Euler–Bernoulli beam theory

EulerBernoulli beam theory1Euler–Bernoulli beam theoryEuler–Bernoulli beam theory (also known asengineer's beam theory or classical beamtheory)[1]

is a simplification of the lineartheory of elasticity which provides a means ofcalculating the load-carrying and deflectioncharacteristics of beams. It covers the case forsmall deflections of a beam which is subjectedto lateral loads only. It is thus a special case ofTimoshenko beam theory which accounts forshear deformation and is applicable for thickbeams. It was first enunciated circa 1750,[2]but was not applied on a large scale until thedevelopment of the Eiffel Tower and theFerris wheel in the late 19th ing these successful demonstrations, itquickly became a cornerstone of engineeringand an enabler of the Second Industrial vibrating glass beam may be modeled as a cantilever beam withacceleration, variable linear density, variable section modulus, some kind ofdissipation, springy end loading, and possibly a point mass at the free onal analysis tools have been developed such as plate theory and finite element analysis, but the simplicity ofbeam theory makes it an important tool in the sciences, especially structural and mechanical yPrevailing consensus is that GalileoGalilei made the first attempts atdeveloping a theory of beams, butrecent studies argue that Leonardo daVinci was the first to make the crucialobservations. Da Vinci lacked Hooke'slaw and calculus to complete thetheory, whereas Galileo was held backby an incorrect assumption he made.[3]The Bernoulli beam is named afterJacob Bernoulli, who made thesignificant discoveries. Leonhard Eulerand Daniel Bernoulli were the first toSchematic of cross-section of a bent beam showing the neutral together a useful theory circa1750.[4]

At the time, science andengineering were generally seen as very distinct fields, and there was considerable doubt that a mathematical productof academia could be trusted for practical safety applications. Bridges and buildings continued to be designed byprecedent until the late 19th century, when the Eiffel Tower and Ferris wheel demonstrated the validity of the theoryon large scales.

EulerBernoulli beam theory2Static beam equationThe Euler-Bernoulli equation describesthe relationship between the beam'sdeflection and the applied load:[5]The curve describes thedeflection of the beam in the

direction at some position (recallthat the beam is modeled as aone-dimensional object). is adistributed load, in other words a forceper unit length (analogous to pressurebeing a force per area); it may be afunction of , , or other that is the elastic modulusBending of an Euler-Bernoulli beam. Each cross-section of the beam is at 90 degrees tothe neutral that is the second moment of area. must be calculated with respect to the centroidal axis perpendicular tothe applied loading. For an Euler-Bernoulli beam not under any axial loading this axis is called the neutral , EI is a constant, so that:This equation, describing the deflection of a uniform, static beam, is used widely in engineering practice. Tabulatedexpressions for the deflection for common beam configurations can be found in engineering handbooks. For morecomplicated situations the deflection can be determined by solving the Euler-Bernoulli equation using techniquessuch as the "slope deflection method", "moment distribution method", "moment area method, "conjugate beammethod", "the principle of virtual work", "direct integration", "Castigliano's method", "Macaulay's method" or the"direct stiffness method".Sign conventions are defined here since different conventions can be found in the literature.[5]

In this article, a righthandedcoordinate system is used as shown in the figure, Bending of an Euler-Bernoulli beam. In this figure, the xand z direction of a right handed coordinate system are shown. Since where , , and areunit vectors in the direction of the x, y, and z axes respectively, the y axis direction is into the figure. Forces acting inthe positive and directions are assumed positive. The sign of the bending moment is positive when the torquevector associated with the bending moment on the right hand side of the section is in the positive y direction (i.e. sothat a positive value of M leads to a compressive stress at the bottom fibers). With this choice of bending momentsign convention, in order to have , it is necessary that the shear force acting on the right side ofthe section be positive in the z direction so as to achieve static equilibrium of moments. To have force equilibriumwith , q, the loading intensity must be positive in the minus z direction. In addition to these signconventions for scalar quantities, we also sometimes use vectors in which the directions of the vectors is made clearthrough the use of the unit vectors, , , and .Successive derivatives of have important meanings where is the deflection in the z direction:•••is the the slope of the the bending moment in the beam.

EulerBernoulli beam theory3•is the shear force in the stresses in a beam can be calculated from the above expressions after the deflection due to a given load has tion of bending moment equationBecause of the fundamental importance of the bending moment equation in engineering, we will provide a shortderivation. The length of the neutral axis in the figure, Bending of an Euler-Bernoulli beam, is . The length ofa fiber with a radial distance, e, below the neutral axis is

The stress of this fiber is

Hooke's Law. The differential force vector,

. Therefore the strain of this fiber iswhere E is the elastic modulus in accordance withThisresulting from this stress is given by,

is the differential force vector exerted on the right hand side of the section shown in the figure. We know that it is inthe direction since the figure clearly shows that the fibers in the lower half are in tension. is the differentialelement of area at the location of the fiber. The differential bending moment vector,

given byThis expression is valid for the fibers in the lower half of expression for the fibers in the upper half of the beam will be similar except that the moment arm vector will bein the positive z direction and the force vector will be in the -x direction since the upper fibers are in the resulting bending moment vector will still be in the -y direction since Therefore weintegrate over the entire cross section of the beam and get for

cross section of the beam the expressionwhere is the second moment of area. From calculus,we know that when

.is small as it is for an Euler-Bernoulli beam. . Thereforethe bending moment vector exerted on the rightassociated with isDynamic beam equationThe dynamic beam equation is the Euler-Lagrange equation for thefollowing actionFinite element method model of a vibration of awide-flange beam (I-beam).

EulerBernoulli beam theory4The first term represents the kinetic energy where is the mass per unit length; the second one represents thepotential energy due to internal forces (when considered with a negative sign) and the third term represents thepotential energy due to the external load . The Euler-Lagrange equation is used to determine the function thatminimizes the functional . For a dynamic Euler-Bernoulli beam, the Euler-Lagrange equation isDerivation of Euler–Lagrange equation for beamsSince the Lagrangian isthe corresponding Euler-Lagrange equation isNow,Plugging into the Euler-Lagrange equation givesor,which is the governing equation for the dynamics of an Euler-Bernoulli Besides deflection, the beam equation describes forces and moments and can thus be used to describe stresses. Forthis reason, the Euler–Bernoulli beam equation is widely used in engineering, especially civil and mechanical, todetermine the strength (as well as deflection) of beams under the bending moment and the shear force cause stresses in the beam. The stress due to shear force is maximumalong the neutral axis of the beam (when the width of the beam, t, is constant along the cross section of the beam;otherwise an integral involving the first moment and the beam's width needs to be evaluated for the particular crosssection), and the maximum tensile stress is at either the top or bottom surfaces. Thus the maximum principal stress inthe beam may be neither at the surface nor at the center but in some general area. However, shear force stresses arenegligible in comparison to bending moment stresses in all but the stockiest of beams as well as the fact that stressconcentrations commonly occur at surfaces, meaning that the maximum stress in a beam is likely to be at the surface.

EulerBernoulli beam theory5Simple or symmetrical bendingFor beam cross-sections that aresymmetrical about a planeperpendicular to the neutral plane, itcan be shown that the tensile stressexperienced by the beam may beexpressed as:Here, is the distance from theneutral axis to a point of interest; andis the bending moment. Note thatthis equation implies that pure bending(of positive sign) will cause zero stressElement of a bent beam: the fibers form concentric arcs, the top fibers are compressedand bottom fibers the neutral axis, positive (tensile)stress at the "top" of the beam, andnegative (compressive) stress at the bottom of the beam; and also implies that the maximum stress will be at the topsurface and the minimum at the bottom. This bending stress may be superimposed with axially applied stresses,which will cause a shift in the neutral (zero stress) m stresses at a cross-sectionThe maximum tensile stress at a cross-section is at thelocation and the maximum compressive stress is atthe location where the height of the cross-section is. These stresses areThe quantities

defined asare the section moduli[5]

and areThe section modulus combines all the important geometricinformation about a beam's section into one quantity. For thecase where a beam is doubly symmetric, and wehave one section modulus .Quantities used in the definition of the section modulus of abeam.

EulerBernoulli beam theory6Strain in an Euler–Bernoulli beamWe need an expression for the strain in terms of the deflection of the neutral surface to relate the stresses in anEuler-Bernoulli beam to the deflection. To obtain that expression we use the assumption that normals to the neutralsurface remain normal during the deformation and that deflections are small. These assumptions imply that the beambends into an arc of a circle of radius (see Figure 1) and that the neutral surface does not change in length duringthe deformation.[5]Let be the length of an element of the neutral surface in the undeformed state. For small deflections, the elementis the not change its length after bending but deforms into an arc of a circle of radius . If

subtended by this arc, then

Let us now consider another segment of the element at a distance above the neutral surface. The initial length ofthis element is . However, after bending, the length of the element becomes. The strain in that segment of the beam is given bywhere is the curvature of the beam. This gives us the axial strain in the beam as a function of distance from theneutral surface. However, we still need to find a relation between the radius of curvature and the beam deflection

.Relation between curvature and beam deflectionLet P be a point on the neutral surface of the beam at a distance from the origin of the

The slope of the beam, i.e., the angle made by the neutral surface with the -axis, at this point iscoordinate ore, for an infinitesimal element , the relation can be written asHence the strain in the beam may be expressed asStress-strain relationsFor a homogenous linear elastic material, the stress is related to the strain by

modulus. Hence the stress in an Euler-Bernoulli beam is given bywhere is the Young'sNote that the above relation, when compared with the relation between the axial stress and the bending moment,leads toSince the shear force is given by , we also have

EulerBernoulli beam theory7Boundary considerationsThe beam equation contains a fourth-order derivative in . To find a unique solution we need fourboundary conditions. The boundary conditions usually model supports, but they can also model point loads,distributed loads and moments. The support or displacement boundary conditions are used to fix values ofdisplacement ( ) and rotations ( ) on the boundary. Such boundary conditions are also called Dirichletboundary conditions. Load and moment boundary conditions involve higher derivatives of

momentum flux. Flux boundary conditions are also called Neumann boundary an example consider a cantileverbeam that is built-in at one end andfree at the other as shown in theadjacent figure. At the built-in end ofthe beam there cannot be anydisplacement or rotation of the means that at the left end bothdeflection and slope are zero. Since noexternal bending moment is applied atthe free end of the beam, the bendingmoment at that location is zero. Inaddition, if there is no external forceapplied to the beam, the shear force atthe free end is also zero.A cantilever representTaking the coordinate of the leftend as and the right end as (the length of the beam), these statements translate to the following set of boundaryconditions (assume is a constant):A simple support (pin or roller) is equivalent to a point force on the beam which is adjusted in such a way as to fixthe position of the beam at that point. A fixed support or clamp, is equivalent to the combination of a point force anda point torque which is adjusted in such a way as to fix both the position and slope of the beam at that point. Pointforces and torques, whether from supports or directly applied, will divide a beam into a set of segments, betweenwhich the beam equation will yield a continuous solution, given four boundary conditions, two at each end of thesegment. Assuming that the product EI is a constant, and defining where F is the magnitude of a pointforce, and where M is the magnitude of a point torque, the boundary conditions appropriate for somefollowed by that derivative. For example,

at the lower boundary of the upper segment, while is the value of

whereat thecommon cases is given in the table below. The change in a particular derivative of w across the boundary as xincreases is denoted by

is the value of

upper boundary of the lower segment. When the values of the particular derivative are not only continuous across theboundary, but fixed as well, the boundary condition is

separate equations (e.g. = fixed).which actually constitutes two

EulerBernoulli beam theory8BoundaryClampSimple supportPoint forcePoint torqueFree endClamp at endSimply supported endPoint force at endPoint torque at endfixedfixedfixedNote that in the first cases, in which the point forces and torques are located between two segments, there are fourboundary conditions, two for the lower segment, and two for the upper. When forces and torques are applied to oneend of the beam, there are two boundary conditions given which apply at that end. The sign of the point forces andtorques at an end will be positive for the lower end, negative for the upper g considerationsApplied loads may be represented either through boundary conditions or through the function whichrepresents an external distributed load. Using distributed loading is often favorable for simplicity. Boundaryconditions are, however, often used to model loads depending on context; this practice being especially common invibration nature, the distributed load is very often represented in a piecewise manner, since in practice a load isn't typicallya continuous function. Point loads can be modeled with help of the Dirac delta function. For example, consider astatic uniform cantilever beam of length with an upward point load applied at the free end. Using boundaryconditions, this may be modeled in two ways. In the first approach, the applied point load is approximated by a shearforce applied at the free end. In that case the governing equation and boundary conditions are:Alternatively we can represent the point load as a distribution using the Dirac function. In that case the equation andboundary conditions areNote that shear force boundary condition (third derivative) is removed, otherwise there would be a are equivalent boundary value problems, and both yield the solutionThe application of several point loads at different locations will lead to being a piecewise function. Use of the

Dirac function greatly simplifies such situations; otherwise the beam would have to be divided into sections, each

with four boundary conditions solved separately. A well organized family of functions called Singularity functions

EulerBernoulli beam theoryare often used as a shorthand for the Dirac function, its derivative, and its c phenomena can also be modeled using the static beam equation by choosing appropriate forms of the loaddistribution. As an example, the free vibration of a beam can be accounted for by using the load function:9where is the linear mass density of the beam, not necessarily a constant. With this time-dependent loading, thebeam equation will be a partial differential equation:Another interesting example describes the deflection of a beam rotating with a constant angular frequency of :This is a centripetal force distribution. Note that in this case, is a function of the displacement (the dependentvariable), and the beam equation will be an autonomous ordinary differential esThree-point bendingThe three point bending test is a classical experiment in mechanics. It represents the case of a beam resting on tworoller supports and subjected to a concentrated load applied in the middle of the beam. The shear is constant inabsolute value: it is half the central load, P / 2. It changes sign in the middle of the beam. The bending momentvaries linearly from one end, where it is 0, and the center where its absolute value is PL / 4, is where the risk ofrupture is the most important. The deformation of the beam is described by a polynomial of third degree over a halfbeam (the other half being symmetrical). The bending moments ( ), shear forces ( ), and deflections ( )for a beam subjected to a central point load and an asymmetric point load are given in the table below.[5]DistributionSimply supported beam with central loadMax. valueSimply supported beam with asymmetric load

at

EulerBernoulli beam theory10Cantilever beamsAnother important class of problems involves cantilever beams. The bending moments ( ), shear forces ( ),and deflections ( ) for a cantilever beam subjected to a point load at the free end and a uniformly distributed loadare given in the table below.[5]DistributionCantilever beam with end loadMax. valueCantilever beam with uniformly distributed loadSolutions for several other commonly encountered configurations are readily available in textbooks on mechanics ofmaterials and engineering ally indeterminate beamsThe bending moments and shear forces in Euler-Bernoulli beams can often be determined directly using staticbalance of forces and moments. However, for certain boundary conditions, the number of reactions can exceed thenumber of independent equilibrium equations.[5]

Such beams are called statically built-in beams shown in the figure below are statically indeterminate. To determine the stresses and deflectionsof such beams, the most direct method is to solve the Euler–Bernoulli beam equation with appropriate boundaryconditions. But direct analytical solutions of the beam equation are possible only for the simplest cases. Therefore,additional techniques such as linear superposition are often used to solve statically indeterminate beam superposition method involves adding the solutions of a number of statically determinate problems which arechosen such that the boundary conditions for the sum of the individual problems add up to those of the originalproblem.(a) Uniformly distributed load q.(b) Linearly distributed load with maximum q0

EulerBernoulli beam theory11(d) Moment M0(c) Concentrated load PAnother commonly encountered statically indeterminate beam problem is the cantilevered beam with the free endsupported on a roller.[5]

The bending moments, shear forces, and deflections of such a beam are listed butionMax. valueExtensionsThe kinematic assumptions upon which the Euler–Bernoulli beam theory is founded allow it to be extended to moreadvanced analysis. Simple superposition allows for three-dimensional transverse loading. Using alternativeconstitutive equations can allow for viscoelastic or plastic beam deformation. Euler–Bernoulli beam theory can alsobe extended to the analysis of curved beams, beam buckling, composite beams, and geometrically nonlinear –Bernoulli beam theory does not account for the effects of transverse shear strain. As a result it underpredictsdeflections and overpredicts natural frequencies. For thin beams (beam length to thickness ratios of the order 20 ormore) these effects are of minor importance. For thick beams, however, these effects can be significant. Moreadvanced beam theories such as the Timoshenko beam theory (developed by the Russian-born scientist StephenTimoshenko) have been developed to account for these deflections

EulerBernoulli beam theory12The original Euler-Bernoulli theory isvalid only for infinitesimal strains andsmall rotations. The theory can beextended in a straightforward mannerto problems involving moderatelylarge rotations provided that the strainremains small by using the vonKármán strains.[6]The Euler-Bernoulli hypotheses thatplane sections remain plane andnormal to the axis of the beam lead todisplacements of the formEuler-Bernoulli beamUsing the definition of the Lagrangian Green strain from finite strain theory, we can find the von Karman strains forthe beam that are valid for large rotations but small strains. These strains have the formFrom the principle of virtual work, the balance of forces and moments in the beams gives us the equilibriumequationswhere is the axial load, is the transverse load, andTo close the system of equations we need the constitutive equations that relate stresses to strains (and hence stressesto displacements). For large rotations and small strains these relations are

EulerBernoulli beam theorywhere13The quantity

bending the extensional stiffness, is the coupled extensional-bending stiffness, and is theFor the situation where the beam has a uniform cross-section and no axial load, the governing equation for alarge-rotation Euler-Bernoulli beam isNotes[1]Timoshenko, S., (1953), History of strength of materials, McGraw-Hill New York[2]Truesdell, C., (1960), The rational mechanics of flexible or elastic bodies 1638-1788, Venditioni Exponunt Orell Fussli Turici.[3]Ballarini, Roberto (April 18, 2003). "The Da Vinci-Euler-Bernoulli Beam Theory?" (/contents/current/webonly/). Mechanical Engineering Magazine Online. . Retrieved 2006-07-22.[4]Seon M. Han, Haym Benaroya and Timothy Wei (March 22, 1999) (PDF). Dynamics of Transversely Vibrating Beams using fourEngineering Theories (/research/vibration/). final version. Academic Press. . Retrieved 2007-04-15.[5]Gere, J. M. and Timoshenko, S. P., 1997, Mechanics of Materials, PWS Publishing Company.[6]Reddy, J. N., (2007), Nonlinear finite element analysis, Oxford University nces•E.A. Witmer (1991-1992). "Elementary Bernoulli-Euler Beam Theory". MIT Unified Engineering Course . 5–114 to 5–164.

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