Introduction to Timoshenko Beam Theory

Transcription

1 Introduction to Timoshenko Beam Theory Aamer Haque Abstract Timoshenko beam theory includes the effect of shear deformation which is ignored in Euler-Bernoulli beam theory. An elementary derivation is provided for Timoshenko beam theory. Energy principles, the stiffness matrix, and Green s functions are formulated. Solutions are provided for some common beam problems. 1. Formulation 1.1. Introduction The goal of beam theory is to simplify the equations of solid mechanics to beams. All loads on a beam act parallel to an axis that is transverse to its longitudinal axis. The length of a beam is much larger than its width and depth. Beam theory provides equations for the deflection and internal forces of the beam. Figure 1 shows a typical deflection of a beam that is loaded from above. We assume that the initial unloaded beam s longitudinal axis coincides with the x axis. The sign convention for the deflection u is that positive deflections have positive y values. The deflection u and angle of deflection θ are assumed to be small. Applied loads are considered positive if they act in the direction of the positive y axis. Euler-Bernoulli beam theory is extensively applied to structural analysis and design. Most design guides and manuals will exclusively use Euler-Bernoulli beams. The derivation of Euler-Bernoulli beam theory is given in: Hibbeler 3, 4, Hjelmstad 5, Kassimali 7, Krenk 9. The problem with Euler-Bernoulli beam theory is that it is inaccurate for deep beams. Deep beams are ones in which the depth is not negligible compared to the length. Either complete solid mechanics or a more accurate beam theory is required for deep beams. Timoshenko beam theory is a simple extension to Euler-Bernoulli beam theory. Shear deformations, which are absent in Euler-Bernoulli beam theory, are included in Timoshenko beam theory. The basic assumptions for Timoshenko beam theory are: The longitudinal axis of the unloaded beam is straight. All applied loads act transverse to the longitudinal axis. All deformations and strains are small. Hooke s law can be used to relate stresses and strains. Plane cross sections, which are initially normal to the longitudinal axis, will remain plane after deformation. The last requirement differs from Euler-Bernoulli theory which requires that the plane cross sections also remain normal to the beam axis after deformation. For Timoshenko beams, plane cross sections will rotate due to shear forces. Figure 1: Deflection of a beam Preprint submitted to Elsevier October 26, 216

2 1.2. Differential Equations Figure 2: Forces on a beam segment This section briefly presents an elementary derivation of the equations of Timoshenko beam theory. A complete derivation from the equations of solid mechanics is provided in Hjelmstad 5 and Wang et al. 16. Figure 2 shows a small segment of the beam. The shear force V and bending moment M are displayed as positive in the figure according to the usual beam convention. The load w is positive when applied upwards. We assume that the applied load is approximately constant for the small beam segment shown in figure 2. Vertical equilibrium requires: V +dv = V +wdx dv = w (1.1) dx For purposes of static equilibrium, the constant load w can be assumed to act as a point load in the middle of the segment. Equilibrium of moments results in: M +dm = M +V dx+ w 2 (dx)2 dm dx = V (1.2) Figure 3: Pure bending deformation The next step in deriving Timoshenko beam theory is to relate the internal forces V and M to deformation. Consider the pure bending of the beam segment shown in figure 3. Pure bending means that the segment is bent only by constant end moments M. The y coordinate is measured from the neutral axis of the beam and is positive upwards. The neutral axis does not change length during the deformation of the beam. The original length of the beam segment is ds. et dθ be the angle between the planar cross sections at the ends of the beam segment. The curvature κ is defined through the equation: κ = 1 R = dθ ds The elongation of a longitudinal fiber in the beam is given by: ds = (R y)dθ = (R y)κds Since strains are assumed to be small, we shall use the engineering strain: ε = ds ds ds 2 = κy

3 Notice that the strain is zero at the neutral axis. Hooke s law gives the stress-strain relation: σ = Eε = κey where σ is the normal stress at the end planes of the segment and E is Young s modulus. The bending moment M is related to normal stress σ through: ˆ ˆ M = σyda = κe y 2 da = κei For beam theory, we need to relate the bending moment M to the deflection u. This is accomplished by substituting the calculus definition of curvature: d M 2 u EI = κ = dx 2 1+ ( ) du 2 3/2 d2 u dx 2 (1.3) dx The approximation is valid because du dx is assumed to be small and hence its square is smaller. Figure 4: Pure shear deformation Now consider the case of pure shear shown in figure 4. The constitutive relation between shear V and angle of shear γ is: V = γ (1.4) where G is the shear modulus and A s is the shear area. A s is usually not equal to the area of the cross section because the distribution of shear stress is not constant along the cross section. Notice that shear deformation does not result in longitudinal strain. Also remember that bending deformation did not cause any shearing strain. Thus bending and shearing deformations are separable and the total angle of deformation is: Using equations (1.1)-(1.5), we arrive at the equations for Timoshenko beam theory: du = θ γ (1.5) dx V (x) = w(x) (1.6) M (x) = V(x) (1.7) θ (x) = M(x) EI(x) (1.8) u (x) = θ(x) V(x) (x) (1.9) These equations can be reduced to the pair of differential equations for < x < : ( d EI dθ ) ( = θ du ) (1.1) dx dx dx ( d θ du ) = w (1.11) dx dx These equations imply that θ and u are two unknown functions whose solutions are to be determined. This is in contrast to Euler-Bernoulli beam theory where a single 4th order differential equation can be written for the displacement u. Equations (1.1) and (1.11) are used to formulate variational and energy principles in the next section. 3

4 Figure 5: Beam deflection using dimensionless distance ξ Figure 6: Internal beam forces using dimensionless distance ξ The equations (1.6)-(1.9) 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 (1.12) V(s)ds+M (1.13) M(s) EI(s) ds+θ (1.14) θ(s)ds ˆ x V(s) (s) ds+u (1.15) 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 (1.16), the solutions (1.12)-(1.15) are now written as: V(ξ) = M(ξ) = θ(ξ) = u(ξ) = f(τ)dτ (1.16) w(τ)dτ +V (1.17) V(τ)dτ +M (1.18) M(τ) EI(τ) dτ +θ (1.19) θ(τ)dτ Henceforth we shall assume prismatic beams with constant EI and. V(τ) (τ) dτ +u (1.2) 4

5 2. Energy Principles 2.1. Principle of Virtual Work Formulation The pair of differential equations (1.1) and (1.11) are the starting point for developing energy principles for Timoshenko beams. We first multiply (1.1) by a function φ and integrate by parts: (EIθ ) φdx = EIθ φ EIθ φ dx = (θ u )φdx (θ u )φdx Remember that M = EIθ. Moving all terms to the left hand side of the equation we have: EIθ φ dx+ φ is interpreted as a virtual angle of rotation of the plane cross section of the beam. We now multiply (1.11) by a function v and integrate by parts: (θ u )φdx Mφ = (2.1) (θ u ) vdx = (θ u )v (θ u )v dx = wvdx wvdx Notice that V = (θ u ). Again moving all terms to the left hand side: (θ u )v dx+ For this equation, v represents a virtual vertical displacement of the beam. Setting equations (2.1) and (2.2) equal to each-other and rearranging: EIθ φ dx+ (θ u )(φ v ) dx = wvdx Vv = (2.2) wvdx+ Mφ Vv (2.3) The left hand side of this equation is internal virtual work and the right hand side is external virtual work. Now let u = θ u T and v = φ v T be the actual and virtual generalized displacements. Also define the generalized force vector f T = M V. Then we can write the Principle of Virtual Work: Internal virtual work δw int and external virtual work δw ext are defined as: δw int (u,v) = δw ext (v) = δw int (u,v) = δw ext (v) (2.4) where ũ T = θ θ u and ṽ = φ φ v T Total Potential Energy The total potential energy for Timoshenko beams is defined as: ũ T EI ṽdx (2.5) w vdx+f T ()v() f T ()v() (2.6) Π(u) = 1 2 δw int(u,u) δw ext (u) (2.7) The mathematical theory of transforming from virtual work equations to total potential energy is given in Hjelmstad 5, Johnson 6. 5

6 Boundary Conditions Essential u = value, V =? or θ = value, M =? or Natural u =?, V = value θ =?, M = value Table 1: Essential vs natural boundary conditions The boundary terms in equation (2.3) provide the essential and natural boundary conditions. Either M or θ can be specified at the end points but both cannot be specified simultaneously. The same argument applies to V and u. In the case that θ or u is specified, the corresponding virtual displacement is not permitted to vary. The possibilities for boundary conditions are summarized in table 2. ocation Type Enforced Value Virtual Displacement x = Essential u() = u v() = x = Natural V() = V v() is free x = Essential θ() = θ φ() = x = Natural M() = M φ() is free x = Essential u() = u v() = x = Natural V() = V v() is free x = Essential θ() = θ φ() = x = Natural M() = M φ() is free Table 2: Essential and natural boundary conditions It should be noted that some combinations of boundary conditions are not permissible since they would violate the existence and uniqueness principles of the differential equations. These issues will not be discussed here but are found in mathematical texts on boundary value problems Function Spaces Solution and variational spaces must be specified so that the integrals in equation (2.3) exist and the boundary conditions are satisfied. If the actual displacement u coincides with the virtual displacement v, then we see that both the displacement function and its derivative are required to be square-integrable. The set of square-integrable vector-valued functions v(x) = φ(x) v(x) Ton, is denoted as: { 2 = v :, R 2 φ(x) 2 dx < and v(x) 2 dx < } The set of functions v(x) in 2 with first derivatives v (x) = φ (x) v (x) T also in 2 is called: H 1 = { v :, R 2 v 2 and v 2} The subset of H 1 which satisfies the essential boundary conditions is notated as Hbc 1. The actual displacements u = T θ u come from H 1 bc. The virtual displacements v must be zero at essential boundary conditions. The function space for virtual displacements is denoted as Hbc 1. Note that H1 bc and H1 bc will be depend on the boundary conditions of the particular beam problem. We always have the following set relations: Hbc 1 H1 bc H1. For the example of a beam clamped at x = and prescribed displacement u() = u at x =, we have: Hbc 1 = { u :, R u H 1 }, θ() =, u() =, u() = u H 1 bc = { v :, R v H 1, φ() =, v() =, v() = } 6

7 Variational Problem The Principle of Virtual Work states that the solution u to the Timoshenko beam problem has the property that internal virtual work equals external virtual work for all admissible virtual displacements. Find u H 1 bc such that δw int (u,v) = δw ext (v) v H 1 bc (2.8) Minimization Problem The Principle of Minimization of Total Potential Energy states that the solution u to the Timoshenko beam problem is the one that minimizes the total potential energy as defined by equation (2.7). Find u H 1 bc such that Π(u) Π(v) v H 1 bc (2.9) Concentrated Applied Moments and Forces Suppose that N concentrated moments M i and point loads p i are applied at x i (,). We allow for the possibility that the applied moment or point load could be zero at x i. The interval, is partitioned into subintervals:, = x i 1,x i N+1 i=1 where x = and x N+1 =. We included the end points of the beam in order to simplify the book-keeping. The Principle of Virtual Work (2.3) for each segment of the beam is: ˆ xi x i 1 EIθ φ dx+ Summing over all segments, we have: ˆ xi x i 1 (θ u )(φ v ) dx = ˆ xi x i 1 wvdx+ Mφ xi x i 1 Vv xi x i 1 EIθ φ dx+ (θ u )(φ v ) dx = + wvdx+m()φ() M()φ() V()v()+V()v() N i=1 M(x + i )φ(x+ i ) M(x i )φ(x i ) N V(x + i )v(x + i ) V(x i )v(x i ) (2.1) i=1 where the plus and minus superscripts indicate right and left hand limits respectively. We associate concentrated moments with jumps in moment: M i = M(x + i ) M(x i ). Similarly, we equate point loads with jumps in shear: p i = V(x + i ) V(x i ). The sign convention for applied concentrated forces is as follows: M i is positive counter-clockwise and p i is positive upwards. Equation (2.1) allows for discontinuities in internal forces and virtual displacements. If all virtual displacements are continuous, then φ(x i ) = φ(x + i ) = φ(x i ) and v(x i) = v(x + i ) = v(x i ). The Principle of Virtual Work simplifies to: EIθ φ dx+ (θ u )(φ v ) dx = wvdx+m()φ() M()φ() V()v()+V()v() N N + M i φ(x i )+ p i v(x i ) (2.11) i=1 i=1 7

8 2.2. Principle of Complementary Virtual Work Formulation et δm(x) be a virtual moment distribution and δv(x) be a virtual shear distribution for the beam. We restrict these distributions to be ones that are in static equilibrium with the virtual applied load δw(x). Multiply both sides of equation (1.8) by δm and integrate by parts: θ δm dx = δm θ θδm dx = M δm EI M δm EI dx dx M δm EI + dx δm θ θδm dx = (2.12) Now multiply equation (1.9) by δv, integrate by parts, and use the fact that δv = δm and δv = δw to get: θδv dx θδm dx (θ u )δv dx = u δv dx = u δv dx = + θδm dx δv u uδv dx = + θδm dx δv u δwudx = V δv dx V δv dx V δv dx V δv dx V δv dx θδm dx δv u + Set equation (2.12) equal to equation (2.13) and rearrange terms: M δm EI V δv dx+ dx = V δv δwudx dx = (2.13) δwudx+ δm θ δv u (2.14) The left hand side is complementary internal virtual work and the right hand side is complementary external virtual work. Equation (2.14) can be written as using matrix and vector notation. et p = M V T and q = δm δv T. Then the Principle of Complementary Virtual Work is: δw int(p,q) = δw ext(q) (2.15) where complementary virtual internal work W int (p,q) and complementary virtual external work W ext(q) are: W int (p,q) = W ext (q) = 1 p T EI 1 qdx δw udx+q T ()u() q T ()u() where u = θ u T is the actual displacement field. Remember that p is related to u through the differential equations (1.6)-(1.9) Complementary Total Potential Energy The complementary total potential energy for the Timoshenko beam is defined as: Π (p) = 1 2 δw int (p,p) δw ext (p) (2.16) 8

9 Function Spaces Suppose the virtual forces coincide with the actual forces in equation (2.14). Then the integrals exist only if the forces are square-integrable functions. However, the virtual forces must also be in equilibrium. We denote the set of virtual forces as 2 eq. The actual forces are derived from displacements which satisfy the essential boundary conditions. The set of actual forces is notated as 2 eqbc. Note the relationship between the sets is: 2 eqbc 2 eq 2. The function spaces are determined by the particular problem being solved. Unfortunately, the 2 eq spaces are more cumbersome to specify than the H1 bc spaces of virtual work methods. This fact helps to explain why complementary virtual work is seldom mentioned in the mathematical theory of continuum mechanics Variational Problem The Principle of Complementary Virtual Work states that the solution u to the Timoshenko beam problem has the property that complementary internal virtual work equals complementary external virtual work for all admissible virtual forces. Find p 2 eqbc such that δw int (p,q) = δw ext (q) q 2 eq (2.17) Minimization Problem The Principle of Minimization of Total Complementary Potential Energy states that the solution p to the Timoshenko beam problem is the one that minimizes the total complementary potential energy as defined by equation (2.16). Find p 2 eqbc such that Π (p) Π (q) q 2 eqbc (2.18) Concentrated Virtual Moments and Forces We now allow for the possibility of N concentrated virtual moments δm i and virtual point loads δp i applied at x i (,). At any location, either of these virtual forces could be zero. The interval, is partitioned into subintervals:, = x i 1,x i where x = and x N+1 =. The Principle of Complementary Virtual Work (2.14) for each segment is: ˆ xi Summing up all the segments: M δm x i 1 EI dx+ ˆ xi x i 1 V δv dx = N+1 i=1 ˆ xi x i 1 δwudx+ δm θ xi x i 1 δv u xi x i 1 M δm EI V δv dx+ dx = + δwudx+δm()θ() δm()θ() δv()u()+δv()u() N i=1 δm(x + i )θ(x+ i ) δm(x i )θ(x i ) N δv(x + i )u(x + i ) δv(x i )u(x i ) i=1 Concentrated virtual moments are associated with jumps in virtual moment: δm i = δm(x + i ) δm(x i ). We also equate virtual point loads with jumps in virtual shear: δp i = δv(x + i ) δv(x i ). The actual displacements θ and u are continuous. Thus the Principle of Complementary Virtual Work becomes: M δm EI V δv dx+ dx = δwudx+δm()θ() δm()θ() δv()u()+δv()u() N N + δm i θ(x i ) δp i u(x i ) (2.19) i=1 i=1 9

10 3. Stiffness Matrix 3.1. Derivation Beam element Figure 7: Sign convention for the beam element Matrix structural analysis requires stiffness matrices for every structural member. Only the derivation of the stiffness matrix is described in this paper. Details of matrix structural analysis are found in: Harrison 2, Kassimali 8, McCormac 1, McGuire et al. 11. The Timoshenko beam element is shown in figure 7. The displacements and end forces are positive in the direction of the positive y axis. End rotations and moments are positive if they are counter-clockwise. This sign convention is demonstrated in figure 7. The stiffness matrix K relates the beam end displacements u to the applied loads f: Ku = f (3.1) where u = u θ u 1 θ 1 T and f = T. F u F M F u1 F M1 Equation (3.1)is explicitly written as: K 11 K 12 K 13 K 14 u K 21 K 22 K 23 K 24 θ K 31 K 32 K 33 K 34 u 1 = K 41 K 42 K 43 K 44 θ 1 F u F M F u1 F M1 (3.2) The most direct way of computing the elements of K is to consider the meaning of the j-th column. K ij is the end force f i that is required to maintain the unit displacement u j = 1 while keeping all other displacements at zero. Thus the procedure to compute the j-th column of K is as follows: 1. Solve the beam equations (1.17)-(1.2) with boundary conditions: { 1 if i = j u i = δ ij = if i j 2. Compute the end shears V, V 1 and end moments M, M 1 from the solution of the beam equations. 3. Translate from the beam sign convention to the element sign convention using: K 1j = V, K 2j = M, K 3j = V 1, K 4j = M 1 This procedure will be applied in the next several pages of this paper Beam equations The general solutions of the beam equations are required in order to compute the stiffness matrix. Since w(ξ) =, the solution of the beam equations (1.17)-(1.2) are: V(ξ) = V (3.3) M(ξ) = V ξ +M (3.4) θ(ξ) = 2 V 2EI ξ2 + M u(ξ) = 3 V ξ3 + 2 M 2EI ξ2 + Boundary conditions are used to determine the values of the constants V, M, θ, and u. We also define the parameter Φ which will be used in forming the stiffness matrix. EI ξ +θ (3.5) ( θ V ) ξ +u (3.6) Φ = 12EI 2 (3.7) 1

11 Unit displacement at left end Figure 8: Unit displacement u = 1 eft BC: u() = 1, θ() = ; Right BC: u(1) =, θ(1) = The left end boundary conditions give: u() = 1 u = 1 and θ() = θ =. Solving for M in (3.5) with θ(1) =, we have: M = V 2 Substituting this expression into (3.6) with u(1) = : The end moments are computed to be: = 3 V 3 V 4EI V +1 GA ( s 3 = 12EI + ) V +1 3 ( 1+ 12EI ) 12EI 2 V = 1 V = 12EI 3 (1+Φ) 1 M = 2 (1+Φ) 1 M 1 = 2 (1+Φ) 1 Translating from beam sign convention to matrix sign convention, the first column of the stiffness matrix is: K 11 V 12 K 21 K 31 = M V 1 = EI 6 (1+Φ) 3 12 K 41 M

12 Unit rotation at left end Figure 9: Unit displacement θ = 1 eft BC: u() =, θ() = 1; Right BC: u(1) =, θ(1) = The left end boundary conditions give: u() = u = and θ() = 1 θ = 1. Solving for M in (3.5) with θ(1) =, we have: Substituting this expression into (3.6) with u(1) = : This value is used to compute M : M = V 2 EI = 3 V 3 V 4EI 2 + V GA ( s 2 = 12EI + 1 ) V ( 1+ 12EI ) 12EI 2 V = 1 2 V = 2 (1+Φ) 1 M 1 is then calculated to be: M = 3EI (1+Φ) 1 EI M = 3EI (1+Φ) 1 EI (1+Φ)(1+Φ) 1 M = EI (4+Φ)(1+Φ) 1 M 1 = V +M M 1 = (1+Φ) 1 EI (4+Φ)(1+Φ) 1 M 1 = EI (2 Φ)(1+Φ) 1 Translating from beam sign convention to matrix sign convention, the second column of the stiffness matrix is: K 12 V 6 K 22 K 32 = M V 1 = EI (4+Φ) 2 (1+Φ) 3 6 K 42 M 1 (2 Φ) 2 12

13 Unit displacement at right end Figure 1: Unit displacement u 1 = 1 eft BC: u() =, θ() = ; Right BC: u(1) = 1, θ(1) = The left end boundary conditions give: u() = u = and θ() = θ =. Solving for M in (3.5) with θ(1) =, we have: M = V 2 Substituting this expression into (3.6) with u(1) = 1: The end moments are computed to be: 1 = 3 V 3 V 4EI V GA ( s 3 1 = 12EI + ) V ( 1 = EI ) 12EI 2 V = 12EI 3 (1+Φ) 1 V M = 2 (1+Φ) 1 M 1 = 3 (1+Φ) 1 Translating from beam sign convention to matrix sign convention, the third column of the stiffness matrix is: K 13 V 12 K 23 K 33 = M V 1 = EI 6 (1+Φ) 3 12 K 43 M

14 Unit rotation at right end Figure 11: Unit displacement θ 1 = 1 eft BC: u() =, θ() = ; Right BC: u(1) =, θ(1) = 1 The left end boundary conditions give: u() = u = and θ() = θ =. Solving for M in (3.5) with θ(1) = 1, we have: Substituting this expression into (3.6) with u(1) = : This value is used to compute M : M = V 2 + EI = 3 V 3 V 4EI + 2 V GA ( s 2 = 12EI + 1 ) V ( 1+ 12EI ) 12EI 2 V = 1 2 V = 2 (1+Φ) 1 M 1 is then calculated to be: M = 3EI (1+Φ) 1 + EI M = 3EI (1+Φ) 1 + EI (1+Φ)(1+Φ) 1 M = EI (2 Φ)(1+Φ) 1 M 1 = V +M M 1 = (1+Φ) 1 EI (2 Φ)(1+Φ) 1 M 1 = EI (4+Φ)(1+Φ) 1 Translating from beam sign convention to matrix sign convention, the fourth column of the stiffness matrix is: K 14 V 6 K 24 K 34 = M V 1 = EI (2 Φ) 2 (1+Φ) 3 6 K 44 M 1 (4+Φ) 2 14

15 3.2. Matrix Form Collecting the information from the previous pages, the stiffness matrix for the Timoshenko beam element is: EI K = 6 (4+Φ) 2 6 (2 Φ) 2 (1+Φ) (3.8) 6 (2 Φ) 2 6 (4+Φ) 2 The equilibrium equation is thus: EI (1+Φ) (4+Φ) 2 6 (2 Φ) (2 Φ) 2 6 (4+Φ) 2 u θ u 1 θ 1 = F u F M F u1 F M1 (3.9) Remember that K is only the element stiffness matrix. The global stiffness matrix is constructed by assembling the stiffness matrices for each element. The right hand side contains the applied forces. Forces occurring along the beam are represented as fixed end shears and moments at the ends. These statically equivalent forces are placed in the right hand side vector. Details concerning this procedure are described in texts on matrix structural analysis such as: Kassimali 8, McCormac 1, McGuire et al. 11, Timoshenko and Young Meaning of Φ The parameter Φ, defined by equation (3.7), appears in every term of the element stiffness matrix. Φ measures the ratio of the bending stiffness to shear stiffness. For a very large shear stiffness, Φ is close to zero. In this case, the solutions for the Timoshenko beam match the solutions to Euler-Bernoulli beam theory. This is expected since shear deformations are ignored for Euler-Bernoulli beams. The relative importance of shear deformation can be explored further by examining the material and shape constants in Φ. We will ignore all numerical constants contained in Φ and focus on dimensional factors instead. For rectangular sections, the moment of inertia and shear area scale according to the following dimensions: I wd 3, A s wd where w is the width and d is the depth of the section. Noting that E and G are usually the same order of magnitude, we have: ( ) 2 d Φ Thus, Φ when d. For I-sections, the analysis is slightly different: I w f t f d 2, A s t w d where w f is the flange width, t f is the flange thickness, and t w is the web thickness. For most sections, t f and t w are much nearer in magnitude than they are to d. Also, w f is usually smaller than d. Thus we can write: Φ w fd 2 < ( ) 2 d We again observe the justification of Euler-Bernoulli beam theory when d. 15

16 4. Green s Functions 4.1. Formulation Figure 12: Download unit point load at ξ = µ The Green s functions are the solutions to the beam equations (1.17)-(1.2) 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) (4.1) 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) = (4.2) if ξ < µ Detailed information on Green s functions and the delta function are found in: Richards and Youn 12, Richtmyer 13, Stakgold 14. 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 (4.3) n+1 These properties allow explicit integration of the solution (1.17)-(1.2). 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 (4.4) M(ξ) = M(ξ) = V(τ)dτ +M H(τ µ)+v dτ +M M(ξ) = (ξ µ)h(ξ µ)+v ξ+m (4.5) θ(ξ) = θ(ξ) = 2 EI θ(ξ) = 2 EI θ(ξ) = 2 2EI 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 (4.6) 16

17 Finally, the deflection is calculated in the form: u(ξ) = u(ξ) = 3 2EI u(ξ) = 3 2EI u(ξ) = 3 θ(τ)dτ V(τ)dτ +u (τ µ) 2 H(τ µ)+v τ M τ + 2EI 2 θ dτ H(ξ µ)+v dτ +u 1 3 (ξ µ)3 H(ξ µ)+ 1 3 V ξ M ξ 2 + 2EI 2 θ ξ + (ξ µ)h(ξ µ) V ξ+u (ξ µ) 3 H(ξ µ)+v ξ M ξ 2 + (ξ µ)h(ξ µ) V ξ+θ ξ +u (4.7) Expressions (4.4)-(4.7) 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) (4.8) G M (ξ,µ) = (ξ µ)h(ξ µ)+v (µ)ξ+m (µ) (4.9) G θ (ξ,µ) = 2 (ξ µ) 2 H(ξ µ)+v (µ)ξ EI M (µ)ξ +θ (µ) (4.1) G u (ξ,µ) = (ξ 3 µ) 3 H(ξ µ)+v (µ)ξ M (µ)ξ 2 + (ξ µ)h(ξ µ) V (µ)ξ+θ (µ)ξ +u (µ) (4.11) 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 Consider the case of N point loads p i at location µ 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 ) (4.12) where q {V,M,θ,u} and G q (ξ,µ) is the corresponding Green s function. For an arbitrary load distribution w(ξ) acting downwards, the total response is given by the integral: q(ξ) = i=1 w(µ)g q (ξ,µ)dµ (4.13) 17

18 4.3. Uniform oads For a uniform load w (weight per unit length) acting downwards, the solution (4.13) becomes: q(ξ) = w G q (ξ,µ)dµ (4.14) 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): Another useful definite integral is: (ξ µ) n H(ξ µ)dµ = ξn+1 n+1, n (4.15) Formulas (4.15) and (4.16) are proved in Haque 1. Using the Green s functions (4.8)-(4.11) and applying the formula (4.15) to (4.14): V(ξ) = w ξ + M(ξ) = w ξ2 +ξ θ(ξ) = w3 2EI u(ξ) = w4 + w2 (1 µ) n dµ = 1 n+1, n (4.16) 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 2 ξ2 ξ V (µ)dµ +w 2 ξ (4.17) M (µ)dµ (4.18) M (µ)dµ +w M (µ)dµ θ (µ)dµ+w θ (µ)dµ (4.19) u (µ)dµ (4.2) 18

19 5. Examples 5.1. Simply-supported Beam Boundary Conditions Figure 13: 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 (4.5) we compute the value of V : = (1 µ)+v V = 1 µ The value of θ is computed using u(1) = in equation (4.7): 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) 2EI G u (ξ,µ) = 3 (ξ µ) 3 H(ξ µ)+(1 µ)ξ 3 +(1 µ) 3 ξ (1 µ)ξ + (ξ µ)h(ξ µ) (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 ) w2 2 ξ(1 ξ) 19

20 5.2. Fixed Beam Boundary Conditions Figure 14: 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 (4.6) and u(1) = in equation (4.7), we get the system of equations: 2 (1 µ) 2 +V + 2 2EI M = 3 (1 µ) 3 +β(1 µ)+(1 β)v + 3 M = where β = / 2. The solution to this system is: For sake of brevity, the Green s functions are: Uniform oad G V (ξ,µ) = H(ξ µ)+v V = 3(1 µ)2 2(1 µ) 3 +2β(1 µ) 1+2β M = (1 µ) 3 +(β 1)(1 µ) 2 β(1 µ) G M (ξ,µ) = (ξ µ)h(ξ µ)+v ξ+m G θ (ξ,µ) = 2 2EI G u (ξ,µ) = 3 1+2β (ξ µ) 2 H(ξ µ)+v ξ M ξ (ξ µ) 3 H(ξ µ)+v ξ M ξ 2 + (ξ µ)h(ξ µ) V ξ ( ) 1 V(ξ) = w 2 ξ M(ξ) = w ξ(1 ξ) θ(ξ) = w3 12EI (ξ 3ξ2 +2ξ 3 ) u(ξ) = w4 24EI ξ2 (ξ 1) 2 w2 2 ξ(1 ξ) 2

21 5.3. Cantilever Beam Boundary Conditions Figure 15: 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 (4.4) we compute the value of V : Then we use M(1) = in equation (4.5)to determine M : The Green s functions are: Uniform oad = 1+V V = 1 = (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 2EI G u (ξ,µ) = 3 (ξ µ) 3 H(ξ µ)+ξ 3 3µξ 2 + (ξ µ)h(ξ µ) ξ 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) w2 2 ξ(2 ξ) 21

22 5.4. Spring-supported Beam Boundary Conditions Figure 16: 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 2EI (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 + (ξ µ)h(ξ µ) (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 ) w2 2 ξ(1 ξ)+ξu 1 +(1 ξ)u 22

23 5.5. End-restrained Beam Boundary Conditions Figure 17: End-restrained beam End Moments The left end displacement boundary condition gives u() = u =. We first use equation (4.5) 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 (4.6) and (4.7): M 1 = 2 (1 µ) 2 + M 1 M +(1 µ)+ 2 β 1 2EI M + M β = 3 (1 µ) 3 + M 1 M +(1 µ)+ 3 M 1 M 1 M + M β The solution of this system of equations for M and M 1 is: M = κ γ µ(1 µ) (1 µ)(2+κ 1 ) Φκ 1 M 1 = κ 1 γ µ(1 µ) (1 µ)(2+κ ) 4 κ 1 2 Φκ where κ = β EI, κ 1 = β 1 EI, Φ = 12EI 2, γ = κ κ 1 +4(κ +κ 1 )+12+Φ(κ κ 1 +κ +κ 1 ) Green s Functions For sake of brevity, the Green s functions (4.8)-(4.11) 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 (ξ µ) 2 H(ξ µ)+v ξ EI M ξ + M β G u (ξ,µ) = (ξ 3 µ) 3 H(ξ µ)+v ξ M ξ 2 + (ξ µ)h(ξ µ) V (µ)ξ+ M ξ β where the values of V and M were computed above for this problem. 23

24 Uniform oad For a uniform load of w, the following integrals are first computed: M (µ)dµ = κ 1 γ 2 + κ 1 12 (1+Φ) M 1 (µ)dµ = κ 1 1 γ 2 + κ 12 (1+Φ) V (µ)dµ = κ 1 1 γ 2 + κ 12 (1+Φ) + κ 1 γ 2 + κ 1 12 (1+Φ) These values are substituted into the uniform load solution: V(ξ) = w ξ + V (µ)dµ M(ξ) = w ξ2 +ξ θ(ξ) = w3 2EI u(ξ) = w4 + w2 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 1 2 ξ2 ξ V (µ)dµ +w 2 ξ M (µ)dµ M (µ)dµ +w M (µ)dµ θ (µ)dµ θ (µ)dµ 24

25 References 1 Aamer Haque. Green s functions for euler-bernoulli beams H.B. Harrison. Computer Methods in Structural Analysis. Civil Engineering and Engineering Mechanics Series. Prentice- Hall, Enlgewood Cliffs, NJ, R.C. Hibbeler. Mechanics of Materials. Prentice Hall, Upper Saddle River, New Jersey, seventh edition, R.C. Hibbeler. Structural Analysis. Prentice Hall, Upper Saddle River, New Jersey, eighth edition, Keith D. Hjelmstad. Fundamentals of Structural Mechanics. Springer, New York, second edition, Claes Johnson. Numerical Solution of Partial Differential Equations by the Finite Element Method. Studentlitteratur, und, Sweden, Aslam Kassimali. Structural Analysis. PWS-Kent, Boston, Aslam Kassimali. Matrix Analysis of Structures. Brooks/Cole, Pacific Grove, CA, Steen Krenk. Mechanics and Analysis of Beams, Columns, and Cables. Springer-Verlag, Berlin, second edition, J.C. McCormac. Structural Analysis using Classical and Matrix Methods. Wiley, New York, fourth edition, W. McGuire, R.H. Gallagher, and R.D. Ziemian. Matrix Structural Analysis. Wiley, New York, 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 Mathematics 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, S.P. Timoshenko and D.H. Young. Theory of Structures. McGraw-Hill, New York, second edition, C.M. Wang, J.N. Reddy, and K.H. ee. Shear Deformable Beams and Plates, Relationships with Classical Solutions. Elsevier, Oxford, 2. 25