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:
> |
dgl:=diff(y(x),x$4)+L*diff(y(x),x$2)+K*y(x)+Q(x)=0; |
 |
(1) |
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(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:
 |
(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$2)(0); C[1]:=diff(y(x),x)(0); C[0]:=y(0); |
![`:=`(C[3], (diff(y(x), `$`(x, 3)))(0))](/view.aspx?SI=5658/Beam_on_an_Elastic_Foundation_7.gif) |
(5) |
![`:=`(C[2], (diff(y(x), `$`(x, 2)))(0))](/view.aspx?SI=5658/Beam_on_an_Elastic_Foundation_8.gif) |
(5) |
![`:=`(C[1], (diff(y(x), x))(0))](/view.aspx?SI=5658/Beam_on_an_Elastic_Foundation_9.gif) |
(5) |
![`:=`(C[0], y(0))](/view.aspx?SI=5658/Beam_on_an_Elastic_Foundation_10.gif) |
(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:
> |
laplace(dgl_eqn,x,s) assuming a >= 0; |

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(6) |
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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]},%); |
![`+`(`*`(`^`(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, `*`...](/view.aspx?SI=5658/Beam_on_an_Elastic_Foundation_15.gif) |
(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)))](/view.aspx?SI=5658/Beam_on_an_Elastic_Foundation_16.gif) |
(9) |
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solution:=invlaplace(%,s,x): |
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Y(x):=simplify(subs({a=1/2,A=1,L=-10,K=100},rhs(%))): |
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:
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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]))))](/view.aspx?SI=5658/Beam_on_an_Elastic_Foundation_17.gif) |
(10) |
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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]))))](/view.aspx?SI=5658/Beam_on_an_Elastic_Foundation_18.gif) |
(11) |
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evalf(solve({Y(1)=0,T(1)=0},{B[2],B[3]}),5); |
![{B[2] = 0.88044e-1, B[3] = -.41161}](/view.aspx?SI=5658/Beam_on_an_Elastic_Foundation_19.gif) |
(12) |
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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.
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plot1:=plot(-0.003635,x=0.4..0.6,color=black): |
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plot2:=plot(-0.004*H(x-1/2),x=0.499..0.501,linestyle=4, color=black): |
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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:
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z(x):=convert(Z(x),`rational`): |
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plot4:=plot(-0.003635,x=0.4..0.6,color=black): |
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plot5:=plot(-0.004*H(x-1/2),x=0.499..0.501,linestyle=4, color=black): |
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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:
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F(x,a):=H(x-(a-epsilon))-H(x-(a+epsilon)); # vertical load |
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(13) |
Concentrated force at a = 1/2 - epsilon and a = 1/2 + epsilon:
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F(x,0.5):=subs({a=0.5,epsilon=0.001},%); |
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(14) |
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F(x,1/2):=convert(F(x,0.5),`rational`); |
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(15) |
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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:
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Z[_max]:=evalf(subs(x=1/2.001,z(x)),4); |
![`:=`(Z[_max], 0.3642e-2)](/view.aspx?SI=5658/Beam_on_an_Elastic_Foundation_26.gif) |
(16) |
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Z[max_]:=evalf(subs(x=1.001/2,z(x)),4); |
![`:=`(Z[max_], 0.3629e-2)](/view.aspx?SI=5658/Beam_on_an_Elastic_Foundation_27.gif) |
(17) |
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Z[max]:=evalf(((Z[_max]+Z[max_])/2),4); |
![`:=`(Z[max], 0.3635e-2)](/view.aspx?SI=5658/Beam_on_an_Elastic_Foundation_28.gif) |
(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:
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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])))](/view.aspx?SI=5658/Beam_on_an_Elastic_Foundation_29.gif) |
(19) |
For K = 100 and L = -10 we arrive at:
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(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](/view.aspx?SI=5658/Beam_on_an_Elastic_Foundation_30.gif) |
(20) |
![`/`(`*`(`+`(z[max][K = 0, L = 0], `-`(z[max][K = 100, L = -10]))), `*`(z[max][K = 0, L = 0])) = `/`(3021, 10000)](/view.aspx?SI=5658/Beam_on_an_Elastic_Foundation_31.gif) |
(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)](/view.aspx?SI=5658/Beam_on_an_Elastic_Foundation_32.gif) |
(22) |
Another example: assuming K = 100 and L = 0 ; then, we arrive at a quotient of:
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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)](/view.aspx?SI=5658/Beam_on_an_Elastic_Foundation_33.gif) |
(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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