Effective Teaching and Learning of the Conjugate Beam ...Fig. 1 An actual beam (i.e., physical beam)...

Session 2468

Proceedings of the 2004 American Society for Engineering Education Annual Conference & ExpositionCopyright 2004, American Society for Engineering Education

Effective Teaching and Learning of the ConjugateBeam Method: Synthesized Guiding Rules

Ing-Chang JongUniversity of Arkansas

Abstract

There are different established methods in Mechanics of Materials for determining deflections ofbeams. No matter which established method is used, one rightfully expects an identical solutionto be obtained for the same problem. Well, not so fast! One will here see a puzzling scenariowhere a certain problem is amenable to solution only by the conjugate beam method, but not byany of the other methods at all. A loaded beam in equilibrium on a simple support is employed asan example of the puzzling scenario, solvable only by the conjugate beam method. The rootcause of such a scenario lies in the fact that the conjugate beam method uses “support condi-tions” while all other methods use “boundary conditions” in the solutions. This paper contributesten synthesized guiding rules for the conjugate beam method to effectively assist in its teachingand learning. Examples having different levels of complexity are included to illustrate the use ofthese 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 engi-neering curricula. Major established methods for determining deflections of beams, as taught insuch a course, may include the following:1-6

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

The conjugate beam method was first derived, defined, and propounded for determining de-flections 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 byWestergaard.1 Additionally, note that this method is one of the established methods for findingdeflections 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 re-garding 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) beknown. 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 connectionsbetween the segments of the beam (e.g., fixed support, roller support, hinge support, internal

2


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

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 thathas a constant flexural rigidity EI, a length of 2L, and a simple support (i.e., a hinge or rollersupport) at its midpoint C as illustrated in Fig. 1. This beam carries a concentrated load P at Aand a couple of moment M = PL # at B. Clearly, the beam is in equilibrium and will deflect. Atthe ends A and B, the boundary conditions (i.e., amounts of slopes and deflections) are notknown, but the support conditions (i.e., free ends) are known. At point C of this beam, both theboundary 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 enoughto allow any of the aforementioned four methods (a) through (d ) to proceed to solve for the de-flections of the beam. On the other hand, we do know the support conditions at points A, B, andC of the beam, which are enough to allow the conjugate beam method to proceed to solve for thedeflections 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 andelementary coverage of the conjugate beam method. Actually, there are two major steps in theconjugate 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 conju-gate beam. In the process, these two steps are to be guided by some ten rules that are synthesizedand 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 issimply the bending-moment diagram of the actual beam divided by the flexural rigidity EIof the actual beam. (This elastic load is upward if the bending moment is positive — to cause top fiberin compression — in beam convention.)

For each existing support condition of the actual beam, there is a corresponding support condi-tion for the conjugate beam. The correspondence is given by rules 3 through 7 listed in Table 1.

3


Table 1 Corresponding support condition for the conjugate beam

Existing support conditionof the actual beam

Corresponding support conditionfor the conjugate beam

Rule 3 Fixed end Free endRule 4 Free end Fixed endRule 5 Simple support at the end Simple support at the endRule 6 Simple support not at the end Unsupported hingeRule 7 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 beamare 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 tothe “shearing force” at the corresponding cross section of the conjugate beam. (This slopeis 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 up-ward 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 ABwith length L and constant flexural rigidity EI, which is acted on by a concentrated force P at itsfree 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 conjugatebeam (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

4


Note in Fig. 3 that the conjugate beam has, by rules 1 through 4, the same length L as the actualbeam, a linearly varying distributed downward elastic load with intensity equal to zero at A andequal to PL/EI at B, a free end at B, and a fixed end at A. Next, we draw in Fig. 4 the free-bodydiagram 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, wereadily find that the reaction “force” Ayc (i.e., the “shearing force” at A of the conjugate beam)and the reaction “moment” MA

c (i.e., the “bending moment” at A of the conjugate beam) have the

following values:

Ayc PL

EI= ↑

2

2 (1)

MAc PL

EI=

3

3 % (2)

By rules 9 and 10 in Sec. III, the slope θA and the deflection yA at the free end A of the actualbeam in Fig. 2 are, respectively, equal to the “shearing force” Ayc and the “bending moment”MA

c 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:

θAPL

EI=

2

2 % (3)

yPL

EIA= ↓

3

3 (4)

Note that yA points downward because MAc causes tension in top fiber of the beam at A. The an-

swers 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

5


Example 2. A Gerber beam (Gerberbalken) of total length 4L has a hinge connection at C andconstant flexural rigidity EI in its segments ABC and CDE. This beam is supported and loadedwith 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 maximumdeflection 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 con-jugate 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 ac-tual beam, a linearly distributed elastic load given by the bending-moment diagram, drawn byparts, of the actual beam divided by the flexural rigidity EI of the actual beam, a free end at A, anunsupported 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 hingesupport, since a beam is usually not subjected to axial loads. Next, we draw in Fig. 8 the free-body diagram for the conjugate beam shown in Fig. 7.


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 thefact the entire conjugate beam subjected to only vertical elastic loads is in equilibrium, we seethat the conjugate beam in Fig. 8 is “statically determinate,” because we can readily write a totalof three independent equations to solve for the three unknowns By , Cy

c , and Eyc appearing in Fig.

8. Readers can readily verify that the solutions obtained are as follows:

BP

y =54

(5)

CPLEIy

c = 1348

2

(6)

EPLEIy

c = 1948

2

(7)

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

VPL

EIBc = − 6

48

2

(8)

VPL

EIDc =

7

48

2

(9)

VPLEIE

c = 1948

2

(10)

VPLEIC

c

l( ) = − 1848

2

(11)

VPL

EICc

r( ) = − 548

2

(12)

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

θBPL

EI= 6

48

2

Z (13)

θDPL

EI= 7

48

2

X (14)

θEPLEI

= 1948

2

X (15)

θC lPLEI

( ) = 18482

Z (16)

θC rPL

EI( ) = 548

2

Z (17)

7


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

MPLEIC

c = − 1448

3

(18)

MPLEID

c = − 1548

3

(19)

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

yPLEIC

= 1448

3

↓ (20)

yPLEID

= 1548

3

↓ (21)

Based on the results obtained in Eqs. (13) through (17) and Eqs. (20) and (21), we can plot inFig. 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 beamare zero at F and G, which are located with the distances

BFL=3

(22)

CGL= 15

6(23)

The maximum deflections occur at F and G, where the slopes of the beam are zero. By comput-ing the “bending moments” at F and G in Fig. 8 and applying rule 10 in Sec. III, we obtain

y MPL

EIF Fc= =

3

54 y

PLEIABmax

( ) =3

54 ↑ (24)

y MPL

EIG Gc= = − +( )126 5

43215 3

yPL

EICEmax( )( ) = +126 5432

15 3 ↓ (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 inFig. 1, which illustrates the case in point described in Sec. II. Applying rules 1, 2, 4, and 6 inSec. 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 theends 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


The free-body diagram for the conjugate beam in Fig. 10 is shown in Fig. 11, where Ayc and Byc

are the unknown reaction “forces” and MAc and MB

c are the unknown reaction “moments” at Aand B, respectively. These four unknowns at the ends A and B are statically indeterminate, butthey may be determined by fully applying rule 8 in Sec. III, which points out that the conjugatebeam of the conjugate beam must also be in static equilibrium. Taking the “flexural rigidity” ofconjugate 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

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 L AL PL

EIAc

yc− + ⋅ =

3 20

2

(26)

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

− + − ⋅ =M L B L PLEIB

cyc

20

2

(27)

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

A BPL

EIPLEIy

cyc+ − − =

2 2

20 (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

M LL L PL

EIL L

A LL L

B LL

M LL L PL

EIAc

yc

yc

Bc

2 5 4 6 3 2 3 2 2 3 2 20

3 3

⋅ + ⋅ ⋅ − ⋅ ⋅ + ⋅ ⋅ − ⋅ − ⋅ ⋅ = (29)

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

APLEIy

c = 4980

2

(30)

BPLEIy

c = 7180

2

(31)

MPLEIA

c = 107240

3

(32)

MPL

EIBc = 93

240

3

(33)

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

VPLEIA

c = 4980

2

(34)

VPL

EIBc = −

71

80

2

(35)

VPL

EICc = 9

80

2

(36)

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

θAPLEI

= 4980

2

X (37)

10


θBPLEI

= 7180

2

Z (38)

θCPL

EI= 9

80

2

X (39)

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

MPLEIA

c = 107240

3

% (40)

MPL

EIBc = 93

240

3

# (41)

The results in Eqs. (40) and (41) together with the sketch in Fig. 11 reveal that both MAc and MB

c

cause the top fiber of the conjugate beam in tension; therefore, they are to be taken as “negativemoments” in beam convention and by rule 10 in Sec. III. According to this rule, the deflectionsat A and B of the actual beam in Fig. 1 are, respectively, as follows:

yPLEIA

= 107240

3

↓ (42)

yPL

EIB= 93

240

3

↓ (43)

Using Eqs. (37) through (39), as well as Eqs. (42) and (43), we depict in Fig. 13 the obtainedsolution 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

CDL= 9

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 DcM

PLEI

= = 8112800

3

yPL

EICBmax( ) = 8112800

3

↑ (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 tosolution only by the conjugate beam method, but not by any other methods. The solutions ob-tained 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 Bof 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 checkof the solutions by regarding the deflected shape of this beam as the elastic curve of two cantile-vered beams: (a) a cantilever beam of length L, fixed at C, and is deflected by a force P fromCA″ to CA′; (b) a cantilever beam of length L, fixed at C, and is deflected by couple of momentM = PL # from CB″ to CB′. From the geometry in Fig. 13, we find the following:

AA BB LPL

EIC″ ″= = =θ 9

80

3

(46)

A A y AAPLEI

PLEI

PLEI

PLEIA

″ ′ = ″− = − = =107240

980

80240 3

3 3 3 3

(47)

θ θ θA C A CPLEI

PLEI

PLEI

PLEI/

= − = − = =4980

980

4080 2

2 2 2 2

(49)

B B y BBPL

EIPL

EIPL

EIPL L

EIML

EIB″ ′ ″= + = + = = =93

240980 2 2 2

3 3 3 2 2( )(48)

θ θ θB C B CPLEI

PLEI

PLEI

PL LEI

MLEI/

( )= − = − − = − = − = −7180

980

2 2 2

(50)

We note that the above values for A A″ ′ , θA C/ , B B″ ′ , and θB C/ all check with those for deflec-tions and slopes of the free ends of cantilever beams loaded with a force P and a moment M attheir 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 be-ing in neutral equilibrium! In other words, there may exist an infinite number of possible con-figurations of deflection for the beam in Fig. 1. Unlike all other methods, the conjugate beammethod can yield a “favored” solution out of a family of possible solutions for the deflections ofa beam in neutral equilibrium, as well as for any beam in stable equilibrium.

12


VII. Concluding Remarks

The method of double integration, method of superposition, method using moment-area theo-rems, and method using Castigliano’s theorem are all well established methods for finding de-flections of beams, but they all require that the boundary conditions of the beams be known orspecified. If not, all of them become helpless. However, the conjugate beam method stands outas the only method that is able to pursue and yield a solution for the deflections of a balancedbeam with a single simple support. In fact, the deflection of any beam in neutral equilibriumcannot 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 neededin the solution by the conjugate beam method is a whole lot different from that needed in the so-lutions by other methods. Consequently, there exist puzzling scenarios where deflections ofbeams in neutral equilibrium are amenable to solution only by the conjugate beam method, butnot by any other methods at all. The root cause of such scenarios lies in the use of support con-ditions versus boundary conditions in the solution.

It is shown in this paper that the solution obtained by the conjugate beam method checks wellanalytically with well-known results found in textbooks. For deflections of beams, the conjugatebeam method — a fabulous method — can work equally well as (or arguably better than) otherestablished methods. Unfortunately, no set of detailed guiding rules for the effective teachingand learning of this method has been found in current textbooks. It is the purpose of this paper toshare mechanics ideas with fellow mechanics educators by contributing ten synthesized guidingrules for the conjugate beam method to effectively assist in its teaching and learning. Should thismethod 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 ofEngineers, Volume XXVI, Number 11, 1921, pp. 369-396.

2. Timoshenko, S., and G. H. MacCullough, Elements of Strength of Materials, Third Edition, D. Van NostrandCompany, Inc., 1949, pp.179-181.

3. Singer, F. L., and A. Pytel, Strength of Materials, Fourth Edition, Harper & Row, Publishers, Inc., 1987, pp. 228-232.

4. Beer, F. P., E. R. Johnston, Jr., and J. T. DeWolf, Mechanics of Materials, Third Edition, The McGraw-HillCompanies, 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 JONGIng-Chang Jong is currently a Professor of Mechanical Engineering at the University of Arkansas. He received aBSCE in 1961 from the National Taiwan University, an MSCE in 1963 from South Dakota School of Mines andTechnology, and a Ph.D. in Theoretical and Applied Mechanics in 1965 from Northwestern University. He served asChair of the Mechanics Division, ASEE, in 1996-97. His research interests are in mechanics and engineering educa-tion.

Effective Teaching and Learning of the Conjugate Beam ...Fig. 1 An actual beam (i.e., physical beam)...

Documents

Transcript of Effective Teaching and Learning of the Conjugate Beam ...Fig. 1 An actual beam (i.e., physical beam)...