Chapter 5

Beams on Elastic Foundation

F F

Winkler foundation elastic foundationno load applied

x

z, w

Winkler Foundation reaction pressure

0Wp k w=

30 [ N/m ]k foundation modulus

floating on liquid (buoyancy)

0 liquid liquidk g= γ = ρ

reaction force intensity (reaction force over unit length)

0Wq k b w k w= − = −

b width of the beam

2[ N/m ]k foundation constant

Basic Equations

x

z,w

V(x+dx)V(x)

M(x) M(x+dx)q(x)

k w(x)

dx

( )WdV q q k w qdx

= − + = −

dM Vdx

=

2

2d M k w qdx

= −

2

2d wM E Idx

= −

3

3d wV E Idx

= −

4

4d wE I k w qdx

+ =

homogeneous equation

4

4 0d wE I k wdx

+ =

4

44 4 0d w w

dx+ β =

4

4kE I

β =

general solution of the homogeneous differential equation

xw eα=

4'''' 0w w− α = ⇔ 4'''' 4 0w w+ β =

4 44α = − β , 2 22 iα = ± β , 2 (1 )i iα = ± ± β = ± ± β

x i xw e e±β ± β=

1 2 3 4( sin cos ) ( sin cos )x xw e C x C x e C x C xβ −β= β + β + β + β

example

sinxw e xβ= β

' (sin cos )xw e x xβ= β β + β

2'' 2 cosxw e xβ= β β

3''' 2 (cos sin )xw e x xβ= β β − β

4'''' 4 sinxw e xβ= − β β or 4'''' 4 0w w+ β =

short-hand notation

(cos sin )xxA e x x−β

β = β + β , sinxxB e x−β

β = β

(cos sin )xxC e x x−β

β = β − β , cosxxD e x−β

β = β

differentials

2 3

2 2 3 31 1 1

2 2x x x

xdD d C d B

Adx dx dxβ β β

β = − = =β β β

2 3

2 2 3 31 1 1

2 2 4x x x

xdA d D d C

Bdx dx dxβ β β

β = − = = −β β β

2 3

2 2 3 31 1 1

2 2x x x

xdB d A d D

Cdx dx dxβ β β

β = = − =β β β

2 3

2 2 3 31 1 1

2 2 4x x x

xdC d B d A

Ddx dx dxβ β β

β = − = − =β β β

distributions

sinx

xB e x−ββ = β cosx

xD e x−ββ = β

ßx

B(ßx

), D

(ßx)

-40

-20

0

20

40

-5 -4 -3 -2 -1 0 1 2 3 4 5

B(ßx)D(ßx)

ßx

B(ßx

), D

(ßx)

-1

0

1

0 1 2 3 4 5

B(ßx)

D(ßx)

Semi-Infinite Beams with Concentrated Loads

wox

z,w

Po

Mo

z,w

xθo

general solution of the homogeneous differential equation

1 2 3 4( sin cos ) ( sin cos )x xw e C x C x e C x C xβ −β= β + β + β + β

boundedness

3 4 3 4( sin cos )xx xw e C x C x C B C D−β

β β= β + β = +

boundary conditions

2

20

( 0)ox

d wM M x E Idx =

= = = −

3

30

( 0)ox

d wP V x E Idx =

− = = = −

2 2

3 4'' 2 2x xw C D C Bβ β= − β + β

3 33 4''' 2 2x xw C A C Cβ β= β + β

( 0) 0xB xβ = = and ( 0) ( 0) ( 0) 1x x xA x C x D xβ β β= = = = = =

232oM E I C= β or

23 2

22

o oM MCkE I

β= =

β

33 42 ( )oP E I C C= β + or

24

2 2o oP MCk kβ β

= −

full solution

2 22 2 2( )o o o

x xM P Mw B D

k k kβ ββ β β

= + −

x x xC D Bβ β β= −

22 2o o

x xP Mw D C

k kβ ββ β

= −

2 32 4' o o

x xP Mw A D

k kβ ββ β

θ = = − +

'' o xo x

P BM E I w M Aβ

β= − = − +β

''' 2o x o xV E I w P C M Bβ β= − = − − β

end force load end moment load

x

z

Po

x

w

π2β

π4β

M

x

x

z

x

w

3π4β

π4β

M

x

Mo

Infinite Beams with Concentrated Loads

x

z,w

Po

symmetric deflection

( ) ( )w x w x= −

'( ) '( )w x w x= − −

'(0) 0w =

MoMoPo /2Po /2

2 3 2 32 ( / 2) 4 4'(0) (0) (0) 0o o o ox x

P M P Mw A Dk k k kβ β

β β β β= − + = − + =

4o

oPM =β

22 ( / 2) 2 ( / 4 ) (2 )

2o o o

x x x xP P Pw D C D Ck k kβ β β β

β β β β= − = −

2o

xPw Ak β

β=

2 3 22 ( / 2) 4 ( / 4 )' ( )o o o

x x x xP P Pw A D A Dk k kβ β β β

β β β βθ = = − + = − −

2

ox

P Bk β

βθ = −

( / 2)'' ( / 4 ) (2 )

4o x o

o x x xP B PM E I w P A B Aβ

β β β= − = − + β = − −β β

4o

xPM Cβ=β

''' ( / 2) 2 ( / 4 ) ( )2o

o x o x x xPV E I w P C P B C Bβ β β β= − = − − β β = − +

2o

xPV Dβ= −

3π4β

w

x

x

z

Po

πβ

θ

x

x

π4β

M

x

V

π2β

Po

x

z,w

Mo

symmetric deflection

( ) ( )w x w x= − −

(0) 0w =

'( ) '( )w x w x= −

PoMo /2

Mo /2Po

2 22 2 ( / 2) 2(0) (0) (0) 0o o o ox x

P M P Mw D Ck k k kβ ββ β β β

= − = − =

2o

oMP β

=

2 22 ( / 2) 2 ( / 2) ( )o o o

x x x xM M Mw D C D Ck k kβ β β β

β β β β= − = −

2

ox

Mw Bk β

β=

2 3 32 ( / 2) 4 ( / 2)' ( 2 )o o o

x x x xM M Mw A D A Dk k kβ β β β

β β β βθ = = − + = − −

3

ox

M Ck β

βθ =

( / 2)'' ( / 2) ( )

2o x o

o x x xM B MM E I w M A B Aβ

β β ββ

= − = − + = − −β

2o

xMM Dβ=

''' ( / 2) 2 ( / 2) ( 2 )2

oo x o x x x

MV E I w M C M B C Bβ β β ββ

= − = − β − β = − +

2o

xMV Aβ

β= −

3π4β

V

x

w

πβ

x

θ

x

π4β

x

M

π2β

Mo

x

z

Mo

Continuous running load on an infinite beam

x

z,w

a bq( )x

x

z,w

q( )x+dxq( )x

x

z,w

q( )x dxdP =

2 xdPdw Ak β

β= ,

2x

dPd Bk β

βθ = − ,

4 xdPdM Cβ=β

, 2 x

dPdV Dβ= −

for running load between x = a and b:

( )2

bx

aw A q x dx

k ββ

= ∫

etc.

Constant running load intensity on an infinite beam

02

bx

a

qw A dxk β

β= ∫

Uniformly distributed load on a semi-infinite beam

x

z

x

z

q0

4

04d wE I k w qdx

+ =

0qwk

=

x

z

x

w

k

R

q0

q0

02

xqRw D

k kββ

= − + so that 02(0) 0qRwk kβ

= − + =

0

2qR =β

0 (1 )x

qw Dk β= − , 0

xq Ak β

βθ = , 0

22x

qM Bβ=β

, 02 xqV Cβ=β

Rx

z

x

w

k

Mw

q0

q0

2022 w

x xM qRw D C

k k kβ βββ

= − + +

2 32 4 w

x xR MA D

k kβ ββ β

θ = −

2

022(0) 0wM qRwk k k

ββ= − + + = and

2 32 4(0) 0wR Mk kβ β

θ = − =

2

02 3

2 20 2 4 w

RqM

⎡ ⎤β − β ⎡ ⎤⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦β − β⎣ ⎦

, 0qR =β

, 022

wqM =β

0 (1 2 )x x

qw D Ck β β= − +

02 ( )x x

q A Dk β ββ

θ = −

02 (2 )

2x x

qM B Aβ β= −β

0 ( )x x

qV C Bβ β= +β

0 (1 )x

qw Ak β= − , 02

xq B

k ββ

θ = , 022

xqM Cβ= −β

, 0x

qV Dβ=β

Uniformly distributed load on an infinite beam

x

z

Qa b

q0

x

z

Q Oξ

dξq0

dξq0

0 ( ) if 02Q xq ddw A

k ββ ξ

= −ξ ξ ≤ and 0 ( ) if 02Q xq ddw A

k ββ ξ

= ξ ξ >

0

0

0( ) ( )

2

b bQ Q x x

a a

qw dw A d A dk β β

ξ=− −

⎡ ⎤β= = −ξ ξ + ξ ξ∫ ∫ ∫⎢ ⎥

⎣ ⎦

0 0

0 0 0 0

1 1( ) ( )2 2

a ba bQ x x x x

q qw A d A d D Dk kβ β β β

⎡ ⎤⎡ ⎤β β ⎛ ⎞ ⎛ ⎞⎢ ⎥= ξ ξ + ξ ξ = − + −∫ ∫⎢ ⎥ ⎜ ⎟ ⎜ ⎟β β⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

0 2

2Q a bqw D D

k β β⎡ ⎤= − −⎣ ⎦

02

Q a bQ

w D Dq a bx k a x b x

β β∂ ∂ ∂⎡ ⎤∂ ∂θ = = − −⎢ ⎥∂ ∂ ∂ ∂ ∂⎣ ⎦

1ax∂

=∂

, 1bx∂

= −∂

02Q a bq A Ak β β

β ⎡ ⎤θ = −⎣ ⎦ , 024

Q a bqM B Bβ β⎡ ⎤= +⎣ ⎦β

, 04Q a bqV C Cβ β⎡ ⎤= −⎣ ⎦β

maximum bending moment

0 0

4Q a bqV C Cβ β⎡ ⎤= − =⎣ ⎦β

short loading span, β ≤ π

only one zero at / 2a b= =

0max / 222

qM Bβ=β

long loading span, 2β > π

dominant zero close to left end 2bβ > π

0 04Q aqV Cβ≈ =β

(cos sin ) 0aaC e a a−β

β = β − β = or 4

a πβ =

0

max / 42 20.08064q qM Bπ≈ ≈β β

Beams of finite length

x

z

a b

Po

general deflection solution

1 2 3 4( ) ( sin cos ) ( sin cos )x xw x e C x C x e C x C xβ −β= β + β + β + β if 0a x− ≤ ≤

1 2 3 4( ) ( sin cos ) ( sin cos )x xr r r r rw x e C x C x e C x C xβ −β= β + β + β + β if 0 x b≤ ≤

boundary conditions

''( ) 0w a− = , '''( ) 0w a− = , ''( ) 0rw b = , '''( ) 0rw b =

(0) (0)rw w= , '(0) '(0)rw w= , ''(0) ''(0)rw w= , '''(0) '''(0) or

Pw wE I

= −

symmetric loading

x

z

Po

/2 /2

( ) ( )w x w x= −

1 2 3 4( ) ( sin cos ) ( sin cos )x xw x e C x C x e C x C xβ −β= β + β + β + β if 0 / 2x≤ ≤

1 2 3 4( ) cosh sin cosh cos sinh sin sinh cosw x D x x D x x D x x D x x= β β + β β + β β + β β

boundary conditions

''( / 2) 0w = , '''( / 2) 0w = , '(0) 0w = , '''(0)2

oPwE I

=

02 cosh cos(0)

2 sinh sinoPw wk

β + β + β= =

β + β

0cosh cos(0)

4 sinh sinoPM M β − β

= =β β + β

0lim2

oPwkβ →∞

β= and 0lim

4oPM

β →∞=

β

00

4lim2 2

o oP Pwk kβ →

β= =

β and 0

0lim 0M

β →=