Conjugate Beam Method

Session 2468

Effective Teaching and Learning of the Conjugate Beam Method: Synthesized Guiding Rules Ing-Chang Jong University of Arkansas

Abstract There are different established methods in Mechanics of Materials for determining deflections of beams. No matter which established method is used, one rightfully expects an identical solution to be obtained for the same problem. Well, not so fast! One will here see a puzzling scenario where a certain problem is amenable to solution only by the conjugate beam method, but not by any of the other methods at all. A loaded beam in equilibrium on a simple support is employed as an example of the puzzling scenario, solvable only by the conjugate beam method. The root cause of such a scenario lies in the fact that the conjugate beam method uses “support conditions” while all other methods use “boundary conditions” in the solutions. This paper contributes ten synthesized guiding rules for the conjugate beam method to effectively assist in its teaching and learning. Examples having different levels of complexity are included to illustrate the use of these rules. The solutions obtained by the conjugate beam method are checked and interpreted. I. Introduction Mechanics of Materials is either a required or an elective course in most undergraduate engineering curricula. Major established methods for determining deflections of beams, as taught in such a course, may include the following: 1-6 (a) (b) (c) (d) (e)

Method of double integration (with or without the use of singularity functions), Method of superposition, Method using moment-area theorems, Method using Castigliano’s theorem, and Conjugate beam method.

The conjugate beam method was first derived, defined, and propounded for determining deflections of beams in 1921 by Westergaard.1 It may well be called a “Westergaard method.” Readers interested in the development of this method are advised to refer to the original paper by Westergaard.1 Additionally, note that this method is one of the established methods for finding deflections of beams in the textbook by Timoshenko/MacCullough2 and that by Singer/Pytel.3 Nevertheless, this method is not easily found in most other textbooks.4,5,6 Solutions using the above methods (a) through (d ) all require that boundary conditions regarding slopes or deflections at two or more different positions of a beam in equilibrium (e.g., zero or a specific slope, zero or a specific deflection, equal slopes, or equal deflections) be known. However, solutions using the above method (e) — conjugate beam method — require, instead, that support conditions regarding the types of support a beam has or the connections between the segments of the beam (e.g., fixed support, roller support, hinge support, internal Proceedings of the 2004 American Society for Engineering Education Annual Conference & Exposition Copyright  2004, American Society for Engineering Education

2 hinge, or free end), if any, be known. The catch here is that the amount of information we know about the boundary conditions of a beam is not necessarily equivalent to that we know about the support conditions of the same beam. This is the root cause that leads to a certain puzzling scenario.

II. A Case in Point: a Beam with a Simple Support and Balanced Loading For example, let it be desired to determine the slopes and deflections of an elastic beam AB that has a constant flexural rigidity EI, a length of 2L, and a simple support (i.e., a hinge or roller support) at its midpoint C as illustrated in Fig. 1. This beam carries a concentrated load P at A and a couple of moment M = PL # at B. Clearly, the beam is in equilibrium and will deflect. At the ends A and B, the boundary conditions (i.e., amounts of slopes and deflections) are not known, but the support conditions (i.e., free ends) are known. At point C of this beam, both the boundary condition (zero deflection) and the support condition (simple support) are known.

Fig. 1 An actual beam (i.e., physical beam) with a simple support and balanced loading Since we know only a single boundary condition at point C of the beam, it is simply not enough to allow any of the aforementioned four methods (a) through (d ) to proceed to solve for the deflections of the beam. On the other hand, we do know the support conditions at points A, B, and C of the beam, which are enough to allow the conjugate beam method to proceed to solve for the deflections of the beam! What happens? Is the conjugate beam method more powerful?

III. Synthesized Guiding Rules for Using the Conjugate Beam Method Westergaard1 propounded a great method, but earlier textbooks2,3 provided mainly brief and elementary coverage of the conjugate beam method. Actually, there are two major steps in the conjugate beam method. The first step is to set up an additional beam, called “conjugate beam,” and the second step is to determine the “shearing forces” and “bending moments” in the conjugate beam. In the process, these two steps are to be guided by some ten rules that are synthesized and inferred by the writer of this paper from the original paper of Westergaard.1 ■

Rule 1: The length of the conjugate beam is the same as the length of the actual beam.

■

Rule 2: The loading diagram showing the elastic loads acting on the conjugate beam is simply the bending-moment diagram of the actual beam divided by the flexural rigidity EI of the actual beam. (This elastic load is upward if the bending moment is positive — to cause top fiber in compression — in beam convention.)

For each existing support condition of the actual beam, there is a corresponding support condition for the conjugate beam. The correspondence is given by rules 3 through 7 listed in Table 1. Proceedings of the 2004 American Society for Engineering Education Annual Conference & Exposition Copyright  2004, American Society for Engineering Education

3 Table 1 Corresponding support condition for the conjugate beam

Rule 3 Rule 4 Rule 5 Rule 6 Rule 7

Existing support condition of the actual beam Fixed end Free end Simple support at the end Simple support not at the end Unsupported hinge

Corresponding support condition for the conjugate beam Free end Fixed end Simple support at the end Unsupported hinge Simple support

The slopes and deflections of the actual beam are obtained by employing the following rules: ■

Rule 8: The actual beam, conjugate beam, and even conjugate beam of a conjugate beam are all in static equilibrium. [Cf. Eq. (29) about the use of conjugate beam of a conjugate beam.]

■

Rule 9: The slope of (the centerline of ) the actual beam at any cross section is equal to the “shearing force” at the corresponding cross section of the conjugate beam. (This slope is positive or counterclockwise if the “shearing force” is positive — to rotate the beam element clockwise — in beam convention.)

■

Rule 10: The deflection of (the centerline of ) the actual beam at any point is equal to the “bending moment” of the conjugate beam at the corresponding point. (This deflection is upward if the “bending moment” is positive — to cause top fiber in compression — in beam convention.)

IV. Illustrative Examples for the Conjugate Beam Method Example 1. Determine the slope θA and deflection yA of the free end A of a cantilever beam AB with length L and constant flexural rigidity EI, which is acted on by a concentrated force P at its free end A as shown in Fig. 2.

Fig. 2 A cantilever beam (actual beam) Solution. According to the rules 1 through 4 in Sec. III, we first draw in Fig. 3 the conjugate beam (i.e., an additional beam) corresponding to the actual beam in Fig. 2.

Fig. 3 Conjugate beam (additional beam) corresponding to the actual beam in Fig. 2 Proceedings of the 2004 American Society for Engineering Education Annual Conference & Exposition Copyright  2004, American Society for Engineering Education

4 Note in Fig. 3 that the conjugate beam has, by rules 1 through 4, the same length L as the actual beam, a linearly varying distributed downward elastic load with intensity equal to zero at A and equal to PL/EI at B, a free end at B, and a fixed end at A. Next, we draw in Fig. 4 the free-body diagram for the conjugate beam shown in Fig. 3.

Fig. 4 Free-body diagram for the conjugate beam in Fig. 3 According to rule 8 in Sec. III, the free body of the conjugate beam is in equilibrium. Thus, we readily find that the reaction “force” Acy (i.e., the “shearing force” at A of the conjugate beam) and the reaction “moment” McA (i.e., the “bending moment” at A of the conjugate beam) have the following values: Acy =

PL2 ↑ 2 EI

(1)

PL3 % (2) 3EI By rules 9 and 10 in Sec. III, the slope θA and the deflection yA at the free end A of the actual beam in Fig. 2 are, respectively, equal to the “shearing force” Acy and the “bending moment” McA at the fixed end A of the conjugate beam in Fig. 3. Thus, we employ the results in Eqs. (1) and (2) to obtain the desired solutions as follows: M cA =

PL2 θA = 2 EI yA =

%

PL3 ↓ 3 EI

(3) (4)

Note that yA points downward because McA causes tension in top fiber of the beam at A. The answers in Eqs. (3) and (4) agree with those obtained by other methods as reported in textbooks.2– 6 The deflections of the cantilever beam and the results in Eqs. (3) and (4) are illustrated in Fig. 5.

Fig. 5 Deflections of the cantilever beam (actual beam) in Fig. 2 Proceedings of the 2004 American Society for Engineering Education Annual Conference & Exposition Copyright  2004, American Society for Engineering Education

5 Example 2. A Gerber beam (Gerberbalken) of total length 4L has a hinge connection at C and constant flexural rigidity EI in its segments ABC and CDE. This beam is supported and loaded with a force P at D as shown in Fig. 6. Determine for this beam (a) the slopes at B, C, D, and E, (b) the deflections at C and D, (c) the maximum deflection between A and B, (d ) the maximum deflection between C and E. [This one is intended to illustrate the solution of a rather challenging problem.]

Fig. 6 A Gerber beam (actual beam) Solution. According to the rules 1, 2, 3, 5, 6, and 7 in Sec. III, we first draw in Fig. 7 the conjugate beam corresponding to the actual beam in Fig. 6.

Fig. 7 Conjugate beam corresponding to the Gerber beam in Fig. 6 Note in Fig. 7 that the conjugate beam has, by these rules, the same total length of 4L as the actual beam, a linearly distributed elastic load given by the bending-moment diagram, drawn by parts, of the actual beam divided by the flexural rigidity EI of the actual beam, a free end at A, an unsupported hinge at B, a simple support (e.g., a roller support) at C, and another simple support (e.g., a hinge support) at E. Notice that a simple support can be either a roller support or a hinge support, since a beam is usually not subjected to axial loads. Next, we draw in Fig. 8 the freebody diagram for the conjugate beam shown in Fig. 7.

Fig. 8 Free-body diagram for the conjugate beam in Fig. 7 Proceedings of the 2004 American Society for Engineering Education Annual Conference & Exposition Copyright  2004, American Society for Engineering Education

6 According to rule 8 in Sec. III, the free body of the conjugate beam in Fig. 8 is in equilibrium. Based on the fact that the bending moment at the unsupported hinge at B must be zero plus the fact the entire conjugate beam subjected to only vertical elastic loads is in equilibrium, we see that the conjugate beam in Fig. 8 is “statically determinate,” because we can readily write a total of three independent equations to solve for the three unknowns By , C yc , and E yc appearing in Fig. 8. Readers can readily verify that the solutions obtained are as follows: 5P 4

(5)

C yc =

13 PL2 48 EI

(6)

E yc =

19 PL2 48 EI

(7)

By =

Using Fig. 8 and Eqs. (5) through (7), we can readily find that the “shearing forces” at B, D, and E, as well as just to the left of C, and just to the right of C in the conjugate beam are, respectively, as follows: VBc = −

6 PL2 48 EI

(8)

VDc =

7 PL2 48 EI

(9)

VEc =

19 PL2 48 EI

(10)

(V ) c C

= −

l

(V ) c C

= −

r

18 PL2 48 EI

(11)

5 PL2 48 EI

(12)

According to rule 9 in Sec. III, the slopes at these locations are, respectively, as follows: θB =

6 PL2 48 EI

Z

(13)

θD =

7 PL2 48 EI

X

(14)

θE =

19 PL2 48 EI

(θC )l

=

18 PL2 48 EI

(θC )r =

5 PL2 48 EI

X

Z

Z

Proceedings of the 2004 American Society for Engineering Education Annual Conference & Exposition Copyright  2004, American Society for Engineering Education

(15) (16) (17)

7 Furthermore, using Fig. 8 and Eqs. (5) through (7), we can readily find that the “bending moments” at C and D in the conjugate beam are, respectively, as follows: M Cc = −

14 PL3 48 EI

(18)

M Dc = −

15 PL3 48 EI

(19)

According to rule 10 in Sec. III, the deflections at these locations are, respectively, as follows: yC =

14 PL3 ↓ 48 EI

(20)

yD =

15 PL3 ↓ 48 EI

(21)

Based on the results obtained in Eqs. (13) through (17) and Eqs. (20) and (21), we can plot in Fig. 9 the slopes and deflections of the Gerber beam in Fig. 6.

Fig. 9 Slopes and deflections of the Gerber beam (actual beam) in Fig. 6 Using Fig. 8 and Eqs. (5) through (7), we find that the “shearing forces” in the conjugate beam are zero at F and G, which are located with the distances BF =

L 3

(22)

15 L (23) 6 The maximum deflections occur at F and G, where the slopes of the beam are zero. By computing the “bending moments” at F and G in Fig. 8 and applying rule 10 in Sec. III, we obtain CG =

yF = M Fc =

yG = MGc = −

PL3 54 EI

(126 + 5 15 ) PL3 432 EI

3

PL (ymax ) AB = 54 EI

( ymax )CE =

↑

(126 + 5 15 )PL3 ↓ 432 EI


(24) (25)

8 V. Solution for the Case in Point by the Conjugate Beam Method Let us return to consider deflections of the beam with a simple support and balanced loading in Fig. 1, which illustrates the case in point described in Sec. II. Applying rules 1, 2, 4, and 6 in Sec. III, we draw in Fig. 10 the corresponding conjugate beam for the actual beam in Fig. 1. Note that this conjugate beam has the same length of 2L as the actual beam, fixed supports at the ends A and B, an unsupported hinge at C, and distributed downward elastic loads as shown.

Fig. 10 Conjugate beam corresponding to the actual beam in Fig. 1

Fig. 11 Free-body diagram for the conjugate beam in Fig. 10 The free-body diagram for the conjugate beam in Fig. 10 is shown in Fig. 11, where Acy and Bcy are the unknown reaction “forces” and McA and MBc are the unknown reaction “moments” at A and B, respectively. These four unknowns at the ends A and B are statically indeterminate, but they may be determined by fully applying rule 8 in Sec. III, which points out that the conjugate beam of the conjugate beam must also be in static equilibrium. Taking the “flexural rigidity” of conjugate beam as 1, we draw in Fig. 12 the conjugate beam of the conjugate beam in Fig. 10.

Fig. 12 Conjugate beam of the conjugate beam in Fig. 10 Proceedings of the 2004 American Society for Engineering Education Annual Conference & Exposition Copyright  2004, American Society for Engineering Education

9 For equilibrium of the preceding conjugate beam in Fig. 11, we write (by rule 8 in Sec. III)

+%Σ MCc = 0 , for just member AC — the left segment of the conjugate beam in Fig. 11: M cA − L Ayc +

L PL2 ⋅ =0 3 2 EI

(26)

+%Σ MCc = 0 , for just member CB — the right segment of the conjugate beam in Fig. 11: − MBc + L Byc −

L PL2 ⋅ =0 2 EI

(27)

+ ↑ Σ Fyc = 0 , for the entire conjugate beam ACB in Fig. 11: Acy + Byc −

PL2 PL2 − =0 2 EI EI

(28)

For equilibrium of the conjugate beam of the conjugate beam, we write (by rule 8 in Sec. III)

+%Σ MCcc = 0 , for the entire conjugate beam of the conjugate beam in Fig. 12: L L L PL3 L L L L L L L PL3 ⋅ MAc L + ⋅ ⋅ − ⋅ ⋅ Acy L + ⋅ ⋅ Byc L − ⋅ MBc L − ⋅ ⋅ =0 2 5 4 6 EI 3 2 3 2 2 3 2 2 EI

(29)

Solving Eqs. (26) through (29) simultaneously for the four unknowns in them, we get Acy =

49 PL2 80 EI

(30)

Byc =

71PL2 80 EI

(31)

MAc =

107 PL3 240 EI

(32)

MBc =

93 PL3 240 EI

(33)

Using Eqs. (30) through (33), we can readily find that the “shearing forces” at A, B, and C in the conjugate beam shown in Fig. 11 are, respectively, as follows: VAc =

49 PL2 80 EI

VBc = −

VCc =

71PL2 80 EI

9 PL2 80 EI

(34) (35) (36)

According to rule 9 in Sec. III, the slopes at A, B, and C in the actual beam in Fig. 1 are, respectively, as follows: θA =

49 PL2 80 EI

X


(37)

10

θB =

71PL2 80 EI

θC =

9 PL2 80 EI

(38)

Z

(39)

X

Furthermore, using Eqs. (30) through (33), we can readily find that the “bending moments” at A and B in the conjugate beam shown in Fig. 11 are, respectively, as follows: MAc =

107 PL3 240 EI

MBc =

93 PL3 240 EI

(40)

%

(41)

#

The results in Eqs. (40) and (41) together with the sketch in Fig. 11 reveal that both MAc and MBc cause the top fiber of the conjugate beam in tension; therefore, they are to be taken as “negative moments” in beam convention and by rule 10 in Sec. III. According to this rule, the deflections at A and B of the actual beam in Fig. 1 are, respectively, as follows: yA = yB =

107 PL3 ↓ 240 EI

(42)

93 PL3 ↓ 240 EI

(43)

Using Eqs. (37) through (39), as well as Eqs. (42) and (43), we depict in Fig. 13 the obtained solution for the slopes and deflections of the actual beam in Fig. 1.

Fig. 13 Slopes and deflections of the actual beam in Fig. 1 Using Fig. 11 and Eqs. (30) and (31), we find that the “shearing force” in the conjugate beam, hence the slope of the actual beam, is zero at D, which is located with the distance CD =

9L 80

(44)

Clearly, a maximum deflection occurs at D, where slope of the beam is zero. By computing the “bending moments” at D in Fig. 11 and applying rule 10 in Sec. III, we obtain the following:

yD = MDc =

81PL3 12800 EI

3

81PL (ymax )CB = 12800 EI

↑


(45)

11 VI. Checking and Interpreting the Results Obtained Deflections of the beam with a simple support and balanced loading in Fig. 1 are amenable to solution only by the conjugate beam method, but not by any other methods. The solutions obtained by this method have been expressed in Eqs. (37) through (39), and (42) through (45). Here, we may analytically check the solutions for the slopes and deflections at the ends A and B of the beam. For ease of reference, let Fig. 13 be repeated here.

Fig. 13 Slopes and deflections of the actual beam in Fig. 1 (repeated) Since we have obtained the slope of the tangent drawn at C, we may perform an analytical check of the solutions by regarding the deflected shape of this beam as the elastic curve of two cantilevered beams: (a) a cantilever beam of length L, fixed at C, and is deflected by a force P from CA″ to CA′; (b) a cantilever beam of length L, fixed at C, and is deflected by couple of moment M = PL from CB″ to CB′. From the geometry in Fig. 13, we find the following: #

AA″ = BB″ = L θC = A″ A′ = yA − AA″ =

θ A /C = θ A − θ C =

107 PL3 9 PL3 80 PL3 PL3 − = = 240 EI 80 EI 240 EI 3 EI PL2 49 PL2 9 PL2 40 PL2 − = = 80 EI 80 EI 80 EI 2 EI

(46) (47) (49)

(PL ) L2 93 PL3 9 PL3 PL3 ML2 + = = = 240 EI 80 EI 2 EI 2 EI 2 EI

(48)

71PL2 9 PL2 PL2 (PL ) L ML − = − = − = − 80 EI 80 EI EI EI EI

(50)

B″ B′ = yB + BB″ =

θB /C = θB − θC = −

9 PL3 80 EI

We note that the above values for A″ A′ , θA /C , B″ B′ , and θB /C all check with those for deflections and slopes of the free ends of cantilever beams loaded with a force P and a moment M at their free ends, respectively, as found in textbooks. The beam in Fig. 1 is not merely in equilibrium. In fact, it should properly be recognized as being in neutral equilibrium! In other words, there may exist an infinite number of possible configurations of deflection for the beam in Fig. 1. Unlike all other methods, the conjugate beam method can yield a “favored” solution out of a family of possible solutions for the deflections of a beam in neutral equilibrium, as well as for any beam in stable equilibrium. Proceedings of the 2004 American Society for Engineering Education Annual Conference & Exposition Copyright  2004, American Society for Engineering Education

12 VII. Concluding Remarks The method of double integration, method of superposition, method using moment-area theorems, and method using Castigliano’s theorem are all well established methods for finding deflections of beams, but they all require that the boundary conditions of the beams be known or specified. If not, all of them become helpless. However, the conjugate beam method stands out as the only method that is able to pursue and yield a solution for the deflections of a balanced beam with a single simple support. In fact, the deflection of any beam in neutral equilibrium cannot be investigated and solved by any methods except the conjugate beam method. This study points out that the fundamental prior knowledge about the condition of a beam needed in the solution by the conjugate beam method is a whole lot different from that needed in the solutions by other methods. Consequently, there exist puzzling scenarios where deflections of beams in neutral equilibrium are amenable to solution only by the conjugate beam method, but not by any other methods at all. The root cause of such scenarios lies in the use of support conditions versus boundary conditions in the solution. It is shown in this paper that the solution obtained by the conjugate beam method checks well analytically with well-known results found in textbooks. For deflections of beams, the conjugate beam method — a fabulous method — can work equally well as (or arguably better than) other established methods. Unfortunately, no set of detailed guiding rules for the effective teaching and learning of this method has been found in current textbooks. It is the purpose of this paper to share mechanics ideas with fellow mechanics educators by contributing ten synthesized guiding rules for the conjugate beam method to effectively assist in its teaching and learning. Should this method be included in the mechanics curriculum? Readers are invited to answer this question. References 1. Westergaard, H. M., “Deflections of Beams by the Conjugate Beam Method,” Journal of the Western Society of Engineers, Volume XXVI, Number 11, 1921, pp. 369-396. 2. Timoshenko, S., and G. H. MacCullough, Elements of Strength of Materials, Third Edition, D. Van Nostrand Company, Inc., 1949, pp.179-181. 3. Singer, F. L., and A. Pytel, Strength of Materials, Fourth Edition, Harper & Row, Publishers, Inc., 1987, pp. 228232. 4. Beer, F. P., E. R. Johnston, Jr., and J. T. DeWolf, Mechanics of Materials, Third Edition, The McGraw-Hill Companies, Inc., 2001. 5. Pytel, A., and J. Kiusalaas, Mechanics of Materials, Brooks/Cole, 2003. 6. Gere, J. M., Mechanics of Materials, Sixth Edition, Brooks/Cole, 2004.

ING-CHANG JONG Ing-Chang Jong is currently a Professor of Mechanical Engineering at the University of Arkansas. He received a BSCE in 1961 from the National Taiwan University, an MSCE in 1963 from South Dakota School of Mines and Technology, and a Ph.D. in Theoretical and Applied Mechanics in 1965 from Northwestern University. He served as Chair of the Mechanics Division, ASEE, in 1996-97. His research interests are in mechanics and engineering education.


Conjugate Beam Method

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