Beams on Elastic Foundation 2

8/13/2019 Beams on Elastic Foundation 2

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This document is part of the notes written by Professor Terje Haukaas and posted at www.inrisk.ubc.ca.

The notes are revised without notice and they are provided as is without warranty of any kind.

You are encouraged to submit comments, suggestions, and questions to [email protected] .

It is unnecessary to print these notes because they will remain available online.

Beams on Elastic Foundation

Beams on elastic foundation, such as that in Figure 1, appear in building foundations,floating structures, beams that rest on a grid of perpendicular beams, and elsewhere. It

also turns out that the governing differential equation is similar to that of cylindricalshells and tapered beams with curved webs. As a result, this material has quite broad

applicability. However, it is important to exercise caution when applying this theory toproblems where the foundation does not behave in a linear fashion, such as soils that

respond nonlinearly to pressure or ships whose waterline area changes with sinkage.

In this document the beam is assumed to be an ordinary Euler-Bernoulli beam, thus the

theory from that document carries over to this one. Accordingly, downwards distributedload on the beam is referred to as q, while it is denoted qzwhen it acts in the opposite

direction, i.e., in the z-direction.

Figure 1: Beam on elastic foundation.

Foundation StiffnessThe foundation stiffness, ks, which is illustrated in Figure 1, is conceptually

straightforward. When the beam displaces downwards, the foundation exerts an upwardsforce, illustrated by the springs in Figure 1. kshas units of force per unit length along the

beam. Another stiffness, denoted k!, is also present, relating to rotation ! of the beam

cross-section around the longitudinal axis of the beam. In other words, k! provides

resistance against torsion of the beam. Techniques to determine ksand k!are described inthe following subsections for different situations.

Stiffness from Buoyancy

An important application of this theory is floating structures. To determine ksfor a beam

that floats in water, suppose the beam cross-section has width equal to bat the locationwhere the cross-section penetrates the water surface. Furthermore, let the mass density of

water be denoted "w, which typically equals 1,025kg/m3for seawater, and let gdenoted

the acceleration of gravity. For an infinitesimal beam length, dx, the change in buoyancy

q

ks

P


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Terje Haukaas University of British Columbia www.inrisk.ubc.ca

Beams on Elastic Foundation Page 2

force due to a vertical displacement !equals the weight of the displaced water, namelyweight density multiplied by volume:

B = !w

"g( ) " b "dx " #( ) (1)

As a result, the stiffness that resists the displacement !, per unit length of the beam, is

obtained by dividing Bby dxand !:

ks = !

w"g"b

kN

m2#

$%

&

'( (2)

which has unit force per unit length squared. The evaluation of ksaccording to Eq. (2) iscorrect as long as the beam width at the waterline does not change. Unfortunately, this

condition is not always satisfied, which is understood from Figure 2. While therectangular beam-cross-section maintains width bregardless of the vertical displacement,

the value of bfor the circular cross-section varies with the heave motion. As a result, ksis

a function of ! for the circular cross-section, which introduces nonlinearity that is

neglected in the subsequent equations.

Figure 2: Heave and roll motion of floating beams.

The fact that the waterline area is only a proxy for displaced water volume becomes even

more apparent when determining k!. To understand this, first consider the classical

approach for determining k!. When the rectangular beam-cross-section in Figure 2 rotatescounterclockwise as indicated by dashed lines, more water is displaced on the left sideand less water is displaced on the right side. As a result, buoyancy forces act against the

rotation, giving rise to k!. By computing the volumes of the shaded triangles in Figure 2the resultant buoyancy force, on either side, is

R = !w"g( ) "

1

2"

b

2"

b

2"#"dx (3)

The distance between the force pair, i.e., the buoyancy forces from each of the shaded

triangles in Figure 2 is 2/3 of b. Thus, the moment from buoyancy that resists the rotationis

T =R !2

3!b ="

w!g!

b3

12!#!dx (4)

b b

!


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As a result, the stiffness that resists the rotation !, per unit length of the beam, is obtained

by dividing Tby dxand !:

k! ="

w#g#

b3

12 (5)

where it is observed that the moment of inertia of the waterline area. Again it is observedthat the derivations do not hold for the circular cross-section in Figure 2. In fact, it iseasily seen that cross-section has zero resistance against rotation. If not easily seen, it is

easily felt when trying to balance on a floating timber log. The problem with applyingEq. (5) to the circular cross-section is the use of movement of the beam at the waterline

as a proxy for displaced water volume. Thus, while Eq. (5) is commonly used, it isimportant to be aware of its limitations.

Stiffness from Soil

Suppose the stiffness, ks, is determined from soil testing. In particular, suppose a vertical

load, P, is placed on an area with dimensions xand y, and that the vertical displacement,

!, is measured. The relationship between the distributed load and the displacement iswritten in terms of a distributed stiffness, kd:

P

x !y= k

d! " # k

d =

P

" !x !y

kN

m3$

%&

'

() (6)

While kd is stiffness per unit area, ks is stiffness per unit length. The sought value isobtained by multiplying kdby the beam width, b:

ks = b ! k

d

kN

m2

"

#$

%

&' (7)

If one assumes that the foundation material is linear elastic, there is no uniquerelationship between the Youngs modulus, E, of the foundation material and the stiffnessks. However, if one imagines that the soil underneath the beam is linear elastic with depth

Lto bedrock then the force-deformation relationship of the soil is

P=EA

L! " (8)

where A=x.yis the area loaded by P. Writing Eq. (8) in the form of Eq. (6) yields

P

x !y= k

d! " # k

d =

E

L

kN

m3$

%&

'

() (9)

and the sought stiffness is, in accordance with Eq. (7):

ks = b ! k

d =

b !E

L

kN

m2

"

#$

%

&' (10)

Stiffness from Girder Grid

Another situation appears when the beam is resting on a grid of closely spaced

perpendicular beams, e.g. joists. Suppose the joists are spaced at xon centre and that their


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stiffness against vertical deflection at the point of intersection with the beam is kb[kN/m].Then the sought stiffness is:

ks =

kb

x

kN

m2

!

"#

$

%& (11)

Differential EquationCompared with the basic Euler-Bernoulli beam theory, it is sufficient to modify theequation for vertical equilibrium to obtain the differential equation for a beam on elastic

foundation. As a result, the following conventions from basic beam bending hold valid:1) Clockwise shear force is positive; 2) Bending moment with tension at the bottom is

positive; 3) Tension stress is positive; 4) The z-axis points upwards, so that upwardsdisplacement, w, is positive; 5) The distributed load, qz, is positive upwards. Figure 3

shows the forces acting on an infinitesimal beam element. The springs that illustrate theelastic foundation exert a downward force when the beam is subjected to an upward

displacement.

Figure 3: Infinitesimal beam element.

Vertical equilibrium yields:

qz!dx"ks!w !dx"dV =0 (12)

Dividing by dxand re-arranging yields

qz =

dV

dx+ ks!w (13)

Substitution of this equation into the Euler-Bernoulli beam theory yields the followingrevised differential equation

EI!d

4w

dx4 + ks!w = qz (14)

dx

VM M+dMV+dV

ks

x

z, w

qz


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Another way of deriving the differential equation is to start with the following basicdifferential equation for beam bending:

EI!d

4w

dx4 =qz (15)

From earlier it is understood that the applied load, qz, plus the elastic foundation yields atotal force on the beam element equal to qzksw. By substituting this total load in theright-hand side of Eq. (15), Eq. (14) is obtained. For convenience, it is rewritten on the

form

d4w

dx4 +

ks

EI!w =

qz

EI (16)

In solving this differential equation it is useful to define a characteristic length,sometimes referred to as the elastic length. To approach the definition it is first noted that

EI/kshas dimension [m4]. As a result, the following definition of the characteristic length

has the dimension of length:

lc !

4 "EI

ks

4 (17)

The convenience of the factor 4 will become apparent later. It is also convenient to work

with the normalized coordinate !instead of the original coordinate, x, along the beam:

! =x

lc

(18)

To transform the differential equation, differentiation with respect to x is related to

differentiation with respect to !by the chain rule of differentiation:

d

dx=

d

d!"

d!

dx=

1

lc

"

d

d! (19)

This yields the following homogeneous version of the transformed differential equation:

d4w

d!4 + 4 "w = 0 (20)

SolutionThe characteristic equation is #

4+4=0 has the four different complex roots (1+i),

(1i), (1+i), and (1i). Consequently, the general solution is:

w(!) = C1"e

#!cos(!)+C2"e

#!sin(!)

"Damped terms"

! "##### $#####

+C3"e

!cos(!)+C

4"e

!sin(!)

"Undamped terms"

! "#### $####

(21)

where the phrase damped terms is employed to identify terms that vanish as !increases

to infinity. This labeling is useful because the solution for a point load must vanish faraway from the point of load application. In fact, only the damped terms appear in many


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practical situations. To shorten the notation under such circumstances, the followingauxiliary functions are defined:

g1= e

!"cos(")

g2= e

!"sin(")

g3=g

1+g

2

g4=g

1!g

2

(22)

These functions are plotted in Figure 4, where it is observed that they decay rapidly with

!. In fact, all functions approach zero once !increases beyond 4.

Figure 4: g-functions.

The auxiliary functions also have the following properties:

dg1

d! ="g

3

dg2

d! =g

4

dg3

d! ="2g

2

dg4

d! ="2g

1 (23)

In short, the general solution to the differential equation, with only the damped termsread:

w(!) = C1"g

1+C

2"g

2 (24)

The bending moment associated with the solution is:

M =EI!d2w

dx2 =

2EI

lc

2C

1g

2"C

2g

1( ) (25)

The shear force is:

V =dM

dx=2EI

lc

3C

1g

4+C

2g

3( ) (26)


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Reference Case 1Consider the beam in Figure 5, where one end is subjected to the forces Voand Mo, whilethe other end is infinitely far away. Both Voand Moare positive, i.e., the shear force is

clockwise and the bending moment gives tension at the bottom. Because the beam isinfinitely long, the solution cannot have contributions from the un-damped terms. As a

result, the solution is given by Eq. (24).

Figure 5: Reference Case 1.

At !=0, g1=g3=g4=1 and g2=0. As a result, Eqs. (25) and (26) yield the follow expression

for the bending moment and shear force at !=0:

M(! = 0) = 2EIlc

2C

2 (27)

V(x = 0) =2EI

lc

3C

1+C

2( ) (28)

Setting those expressions equal to Moand Vo, respectively, yields:

C1 =

lc

2

2EIlcVo!M

o( )

C2

=

lc

2

2EI

Mo

(29)

Hence, the solution reads:

w =! l

c

2

2EIM

og

4+ l

cV

og

1( )

" =lc

2EI2M

og

1+ l

cVog

3( )

M =Mog

3+ l

cV

og

2

V =!2

lc

Mog

2+V

og

4

(30)

Reference Case 2Consider the infinitely long beam in Figure 6, with a point load applied at !=0.

Immediately to the left of the point load the shear force is P/2, while immediately to theright it is P/2. That is, with reference to the previous case, Vo=P/2. Furthermore, the

rotation at the point load is zero:

!

Mo

Vo

EI

ks


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!(0) =lc

2EI2M

o+ l

cV

o( ) = 0 (31)

Substituting Vo=P/2 and solving for Moyields

Mo =

lcP

4 (32)

where it is noted that the bending moment at the point load is the same as that of a simply

supported beam with length lcloaded at midspan. In summary, the solution from Case 1is applicable also here, with

Vo =!

P

2 (33)

and

Mo =

lcP

4

(34)

Figure 6: Reference Case 2.

!

EI

ks

P

Beams on Elastic Foundation 2

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