Conjugate Beam Method

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Introduction

In the field of Structural Analysis, the deformation of beams under various loading is studied to determine their slope and deflection via a variety of geometric methods. The conjugate beam method is one of these geometric methods. As "Conjugate" means 'related or reciprocal'[1], this method focuses on replacing a real beam with an reciprocal imaginary beam for analysis.

First published by Westergaard in 1921[2], the development or inspiration of "using a conjugate beam" is often accredited to the works of Breslau (1865) and Mohr (1868). The method is an extension of Moment Area Theorem and provides an alternate approach to solve for the slope and deflection in determinate beams. In addition to the moment area theorems, the theory behind the conjugate beam is based around the analogy of the relationships between load,shear, and bending moment and the similar relationships between $ \frac{M}{EI} $, slope and deflection[3]. In other words, the mathematics of proceeding from load to bending moment is near identical to that of proceeding from $ \frac{M}{EI} $ to deflection, meaning the straightforward process of analyzing load, shears and moments on a beam can be adapted for analyzing curvature, slope and deflection on a 'conjugate' beam.

Although it often requires the same level of effort or calculation as the Moment Area Theorem , the Conjugate Beam Method's calculation essentially only involve application of the equations of statics, making this method easier to follow and a favourite among engineering students.

Theory

The main principle being explored with the conjugate beam method is the analogous relationships between load,shear and moments versus the relationships between curvature ($ \frac{M}{EI} $), slope and deflection[3]. As seen in Table 1, both sets are all related by integrals of one another. Due to these similarities, if curvature could be treated as a load, the process to find slope and deflection should be the same as that of finding shear and moments for a load. This is where the conjugate beam method comes in : a "conjugate" (imaginary) beam is created and given the values of $ \frac{M}{EI} $ as a loading (paying close attention to sign convention). For this conjugate beam, the boundary conditions must be changed from the real beam so that shears represent slopes and moments represent deflections. This is done by replacing the real supports with their conjugate counterparts (see Table 2 for details). The idea is to create new support conditions so that the properties of shear and moments on the real beam are the same for slope and deflection on the conjugate beam[4]. With the conjugate beam constructed and $ \frac{M}{EI} $ placed as its respective loading conditions, the beam can now be solved the same as a regular beam, but with the shears and moments found now representing slopes and curvature. See the example below and The Method of Analysis for further details.


Table 1: Relationship between Load-Shear-Bending Moments and Curvature-Slope-Deflection[3]
Load-Shear-Bending Moment Curvature-Slope-Deflection
$ \frac{dS}{dx} = w $ $ \frac{d\theta}{dx} = \frac{M}{EI} $
$ \frac{dM}{dx} = S $ $ \frac{dy}{dx} = \theta $
$ \frac{d^2M}{dx^2} = w $ $ \frac{d^2y}{dx^2} = \frac{M}{EI} $


Rules for Application

In order to apply the conjugate beam method, these conditions must be applied[4]:

Rule 1 The original beam must be statically determinate
Rule 2 The conjugate beam and original beam must be made the same length
Rule 3 The conjugate beam must be in static equilibrium
Rule 4 No other loading should be applied to the conjugate beam other than the elastic weight, $ \frac{M}{EI} $
Rule 5 No other support or boundary conditions are involved with the conjugate beam other than the support reactions obtained from Table 2


Conjugate Supports

To determine the conjugate beam to be used for analysis, the relationship between loads and curvature previously discussed in the theory is applied. That is, that the slope and deflection at a support or connection on the real beam is equal to the shear and moment at the same point on the conjugate beam respectively. Therefore, each support or connection on the real beam will be replaced with the conjugate support or connection that will follow this relationship. For example, at a fixed support both the slope and deflection are zero, meaning the shear and moment on the conjugate beam must both be zero, as such, the fixed support is replaced by a free end in the conjugate beam as a free end experiences a zero shear and moment.[3] Below, Table 2 outlines the supports and connections commonly used when determining the conjugate beam.


Table 2: Real supports and their conjugate counterparts[3]
Real Beam Conjugate Beam
Type of Support Slope and Deflection Shear and Bending Moment Type of support
Simple End Support [5]
$ \theta\;\neq\;0\\\triangle\;=\;0 $ $ S\;\neq\;0\\M\;=\;0 $
Simple End Support [5]
Fixed Support[5]
$ \theta=0\\\triangle=0 $ $ S=0\\M=0 $
Free End[5]
Free End[5]
$ \theta\;\neq\;0\\\triangle\;\neq\;0 $ $ S\;\neq\;0\\M\;\neq\;0 $
Fixed Support[5]
Simple Interior Support[5]
$ \theta\;\neq\;0\;and\\continuous\\\triangle\;=\;0 $ $ S\;\neq\;0\;and\\continuous\\M\;=\;0 $
Internal Hinge[5]
Internal Hinge[5]
$ \theta\;=\;0\;and\\discontinuous\\\triangle\;\neq\;0 $ $ S\;\neq\;0\;and\\discontinuous\\M\;\neq\;0 $
Simple Interior Support[5]
Interior Hinge[5]
$ \theta\;\neq\;0\;and\\discontinuous\\\triangle\;\neq\;0\;and\\continuous $ $ S\;\neq\;0\;and\\discontinuous\\M\;\neq\;0\;and\\continuous $
Interior Hinge[5]

Method of Anaylsis

  1. Draw the $ \frac{M}{EI} $ diagram for the real beam. For a detailed explanation of how this is achieved, see Beam Analysis
  2. Redraw the real beam as the imaginary beam replacing its real supports with conjugate supports. Refer to Table 2 to determine how this is achieved.
  3. Apply the $ \frac{M}{EI} $ diagram as loading on the conjugate beam, paying close attention to the magnitude and direction. The negative portion of the $ \frac{M}{EI} $ diagram is applied as a downward force and the positive portion of the $ \frac{M}{EI} $ diagram is applied as an upward force.
  4. Solve the support reactions on the conjugate beam. For a detailed explanation of how this is achieved, see Calculation of Support Reactions


Once the conjugate beam with its loading has been determined, any slope and/or deflection at any point on the real beam can now be determined. There are two different methods of achieving this.

Method 1
Draw the slope and deflection diagram of the curvature loading on the conjugate beam. These diagrams are drawn similar to shear and moment diagrams. The slope diagram is the integral of the curvature diagram and the deflection diagram is the integral of the slope diagram. Once these have been found, the slope and deflection can be found at any point.
Method 2
Find the internal shear and moment at the point in question on the conjugate beam. Because of the relationship between the load-shear-bending moment and curvature-slope-deflection, the internal shears and moments calculated on the conjugate beam are the slopes and deflection respectfully on the real beam.

Example

Find the slope and deflection of point A in Figure (a) using Method 1. The beam is aluminium therefore: E = 70 GPa and I = 90 x 106 mm4.


Figure a.PNG

As stated in the Method of Analysis, the first step is to solve the support reactions. As the beam has a hinge, it can be separated into two section: section 1-A and section A-2.

By first looking at section A-2, the vertical reaction can be found by taking a moment at A, with the counterclockwise direction as the positive direction.


$ \sum M_a = 0 $

$ F_2\times5 - 20\times4 = 0 $

$ F_2 = 16 kN $


Looking at the entire structure: by summing the forces in the y-direction, $ F_1 $ can be found; by taking the moment at Support 1, $ M_1 $ can be found.


$ \sum F_y = 0 $

$ -10 - 20 + 16 + F_1 = 0 $

$ F_1 = 14 kN $


$ \sum M_1 = 0 $

$ -10\times3 - 20\times8.5 = F_2\times9.5 - M_1 $

$ M_1 = - 48 kNm $


The completed diagram of the support reactions as well as the shear and curvature diagram can be found below in Figure (b), (c) and (d) respectively. For the method on how the shear and curvature diagram was calculated, refer to Beam Analysis.

Figure bcd.PNG


The deflected shape in Figure (e) below shows the slope and deflection for point A. Because there is a hinge at A, the slope will be discontinous thus it must be found on the left and right side of A.

Figure e.PNG


Figure (f) shows the imaginary beam with the conjugate supports. The fixed end becomes a free end, the internal hinge becomes a fixed hinge and the hinge becomes a roller. Furthermore, the curvature diagram of the real beam is applied as loading on the conjugate beam. The negative portion is applied as a negative load and vice versa.

Figure f.PNG


Figure (g) below shows the imaginary beam with the conjugate supports. The fixed end becomes a free end, the internal hinge becomes a fixed pin support and the hinge support becomes a roller to make the structure statically determinate. Figure (h) and (i) show the slope and deflection diagram for the conjugate beam.

The support reactions for the conjugate beam are calculated using methods described in Calculation of Support Reactions. From this, the slope and deflection diagrams (Figure (h) and (i)) can be found using Beam Analysis.

The following equations in Table 3 below were used to determine key values on seen in Figure (h) and (i). The equation of each different loading pattern on the conjugate beam (Figure (g)) was determined - by simply using a $ y = mx + b $ approach - and then integrated once for the slope diagram and a twice for the displacement diagram.


Table 3: Intermediate equations
Distance along beam Slope equation Displacement equation
$ 0\leq x \leq3\ $ $ \theta=7x^{2}-48x $ $ y=2.33x^{3}-24x^{2} $
$ 3\leq x \leq4.5\\ where \ p=x-3 \\ therefore \ 0\leq p \leq1.5 $ $ \theta=2p^{2}-6p $ $ y=\frac{2}{3}p^{3}-3p^{2} $
$ 4.5\leq x \leq8.5 \\ where \ p=x-4.5 \\ therefore \ 0\leq p \leq4 $ $ \theta=2p^{2} $ $ y=\frac{2}{3}p^{3} $
$ 8.5\leq x \leq9.5\\ where \ p=x-8.5 \\therefore \ 0\leq p \leq1 $ $ \theta=-8p^{2}+16p $ $ y=-2.67p^{3}+8p^{2} $


Figure ghi.PNG


To determine the slope and deflection at point A, the slope and deflection diagrams are used. Since the slope at point A is discontinuous, it must be found on both the left and right side of A. So by looking at Figure (h), the slope at A is $ \frac{85.5}{EI} $ and $ \frac{39.8}{EI} $. Once divided by EI, the slopes can be found.


$ \theta _{L}= \frac{+85.5 \ kN\cdot m^{2}}{EI} = \frac{85.5 \ kNm^{2}}{(70 \times \ 10^{6} \ kN/m^{2})(90 \times \ 10^{-6} \ m^{4}) } = 0.01357 \ rad = 0.77758 ^{\circ} = 0.78^{\circ} $ ~ANS

$ \theta _{R}= \frac{+39.8 \ kN\cdot m^{2}}{EI} =\frac{39.8 \ kN\cdot m^{2}}{(70 \times \ 10^{6} \ kN/m^{2})(90 \times \ 10^{-6} \ m^{4}) } = 0.0063 \ rad = 0.36^{\circ} $ ~ANS


Since the deflection is continuous for point A, it can be directly taken as $ \frac{-279}{EI} $ from Figure (i). Once divided by EI, the deflection can be found.


$ y_{A} = \frac{-279 \ kN\cdot m^{3}}{EI} =\frac{-279 \ kN\cdot m^{3}}{(70 \times \ 10^{6} \ kN/m^{2})(90 \times \ 10^{-6} \ m^{4}) } = 0.04428 \ m = 44.3 \ mm $ ~ANS


For a more complex example, check out Professor Erochko's video on YouTube

Note: All the images in the example were made by [5]

References

  1. "conjugate." Merriam-Webster.com. Merriam-Webster, 2013. Web. 23 November 2013..
  2. H. M. Westergaard, “Deflections of Beams by the Conjugate Beam Method,” Journal of the Western Society of Engineers, Vol. XXVI, No. 11, 1921, pp. 369-396.
  3. 3.0 3.1 3.2 3.3 3.4 Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.). Stamford, CT: Cengage Learning.
  4. 4.0 4.1 Int. J. Engng. Ed., Vol. 26, No. 6, pp. 1422-1427, 2010.
  5. 5.00 5.01 5.02 5.03 5.04 5.05 5.06 5.07 5.08 5.09 5.10 5.11 5.12 Bennet, I., Student, Carleton University, 2013