Lecture 5: Solution Method for Beam Deflections

Transcription

1 Structura Mechanics.080 Lecture 5 Semester Yr Lecture 5: Soution Method for Beam Defections 5.1 Governing Equations So far we have estabished three groups of equations fuy characterizing the response of beams to different types of oading. In Lecture reations were estabished to cacuate strains from the dispacement fied. where ɛ (x) = du + 1 ɛ(x, z) = ɛ (x) + zκ (5.1) ( ), κ = d w (5.) The above geometrica reation are independent on equiibrium and appy to any kind of materias. The second set of equations, derived in Lecture 3, is the equiibrium requirement dv + q(x) = 0 force equiibrium (5.3) dm V = 0 moment equiibrium (5.4) where V = V + N is the effective shear. (5.5) dn = 0 (5.6) Eiminating V and V between the above equations, the beam equiibrium equation was obtained (See Eq. (3.74)) d M + N d w + q = 0 (5.7) The derivation of the equiibrium is vaid for a types of materias. In the theory of moderatey arge defections, the equiibrium is couped with the kinematics. The third group of equation define the materia behavior and reates the generaized strains to generaized forces N = EAɛ (5.8) M = EIκ (5.9) Independence of geometry and equiibrium on constitutive equation aows to deveop the genera framework of a sover in the Finite Eement codes. The constitutive equations can then be added as a user Defined Subroutines. Let s consider first the infinitesima deformations (sma rotations for which the term ( ) 1 vanish in Eq. (5.) and the term d w 5.4 and 5.8) one obtains = 0 in Eq. (5.7). Then from Eqs. (5., EA d u = 0 (5.10) 5-1

2 Structura Mechanics.080 Lecture 5 Semester Yr Eiminating the curvature and bending moments between Eqs. (5., 5.7 and 5.9), the beam defection equation is obtained EI d4 w = q(x) (5.11) 4 The concentrated oad P can be treated as a specia case of the distributed oad q(x) = P δ(x x 0 ), where δ is the Dirac deta function. Let s consider first Eq. (5.4) for the axia dispacement. The boundary conditions in the x-direction are (N N)δu = 0 (5.1) The genera soution for u(x) is du = D 1, u = D 1 x + D 0 (5.13) There are two integration constants, and two boundary conditions are needed. There are ony four combinations of boundary conditions: 1. Beam restricted from axia motion, see Fig. (5.1). This gives rise to the soution of two agebraic equation u(x = 0) = u(x = ) = 0 (5.14) 0 = D 0 + D 1 0 (5.15a) 0 = D 0 + D 1 (5.15b) which gives D 0 = D 1 = 0 and u(x) = 0. This is a trivia case, for which the axia force N = EA du vanishes as we. u = 0 x u = 0 Fixed (Kinematic) N = 0 N = 0 Siding (Static) N = 0 u = 0 Mixed Figure 5.1: Three combinations of in-pane boundary conditions for u(x). 5-

3 Structura Mechanics.080 Lecture 5 Semester Yr. Beam aowed to side in the x-direction on both ends. N = N = 0 at x = 0 and x = (5.16) The axia force is proportiona to du. From Eq. (5.13) we can see that the gradient of u is zero aong the entire beam. So, if N = 0 or du vanishes at one end, say x = 0, D 1 = 0 and automaticay N = 0 is satisfied at the other end, x =. The integration constant D 0 is undetermined meaning that the rigid body transation of the entire beam is aowed. 3. In order to prevent the rigid body transation, one end of the beam, say x = 0, must be fixed against motion in the x-direction. Thus N = 0 or du = 0 at x = 0 (5.17a) u = 0 at x = (5.17b) which are precisey the boundary conditions for the third case. From Eq. (5.13) we get D 1 = 0 D 1 + D = 0 D = 0 (5.18a) (5.18b) Now, the axia dispacement vanishes, u(x) = 0 but the rigid body transation is eiminated. For a the above three cases of kinematic static and mixed boundary conditions, the axia force was zero. 4. If one end of the beam (bar) is oaded by a given force N and the other one is fixed, the boundary conditions (BC) are N = N, EA du = 0 at x = 0 u = 0 at x = (5.19) and the soution is D 1 = N EA, D = u(x) = N EA (5.0) N ( x) (5.1) EA The case in which the noninear term is retained in Eq. (5.) is much more interesting. This wi be deat with in the section on moderatey arge defection of beams. 5-3

4 Structura Mechanics.080 Lecture 5 Semester Yr We now turn our attention to the soution of the beam defection, Eq. (5.11). This is the fourth-order inear inhomogeneous equation which requires four boundary conditions. There are four types of boundary conditions, defined by (M M)δw = 0 (5.a) (V V )δw = 0 (5.b) For the sake of iustration, we seect a pin-pin BC for a beam oaded by the uniform ike oad q, Fig. (5.). Figure 5.: Pin support aows for rotation but not for vertica transation. The bending moment is proportiona to the curvature. Eq. (5.11) is then subjected to the foowing boundary conditions: w(x = 0) = w(x = ) = 0 d w = d w x=0 = 0 x= (5.3a) (5.3b) Let s integrate the differentia equation four times with respect to x: d 3 w 3 = qx EA + C 1 (5.4a) d w = qx EA + C 1x + C (5.4b) = qx3 EA6 + C 1x + C x + C 3 (5.4c) = qx4 EA4 + C 1x 3 + C x + C 3 x + C 4 6 (5.4d) Substituting the BC into the genera soutions, one gets 0 = C (5.5a) 0 = q3 EA + C 1 + C (5.5b) 0 = C 4 (5.5c) 0 = q4 4EA + C C + C 3 + C 4 (5.5d) 6 5-4

5 Structura Mechanics.080 Lecture 5 Semester Yr The soution of the above system is C 1 = q C = 0 C 3 = q3 1 C 4 = 0 (5.6a) (5.6b) (5.6c) (5.6d) The oad-dispacement reation becomes w(x) = qx 4EA (3 x + x 3 ) (5.7) Differentiating Eq. (5.7) twice, the expression for the bending moment is M(x) = qx ( x) (5.8) and differentiating again, the shear force becomes V (x) = dm = q ( x) (5.9) Pots of the normaized bending moments and shear forces are shown in Fig. (5.3). Figure 5.3: Paraboic distribution of the bending moment and inear variation of the shear force. The shear force V = EI d3 w 3 sope is seen to vanish at the mid-span of the beam. Aso the is zero at this ocation. We have proved that at the symmetry pane V (x = ) = 0 (5.30a) = 0 (5.30b) x= Inversey, if the probem is symmetric, that Eq. (5.30) must hod at the symmetry pane. As an aternative formuation, one can consider a haf of the beam with the symmetry BC. Can you sove the above probem and compare it with soution of the pin-pin beam, Eq. (5.7)? 5-5

6 Structura Mechanics.080 Lecture 5 Semester Yr / M = 0 w = 0 x V = 0 =0 Figure 5.4: Simpy-supported pate with symmetry boundary conditions. It shoud be mentioned that the pin-pin supported beam is a staticay determinate structure. Therefore the distribution of shear forces and bending moments can be determined from the equiibrium equation aone. Can you do it and get correcty the signs? The purpose of Lecture 5 is to present properties of the governing equations and soutions. interested students are referred to end chapter of probem sets where many beams with different oading and BC are considered. Aso the recommended reference book and monographs present soution to some common beam probems. 5. Genera Properties of the Beam Governing Equation: Genera and Particuar Soutions Reca from the Cacuus that soution of the inhomogeneous, inear ordinary differentia equation is a sum of the genera soution of the homogeneous equation w g and the particuar soution of the inhomogeneous equation w p. The property of homogeneity means that f(ax) = Af(x). The homogeneous counterpart of Eq. (5.11) is EI d4 w 4 = 0 or d 4 w 4 = 0 (5.31) and its soution, obtained by four integrations is the third order poynomia w g (x) = C 1x C x + C 3 x + C 4 (5.3) The particuar soution w p of the beam defection equation, Eq. (5.11) depends on the oading, but not the boundary conditions. For the uniformy oaded beam the particuar soution is the first term in Eq. (5.3d). As an iustration, consider the same pin-pin supported beam oaded by the trianguar ine oad q(x) = q 0 x, 0 < x < (5.33) where q 0 is the oad intensity at mid-span x = /. The particuar soution of this probem, satisfying the governing equation is w p = q 0x 5 60EI 5-6 (5.34)

7 Structura Mechanics.080 Lecture 5 Semester Yr Then, the fu soution is w(x) = w g + w p. Beam oaded by concentrated forces (or moments) requires specia consideration. Continuity requirements A sudden change in the beam cross-section or oading may produce a discontinuous soution. What quantities may suffer a jump and what must be continuous? w Figure 5.5: The dispacement and sope discontinuities are not aowed in beams. In mechanics the discontinuity of a given function is denoted by a square bracket [f(ξ)] = f(ξ + ) f(ξ ) (5.35) where ξ + and ξ denote the vaues of the argument on the right and eft hand of a discontinuity. In the quasi-static theory of beam [w] = 0 (5.36a) [ ] = 0 (5.36b) The discontinuity in the vertica dispacement means separation so of course it may not occur. Why then sopes must be continuous for eastic beams? This is simpe. A change of sope is caed a curvature. A jump in the sope gives an infinite curvature, and thus an infinite bending moments. Such a situation is impossibe, because the beam cross-section wi go into pastic range, and the beam wi no onger stay eastic. Quantities that can be discontinuous are Bending monents [M] = M (5.37a) Shear force [V ] = V (5.37b) As an iustration, consider a pin-pin supported beam oaded at mid-span by a point force P. As mentioned earier, the point oad can be considered as a imiting case of a continuous ine oad with the hep of the Dirac deta function q(x) = P δ(x ), where δ(x ) = 1 (5.38) Even though techniques have been deveoped to dea with singuarity functions for beams, they require to use the apparatus of the mathematica theory of distribution. This 5-7

8 Structura Mechanics.080 Lecture 5 Semester Yr w = 0 M = 0 x P w = 0 M = 0 w = 0 M = 0 / P V = P =0 Figure 5.6: Symmetric oading of a beam by a concentrated force. is not the avenue that we wi take. instead, a symmetry boundary condition wi be imposed. Now, the concentrated oad is not appied inside the beam 0 < x <, governed by the inhomogeneous differentia equation, but at the boundary. Each haf of the beam is carrying haf of the oad. Therefore, the boundary conditions are at x = 0 w = 0 (5.39a) at x = d w = 0 V = P = 0 (5.39b) (5.39c) (5.39d) Because the oading is appied on the boundary, the differentia equation becomes homogeneous. The soution of Eq. (5.31) is given by the third order poynomia, substituting the above BC into the soution given by Eq. (5.3), a system of four inear agebraic equations is obtained, where the soution is The defection ine is given by C 1 = D EI, C = 0, C 3 = w(x) = and the centra defection (something to remember) is P 16EI, C 4 = 0 (5.40) P x 48EI (3 4x ) (5.41) w 0 = w(x = ) = p3 48EI (5.4) The pot of the distribution of bending moment and shear forces aong the ength of the beam determined from the cacuated defection ine is shown in Fig. (5.7). Note that the jump in the interna shear force is equa to the appied force [V ] = V right (x = ) V eft(x = ) = P (5.43) 5-8

9 Structura Mechanics.080 Lecture 5 Semester Yr P P V(x) M(x) Figure 5.7: Bending moment is continuous at the mid-span, but the shear force is not. P If the point oad is not appied at the mid-span but at an arbitrary distance x = a, the beam must be divided into two parts 0 < x < a, a < x <, and each part must be soved independenty. First segment 0 < x < a w I (x) = C 1x 3 6 Second segment a < x < w II (x) = C 5x C x + C 6x + C 3 x + C 4 (5.44a) + C 7 x + C 8 (5.44b) This gives rise to eight integration constants, four for each side. Woud there be enough conditions to determine these constants? The answer is YES.There are two boundary conditions at x = 0, four continuity conditions at x = a, given by Eqs. ( ) and, again, two boundary conditions at x =. In summary BC, x = 0 Continuity, x = a BC, x = w = 0 [w] [ = ] 0 w = 0 M = 0 = 0 M = 0 [M] = 0 [V ] = P (5.45) Note that there is no concentrated bending moment appied M = 0 so that the bending moment fied is continueous across x = a. The concentrated force produces a jump in the distribution of the shear forces, so V = P. We eave it to the reader to appy the above condition and sove the probem. More on this probem can be found in two sections of this notes: Probem Sets and Recitations. The method of superposition says that the defections and sopes of the beam subjected to a system of oads are equa to the sum of those quantities due to individua oads. In other words the individua resuts may be superimposed to determine a combined response, hence the term method of superposition. This is a very powerfu and convenient method since soutions for many support and oading conditions are readiy avaiabe in various engineering handbooks. Using the principe of superposition, we may combine these soutions to obtain a soution for more compicated oading conditions. As an exampe, consider a camped-camped beam oaded by a uniform ine oad q and concentrated force at the center P. The defection formuas for the two individua oading 5-9

10 Structura Mechanics.080 Lecture 5 Semester Yr are w uniform = qx ( x) (5.46a) 4EI w point = P x (3 4x) 48EI (5.46b) The soution for both oads acting together is w tota = w uniform + w point (5.47) 5.3 Staticay Determined Beams Beam for which the distribution of bending moments and shear forces can be determined from the equiibrium aone are caed staticay determinate beams. For such beams M(x) and V (x) are known and determination of beam defection wi be a much easier task. Combining Eq. (5.9) with Eq. (5.) one ends up with the foowing second order inear differentia equation EI d w = M(x) (5.48) The bending moment, which by itsef shoud satisfy the second order differentia equation, Eq. (5.7) shoud now obey two stress boundary conditions at the beam ends. The static boundary conditions are indicated in Fig. (5.8) for a pin-pin supported and cantiever beam. Figure 5.8: The static boundary conditions for a fu and haf of a beam. Determination of bending moment and shear force diagrams is the subject of eementary courses in statics, and the genera procedure is not expained here. In the case of the simpy supported beam with a point oad at the mid-span, the bending moments P x M(x) =, 0 < x < P ( x), < x < The bending moment vanishes at x = 0 and x =. The corresponding shear force V = dm is P V (x) =, 0 < x < P, < x < 5-10 (5.49) (5.50)

11 Structura Mechanics.080 Lecture 5 Semester Yr At the beam center [V ] = P [ P ] = P (5.51) Because of shear force discontinuity at the beam center, the soution wi be sought for a haf of the beam. Each haf of the beam is carrying haf of the oad. We have shown that the bending moment distribution satisfy two satin boundary condition. Therefore the differentia equation (5.49) is subjected ony to two kinematic boundary conditions w = 0 x =0 Integrating Eq. (5.48) twice one gets Figure 5.9: Symmetry boundary condition. EIw = P x3 1 + C 1x + C (5.5) The two integration constants, determined from the boundary conditions w(0) = 0, 0, are x= C 1 = P 16, C = 0 (5.53) and the defection ine of the beam is given by w(x) = P x 48EI (3 4x ) 0 < x < (5.54) The second haf of the beam is the mirror refection, by symmetry. In particuar, the centra defection w o = w(x = ) is expressed by a input parameters of the beam as = w o = P 3 48EI (5.55) It wi be hepfu to remember the above formua for the rest of your professiona ife. In summary, determination of defections of staticay determinate beams is much easier than its staticay indeterminate counterparts. The governing equation is of the second order, and for symmetric probems there are ony two integration constants. 5-11

12 Structura Mechanics.080 Lecture 5 Semester Yr A a P - a B R A R B Figure 5.10: Beam under off-center point oad. 5.4 Continuity Conditions, an Exampe In Section 5.4 the continuity requirements were formuated, but the system of eight agebraic equations was not soved. Here a compete soution wi be presented for a beam oaded by a point force acting at an arbitrary ocation x = a. The reaction forces are cacuated from moment equiibrium: R A = P a R B = P a (5.56a) (5.56b) The sum of the reaction forces is equa to P. The corresponding bending moments and shear forces are P ( a)x P ( a) R A x =,, 0 < x < a M(x) = V (x) = P a( x) R B ( x) =, P a (5.57), a < x < The jump in the shear force across the discontinuity point x = a is [V ] = V + V = P ( a) ( P a ) = P (5.58) The bending moments are continuous on both sides, [M] = 0. Therefore the static continuity conditions are automaticay satisfied at x = a. The kinematic continuity conditions, formuated in Eq. (5.36) require dispacements and sopes to be continuous. Integrating the governing equations (5.48) with (5.57) in two regions gives EIw I = P ( a)x3 6 + C 1 x + C 0 < x < a (5.59a) EIw II = P a ( x x3 6 ) + C 3x + C 4 a < x < (5.59b) The four integration constants are found from two boundary condition and two continuity condition w(0) = w() = 0, w I (a) = w II I (a), = II x=a (5.60) x=a 5-1

13 Structura Mechanics.080 Lecture 5 Semester Yr This gives rise to the system of four inear inhomogeneous agebraic equations for C 1, C, C 3, and C 4 C = 0 P a + C 3 + C 4 = 0 3 P ba 3 + C 1 a = P a ( ) a 6 a3 (5.61) + C 3 a + C 4 6 P ba + C 1 = P a (a 1 ) a + C 3 A simpe probem has ed to a quite compex agebra. Now, you understand why the previous exampe with eight unknown coefficients was ony formuated but not soved. The soution to the system (5.61) is C 1 = P a(a 3a + ) 6 C = 0 C 3 = P a(a + ) 6 C 4 = P a3 6 and the fina soution of unsymmetricay oaded beam is w I (x) = P x [ a 3 3a x + a( + x ) ] 6EI w II (x) = P a( x) [ a + x( + x) ] 6EI (5.6a) (5.6b) (5.6c) (5.6d) 0 < x < a (5.63a) a < x < (5.63b) One can easiy check that the continuity conditions are met at x = a. The above exampe teaches us that symmetry in nature and engineering not ony means beauty, but aso brings simpicity. 5-13

14 MIT OpenCourseWare / 1.573J Structura Mechanics Fa 013 For information about citing these materias or our Terms of Use, visit: