CHAPTER 6 Beam theory blahLinear beam theoryBendingetcEx: apply load st. the moment is zero at the center of the beamEx: deflection of a beam under its own weightEx: residual stress induced bendingCaveats: only works for small deflectionsMoments of common cross-sectionsExample: moment of an I-beamBi-metal and composite beamsAnticlastic curvature and bending of platesTorsionReferencesProblemsGreat Events of the Twentieth Century 73CHAPTER 6 Beam theory blahSynopsisMoment/curvature relationship ,Bending under end loadingAxial compression:Torsional deflectionMatrix representation of end deflection vs. applied force/momentκ1ρ---MEI------== Ia3b12--------=yx()x26EI------- --x 3L–()Fx22EI------ ---M+=yL()L33EI---------FL22EI---------M+=KxLEA-------F=θMLJG--------=Beam theory blah74 Great Events of the Twentieth CenturyLinear beam theorySolving for the 3D deformation of an arbitrary object under arbitrary loading condi-tions is the domain of the theory of elasticity, and in general a nasty problem. Wereduce the problem to a single dimension, where all quantities on the beam varyonly as functions of a single variable such as arc length. This is the fundamentalassumption of beam theory, which is tractable for hand analysis, and very often justas accurate for MEMS problems as the full three dimensional theory of elasticity.A further assumption is that the deflections are “small”. Let us start by assumingthat the deflections are infinitesimal, and then see what the limits to this linearbeam theory are in the next chapter.We will also assume that the cross-sections of the beams vary smoothly in shapealong the length of the beam, and that the cross-sectional dimensions are smallcompared to the length of the beam.The goal of this chapter will be to develop the relationship between forces andmoments applied at the ends of a beam and the resulting deflections. Since this is alinear theory, the resulting relationship will take the form of a matrix.BendingFind yourself a big rubber eraser and draw a cartesian grid on one side. Now bendthe eraser by pushing your thumbs up in the middle and pulling down on the ends.You should see something like what’s in Figure xxx. Your fingers are doing aGreat Events of the Twentieth Century 75Bendingdecent job of applying a uniform moment along the length of the eraser, and it isdoing its best to bend to a constant radius of curvature.A horizontal line midway between the top and the bottom of the eraser is known asthe neutral axis because it will stay the same length before and after the bending,whereas lines above the midpoint will get longer, and lines below the midpoint willget shorter.We define a coordinate system on the undeformed beam (eraser) where x is the dis-tance along the length of the beam and z is the vertical distance above the neutralaxis of the beam. If we take the thickness of the beam to be a, then we see that zvaries from -a/2 to a/2.From geometrical arguments we can show that the strain as a function of positionalong the beam is(EQ 27)where we are allowing the radius to vary as a function of position along the lengthof the beam.Recalling that the stress and strain are related by the Young’s modulus, we canwrite an expression for the stress (in the x direction) as a function of position alongthe beam(EQ 28)FIGURE 23. Deformation of a cartesian grid under a uniform torque.M0M0xzaLrhoε xyz,,()εxz,()zρ x()-----------==σxxxyz,,()Ezρ x()-----------=Beam theory blah76 Great Events of the Twentieth CenturyIntegrating over the cross-section of the beam we see that(EQ 29)Which is the moment/curvature relationship for beams, and is usually written(EQ 30)where is the radius of curvature, M is the moment, and(EQ 31)is the moment of inertia of the cross-section. The cross-sectional geometry canvary slowly as a function of position along the beam, making I a function of x. Inprinciple, E can also be a function of x. Discontinuities or rapid changes in thecross section require more detailed modeling.The curvature of the beam is approximately the second derivate of displacement ywith respect to a fixed coordinate system x(EQ 32)from which we see that(EQ 33)Example: Calculate the shape of a cantilevered beam with a pure moment applied atthefreeendofthebeam.Soln: The cantilevered beam implies a boundary condition ofMx() zσxxbzda2---–a2---∫zEzρ x()-----------bzda2---–a2---∫Eρ x()-----------bz33----a2---–a2---Ea3b12ρ x()-----------------== ==1ρ x()-----------Mx()EI x()-------------=ρIa3b12--------=κx22ddy1xddy32/+12/----------------------------------x22ddy≈=x22ddy Mx()EI------------≈Great Events of the Twentieth Century 77Bending(EQ 34)We can calculate the moment at any point along the beam using a freebody diagramwhere we break the beam at a location x along its length. In order to maintain staticequilibrium we must have the sum of all forces and moments on a body equal tozero. This implies that across a virtual break at location x, we would need to apply-M0 on the left end of end of the end body, which implies that a moment M0 mustbe applied on the right end of the remaining cantilever. So(EQ 35)Now we have an ordinary differential equation with a boundary condition. We canintegrate the ODE directly, yieldingSubsituting the boundary conditions shows that the constants of integration arezero. Note that the actual shape of the deflected beam will be a section of a circle (acurve with a constant radius of curvature). The expression that we derived is anaccurate approximation to a circle for x much less than . The error between theexpression that we calculated and the true shape is due to the approximation that theradius of curvature was equal to the reciprocal of the second derivative.Often we are just interested in the deflection of the endpoint of the beam, which isy 0() 0xddy0(), 0==M0xL-xM0FIGURE 24. Freebody diagram for transmission of moment along the beamat location x.Mx() M0=xddyM0EI-------xC1+=yx()M0EI-------x22-----C1xC2++=ρyL()MEI------L22-----θ L();xddyL()MEI------L===Beam theory blah78 Great Events of the Twentieth Centurywhich gives us two spring constantsExample: Consider an isotropic beam of length L and cross section a by b. Com-pare the angular deflection when a moment is applied axially or transversely on thebeam.Soln: Since we are in the linear region we can compare deflections simply by look-ingataratioofthetwospringconstants.Example: A cantilevered beam has a rigid body bolted on the free
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