2.080 Structural Mechanics Lecture 5 ... - MIT OpenCourseWare

1 Structural Lecture 5 Semester YrLecture 5: Solution Method for Beam Governing EquationsSo far we have established three groups of equations fully characterizing the response ofbeams to different types of loading. In Lecture 2 relations were established to calculatestrains from the displacement field. (x,z) = (x) +z ( )where (x) =dudx+12(dwdx)2, = d2wdx2( )The above geometrical relation are independent on equilibrium and apply to any kind secondset of equations, derived in Lecture 3, is the equilibrium requirementdV dx+q(x) = 0 force equilibrium( )dMdx V= 0 moment equilibrium( )whereV =V+Ndwdxis the effective shear.

2 ( )dNdx= 0( )EliminatingVandV between the above equations, the beam equilibrium equation wasobtained (See Eq. ( ))d2 Mdx2+Nd2wdx2+q= 0( )The derivation of the equilibrium is valid for all types of materials. In the theory ofmoderately large deflections, the equilibrium is coupled with the third groupof equation define the material behavior and relates the generalizedstrains to generalized forcesN=EA ( )M=EI ( )Independence of geometry and equilibrium on constitutive equation allows to develop thegeneral framework of a solver in the Finite Element codes.

3 The constitutive equations canthen be added as a user Defined s consider first the infinitesimal deformations (small rotations for which the term12(dwdx)2vanish in Eq. ( ) and the termd2wdx2= 0 in Eq. ( ). Then from Eqs. ( , and ) one obtainsEAd2udx2= 0( )5-1 Structural Lecture 5 Semester YrEliminating the curvature and bending moments between Eqs. ( , ), the beamdeflection equation is obtainedEId4wdx4=q(x)( )The concentrated loadPcan be treated as a special case of the distributed loadq(x) =P (x x0), where is the Dirac delta s consider first Eq.)

4 ( ) for the axial displacement. The boundary conditions inthex-direction are(N N) u= 0( )The general solution foru(x) isdudx=D1,u=D1x+D0( )There are two integration constants, and two boundary conditions are needed. There areonly four combinations of boundary conditions:1. Beam restricted from axial motion, see Fig. ( ).u(x= 0) =u(x=l) = 0( )This gives rise to the solution of two algebraic equation0 =D0+D1 0( )0 =D0+D1l( )which givesD0=D1= 0 andu(x) = 0. This is a trivial case, for which the axialforceN=EAdudxvanishes as u = 0 u = 0 Fixed (Kinematic) N = 0 N = 0 Sliding (Static) Mixed N = 0 u = 0 Figure : Three combinations of in-plane boundary conditions foru(x).

5 5-2 Structural Lecture 5 Semester Yr2. Beam allowed to slide in thex-direction on both ends. N=N= 0 atx= 0 andx=l( )The axial force is proportional todudx. From Eq. ( ) we can see that the gradientofuis zero along the entire beam. So, if N= 0 ordudxvanishes at one end, sayx= 0,D1= 0 and automatically N= 0 is satisfied at the other end,x=l. The integrationconstantD0is undetermined meaning that the rigid body translation of the entirebeam is In order to prevent the rigid body translation, one end of the beam, sayx= 0, mustbe fixed against motion in thex-direction.

6 Thus N= 0 ordudx= 0 atx= 0( )u= 0 atx=l( )which are precisely the boundary conditions for the third case. From Eq. ( ) wegetD1= 0( )D1l+D2= 0 D2= 0( )Now, the axial displacement vanishes,u(x) = 0 but the rigid body translation all the above three cases of kinematic static and mixed boundary conditions, theaxial force was If one end of the beam (bar) is loaded by a given force Nand the other one is fixed,the boundary conditions (BC) areN= N,EAdudx= 0atx= 0u= 0atx=l( )D1= NEA,D2= NlEA( )and the solution isu(x) = NEA(l x)( )The case in which the nonlinear term is retained in Eq.

7 ( ) is much more will be dealt with in the section on moderately large deflection of Lecture 5 Semester YrWe now turn our attention to the solution of the beam deflection, Eq. ( ). This isthe fourth-order linear inhomogeneous equation which requires four boundary are four types of boundary conditions, defined by(M M) w = 0( )(V V) w= 0( )For the sake of illustration, we select a pin-pin BC for a beam loaded by the uniformlike loadq, Fig. ( ).Figure : Pin support allows for rotation but not for vertical bending moment is proportional to the curvature.

8 Eq. ( ) is then subjected tothe following boundary conditions:w(x= 0) =w(x=l) = 0( )d2wdx2 x=0=d2wdx2 x=l= 0( )Let s integrate the differential equation four times with respect tox:d3wdx3=qxEA+C1( )d2wdx2=qx2EA2+C1x+C2( )dwdx=qx3EA6+C1x22+C2x+C3( )dwdx=qx4EA24+C1x36+C2x22+C3x+C4( )Substituting the BC into the general solutions, one gets0 =C2( )0 =ql32EA+C1l+C2( )0 =C4( )0 =ql424EA+C1l36+C2l22+C3l+C4( )5-4 Structural Lecture 5 Semester YrThe solution of the above system isC1= ql2( )C2= 0( )C3=ql312( )C4= 0( )The load-displacement relation becomesw(x) =qx24EA(l3 2lx2+x3)( )Differentiating Eq.

9 ( ) twice, the expression for the bending moment isM(x) =qx2(l x)( )and differentiating again, the shear force becomesV(x) =dMdx=q2(l 2x)( )Plots of the normalized bending moments and shear forces are shown in Fig. ( ).Figure : Parabolic distribution of the bending moment and linear variation of the shear forceV=EId3wdx3is seen to vanish at themid-spanof the beam. Also theslopedwdxis zero at this location. We have proved that at the symmetry planeV(x=l2) = 0( )dwdx x=l2= 0( )Inversely, if the problem is symmetric, that Eq.

10 ( ) must hold at the symmetry an alternative formulation, one can consider a half of the beam with the symmetry you solve the above problem andcompareit with solution of the pin-pin beam, Eq.( )?5-5 Structural Lecture 5 Semester YrM = 0 w = 0 x V = 0 dwdx=0l/2 Figure : Simply-supported plate with symmetry boundary should be mentioned that the pin-pin supported beam is a statically determinatestructure. Therefore the distribution of shear forces and bending moments can be deter-mined from the equilibrium equation you do it and get correctly the signs?