Beams on Elastic Foundation

Lecture notes: Structural Analysis II2011 S. Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 111 Beams on elastic foundation I. Basic concepts. The beam lies on elastic foundation when under the applied external 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) Figure 1 Beam on elastic foundation Because of the external 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(x) at this point via the constant k: R(x)=k·y(x). 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 0 [kN/m 2 /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 equal to k 0 ·b in the foundation, therefore the intensity of distributed reaction (per unit length of the beam) will be: R(x)= b·k 0 ·y(x)= k·y(x), where k= k 0 ·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 equation of equilibrium of a beam on elastic foundation Consider an infinitely small element enclosed between two vertical cross sections at distance dx apart on the beam into consideration (Fig. 2). Assume that this element was taken from a portion F M q R(x) x y A B plane of symmetry cross section b Table of contents Lecture notes: Structural Analysis II 2011 S. Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 112 where the beam was acted upon by a distributed loading q(x). The internal forces that arise in section cuts are depicted in Fig. 2. Figure 2 Differential element of length dx Considering the equilibrium of the differential element, the sum of the vertical forces gives: ( ) ( ) 0 ( ) ( k y x V Q Q dQ R x dx q x dx ⋅ Σ = − + + ⋅ − ⋅ = ) 0; dQ k y q dx = ⋅ − . Considering the equilibrium of moments along the left section of the element we get: ( ) 2 2 0 0 2 2 dx dx M dM Q dQ dx q r Σ = − + ⋅ − + = ; dM q dx = . Using now the well known differential equation of a beam in bending: 2 2 d y M EI dx = − ’ it can be written: 2 4 2 4 dQ d M d y EI dx dx dx = = − . Finally, from the summation of the vertical forces 0 V Σ = : 4 4 d y EI k y q dx − = ⋅ − ; 4 ( ) 4 4 IV k q y y EI EI + ⋅ = α x . In the above equation the parameter α includes the flexural 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/α is referred to as the so called characteristic length. Therefore, x ⋅ α will be an absolute number. The differential equation of equilibrium of an infinitely small element becomes: q(x) dx R(x) Q+dQ M+dM Q M 0 0 0 Lecture notes: Structural Analysis II 2011 S. Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 113 4 ( ) 4 IV q x y y EI + ⋅ = α , 4 4 k EI = α . The solution of this differential equation could be expressed as: 0 ( ) ( ) ( ) y x y x v x = + , where 0 ( ) y x is the solution of homogeneous differential equation , is a particular integral corresponding to q(x). 4 4 IV y y + ⋅ = α 0 ( ) v x 1. Solution of homogeneous differential equation The characteristic equation of the differential equation under consideration is: 4 4 4 0 r + = α . After simple transformations this characteristic equation could be presented as: ( ) ( ) ( ) ( ) 2 2 4 2 4 2 2 2 2 2 2 2 2 2 2 r r i r i r i ⎡ ⎤ ⎡ + = − ⋅ = − ⋅ + ⋅ ⎢ ⎥ ⎢ ⎣ ⎦ ⎣ α α α α 0 ⎤ = ⎥ ⎦ , ( ) ( ) 2 2 2 2 2 2 2 2 2 2 r i r i ⎡ ⎤ ⎡ + − ⋅ − + + ⋅ − = ⎢ ⎥ ⎢ ⎣ ⎦ ⎣ α α α α α α 0 ⎤ ⎥ ⎦ ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 r i i r i i ⎡ ⎤ ⎡ + − ⋅ + + + ⋅ + = ⎢ ⎥ ⎢ ⎣ ⎦ ⎣ α α α α α α 0 ⎤ ⎥ ⎦ 2 ⎤ ⎥ ⎦ ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 0 r i r i r i i r i i ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ + − ⋅ + + ⋅ = − − ⋅ − + ⋅ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ α α α α α α α α ( ) ( ) ( ) ( ) 0 r i i r i i r i i r i i − − ⋅ + − ⋅ − + ⋅ + + ⋅ = ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ α α α α α α α α ⎤ ⎦ , wherefrom the roots of the above characteristic equation are: ( ) ( ) ( ) ( ) 1 2 3 4 , , , r i r i r i r i = + ⋅ = − ⋅ = − + ⋅ = − − ⋅ α α α α α α α α 4 . The general solution of homogeneous differential equation takes the form: 1 2 3 0 1 2 3 4 ( ) r x r x r x r x y x A e A e A e A e ⋅ ⋅ ⋅ ⋅ = ⋅ + ⋅ + ⋅ + ⋅ ; ( ) ( ) ( ) ( ) 0 1 2 3 4 ( ) i x i x i x i x y x A e A e A e A e + ⋅ ⋅ − ⋅ ⋅ − + ⋅ ⋅ − − ⋅ ⋅ = ⋅ + ⋅ + ⋅ + ⋅ α α α α α α α α ; 0 1 2 3 4 ( ) x i x x i x x i x x i x y x A e e A e e A e e A e e − − − = ⋅ + ⋅ + ⋅ + ⋅ − α α α α α α α α x x . Using the well known Euler’s expressions: cos sin , cos sin . i x i x e x i e x i − = + ⋅ = − ⋅ α α α α α α the solution takes the form: ( ) ( ) ( ) ( ) 0 1 2 3 4 ( ) cos sin cos sin cos sin cos sin x x x x y x A e x i x A e x i x A e x i x A e x i x − − = ⋅ + ⋅ + ⋅ − ⋅ + ⋅ + ⋅ + + ⋅ − ⋅ α α α α α α α α α α α α After simple regrouping of the members the expression becomes: ( ) ( ) ( ) ( ) 3 4 1 2 0 3 4 3 4 1 2 1 2 ( ) cos sin cos sin x x B B B B y x e A A x iA iA x e A A x iA iA x − ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ = + + − + + + − ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ α α α α α α . Lecture notes: Structural Analysis II 2011 S. Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 114 By introducing the new constants B 1 -B 4 where ( ) 1 3 4 B A A = + , ( ) 2 3 4 B iA iA = − , ( ) 3 1 2 B A A = + and ( 4 1 2 ) B iA iA = − the solution can be written in a more convenient form: [ ] [ ] 0 1 2 3 4 ( ) cos sin cos sin x x y x e B x B x e B x B x − = + + + α α α α α α Taking into account the expressions: ( ) 1 ch 2 x x x e e − = + α α α and ( ) 1 sh 2 x x x e e − = − α α α , respectively: ch sh x e x = + x α α α and ch sh x e x − = − x α α α , the equation of elastic line takes the form: ( )[ ] ( )[ ] 0 1 2 3 ( ) ch sh cos sin ch sh cos sin 4 y x x x B x B x x x B x B x = − + + + + α α α α α α α α 2 4 , and after regrouping of the members of the expression: ( ) ( ) ( ) ( ) 2 4 1 3 0 1 3 2 4 1 3 ( ) ch cos sin sh cos sin C C C C y x x x B B x B B x x B B x B B ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ = + + + + − + + − + ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ α α α α α α By introducing the new constants C 1 -C 4 the final solution reads: [ ] [ ] 0 1 2 3 4 ( ) ch cos sin sh cos sin y x x C x C x x C x C = + + + x α α α α α α . 1.1 Derivation of the integration constants C 1 -C 4 By differentiation of the above equation we get: [ ] [ ] [ ] [ 0 1 2 1 2 3 4 3 4 ( ) sh cos sin ch sin cos ch cos sin sh sin cos . y x x C x C x x C x C x ] x C x C x x C x C x ′ = ⋅ + + ⋅ ⋅ − + + + ⋅ + + ⋅ ⋅ − + α α α α α α α α α α α α α α α α After regrouping of the members about integration constants C i the first derivative takes the form: ( ) ( ) ( ) ( 0 1 2 3 4 ( ) sh cos ch sin sh sin ch cos ch cos sh sin ch sin sh cos . y x C x x x x C x x x x C x x x x C x x x x ′ = ⋅ ⋅ − ⋅ + ⋅ ⋅ + ⋅ + + ⋅ ⋅ − ⋅ + ⋅ ⋅ + ⋅ α α α α α α α α α α α α α α α α α α α α ) By differentiation of the first derivative 0 ( ) y x ′ the second derivative of the elastic line is: ( ) ( ) ( ) 2 0 1 2 2 2 3 2 4 ( ) ch cos sh sin sh sin ch cos ch sin sh cos sh cos ch sin sh cos ch sin ch sin sh cos sh sin ch cos ch cos sh y x C x x x x x x x x C x x x x x x x x C x x x x x x x x C x x x x x x x ′′ = ⋅ ⋅ − ⋅ − ⋅ − ⋅ + ⋅ ⋅ + ⋅ + ⋅ − ⋅ + ⋅ ⋅ − ⋅ − ⋅ − ⋅ + ⋅ ⋅ + ⋅ + ⋅ − α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α α ( ) sin . + + + x ⋅ α Finally, for the second derivative we have: 2 2 2 0 1 2 3 2 4 ( ) 2 sh sin 2 sh cos 2 ch sin 2 ch cos . y x C x x C x x C x x C x x ′′ = − ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − ⋅ ⋅ ⋅ + + ⋅ ⋅ ⋅ α α α α α α α α α α α α After differentiation of the second derivative the third derivative of 0 ( ) y x becomes: ( ) ( ( ) ( 3 3 0 1 2 3 3 3 4 ( ) 2 ch sin sh cos 2 ch cos sh sin 2 sh sin ch cos 2 sh cos ch sin . y x C x x x x C x x x x C x x x x C x x x x ′′′ = − ⋅ ⋅ + ⋅ + ⋅ ⋅ − ⋅ − − ⋅ ⋅ + ⋅ + ⋅ ⋅ − ⋅ α α α α α α α α α α α α α α α α α α α α ) ) Lecture notes: Structural Analysis II 2011 S. Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 115 Knowing that ( ) dy x dx = ϕ , 2 2 d y M EI dx = − and 3 3 d y Q EI dx = − we can obtain the general expressions for the slope of the deflected line ( ) x ϕ , for the bending moment ( ) M x and for the shear force at any point of distance ( ) Q x x at the beam axis. Taking in these equations , bearing in mind that , , 0 x = sin 0 0 = sh 0 0 = cos 0 1 = , ch 0 1 = and cos 0 ch 0 1 ⋅ = , we get the initial parameters of the left end of the beam as follows: 0 0 (0) y y = = 1 C 3 C ; 0 0 2 (0) y C ′ = = ⋅ + ⋅ ϕ α α ; 2 0 0 4 ( ) 2 M y x C EI ′′ = − = ⋅ α ; 3 3 0 0 2 ( ) 2 2 Q y x C C EI ′′′ = − = ⋅ − ⋅ α α 3 . After simple transformations: 0 1 y C = ; 0 2 3 C C = + ϕ α ; 0 4 2 2 M C EI − = ⋅ α ; 0 2 3 3 2 Q C C EI − = ⋅ α − . Now expressing the constants C 1 -C 4 as unknowns, from the above system of equations we have: 1 0 C y = ; 0 0 2 3 2 4 Q C EI = − ⋅ ϕ α α ; 0 0 3 3 2 4 Q C EI = + ⋅ ϕ α α 0 4 2 2 M C EI = − ⋅ α . Substituting these results in the above expression for the solution of homogeneous differential equation 0 ( ) y x we get: 0 0 0 0 0 0 0 3 3 ( ) ch cos sin sh cos sin . 2 2 4 4 2 Q Q y x x y x x x x x EI EI EI ϕ ϕ α α α α α α α α α α ⎡ ⎤ ⎡ ⎛ ⎞ ⎛ ⎞ = + − + + − ⎢ ⎥ ⎢ ⎜ ⎟ ⎜ ⎟ ⋅ ⋅ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ ⎣ 2 M α ⎤ ⎥ ⋅ ⎦ After regrouping of the members about the initial parameters the solution becomes: ( ) ( ) 0 0 0 0 0 3 2 ch sin sh cos ( ) ch cos 2 ch sin sh cos sh sin . 4 2 x x x x y x x x y x x x x Q M x x EI E α α α α ϕ α α α α α α α α α α α ⋅ + ⋅ = ⋅ ⋅ + + ⋅ − ⋅ ⋅ ⎛ ⎞ ⎛ + − + ⎜ ⎟ ⎜ ⋅ ⋅ ⎝ ⎠ ⎝ I ⎞ − ⎟ ⎠ Lecture notes: Structural Analysis II 2011 S. Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 116 By introducing the new functions ( ) A x α , ( ) B x α , ( ) C x α and ( ) D x α in such a way that the following substitutions e valid: ar ( ) ch cos A x x = ⋅ x α α α , ( ) ( ) ch sin sh cos 2 x x x x B x ⋅ + ⋅ = α α α α α , ( ) sh sin 2 x x C x ⋅ = α α α and ( ) ( ) ch sin sh cos 4 x x x x ⋅ − ⋅ D x = α α α α , the equation of elastic line becomes: α ( ) ( ) ( ) ( ) 0 0 0 0 2 3 ( ) M Q y x A x y B x C x D x 0 EI E = ⋅ + − − I ⋅ ⋅ ϕ α α α α α α α . The functions ( ) A x α , ( ) B x α , ( ) C x α and ( ) D x α are known as Krilov’s functions, and the following expressions are true for their first derivatives: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 4 ; ; ; A x D x B x A x C x B x D x C x ′ ′ ′ ′ = − ⋅ = ⋅ = ⋅ = ⋅ α α α α α α α α α α α α . Recall that the solution of original differential equation is: In that respect the equation of elastic line of a beam on elastic foundation get the form: 0 ( y x ( ) ) ( ) y x v x = + , where ( ) v x is a particular integral corresponding to the applied loads. ( ) ( ) ( ) ( ) 0 0 0 0 2 3 ( ) ( ) M Q y x A x y B x C x D x v x EI EI = ⋅ + − − + ⋅ ⋅ ϕ α α α α α α α Using the above relations for first derivative of the Krilov’s functions and fi quation of elastic line we can obtain general expressions f e slope of the deflected li e ( ) nal e or th n x ϕ , for the bending moment ( ) M x and for the shear force ( ) Q x at any point of distance x of the beam axis. These relationships are as follows: ( ) ( ) ( ) ( ) 0 0 0 0 2 3 ( ) ( ) M Q y x A x y B x C x D x v x EI EI = ⋅ + − − + ⋅ ⋅ ϕ α α α α α α α , ( ) ( ) ( ) ( ) ( ) 0 0 0 0 2 3 ( ) 4 ( ) M Q x y x D x y A x B x C x v x EI EI ′ ′ = = − ⋅ ⋅ + ⋅ − ⋅ − ⋅ + ⋅ ⋅ ϕ ϕ α α α α α α α α α α α , ( ) ( ) ( ) ( ) ( ) 2 2 2 0 0 0 2 2 0 3 ( ) 4 4 ( ), M M x EIy x EI C x y EI D x EI A x EI Q EI B x EI v x EI ϕ α α α α α α α α α α α ′′ = − = ⋅ ⋅ + ⋅ + ⋅ ⋅ ′′ + ⋅ − ⋅ ⋅ ( ) ( ) ( ) ( ) ( ) 3 3 3 0 0 0 2 3 0 3 ( ) 4 4 4 ( ). M Q x EIy x EI B x y EI C x EI D x EI Q EI A x EI v x EI ϕ α α α α α α α α α α α ′′′ = − = ⋅ ⋅ + ⋅ − ⋅ ⋅ ′′′ + ⋅ − ⋅ ⋅ Lecture notes: Structural Analysis II 2011 S. Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 117 The same equations written in matrix form are: ( ) ( ) ( ) 0 0 0 2 2 2 2 2 3 3 3 3 0 3 ( ) ( ) 4 ( ) 4 4 ( ) 4 4 4 y A B C D y x v x D A B C x v x M M x ( ) EI C EI D EI A EI B EI v x EI Q x EI v x EI B EI C EI D EI A Q ⎧ ⎫ ⎪ ⎪ − − ⎡ ⎤ ⎧ ⎫ ⎪ ⎪ ⎧ ⎫ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ − ⋅ ⋅ − ⋅ − ⋅ ′ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ = + ⎨ ⎬ ⎨ ⎬ ⎨ ⎢ ⎥ ′′ ⋅ ⋅ ⋅ ⋅ − ⋅ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⋅ ⎪ ⎪ ⎪ ⎪ ⎪ ′′′ − ⋅ ⎢ ⎥ ⎩ ⎭ ⎩ ⎭ ⋅ ⋅ − ⋅ ⋅ ⎣ ⎦ ⎪ ⎪ ϕ α α α α α ϕ α α α α α α α α α ⎪ EI ⎪ ⎪ ⋅ ⎩ ⎭ α the above expressions the initial integral constants C 1 -C 4 are replaced by the 0 0 0 , , y M ⎪ ⎬ ⎪ In ϕ and 0 Q quantities, called initial parameters. This representation is known as the m ) and EI multiple values of slo deflection line ( ethod of initial conditions. It is more convenient to express EI multiple values of the transverse displacements ( ( ) EIy x pe of ( ) EI x ϕ ). If the following relations ar ; e valid: 0 0 V ( ), V EIy x EIy = = 0 0 ( ), EI x EI Φ = Φ = ϕ ϕ ; V ( ), ( ), M ( ), Q ( ) EIv x EIv x EIv x EIv x ′ ′′ ′′′ = Φ = = − = − ; nowns of any section of the beam axis expressed by the initial parameters finally get the form: nowns of any section of the beam axis expressed by the initial parameters finally get the form: The basic unk The basic unk ( ) ( ) ( ) ( ) 2 3 0 0 2 0 2 0 3 2 V V V 4 M 4 4 Q 4 4 4 B C D A x B C D A x M M x B C D A Q Q x B C D A ⎡ ⎤ − − ⎢ ⎥ ⎢ ⎥ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ − ⋅ − − Φ Φ Φ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ = + ⎨ ⎬ ⎨ ⎬ ⎨ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎢ ⎥ − ⎢ ⎥ ⎣ ⎦ α α α α α α α α α α α α ⎪ ⎬ ⎪ ⎪ moment , concentrated vertical force F and uniformly distributed load q(x) Figure 3 Beam on elastic foundation – different types of loading III. Derivation of particular integrals corresponding to concentrated M F M q x y 0 1 2 3 4 5 k n f m x y 0 φ 0 M 0 Q 0 M(x) Q(x) y(x) φ(x) Lecture notes: Structural Analysis II 2011 S. Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 118 Let us assume that initial parameters 0 0 0 , , y M ϕ and 0 Q are known. Then we can proceed from the left end of the beam (Fig. 3) toward the right along the unloaded portion 0-1 until we arrive at e point 1 where the concentrated moment is applied. . Particular integral owe to the concentrated moment M atrix. In accordance with this column the influence of concentrated moment can be expressed as: th 1 The concentrated moment M must have an effect to the right of section 1 similar to the effect, which the initial moment M 0 had on portion 0-1. The influence of M 0 is given by the third column of the above m ( ) ( ) ( ) ( ) 2 V( ) , ( ) , M , Q 4 . C m x M B m x M A m M D m M α α α α α α α = − Φ = − = ⋅ = − ⋅ 2. Particular integral due to the concentrated force F rce). Thus, the influence of F could be obtained by the forth column of the above matrix, or: In a similar way we can find the influence of concentrated force F. It is the same as the influence of Q 0 to the portion 0-2 of the beam, taken with reverse sign (because concentrated force is opposing to the initial shear fo ( ) ( ) ( ) ( ) 3 2 V( ) , ( ) , M( ) , Q( ) . D f x F C f x F B f x F x A f F α α α α α α α = Φ = = − = − ⋅ 2. Particular integral corresponding to the distributed load q The d rtion 3-5 of the beam into consideration is the me as the effect of the force F (Fig. 4), namely: istributed load q can be regarded as consisting of infinite small concentrated forces such as q q dx ⋅ in Fig. 4. The effect of this force for the po sa ( ) 3 ( ) V( ) q q D n x x q dx − = ⋅ α α . Lecture notes: Structural Analysis II 2011 S. Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 119 Figure 4 Derivation of particular integral due to distributed loading centrated forces belonging to the distributed load can be expressed as the following integral: The effect of all infinite small con ( ) ( ) ( ) 3 3 0 0 ( ) ( ) V( ) n k n k q q q q D n x D n x x q dx q d n x − − − − = ⋅ = − ⋅ − ∫ ∫ α α Bearing in mind that α α ( ) ( ) 4 A x D ′ = − ⋅ x α α α it can be written that: ( ) ( ) 4 A x dx ′ ⋅ = ∫ ∫ α α D x dx − ⋅ ⋅ α , respectively ( ) ( ) / 4 D x dx A x ⋅ = − ∫ α α α . our case In ( ) ( ) ( ) ( ) ( ) / 4 q q q D n x d n x A n x − ⋅ − = − − ∫ α α α . Finally, for the influence of distributed load on the transverse displacements we get: ( ) ( ) ( ) ( ) ( ) ( ) 3 3 4 0 ( ) ( ) ( 0) V( ) 4 4 4 n k q A n x A n n k A n q q q x A n A k − − − − − + + − = − = − = − − α α α α α α α α α α ( ) ( ) ( ) 4 V( ) 4 q x A n A k = − − α α α . particular integrals can be derived by differentiation of the above equation - recall the relations The other 4 ; ; ; A D B A C B D C ′ ′ ′ ′ = − ⋅ = ⋅ = ⋅ = ⋅ α α α α , or: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 2 ( ) V( ) ; M( ) ( ) ; Q( ) M( ) . q x x D n D k q x x C n C k q x x B n B k ′ Φ = = − ′ = −Φ = − − ′ = = − − α α α α α α α α α q x y 0 3 4 5 k n x y 0 φ 0 M 0 Q 0 M(x) Q(x) y(x) φ(x) q dx q x q q dx ⋅ k Lecture notes: Structural Analysis II 2011 S. Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 120 IV. Classification of the beams according to their stiffness The term l ⋅ α , where l is the beam length, characterizes the relative stiffness of the beam on elastic foundation. According to the values of l ⋅ α the beams can be classified into three groups: 1) Short beams for which: 0.5 l ⋅ < α ; 2) Beams of medium length: 0.5 5 l ≥ ⋅ ≤ α ; 3) Long beams: 5 l ⋅ > α . 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 approximations 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 ⋅ α 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 α , ( ) B l α , ( ) C l α and ( D l) α are practically zero which greatly simplifies the computations. V. Numerical example In the following numerical example 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 3·10 7 kN/m 2 (concrete) and the modulus of the foundation is k 0 =50000 kN/m 2 /m. The Wikler’s constant or constant of the foundation is: k= k 0 ·b=50000·1.1=55000. Figure 5 Numerical example: geometrical dimensions and mechanical properties 1 4 5 10 250 100 200 1.1 0 . 5 Cross section E=3·10 7 kN/m 2 ; k 0 =50000 kN/m 2 /m f=9 m=6 n=5 Lecture notes: Structural Analysis II 2011 S. Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 121 The main characteristics of the beam into consideration are: 3 3 7 2 1 4 4 1 1 1.1 0.5 0.011458 ; 12 12 3 10 0.011458 343750 . ; 55000 0.44721 ; 4 4 343750 0.44721 10 4.472. 4 I b h m EI kN m k m EI l α α − = ⋅ = ⋅ = = ⋅ ⋅ = = = = ⋅ ⋅ = ⋅ = The beam is of a medium length according to its stiffness l ⋅ α , so the method of initial conditions is applicable. Next 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 M 0 and the shear force Q 0 are equal to zero, the vertical deflection y 0 and rotation φ 0 are the unknown initial parameters which should be determined using the boundary conditions of the right end. These boundary conditions are: ( ) 0; ( ) 0. M l Q l = = Using the expressions for bending moment and shear force in terms of initial parameters and accounting for the influence of loading the following equations are compounded: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 0 0 2 3 2 0 0 4 ( ) V 4 ( ) 0, 4 ( ) V 4 ( ) 4 0 B f q M l C l D l A m M F C n C k q Q l B l C l D m M A f F B n B k α α α α α α α α α α α α α α α α α α α α = ⋅ + ⋅ Φ + ⋅ − − − = = ⋅ + ⋅ Φ − ⋅ − ⋅ − − . = For x=l the distances m, f , n and k are as shown in Fig. 5 or m =6, f =9, n =5 and k =0. In this case the above equations become: ( ) ( ) 0 0 0 0 0 0 0 0 17.004 V 14.358 6.5921 100 44.042 250 1000 1.81928 0 0, 9.4692 V 17.004 4.3782 100 17.7663 250 447.21 0.43394 0 0. 17.004 V 14.358 8532.07, 9.4691 V 17.004 3810.01. − ⋅ − ⋅ Φ − ⋅ + ⋅ − − = − ⋅ − ⋅ Φ − ⋅ + ⋅ − − = − ⋅ − ⋅ Φ = − − ⋅ − ⋅ Φ = − Wherefrom: 0 0 V 590.03 104.51 = Φ = − Having the initial parameters available, for the given external loadings, we can obtain any parameter of an arbitrary section of the beam axis using the expressions below: Lecture notes: Structural Analysis II 2011 S. Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 122 ( ) ( ) ( ) ( ) 2 3 0 0 2 0 2 0 3 2 ( ) ( ) ( ) ( ) V V V ( ) ( ) 4 ( ) ( ) M ( ) 4 ( ) 4 ( ) ( ) Q 4 ( ) 4 ( ) 4 ( ) ( ) V M B x C x D x A x x B x C x D x A x x M M x B x C x D x A x Q Q x B x C x D x A x α α α α α α α α α α α α α α α α α α α α α α α α α α α α ⎡ ⎤ − − ⎢ ⎥ ⎢ ⎥ ⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ − ⋅ − − Φ Φ Φ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ = + ⎨ ⎬ ⎨ ⎬ ⎨ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎢ ⎥ − ⎢ ⎥ ⎣ ⎦ Φ ( ) ⎪ ⎬ ⎪ ⎪ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 4 3 2 3 2 2 4 Q 4 ( ) ( ) q D f A n A k C m F M q C f D n D k B m F M q C n C k B f A m M F q D m M B n B k A f F V x R x k y x k EI α α α α α α α α α α α α α α α α α α α α α α α α α α ⎧ ⎫ ⎧ ⎫ − − ⎪ ⎪ ⎧ ⎫ ⎪ ⎪ − ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ − ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ − = + + ⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ − − ⋅ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ − ⎩ ⎭ ⎪ ⎪ ⎪ ⎪ ⎪ − ⋅ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ − − − ⋅ ⎩ ⎭ ⎪ ⎪ ⎩ ⎭ = ⋅ = ⋅ ⎪ It should be pointed out that for every different section of the beam, with a single abscissa x, the values of m, f, n and k are different and depend on the distance x. 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 x V(x)=EIy(x) Φ(x)=EIφ(x) M(x) Q(x) R(x) 0 590.035 -104.514 0.000 0.000 94.406 0.9999 481.727 -119.549 44.395 85.922 77.076 1 481.727 -119.549 44.395 -164.078 77.076 2 364.279 -93.929 -84.426 -96.754 58.285 3 326.363 29.653 -153.863 -43.162 52.218 3.9999 437.891 195.983 -169.054 15.748 70.063 4 437.891 195.983 -69.054 15.748 70.063 5 662.582 244.093 -12.626 103.125 106.013 6 899.391 219.197 49.981 28.431 143.903 7 1090.916 162.960 55.849 -11.586 174.547 8 1228.738 116.131 35.530 -25.386 196.598 9 1331.329 93.128 11.337 -20.275 213.013 10 1421.503 89.150 0.000 0.000 227.440 Lecture notes: Structural Analysis II 2011 S. Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 123 The diagrams of required 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. Figure 6 (a) Vertical deflections; (b) slope of the deflected line; (c) bending moment diagram; (d) shear force diagram V ( EIy x = ) + 5 9 0 . 0 3 4 8 1 . 7 3 3 6 4 . 2 8 3 2 6 . 3 6 4 3 7 . 8 9 6 6 2 . 5 8 8 9 9 . 3 9 1 0 9 0 . 9 2 1 2 2 8 . 7 4 1 3 3 1 . 3 3 1 4 2 1 . 5 0 (a) 2 4 4 . 0 9 ( ) EI x Φ = ϕ 2 9 . 6 5 3 + 1 0 4 . 5 1 1 1 9 . 5 5 9 3 . 9 2 9 1 9 5 . 9 8 2 1 9 . 2 0 1 6 2 . 9 6 1 1 6 . 1 3 9 3 . 1 2 8 8 9 . 1 5 (b) M 4 4 . 3 9 5 8 4 . 4 2 6 1 5 3 . 8 6 1 6 9 . 0 5 4 1 2 . 6 2 6 6 9 . 0 5 4 4 9 . 9 8 1 5 5 . 8 4 9 3 5 . 5 3 0 1 1 . 3 3 7 + (c) Q 1 1 . 5 8 6 2 5 . 3 8 6 2 0 . 2 7 5 + 2 8 . 4 3 1 1 0 3 . 1 2 5 1 5 . 7 4 8 4 3 . 1 6 2 9 6 . 7 5 4 8 5 . 9 2 2 1 6 4 . 0 7 8 (d) Lecture notes: Structural Analysis II 2011 S. Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 124 Figure 7 Diagram of vertical reactions Verification In order to verify the obtained results we shall check the equilibrium of the vertical forces using the condition . 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: 0 V Σ = 2 94.406 1 2 77.076 58.285 2 52.218 70.063 2 106.01 143.90 2 174.55 196.60 3 2 227.44 2 213.01 1250.33 2 A ⎛ = + ⋅ + + ⋅ + + ⋅ + + ⋅ + ⎜ ⎝ ⎞ + ⋅ + = ⎟ ⎠ + 0 1250.33 250 200 5 1250.33 1250 V Σ = − − ⋅ = − The numerical error is 0.026%. 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. R + 9 4 . 4 0 6 7 7 . 0 7 6 5 8 . 2 8 5 5 2 . 2 1 8 7 0 . 0 6 3 1 0 6 . 0 1 1 4 3 . 9 0 4 . 5 5 6 . 6 0 0 1 4 4 1 7 1 9 2 1 3 . 2 2 7 . Lecture notes: Structural Analysis II 2011 S. Parvanova, University of Architecture, Civil Engineering and Geodesy - Sofia 125 References HETENYI, M. Beams on elastic foundation. Waverly press, Baltimore, 1946 КАРАМАНСКИ Т. Д. И КОЛЕКТИВ. Строителна механика. Издателство „Техника” София, 1988