276 Calculus and Structures

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1 76 Calculus and Structures

2 CHAPTER THE CONJUGATE BEA ETHOD Calculus and Structures 77 Copyright

3 Chapter THE CONJUGATE BEA ETHOD.1 INTRODUCTION To find the deflection of a beam you must solve the equation, d y dx. Two integrations will solve this problem as was shown in Chapter 19. But there is another method that avoids having to carry out this involved computation when the beam that we are considering is a free standing beam. First solve for the bending moment of the beam and then consider a new beam problem called the conjugate beam problem in which the old bending moment divided by is taken to be the new load (x) on the beam. The bending moment of this new beam problem turns out to be the deflection.. THE CONJUGATE BEA ETHOD et s see how the conjugate beam method works for the familiar free standing beam problem in which a constant density of 1 lb/ft acts on a 1 foot beam as shown in Fig. 1 1lb/ft Fig. 1 By an analysis that we have carried out several times in this book, V 1x, x 1 where V when x =, and (1) x x, x 1 where max 1 at x =. The bending moment diagram is shown in Fig.. 78 Calculus and Structures

4 Section. 1 = = x x 1 X R 1 R Fig. However we now consider the bending moment,, divided by to be the load x x of a new beam problem. We can then compute the bending moment of this new beam in which case its bending moment will represent the deflection of the original beam problem shown in Fig. 1. However, we must first determine the total weight W acting on the beam in order to compute the resistance forces R 1 and R. 1 1 W ( x) dx (x x ) dx 3 1 (x x ) Therefore, R1 R W () 6 We are generally only interested in the maximum bending moment which we can compute by statics. Consider this conjugate beam redrawn in Fig = x x 1 X 6 6 Fig. 3 Calculus and Structures 79

5 Chapter THE CONJUGATE BEA ETHOD To find the bending moment at x = calculate the moment about point O at x = (see Fig. 4). 8 ( ) 6 6 V= Fig. 4 Since is a maximum at x =, V = at that point so that the sum of the moment about O is, ( )() or 13 (3) 6 8 Note that we have used the fact that the centroid of a right parabola is =. x c where 8 This is the identical result that we obtained in Chapter 19 with a great deal of more effort. Problem: Apply the conjugate beam method to the following beam (Fig. ), 1 lb Fig. 8 Calculus and Structures

6 Section. where the usual analysis yields, V x (4a), x 1 and, x x (4b) x, x 1 The bending moment diagram is shown in Fig R = PUTTING IT A TOGETHER R = 6 Fig. 6 et us now test the theory that we have developed throughout this book. Consider the beam in Fig. 7 in which a person weighing W lbs. with a carriage of a inches sits upon a beam of length in., width b, and depth d. We shall assume that the beam is free standing, and we wish to determine the shear force, bending moment, maximum stress on the beam, and maximum deflection. From the maximum stress we will predict if the beam will rupture. a Fig. 7 Calculus and Structures 81

7 Chapter THE CONJUGATE BEA ETHOD We have solved many problems of this type before. However, now we are not specifying numbers but developing the solution in terms of the parameters, W, a, and. The maximum stress, max, is determined by Eq. 3 of Chapter, while the maximum deflection is determined by the conjugate beam method. I will give the results and leave it to the student to verify the results as an exercise. First we compute V and. W V, x a (4a) = W ( x), a x a 4a W =, a x W x x a (4b) W W ( a) [( x) 4a, a x a 4 16a = W ( x), a x The maximum bending moment clearly occurs at / and has the value, max W ( a) ( ). 4 From Eq. 3 of Chapter, the maximum stress of a beam of width b and depth d is, max 6 max 3W ( a) d b d b Finally, the maximum depth is determined from the conjugate beam method. Again, the maximum deflection, y max, occurs at the midpoint of the beam, /. Its value is the bending moment obtained from a density distribution of,. 8 Calculus and Structures

8 Section.3 d 3 b For a wooden beam with rectangular cross section from Eq. 4 of Chapter 19, I. 1 and letting E = 1,9, lbs. /in, The maximum deflection occurs at the midpoint of the beam by the statics method applied to the free body diagram in Fig. 8. Note: V= when x= x c m The equation of moments about point O at x = in Fig. 8 is, m Fig. 8 y m max x, (7) c where m is the weight of the conjugate beam in which the density, of m is the integral of the density over the interval [, ], i.e., m ( ) dx (8) where is given by Eq. 4b. After a lengthy calculation we have found that,. The value W m (9) 16 Unfortunately it is very complicated to compute the center of gravity for this portion of the beam. Instead we will compute an approximate value. The portion of the beam over the interval [,/] has two extreme positions: a) a = in which case the moment curve is a straight line over the interval x /. In this case x ( c ) (see Fig. 9b) ; or 3 b) when a = / in which case the moment curve is a parabola over the interval x / with a maximum value at x = /. In this case x ( c ) (see Fig. 9c). 8 If neither of these extreme cases hold, the interval x ( a) will be a straight line while the remaining part of the interval, ( a) x /, will be a parabola (see Fig. 9c). We will take the approximation to the center of gravity, Calculus and Structures 83

9 Chapter THE CONJUGATE BEA ETHOD x c ( a) a a (1) 4 x c = 3 ( ) Fig. 9a x c = 8 ( ) Fig. 9b - a x c 8 - a x c= 4 Fig. 9c 84 Calculus and Structures

10 Section.3 You will see from Eq. 1 that for extreme cases a) and x ( c ) 3 a and x c and; b) a = 8. This value of x c was used in Eq. 7 to compute y max. The result is, y max W (8 a) (11) 3 3Ebd.4 CONCUSION This method works whenever the beam is supported by hinges and rollers, i.e., a free standing beam. However, the method can also be adapted to solve for the deflection of cantilever beams or beams with hinges. However, we will not consider these beams in this book. Calculus and Structures 8

Calculus and Structures

Calculus and Structures CHAPTER 8 SHEAR FORCE AND BENDING MOMENTS FOR BEAMS WITH CONTINUOUS FORCES Calculus and Structures 11 Copyright Chapter 8 CONTINUOUS FORCE 8.1 INTRODUCTION The last section was