Barber J.R. Intermediate Mechanics of Materials

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7.1 The governing equation 355

V(z)

V(z+δz)

M(z)

M(z+δz)

p(z)δz

w(z)δz

δz

(b)

Figure 7.3: (a) A beam supported on a Winkler foundation, (b) equilibrium of a beam

element

Equilibrium of vertical forces therefore requires that the shear force V satsify the

equation

V (z +

z) −V(z) −w(z)

z − p(z)

z = 0

and hence

V (z +

z) −V (z)

= w(z)+ p(z) .

Allowing

z →0 and using (7.1), we then obtain

= w(z)+ p(z) = w(z)+ ku(z) . (7.2)

As in §6.1, moment equilibrium requires

= V (7.3)

and the bending moment M and the displacement u are related by the bending equa-

tion (1.17)

M = −EI

. (7.4)

Eliminating M,V between equations (7.2–7.4), we obtain

+ ku = −w , (7.5)

which is the governing equation for an elastic beam supported by a Winkler founda-

tion.

7.1.1 Solution of the governing equation

The applied load w will generally be a known function of z and hence (7.5) consti-

tutes an inhomogeneous ordinary differential equation for the unknown displacement

u(z). Once this equation is solved, the bending moment can be recovered by substi-

tution in (7.4).

beam

u(z)

w(z) per unit length

(a)

356 7 Beams on Elastic Foundations

The general solution of (7.5) can be written as the sum of a particular solution

and the general homogeneous solution. The particular solution is any function that

satisﬁes (7.5) and the homogeneous solution is the general solution of the corre-

sponding homogeneous equation

+ ku = 0 . (7.6)

Since (7.6) is a fourth order equation, its solution will contain four arbitrary constants

which permit the satisfaction of two boundary conditions at each end of the beam, as

in elementary beam problems.

The particular solution and the homogeneous solution can be given a physical in-

terpretation. The particular solution corresponds to a state in which the correct load

w(z) is applied, without regard to the end conditions, which will therefore gener-

ally not be those required in the problem. Hence, we must superpose forces and/or

moments at the ends so as to restore the correct end conditions. This is the func-

tion of the general homogeneous solution, which describes all possible states of the

same beam loaded at the ends only — after all, (7.6) is simply (7.5) with w(z) equal

to zero. We shall ﬁrst examine the homogeneous solution in some detail, since it

affords considerable insight into the general behaviour of beams on elastic founda-

tions. In particular, we shall ﬁnd that the deformation tends to be quite localized near

the loads, in contrast to the behaviour of beams on discrete supports.

7.2 The homogeneous solution

If the beam has no distributed load w(z), the displacement is given by equation (7.6),

whose form suggests the existence of solutions of the form

u(z) = Ae

, (7.7)

where A,b are constants. Substitution into (7.6) yields

EIb

+ kAe

= 0 ,

showing that (7.7) will be a solution of (7.6) if and only if

= −

. (7.8)

This equation has no real roots, but it has four complex roots which can be written

b = (±1 ±i)

where

4EI

. (7.9)

This solution is easily veriﬁed by substitution back into (7.8).

7.2 The homogeneous solution 357

It follows that the general solution of the homogeneous equation (7.6) can be

written

u(z) = A

(1+i)

+ A

(1−i)

+ A

(−1+i)

+ A

(−1−i)

, (7.10)

where A

are four independent complex constants. However, the displace-

ment must be a real function, so we must have A

, where the overbar

denotes the complex conjugate. After some algebraic manipulations, the most gen-

eral real function of the form (7.10) can be written

u(z) = B

cos(

z) + B

sin(

z) + B

−

cos(

z) + B

−

sin(

z) . (7.11)

An alternative form which is sometimes more convenient for beams of ﬁnite length

u(z) = C

cosh(

z)cos(

z) +C

sinh(

z)sin(

z) +C

cosh(

z)sin(

sinh(

z)cos(

z) . (7.12)

Notice that (7.11) and (7.12) are equivalent because of the identities

cosh(

z) =

+ e

−

; sinh(

z) =

−e

−

In fact, (7.12) can be obtained from (7.11) by writing

−C

; B

−C

The reader uncomfortable with the complex algebra leading to equations (7.11, 7.12)

is encouraged to verify them by direct substitution into (7.6).

7.2.1 The semi-inﬁnite beam

Figure 7.4 shows a semi-inﬁnite beam, z> 0, loaded by a force F

and a moment M

at the end z = 0. The displacement must be bounded as z →∞ and hence only the

exponentially decaying terms should be retained from (7.11), giving

u(z) = B

−

cos(

z) + B

−

sin(

z) . (7.13)

u(z)

Figure 7.4: The semi-inﬁnite beam loaded only at the end

358 7 Beams on Elastic Foundations

A more rigorous argument showing that the exponentially growing terms should be

dropped can be obtained by considering a beam of ﬁnite length L, where

L ≫1.

In this case, all four terms in (7.11) should be retained in order to satisfy the two

boundary conditions at each end, but it will be found that the coefﬁcients B

→0

L →∞, for any boundary conditions on the end z = L. We shall consider ﬁnite

beam problems further in §7.6 below.

Progressive derivatives of equation (7.13) with respect to z provide corresponding

expressions for the slope

, bending moment M, and shear force V as functions of z.

We have

≡

= −B

−

[cos(

z) + sin(

z)]

−

[cos(

z) −sin(

z)] (7.14)

M = −EI

= −2B

−

sin(

+2B

−

cos(

z) (7.15)

V =

= −2B

−

[cos(

z) −sin(

z)]

−2B

−

[cos(

z) + sin(

z)] . (7.16)

It is convenient to deﬁne the four functions

(x) = e

−x

cosx ; f

(x) = e

−x

sinx ; f

(x) = e

−x

(cosx + sinx) ;

(x) = e

−x

(cosx −sinx) , (7.17)

which permit equations (7.13–7.16) to be written in the compact form

u(z) = B

(

z) + B

(

z) (7.18)

(z) = −B

(

z) + B

(

z) (7.19)

M(z) = −

(

z) +

(

z) (7.20)

V (z) = −

(

z) −

(

z) , (7.21)

where we have used (7.9) to eliminate the ﬂexural rigidity EI in favour of the mod-

ulus k in equations (7.20, 7.21). It can be veriﬁed by substitution that the functions

, f

satisfy the relations

( f

+ f

) ; f

( f

− f

) ; f

= f

+ f

; f

= f

− f

(7.22)

and

d f

= −f

;

d f

= f

;

d f

= −2 f

;

d f

= −2 f

. (7.23)

Equations (7.18–7.21) can be used to determine the constants B

for any two

boundary conditions at the end z=0. For the case illustrated in Figure 7.4, we have

7.2 The homogeneous solution 359

= V (0) = −

−

; M

= M(0) =

and hence

= −

−

; B

Substituting into (7.18–7.21) and using (7.22) to simplify the resulting expres-

sions, we obtain

u(z) = −

(

z) −

(

z) (7.24)

(z) =

(

z) +

(

z) (7.25)

M(z) =

(

z) + M

(

z) (7.26)

V (z) = F

(

z) −2M

(

z) . (7.27)

Example 7. 1

A semi-inﬁnite steel beam of second moment of area I

=0.5 ×10

is supported

on an elastic foundation of modulus k = 10 MPa and loaded by a downward force

=10 kN at the end, as shown in Figure 7.5. Find the slope and deﬂection at the free

end and the location and magnitude of the maximum bending moment (E

steel

= 210

GPa).

Figure 7.5: The semi-inﬁnite beam loaded by an end force

We ﬁrst use equation (7.9) to calculate

4EI

10 ×10

4 ×210 ×10

×0.5 ×10

−6

= 2.209 m

−1

The complete solution can be obtained by substituting F

= 10 kN, M

= 0 into

equations (7.24–7.27). In particular, the end displacement and slope are

u(0) = −

= −



2 ×2.209 ×10 ×10

10 ×10



= −4.4 ×10

−3

— i.e. 4.4 mm downwards, and

u(z)

360 7 Beams on Elastic Foundations

(0) =

2 ×2.209

×10 ×10

10 ×10

= 9.8 ×10

−3

radians (0.56

) .

To ﬁnd the maximum bending moment, we ﬁrst note that

= V = −



−



(

z) = F

−

[cos(

z) −sin(

z)] ,

using (7.24). The maximum bending moment occurs when dM/dz= 0 and hence

cos(

z) −sin(

z) = 0 .

This equation has roots at

z =

,...

In view of the exponential decay, the magnitude of M will be greatest at the ﬁrst root,

where

z =

4 ×2.209

= 0.356 m .

At this point, we have

max

−

√

= 0.322

0.322 ×10 ×10

2.209

= 1457 Nm .

Figure 7.6 shows the deformed shape of the beam and the bending moment as a

function of z. Notice that the disturbance decays rapidly with z, but the trigonometric

functions in equations (7.13–7.16) cause the decay to be oscillatory, so there are

actually regions in which the beam is displaced upwards

by the downward force F

(b)

Figure 7.6: (a) Deformed shape of the beam and (b) the bending moment distribution

If the beam is simply resting on the support, rather than being attached to it, an upward

displacement will cause the beam to separate from the foundation and the restoring force

p(z) will then be locally zero, rather than being given by equation (7.1). A foundation of

this kind that exerts a restoring force only for displacements in one direction is known as a

unilateral support. However, in problems with downward loading, the weight of the beam

will probably be sufﬁcient to maintain contact in the upward displacement regions.

original centreline

(a)

7.3 Localized nature of the solution 361

7.3 Locali zed nature of the solution

The rapid decay of the disturbance observed in the above example is typical of end

loading problems for beams on elastic foundations. Figure 7.7 shows the four func-

tions f

, f

, of equations (7.17) and all have esentially decayed to negligible

values at x= 4. In view of equations (7.18–7.21), we conclude that for any end load-

ing of the beam, the deformation will be restricted to a region at the end deﬁned

approximately by

0 < z < 4l

, (7.28)

where

≡

4EI

(7.29)

has the dimensions of length and functions as a characteristic decay length for the

beam/support system. For Example 7.1, the decay length was l

= 1/2.209 = 0.453

Figure 7.7: The functions f

, f

, of equations (7.17)

In any practical problem involving a beam on an elastic foundation, the very

ﬁrst step should always be to calculate l

from the properties of the beam and the

foundation. Comparison of this length with other length scales in the problem then

provides important information about the behaviour of the system and can often be

used to justify a simpler solution procedure. If the length of the beam L ≫l

(i.e. if

L ≫1), it follows that end loads on the beam will have only local effects and there

will be a central region, distant from the ends, where the beam will be unaffected

by the end conditions. This condition is often satisﬁed in practice and it simpliﬁes

the resulting problem, since effects at the two ends can then be analyzed separately,

using equations (7.18–7.21).

Similar localization occurs when loads are applied over a restricted central region

of the beam. If the loads are at least 4l

from either end of the beam, their effects will

not be inﬂuenced by conditions at the ends and it is sufﬁcient to solve the simpler

problem in which the same loads are applied to an inﬁnite beam (see §7.4 below).

362 7 Beams on Elastic Foundations

non-local displacements

Figure 7.8: Estimation of the signiﬁcance of non-local effects

We can also use an estimate of l

to determine whether a more realistic ‘non-

local’ foundation can be approximated by an equivalent Winkler foundation in a

particular application (see Figure 7.1 and the associated discussion). Suppose a uni-

form load is applied to the non-local foundation over an extended region, as shown

in Figure 7.8. The Winkler approximation is reasonable if the regions of signiﬁcant

non-local displacements adjacent to the load extend a distance signiﬁcantly smaller

than l

Figure 7.9: A beam supported on discrete springs with spacing d

Another case of some interest concerns a beam supported on a set of discrete

springs, as shown in Figure 7.9. If the springs each have stiffness k

and the distance

between adjacent springs is d, the effective modulus of the foundation (the stiffness

associated with a unit length) will be

k =

, (7.30)

giving

4EId

. (7.31)

We can then argue that representation of the discrete springs as a continuous founda-

tion will be reasonable if d ≪ l

and hence

4EI

≪ 1. (7.32)

7.4 Concentrated force on an inﬁnite beam

In view of the preceding discussion, it is of interest to consider the problem of Fig-

ure 7.10 (a), in which an inﬁnite beam on an elastic foundation is loaded by a con-

centrated force F

. We can locate the origin at the point of application of the force

without loss of generality.

stiffness k

7.4 Concentrated force on an inﬁnite beam 363

(a) (b)

Figure 7.10: (a) Inﬁnite beam loaded by a concentrated force, (b) equilibrium of an

inﬁnitesimal element at O

Strictly speaking, this is an inhomogeneous problem in which the load w(z) =

(z), where

(z) is the Dirac delta function. However, we can reduce the prob-

lem to a homogeneous problem for the semi-inﬁnite beam z > 0 using a symmetry

argument.

The problem is symmetrical about O and hence the slope

must be zero at O.

Furthermore, equilibrium of an inﬁnitesimal beam element immediately under the

load [Figure 7.10 (b)] shows that the shear force immediately to the right of the force

must be F

/2. Thus the region z> 0 has end conditions

(0) = 0 ; V (0) =

Substituting into equations (7.19, 7.21), we obtain the two simultaneous equations

−B

+ B

= 0 ; −

−

for B

, with solution

= B

= −

It follows from (7.18, 7.20) that

u(z) = −

−

[cos(

z) + sin(

z)] = −

(

z) (7.33)

M(z) = −

−

[cos(

z) −sin(

z)] = −

(

z) (7.34)

in z>0. The symmetry of the problem requires that the same solution can be used in

z< 0 with z replaced by −z. Thus, expressions that are correct for all values of z can

be written

u(z) = −

(

|z|) ; M(z) = −

(

|z|) . (7.35)

The slope

and shear force V are odd functions of z (e.g. V(z) = −V (−z)) and can

be written

(z) =

sgn(z) f

(

|z|) ; V (z) =

sgn(z) f

(

|z|) , (7.36)

364 7 Beams on Elastic Foundations

using (7.3, 7.21), where sgn(z)=1 for z>0 and −1 for z< 0.

The maximum displacement and bending moment occur under the load at z = 0

and are

max

= u(0) = −

; M

max

= M(0) = −

. (7.37)

The deformed shape of the beam is shown in Figure 7.11. As in Example 7.1 [Figure

7.6 (a)], the decay of the disturbance is oscillatory, so there are regions where the

downward force causes an upward (albeit small) displacement.

Figure 7.11: Deformation of the beam under a concentrated force

7.4.1 More general loading of the inﬁnite beam

The preceding solution can be extended to problems involving several concentrated

forces, using superposition. For example, if forces F

act at the points z =

respectively, as shown in Figure 7.12, the resulting displacement will be

u(z) = −

(

|z −a

|) −

(

|z −a

|) −

(

|z −a

|) ,

from (7.35). A similar expression can be written for the moment M(z).

Figure 7.12: Inﬁnite beam with several concentrated forces

The same method can be extended to the case of a distributed load w(z) in a <

z <b by considering it to consist of a set of concentrated forces

w(z

′

)

′

, as shown

in Figure 7.13.

Notice that this is another case where we need to introduce a second variable z

′

for the

purposes of integration (cf §6.1.4).