Green s Functions for Euler-Bernoulli Beams

Transcription

1 Green s Functions for Euler-Bernoulli Beams Aamer Haque Abstract A Green s function for the Euler-Bernoulli beam is the response of a dependent variable to a unit load. The form of the Green s functions for shear, bending moment, angle of deflection, and deflection are derived. Specific Green s functions are provided for simple beams, fixed beams, cantilever beams, beam on spring supports, and moment-restrained beams. Responses for uniform loading are computed using the Green s functions. 1. Formulation 1.1. Differential Equations Figure 1: Beam deflection Figure 2: Internal beam forces and free body diagram Consider the deflected beam displayed in figure 1. The deflection u(x) of the beam is assumed positive upwards and this convention is also used for applied loads. The 4th order differential equation for deflection u(x) of Euler-Bernoulli beams is: d 2 dx 2 EI d2 u dx 2 = w, < x < (1.1) where EI(x) is the flexural rigidity and w(x) is the applied load. Four boundary conditions are required to determine the solution of (1.1). The assumptions for validity of (1.1) and its derivation are found in texts on structural mechanics such as: Hibbeler 1, Kassimali 2, Krenk 3. The 4th order differential equation can be written as the following series of 1st order equations: V (x) = w(x) (1.2) M (x) = V(x) (1.3) θ (x) = M(x) EI(x) (1.4) u (x) = θ(x) (1.5) V(x) is shear force in the beam, M(x) is the bending moment, and θ(x) is the angle of deflection. The standard sign convention for beams is assumed. Figure 2 shows positive shears and bending moments. Preprint submitted to Elsevier January 14, 216

2 The 1st order equations (1.2)-(1.5) can be integrated to yield the following solutions: V(x) = M(x) = θ(x) = u(x) = ˆ x ˆ x ˆ x ˆ x w(s)ds+v (1.6) V(s)ds+M (1.7) M(s) EI(s) ds+θ (1.8) θ(s)ds+u (1.9) The constants of integration V, M, θ, and u are determined using the boundary conditions for the particular problem being solved. The solution to the beam equations is more conveniently written using the dimensionless spatial variable ξ = x/. We then reformulate functions f(x) as f(ξ). Since this involves reformulation of the function instead of mere replacement of the dependent variable, we notate this process as: f(x) f(ξ). Integrals of f(x) are transformed in the following manner: ˆ x τ = s, dτ = ds f(s)ds Using (1.1), the solutions (1.6)-(1.9) are now written as: V(ξ) = M(ξ) = θ(ξ) = u(ξ) = f(τ)dτ (1.1) w(τ)dτ +V (1.11) V(τ)dτ +M (1.12) M(τ) EI(τ) dτ +θ (1.13) θ(τ)dτ +u (1.14) Henceforth we shall assume prismatic beams with constant flexural EI rigidity along its length. Figure 3: Beam deflection using dimensionless distance ξ Figure 4: Internal beam forces using dimensionless distance ξ 2

3 1.2. Green s Functions Figure 5: Download unit point load at ξ = µ The Green s functions are the solutions to the beam equations (1.11)-(1.14) for a downward unit point load acting at location x = a. The point load is represented as a delta function: w(x) = δ(x a). In terms of the dimensionless variable ξ, we set µ = a/. The delta function obeys the scaling property: δ(x a) = δ((ξ µ)) = 1 δ(ξ µ) (1.15) Thus we write: w(ξ) = δ(ξ µ)/. We also require the fact that the integral of the delta function is the Heaviside function: ˆ { ξ 1 if ξ µ δ(τ µ)dµ = H(ξ µ) = (1.16) if ξ < µ Detailed information on Green s functions and the delta function are found in: Richards and Youn 4, Richtmyer 5, Stakgold 6. Successive integrals of the Heaviside function are given by the formula: (τ µ) n H(τ µ)dτ = (ξ µ)n+1 H(ξ µ), n (1.17) n+1 These properties allow explicit integration of the solution (1.11)-(1.14). The shear force is computed to be: The bending moment is given as: The angle of deflection can be simplified: V(ξ) = V(ξ) = w(τ)dτ +V 1 δ(τ µ)dτ +V V(ξ) = H(ξ µ)+v (1.18) M(ξ) = M(ξ) = V(τ)dτ +M H(τ µ)+v dτ +M M(ξ) = (ξ µ)h(ξ µ)+v ξ+m (1.19) θ(ξ) = θ(ξ) = 2 EI θ(ξ) = 2 EI θ(ξ) = 2 M(τ) dτ +θ EI (τ µ)h(τ µ)+v τ + 1 M dτ +θ 1 2 (ξ µ)2 H(ξ µ)+ 1 2 V ξ M ξ +θ (ξ µ) 2 H(ξ µ)+v ξ M ξ +θ (1.2) 3

4 Finally, the deflection is calculated in the form: u(ξ) = u(ξ) = 3 u(ξ) = 3 u(ξ) = 3 θ(τ)dτ +u (τ µ) 2 H(τ µ)+v τ M τ + 2 θ dτ +u 1 3 (ξ µ)3 H(ξ µ)+ 1 3 V ξ M ξ θ ξ +u (ξ µ) 3 H(ξ µ)+v ξ M ξ 2 +θ ξ +u (1.21) Expressions (1.18)-(1.21) are the Green s functions for the Euler-Bernoulli beam. et G q (ξ,µ) denote the Green s function for the value of quantity q at location ξ due to the unit downward load at location µ. We let q {V, M, θ, u} and list all the corresponding Green s functions: G V (ξ,µ) = H(ξ µ)+v (µ) (1.22) G M (ξ,µ) = (ξ µ)h(ξ µ)+v (µ)ξ+m (µ) (1.23) G θ (ξ,µ) = 2 (ξ µ) 2 H(ξ µ)+v (µ)ξ M (µ)ξ +θ (µ) (1.24) G u (ξ,µ) = 3 (ξ µ) 3 H(ξ µ)+v (µ)ξ M (µ)ξ 2 +θ (µ)ξ +u (µ) (1.25) Notice that the constants of integration V, M, θ, and u depend on the boundary conditions and are functions of the location µ of the point load Arbitrary oads First consider the case of N point loads p i. By the principle of superposition, the total response for quantity q is the sum of responses due to each point load p i. Thus the solution to the Euler-Bernoulli beam equations is: q(ξ) = N p i G q (ξ,µ i ) (1.26) i=1 where q {V,M,θ,u} and G q (ξ,µ) is the corresponding Green s function. To generalize to an arbitrary load distribution w(x), we transform from the original spatial coordinate x to the dimensionless coordinate µ. A uniform discretization of x is given by: The corresponding uniform discretization of µ is: x = N, x i = i x, i =,...,N µ = 1 N, µ i = i µ, i =,...,N We define w i to be the applied load on the interval x i 1,x i. This value is approximated by using the value of w(x) at the midpoint of the interval and multiplying by the length of the interval. Thus an increment of distributed load is approximated by a point load. The locations of the midpoints are: ˆx i = 1 ( 2 (x i 1 +x i ) = i 1 ) x, i = 1,...,N 2 ˆµ i = 1 ( 2 (µ i 1 +µ i ) = i 1 ) µ, i = 1,...,N 2 The approximate value for w i is: w i = w(ˆx i ) x w i = w(ˆµ i ) µ Notice that the factor of is necessary when changing to the non-dimensional coordinate µ. Using the principle of superposition (1.26) and evaluating the Riemann sum using midpoint rule we arrive at: q(ξ) = lim N i=1 N w i G(ξ, ˆµ i ) = lim q(ξ) = N i=1 N w(ˆµ i )G(ξ, ˆµ i ) µ w(µ)g q (ξ,µ)dµ (1.27) 4

5 1.4. Uniform oads For a uniform load of w (weight per unit length), the solution (1.27) becomes: q(ξ) = w G q (ξ,µ)dµ (1.28) In order to proceed further, we require the evaluation of certain types of definite integrals. The first of which involves non-negative integer powers of ξ µ multiplied by H(ξ µ): (ξ µ) n H(ξ µ)dµ = ξn+1 n+1, n (1.29) Formula (1.29) is proved by changing variables. et z = ξ µ, then dz = dµ and we have: (ξ µ) n H(ξ µ)dµ = = = = ξ 1 ξ 1 = zn+1 z n H(z)dz z n H(z)dz z n H(z)dz + z n dz n+1 = ξn+1 n+1 ξ ˆ ξ 1 z n H(z)dz We arrive at the fourth line because ξ 1 and thus H(z) = for the second integral in the third line. Another useful definite integral is: (1 µ) n dµ = 1 n+1, n (1.3) This expression is easily proved by letting z = 1 µ and dz = dµ. We now compute the integral as: (1 µ) n dµ = = ˆ 1 = zn+1 = z n dz z n dz n+1 1 n+1 Using the Green s functions (1.22)-(1.25) and applying the formula (1.29) to (1.28): V(ξ) = w ξ + M(ξ) = w ξ2 +ξ θ(ξ) = w3 u(ξ) = w4 V (µ)dµ V (µ)dµ +w 13 2 ξ3 +ξ V (µ)dµ+ 2 ξ 14 3 ξ4 +ξ V (µ)dµ+ 3 ξ2 1 (1.31) M (µ)dµ (1.32) M (µ)dµ +w M (µ)dµ +w 2 ξ θ (µ)dµ (1.33) θ (µ)dµ+w u (µ)dµ (1.34) 5

6 2. Examples 2.1. Simply-supported Beam Boundary Conditions Figure 6: Simply-supported beam Green s Functions The left end boundary conditions give: u() = u = and M() = M =. The right end boundary conditions are used to determine θ and V. Using M(1) = in equation (1.19) we compute the value of V : = (1 µ)+v V = 1 µ The value of θ is computed using u(1) = in equation (1.21): The Green s functions are: = 3 (1 µ) 3 +(1 µ) +θ = 2 (1 µ) 3 (1 µ) θ G V (ξ,µ) = H(ξ µ)+(1 µ) G M (ξ,µ) = (ξ µ)h(ξ µ)+(1 µ)ξ G θ (ξ,µ) = 2 (ξ µ) 2 H(ξ µ)+(1 µ)ξ (1 µ) 3 (1 µ) G u (ξ,µ) = 3 (ξ µ) 3 H(ξ µ)+(1 µ)ξ 3 +(1 µ) 3 ξ (1 µ)ξ Uniform oad ( ) 1 V(ξ) = w 2 ξ M(ξ) = w2 2 ξ(1 ξ) θ(ξ) = w3 24EI (1 6ξ2 +4ξ 3 ) u(ξ) = w4 24EI (ξ 2ξ3 +ξ 4 ) 6

7 2.2. Fixed Beam Boundary Conditions Figure 7: Fixed beam Green s Functions The left end boundary conditions give: u() = u = and θ() = θ =. The right end boundary conditions are used to determine M and V. Using θ(1) = in equation (1.2) and u(1) = in equation (1.21), we get the system of equations: The solution to this system is: The Green s functions are: Uniform oad 2 (1 µ) 2 +V + 2 M 3 (1 µ) 3 +V + 3 M V = 3(1 µ) 2 2(1 µ) 3 M = (1 µ) 3 (1 µ) 2 G V (ξ,µ) = H(ξ µ)+3(1 µ) 2 2(1 µ) 3 = = G M (ξ,µ) = (ξ µ)h(ξ µ)+3(1 µ) 2 ξ 2(1 µ) 3 ξ + (1 µ) 3 (1 µ) 2 G θ (ξ,µ) = 2 (ξ µ) 2 H(ξ µ)+3(1 µ) 2 ξ 2 2(1 µ) 3 ξ 2 +2(1 µ) 3 ξ 2(1 µ) 2 ξ G u (ξ,µ) = 3 (ξ µ) 3 H(ξ µ)+3(1 µ) 2 ξ 3 2(1 µ) 3 ξ 3 +3(1 µ) 3 ξ 2 3(1 µ) 2 ξ 2 ( ) 1 V(ξ) = w 2 ξ M(ξ) = w ξ(1 ξ) θ(ξ) = w3 1 (ξ 3ξ2 +2ξ 3 ) u(ξ) = w4 24EI ξ2 (ξ 1) 2 7

8 2.3. Cantilever Beam Boundary Conditions Figure 8: Cantilever beam Green s Functions The left end boundary conditions give: u() = u = and θ() = θ =. The right end boundary conditions are used to determine M and V. Using V(1) = in equation (1.18) we compute the value of V : = 1+V V = 1 Then we use M(1) = in equation (1.19)to determine M : The Green s functions are: Uniform oad = (1 µ)+1+m M = µ G V (ξ,µ) = H(ξ µ)+1 G M (ξ,µ) = (ξ µ)h(ξ µ)+ξ µ G θ (ξ,µ) = 2 (ξ µ) 2 H(ξ µ)+ξ 2 2µξ G u (ξ,µ) = 3 (ξ µ) 3 H(ξ µ)+ξ 3 3µξ 2 V(ξ) = w(1 ξ) M(ξ) = w2 2 (1 ξ)2 θ(ξ) = w3 (3ξ 3ξ2 +ξ 3 ) u(ξ) = w4 24EI ( 6ξ 2 4ξ 3 +ξ 4) 8

9 2.4. Spring-supported Beam Boundary Conditions Figure 9: Spring-supported beam Green s Functions The left end moment boundary condition gives M() = M =. Using M(1) = we compute the value of V : The value of V(1) is thus: = (1 µ)+v V = 1 µ V(1) = µ The spring equations provide the value of the end displacements: u() = 1 k (1 µ) u(1) = µ k 1 Thus u = (1 µ)/k and u 1 = µ/k 1. Notice that k i implies u i for i = 1,2. The value of θ is computed using u(1) = u 1 : The Green s functions are: u 1 = 3 (1 µ) 3 +(1 µ) +θ +u θ = 2 (1 µ) 3 (1 µ) + 1 (u 1 u ) G V (ξ,µ) = H(ξ µ)+(1 µ) G M (ξ,µ) = (ξ µ)h(ξ µ)+(1 µ)ξ G θ (ξ,µ) = 2 (ξ µ) 2 H(ξ µ)+(1 µ)ξ (1 µ) 3 (1 µ) + 1 (u 1 u ) G u (ξ,µ) = 3 (ξ µ) 3 H(ξ µ)+(1 µ)ξ 3 +(1 µ) 3 ξ (1 µ)ξ +ξu 1 +(1 ξ)u It is seen that Green s functions for θ and u are the sum of the corresponding Green s functions for simple beams and a linear displacement field connecting the ends Uniform oad ( ) 1 V(ξ) = w 2 ξ M(ξ) = w2 2 ξ(1 ξ) θ(ξ) = w3 24EI (1 6ξ2 +4ξ 3 )+ 1 (u 1 u ) u(ξ) = w4 24EI (ξ 2ξ3 +ξ 4 )+ξu 1 +(1 ξ)u 9

10 2.5. End-restrained Beam Boundary Conditions Figure 1: End-restrained beam End Moments The left end displacement boundary condition gives u() = u =. We first use equation (1.19) to derive an expression for V in terms of the end moments: M = M() and M 1 = M(1). M 1 = (1 µ)+v +M V = M 1 M +(1 µ) This expression and the end moment-angle relations are incorporated into equations (1.2) and (1.21): M 1 = 2 (1 µ) 2 + M 1 M +(1 µ)+ 2 β 1 M + M β = 3 (1 µ) 3 + M 1 M +(1 µ)+ 3 M + M β The solution of this system of equations for M and M 1 is: where M = κ γ µ(1 µ)(1 µ)(2+κ 1)+2 M 1 = κ 1 γ µ(1 µ)(1 µ)(2+κ ) 4 κ κ = β EI, κ 1 = β 1 EI, γ = κ κ 1 +4(κ +κ 1 ) imiting Cases The limiting cases for zero and infinite rotational stiffness are given below. Zero stiffness corresponds to a pin or roller end. Infinite stiffness implies a fixed end connection. Case κ, κ 1 : M, M 1 Case κ, κ 1 > : Case κ, κ 1 : Case κ >, κ 1 : Case κ >, κ 1 : M, M, M 1 κ 1µ(1 µ)(1+µ) 2κ 1 +6 M 1 µ(1 µ)(1+µ) 2 M κ µ(1 µ)(2 µ), M 1 2κ +6 M κ µ(1 µ) 2, M 1 µ(1 µ)(1 µ)(2+κ ) 4 κ κ +4 κ +4 Case κ, κ 1 : M µ(1 µ)(2 µ), M 1 2 Case κ, κ 1 > : Case κ, κ 1 : M µ(1 µ)(1 µ)(2+κ 1)+2, M 1 κ 1µ 2 (1 µ) κ 1 +4 κ 1 +4 M = µ(1 µ) 2, M 1 µ 2 (1 µ) We can show that this expression for M is identical to the one obtained for the fixed end beam: M µ(1 µ) 2 = (1 µ) 2 (1 µ) 1 = (1 µ) 3 (1 µ) 2 1

11 Green s Functions For sake of brevity, the Green s functions (1.22)-(1.25) are: G V (ξ,µ) = H(ξ µ)+v G M (ξ,µ) = (ξ µ)h(ξ µ)+v ξ+m G θ (ξ,µ) = 2 G u (ξ,µ) = 3 (ξ µ) 2 H(ξ µ)+v ξ M ξ where the values of V and M were computed above for this problem Uniform oad For a uniform load of w, the following integrals are first computed: V (µ)dµ = κ 1 γ + M (ξ µ) 3 H(ξ µ)+v ξ M ξ 2 + M β ξ These values are substituted into the uniform load solution: V(ξ) = w ξ + V (µ)dµ M(ξ) = w ξ2 +ξ θ(ξ) = w3 u(ξ) = w4 M (µ)dµ = κ ( 1 γ 2 + κ ) 1 12 M 1 (µ)dµ = κ ( 1 1 γ 2 + κ ) 12 ( κ ) + κ 12 γ ( κ 1 12 θ (µ)dµ = 1 1 M (µ)dµ β ˆ V (µ)dµ +w 13 2 ξ3 +ξ V (µ)dµ+ 2 ξ 14 3 ξ4 +ξ V (µ)dµ+ 3 ξ2 M (µ)dµ ) M (µ)dµ +w β M (µ)dµ +w 2 ξ θ (µ)dµ θ (µ)dµ 11

12 References 1 R.C. Hibbeler. Structural Analysis. Prentice Hall, Upper Saddle River, New Jersey, eighth edition, Aslam Kassimali. Structural Analysis. PWS-Kent, Boston, Steen Krenk. Mechanics and Analysis of Beams, Columns, and Cables. Springer-Verlag, Berlin, second edition, J Ian Richards and Heekyung K. Youn. Theory of Distributions, A Nontechnical Introduction. Cambridge University Press, Cambridge, Robert D. Richtmyer. Principles of Advanced Mathematical Physics, volume 1 of Texts and Monographs in Physics. Springer-Verlag, Berlin, Ivar Stakgold. Green s Functions and Boudnary Value Problems. Wiley-Interscience, New York,