortened or
elongated by yx εε , and the solution in Eq. 4, 8, 18 remains the same and only is effected by xi
and yi as in Eq. 19, where:
∑∫
∑∫−
=
−
=
+
+
−−=
−−=
1
0
1
0
1
1
sin)sin(
cos)sin(
n
i
x
xi
ii
y
n
i
x
xi
ii
x
dsEA
P
dsEA
P
i
i
i
i
θαθ
ε
θαθ
ε
………………………………………………….. (46)
22
Finale note on extensibility and large deflections: extensibility may have a minor effect on the
moments however it can affect the buckling deflection criterion. In general when loading a beam
the moment and axial load reduces with time however the deflection increases with time until the
final iii uyx or , occurring at tfinal . In this case the safety factor on the stresses must account for
the dynamic problem of loading and reloading and care must be taken when using large
deflections in design.
Comparison with current methods for large deflections: It would be very difficult to draw
conclusions from one or two examples when comparing the exact solutions with any
approximate method including finite elements. Thus, comparison is left out to a more in-depth
study in a different article. The finding in this paper stands alone on its own two feet, is complete
and it is a bench mark.
Conclusion:
The closed form and general solution of non-linear differential equation of Bernoulli-Euler beam
theory is solved numerically for general loading function for a non-prismatic beam and can be
approximated for a non-prismatic curved beam when the presented solution of curvilinear beam
is not used. In some cases it is solved in closed form for prismatic and non-prismatic beam. In
general the Elastica, as called by Timoshenko and Gere [2], is solved.
23
References:
1. Bisshopp, K. E., and Drucker, D. C., “Large Deflections of Cantilever Beams,” Quarterly
of Applied Mathematics, Vol. 3, 1945, pp. 272-275
2. Timoshenko, S. P. and Gere, J. M., “Theory of Elastic Stability” 1961, McGraw-Hill
Book Company, New York, pp. 76-82 and pp. 55-57
3. Rohde, F. V., “Large Deflection of Cantilever Beam with a Uniformly Distributed Load,”
Quarterly of Applied Mathematics, Vol. 11, 1953, pp. 337-338
4. Lau, John H. “Large Deflection of Cantilever Beams,” Journal of Engineering Mechanics
Division, Proceedings of the American Society of Civil Engineers, Vol. 107, NO. EM1,
February, 1981. pp. 259-264
5. Scott, E. J. and Carver, D. R. “On the Nonlinear Differential Equation for Beam
Deflection,” ASME Applied Mechanics Division June, 1955, pp. 245-248.
6. Jong-Dar Yau "Closed-Form Solution of Large Deflection for a Guyed Cantilever
Column Pulled by an Inclination Cable" Journal of Marine and Technology, Vol. 18, No.
1, 2010, pp 120-136
7. Nellis, J. William "Tables of Elliptic Integrals" NASA Contractor Report NASA CR-289
Prepared under Grant No. NsG-293 by IOWA State University, Ames, Iowa for
NATIONAL AERONAUTICS AND SPACE ADMINISTRATION . WASHINGTON,
D. C. AUGUST 1965
8. B. C. Carlson "Three Improvements in Reduction and Computation of Elliptic Integrals"
Journal of Research of National Institute of Standards and Technology, Volume 107,
Number 5, September-October 2002, pp 413-418
24
APPENDIX A
General algorithm solution of solving for xi for application example Case I:
From Eq. 27 let the function:
∫∑
+
+
−+−−
=Φ
=
c
CxxQxxPEI
dxx
i
j
jjjj
i
i2
0
2 1)()(5.01
1
)( ……………..…..………. (47)
When setting 0=∂
Φ∂
x
i there is no solution and the function )(xiΦ is completely odd but
translated, increasing and crossing the x axis once. Thus there is only one root xi*. To proceed to
find the inflection points for 02
2
=∂
Φ∂
x
i rewrite Eq. 47 to be:
[ ]∫ ++−−
=Φ c
BAxC
dxx
iii
i22)(1
)( …………………………………….........………..….. (48)
Where:
∑
∑
∑
∑
=
=
=
=
=
+−+−=
−
=
i
j
j
i
i
i
j
jjjj
i
iii
i
j
j
i
j
jjj
i
PEI
C
CxQxPEI
CAB
P
QxP
A
0
0
22
0
0
5.01
1)5.0(1
)(
……………………………......……………….. (49)
And from Eq. 25:
25
−+−−= ∑
−
=−
1
0
2
1
)()(5.01
1n
j
jjjj
n
xLQxLPEI
C ………………………......………………… (50)
Thus the inflection points are only three and they are:
i
i
ii
ii
C
BAx
Ax
−±=
=
3,2
1
…………………………………………………………………………… (51)
The function becomes:
[ ][ ] cdxBAxC
k
kc
BAxC
dxx
k
k
iii
iii
i +
+−
⋅⋅⋅⋅⋅−⋅⋅⋅⋅⋅
+=++−−
=Φ ∫ ∑∫∞
=
)(2642
)12(5311
)(1
)(1
22
22
………….………………………………. (52)
The slopes and the deflections become:
[ ]1
22
2
for
)(1
)()( +≤≤
+−−
+−= ii
iii
iiii xxx
BAxC
BAxCxy& …………………..………….. (53)
[ ][ ]
[ ]
[ ][ ] i
x
A
iii
iii
i
k
k
iiiiiii
iii
iii
iiii
C
BAzC
dzBAzC
CdxBAxCk
kAxBAxC
xxxC
BAxC
dxBAxCxy
i
2
)(1
)(
2 )(2642
)12(531)()(
3
1
for 2
)(1
)()(
22
2
1
1223
122
2
++−−
+−=
+
+−
⋅⋅⋅⋅⋅−⋅⋅⋅⋅⋅
+−+−=
≤≤++−−
+−=
∫
∫ ∑
∫
∞
=
+
+
…………………………………………… (54)
And find
i
x
Ai
x
Aii
L
Ann
CdxxydxxyC
dxxyC
i
i
i
i
n
2 )( )(2
)(2
1
1
1
1
1
1
1
+−=
−=
∫∫
∫−
−
−
−
−−
−
&&
&
…………..……………………………….……… (55)
And let
26
[ ][ ] i
x
xk
k
iiii
x
x
iii
ii LdxBAxCk
kL
BAxC
dxx
i
i
i
i
−
+−
⋅⋅⋅⋅⋅
−⋅⋅⋅⋅⋅+=−
+−−= ∫ ∑∫
++∞
=
)(2642
)12(5311
)(1
)(11
1
22
22
ψ
…………………………………………. (56)
For the numerical procedure, use Newton-Raphson method for systems of nonlinear algebraic
equations. The following is the result for the Jacobian matrix. For a given function )( ii xψ from
Eq. 56 the derivative with respect to xm for m = 1 to n is:
[ ]
[ ]
[ ][ ][ ][ ]
∫
∑ ∫
+
+
+
+
+−−
+−++
+−−
−−−
∂∂
=
+−
+⋅⋅⋅⋅⋅−⋅⋅⋅⋅⋅
+
−++−−
−−−
∂∂
=∂
∂
=
=
∞
=
+
=
=
1
1
1
1
2
322
22
22
1
22
122
)(1
)(1
2)(1
1
2)(
)(2
12
2642
)12(531
)(2)(1
1
2)(
)(
i
i
i
i
i
i
i
i
x
x
iii
iii
i
mi
xx
xxiiii
mi
imi
m
k
x
x
k
iii
i
mi
ii
i
mi
xx
xxiiii
miimi
mm
ii
dx
BAxC
BAxC
B
B
BAxCB
BAxA
x
x
dxBAxCB
Bk
k
k
xxB
B
BAxCB
BAxA
x
x
x
xψ
………………………………………………………. (57)
Where:
[ ]
[ ] imQxLPEI
B
imQxLPEIEI
QxPCAAB
imA
im
P
PA
mmm
n
mi
mmm
ni
mmmimiimi
mi
i
j
j
m
mi
>+−=
≤+−+−
+−=
>=
≤=
∑=
)(1
and for )(1
2
0
and for
0
……….………….. (58)
27
Thus, choose ∑=
=i
j
ii Lx0
for the initial condition and the Newton-Raphson method requires the
updated { } { } [ ] { }ψ1−−= Jxx . Where,{ }x is the updated vector of ix , { }x is the old vector of ix ,
[ ]J is the Jacobian matrix evaluated ix and { }ψ is the vector of function of Eq. 56 evaluated ix .
Example:
In some cases this solution is the exact solution when the loads are actually point loads and
moment for a beam. Setting up the solution for two point loads and two moments on a beam that
is of two moment of inertia, see Fig. 8, and substituting yield,
[ ]
[ ]
[ ] ( )111
1
10000
1
00
1000
00
2
00
0
0
2
00
0
00
0
000
1100
2
11
2
00
1
)(1
)(1
0 1
15.01
5.0
)()()(5.0)(5.01
1
QxLPEI
BQxLPEI
B
AA
CxQxPEI
CAB
EI
PC
P
QxA
xLQxLQxLPxLPEI
C
+−=+−=
==
+−+−=
=
−=
−+−+−+−−=
[ ]
[ ] [ ]
1)(5.05.01
)(5.0
)(
1
1
1101
1
001
10
111
10
0
01
1100
2
11
2
00
1
1
2
11
0
101
10
101100
1
ALEI
PBAL
EI
PB
PP
PA
PP
PA
CxQxQxPxPEI
CAB
EI
PPC
PP
QQPxPxA
−=−=
+=
+=
++−++−=
+=
+
+−+=
28
Let H consider the variables in Eq. 57
=
=∂
∂
=∂
∂
),,,,,,(),,,,,,(
),,,,,,(),,,,,,(
),,,,,,(
),,,,,,(
111111110101011111011
1010000101000000001000
111111011
000001000
BACBAxxHBACBAxxH
BACBAxxHBACBAxxHJ
BACBAxxHx
BACBAxxHx
m
mmm
m
mmm
m
ψ
ψ
And an exact numerical solution is obtained.
If for example the beam has to be divided to small increments due to the load function or the
moment of inertia function, a less computational analysis may be selected as in Eq. 28.
29
APPENDIX B
Finding coefficients C1i with no discontinuity in the moment of inertia:
In Application for Case I, II, and III – (Numerical Solution for Any Load Function Non-
Prismatic Beam including the examples), the moment of inertia at the joints were imposed equal.
In the following equations this assumption will be shown valid and in Appendix C the derivation
for discontinues beam for abrupt changes of the moment of inertia will be derived.
To start with the closed form solution of Eq.1 through Eq 18 will be used and the moment of
inertia is taken as:
∞≠=⇒
≤≤=
≤≤=
≤≤=
∑
∑
∑
=
=
=
rrn
uuuauEI
yyyayEI
LxxaxEI
r
n
n
n
r
n
n
n
r
n
n
n
and ,........,2 ,1
0 interval in the defined III Casefor )(
1
0 interval in the defined II Casefor )(
1
0 interval in the defined I Casefor )(
1
0
0
0
0
0
………………………………. (59)
The following proof is for Case I. All other cases can be done with a similar proof.
From Eq. 3 the integral term for each segment becomes
1
00
for 1 )(1)(1)( +==
≤≤+
+−
=+=+ ∫ ∑∫ ∑ iii
i
j
jjj
nr
n
niii xxxCdxQxxPxaCdxxfCxZ
……………………………...……………..….. (60)
30
Using integration by parts starting with
yeilds, )( and 00
∑∑==
+−==i
j
jjj
nr
n
n QxxPdvxau
dxxxQxxP
xnaxxQxxP
xaxZi
j
jj
jjnr
n
n
i
j
jj
jjnr
n
ni )(2
)( )(
2
)( )(
0
2
1
10
2
0
−+
−
+
−+
−
= ∑∫ ∑∑∑
=
−
===
……………………………………………………. (61)
Continuing the integration by parts on each integral leads to
∑ ∑∑=
+=
++−
=
≤≤
+
−+
+
−
−=
r
0k
1
0
12
for )!1(
)(
)!2(
)(
)!(
!)( ii
i
j
k
jj
k
jjknr
kn
ni xxxk
xxQ
k
xxPxa
kn
nxZ
……..………………………… (62)
At xi Eq. 3 for to consecutive segments at the joint becomes:
[ ]
[ ] 1)(1
1)()(
and 1)(1
1)()(
2
11
111
2
−−
−−−
+−
+=
+−
+=
iii
iiiii
iii
iii
ii
CxZ
CxZxy
CxZ
CxZxy
&
&
………………………………………………………….. (63)
Apply compatibilities yields:
ii
xxxyxy iiiii
Bm. Seg. to1 Bm. Seg.
at )()( 1
−
==− && ………………………………………………… (64)
Thus from Eq. 63
iiiiii CxZCxZ 1)(1)( 11 +=+ −− ……………………………………………………………… (65)
Substituting in Eq. 62 yields;
∑ ∑∑=
−
=
++−
=−
+
−+
+
−
−==
r
0k
1
0
12
1 )!1(
)(
)!2(
)(
)!(
!)()(
i
j
k
jij
k
jijkn
i
r
kn
niiiik
xxQ
k
xxPxa
kn
nxZxZ ……..…... (66)
31
Or,
ii CC 11 1 =− ……………………….………………………………………………………. (67)
Thus:
11111 1210 CCCCC n ==⋅⋅⋅⋅⋅⋅=== − ….…………………………………………………… (68)
Thus, the coefficients C1i with no discontinuity in the moment of inertia is correct and the
moment of inertia of the segment for Eq. 60 can be approximated as:
∑∑∫ +
=+
=++ ===
+
1
0
1
01
1 and where)(
1
i
j
ji
i
j
ji
i
l
l
i LlLlL
xIEI
i
i ….…….……………………………… (69)
32
APPENDIX C
Finding coefficients C1i with discontinuity in the moment of inertia:
Application for Case I, II, and III are all similar and only Case I will be addressed. By using Eq.
64 and E. 65 in Eq. 22 the Zi can be written at xi as:
[ ]
[ ]∑
∑
−
=
−
=−−
−+−=
−+−=
1
0
2
1
0
2
1
1
)()(5.01
and
)()(5.01
i
j
jijjij
i
i
i
j
jijjij
i
i
xxQxxPEI
Z
xxQxxPEI
Z
…………………………………………… (70)
Thus from Eq. 65 yeilds;
[ ]
[ ] 1
1
0
2
1
1
0
2
1
1
1)()(5.011
1
or
1)()(5.011
1
−
−
=−
−
=−−
+−+−
−=
+−+−
−=
∑
∑
i
i
j
jijjij
ii
i
i
i
j
jijjij
ii
i
CxxQxxPEIEI
C
CxxQxxPEIEI
C
…………………………….. (71)
Starting with Eq. 25 yields;
−+−−= ∑
−
=−−
1
0
2
1
1 )()(5.01
1n
j
jjjj
n
n xLQxLPEI
C . ………………………………………… (72)
Thus, all of the C1i can be found consecutively from Eq. 71 and Eq. 30 becomes:
2
0
2
1
1)()(5.01
1
+
−+−−
−=
∑=
+
i
i
j
jijjij
i
ii
i
CxxQxxPEI
xxL …………………………………. (73)
So Eq. 31 becomes:
[ ]20
01
0
11 C
xxL
−
−= …………………………………………………………………………… (74).
33
So the approximate numerical method with Eq. 28 starts by guessing C10 then find (x1 – x0) from
Eq. 74 and Eq. 71 with i = 1yields;
[ ] 0010
2
010
10
1 1)()(5.0 11
1 CxxQxxPEIEI
C +−+−
−= …………………………………….. (75)
So C11 can be found. By using Eq. 73 for i = 1yields;
[ ]2
1010
2
010
1
121
1)()(5.01
1
+−+−−
−=
CxxQxxPEI
xxL …………………………………….. (76)
Now find (x2 – x1) using Eq. 74, Eq. 75 and Eq. 76. , then find (x2 – x0) = (x2 – x1) + (x1 – x0) and
substitute in Eq. 71 yields;
[ ] 1020
2
020121
2
121
01
2 1)()(5.0)()(5.0 11
1 CxxQxxPxxQxxPEIEI
C +−+−+−+−
−=
…………………………………………………………………………… (77).
So all of the C1i then find (xi+1 – xi) can be found using Eq. 71 and Eq. 73. Now find (xn – x0)
from
)(...........)()()( 012110 xxxxxxxx nnnnn −++−+−=− −−− …………………………………. (78)
If (xn – x0) = (L – x0) then x0 = L - (xn – x0), and all of x1 , x2 , ……. , xn-1 can be found. By
checking the end condition of Eq. 25 or Eq. 72 with Eq. 71
34
[ ] 2
1
0
1
2
1
12
1
1
0
2
1
1
1
1
0
2
1
1)()(5.011
1
ifcheck
0 since )()(5.01
1
or
0 1)()(5.01
0)(
−
−
=−−
−−−
−
=−−
−
−
=−
+−+−
−=
==
−+−−=
=+
−+−⇒=
∑
∑
∑
n
n
j
jnjjnj
nn
n
nn
n
j
jjjj
n
n
n
n
j
jjjj
n
n
CxxQxxPEIEI
C
QPxLQxLPEI
C
CxLQxLPEI
Ly&
………………… (79)
If it is not satisfied update C10 with numerical analysis until all the variables are found. For the
deflection from Eq. 4 yields:
12
0
2
0
2
for 2
1)()(5.01
1
1)()(5.01
)( +
=
=≤≤+
+
−+−−
+
−+−
= ∫∑
∑iii
i
i
j
jjjj
i
i
i
j
jjjj
i
i xxxCdx
CxxQxxPEI
CxxQxxPEI
xy
………………………....…………………. (80)
To find C2i assume compatibility and enforce:
.....
2 find and )()(
2 find and 0)()(
2111
1
etc
Cxyxy
CLyxy
nnnnn
nnnn
−−−−
−
=
==
And the solution is found numerically.
For using Newton-Raphson method in Appendix A, replace Bi in Eq. 43 by
35
∑
∑
∑
∑
=
=
=
=
=
+−+−=
−
=
i
j
j
i
i
i
i
j
jjjj
i
iii
i
j
j
i
j
jjj
i
PEI
C
CxQxPEI
CAB
P
QxP
A
0
0
22
0
0
5.01
1)5.0(1
)(
………………………..……………………(81)
Where C1i is from Eq. 71 and from Eq. 79
0 since )()(5.01
11
0
2
1
1 ==
−+−−= ∑
−
=−− nn
n
j
jjjj
n
n QPxLQxLPEI
C …………….….. (82)
So all of the C1i can be calculated from the vector {x} and update { } { } [ ] { }ψ1−−= Jxx .
Where,{ }x is the updated vector of ix , { }x is the old vector of ix , [ ]J is the Jacobian matrix
evaluated ix and { }ψ is the vector of function of Eq. 50 evaluated ix .
36
APPENDIX D
Converting the Equations for Case I application to Elliptic functions:
Eq. 26 and Eq. 27 in Case I applications can be re-written as in Eq. 48 yields:
[ ]
[ ]22
2
22
)(1
)(
)(1
1
iii
iii
i
x
x
iii
i
BAxC
BAxCy
BAxC
dxL
i
i
+−−
+−=
+−−= ∫
+
&
……………………………..…………………………… (83)
Where iii BCA and , are define in Eq. 49
Re-writing Eq. 77 to become:
[ ]∫+
+−−−−=
1
22 1)1()(1
i
i
x
x
iii
i
BAxC
dxL …………………………………………………. . (84)
Now let
φ
φφ
22 cos)1()(
and
sin1ith wcos1)(
iii
iiiii
BAxC
BdxCBAxC
−=−
−−=−=−
………………………… (85)
[ ]∫+
+−−−
−−=
12
1 22 1sin)1(1
sin1 i
i
d
BC
BL
ii
ii
φ
ϕφ
φ
φ ………………………………………….. (86)
Now multiply the square in the denominator in Eq. 86 and rearrange yield:
∫+
−−
−=12
1
2
2
sin2
11
2
1 i
i
ii
i
B
d
CL
φ
φ
φ
φ ………………………………………………… (87)
Where:
37
( ) ( )
( ) ( )
−=
−
−=
−=
−
−=
+−
+−
+
−−
ii
i
i
ii
i
i
i
ii
i
i
ii
i
ii
Axp
CAx
B
C
Axp
CAx
B
C
1
1
1
1
1
11
2/cos
1cos2
and
2/cos
1cos1
φ
φ
……………………………… (88)
Where
2
1 i
i
Bp
−= ……………………………………..………………………………………….. (89)
And Eq. 87 becomes:
∫+
−−=
12
1 22 sin12
1 i
i
ii
i
p
d
CL
φ
φ φ
φ ………………………………………………………….. (90)
Which can be expressed in Elliptic Integral as follows:
[ ])1,()2,(2
11 iiii
i
i pFpFC
L φφ −= + ……………………………………...………………. (91)
Where the function F is the elliptic integral of the first kind.
Similarly Eq. 83 becomes:
[ ]22
2
1)1()(1
1)1()()(
+−−−−
+−−−=
iii
iiii
BAxC
BAxCxy& ………………………………….………………… (92)
Now substitute Eq. 89 in 92 yields:
i
i
i
ii
ii
AC
px
pp
pxy
+=
−
+−=
φ
φφ
φ
cos2/
where
sin1sin2
1sin2)(
22
22
&
…………………………………………………………… (93)
The deflection between the joints becomes:
38
[ ]dx
BAxC
BAxCyydxy
i
i
i
i
x
x
iii
iiiii
y
y
1)1()(1
1)1()(11
22
2
1 ∫∫++
+−−−−
+−−−=−= +& ………………………………. (94)
Eq. 94 can be re-written and using Eq.84 yields:
[ ] [ ]
[ ] i
x
x
iii
iii
x
x
iii
x
x
iii
iiiii
Ldx
BAxC
BAxC
dx
BAxC
dx
BAxC
BAxCyy
i
i
i
i
i
i
++−−−−
−−−=
+−−−−+
+−−−−
−−−=−
∫
∫∫
+
++
+
1)1()(1
)1()(
1)1()(1
1
1)1()(1
)1()(
1
11
22
2
2222
2
1
…………………………..……………. (95)
Now substitute Eq. 89 in 95 yields:
)1,()2,(2
sin12
2
sin1 sin1
1
2
2
sin1
sin
2/
1
sin1
sin1
1
2
1
22
2
1
222
1 22
2
1 22
22
2
1 22
22
1
1
11
1
1
iiii
i
ii
i
i
i
ii
i
i
i
i
i
ii
i
i
i
iii
pEpEC
LdpC
L
dpdpC
L
dp
p
CL
dpp
p
C
BLyy
i
i
i
i
i
i
i
i
i
i
φφφφ
φφφφ
φφ
φ
φφ
φ
φ
φ
φ
φ
φ
φ
φ
φ
φ
φ
−−−=−−−=
−−
−+=
−+=
−
−+=−
+
+
∫
∫∫
∫
∫
+
++
+
+
……… (96)
Where the function E is the elliptic integral of the second kind.
39
APPENDIX E
Other End Conditions:
End Condition #1: This condition has a pin at bottom end and rotational fixed at the top end but
free to translate at the top as in Fig. 9.
Thus the moment Q0 at the tip of the column makes 0@
0==
=xydx
dy Now at y = yn = 0 and x = L
fn(L) = 0 since the moment at the bottom is zero or Eq. 21 becomes:
( )∑
∑
∑
−
=
−
=
−
=−−
−
+−−−−=
=+−+−+
=
+−==
1
1
000
1
1
000
1
011
1
)()(
or
0)()(
or
0)(1
)(
n
j
jjj
n
j
jjj
n
j
jjj
nn
nn
QxLPxLPQ
QxLPxLPQ
QxLPEIEI
MLf
……………….……..……….. (97)
Let
∑=
=i
j
ji PR0
…………………………………………………….…………………….. (98)
Then from Eq. 49
[ ]
+−−−−== ∑−
=
1
1
0
00
0
0 )()(1 n
j
jj QxLPxLPRR
QA …………………………….…….. (99)
Thus 0
2
00
0@
)( 83 Eq. from 0 BACdx
dy
xy
−=→===
And Eq. 24 does not apply. So rewriting Eq. 23 using Eq. 83 to
40
1
2
11
2
2
1
2
11
)()(
or
)()(
−−−
−−−
+−+−−=
+−=+−
iiiiiiii
iiiiiiii
BAxCAxCB
BAxCBAxC
…………………………………………… (100)
Therefore by substituting Ai , Ai-1 , Bi-1 in Eq. 99 then Bi is found and iii p and 1 , 2 1 φφ + are
found from Eq. 88 and Eq. 89 and the problem is solved using the elliptical integral.
Note: Eq. 56 and 57 can be written as:
i
ii
ii Lp
d
Cx
i
i
−−
−= ∫+12
1 22 sin12
1)(
φ
φ φ
φψ …………………………………………….. (101)
[ ]
[ ]
[ ]( )[ ]
∫
∑ ∫
+
+
+
+
−
+++
+−−
−−−
∂∂
=
+−
+⋅⋅⋅⋅⋅−⋅⋅⋅⋅⋅
+
−++−−
−−−
∂∂
=∂
∂
=
=
∞
=
+
=
=
1
1
1
1
2
1
2
3222
222
222
1
22
122
sin1sin
sincos1
24)(1
1
2)(
)(2
12
2642
)12(531
)(2)(1
1
2)(
)(
i
i
i
i
i
i
i
i
d
p
B
CpB
B
BAxCB
BAxA
x
x
dxBAxCB
Bk
k
k
xxB
B
BAxCB
BAxA
x
x
x
x
i
i
iii
mi
xx
xxiiii
mi
imi
m
k
x
x
k
iii
i
mi
ii
i
mi
xx
xxiiii
mi
imi
mm
ii
φ
φφ
φφ
φ
ψ
………………………………………………………… (102)
End Condition #2 Same as condition #1 except fixed at x = L as in Fig 10.
Using Eq. 93 at i = n, xn = L Lxdx
dy
=
=→@
0 yields:
0)( 1
2
11 =+− −−− iii BALC ……………………….…………………….……………… (103)
For fn(L) = 0 since the total moment at the bottom is zero implies
41
( ) n
n
j
jjj
n
n
j
jjj
n
j
njjj
nn
nn
QQxLPxLPQ
QQxLPxLPQ
QQxLPEIEI
MLf
++−−−−=
=−+−+−+
=
−+−==
∑
∑
∑
−
=
−
=
−
=−−
−
1
1
000
1
1
000
1
011
1
)()(
or
0)()(
or
0)(1
)(
………………………….…………….(104)
Thus the procedure is to pick Qn and find Q0 from Eq. 97 and use procedure in condition #1
above to solve for yi for given Li and find all Bi then iii p and 2 , 1 1 φφ + are from Eq. 82 and
Eq. 83 and check if Eq. 96 is satisfied if not update Qn and the problem is solved using the
elliptical integral.
End Condition #3 Pined both ends as in Fig 11.
Subdividing the column into two parts Part A and Part B at point E at line a – a where the slope
0=dx
dy in which it is to be found. Separate the loads of Part A and Part B and solve for y0A for
Part A using a straight cantilever column fixed at point E. Then solve Part B using end condition
#1 (pin at bottom end and rotational fixed at the top end but free to translate at the top.) and find
y0B . Check and see if y0A = y0B . If y0A not equal to y0B then move point E at line a – a up if y0A >
y0B or down if y0A < y0B until point E is found.
End Condition #4 Fixed both ends.
This condition is similar to end condition #3 where Part A and B are as condition #2.