Exact Solution for Large Deflection on Beams · PDF fileExact and Numerical Solutions for Large Deflection of Elastic Non-Prismatic Beams by Farid A. Chouery, ... The solution for
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1
Exact and Numerical Solutions for Large Deflection of Elastic
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:
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.
10
0 since )()(5.01
11
or
0 1)()(5.01
0)(
1
0
2
11
1
1
0
2
1
nn
n
jjjjj
nn
n
n
jjjjj
nn
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
jjjjj
i
i
jjjjj
ii 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
2 dxxyLi
i
x
x ii or
1
2
0
2 1)()(5.01
1
i
i
x
xi
jjjjj
i
i
CxxQxxPEI
dxL ……………….....…….………. (27)
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
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
P0
P1
P2
Pi
Pi+1
L0 L1 Li
Q0 Q1 Qi
Q2
Qi+1
P0
P1 P2 Pi Pi+1
Q0
Q1
Qi
Q2
Qi+1
x2
xi
x1
x0
xn = L
xi+1
x
x
y
Segment Beam i Length = Li
y0 y1
y
17
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
sin and cos
where,
)()(sin)(cos)(),(
or
sin)(cos)(),(
00
0
0
0
0
0
000
0000
i
j i
ji
i
j i
jii
j i
j
i
j i
jj
ii
j i
j
i
j i
jj
i
iiii
i
jjj
i
jiii
EI
Pb
EI
Pa
EI
PEI
yP
y
EI
PEI
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
18
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.
FIG.4 – Fishing Pole Example (a)
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.
y
P0
P1
L0, I0
P0
P1
x1
x0
xn = L
x
x
L1, I1
α
α
α
P0 sinα
P0 cosα
19
(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:
FIG.5 – Bow and Arrow Example (c)
2)(
)()(
2)(
1
2cot)(
2)(
1),(
20
2
000
0
000
000
P
xLL
xLyy
Pxx
EI
Pyy
Pxx
EIybxah
y0
x0
L - x0
L
20
2 )( xLL
x y
0.5P
0.5P
P
β
20
20
2
0
00
20
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
ii
ii
ii
i
iiii
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,
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:
21
20
20
0
20
20
0
00
20
20
002
02
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
ii
dd
d
ii
ii
dd
d
dd
d
iii
xy
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,
β
y
x
y0 x0P
xd
D
L,I
yd
22
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
2xt
xR
xR
xEI
xM
y
y
………………………………………… (42)
And solution is as Eq. 3 and Eq. 4 by replacing f(x) by t(x).
23
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.
Fig. 7 Extensibility
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
iii
1
)sin( ……………………………………………….……………… (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
x
y
θi
αi
90 - αi
Pi
Li
180 – (90 - αi) – θi = 90 – (θi - αi)
24
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:
1
sincos
sin)sin(
ˆˆ
1
1sincos
/1
1sincos
sincos
sincos
sincoscossin
cos)sin(
ˆˆ
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
jjji
i
j
x
xj
jji
i
jjii
j
j
x
xj
jjji
j
j
x
xj
jjj
i
j
j
x
xj
jjj
i
j
j
x
xj
jjj
i
j
j
x
xj
jjji
i
j
x
xj
jji
i
jjii
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
…………………. (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 .
25
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
x i
x
xi
iiiix
x i
i
x
xi
i
i
i
i
ii
x
x i
i
x
xi
iiix
x i
i
x
xi
iiix
x i
x
xi
iix
x ii
x
x iii
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
i
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 , 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
iiy
n
i
x
xi
iix
dsEA
P
dsEA
P
i
i
i
i
………………………………………………….. (46)
26
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.
27
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
28
APPENDIX A
General algorithm solution of solving for xi for application example Case I:
From Eq. 27 let the function:
c
CxxQxxPEI
dxx
i
jjjjj
i
i 2
0
2 1)()(5.01
1
)( ……………..…..………. (47)
When setting 0x
i there is no solution and the function )(xi 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
x
i rewrite Eq. 47 to be:
cBAxC
dxx
iii
i 22)(1)( …………………………………….........………..….. (48)
Where:
i
jj
ii
i
jjjjj
iiii
i
jj
i
jjjj
i
PEI
C
CxQxPEI
CAB
P
QxP
A
0
0
22
0
0
5.01
1)5.0(1
)(
……………………………......……………….. (49)
And from Eq. 25:
29
1
0
2
1
)()(5.01
1n
jjjjj
n
xLQxLPEI
C ………………………......………………… (50)
Thus the inflection points are only three and they are:
i
iii
ii
C
BAx
Ax
3,2
1
…………………………………………………………………………… (51)
The function becomes:
cdxBAxC
k
kc
BAxC
dxx
k
k
iii
iii
i
)(2642
)12(5311
)(1)(
1
22
22
………….………………………………. (52)
The slopes and the deflections become:
122
2
for )(1
)()(
ii
iii
iiii xxx
BAxC
BAxCxy …………………..………….. (53)
i
x
A
iii
iii
ik
k
iiiiiii
iii
iii
iiii
CBAzC
dzBAzC
CdxBAxCk
kAxBAxC
xxxCBAxC
dxBAxCxy
i
2)(1
)(
2 )(2642
)12(531)()(
3
1
for 2)(1
)()(
22
2
1
1223
122
2
…………………………………………… (54)
And find
i
x
A i
x
A ii
L
A nn
CdxxydxxyC
dxxyC
i
i
i
i
n
2 )( )(2
)(2
1
1
1
1
1
1
1
…………..……………………………….……… (55)
And let
30
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
…………………………………………. (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 x 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
miimi
m
k
x
x
k
iiii
mi
iii
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
x
………………………………………………………. (57)
Where:
imQxLPEI
B
imQxLPEIEI
QxPCAAB
imA
imP
PA
mmmn
mi
mmmni
mmmimiimi
mi
i
jj
mmi
)(1
and for )(1
2
0
and for
0
……….………….. (58)
31
Thus, choose
i
jii Lx
0
for the initial condition and the Newton-Raphson method requires the
updated 1 Jxx . Where, x is the updated vector of ix , x is the old vector of ix ,
J is the Jacobian matrix evaluated ix and 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,
1111
100001
00
1000
00200
00
200
0
00
0
000
11002
112
001
)(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
)(
11
1101
1
001
10
111
10
001
1100211
200
11
211
0
101
10
1011001
ALEI
PBAL
EI
PB
PP
PA
PP
PA
CxQxQxPxPEI
CAB
EI
PPC
PP
QQPxPxA
32
Let H consider the variables in Eq. 57
),,,,,,(),,,,,,(
),,,,,,(),,,,,,(
),,,,,,(
),,,,,,(
111111110101011111011
1010000101000000001000
111111011
000001000
BACBAxxHBACBAxxH
BACBAxxHBACBAxxHJ
BACBAxxHx
BACBAxxHx
m
mmmm
mmmm
And an exact numerical solution is obtained.
FIG.8 - Example
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.
P0
P1
L0, I0
Q0 Q1
P0
P1Q0
Q1
x1
x0
xn = L
x
x
y
L1, I1
33
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
nn
r
n
nn
r
n
nn
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
00
00
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
100
for 1 )(1)(1)(
iii
i
jjjj
nr
nniii xxxCdxQxxPxaCdxxfCxZ
……………………………...……………..….. (60)
34
Using integration by parts starting with
yeilds, )( and 00
i
jjjj
nr
nn QxxPdvxau
dxxxQxxP
xnaxxQxxP
xaxZi
jjj
jjnr
nn
i
jjj
jjnr
nni )(
2
)( )(
2
)( )(
0
21
10
2
0
……………………………………………………. (61)
Continuing the integration by parts on each integral leads to
r
0k1
0
12
for )!1(
)(
)!2(
)(
)!(
!)( ii
i
j
kjj
kjjkn
r
knni xxx
k
xxQ
k
xxPxa
kn
nxZ
……..………………………… (62)
At xi Eq. 3 for to consecutive segments at the joint becomes: