Beams on elastic foundation

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1 Beams on elastic foundation I Basic concepts The beam lies on elastic foundation when under the applied eternal loads, the reaction forces of the foundation are proportional at every point to the deflection of the beam at this point This assumption was introduced first by Winkler in 1867 Consider a straight beam supported along its entire length by an elastic medium and subjected to vertical forces acting in the plane of symmetry of the cross section (Fig 1 A F B cross section b y R( Figure 1 Beam on elastic foundation plane of symmetry Because of the eternal loadings the beam will deflect producing continuously distributed reaction forces in the supporting medium The intensity of these reaction forces at any point is proportional to the deflection of the beam y( at this point via the constant k: R(=k y( The reactions act vertically and opposing the deflection of the beam Hence, where the deflection is acting downward there will be a compression in the supporting medium Where the deflection happens to be upward in the supporting medium tension will be produced which is not possible In spite of all it is assumed that the supporting medium is elastic and is able to take up such tensile forces In other words the foundation is made of material which follows Hooke s law Its elasticity is characterized by the force, which distributed over a unit area, will cause a unit deflection This force is a constant of the supporting medium called the modulus of the foundation k [kn/m /m] Assume that the beam under consideration has a constant cross section with constant width b which is supported by the foundation A unit deflection of this beam will cause reaction eual to k b in the foundation, therefore the intensity of distributed reaction (per unit length of the beam will be: R(= b k y(= k y(, where k= k b is the constant of the foundation, known as Winkler s constant, which includes the effect of the width of the beam, and has dimension kn/m/m II Differential euation of euilibrium of a beam on elastic foundation Consider an infinitely small element enclosed between two vertical cross sections at distance d apart on the beam into consideration (Fig Assume that this element was taken from a portion 11 S Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 111

2 where the beam was acted upon by a distributed loading ( The internal forces that arise in section cuts are depicted in Fig ( d Q QdQ d R( Figure Differential element of length d Considering the euilibrium of the differential element, the sum of the vertical forces gives: Σ V = Q Q dq R( d ( d= ; dq k y d = ( ky ( Considering the euilibrium of moments along the left section of the element we get: d d Σ = d ( Q dq d r = ; d d = Using now the well known differential euation of a beam in bending: d y = d EI it can be written: dq d d y = = EI d d d Finally, from the summation of the vertical forces Σ V = : d y EI = k y ; d IV ( k y y = EI EI In the above euation the parameter includes the fleural rigidity of the beam as well as the elasticity of the foundation This factor is called the characteristic of the system with dimension length -1 In that respect 1/ is referred to as the so called characteristic length Therefore, will be an absolute number The differential euation of euilibrium of an infinitely small element becomes: 11 S Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 11

3 IV ( k y y =, = EI EI The solution of this differential euation could be epressed as: y( = y ( v(, where y ( is the solution of homogeneous differential euation y y =, v ( is a particular integral corresponding to ( 1 Solution of homogeneous differential euation The characteristic euation of the differential euation under consideration is: r = After simple transformations this characteristic euation could be presented as: ( ( ( ( ( ( ( ( r = r i = r i r i =, r i r i = r i i r i i = r ( i r ( i r i ( i r i ( i = = r i( i r i( i r i( i r i( i =, wherefrom the roots of the above characteristic euation are: r = i, r = i, r = i, r = i ( ( ( ( 1 The general solution of homogeneous differential euation takes the form: r1 r r r ( = 1 ; ( i ( i ( i ( i ( = 1 i i i i ( = 1 y A e A e A e A e y A e A e A e A e y A e e A e e A e e A e e Using the well known Euler s epressions: i e = cos i sin, i e = cos i sin the solution takes the form: y ( = A e cos i sin A e cos i sin A e cos i sin ( ( ( 1 ( cos sin A e i After simple regrouping of the members the epression becomes: y( e ( A A cos ( ia ia sin = e ( A1 A cos ( ia1 ia sin B1 B B B ; IV 11 S Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 11

4 By introducing the new constants B 1 -B where B = ( A A, B = ( ia ia, B = ( A A 1 1 ( = 1cos sin cos sin and B = ( ia ia the solution can be written in a more convenient form: [ ] [ ] y e B B e B B Taking into account the epressions: 1 ch = ( e e 1 and sh = ( e e, respectively: e = ch sh and e = ch sh, the euation of elastic line takes the form: y ( = ch sh B cos B sin ch sh B cos B sin, ( [ ] ( [ ] 1 1 and after regrouping of the members of the epression: y( = ch cos ( B1 B sin ( B B sh cos ( B1 B sin ( B B C1 C C C By introducing the new constants C 1 -C the final solution reads: y ( = ch C cos C sin sh C cos C sin [ ] [ ] 1 11 Derivation of the integration constants C 1 -C By differentiation of the above euation we get: y ( = sh[ C1cos Csin] ch [ C1sin Ccos] ch[ Ccos Csin] sh [ Csin Ccos ] After regrouping of the members about integration constants C i the first derivative takes the form: y ( = C1( sh cos ch sin C( sh sin ch cos C ch cos sh sin C ch sin sh cos ( ( By differentiation of the first derivative y ( the second derivative of the elastic line is: y ( = C ch cos sh sin sh sin ch cos 1 ( ( ch sin sh cos sh cos ch sin ( sh cos ch sin ch sin sh cos ( sh sin ch cos ch cos sh C C C sin Finally, for the second derivative we have: y ( = C sh sin C sh cos C ch sin 1 C ch cos After differentiation of the second derivative the third derivative of y ( becomes: ( ( ( ( 1 y ( = C ch sin sh cos C ch cos sh sin C sh sin ch cos C sh cos ch sin 11 S Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 11

5 dy Knowing that ( d = ϕ, d y d y Q = and = we can obtain the general epressions d EI d EI for the slope of the deflected line ϕ (, for the bending moment ( and for the shear force Q( at any point of distance at the beam ais Taking in these euations =, bearing in mind that sin =, sh =, cos = 1, ch = 1 and cos ch = 1, we get the initial parameters of the left end of the beam as follows: y( = y = C 1 ; y ( = ϕ = C C ; y ( = = C; EI Q y ( = = C C EI After simple transformations: y = C1; ϕ = C C; = C ; Q = C C Now epressing the constants C 1 -C as unknowns, from the above system of euations we have: C1 = y; ϕ Q C = ; ϕ Q C = C = Substituting these results in the above epression for the solution of homogeneous differential euation y ( we get: ϕ Q ϕ Q y( = chycos sin sh cos sin EI EI EI After regrouping of the members about the initial parameters the solution becomes: ( ch sin sh cos ϕ y ( = ch cos y ( ch sin sh cos Q sh sin EI EI 11 S Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 115

6 By introducing the new functions A(, B (, C( and D( following substitutions are valid: A =, ( ch cos ( ch sin sh cos ( = B, sh sin C( = and ( ch sin sh cos D( =, the euation of elastic line becomes: ϕ Q y( = A( y B( C( D( The functions A(, B(, C( and D( in such a way that the are known as Krilov s functions, and the following epressions are true for their first derivatives: A = D ; B = A ; C = B ; D = C ( ( ( ( ( ( ( ( Recall that the solution of original differential euation is: y ( = y ( v (, where v ( is a particular integral corresponding to the applied loads In that respect the euation of elastic line of a beam on elastic foundation get the form: ϕ Q y ( = A ( y B( C( D( v ( EI EI Using the above relations for first derivative of the Krilov s functions and final euation of elastic or th n ϕ, for the bending line we can obtain general epressions f e slope of the deflected li e ( moment ( and for the shear force ( Q at any point of distance of the beam ais These relationships are as follows: ϕ Q y ( = A ( y B( C( D( v (, EI EI ϕ Q ϕ( = y ( = D ( y A( B( C( v (, EI EI ϕ ( = EIy ( = EI C ( y EI D ( EI A( Q EI B ( EI v (, ϕ Q ( = EIy ( = EI B ( y EI C ( EI D( Q EI A ( EI v ( 11 S Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 116

7 The same euations written in matri form are: y y ( A B C D ϕ v ( ( D A B C v( ϕ = ( EI C EI D EI A EI B EI v ( Q( EI v ( EI B EI C EI D EI A Q In the above epressions the initial integral constants C 1 -C are replaced by the y, ϕ, and Q uantities, called initial parameters This representation is known as the method of initial conditions It is more convenient to epress EI multiple values of the transverse displacements ( EIy( and EI multiple values of slo pe of deflection line ( EIϕ ( If the following relations are valid: V = EIy(, V = EIy ; Φ= EIϕ(, Φ = EIϕ ; V = EIv(, Φ= EIv (, = EIv (, Q = EIv ( ; The basic unk nowns of any section of the beam ais epressed by the initial parameters finally get the form: B C D A V( V V B C ( D A Φ Φ Φ = ( B Q( C D A Q Q B C D A III Derivation of particular integrals corresponding to concentrated, concentrated vertical force F and uniformly distributed load ( moment Q F ( φ( φ 1 5 y k n f y Figure Beam on elastic foundation different types of loading m y( Q( 11 S Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 117

8 Let us assume that initial parameters y, ϕ, and Q are known Then we can proceed from the left end of the beam (Fig toward the right along the unloaded portion -1 until we arrive at the point 1 where the concentrated moment is applied 1 Particular integral owe to the concentrated moment The concentrated moment must have an effect to the right of section 1 similar to the effect, which the initial moment had on portion -1 The influence of is given by the third column of the above matri In accordance with this column the influence of concentrated moment can be epressed as: C( m V( =, B( m Φ ( =, = A m, ( ( Q= D m Particular integral due to the concentrated force F In a similar way we can find the influence of concentrated force F It is the same as the influence of Q to the portion - of the beam, taken with reverse sign (because concentrated force is opposing to the initial shear force Thus, the influence of F could be obtained by the forth column of the above matri, or: ( f D V( = F, C( f Φ ( = F, B( f ( = F, Q( = A f F ( Particular integral corresponding to the distributed load The distributed load can be regarded as consisting of infinite small concentrated forces such as d in Fig The effect of this force for the portion -5 of the beam into consideration is the same as the effect of the force F (Fig, namely: ( ( D n V( = d 11 S Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 118

9 d Q ( φ( φ 5 y d k y( Q( n y Figure Derivation of particular integral due to distributed loading The effect of all infinite small concentrated forces belonging to the distributed load can be epressed as the following integral: n k ( ( n k ( ( D n D n V( = d = d n A ( = D( ( D( d, Bearing in mind that A d= respectively D( d = A( / ( it can be written that: our case D ( n d ( n = A ( n / In ( ( Finally, for the influence of distributed load on the transverse displacements we get: ( ( n k ( ( ( ( A n A n n k A n V( = = = A n A k V( = A n A k ( ( ( ( ( ( The other particular integrals can be derived by differentiation of the above euation - recall the relations A = D; B = A; C = B; D = C, or: Φ ( = V( = ( D( n D( k ; ( = Φ ( = ( C( n C( k ; Q( = ( = ( B( n B( k 11 S Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 119

10 IV Classification of the beams according to their stiffness The term l, where l is the beam length, characterizes the relative stiffness of the beam on elastic foundation According to the values of l the beams can be classified into three groups: 1 Short beams for which: l < 5 ; Beams of medium length: 5 l 5 ; Long beams: l > 5 For beams belonging to the first group the bending deformations of the beam can be neglected in the most practical cases These deformations are negligible small compared to the deformations produced in the foundation Therefore such beams can be considered as absolutely rigid For the beams belonging to the second group the loads applied at the one end of the beam have a finite and not negligible effect to the other end In this case an accurate computation of the beam is necessary and no approimations are possible For these beams the method of initial conditions is very suitable and the obtained results are accurate The beams belonging to the third group have a specific feature l such that the counter effect of the end conditioning terms (forces and displacements on each other is negligible When investigate the one end of the beam we can assume that the other end is infinitely far away The forces applied at the one end of the beam have a negligible effect to the other end In other words the Krilov s functions A( l, B( l, C( l and D( l are practically zero which greatly simplifies the computations V Numerical eample In the following numerical eample we shall construct the diagrams of vertical displacements, the slope of the deflection line, bending moment and shear force and the diagram of vertical reactions in the foundation All the diagrams will be found by the method of initial conditions Consider a beam on elastic foundation with free ends The geometrical dimensions, mechanical properties and loadings are shown in Fig 5 The modulus of elasticity of material of the beam is 1 7 kn/m (concrete and the modulus of the foundation is k =5 kn/m /m The Wikler s constant or constant of the foundation is: k= k b=5 11= Cross section 5 1 f= m=6 n=5 E= 1 7 kn/m ; k =5 kn/m /m Figure 5 Numerical eample: geometrical dimensions and mechanical properties 11 S Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 1

11 The main characteristics of the beam into consideration are: I = b h = = m ; EI = = 75 kn m ; k 55 1 = = = 71 m ; EI 75 l = 71 1 = 7 The beam is of a medium length according to its stiffness l, so the method of initial conditions is applicable Net we should determine the initial parameters of the left end using the boundary conditions of the right end of the beam Obviously for the free left end the bending moment and the shear force Q are eual to zero, the vertical deflection y and rotation φ are the unknown initial parameters which should be determined using the boundary conditions of the right end These boundary conditions are: l ( = ; Ql ( = Using the epressions for bending moment and shear force in terms of initial parameters and accounting for the influence of loading the following euations are compounded: B( f ( l = C( l V D( l Φ A( m F ( C( n C( k =, Q( l = B( l V C( l Φ D( m A( f F ( B( n B( k = For =l the distances m, f, n and k are as shown in Fig 5 or m =6, f =9, n =5 and k = In this case the above euations become: 17 V 158 Φ ( =, 969 V 17 Φ = ( 17 V 158 Φ = 857, 9691 V 17 Φ = 811 Wherefrom: V = 59 Φ = 151 Having the initial parameters available, for the given eternal loadings, we can obtain any parameter of an arbitrary section of the beam ais using the epressions below: 11 S Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 11

12 B( C( D( A( V( V V B( C( ( D( A( Φ Φ Φ = ( B( Q( C( D( A( Q Q B( C( D( A( D ( ( f C m ( A( n A( k F V C Φ ( ( f ( D( n D( k B m F = ( B( f ( ( ( C n C k Q A m F D( m A( f F ( B( n B( k V( R( = k y( = k EI It should be pointed out that for every different section of the beam, with a single abscissa, the values of m, f, n and k are different and depend on the distance The influence of different loading appears when the section into consideration is on the right of this loading The obtained results for vertical displacements, the slope of the deflection line, bending moments, shear forces and the vertical reactions, for different sections of the beam, are calculated and given in table 1 Table 1 V(=EIy( Φ(=EIφ( ( Q( R( S Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 1

13 The diagrams of reuired displacements and internal forces of the beam are presented in Fig 6 The diagram of continuously distributed reaction forces in the foundation is depicted in Fig 7 (a V = EIy( Φ = EIϕ ( (b (c (d Q Figure 6 (a Vertical deflections; (b slope of the deflected line; (c bending moment diagram; (d shear force diagram 11 S Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 1

14 R Figure 7 Diagram of vertical reactions Verification In order to verify the obtained results we shall check the euilibrium of the vertical forces using the condition Σ V = In order to do that, we should find the resultant force of distributed vertical reaction in the foundation, presented in Fig 7 In other words we have to calculate the area of the corresponding diagram of R Using a numerical integration the area of the diagram of vertical reactions is: 96 A = = 15 Σ V = = The numerical error is 6% The error is due to numerical integration and will decrease if we calculate the values of vertical reactions in more sections of the beam, respectively if we decrease the step between two neighboring values of the diagram 11 S Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 1

15 References HETENYI, Beams on elastic foundation Waverly press, Baltimore, 196 КАРАМАНСКИ Т Д И КОЛЕКТИВ Строителна механика Издателство Техника София, S Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 15

7.4 The Elementary Beam Theory

7.4 The Elementary Beam Theory In this section, problems involving long and slender beams are addressed. s with pressure vessels, the geometry of the beam, and the specific type of loading which will be