1

Exact and Numerical Solutions for Large Deflection of Elastic

Non-Prismatic Beams

by

Farid A. Chouery, PE, SE1 ©2006, 2007, 2008 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, US Army Corps of Engineers Seattle DistrictAnd FAC Systems Inc. 6738 19th 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,

)()(1

32

xfy

y

………………………………………………….…………… (2)

Let

2

2

cos and

cos

1)(1tan

yyy

Where is a new variable, then substitute inq. 2 yields,

11 )(sinor )( cos )(cos CdxxfCdxxfdxf

2

12

12 )(1sin1cos

Cdxxf

Thus,

21

1

)(1

)(

cos

sin

Cdxxf

Cdxxfy

………………………………………………...……. (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,

)()(1

32

ygy

y

……………………………………………….……………… (5)

Let 322

11

1

x

x

dx

dy

dy

xd

xdx

xd

xy

xy

By replacing yields,

)()(1

32

ygx

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)

Case III:

5

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,

)()(1

32

byaxhy

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

2

2

22

22

2

22

22

22

22

cos1

tan1

ba

abv

ba

ab

ba

abv

…………………………………………………..…….. (15)

Where is a new variable, then substitute inq. 14 yields;

2122

12222

1cos

sin cos

Cduubah

Cduubahubah

………………....………….. (16)

7

22

22

2

122

1222

22

22

22

222

22

22

1

1

tan1

ba

ba

Cduubah

Cduubah

ba

ba

ba

ab

ba

abv

………………………………..……………. (17)

222

22

2

122

1222

22

22

1

1 Cu

ba

badu

Cduubah

Cduubah

ba

bav

………………..... (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

jjjjn

ii

i

jjjji

xxxQxxPM

xxxQxxPM

xxxQxxPQxxPM

xxxQxxPM

1

1

01

10

211110001

100000

for )(

for )(

for )( )(

for )(

………….........……………… (19)

Where, all xi are unknown.

FIG.1 – Cantilever Beam Analysis – Non-Prismatic Beam

P0

P1 P2

PiPi+1

L0 L1 Li

Q0 Q1 QiQ2

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

9

Point Loads are in the y direction

Now the moment of inertia:

i

jiTiTii LLLII

0

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

jjjj

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

jjjjj

i

i

i

jjjjj

ii xxx

CxxQxxPEI

CxxQxxPEI

xy

………………………..…....…………… (22)

Apply compatibilities yield:

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:

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

solution), so:

)()( 1 iiii xyxy ………………………………………………………...………………….. (28)

Thus,

11

222 )(1 iiiiii xyxyxL ……………………………......……………………… (29)

Or

2

0

2

1

1)()(5.01

1

CxxQxxPEI

xxL

i

jjijjij

i

iii …………………......………….. (30)

Thus, at i = 0 yields:

111)()(5.0

11

2

01

2

0002

0000

010

C

xx

CxxQxxPEI

xxL

……………...………. (31)

At i = 1 yields

1)()(5.01

)()(5.01

12

1112

1111

1002

1001

121

CxxQxxPEI

xxQxxPEI

xxL

2

1012

1011

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.

At i = 2 yields

1)()(5.01

)()(5.01

12

1222

1222

0102

0202

232

CxxQxxPEI

xxQxxPEI

xxL

………………......…………………………… (33)

12

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 000

1 )( xLxxxx n

n

iii

, 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

jjjjj

i

i

jjjjj

ii 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.

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.

13

FIG.2 – Cantilever Beam Analysis – Non-Prismatic Beam

Point Loads are in the x direction

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

yiii

xi

xiii

i

yii

i

xii

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:

P0 P2 Pi Pi+1

L0 L1 Li

Q0 Q1 Q2

Qi+1

P0

P1

P2

PiPi+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

P1

Qi

14

i

yii

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)

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

ii

iii

i

iii

i

i

i

iiiii v

b

bav

a

bau

u

y

u

xuyxL

………………………..………………………… (37)

Or

2222

1

)(1)(1

2)(

i

ii

i

iiiiiii b

uv

a

uvbauuL

……..…………………..………….. (38)

15

Which is again for a given C1 ui‘s can be found from Eq. 38. When using Eq.10:

021

21

10

1

01 )( uL

ba

auuuu

nn

nn

n

iii

. Thus, all ui‘s can be obtained by back substitutions

with checking if C1 satisfies:

0

1

1

)(1

)(1

or

0)()(

21

21

1

21

21

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

Lba

aL

ba

avuv

nn

n

nn

nnnn 2

12

1

1

21

21

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 .

16

FIG.3 – Cantilever Beam Analysis – Non-Prismatic Beam

Point Loads are in any direction

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:

1)(1

1)()(

and 1)(1

1)()(

211

111

2

iii

iiiii

iii

iiiii

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

kjij

kjijkn

i

r

knniiii k

xxQ

k

xxPxa

kn

nxZxZ ……..…... (66)

35

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

01

011 and where

)(1

i

jji

i

jji

i

l

li LlLl

L

xIEI

i

i ….…….……………………………… (69)

36

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

11

)()(5.01

and

)()(5.01

i

jjijjij

ii

i

jjijjij

ii

xxQxxPEI

Z

xxQxxPEI

Z

…………………………………………… (70)

Thus from Eq. 65 yeilds;

1

1

0

2

1

1

0

2

11

1)()(5.011

1

or

1)()(5.011

1

i

i

jjijjij

iii

i

i

jjijjij

iii

CxxQxxPEIEI

C

CxxQxxPEIEI

C

…………………………….. (71)

Starting with Eq. 25 yields;

1

0

2

11 )()(5.0

11

n

jjjjj

nn xLQxLP

EIC . ………………………………………… (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

jjijjij

i

iii

CxxQxxPEI

xxL …………………………………. (73)

So Eq. 31 becomes:

20

010

11 C

xxL

…………………………………………………………………………… (74).

37

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;

00102

01010

1 1)()(5.0 11

1 CxxQxxPEIEI

C

…………………………………….. (75)

So C11 can be found. By using Eq. 73 for i = 1yields;

2

10102

0101

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;

10202

0201212

12101

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

38

2

1

01

21

121

1

0

2

11

1

1

0

2

1

1)()(5.011

1

ifcheck

0 since )()(5.01

1

or

0 1)()(5.01

0)(

n

n

jjnjjnj

nnn

nn

n

jjjjj

nn

n

n

jjjjj

nn

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

jjjjj

i

i

i

jjjjj

ii 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

39

i

jj

ii

i

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

)(

………………………..……………………(81)

Where C1i is from Eq. 71 and from Eq. 79

0 since )()(5.01

11

0

2

11

nn

n

jjjjj

nn QPxLQxLP

EIC …………….….. (82)

So all of the C1i can be calculated from the vector {x} and update 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. 50 evaluated ix .

40

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

iiii

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

22 cos)1()(

and

sin1ith wcos1)(

iii

iiiii

BAxC

BdxCBAxC

………………………… (85)

12

1 22 1sin)1(1

sin1 i

i

dBC

BL

ii

ii

………………………………………….. (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

………………………………………………… (87)

Where:

41

iii

iii

i

ii

iii

iii

i

ii

Axp

CAx

B

C

Axp

CAx

B

C

11

11

1

11

2/cos

1cos2

and

2/cos

1cos1

……………………………… (88)

Where

2

1 ii

Bp

……………………………………..………………………………………….. (89)

And Eq. 87 becomes:

12

1 22 sin12

1 i

iii

ip

d

CL

………………………………………………………….. (90)

Which can be expressed in Elliptic Integral as follows:

)1,()2,(2

11 iiii

i

i pFpFC

L ……………………………………...………………. (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

cos2/

where

sin1sin2

1sin2)(

22

22

…………………………………………………………… (93)

The deflection between the joints becomes:

42

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

LdxBAxC

BAxC

dxBAxC

dxBAxC

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

iiiii

ii

i

i

i

ii

i

i

i

i

i

ii

i

i

iiii

pEpEC

LdpC

L

dpdpC

L

dp

p

CL

dpp

p

C

BLyy

i

i

i

i

i

i

i

i

i

i

……… (96)

Where the function E is the elliptic integral of the second kind.

43

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.

Fig. 9 Pin – Rotational Fixed Column

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:

y0

y

x

Q0

Ri

RiAi

x0

44

1

1000

1

1000

1

011

1

)()(

or

0)()(

or

0)(1

)(

n

jjjj

n

jjjj

n

jjjj

nn

nn

QxLPxLPQ

QxLPxLPQ

QxLPEIEI

MLf

……………….……..……….. (97)

Let

i

jji PR

0

…………………………………………………….…………………….. (98)

Then from Eq. 49

1

10

00

00 )()(

1 n

jjj QxLPxLP

RR

QA …………………………….…….. (99)

Thus 02

000@

)( 83 Eq. from 0 BACdx

dy

xy

And Eq. 24 does not apply. So rewriting Eq. 23 using Eq. 83 to

12

112

21

211

)()(

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 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)(

…………………………………………….. (101)

45

1

1

1

1

2

12

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

miimi

m

k

x

x

k

iiii

mi

iii

mi

xx

xxiiii

miimi

mm

ii

………………………………………………………… (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)( 12

11 iii BALC ……………………….…………………….……………… (103)

For fn(L) = 0 since the total moment at the bottom is zero implies

n

n

jjjj

n

n

jjjj

n

jnjjj

nn

nn

QQxLPxLPQ

QQxLPxLPQ

QQxLPEIEI

MLf

1

1000

1

1000

1

011

1

)()(

or

0)()(

or

0)(1

)(

………………………….…………….(104)

46

Fig. 10 Fixed – Rotational Fixed Column

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 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

0dx

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.

y0

y

x

Q0

Ri

RiAi

x0

Qn

47

Fig 11 – Pined both ends.

End Condition #4 Fixed both ends.

This condition is similar to end condition #3 where Part A and B are as condition #2.

x

RiB

RiAiB

RiA

RiAiA PART A

PART B

y0A

y0B

E a a