Solution Methods for Beam and Frames on Elastic Fou ndation Using the Finite Element Method

Spôsoby riešenie nosníkov a rámov na pružnom podkla de pomocou metódy kone čných prvkov

Roland JANČO1

Abstract: This paper contains numerical methods for the solution of plane beams and frames on elastic foundation. In the first case, the solution is based on beam elements BEAM54 (ANSYS software) and the derivation of the stiffness matrix of the element is presented. The second approach uses a beam element with a contact element with the description of the derivation of the stiffness matrix for the frame on elastic foundation. Both solutions are compared with theoretical solutions. In conclusions, the influence of the number of divisions of the beam element on the accuracy of the solution is shown.

Abstrakt: Príspevok obsahuje numerické spôsoby riešenia rovinných nosníkov a rámov na pružnom podklade. V prvom prípade na riešenie je použitý nosníkový prvok BEAM54 (program ANSYS) a odvodení matice tuhosti tohto prvku je uvedené. Druhý prístup je použitím nosníkového prvku s kontaktným prvkom aj s odvodením matice tuhosti. Oba spôsoby riešenia sú porovnané s teoretickým riešením. V závere je ukázaný aj vplyv delenia nosníka na prvku na presnosť riešenia.

Keywords: elastic (Winkler's) foundation, Finite Element Method, beam element, mechanical contact

Klíčová slova: pružný (Winklerov) podklad, metóda konečných prvkov, nosníkový prvok, mechanický kontakt

1. Introduction

Solutions of the frames and beams on elastic foundation often occur in many practical cases. For example, solution of building frames and constructions, buried gas pipeline systems and in design of railway tracks for railway transport, etc. Solution of beam on elastic foundation is a statically indeterminate problem of mechanics. In this case, we have the beam with elastic foundation along the whole length and width or only over some part of the length or width. From the theoretical point of view, the exact solution of this type of problems exist only for some type of beams. Detailed explanation of theoretical solution can be found in [2] and [3].

Not all problems can be solved by theoretical approach because theoretical solution is very complicated. In solution of this type of problems, it is possible to use the numerical methods. In this paper, the Finite Element Method approach for the solution of 2D beam and frame on elastic foundation is described.

1 M.Sc. Roland JANČO, Ph.D. ING-PAED IGIP, Slovak University of Technology in Bratislava, Faculty of Mechanical Engineering, Institute of Applied Mechanics and Mechatronics, Nám. slobody 17, 812 31 Bratislava, Slovak republic, roland.janco@stuba.sk.

2. Theoretical background for 2D beam on elastic foundation

The Winkler's foundation model is easy to formulate using energy concepts. The Winkler's model postulates a foundation to resists only displacement normal to its surface, with resisting pressure p Kv= /Pa/, where K /Nm-3/ is the foundation modulus and v /m/ is lateral deflection at the foundation surface. This model deflects only where load is applied. The surrounding foundation is utterly unaffected (Fig. 1a). An area dA of the foundation surface acts like a linear spring of stiffness C p dA / v Kv dA / v K dA= = = /Nm-1/ Strain energy in a linear

spring is 21

2Cv . Now considering a structural element, perhaps a plate bending element or one

face of a 3D solid element that has an area A /m2/ in contact with the foundation. Lateral deflection of area A normal to the foundation, is [ ]{ }v ,= f fN d where { }fd contains D.O.F. of

element nodes in contact with foundation. Strain energy U in foundation over area is

{ } [ ]{ }21 1 1

2 2 2TTU Kv dA v Kv dA= = =∫ ∫ f f fd K d (1)

in which the Winkler's foundation stiffness matrix for the element is

[ ] [ ] [ ]TK dA= ∫f f fK N N . (2)

Fig. 1 Deflection of elastic foundations under uniform pressure q applied directly to the

foundation. (a) Winkler's foundation. (b) Elastic solid foundation [1].

3. Solution base on 2D beam element with elastic foundation properties (1. method of solution)

When we consider the beam element with Winkler's foundation, than matrix [ ]fN is

shape function matrix in the eqn. (2), which is identical to the shape function matrix of the beam without elastic foundation. Components of matrix [ ]fN are

2 3

1 2 3

3 21

L L

x xN = − + ,

2 3

2 2

2

L L

x xN x= − + , (3)

2 3

3 2 3

3 2

L L

x xN = − ,

2 3

4 2L L

x xN = − +

and b dA dx= , where b is the width of the beam face contact with the foundation.

(a) (b)

We can input eqn. (3) into (2), and get

[ ]

2 2

2 3 2 3

2 2

2 3 2 3

13bL 11bL 9bL 13bL

35 210 70 420

11bL bL 13bL bL

210 105 420 140

9bL 13bL 13bL 11bL

70 420 35 210

13bL bL 11bL bL

420 140 210 105

K K K K

K K K K

K K K K

K K K K

−

−

= − − − −

fK (4)

The stiffness matrix of beam without shear deformation can obtain the formal approach using equation

[ ] EI dx,= ∫T

bK B B (5)

where B is the strain-displacement matrix, which is defined for beam by 2

2

d

dx= N

B .

After mathematical solution of equation (5) using eqn. (3), we obtain the stiffness matrix for beam considering only bending moment and transversal load at the nodes

[ ]2 2

3

2 2

12 6L 12 6L

6L 4L 6L 2L

12 6L 12 6LL

6L 2L 6L 4L

EI

− − = − − − −

bK . (6)

If we consider axial loading, stiffness matrix for the beam becomes

[ ]

3 2 3 2

2 2

3 2 3 2

2 2

0 0 0 0L L

12 6 12 60 0

L L L L6 4 6 2

0 0L L L L

0 0 0 0L L

12 6 12 60 0

L L L L6 2 6 4

0 0L L L L

AE AE

EI EI EI EI

EI EI EI EI

AE AE

EI EI EI EI

EI EI EI EI

− − − = −

− − − −

bK . (7)

Resulting stiffness matrix for beam on elastic foundation is the sum of matrix [ ]fK (eqn.

4) and [ ]bK (eqn. 7), which is following

[ ] [ ] [ ]

2 2

3 2 3 2

2 3 2 3

2 2

2

3 2 3

0 0 0 0L L

12 13bL 6 11bL 12 9bL 6 13bL0 0

L 35 L 210 L 70 L 420

6 11bL 4 bL 6 13bL 2 bL0 0

L 210 L 105 L 420 L 140

0 0 0 0L L

12 9bL 6 13bL 12 10 0

L 70 L 420 L

AE AE

EI K EI K EI K EI K

EI K EI K EI K EI K

AE AE

EI K EI K EI

= + =

−

+ + − + −

+ + − + −=

−

− + − + +

bef f bK K K

2

2

2 3 2 3

2 2

3bL 6 11bL

35 L 210

6 13bL 2 bL 6 11bL 4 bL0 0

L 420 L 140 L 210 L 105

K EI K

EI K EI K EI K EI K

− − − − − − +

(8) Hence, in ANSYS software [4], the element BEAM54 has a stiffness matrix derived from eqn. (8).

4. Solution base on 2D beam and contact elements (2. method of solution)

The second method of solving beam and frames is by using a beam and contact element. Stiffness matrix for beam without shearing deformation and axial loading is given in eqn. (6). Contact will be simulated by spring element between rigid ground and beam in the fig. 2. Stiffness matrix for spring element is as follows

spring

C C

C C

− = − K , (9)

where C is stiffness of spring.

Fig. 2. Beam and contact element. Global stiffness matrix for beam and spring element is given by combining eqn. (6) and eqn. (9), which is

[ ]

2 2

2 23

12 6L 12 6L 0

6L 4L 6L 2L 0 0

12 6L 12 6L 0

6L 2L 6L 4L 0 0L

0 0 0 0

0 0 0 0

M M

M M

M M

M M

C C

C CEI

C C

C C

+ − − − − − + − −

= − −

−

bcK , (10)

where MC is the modified stiffness of spring solved by the following equation

3L

MC CEI

= . (11)

5. Verification of numerical solution

Consider the beam of elastic foundation in fig. 3, where the length of one-half of beam is 1.8 m, this is the parameterL . Beam is made from steel, which has the Young’s modulus

52x10 MPaE = with rectangular cross-section area by parameters b 200 mm= and h 400 mm= . Foundation modulus K is equal 1×108 Nm-3 and applied force F is equal 1×105 N

Fig. 3. Beam on elastic foundation.

Theoretical solution

The theoretical solution of beam on elastic foundation in Fig. 3. is described in literature [1, chapter 9]. Solution from literature is in Tab. 1.

4

40

d vEI kv

dx+ = 4

4

k

EIω =

ωx -ωx1 2 3 4e (A cos x+A sin x)+e (A cos x+A sin x)v( x ) ω ω ω ω=

2ωL 2ωL 2ωL1A B [1 2e +e cos(2 L) e sin(2 L)]ω ω= + − 2ωL 2ωL

3A Be [2 e +cos(2 L) sin(2 L)]ω ω= + +

2ωL 2ωL2A B [-1 e cos(2 L) e sin(2 L)]ω ω= + + 2ωL 2ωL

4A Be [e cos(2 L) sin(2 L)]- ω ω= +

3 4ωL 2ωL

FB

8 [ 1 e 2e sin(2 L)]EIω ω=

− + +

( )2

2 ωx -ωx2 1 3 42

2 e [A cos x A sin x] e [A sin x A cos x]o

d vM EI EI - -

dxω ω ω ω ω= − = − +

33 ωx

2 13

-ωx3 4

2 (e [A (cos x sin x) A (cos x sin x)]

e [A (cos x sin x) A (cos x sin x)])

d vT EI EI - -

dx

- -

ω ω ω ω ω

ω ω ω ω

= − = − + +

+ + (0 L)x ,∈

Tab.1 Theoretical solution for beam on elastic foundation in fig.3.

From the given parameters using equations in Tab. 1 the maximum deflection of beam is 0.00145526 m in x = 0, minimum deflection is 0.00128961 m in x = L, maximum bending moment is Mo (x = 0) = 44045.7 Nm and minimum bending moment is Mo (x = L = 1.8 m) = 0 Nm. Maximum transversal load is T (x = 0) = - 50000 N. Minimum transversal load is T (x = L = 1.8 m) = 0.

Numerical solution using BEAM54 (1. method of solution)

For numerical solution we use the program environment ANSYS, which includes a special 2D element. This element is BEAM54. Properties and the characteristics of cross-sectional area are entered in the real constants. In Fig. 4 the FEM model shows the numbering of nodes and elements. The theoretical solution of this model using the principle of FEM can be entered using the eqn. (6) and boundary conditions. Boundary conditions are as follows: the displacement of all nodes in the direction of x is equal to zero. At point number two the force F in the y direction is applied. Theoretical solution written in matrix form is as follows:

Fig. 4 FE model using BEAM54.

11 1 2 1 3 1 4 1

1 2 2 2 1 4 2 4 1

1 3 1 4 11 1 3 1 4 2

1 4 2 4 2 2 1 4 2 4 2

1 3 1 4 11 1 2 3

1 4 2 4 1 2 2 2 3

K K K K 0 0 0

K K K K 0 0 0

K K 2K 0 K K F

K K 0 2K K K 0

0 0 K K K K 0

0 0 K K K K 0

, , , ,

, , , ,

, , , , ,

, , , , ,

, , , ,

, , , ,

v

v

v

θ

θ

θ

− − − = − − −

−

, (12)

where 11 3

12 13bLK

L 35,

EI K= + , 2

1 2 2

6 11bLK

L 210,

EI K= + , 1 3 3

12 9bLK

L 70,

EI K= − + ,

2

1 4 2

6 13bLK

L 420,

EI K= − , 3

2 2

4 bLK

L 105,

EI K= + , 3

2 4

2 bLK

L 140,

EI K= − , b

kK = .

In fig. 5 the deflection of beam is shown. The bending moment diagram is shown in Fig. 6 and transversal load diagram is shown in Fig. 7.

Fig. 5 Diagram of the deflection of beam in meters.

Fig. 6 Diagram of the bending moment in Nm.

Fig. 7 Diagram of the transversal load in N.

Numerical solution using BEAM54 (k=0) and CONTAC52 (2. method of solution)

The second method of solution is using the beam and the contact element. The model is shown in Fig. 8 with numbering of nodes and elements. Matrix form of the theoretical solutions using FEM approach, where the stiffness matrix for one element is given by eqn. (10) is as follows

Fig. 8 FE model using BEAM and CONTACT element.

1

2 21

2

2 2 232

3

2 23

12 6L 12 6L 0 0 0

6L 4L 6L 2L 0 0 0

12 6L 24 0 12 6L F

6 2L 0 8L 6L 2L 0L

0 0 12 6L 12 6L 0

0 0 6L 2L 6L 4L 0

M

M

M

vC

vCEI

L

vC

θ

θ

θ

+ − − − − + − − = − − − + −

−

, (13)

where 3L

MC CEI

= . Stiffness of spring in contact element is followingcontactn

KC = , where contactn

is number of contact element. Result of the solution for deflection is shown in Fig. 9. The bending moment diagram is shown in Fig. 10 and transverse load diagram is shown in Fig. 11.

Fig. 9 Diagram of the deflection of beam in meters.

Fig. 10 Diagram of the bending moment in Nm.

Fig. 11 Diagram of the transversal load in N.

Conclusions

In this paper, the solution of plane beams and frames using the Finite Element Method is

shown. Two approaches are shown for the numerical solution of beams. The first approach is using the BEAM54 element, where the element stiffness matrix is according to equation (8). This approach can be used when the foundation is without compression resistance. If we consider compression resistance, we have to applied the contact element approach (for example. CONTACT52), where compression resistance is input by a gap. In our example, the gap is equal to zero. Because the ANSYS software contains beam elements with shear deformation, only BEAM54 is without shear deformation. For verification an example of contact was considering using the element BEAM54, when the stiffness of elastic foundation is equal to zero.

Of course the accuracy of the result for the numerical methods is influenced by the number of elements over the length of beam. The verification examples used only one element over the length of beam L. Influence of the number of divisions in both the approaches is illustrated in the Fig. 12, where deflection of the beam is shown. In Fig. 13 a diagram of bending moment is shown and in Fig. 14 the transverse load is shown.

0,08

0,09

0,10

0,11

0,12

0,13

0,14

0,15

0 200 400 600 800 1000 1200 1400 1600 1800 2000

Number of division (-)

De

fle

ctio

n (

m)

Max. deflection using Beam and Contact (2. method)Min. deflection using Beam and Contact (2. method)Max. deflection (theoretical solution)Min. deflection (theoretical solution)Max. deflection using Beam54 (1. method)Min. deflection using Beam54 (1. method)

x 10-2

Fig. 12 The influence of number of element divisions along the length L on the minimum and maximum deflection.

42000

44000

46000

48000

50000

52000

54000

56000

0 200 400 600 800 1000 1200 1400 1600 1800 2000

Number of division (-)

Be

nd

ing

mo

me

nt

(Nm

)

Max. bending moment using Beam and Contact (2. method)

Max. bending moment (theoretical solution)

Max. bending moment using BEAM54 (1. method)

Fig. 13 The influence of the number of elements division along

the length L on the bending moment.

30000

35000

40000

45000

50000

55000

0 200 400 600 800 1000 1200 1400 1600 1800 2000

Number of division (-)

Tra

nsv

ers

e f

orc

e (

N)

Max. Transverse force using Beam and Contact (2. method)

Max. Transverse force (theoretical solution)

Max. Transverse force using Beam54 (1. method)

Fig. 14 The influence of number of elements division along the length L on maximum

transverse force It is shown in Fig. 12 to 14, the difference between solutions using the element BEAM54

with given foundation stiffness and using the BEAM54, when the stiffness of foundation is zero

in the real constants, and elastic foundation is modelled by contact elements CONTAC52. Both solutions are compared with theoretical solution. Reasons of difference are shown in Fig. 1, where the solution without contact is shown in Fig. 1a and with the contact in Fig.1b.

Based on the above information, it is possible to create the program code for the solution of plane beams on elastic foundations, for example in the program environment MATLAB. Practical solutions of problem related to gas pipelines using the contact elements is explained in Chapter 16 of Literature [3].

Acknowledgements

The paper was supported by grant from Grant Agency of VEGA no. 1/0100/08.

References

[1] Cook, R.D., Malkus D.S., Plesha, M.E., Witt, R.J.: Concepts and Applications of Finite Element Analysis, Fourth Edition, WILEY, ISBN 0-471-35605-0, 2002, pp.719

[2] Frydrýšek, K.: Nosníky a rámy na pružném podkladu 1, Faculty of Mechanical Engineering, VŠB - Technical University of Ostrava, ISBN 80-248-1244-4, Ostrava, Czech Republic, 2006, pp.463.

[3] Frydrýšek, K. Jančo, R: Nosníky a rámy na pružném podkladu 2, Faculty of Mechanical Engineering, VŠB - Technical University of Ostrava, ISBN 978-80-248-1743-9, Ostrava, Czech Republic, 2008, pp.516.

[4] ANSYS Theoretical manual

Reviewer: doc. Ing. Karel FRYDRÝŠEK, Ph.D., ING-PAED IGIP, VŠB - Technical University of Ostrava, Czech Republic