of 26

• date post

02-Dec-2014
• Category

## Documents

• view

125

12

TAGS:

Embed Size (px)

### Transcript of Conjugate Beam

UNIT 2

CONJUGATE BEAM METHODOBJECTIVES At the end of this unit, students are supposed to; 1. Understand Mohrs first and second conjugate beam theorem. 2. Be able to convert real beams into conjugate beams 3. Be analyse beams and frames using the conjugate beam method4.

2.1 INTRODUCTIONIn the course module BT 254, we were introduced to the concept of Moment Area Method as a means of calculating the slope and the deflection of beams at various sections. The concept was based on the theorems by Castigliano, which stated that; the change in slope between two points on a straight member under flexure is equal to the area of the diagram between those two points. That is, the slope between the point A and B on beam is described mathematically as;

Thus, in finding the magnitude of one of the slope, say slope at A, the value of the slope at B should be known before hand or should be a point on the beam with a slope value of zero.

2.2 DERIVATION OF THEOREMThe conjugate beam theorem is derived from the moment area theorems. It has the advantage of being able to calculate for the slope in a member even when they are no points on the beam where the slopes are zero. To illustrate the theorem, we will consider the simply supported beam which is uniformly loaded as shown in Fig. 2.0(a) below. The the elastic curve for the beam is show in Fig. 2.0(b) and (c) respectively. diagram and

CONJUGATE BEAM METHODW A x LFig 2.1(a)

W C B

RA

Fig 2.1(b)

RB

x AA

C C2 C1B

B

Fig 2.1(c)

From Fig 1.0 the change in slope between A and C is given by;

The slope at C is thus found as

It can be seen that,

Therefore It can be seen that deflection at C is;

12 | P a g e

CONJUGATE BEAM METHOD

Lets look at a case of a similar beam loaded with the

diagram and having the span equal to

the above considered beam. The reaction at the point A would be given by,

And the shear force at C would be given by,

It can be observed that the slope at C, in the beam earlier considered is equal to the shear diagram. And that the

force in the latter beam considered which was loaded with the

deflection at C in the earlier beam is equal to the bending moment in the latter beam with the diagram as its load. The latter beam considered is the conjugate beam. The about discussion is the result of two theorems; Theorem I: The rotation/slope at a point in a beam is equal to the shear force in the conjugate beam Theorem II: The deflection in a beam is equal to the bending moment in the conjugate beam. The generalizations above are made on the premise that the beam is simply supported. Students are being advised, however, to make suitable changes considering the boundary conditions for different beams to see if the above theorem would hold. The conjugate beam can thus be defined as; An imaginary beam with a span which is equal to the span of the original beam and is loaded with the diagram of the original beam, in a way that the shear force and the bending moments at a section of the conjugate beam represents slope and the deflection in the original beam. It is worth noting that in the original beam with a fixed end, the slope and the deflection is zero. Thus in the corresponding conjugate beam, the shear and the bending moment would be equally zero. These conditions exist in a beam with a free end, as such; a fixed support in an

13 | P a g e

CONJUGATE BEAM METHODoriginal beam connotes a free end/support in a conjugate beam. The opposite holds in both cases. As in an original beam with a free end, the slope and the deflection at end is note equal to zero. Similarly, in the conjugate beam, the shear and the bending moment at the support would not be zero. This indicates a support which is fixed; since its only a fixed support that resists both rotation and translation. In this case a free end/support in an original beam will give e fixed end/support in the conjugate beam. There are various end conditions and there corresponding end conditions in the conjugate beam. These are illustrated in Table 2.1 below. Also original beams and their corresponding conjugate beams are illustrated in Table 2.2.

SING CONVENTION: The sign convention that has been adopted for the purpose of this chapter is as followed. 1. Sagging moment are positive 2. Left side upwards force or right side downwards force () gives positive shear, Thus positive shear gives clockwise rotation and positive moment gives downward deflection

CONJUGATE BEAM METHODTable 2.1 SI NO ORIGINAL BEAM CONJUGATE BEAM

0 1

0

0

0

Simply supported or Roller end

Simply supported or Roller end

0 2 Hinged end

0

0 Hinged end

0

3

0

0

0

0

Fixed end

Free end

0 0 4 Free end

0 0 Fixed end

0 0 5 Interior support

0 0

Interior hinge

0 0 6 Interior hinge

0 0

Interior support

15 | P a g e

CONJUGATE BEAM METHODTable 2.2 SI NO ORIGINAL BEAM CONJUGATE BEAM

1

2

3

4

5

6

7

8

16 | P a g e

CONJUGATE BEAM METHODconjugate beam. As would be realised from Table 2.1 and 2.2, the reaction in the conjugate beam do sometimes differ from that of the original beam. Again the condition at the support of the original beam differs from the conditions at the support of the conjugate beam. Whereas the condition that might exist at the support of an original beam would be either a translation or movement in any of the two perpendicular plane which is termed as the deflection in that particular plane and a rotation about a given point the same when measured in the conjugate beam is reckoned as the shear and the moment respectively. That is to say that whenever the slope of the original beam is measures or calculated for, the same passes as the shear in the conjugate beam. And whenever the deflection of the original beam is measured it passes as the moment in the conjugate beam. At this moment it is worth saying that a slope in the original beam is equal to a shear in the conjugate beam. And a deflection in the original beam is a moment in the conjugate beam. To throw a little more light on this lets consider Table 2.1. The table has two main columns, one corresponding to original beam configurations and their respective end conditions and the other, conjugate beam configuration and their end conditions for their respective original beams in the first column. It is worth noting these facts, considering loadings in the y-plane only; a. A simple support does not allow for translation in that plane (y-plane), hence deflection in the y-plane is zero. But then the beam can rotate about that reaction since a simple support does not have the capacity to resist rotation. In which case there is going to be a slope between the beam at its original position and the beam in its deflected position. This can be summarised that at a simple support deflection is zero but slope is not equal to zero. b. A free end of a beam does allow for both translation in the y-plane and rotation about that point. Hence at the free end of a beam there exists a slope as much as there is a rotation. Thus in summary, at the free end of a beam slope is not equal to zero and deflection is not equal to zero c. The fixed end of a beam or a beam with fixed supports has the capacity to resist both rotation and translation within any plane. Since the beam cannot translate at that support, deflection at that support is zero. Because the beam does not allow for rotation at that support as well, the deflected beam would lie parallel the profile of the beam before deflection, in which case slope would be equal to zero. It can thus be said of the fixed support that it has a zero slope and a zero deflection.

2.4 PROCEDUREIn solving problems related to the conjugate beam method, the following step should be followed, 1. Solve for the reactions of the original beam; that is determining the values of the