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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;

 

`:=`(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;

 

`:=`(L, `/`(`*`(l), `*`(EI))) (2)

 

`:=`(K, `/`(`*`(beta), `*`(EI))) (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;

 

 

`:=`(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);

 

`:=`(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);

 

`:=`(C[3], (diff(y(x), `$`(x, 3)))(0)) (5)

 

`:=`(C[2], (diff(y(x), `$`(x, 2)))(0)) (5)

 

`:=`(C[1], (diff(y(x), x))(0)) (5)

 

`:=`(C[0], y(0)) (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;

 

`+`(`*`(`^`(s, 4), `*`(laplace(y(x), x, s))), `-`(((`@@`(D, 3))(y))(0)), `-`(`*`(s, `*`(((`@@`(D, 2))(y))(0)))), `-`(`*`(`^`(s, 2), `*`((D(y))(0)))), `-`(`*`(`^`(s, 3), `*`(y(0)))), `*`(L, `*`(`^`(s, ...
`+`(`*`(`^`(s, 4), `*`(laplace(y(x), x, s))), `-`(((`@@`(D, 3))(y))(0)), `-`(`*`(s, `*`(((`@@`(D, 2))(y))(0)))), `-`(`*`(`^`(s, 2), `*`((D(y))(0)))), `-`(`*`(`^`(s, 3), `*`(y(0)))), `*`(L, `*`(`^`(s, ...
(6)

 

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

 

`+`(`*`(`^`(s, 4), `*`(laplace(y(x), x, s))), `-`((diff(y(x), `$`(x, 3)))(0)), `-`(`*`(s, `*`((diff(y(x), `$`(x, 2)))(0)))), `-`(`*`(`^`(s, 2), `*`((diff(y(x), x))(0)))), `-`(`*`(`^`(s, 3), `*`(y(0)))...
`+`(`*`(`^`(s, 4), `*`(laplace(y(x), x, s))), `-`((diff(y(x), `$`(x, 3)))(0)), `-`(`*`(s, `*`((diff(y(x), `$`(x, 2)))(0)))), `-`(`*`(`^`(s, 2), `*`((diff(y(x), x))(0)))), `-`(`*`(`^`(s, 3), `*`(y(0)))...
(7)

 

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

 

`+`(`*`(`^`(s, 4), `*`(laplace(y(x), x, s))), `-`(B[3]), `-`(`*`(s, `*`(B[2]))), `*`(L, `*`(`^`(s, 2), `*`(laplace(y(x), x, s)))), `*`(K, `*`(laplace(y(x), x, s)))) = `*`(A, `*`(exp(`+`(`-`(`*`(s, `*`... (8)

 

> readlib(isolate)(%,laplace(y(x),x,s));

 

laplace(y(x), x, s) = `/`(`*`(`+`(`*`(A, `*`(exp(`+`(`-`(`*`(s, `*`(a))))))), B[3], `*`(s, `*`(B[2])))), `*`(`+`(`*`(`^`(s, 4)), `*`(L, `*`(`^`(s, 2))), K))) (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);

 

 

`:=`(Y(1), `+`(`+`(0.2342e-1, `-`(`*`(0., `*`(I)))), `*`(`+`(.2471, `*`(0., `*`(I))), `*`(B[3])), `*`(`+`(.8892, `*`(0., `*`(I))), `*`(B[2])))) (10)

 

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

 

`:=`(T(1), `+`(`+`(.1510, `-`(`*`(0., `*`(I)))), `*`(`+`(.8892, `*`(0., `*`(I))), `*`(B[3])), `*`(`+`(2.442, `*`(0., `*`(I))), `*`(B[2])))) (11)

 

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

 

{B[2] = 0.88044e-1, B[3] = -.41161} (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);

 

Plot_2d  

 

 

 

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);

 

Plot_2d  

 

 

 

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

 

`:=`(F(x, a), `+`(H(`+`(x, `-`(a), epsilon)), `-`(H(`+`(x, `-`(a), `-`(epsilon)))))) (13)

 

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

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

 

`:=`(F(x, .5), `+`(H(`+`(x, `-`(.499))), `-`(H(`+`(x, `-`(.501)))))) (14)

 

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

 

`:=`(F(x, `/`(1, 2)), `+`(H(`+`(x, `-`(`/`(499, 1000)))), `-`(H(`+`(x, `-`(`/`(501, 1000))))))) (15)

 

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

 

Plot_2d  

 

 

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

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

 

`:=`(Z[_max], 0.3642e-2) (16)

 

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

 

`:=`(Z[max_], 0.3629e-2) (17)

 

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

 

`:=`(Z[max], 0.3635e-2) (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];

 

`:=`(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);

 

`/`(`*`(`+`(z[max][K = 0, L = 0], `-`(z[max][K = 100, L = -10]))), `*`(z[max][K = 0, L = 0])) = .3021 (20)

 

> convert(%,`rational`);

 

`/`(`*`(`+`(z[max][K = 0, L = 0], `-`(z[max][K = 100, L = -10]))), `*`(z[max][K = 0, L = 0])) = `/`(3021, 10000) (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);   

`:=`(quotient, `/`(`*`(z[max][K = 0, L = 0]), `*`(z[max][K = 100, L = -10])) = 1.435) (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);

 

`:=`(Quotient, `/`(`*`(z[max][K = 100, L = -10]), `*`(z[max][K = 100, L = 0])) = .8324) (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. 

>  

 

 

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