Chapter 4 - Beam Deflections - Weebly

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Chapter 4Beam DeflectionsThis chapter will discuss various methods to determine the deflection and slope atthe specific points in determinate beam. The methods include the DoubleIntegration method and Macaulay method as well as Moment Area method.After successfully completing this chapter you should be able to: Develop the general equation for the elastic curve of a deflected beam byusing double integration method and area-moment method.State the boundary conditions of a deflected beamDetermine the deflections and slopes of elastic curves of simply supportedbeams and cantilever beams.4.1IntroductionWhen a structure is placed under load it will bend, deflect or displace. Thedeflection will depend on the following factors:1.2.3.4.Geometry of the structure, including shape and flexural rigidity ofmember.Flexibility/rigidity of the material used.Restraint of the supports.Load pattern.In the design of structures the primary requirement is to ensure that the structureor structural component can adequately resist the loading to which it is beingsubjected. This aspect of design is concerned with designing for strength.However there are other aspects of design that are important, and in the case ofthe design of beams another consideration is the value of the vertical deflectionsthat will occur when such beam is loaded. This chapter is intended as anintroduction to the analytical techniques used for calculating deflections in beamsand also for calculating the rotations at critical locations along the length of abeam.4.2Deflection of BeamsThe deformation of a beam is usually expressed in terms of its deflection from itsoriginal unloaded position. The deflection is measured from the original neutralsurface of the beam to the neutral surface of the deformed beam. The97

configuration assumed by the deformed neutral surface is known as the elasticcurve of the beam as shown in Figure 4.1.Figure 4.1: Elastic curveNumerous methods are available for the determination of beam deflections.These methods include:1.2.3.4.5.Double Integration MethodMoment Area MethodStrain Energy Method (Castigliano’s Theorem)Conjugate Beam MethodMethod of SuperpositionOf these methods, the first two are the ones that are commonly used. Therefore,this chapter will be only focus on the first two methods.4.3Double Integration MethodThe Double Integration Method, also known as Macaulay’s Method is a powerfultool in solving deflection and slope of a beam at any point because we will beable to get the equation of the elastic curve.In calculus, the radius of curvature of a curve y f(x) is given byThe radius of curvature of a beam is given asDeflection of beams is so small, such that the slope of the elastic curve dy/dx isvery small, and squaring this expression the value becomes practically negligible,hence98

Thus, EI / M 1 / y''If EI is constant, the equation may be written as:where x and y are the coordinates shown in the Figure 4.1 of the elastic curve ofthe beam under load, y is the deflection of the beam at any distance x. E is themodulus of elasticity of the beam, I represent the moment of inertia about theneutral axis, and M represents the bending moment at a distance x from the endof the beam. The product EI is called the flexural rigidity of the beam.The first integration y' yields the slope of the elastic curve and the secondintegration y” gives the deflection of the beam at any distance x. The resultingsolution must contain two constants of integration since EI y" M is of secondorder. These two constants must be evaluated from known conditions concerningthe slope deflection at certain points of the beam. For instance, in the case of asimply supported beam with rigid supports, at x 0 and x L, the deflection y 0,and in locating the point of maximum deflection, we simply set the slope of theelastic curve y' to zero.4.3.1 Boundary ConditionsGenerally, the deflections is known as y-values and slopes is known asdy. Thedxvalues are called boundary conditions, which normally are:1.2.For simply supported beams:(i)At the x-values of the two supports, deflection is zero, i.e. y 0.(ii)If the point (i.e. the x-value) of maximum deflection is known,dythen at the x-value of the point, the slope is zero, i.e. 0.dxFor cantilever beams, at the x-value of the built-in end:(i)The deflection is zero, i.e. y 0.dy(ii)The slope is zero, i.e. 0.dx99

Example 4.1Determine the maximum deflection δ in a simply supported beam of length Lcarrying a concentrated load P at midspan.Solution:At x 0, y 0, therefore, C2 0At x L, y 0Thus,Maximum deflection will occur at x ½ L (midspan)100

The negative sign indicates that the deflection is below the undeformed neutralaxis.Therefore,Example 4.2Find the equation of the elastic curve for the cantilever beam shown in Fig. E4.2;it carries a load that varies from zero at the wall to wo at the free end. Take theoriginatthewall.Figure E4.2Solution:By ratio and proportion101

At x 0, y' 0, therefore C1 0At x 0, y 0, therefore C2 0Therefore, the equation of the elastic curve isExample 4.3Compute the value of EI δ at midspan for the beam loaded as shown in Fig. E4.3.If E 10 GPa, what value of I is required to limit the midspan deflection to 1/360of the span?Figure E4.3Solution:102

At x 0, y 0, therefore C2 0At x 4 m, y 0Therefore,At x 2 m (midspan)Maximum midspan deflectionThus,103

orExample 4.4For the beam loaded as shown in Fig. E4.4, determine (a) the deflection andslope under the load P and (b) the maximum deflection between the supports.Figure E4.4Solution:104

At x 0, y 0, therefore C2 0At x a, y 00 -[ b / (6a) ] Pa3 aC1C1 (ab/6)PTherefore,Part (a): Slope and deflection under the load PSlope under the load P: (note x a b L)Deflection under the load P: (note x a b L)105

Part (b): Maximum deflection between the supportsThe maximum deflection between the supports will occur at the pointwhere y' 0.At y' 0, ⟨ x - a ⟩ do not exist thus,At,106

Example 4.5Determine the value of EIy midway between the supports for the beam loaded asshown in Fig. E4.5.Figure E4.5Solution:At x 0, y 0, therefore C2 0At x 6 m, y 00 50(63) - 900(42) - (25/3)(24) 6C1107

C1 5600/9 N·m3Therefore,At x 3 m4.4Moment Diagrams by PartsThe moment-area method of finding the deflection of a beam will demand theaccurate computation of the area of a moment diagram, as well as the moment ofsuch area about any axis. To pave its way, this section will deal on how to drawmoment diagrams by parts and to calculate the moment of such diagrams aboutaspecifiedaxis.4.4.1 Basic Principles1. The bending moment caused by all forces to the left or to the right of anysection is equal to the respective algebraic sum of the bending momentsat that section caused by each load acting separately.2. The moment of a load about a specified axis is always defined by theequation of a spandrelwhere n is the degree of power of x.The graph of the above equation is as shown below108

and the area and location of centroid are defined as follows.4.4.2 Cantilever LoadingsA area of moment diagramMx moment about a section of distance xbarred x location of centoidDegree degree power of the moment diagramCouple or Moment LoadDegree: zeroConcentrated LoadDegree: firstUniformly Distributed LoadDegree: second109

Uniformly Varying LoadDegree: thirdExample 4.6For the beam loaded as shown in Fig. E4.6, compute the moment of area of theM diagrams between the reactions about both the left and the right reaction.Figure E4.6Solution:Moment diagram by parts can be drawn in different ways; three are shown below.110

1st Solution:2nd Solution:111

Example 4.7For the beam loaded as shown in Fig. E4.7, compute the moment of area of theM diagrams between the reactions about both the left and the right reaction. (Hint:Draw the moment diagram by parts from right to left.)Figure E4.7112

113

4.5Area-Moment MethodAnother method of determining the slopes and deflections in beams is the areamoment method, which involves the area of the moment diagram.4.5.1 Theorems of Area-Moment MethodTheorem I:The change in slope between the tangents drawn to the elastic curve at any twopoints A and B is equal to the product of 1/EI multiplied by the area of themoment diagram between these two points.Theorem II:The deviation of any point B relative to the tangent drawn to the elastic curve atany other point A, in a direction perpendicular to the original position of the beam,is equal to the product of 1/EI multiplied by the moment of an area about B of thatpart of the moment diagram between points A and B.and114

Rules of Sign:1. The deviation at any point is positive if the point lies above the tangent,negative if the point is below the tangent.2. Measured from left tangent, if θ is counterclockwise, the change of slope ispositive, negative if θ is clockwise.Example 4.8The cantilever beam shown in Fig. E4.8 has a rectangular cross-section 50 mmwide by h mm high. Find the height h if the maximum deflection is not to exceed10 mm. Use E 10 GPa.Figure E4.8Solution:115

Example 4.9For the cantilever beam shown in Fig. E4.9, what is the force (P) will cause zerodeflection at A?Figure E 4.9Solution:116

Example 4.10Compute the deflection and slope at a section 3 m from the wall for the beamshown in Figure E4.10. Assume that E 10 GPa and I 30 106 mm4.Figure E4.10Solution:117

Therefore:4.6Deflections in Simply Supported Beams Using Area-Moment MethodThe deflection δ at some point B of a simply supported beam can be obtained bythe following steps.1. Compute2. Compute3. Solve δ by ratio and proportion (see figure above).118

Example 4.11Compute the midspan value of EIδ for the beam shown in Fig. E4.11. (Hint: Drawthe M diagram by parts, starting from midspan toward the ends. Also takeadvantage of symmetry to note that the tangent drawn to the elastic curve atmidspanishorizontal.)Figure E4.11Solution:By symmetry:From the figure,119

ThusExample 4.12A simple beam supports a concentrated load placed anywhere on the span, asshown in Fig. E4.12. Measuring x from A, show that the maximum deflectionoccurs at x [(L2 - b2)/3].Figure E4.12Solution:120

From the figure:121

Tutorial ProblemsP4.1 A simply supported beam is loaded as shown in Fig. P4.1. Determine theEIy value of the beam at point C.Figure P4.1Answer:-461.33 Nm3P4.2 A simply supported beam, of diameter 12 mm, is loaded as shown in Fig.P4.2. Using double integration method, determine the location of themaximum deflection and magnitude of the deflection. Modulus of elasticityis 200 GPa for the beam material.Figure P4.2Answer:-1.1263 mm at 0.9 mP4.3 A cantilever beam AD carries a moment and a uniformly distributed loadas shown in Fig. P4.3. Determine the deflection at the free end A. Modulusof elasticity is 200 GPa and I is 491 x 10-8 m4 for the beam.Figure P4.3122

Answer:2.35 mmP4.4 A cantilever beam is loaded as illustrated in Fig. P4.4. Determine thedeflection at the free end. E 200 GPa, I 2.5 x 10-6 m4.Figure P4.4Answer:4.99 mm123

Integration method and Macaulay method as well as Moment Area method. 98 configuration assumed by the deformed neutral surface is known as the elastic curve of the beam as shown in Figure 4.1. Figure 4.1: Elastic curve Numerous methods are available for the determination of beam deflections. These methods include: 1. Double Integration Method 2. Moment Area Method 3. Strain Energy Method .