Application Center - Maplesoft

# Beam on an Elastic Foundation

You can switch back to the summary page by clicking here.

Beam on an Elastic Foundation

Univ.-Prof. Dr.-Ing. habil. Josef  BETTEN
RWTH University Aachen
Mathematical Models in Materials Science and Continuum Mechanics
Augustinerbach 4-20
D-52056  Aachen,  Germany
betten@mmw.rwth-aachen.de

Abstract

This MAPLE worksheet is concerned with the deflection curve of a statically indeterminate

beam rigidly clamped at both ends. The beam may be loaded by a concentrated force F in

vertical and a load L in longitudial direction. The elastic foundation is characterised by the

parameter K. This problem can easily solved by applying the LAPLACE transformation to

the differential equation of the problem.

Keywords:  Elastic foundation; Deflection of  beams; LAPLACE transformation

Differential Equation

The deflection curve y(x) of  a beam on an elastic foundation is the solution of  the following

ordinary differential equation with constant coefficients:

 > restart:

 > dgl:=diff(y(x),x\$4)+L*diff(y(x),x\$2)+K*y(x)+Q(x)=0;

 (1)

 > L:=l/EI; K:=beta/EI;

 (2)

 (2)

The parameters L and K are expressing  the longitudial loading and the elstic foundation,

respectively, where EI is the flexural rigidity and beta is known as the WINKLER elastic

foundation number. The vertical load  Q(x)  is:

 > Q(x):=q(x)/EI;

 (3)

In the following example Q(x)  is assumed to be a concentrated force F = A*EI

at  x = a , which can be expressed by the DIRAC delta function. Thus, we start from

the following differential equation:

 > restart: alias(delta=Dirac, H=Heaviside, th=thickness):

 > dgl_eqn:=diff(y(x),x\$4)+L*diff(y(x),x\$2)+K*y(x)=A*delta(x-a);

 (4)

Boundary Conditions

To calculate the deflection curve of a statically indeterminate beam rigidly clamped at both ends,

we have to take the following boundary conditions into consideration:

 > C[3]:=diff(y(x),x\$3)(0); C[2]:=diff(y(x),x\$2)(0);          C[1]:=diff(y(x),x)(0); C[0]:=y(0);

 (5)

 (5)

 (5)

 (5)

Where  C[0] = C[1] = 0. The constants C[2] and C[3] are determined later.

LAPLACE Transformation

Using the MAPLE inttrans package we arrive at the following solution:

 > with(inttrans):

 > laplace(dgl_eqn,x,s)  assuming a >= 0;

 (6)

 > subs({D(D(D(y)))(0)=C[3],D(D(y))(0)=C[2],             D(y)(0)=C[1],y(0)=C[0]},%);

 (7)

 > subs({C[0]=0,C[1]=0,C[2]=B[2],C[3]=B[3]},%);

 (8)

 (9)

 > solution:=invlaplace(%,s,x):

 > Y(x):=simplify(subs({a=1/2,A=1,L=-10,K=100},rhs(%))):

 > Y(x):=evalf(%):

Note that the commands of the solution  Y(x)  are executed but the outputs are not printed, since                                            the input lines are ending  by colons instead of semicolons. The constants  B[2]  and  B[3]  can                                             be determined by considering the boundary conditions at  x = 1:

 > Y(1):=evalf(subs(x=1,Y(x)),4);

 (10)

 > T(1):=evalf(subs(x=1,diff(Y(x),x)),4);

 (11)

 > evalf(solve({Y(1)=0,T(1)=0},{B[2],B[3]}),5);

 (12)

 > Z(x):=subs({B[2]=0.088044,B[3]=-0.41161},Y(x)):

Again this solution is not printed, since the input command is ended by a colon instead

of a semicolon.

 > plot1:=plot(-0.003635,x=0.4..0.6,color=black):

 > plot2:=plot(-0.004*H(x-1/2),x=0.499..0.501,linestyle=4,             color=black):

 > plot3:=plot(-Z(x),x=0..1,color=black,th=3,title=                                 "Deflection # parameters: a = 1/2, A = 1, L = -10,  K = 100"):

 > plots[display](plot1,plot2,plot3);

Representation with rational numbers:

 > z(x):=convert(Z(x),`rational`):

 > z(x):=simplify(%):

 > plot4:=plot(-0.003635,x=0.4..0.6,color=black):

 > plot5:=plot(-0.004*H(x-1/2),x=0.499..0.501,linestyle=4,             color=black):

 > plot6:=plot(-z(x),x=0..1,color=black,th=3,title=                      "Deflection # parameters: a = 1/2, A = 1, L = -10, K = 100"):

 > plots[display](plot4,plot5,plot6);

The concentrated force  F = A*EI  at  x = a  can be expresed by the DIRAC delta function, mentioned before.

Alternatively, we can use the HEAVISIDE function:

 > F(x,a):=H(x-(a-epsilon))-H(x-(a+epsilon)); # vertical load

 (13)

Concentrated force at  a = 1/2 - epsilon  and  a = 1/2 + epsilon:

 > F(x,0.5):=subs({a=0.5,epsilon=0.001},%);

 (14)

 > F(x,1/2):=convert(F(x,0.5),`rational`);

 (15)

 > plot(F(x,1/2),x=0..1,scaling=constrained,color=black,th=3,          title="Single load with parameter  A = 1");

Maximum deflection at  a  =  1/2  with  A = 1,  L = -10,  K = 100:

 > Z[_max]:=evalf(subs(x=1/2.001,z(x)),4);

 (16)

 > Z[max_]:=evalf(subs(x=1.001/2,z(x)),4);

 (17)

 > Z[max]:=evalf(((Z[_max]+Z[max_])/2),4);

 (18)

 >

The dimensionless maximum of the deflection of a beam with  L = K = 0  is  1/192 , as

shown in: BETTEN, J., Creep Mechanics, Third Edition, Springer-Verlag, Berlin /

Heidelberg / New York, 2008. This value is compared with the corresponding value

for L and K not equal to zero in the following:

 > relative_maximum_deflection:=                (z[max][K=0,L=0]-z[max][K,L])/z[max][K=0,L=0];

 (19)

For  K = 100  and  L = -10   we arrive at:

 > (z[max][K=0,L=0]-z[max][K=100,L=-10])/z[max][K=0,L=0]=        evalf(((1/192-0.003635)*192),4);

 (20)

 > convert(%,`rational`);

 (21)

 >

The length of the beam may be 10 m. Then, we arrive for  L = -10  and  K = 100 at a deflection

maximum of  3.64 cm. The corresponding value for  L = K = 0  is (1/192)*10^3 cm = 5,21 cm.

The quotient is:

quotient:=           evalf(z[max][K=0,L=0]/z[max][K=100,L=-10]=(1/192)/Z[max_],4);

 (22)

 >

Another example: assuming  K = 100  and  L = 0 ; then, we arrive at a quotient of:

 > Quotient:=evalf(z[max][K=100,L=-10]/z[max][K=100,L=0]=      0.003635/0.004367,4);

 (23)

This value demonstrates the influence of  a longitudinal force (L = -10) on the deflection

maximum of  a beam on an elastic foundation, i.e., because of  L the deflection decreases.

An increasing positive parameter L leads to a  critical load in horizontal direction.

 >

Legal Notice: The copyright for this application is owned by the author(s). Neither Maplesoft nor the author are responsible for any errors contained within and are not liable for any damages resulting from the use of this material. This application is intended for non-commercial, non-profit use only. Contact the author for permission if you wish to use this application in for-profit activities.