Beams on Elastic Foundation

Transcription

1 Professor Terje Haukaas University of British Columbia, Vanouver Beams on Elasti Foundation Beams on elasti foundation, suh as that in Figure 1, appear in building foundations, floating strutures, beams that rest on a grid of perpendiular beams, and elsewhere. It also turns out that the governing differential equation is similar to that of ylindrial shells and tapered beams with urved webs. As a result, this material has quite broad appliability. However, it is important to exerise aution when applying this theory to problems where the foundation does not behave in a linear fashion, suh as soils that respond nonlinearly to pressure or ships whose waterline area hanges with sinkage. In this doument the beam is assumed to be an ordinary Euler-Bernoulli beam, thus the theory from that doument arries over to this one. Aordingly, downwards distributed load on the beam is referred to as q, while it is denoted q z when it ats in the opposite diretion, i.e., in the z-diretion. P q Figure 1: Beam on elasti foundation. Foundation Stiffness The foundation stiffness,, whih is illustrated in Figure 1, is oneptually straightforward. When the beam displaes downwards, the foundation exerts an upwards fore, illustrated by the springs in Figure 1. has units of fore per unit length along the beam. Another stiffness, denoted k φ, is also present, relating to rotation φ of the beam ross-setion around the longitudinal axis of the beam. In other words, k φ provides resistane against torsion of the beam. Tehniques to determine and k φ are desribed in the following subsetions for different situations. Stiffness from Buoyany An important appliation of this theory is floating strutures. To determine for a beam that floats in water, suppose the beam ross-setion has width equal to b at the loation where the ross-setion penetrates the water surfae. Furthermore, let the mass density of water be denoted ρ w, whih typially equals 1,05kg/m 3 for seawater, and let g denoted the aeleration of gravity. For an infinitesimal beam length, dx, the hange in buoyany fore due to a vertial displaement Δ equals the weight of the displaed water, namely weight density multiplied by volume: B = ( ρ w g) ( b dx Δ) (1) As a result, the stiffness that resists the displaement Δ, per unit length of the beam, is obtained by dividing B by dx and Δ: Beams on Elasti Foundation Updated Marh 10, 014 Page 1

2 Professor Terje Haukaas University of British Columbia, Vanouver = ρ w g b kn m () whih has unit fore per unit length squared. The evaluation of aording to Eq. () is orret as long as the beam width at the waterline does not hange. Unfortunately, this ondition is not always satisfied, whih is understood from Figure. While the retangular beam-ross-setion maintains width b regardless of the vertial displaement, the value of b for the irular ross-setion varies with the heave motion. As a result, is a funtion of Δ for the irular ross-setion, whih introdues nonlinearity that is negleted in the subsequent equations. ϕ b b Figure : Heave and roll motion of floating beams. The fat that the waterline area is only a proxy for displaed water volume beomes even more apparent when determining k φ. To understand this, first onsider the lassial approah for determining k φ. When the retangular beam-ross-setion in Figure rotates ounterlokwise as indiated by dashed lines, more water is displaed on the left side and less water is displaed on the right side. As a result, buoyany fores at against the rotation, giving rise to k φ. By omputing the volumes of the shaded triangles in Figure the resultant buoyany fore, on either side, is R = ( ρ w g) 1 b b φ dx (3) The distane between the fore pair, i.e., the buoyany fores from eah of the shaded triangles in Figure is /3 of b. Thus, the moment from buoyany that resists the rotation is T = R 3 b = ρ g b3 φ dx (4) w 1 As a result, the stiffness that resists the rotation φ, per unit length of the beam, is obtained by dividing T by dx and φ: k φ = ρ w g b3 1 where it is observed that the moment of inertia of the waterline area. Again it is observed that the derivations do not hold for the irular ross-setion in Figure. In fat, it is (5) Beams on Elasti Foundation Updated Marh 10, 014 Page

3 Professor Terje Haukaas University of British Columbia, Vanouver easily seen that ross-setion has zero resistane against rotation. If not easily seen, it is easily felt when trying to balane on a floating timber log. The problem with applying Eq. (5) to the irular ross-setion is the use of movement of the beam at the waterline as a proxy for displaed water volume. Thus, while Eq. (5) is ommonly used, it is important to be aware of its limitations. Stiffness from Soil Suppose the stiffness,, is determined from soil testing. In partiular, suppose a vertial load, P, is plaed on an area with dimensions x and y, and that the vertial displaement, Δ, is measured. The relationship between the distributed load and the displaement is written in terms of a distributed stiffness, k d : P x y = k Δ k = P N d d Δ x y m 3 (6) While k d is stiffness per unit area, is stiffness per unit length. The sought value is obtained by multiplying k d by the beam width, b: N = b k d m (7) If one assumes that the foundation material is linear elasti, there is no unique relationship between the Young s modulus, E, of the foundation material and the stiffness. However, if one imagines that the soil underneath the beam is linear elasti with depth L to bedrok then the fore-deformation relationship of the soil is P = EA L Δ (8) where A=x. y is the area loaded by P. Writing Eq. (8) in the form of Eq. (6) yields P x y = k d Δ and the sought stiffness is, in aordane with Eq. (7): = b k d = b E L k = E N d L m 3 (9) N m (10) Stiffness from Girder Grid Another situation appears when the beam is resting on a grid of losely spaed perpendiular beams, e.g. joists. Suppose the joists are spaed at x on entre and that their stiffness against vertial defletion at the point of intersetion with the beam is k b [N/m]. Then the sought stiffness is: = k b x N m (11) Beams on Elasti Foundation Updated Marh 10, 014 Page 3

4 Professor Terje Haukaas University of British Columbia, Vanouver Bending Differential Equation Compared with the basi Euler-Bernoulli beam theory, it is suffiient to modify the equation for vertial equilibrium to obtain the differential equation for a beam on elasti foundation. As a result, the following onventions from basi beam bending hold valid: 1) Clokwise shear fore is positive; ) Bending moment with tension at the bottom is positive; 3) Tension stress is positive; 4) The z-axis points upwards, so that upwards displaement, w, is positive; 5) The distributed load, q z, is positive upwards. Figure 3 shows the fores ating on an infinitesimal beam element. The springs that illustrate the elasti foundation exert a downward fore when the beam is subjeted to an upward displaement. q z z, w M V x V+dV M+dM dx Figure 3: Infinitesimal beam element. Vertial equilibrium yields: q z dx w dx dv = 0 (1) Dividing by dx and re-arranging yields q z = dv dx + w (13) Substitution of this equation into the Euler-Bernoulli beam theory yields the following revised differential equation EI d 4 w dx 4 + w = q z (14) Another way of deriving the differential equation is to start with the following basi differential equation for beam bending: EI d 4 w dx 4 = q z (15) Beams on Elasti Foundation Updated Marh 10, 014 Page 4

5 Professor Terje Haukaas University of British Columbia, Vanouver From earlier it is understood that the applied load, q z, plus the elasti foundation yields a total fore on the beam element equal to q z w. By substituting this total load in the right-hand side of Eq. (15), Eq. (14) is obtained. For onveniene, it is rewritten on the form d 4 w dx 4 + EI w = q z EI In solving this differential equation it is useful to define a harateristi length, sometimes referred to as the elasti length. To approah the definition it is first noted that EI/ has dimension [m 4 ]. As a result, the following definition of the harateristi length has the dimension of length: (16) l 4 EI 4 EI = l 4 4 (17) The onveniene of the fator 4 will beome apparent later. It is also onvenient to work with the normalized oordinate ξ instead of the original oordinate, x, along the beam: ξ = x l (18) To transform the differential equation, differentiation with respet to x is related to differentiation with respet to ξ by the hain rule of differentiation: d dx = d dξ dξ dx = 1 d l dξ Hene, the fourth-order derivative with respet to x equals the fourth-order derivative with respet to ξ divided by l 4, whih yields the following homogeneous version of the transformed differential equation, now formulated in the oordinate ξ: (19) d 4 w + 4 w = 0 (0) 4 dξ General Solution The harateristi equation is γ 4 +4=0 has the four different omplex roots (1+i), (1 i), ( 1+i), and ( 1 i). Consequently, the general solution is: w(ξ) = C 1 e ξ os(ξ) + C e ξ sin(ξ) + C 3 eξ os(ξ) + C 4 e ξ sin(ξ) "Damped terms" "Undamped terms" (1) where the phrase damped terms is employed to identify terms that vanish as ξ inreases to infinity. This labeling is useful beause the solution for a point load must vanish far away from the point of load appliation. In fat, only the damped terms appear in many pratial situations. To shorten the notation under suh irumstanes, the following auxiliary funtions are defined: g 1 = e ξ os(ξ) () Beams on Elasti Foundation Updated Marh 10, 014 Page 5

6 Professor Terje Haukaas University of British Columbia, Vanouver g = e ξ sin(ξ) (3) g 3 = g 1 + g (4) g 4 = g 1 g (5) These funtions are plotted in Figure 4, where it is observed that they deay rapidly with ξ. In fat, all funtions approah zero one ξ inreases beyond 4. g x g x g x g x Figure 4: g-funtions. The auxiliary funtions also have the following properties: dg 1 dξ = g 3 dg dξ = g 4 dg 3 dξ = g dg 4 dξ = g 1 (6) These relationships are helpful for deriving the bending moment, et. from the general solution. The starting point is the general solution with only the damped terms, whih reads: Differentiation yields the rotation: w(ξ) = C 1 g 1 + C g (7) θ(ξ) = dw dx = 1 dw l dξ = 1 ( C l 1 g 3 C g 4 ) (8) Another differentiation yields the bending moment: M = EI d w dx dθ = EI dx = EI 1 dθ l dξ = EI ( C l 1 g C g 1 ) (9) Beams on Elasti Foundation Updated Marh 10, 014 Page 6

7 Professor Terje Haukaas University of British Columbia, Vanouver Yet another differentiation yields the shear fore: V = dm dx = 1 dm l dξ = EI 3 ( C l 1 g 4 + C g 3 ) (30) Beam with End Fores As a referene ase for the analysis of several other problems it is useful to onsider the beam shown in Figure 5, where one end is subjeted to the fores V o and M o, while the other end is infinitely far away. V o M o EI! Figure 5: Beam with end-fores. Notie that both V o and M o are positive, i.e., the shear fore is lokwise and the bending moment gives tension at the bottom. Beause the beam is infinitely long in one diretion, the solution annot have ontributions from the un-damped terms. As a result, the solution is given by Eq. (7). Beause g 1 =g 3 =g 4 =1 and g =0 at ξ=0, Eqs. (9) and (30) yield the follow expression for the bending moment and shear fore at ξ=0: M(ξ = 0) = EI l C (31) V (x = 0) = EI 3 ( C l 1 + C ) (3) Setting those expressions equal to M o and V o, respetively, yields: C 1 = l ( EI l V + M o o ) (33) C = l EI M (34) o Substitution of Eqs. (33) and (34) into Eq. (7) to (30) yields the following solution for this loading ase: w = l EI ( M o g 4 + l V o g 1 ) (35) θ = l ( EI M g + l V g o 1 o 3) (36) M = M o g 3 + l V o g (37) Beams on Elasti Foundation Updated Marh 10, 014 Page 7

8 Professor Terje Haukaas University of British Columbia, Vanouver V = l M o g +V o g 4 (38) Beam with Point Load The previous solution is now utilized to analyze the infinitely long beam in Figure 6, whih has a point load applied at ξ=0. P EI! Figure 6: Beam with point load. Immediately to the left of the point load the shear fore is P/, while immediately to the right it is P/. That is, with referene to the previous ase, V o = P/. Furthermore, the rotation at the point load is zero: θ(0) = Substituting V o = P/ and solving for M o yields l ( EI M + l V o o ) = 0 (39) M o = P l 4 where it is noted that the bending moment at the point load is the same as that of a simply supported beam with length l loaded at midspan. In fat, that is the reason that the integer 4 was introdued in the definition of l in Eq. (17). In short, the solution for the beam in Figure 6 is given by Eqs. (35) to (38) with (40) V o = P (41) and M o = l P 4 That yields the following solution for a beam on an elasti foundation with a point load: w = Pl 3 8EI θ = Pl 4EI (4) ( g 4 g 1 ) (43) ( g 1 g 3 ) (44) M = Pl ( 4 g g 3 ) (45) Beams on Elasti Foundation Updated Marh 10, 014 Page 8

9 Professor Terje Haukaas University of British Columbia, Vanouver V = P ( g + g 4 ) (46) This solution, as the previous ones, is defined only for positive ξ-values. However, it is readily plotted on both sides of the origin by multiplying Eqs. (44) and (46) by the sign of ξ, namely ξ/ ξ. This is done beause the rotation and the shear fore are asymmetri funtions. For example, from the disussion before Eq. (39) it is lear that the shear fore is positive where ξ is negative, and vie versa. Another ommon plotting onvention is to multiply Eq. (45) by ( 1) to get the moment diagram on the tension side of the beam. Cluster of Point Loads If more than one point load is present, suh as the situation in Figure 7, the solution is obtained by superposition. Assuming that the beam is suffiiently long for the solution to dampen out before the beam end is reahed, the solution in Eqs. (43) to (46) is employed. For example, for the situation in Figure 7 the bending moment is: M = P 0 l 4 P ( g 3 (ξ) g (ξ)) + i l ( g 4 3 (ξ ξ i ) g (ξ ξ i )) (47) ξ 1 ξ i=1 P 0 P 1 P ξ EI Figure 7: Beam with point load luster. Distributed Load Above, the solutions w, θ, M, and V were established for a beam with a point load. That solution an also be employed in the analysis of beams with distributed load. The point load that was previously applied at ξ=0 is now applied at the generi loation ξ=. Moreover, the point load, P, is onsidered to be one infinitesimal part of a distributed load, i.e., P=q. dx. Beause dx=l. dξ the load to be inserted into the previous solutions w(ξ), θ(ξ), M(ξ), and V(ξ) is P=q. l. dξ. On that basis, the solution for a distributed load between ξ=a and ξ=b is w(ξ) = b a w(ξ )d b = ql 3 8EI l g ( (ξ ) g (ξ ) 4 1 )d (48) et. for the θ(ξ), M(ξ), and V(ξ), assuming onstant load intensity, q. a Beams on Elasti Foundation Updated Marh 10, 014 Page 9

10 Professor Terje Haukaas University of British Columbia, Vanouver Torsion Differential Equation The differential equation for St. Venant torsion, without resistane from an elasti foundation, is GJ φ ''+ m x = 0 (49) The torsional resistane from an elasti foundation is a distributed torque equal to k φ. φ, whih implies that the effetive distributed torque is equal to m x k φ. φ. That leads to the following modified differential equation for St. Venant torsion: φ '' k φ GJ φ + m x GJ = 0 (50) Now onsider the homogeneous version of this equation, i.e., let m x =0, and define the auxiliary quantity l = GJ k φ (51) whih is the harateristi length in torsion. This definition gives a length-measure beause k φ is in units of fore. As done previously for bending, the normalized oordinate ξ is defined as in Eq. (18): ξ=x/l, with resulting differentials given in Eq. (19). This means that the homogeneous version of Eq. (50) is d φ dx 1 l φ = 1 l d φ dξ 1 l φ = d φ dξ φ = 0 (5) General Solution The harateristi equation is γ +γ=0 has the two roots γ=0 and γ=1. Consequently, the general solution is: The torque assoiated with this solution is φ = C 1 e ξ + C e ξ (53) T = GJ dφ dx = GJ 1 ( C l 1 e ξ + C e ) ξ (54) Beam with End Torque As a referene ase for the analysis of other problems, onsider a beam like the one in Figure 5, with a torque, T o, applied at the free end. Beause the beam is infinitely long in one diretion, the solution annot have ontributions from the un-damped term. As a result, C 1 =0, and the equation that determines C is: whih yields T o = T (ξ = 0) = GJ 1 l C (55) Beams on Elasti Foundation Updated Marh 10, 014 Page 10

11 Professor Terje Haukaas University of British Columbia, Vanouver whih in turn means that the solution is C = T o l GJ φ = T o l GJ e ξ (56) (57) Rigid Beam For ases with long harateristi length relative to the beam length, L, it may be appropriate to onsider the beam as a rigid body as it rotates under applied torque, T o. In that ase the rotation is simply the onstant φ o = T o k φ L (58) and the torque in the beam inreases linearly from zero at the free end to T (x) = k φ φ o x = T o x L (59) at a distane x from the free end. Beams on Elasti Foundation Updated Marh 10, 014 Page 11