Governing Equations For Beams

Beams II -- Deflections: 1Governing Equations for BeamsWhat is the relationship between a beam'sloading and it's deflection?In this stack we will derive this relationship,and also establish relationships between the load,shear and moment at any point along the beam.This stack concludes with a brief outline of twomethods used to solve for deflections in beams.1Consider the loaded beam shownabove. The function w(x) denotes thevalue of the distributed load at any point x.2Hide TextHide TextA Section of the BeamTo derive the load, shear,moment relationships we will focuson a short length of the beam takenfrom anywhere along the beam.3Hide TextHere is that short element of the beamshown with the portion of the load that isacting on it.Does this figure appear to satisfy staticequilibrium?4Hide Text

Beams II -- Deflections: 2Internal Forces on the SectionSomething on sign convention for internal moments andshears!In order to keep the short length ofbeam in equilibrium, we must include theinternal forces acting on the cut of thebeam.For beam bending we will ignore theaxial force acting on this cut, and focus onthe internal shear and moment.56Hide TextHide TextHopefully the picture on the previous card appearedincomplete. This figure shows the internal forces which acton the right face of the cut.In general the shear and moment on the right face willnot be the same as that on the left face. The shear willhave changed by some amount equal to the change in shearwith respect to x (dV/dx) times the length of the element,Taking a closer look at the short length of the beam, we nowdefine a local coordinate system for the beam. The x coordinatestarts at one end of the beam (usually the left end) and runsalong the "neutral axis" of the beam. The y coordinatemeasures the perpendicular distance from the neutral axis.7Hide TextNeutral Axis ?8Hide Text

Beams II -- Deflections: 3The three components which contribute tovertical force equilibrium are the shear on the leftcut, the shear on the right cut, and the distributedload. Notice that the distributed load must bemultiplied times the length over which it actsbefore it can be treated as a force.Let's begin our examination ofthis free body diagram by summingthe forces in the vertical direction.910Hide TextHide TextDividing both sides by the length of the element, dx, we areleft with the relationship:The V(x) and – V(x)terms cancel each other.The change in shear along the beam isequal to the distributed load acting on thebeam !(with a negative sign to be consistent with our sign convention).11Hide Text12Hide Text

Beams II -- Deflections: 4We continue our quest byapplying moment equilibrium tothe free body diagram.This is an important result, so let'sstore it at the bottom of the screen.13Hide Text14Hide TextWe sum the moments about a point locatedon the neutral axis at the left cut. See if youcan determine where each component in theequilibrium equation above has come from.15Hide TextWe can eliminate like terms.16Hide Text

Beams II -- Deflections: 5And then divide through by dx .We can eliminate like terms.1718Hide TextHide TextFinally, we take the limit of theexpression as dx goes to zero.And then divide through by dx .19Hide Text20Hide Text

Beams II -- Deflections: 6And we are left with arelationship between shear forceand change in moment.Finally, we take the limit of theexpression as dx goes to zero.2122Hide TextHide TextFormally, we saythat the shear in a beam isequal to the change inmoment along the lengthof the beam.These are two important relationships betweenshear, moment, and external load. We will refer back tothem often in our discussion of stresses in beams.We arrived at these relationships by applyingequilibrium to the free body diagram shown above.23Hide Text24Hide Text

Beams II -- Deflections: 7Here are the shear andmoment diagrams from aprevious example. Confirmfor yourself that the changein the shear equals theloading, and that thechange in the momentequals the shear.We now have relationships between theshear, moment and the externally appliedload. However, we are not yet done.Remember, that our goal was to establish alink between externally applied load and thedisplaced shape of the beam.2526Hide TextCombining the two equations we see that thesecond derivative of the moment with respect to xequals minus the externally applied load.27Hide TextHide TextAt this point we need to recall the relationship between the momentin a beam and the displaced shape, υ(x).If you don't remember how we arrived at this relationship, you maywish to review this information by clicking the M EI v'' button.28Hide TextM EI v"

Beams II -- Deflections: 8We can further simplify the load-displacementrelationship if we know that E and I are constant alongthe length of the beam.We can combine these two equationsto arrive at the coveted load-displacementrelationship for beams.29Hide Text30Hide TextThe Governing Differential EquationThis last equation is the one most commonlyreferred to as the governing equation for beams.Never use it blindly. Always confirm that I for thebeam is constant.31Hide TextThe Whole StoryThe governing equation for beam deflections, shown at the top, is a fourth orderdifferential equation. The four integrations needed to calculate the deflections of the beamare shown below the governing equation. Note the result of each integration is related to aparticular property of the beam's internal loading or shape. Refer back to this figure if youare unsure at what step the beam equation must satisfy a certain boundary condition.32Hide Text

Beams II -- Deflections: 9Determining Displaced Shapes: 1Determining Displaced Shapes: 2There are two ways we can use the previously derived relationships to calculate abeams displaced shape from its loading. The first method is outlined here.The procedure begins by determining the function which defines moment in thebeam as a function of position, M(x). To do this, use your favorite (or the easiest)method to calculate the moment diagram for the beam. Note that (1) this method isonly appropriate for statically determinate beams, and (2) if you have point loads onthe beam, the function M(x) will have kinks.Once you establish M(x), integrate the function twice. Don't forget the integrationconstants that come with each indefinite integration!Finally apply the two displacement boundary conditions for the beam. For33The second method for determining beam deflections involvesintegrating the "beam equation" four times. The four integrations willresult in four unknown integration constants. To solve for theconstants, apply the four boundary conditions for the beam.The four integration constants come from the fact that every beamhas two ends, and each end has two boundary conditions.34Hide TextHide TextBoundary ConditionsIt was just stated that we canread two boundary conditions fromany end of a beam. Shown at theright are the three common endconditions for beams; fixed, simple,35Hide TextBoundary ConditionsAt a fixed support we know that thedeflection of the beam is zero and the slope ofthe beam is zero.36Hide Text

Beams II -- Deflections: 10Boundary ConditionsBoundary ConditionsAt a simple support, the deflectionof the beam is zero and the moment inthe beam is zero.37At a free end, themoment in the beam iszero, and the shear in thebeam is zero.38Hide TextHide TextOther Supports/ConnectionsSummaryThe key to determining displacements in beams lies inintegrating simple differential equations: the tedious part ishandling boundary conditions and non-continuous loads. Insubsequent courses you will learn many methods for findingdisplacements more conveniently, but you already have thecomplete theory under your belt: EIv'''' w(x).Other supports which you mightencounter when analyzing a beam are thepin support in the middle of a beam, andthe internal hinge.At an internal pin support the deflectionmust be zero. At an internal hinge, themoment must be zero.39Hide Text40Hide Text

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deflection of the beam is zero and the slope of the beam is zero. Beams II -- Deflections: 9 . At a simple support, the deflection of the beam is zero and the moment in the beam is zero. Boundary Conditions Hide Text 38 At a free end, the moment in the beam is zero, and