工程力学英文版课件13DeflectionofBeamsDuetoBending.ppt

1、1,Deflection of Beams Due to Bending,Engineering Mechanics,2,151 Slope and Deflection of Beams 152 Deflection of Beams by Integration 153 Method of Superposition 154 Statically Indeterminate Beams,Bending and Shearing Stresses in Beams,Deflection of Beams Due to Bending,3,15-1 Slope and Deflection o

2、f Beams,Introduction,Having studied the stresses set up in bending, we now turn to the equally important aspect of beam stiffness. In many structural elements, such as floor joists or aircraft wings, the limiting constraint on the design is stiffness. Any design which is stiff enough will be strong

3、enough.,Deflection of Beams Due to Bending,4,The x axis extends positive to the right, along the initially straight longitudinal axis of the beam. The w axis extends positive upward from the x axis. It measures the deflection of the centroid on the cross-sectional area of the element.,Deflection of

4、Beams Due to Bending,5,The slope is the turning angle of the cross-sectional area. The positive slope angle will be measured counterclockwise from the x axis when x is positive to the right. The slope can be determined from dw/dx.,Deflection of Beams Due to Bending,6,It is important that we should b

5、e able to calculate the deflection of a beam of given section, since for given conditions of span and load it would be possible to adopt a section which would meet a strength criterion but would give an unacceptable deflection. The total deflection of a beam is due to a very large extent to the defl

6、ection caused by bending, and to a very much smaller extent to the deflection caused by shear.,Deflection of Beams Due to Bending,7,15-2 Deflection of Beams by Integration,From the previous chapter, we have: If the material is homogeneous and behaves in a linear-elastic manner, then Hookes law appli

7、es, . Also, since the flexure formula applies, . Combining these equations and substituting into the above equation, we have: (1),Deflection of Beams Due to Bending,8,The elastic curve for a beam can be expressed mathematically as w = f (x). To obtain this equation, we must first represent the curva

8、ture in terms of w and x. In most calculus books it is shown that this relationship is,Deflection of Beams Due to Bending,Substituting into equation (1), we get (2),9,Most engineering design codes specify limitations on deflections for tolerance or esthetical purposes, and as a result the elastic de

9、flections for the majority of beams and shafts form a shallow curve. Consequently, the slope of the elastic curve which is determined from dw/dx will be very small, and its square will be negligible compared with unity. Using this simplification, equation (2) can now be written as (3),Deflection of

10、Beams Due to Bending,10,If EI is constant, rewriting equation (3), we have (4) Integrating equation (4) twice yields,Deflection of Beams Due to Bending,11,Therefore, the equation of the slope and elastic curve for the beam are,Deflection of Beams Due to Bending,Where c1 and c2 are constant of integr

11、ation. They will be determined by using the boundary and continuity conditions.,12,Boundary and Continuity Conditions,The constants of integration are determined by evaluating the functions for shear, moment, slope, or displacement at a particular point on the beam where the value of the function is

12、 known. These values are called boundary conditions. If a single x coordinate cannot be used to express the equation for the beams slope or the elastic curve, the continuity conditions must be used to evaluate some of the integration constants.,Deflection of Beams Due to Bending,13,Example 1 The can

13、tilevered beam shown in figure is subjected to a vertical load P at its end. Determine the equation of the elastic curve. EI is constant.,Moment Function. From the free-body diagram, with M acting in the positive direction, we have,Deflection of Beams Due to Bending,14,Slope and Elastic Curve. Apply

14、ing equation (4) and integrating twice yields,(a),(b),(c),Deflection of Beams Due to Bending,15,Using the boundary conditions dw / dx = 0 and w = 0 at x = L, equation (b) and (c) become,Thus, c1= PL2/2 and c2= -PL3/3. Substituting these results into equation (b) and (c), we get,Deflection of Beams D

15、ue to Bending,16,Maximum slope and deflection occur at A (x = 0), for which,Deflection of Beams Due to Bending,17,Example 2 The simply supported beam shown in figure is subjected to the concentrated force P. Determine the maximum deflection of the beam. EI is constant.,Elastic Curve. Two coordinates

16、 must be used, since the moment becomes discontinuous at P.,Here we will take x1 and x2, having the same origin at A, so that,Deflection of Beams Due to Bending,18,Moment Function From the free-body diagrams shown in figure,Deflection of Beams Due to Bending,19,(2-1),(2-2),Slope and Elastic Curve. A

17、pplying equation (4) for M1 and integrating twice yields,Deflection of Beams Due to Bending,20,(2-3),(2-4),Likewise for M2,Deflection of Beams Due to Bending,21,The four constants are evaluated using two boundary conditions, namely, x1= 0, w1= 0 and x2= 3a, w2= 0. Also, two continuity conditions mus

18、t be applied at B, that is, dw1/dx1 = dw2/dx2 at x1=x2=2a and w1=w2 at x1=x2=2a. Substituting these conditions into equations (2-1)-(2-4), and solving these equations we get,Deflection of Beams Due to Bending,22,(2-5),(2-8),Thus equations (2-1)-(2-4) become,(2-6),(2-7),Deflection of Beams Due to Ben

19、ding,23,By inspection of the elastic curve, the maximum deflection occurs at somewhere within region AB. Here the slope must be zero. From equation 2-5,Substituting into equation 2-6,Deflection of Beams Due to Bending,24,PROCEDURE FOR ANALYSIS,The following procedure provides a method for determinin

20、g the slope and deflection of a beam using the method of integration.,Draw an exaggerated view of the beams elastic curve. Recall that zero slope and zero displacement occur at all fixed supports, and zero displacement occurs at all pin and roller supports.,Support Reactions,Deflection of Beams Due

21、to Bending,25,Establish the x and w coordinate axes. The x axis must be parallel to the undeflected beam and can have an origin at any point along the beam, with a positive direction to the right. If several discontinuous loads are present, establish x coordinates that are valid for each region of t

22、he beam between the discontinuities. Choose these coordinates so that they will simplify subsequent algebraic work. The positive w axis should be directed upward.,Deflection of Beams Due to Bending,26,For each region in which there is an x coordinate, express the internal moment M as a function of x

23、. In particular, always assume that M acts in the positive direction when applying the equation of moment equilibrium to determine M = f (x).,Moment Function,Deflection of Beams Due to Bending,Provided EI is constant, apply the moment equation. It is important to include a constant of integration.,S

24、lope and Elastic Curve,27,The constants are evaluated using the boundary conditions for the supports and the continuity conditions that apply to slope and deflection at points where two functions meet. Once the constants are evaluated and substitute back into the slope and deflection equations, the

25、slope and deflection at specific points on the elastic curve can then be determined. Realize that positive values for slope are counterclockwise, and positive deflection is upward.,Deflection of Beams Due to Bending,28,15-3 Method of Superposition,In cases where a beam is subjected to several concen

26、trated forces, couple moments, and distributed loads, the method of integration would be very complicated. If the two requirements, i.e., the load is linearly related to the deflection, and the load is assumed not to change significantly the original geometry of the beam, are satisfied, the principl

27、e of superposition can be applied.,Deflection of Beams Due to Bending,29,As a result, the deflections for a series of separate loadings acting on a beam may be superimposed. For example, if w1 is the deflection for one load and w2 is the deflection for another load, the total deflection for both loads acting together is the algebraic sum w1+w2.,Deflection of Beams Due to Bending,30,15-4 Statically Indeterminate Beams,The additional support reactions on the beam or shaft that are not needed to keep it in stable equilibrium are called redundant. The number of the redundant is ref