Theory and Analysis of Continuous Beams

Continuous beams March: 2012

1

A B 1N/m C D

6m

RA RB RC

0

5N 2N

24m 30m

THEORY AND ANALYSIS

OF

CONTINUOUS BEAMS

1Phumlani G. Nkosi

1 Phumlani G. Nkosi, BSc. (Mathematical and statistical sciences: Mathematical Modelling), BSc Hons

(Technology Management), DP( Mechanical engineering), B.Tech. Mechanical Engineering student UNISA


2

Table of Contents

Introduction ............................................................................................................................................ 3

Analysis of continuous beams .............................................................................................................. 3

Claperon’s Theorem of Three Moments ............................................................................................ 4

Bending moments ............................................................................................................................. 4

Reactions at the support ...................................................................................................................... 8

Steps in solving continuous beams ........................................................................................................ 9

Analysis ................................................................................................................................................. 12

Summary ............................................................................................................................................. 23

References ............................................................................................................................................ 23

APPENDIX A .......................................................................................................................................... 24


3

CONTINUOUS BEAMS

Introduction

Before we elaborate more on continuous beam let us first understand what is meant by the term

beam and bending moments in structural engineering. A beam is a horizontal structural element that

is capable of withstanding load primarily by resisting bending, bending moments are the bending

forces induced to the material as a results of external load/s and due to its own weight over the span.

We are now ready to define a continuous beam; A Continuous beam is a beam that is supported by

more than two supports as shown in figure 1.

Figure 1: Continuous beam subjected to point load

These beams (Continuous beams) are statistically indeterminate and are known as Redundant or

Indeterminate Structures as they cannot be analysed by making use of basic equilibrium, i.e.

∑ �� = 0, ∑ �� = 0 � ∑ � = 0 ,

As opposed to determinate structures where by the above conditions are applicable.

Continuous are mainly used in high risk structures like bridges and buildings as the supports are

strong enough to withstand heavy loads. The following are some of the advantages and

disadvantages of continuous beams (just to name the few),

Advantage/s:

1. Has more vertical load capacity – can support a very heavy load/s

2. Deflection at the middle of the span is minimal as opposed to simple supported beams.

Disadvantage/s:

Difficult analysis and design procedure

Analysis of continuous beams

As stated from the introduction that continuous beams are difficult to analyse one method (not the

only method) of analysing them is by making use of Claperon’s Theorem this theorem is also known

as Theorem of Three Moments and it is used as follows.

Two consecutive spans of the continuous beam are considered at simultaneously, each span is

treated individually (see figure 2a) as a simple supported beam with external loads and two end

support moments. For each central support, one equation is written in terms of three moments (EQ 1).

Thus we get as many equations as there unknowns; each will have only three unknowns.

The term three moments refers to the unknown moments at the central support and at the two ends

of any pair of adjacent spans. The theorem equation is derived by considering the deflection and the


4

slope of the beam; we will not derive the theorem, we will just give the question and show how it is

applied in determining the bending moments acting on the beam for analysis.

(The derivation can be found on Strength of Materials for Technologists by JG Drotsky ).

Claperon’s Theorem of Three Moments

Consider the continuous beam figure 2(a) carrying a load �� supported at � , � � �

By load we mean any external force such as Transverse load (concentrated point load or uniform

distributed load) and bending moments

Assume that the beam:

1. Has Uniform cross section

2. Has Uniform Modulus of rigidity and

3. The supports are at the same level

Therefore the Claperyon’s equation is given by:

− �� × �� − �� + �� − �� × �� = � �� + �� ! 1

Where:

�" = The Left hand side (RHS) moment due to the support �

�# = The central moment due to the central support �

�$ = The right hand side (RHS) moment due to the support �

�" = The length of the segment ��

�$ = The length of the segment ��

%" = The Area of the segment ��

%$ = The Area of the segment ��

�̅" = The length from the left support � to the centroid of area %"

�̅$ = The length from the left support � to the centroid of area %$

Bending moments

Before continuing with this section it is important that you first go through appendix A for revision on

simple supported beams

Knowing what bending moments are acting on the beam will help us to able to calculate the Area/s as

well as the net bending moment for segment/span

Before applying the equation of the three moments let us first examine what bending moments the

beam on figure 2(a) is subjected to:


5

The beam is subjected to the bending moments coursed by the external load figure 2(b), the bending

moments coursed by the central support � figure 2 (c) and bending moments coursed by the support � and � at the end figure 2 (d) which they are assume to be equal to zero unless the beam is fixed at

either both supports (Fixed ends Beam, or at one End (Proped beam see figure 6(a)) of is

overhanging see figure 4

Before we can express the above statement mathematically it is important to know that we keep sign

conversion the same that is if the beam is compressed at the top (i.e. above neutral axis) and

elongated at the bottom (i.e. subjected to tensile force) therefore the bending moments is positive,

now we can express the above mathematically:

Total bending moments = Bending moment due to load + bending moments due to supports

� = �'�� + �� + �� + �� 2

Taking into consideration the sign conversion, the net bending moment is given by

�()* = �'�� − �� + �� + �� 3

For the simple beam with three supports as shown in figure 2 (a),

�� = �� = + 4

∴ �()* = �'�� − �� 5


6

Figure 2: (a) continuous beam supporting a load w(x), (b) Bending moment diagram due to external load w(x), (c) Bending moment diagram due to central support C, (d) Bending moment due to end supports , L and R

Let us now illustrate the above with the following example, figure 3 show the concentrated load P

acting on the mid-span of LC and uniform distributed load W acting on the span CR, let LC = a and

CR = b. Determine the bending moments.

The procedure is as follows:

We treat each span individually as simple supported beam with external loads and two end support moments:

The table below show the Areas (A), the centrionds ( �̅ ) for each span

Span Area A Distance � Comment

LC %" = -.8 �̅" = 2 Simple supported beam with the load acting at the centre( Appendix A, Figure A2)

CR %$ = 12312 �̅$ = 22 Simple supported beam with uniform distributed load( Appendix A, Figure A1)

Table 1

We can now use the three moment’s equation (EQ1) and substitute the values on table 1 to

determine the central bending moment.

The equation is given by:


7

− �" × �" − 2�#��" + �$� − �$ × �$ = 6 6%"�̅" �" + %$�̅$�$ 7 From EQ 4:

�" = �$ = 0

∴ − 2��# = 6 8-.8 × 2 + 12312 × 222 9

∴ �# = −3� ;-.16 + 12324 = From the above we can now able to calculate the net bending moment at the middle of each span, the

net bending at the mid-span is the difference between the maximum bending moment for a simple

beam (figure 3(b)) and the bending moments due to the supports (see figure 3 (c)) :

For the span LC the net bending moment is:

�>?@ "# = -4 − 12 ��" + �#�

∴ �>?@ "# = -4 − �#2

And for the span CR

�>?@ #$ = -2.8 − 12 ��# + �$�

∴ �>?@ #$ = -2.8 − �#2

And for the middle of the beam

�>?@ AB = 0 − �#

∴ �>?@ AB = − �#

We can now draw the bending moment diagram using the above results see figure 3 (d) or 4(c) for the

bending moment diagram


8

Figure 3

Reactions at the support

Having calculated the bending moments above we can now determine the reactions by either taking

moment at the supports or using the relationship between the bending moment and the shear force,

we will only used the moment theorem, once again we treat each span individually as simple

supported beam with external loads and two end support moments

Reaction RL:

Consider span LC as a simple supported beam:

Treat the moments at L as redundant (i.e. the moment at the support that we are calculation the

reaction)

C �# = 0

�" × + �# = - × 2 ∴ �" = D- − 2�#2 E

Reaction RC:

P

L C R

(a)

RL RC RR

= =

(b) AL AR

+ +

ML MR

(c)

MC

= =

(d)

W

L

a b


9

Consider the entire beam; and treat the moments at the support L and C as redundant, therefore

C �B = 0

�" × � + �#2 + �B = - × F2 + 2G + 2 + 22

D- − 2�#2 E × � + �#2 = - × D + 222 E + 2.2 ∴ �# = - × D + 2222 E + 2.22 − D- − 2�#2 E × �

Reaction RD:

RD is determined b using the equilibrium equation:

Thus C HI = 0

∴ �" + �# + �B = 0

∴ �B = − �" − �#

After determining the reactions the SF and the BM diagrams can be drawn as shown on figure 4(b)

and (c)

Steps in solving continuous beams

� Step 1

Three condition need to be taken into account before start analysing the beam, namely:

I. Simple supported beam:

If the beam is simple supported at both ends as shown in figure 3(a), set the bending

moments at the supports equal to zero and start analysing the beam as explained

below.

II. Overhanging beam

If the beam is supporting an overhanging load on either both or one side as shown if

figure

5(a), start by taking moments about support/s supporting the overhang load to

determine the induced moment at that support/s, see figure 5(b), if it is only one side

that is overhanging set the other side bending moment at the support equal to zero

and continue analysing the beam as explained below

III. Fixed ends beam

If the beam is fixed in one or both ends as shown if figure 6(a) first release the fixed

end/s as shown


10

in figure 6(b) set, the released length equal to zero, hence the area for this span is

zero then set the bending moments at the far end supports equal to zero and start

analysing the beam as explained below.

Figure 4: (a) continuous beam, (b) SF diagram, (c) BM diagram

P

L C R

(a)

RL RC RR

RL

0

(b)

x

- RR

Mmax

0

ML MR

(c)

MC

BM Diaram

W

a b

L

SF Diagram


11

Figure 5: (a) Simple beam with over-hanging load, (b) a beam is reduced to simple supported ends with no over-hanging

load.

Figure 6: (a) continuous beam with the fixed end to the right, (b) The fixed end has been released with the span CD

� Step 2

Treat each span individual as simple supported beam taking load/s acting on the span into

consideration.

� Step 3

Draw a free body diagram (figure 3 (b)) representing the bending moments endued by the

applied load/s on the span.

� Step 4

Determine the Area(s) /area moments for the load/s acting on simple supported beam.

� Step 5

Use the three moment theorem equation (EQ1) by considering the consecutive spans (two

spans and three supports) to estimate the bending moments induced by the supports.

� Step 6

Determine the reactions on each support using the moment’s equilibrium equation.

� Step 7

Draw the shear force (SF) and the bending moments (BM) diagram.

A B w N/m C D

(a)

c

A B w N/m C

(b)

MA= 0

a b

MC = Wc2/2

a b

A B w N/m 'C

(a)

a b

A B w N/m C

(b) D

MA = 0 MD = 0

a b c= 0


12

Analysis

The following examples will better clarify the steps above:

Example 1

A 60m long beam supports a uniform distributed loan (u.d.l.) of 1N/m throughout its entire length AD,

with

AB = 24m, BC = 30m and CD = 6m. In addition a concentrated load of 5 N acts at the mid-point of AB

and a load of 2 tons acts at D. See figure 7

Determine the bending moments and the reaction forces ant the supports. Draw the S.F. and the B.M.

diagram.

Figure 7

Step 1 �J = 0 C �# = 0

�# = −2 × 6 + �1�6 D62E �# = −30KL

Step 2

Figure 8(a) shows that we have three consecutive spans/segments i.e. AB, BC and CD. For segment:

AB

Segment AB of a beam is subjected to two loads, i.e. the point load with the magnitude of 5N

acting on the mid-span and the u.d.l. of 1Nm over the entire span.

BC

This segment is only subjected to u.d.l. of 1Nm,

CD

This segment is subjected to two loads, i.e. the u.d.l and the point load with the magnitude of

2N acting at D

Step 3

Draw a free body diagram representing the bending moments endued by the applied load on the

span.

See figure 8(b)

A B 1N/m C D

6m

RA RB RC

24m 30m

2N5N


13

Step 4

After separating and identifying the load/s acting on each span, we can now calculate the area/s

under each curve, see Appendix A

Note: If the spun is subjected to two or more loads therefore the total area moment for span will be

sum of the area moments for each load (for this example: %MN�̅MN = �%�̅�OPQ>@ RPST + �%�̅�UTR)

Calculation of area moment

Span AB %MN�̅MN = �%�̅MN�OPQ>@ RPST + �%�̅MN�UTR For point load:

Method 1

�%�̅�OPQ>@ RPST = -�.MN8 × �MN 2

�%�̅�OPQ>@ RPST = 5 × 24.8 × 24 2 = 4320

Method 2

%WX = 2ℎ2

�%�̅�OPQ>@ RPST = D12 × Z-�4 E × 2Z 3 + D12 × Z-�4 E × DZ3 + ZE

1ℎXWX: Z = �2 � � = �MN

�%�̅�OPQ>@ RPST = \12 × 12 × D5 × 244 E] × D2 × 12 3 E + \12 × 12 × D5 × 244 E] × D123 + 12E

�%�̅�OPQ>@ RPST = 4320

For u.dl. load

�%�̅�UTR = 1�3MN12 × �2

�%�̅�UTR = 1 × 24312 × 242

�%�̅�UTR = 13824

%MN�̅MN = 4320 + 13824 %MN�̅MN = 18144


14

Span BC

%N#�̅N# = 1�3N#12 × �N#2

%N# �̅N# = 1 × 30312 × 302

%N# �̅N# = 33750

Step 5

Applying Equation 1 to the span A B C

− �M × �MN − 2�N��MN + �N# � − �# × �N# = 6 6%MN�̅MN �MN + %N# �̅N#�N# 7 0 × 24 − 2�N�54� − �−30� × 30 = 6 618114 24 + 3375030 7 − 108�N + 900 = 6 × 1881 ∴ �N = 96.2 KL

The Bending Moment at mid-point of AB

�>?@ MN = -�MN4 + 1�.MN8 − �N2

�>?@ MN = 5 × 244 + 1 × 24.8 − 96.22

∴ �>?@ N# = 53.9KL

The Bending Moment at the mid-point of BC

�>?@ N# = 1�.N#8 − 12 ��N + �#�

�>?@ N# = 1 × 30.8 − 12 �96.2 + 30�

∴ �>?@ N# = 49.4KL

Step 6

Determine the reactions on each support using the moment’s equilibrium equation.

�� For span AB C �N = 0


15

�# × 24 + �N = 5 × D242 E + 1 × 24 × D242 E

�M = D288 + 60 − 96.224 E = 10.5 K

�a For span ABC C �# = 0

54 × �M + 30 × �N × + �# = +5 D242 + 30E + 1 × 54 × D542 E 54 × 10.5 + 30�N + 30 = 5 × 42 + 54 × 27

�N = D1668 − 59730 E

�N = 35.7K �� C HI = 0

�M + �N + �# = 5 + 2 + 1 × 60

�# = 67 − 10.5 − 35

∴ �# = 20.8 K

Step 7

Draw the SF and the BM diagrams figure 8 (c) and (d)

From the Shear Force diagram at the maximum bending moment occurs at:

�b = 10.5 D 1210.5 + 1.5E

�b = 10.5 L cWdL %

And at

�. = 17.2 D 3017.2 + 12.8E

�. = 17.2 L cWdL �


16

The maximum bending moment is given by the area under the curve from the shear force diagram

At x = 10.5m the maximum bending moment is

�eSf = �M + 12 2ℎ �M = 0

∴ �eSf = 12 × 10.5 × 10.5 = 55.3 KL

At x = 17.2m the maximum bending moment is:

�eSf = 12 2ℎ − �N

∴ �eSf = 12 × 17.2 × 17.2 − 96.2 = 51.72 � gh 52 KL�

The diagram for the SF and BM is shown below:


17

Figure 8: (a) continuous beam, (b) free body diagram for BM induced by external load, (c) SH diagram, (d) BM diagram

The bending moment diagram can also be drawn as shown below, figure 9

A B 1N/m C D

6m

RA RB RC

(b)

MB

17.2

10.5 SF DIAGRAM

F '8

0 '2

'1.5

'6.5

x1 = 10.5 x2 = 17.2

12.8

'18.5

(d) '55.1 53.9 '52

'49.4 BM DIAGRAM

0

M

'30

'96.2

24m 30m

L

2N5N

1

2 3


18

Figure 9: BM Diagram

Example 2

This question is taken directly from UNISA past exam paper (January/February 2010)

Solution

Let: �M = 0

�B = −10 × 1 = −10iKL

Areas

Consider span AB

%MN = 1�3 12 = 1 × 5312 = 12512

%MN�̅ = 12512 × 52 = 62524

Consider the span BC

Since the concentrated does not act at the mid-span: �̅N# ≠ ". , the figure below represent the BM

diagram induced by the concentrated point load, figure 10

49.4

96.2

53.9

M

30

0

BM Diagram


19

Figure 10: BM diagram for the point load

ℎ = - × �b × �.�

Where: � = �b + �.

∴ ℎ = 20 × 2 × 1 3 = 403

Method 1:

%WX %k = %MN = 1 2 × 2ℎ = 12 × 3 × 403 = 20

Moments about B

%MN�̅N = %WX %k × �� + �b�3 = 20 × 53 = 1003

Moments about C

%MN�̅N = %WX %k × �� + �.�3 = 20 × 43 = 803

Method 2:

%WX 1 = 1 2 × 2ℎ = 12 × 2 × 403 = 403

%WX 2 = 1 2 × 2ℎ = 12 × 1 × 403 = 203

Moments about B:

%MN�̅N = 403 × 2 3 × 2 + 303 D13 × 1 + 2E = 1003

L1 =2m L2=1m

12


20

Moments about C:

%MN�̅# = 403 D13 × 2 + 1E + 203 × 2 3 × 1 = 803

Note: �l)m �a = �l)mn + �l)m� = o+p + �+p = �+

Consider the span CD

%#B = 1�3 12 = 3 × 4312 = 16

%#B�̅ = 16 × 42 = 32

Consider span AB and BC:

− �M × �MN − 2�N��MN + �N#� − �# × �N# = 6 6%MN�̅M �MN + %N# �̅N�N# 7 0 × 5 − 2�N�8� + �# × 3 = 6 662524 × 15 + 1003 × 137

−16�N − 3 �# = 97.2

�# = −�32.64 + 5.33�N�

Consider span BC and CD:

− �N × �N# − 2�#��N# + �#B� − �B × �#B = 6 6%N#�̅# �N# + %#B�̅B�#B 7 −3�N − 2�#�7� + 10 × 4 = 6 6803 × 15 + 32 × 147

−3�N − 14�# = 101.33 − 40

−3�N − 14q−�32.64 + 5.33�N�r = 101.33 − 40

∴ �N = D61.33 − 456.9671.62 E = −5.52 iKL

∴ �# = 32.64 − 5.33�−5.52� = −3.22 iKL

Reactions:

Consider span AB: C �N = 0

�M × 5 + �N = 1 × 5 × D52E

∴ �M = D12.5 − 5.525 E = 1.4 iK

Consider span ABC:


21

C �# = 0

�M × 8 + �N × 3 + �# = 1 × 5 × D52 + 3E + 20 × 1

∴ �N = D47.5 − 14.423 E = 11.03 iK

Consider span ABCD:

C �B = 0

�M × 12 + �N × 7 + �# × 4 + �B = 1 × 5 × D52 + 7E + 20 × �1 + 4� + 3 × 4 × 42

∴ �# = D171.5 − 104.14 E = 16.85 iK

And �B

C HI = 0

�M + �N + �# + �B = 1 × 5 + 20 + 3 × 4 + 10 ∴ �B = 17.72 iK

Maximum bending moment:

From the SF diagram the maximum bending moments occurs at:

�b = 1.4 D 51.4 + 3.6E = 1.4L cWdL %

And at

�. = 4.2 D 412E = 1.43L cWdL �

The SF and the BM diagram is shown below figure11 (c) and (d)

From here one can easily calculate the maximum bending moments when V =0 and the resultant

moments at the mid-span AB and CD.

From the results above one can also draw the bending moments like one in figure 9 for the this

problem


22

Figure 11: (a) continuous beam, (b) free body diagram, (c) SF Diagram, (d) BM Diagram

20KN 10KN

1m 1m

(a) A B C D E

(b) 0

MB

MC

MD

SF DIAGRAM 10

7.43

F 4.28

(c) 0 1.4

1.4m 1.43m

3.6

7.72

12.57

M

BM DIAGRAM

(d) 0.98

0 0.37 0.4

0.61

3.2

5.5

10

9.36

1KN/m 3KN/m

5m 3m 4m


23

Summary

Continuous beams have been employed for ages as they are most reliable for the civil and

engineering structures with high level of risk to collapse e.g. bridges, their complex structure make

them difficult to analyse during the design phase, one way of analysing them is by making use of

Claperon’s Theorem of Three Moments, whereby for each inter mediate support an equation is

developed in terms of three moments, thus we get as many equation as there are unknowns in order

to analysis the beam, this method was demonstrated in this paper by making used of examples.

References

q1r JG Drotsky, Strenght of materials for technologist. 1997. Tech Books

q2r H.P. Gavin, The three-moment equation for continuous beam analysis, Uncertainty, Design and

Optimization. CE130L. Spring 2009.Duke University.

q3r Dr. Amlan K. Sengupta, Prof. Devdas Menon, Pre-stressed Concrete Structures, Indiana Institute

of Technology Madras

s4t Kharagpur, The three moment’s equation for continuous beams analysis, Version 2 IIT


24

APPENDIX A

Revision on simple supported beams

Figure A1: Simple beam uniform distributed load

The area under the curve for a bending moment is calculated as follows:

Consider the diagram below:

The bending moment at any point along the span is given by:

�f = h = 1�2 �� − ��

Consider the strip for the above figure:

The Area (Audl) of a strip is given by:

%UTR = 2ℎ = h��

The total area is:

%uP@SR = v 1�2 �� − ��"w

R R

SF Diagram

BM Diagram

L/2 L/2

W

L

Mmax = WL2/8

Vmax = R = WL/2

L

x

y

y

x

dx

W


25

%uP@SR = v 1��2 �� − v 1�.2"w ��"

w

%uP@SR = x1��.4 − 1�36 x �00

%uP@SR = 31�3 − 21�312

∴ %uP@SR = 1�312

The centriod:

�̅ = �2

Figure A2: Simple beam concentrated load at the centre

The area under the curve for a bending moment is calculated as follows:

Consider the diagram below:

P

L/2 L/2

R R

SFD

BMD

Mmax = PL/4

L

Vmax = R = P/2

x

y

y1 y2

y

x

dx


26

From the above figure it can be seen that are two curves, thus:

hb = -�2 cdW 0 ≤ � ≤ �2

And

h. = -�� − ��2 cdW �2 ≤ � ≤ �

The Area (Apl) of a strip is given by:

%OR = h�� = hb�� + h.��

The total area is:

%uP@SR = v hb�� + v h.��"".

".w

%uP@SR = v -�2 �� + v -�� − ��2 ��"".

".w

%uP@SR = z-�.4 x �200 + z-��2 { �0 �2 − z-�.4 x �0�2 %uP@SR = -�.16 + -�.2 − -�.4 − -�.4 + -�.16

∴ %uP@SR = -�.8

The area can also be calculated from the bending moment diagram as follows:

% = 12 × 2ℎ

∴ % = 12 × � × -�4 = -�.8

The centriod:

�̅ = �2

Figure A2: Simple beam concentrated load at any point


27

The area for r the above figure can be calculated as follows:

% = 12 × 2ℎ

% = 12 × � × -2� = -22

The centriod:

�̅ = �� + �3

P

a

a

Mmax = Pab/L

b

L

b

L

V2 = Pa/L

V1 = Pb/L

Documents

Theory and Analysis of Continuous Beams