The partially supported elastic beam

Journal of Elasticity 13 (1983) 71-82 © 1983 Ma.rtinus Nijhoff Publishers, The Hague. Printed in The Netherlands

The partially supported elastic beam

FABIO GASTALDI* and DAVID KINDERLEHRER

Instituto di Analisi Numerica del CN.R., Corso Carlo Alberto 5, 27100 Pavia, Italy School o f Mathematics, University o f Minnesota, 206 Church St. S.E., Minneapolis, MN55455, U.S.A.

(Received: April 10, 1981)

I. Introduction

An elastic beam resting on a horizontal table and partially extending beyond it is subjected to a known force per unit length, like gravity. Constrained to lie on or above the table in the supported region, the displacement of the beam resolves a variational inequality or a uni- lateral problem. Our intention here is to derive the solution of this problem by means of variational inequalities. Since obtaining the solution is a well known classroom exercise in elementary mechanics (cf. Crandall, Dahl, and Lardner [3] p. 524), the object of this note is to illustrate the variational methods. This will entail applying them in the most elementary situation and providing the appropriate mechanical interpretations where we are able. Figure 1 depicts the beam in its natural configuration and Figure 2 a virtual displacement. Two possible equilibrium configuration are shown in Figures 3 and 4.

As a result we offer several observations which are not generally found in textbooks. These include existence, uniqueness, and smoothness properties as well as the explicit calcu- lation of the solution. The question of smoothness is connected to the existence of concen- trated support forces. The principles we adopt here may be utilized equally well in many other situations, for example in the case of an arbitrary force or a beam of varying density.

Our principal sources for this subject have been the books of Duvaut and Lions [4] and Kinderlehrer and Stampacchia [6]. We also mention [1] and the conference proceedings [2]. For discussions of various problems in solid mechanics from our point of view we cite [51, [9] ,and [10].

We wish to thank Roger Fosdick for suggesting this question and for his interest in our work.

Suppose the beam in its natural configuration occupies the interval [a, b] , a < 0 < b, of the x-axis and the table occupies [a, 0]. Assuming the beam to be a homogenous one di- mensional elastic material we denote by y = u (x) its vertical displacement (the horizontal

*Requests for reprints to F. Gastaldi.

71

72 F. Gastaldi and D. Kinderlehrer

H/1/1/

O

Y

x b

Figure 1

II///11

0

Y r

x b

Figure 2

1/1/H/

0 x

Figure 3

Y

b < lal <_ ~ b

Y

0 X

Figure 4

b < lal

b

The partialty supported elastic beam 73

displacement is negligible). After suitable normalization, the potential energy of a virtual displacement y = v(x), a < x <= b, of the beam is given by

' - . 0 ~'(e -- ~

where f i s proportional to the applied force. We shall suppose in the sequel that gravity is the only active force so

f = constant < 0.

In this context a virtual displacement v(x) is admissible provided v(x) => 0 for a -< x _<- 0. We set

~ = {y = v ( x ) ; a<=x<=b:v(x)>=O for a<=x<O}

omitting for the moment the regularity hypotheses we shall impose shortly. In equilibrium the configuration y = u (x), a _-< x < b, o f the beam is given by the dis-

placement which renders the functional E(v) a minimum, namely,

E(u) = mi%~l~E(v) . (1.2)

To give a more precise statement of (1.2), let us introduce a few notations. With ~2 = (a, b), by

Hm([2), m = 1 ,2 . . . . .

we mean the space of functions {'(x) such that ~', (d/dx)~ . . . . . (dra/d2cra)~ are square integrable on [2 which we endow with the norm

,~,~,m,., = f ¢(x)'+ ~ ~-~ ~(~ ~ . F/ 1

As usual, L 2 ([2) denotes functions square integrable on ~2.

Observe that if ~ E H ' n ( ~ ) then [ ' E C " - ~ ( ~ ) (i.e., ~" is ( m - 1) - times continuously differentiable in the closed interval [a, b] ). We now redefine I~ as

~ = { v ~ H 2 ( ~ ) : ~ ( x ) > O for a < x < O } (1.3)

and state our problem:

Problem 1.1 To find u ~ K : E ( u ) = min~¢l~E(v ).

Given a solution u ~ ~ and any v ~ l~, the convex combination wt = u + t ( v - - u ) ~ ~ , 0 =< t <= 1, so ),(t) -- E(wt) has a minimum at t = 0. Thus

0__< - - - j .

or, with the notat ion

a(w, ~) = f a w"~"~,

u is a solution of


Problem 1.2 T o f i n d u E K : a ( u , v - - u ) > = f ~ f ( v - - u ) d x f o r a l l v E K . This is a variational inequality (cf. Stampacchia [8] or [6] pp. 1, 2). Conversely, since

?~(t) is a convex function, if u is a solution o f Problem 1.2 it is also a solution o f Problem 1.1.

A suggestive way to describe Problem 1.2 is in terms of its complementarity conditions,

or the natural conditions which prevail in [2 and at its boundary. Let us give a formal deri-

vation which will be justified once our regularity statement, Theorem 2.3, is known.

To begin, integrating by parts twice in the variational inequality we obtain

f uiV(v-u)ax+ [u"(v-u)'-u'"(v-u)]l~>= f f ( v -u )ax forall v E K . (1.4)

Setting v = u + ~" with ~" E C~*(I2) and ~" => 0 on [a, 0 ] , we find that the boundary terms vanish and we infer that u iv - - f > 0 on [2 and u iv = f o n (0, b) because ~" is not constrained there. By the same procedure we also derive that u iv = f on any open subset of (a, 0)where u > 0. So we can write the complementarity conditions in [2

u=>0 on [a, 0 ] , (1.5)

u i v - f > O and u(u i v - f ) = 0 on (a,b).

Choosing v = u + ~" where ~" is smooth on [a, b] and identically zero on [a, b/2], we have from (1.4), (1.5),

~/i~ (ui~ - / ) f a x 0 < + u"(b) ~'(b) -- u'"(b) ~ (b) . . . . . . = = u ( b ) ~ ( b ) - - u (b)~(b).

Since ~'(b), ~"(b) are arbitrary, this inequality is true if and only if

u"(b) = u"(b) = 0. (1.6)

Then choosing first v = 0 and then v = 2u in (1.4), we find

0 = j " (u iv - - / ) u d x + u"(a)u(a) --u"(a)u'(a),

0 = u'"(a) u(a) -- u"(a)u'(a). (1.7)

Now (1.4) becomes

(u i~ - D vax + u"(a) v(a) - u"(a) v'(a). (1.8) 0 <=

At this point we choose a smooth v E K which is identically zero except in a small neighborhood o f x = a. Then for some e > 0 we have that

- - e < u " (a) v (a ) - - u" (a) z/ (a). (1.9)

This can hold in particular for v with v(a) > 0 and v'(a) = 0 only if u"'(a) ~- O. But this implies that u"(a) = 0 since we can choose v with v(a) > 0, fixed, and v'(a) arbitrary in (1.9). From (1.7), u(a)u"(a) = 0. We summarize the conditions at x = a:

u(a) =>_ 0, u"'(a) >= o, u(a)u"'(a) = o, (1

~"(a) = 0.

The partially supported elastic beam 75

Note that the first three conditions are themselves a complementarity system. The formal derivation we have given shows that a solution u o f Problem 1.2 satisfies

(1.5), (1.6), (1.10). We point out that (1.5) must be understood in a weak sense since u ~ is not a function (it will be shown to be a measure). On the other hand, while (1.6) is easily justified, u"(a) and u"(a) in (1.10) as yet have no meaning. In the sequel we shall verify

these conditions and exploit them to analyze the solution. Let us note that one cannot expect Problem 1.2 always to admit a solution. Suppose, for

example, that b > lal and set v(x) = ~x, o~ < 0, in (1.1). One finds that

E(v) = - fvdx = - - f ( b - - a ) v = --o. as a--*--o*,

so E(v) does not have a minimum in K. Certainly this is what we expect to occur on physical

grounds, but it illustrates that the existence theorem will have to contain some compatibility conditions.

2. Existence, uniqueness, and regularity of the solution

We shall always assume that b < lal.

Theorem 2.1 There exists a solution o f Problem 1.2

Proof. It is easily checked that the theorem of Lions and Stampacchia may be applied

([7] Theorem 5.1, of. also [6] Theorem 2.3, p. 91). The most significant hypothesis of this theorem for us to check is the compatibility condition full'died by the data, which is that

S f ~ ( x ) d x < O for all E K , ~ 0, and a(rt, r/) = 0. (2.1) 7/

Since a(r/, 7/) = 0 if and only if 17(x) = ax +/3 is linear, we obtain from (2.1) the condition

Now n E K and (a + b)/2 < 0 imply that r/((a + b)/2) > 0 unless r / = 0. Q.E.D. Let us observe that the more familiar existence and uniqueness theorem for variational

inequalities with coercive forms is not appropriate here because the kernel of a ( - , - ) is not

{0}. Sometimes this fact can jeopardize both existence and uniqueness, for instance in the case o f Neumann boundary conditions. However in our case uniqueness also holds.

Theorem 2.2 The solution u of Problem 1.2 is unique.

Proof. Suppose that u and w are two solutions. Setting v = w in the variational inequality satisfied by u and v = u in the variational inequality satisfied by w we have


a(u, w-u) >= ~ f(w-u)ax and

,(w, u-w) >= J'. : (u-

Adding these gives that

O < = a ( u - w , u - w ) <O,

so a(u -- w, u -- w) = 0 and there is a linear function rl(x) such that w = u + ~7. Again using the variational inequality :

= a(u, rl) = a(u, w--u)>= f~ f ( w - - u ) d . x = fs~ f~?dx 0

and

f. fo 0 = a(w, rl) .= a(w, u -- w) > f (u -- w)dx = -- frld.x,

whence

(b --a)fl? f---~- + b = .f~ f~lclx = O.

Thus writing r/(x) = ot(x -- (a + b)/2) +/~, we have found that/~ = 0 and u((a + b)/2) = w((a + b)/2) .

Suppose a : / : 0 , say a < 0 . If u((a + b)/2) = w((a + b)/2) = 0 , then u, w have local minima at x = (a + b)/2 so

= - - = ~ = o r < 0 ,

contrary to our assumption. Thus w((a + b)/2) = u((a + b)/2) > 0, and both u and w are solutions o f [-iv _ f = 0 near x = (a + b)/2, say Ix -- (a + b)/21 < 6, by (1.5). Now r/(x) > 0

for x < (a + b)[2 so

<a+b. w ( x ) = u (x ) + r/(x) > r/(x) > 0 for x 2

Thus w iv - - f = 0 for a < x < ½ (a + b) + 8. On the other hand rl(x) < 0 for x > (a + b)/2 so

a + b u(x) = w(x)-n(x)>=-n(x)>O for x > ~

2

Again by the complementarity conditions (1.5),

w i V - - f = u t V - - f = 0 for ½ ( a + b ) < x < b .

Since w(a) > 0, we may conclude rigorously that w'"(a) = w"(a) = 0. Consequently

w iv = f in I2,

w"(a) = w"(a) = w"(b) = w"(b) = O.


Integrating by parts gives, for any ~" E H 2 ( ~ ) ,

a(w , D = f ~ W"~'~

Now choose ~" - 1 above, which gives

0 = a(w, 1) = S a f d x < 0 ,

a contradiction. Thus ~ (x ) -- 0. Q.E.D. Tumingnow to the regularity of the solution, both the methods discussed in [6] (Chapter

VII, §9) or [9] are available to us but the one based on penalization has a number of technical difficulties. We only consider the technique based on measure.

Theorem 2.3 The solution u o f Problem 1.2 has a third derivative u" which is a function o f bounded variation.

Proof. For any ~ E C~'(~2), [ > 0, the function v = u + ~ E K. Thus

a(u,~) faI'~'dx > 0 for ~" > - = = o, ~ e c ~ ( a ) ,

and the functional

is linear, non-negative and continuous with respect to convergence in C~0(~2). Thus, by

Schwartz's extension of the Riesz representation theorem, there is a measure/z satisfying

S . ~d , = a(u,~)-- S . f fdx . (2.2)

In particular this means that u iv is a (signed) measure and

d/~ = u iv - - f in the sense of distributions on ~2. (2.3)

Since u"'(b) = O,

u " ( x ) = - u ( l x , b]) +fx is a function of bounded variation. Q.E.D.

The complementar i ty conditions (1.10) at x = a may be verified rigorously using this theorem. This elementary exercise is left to the reader.

3. Expficit determination of the solution

Recall that b < lal.

78

Property 3.1 Let u be the solution o f Problem l.2 and set

~ = { x E [a, 0 ) : u ( x ) = 0}.

Then

( i ) I = [a,a'l (ii) I = {a}

( ia) ,~ = ~

F. Gastaldi and D. Kinderlehrer

for some a' ~ O,

or

a n d u ( x ) > O for a<=x<O.

Proof. We first dispose of the possibility that I = [a, 0), namely that u (x) = 0 for a < x <

O. In this case, recalling that u is twice continuously differentiable in I2, we have u (0) =

u'(O) = u"(O) = 0 and thus u is a solution o f

u ~v = / < 0 for O < x < b ,

u(O) = u'(O) = u"(o) = o,

u"(b) = u"'(b) = O.

These five boundary conditions are incompatible. Indeed in the present situation, where f i s

constant, u" = ½f(x " b ) ~ , 0 < x < b, which does not vanish at x = 0. So ! is properly

contained in [a, 0).

The set I is closed in [a, 0), because u is continuous, so its complement is the union of

open intervals. Suppose that (Xo, x l ) C [a, 0) - - I with x l E 1 and xo E 1 or Xo = a as the

case may be. Then,

u > 0 in (Xo,Xl)

and u attains a minimum at x l and Xo i fxo > a. Thus,

u iv = f in (Xo,X~), (3.1)

u ( x , ) = u ' (x , ) . = O, (3.2)

U(Xo) = u'(xo) = 0 i f Xo > a , while (3.3) I t l ~

U(Xo)U (Xo) - 0 and u"(Xo) = 0 i f Xo = a.

Let

fO Xo =<x =<x~, w(x)

t u ( x ) otherwise

Now w ( x ) is a non-negative C 1 function with bounded second derivative, so w E K. We shall

calculate that E ( w ) < E ( u ) , so by uniqueness of the minimum u, w = u and (Xo, x t ) is

empty.

From (1.1),

1 =


= E ( u ) - - f X,u,Vuax. + ~' fudx + (u"'u -- u u )Ix0 d X o 0

Because of the conditions (3.2), (3.3), the endpoint terms above all vanish. In view of (3.1), we obtain

1 X 1

~(w) = e (u) + Jxo fuax =< E(u). 2

This shows that the complement o f I is connected in [a, 0) and that x = a E I i f / 4 : 0.

Q.E.D. Property 3.1 allows us three possible expressions for the measure dta o f (2.2) and (2.3).

These are

(i) I = [ a , a ' ] , a < a ' <O: d# = --fsoId.x + Fo,5,; + Fo6o,

(ii) I = {a} : do = Fa6a + Fo6o,

(iii) I = 0 : d# = Fo6o,

where

{[u '"~y = u ' " ( y + ) - - u ' " ( y - ) if y = a ' or y = 0,

Fy = u"(a+) if y = a,

and in physical language represents concentrate densities o f support forces, 6y is the Dirae

measure at y , and ~0x is the characteristic function o f the interval I. Note that in eases (i) and (iii), u'"(a ÷) = b.

Property 3.2 Let u be the solution o f Problem 1.2. Then

u(a) = u(O) = 0 and [[u'"]o ~ O.

In particular, 14= 0 and case Oil) does not occur.

Proof. First suppose that [u ' " ]o = O. Consider case (i) above, we then have that

f1 ~u ~a'¢(a ) a ~du = - - f ~ c ~ + . . . .

= a ( u , ~ ) - - f f~dx .

Choosing ~" = p + qx, a linear function, we obtain

fa b m t o = . f ~ a x + ~u L ' ~ ( a )

= f ( b - - a ' ) ~ ( a ' - - 2 b ) + [u'"~d~(a')


which is not possible for arbitrary linear functions since ½ (a' + b) ~ a' . Thus, ~u"']lo :~ 0. This precludes that u ( 0 ) > 0 since in that case u tv = f in a neighborhood of x = 0, u is

smooth there, and thus ~u"]o = u"(O ÷) - - u ' " ( 0 - ) = 0. Consequently u ( 0 ) = 0. Case (ii) is treated in the same manner.

In case (iii), for ~ = p + qx,

Fop = ¢au = - /~ax = - - / ( b - - a ) ¢ - - 7 -

which cannot hold for arbitrary linear functions unless b = - a, contrary to hypothesis, whether or not Fo = 0. Thus case (iii) never occurs. In particular, i f u ( a ) > 0, then case (iii) prevails. So u(a) = 0. Q.E.D.

Property 3.3 Le t u be the solution o f Problem 1.2 and set

I = { x E [ a , O ) : u ( x ) = 0} = [a,a'].

Either

a' > a and a' = --vC2b or a' = a and lal < x / 2 b .

Proof. We first derive a representation o f u " . Since u"(b) = u" (b ) = O,

u" (x ) = ½ f ( x - - b ) 2 for 0 < x < b .

For our measure lz we have the representation

fa ~du = - fa Ao~dx + Yo,~(a') + Fd'(O),

a formula we can use in either case (i) or (ii) if we agree that I = {a} and a ' = a in case (ii). This means that

a(u, ~) = f~_x:¢~ + Fa,~(a' ) + Fo~'(0),

Choosing [ = I and ~" = x in (3.4), we see that

0 = f ( b - - a ' ) + F d + F o ,

0 = ~ ( b 2 - - a ' 2 ) + F a , a '.

Solving these equations for Fa, , Fo and using that

u " ( 0 - ) = - - Fo + u'"(0 +) = - - F o - - f b and

u"(a '+) = Fa, ,

we obtain

u"(x) = -~(x

for all ~" E t t2 (~2). (3.4)

(3.5)

(3.6)

a ' < x < 0 .

The partially supported elastic beam

Finally, since

0 F

f n ~xu (x)dx = --a'u'(a'),

it readily follows that

a' f ( a ' ) 2 u'(a') = }

81

Consequently, if a'>a, then u ' ( a ' ) = 0 so a' =--.qt-2b>a. If a ' = a , then u'(a)>O so

lal < ' v~b . Q.E.D. Note that (3.5) and (3.6) are the equations which express balance of force and moment.

The function u(x) may now be calculated on (a', O) by integrating

if a > a ,

- - - - i f a' = a. u"(x) = x - a ) a

We note that in the exceptional case b = -- a, there is a family o f solutions given by

t f ( x - - b ) 2 f o r O < = x < = b ,

u"(x) =

f ( x + b) for --b<x<O.

Upon integration, one imposes the conditions u(O) = 0 and u(x) > 0 f o r x < 0 which remain

valid in this context.

A c k n o w l e d g e m e n t

Supported in part by the N.S.F. and the C.N.R. The first author wishes to acknowledge the

hospitality o f the University o f Minnesota during the preparation of this paper.

REFERENCES

[ 1] Baiocchi, C. and Capelo, A., Disequazioni variazionali e quasivariazionali, Applicazioni a problemi di frontiera fibera, I, II, Pitagora, Bologna (1978).

[2] Cottle, R., Giannessi, F., and Lions, J. L., Variational inequalities and complementarity problems, Wiley, Chichester - New York (1980).

[3] Crandall, S., Dald, N., and Lardner, T., An introduction to the mechanics of solids, McGraw Hill, New York (1978).


[4] Duvaut, G., and Lions, J. L., Inequalities in mechanics and physics, Springer, New York. 1976. [5] Kalker, J. J., Aspects of contact mechanics, in Proc. Int. Union of Theoretical and Appl. Mechanics,

(de Pater and Kalker, eds) Delft. Univ. Press (1975) 1-25. [6] Kinderlehrer, D. and Stampacchia, G., An introduction to variational inequalities and their appli-

cations, Academic Press, New York, (1980). [7] Lions, J. L. and Stampacchia, G., Variational inequalities, Comm. Pure and Appl. Math. 20 (1967)

493-519. [8] Stampacchia, G., Formes bilineaires coercitives sur les ensembles convexes, C. R. Acad. Sci. Paris,

258 (1964) 4413-4416. [ 9 ] - - , Su una disequazione variazionale legata al comportamento elastoplastico delle travi

appoggiate agli estremi, Boll. Un. Mat. Ital. 11 (4) (1975), 444-454. [ 10] Villaggio, P., Monodimensional solids with constrained solutions, Meccanica 2 (1967) 65-68.

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The partially supported elastic beam