FIXED BEAMS CONTINUOUS BEAMS

INTRODUCTION

A beam carried over more than two supports is known as a continuous beam. Railway bridges are common examples of continuous beams. But the beams in railway bridges are subjected to travelling loads in addition to static loads. We will only consider the effect of static, concentrated, and distributed loads for the analysis of reactions and support moments. Figure 1 shows a beam ABCD, carried over three spans of lengths L1, L2, and L3, respectively. End A of the beam is fixed, while end D is simply supported. At the end D support moment will be zero, but at end A, supports B and C there will be support moments in the beam, to be determined.

Figure 1 Continuous beam

Prof. Clapeyron has provided a theorem showing the relationship

between three support moments of any two consecutive spans

of a continuous beam and the loads applied on these two spans.

This theorem is generally known as the “Clapeyron’s theorem

of three moments”.

Engineers are generally responsible with the job of analyzing

support moments, support reactions, and bending moments in a

continuous beam for the optimum design of beam sections.

CLAPEYRON’S THEOREM OF THREE MOMENTS

This theorem provides a relationship between three moments of two

consecutive spans of a continuous beam with the loading arrangement on these

spans.

Let us consider a continuous beam A′ ABCC′ supported over five supports, and

there are four spans of the beam with lengths L1′ L1, L2, and L2′, respectively,

as shown in Figure 2. In this beam let’s consider consecutive spans AB and BC

carrying a uniformly distributed loading (udl) of intensities w1 and w2,

respectively, as shown in Figure 2.

Figure 2 Continuous beam

CLAPEYRON’S THEOREM OF THREE MOMENTS

Let’s say that support moments at A, B, and C are MA, MB, MC, respectively. If

the bending moment on spans AB and BC is positive, then support moments

will be negative.

Now let’s take two spans AB and BC independently and draw the bending

moment (BM) diagram of each considering simple supports at ends. Maximum

bending moment of AB will occur at its centre and be equal to: (Fixed beams,

distributed loading, SS case, page 24)

Figure 2 Continuous beam

8

2

11Lw

Similarly, maximum bending moment at the centre of span BC will be

as shown in Figure 3.

Figure 3 BM diagrams over two supports

8

2

22Lw

Span AB (Independently)

Origin at B, x positive towards left, bending moment at any section X-X is

Support moments MA, MB, MC are shown in the diagram. Bending moment at this section due to support moments (using the top triangle)

222

2

11111

xwx

LwxxLwM x

xM BM

AMxxL 1

AABBAx

ABAx

ABAx

MLx

MMLx

MMM

MML

xLMM

L

MM

xL

MM

11

1

1

11

BABx MMLx

MM 1

Resultant bending moment at the section (when AB is a part of a continuous beam)

or

Integrating this equation, we get

where C1 is a constant of integration.

At support B, where x = 0, slope (say)

EIiB = 0–0+0+0+C1

or constant of integration, C1 = +EIiB

so,

BABxxx MMLx

MxwxLw

MMM 1

2

111

22

BAB MMLx

MxwxLw

dx

ydEI

1

2

111

2

2

22

Now

Integrating this, we get

where C2 is another constant of integration.

At support B, x = 0, deflection y = 0,

so, 0 = 0 − 0 + 0 + 0 + 0 + C2

Constant C2 = 0

Finally

At the support A, x = L1, deflection y = 0, so,

or

Similarly considering span BC, origin at B, x positive towards right and proceeding in the same manner as before, we can write the equation

The slope i′B = –i′B because in portion AB, x is taken positive towards left

and in portion BC, x is taken positive towards right.

Adding and we get

This is a well-known “Clapeyron’s theorem of three moments” (for

support moments having three consecutive supports for a continuous

beam carrying uniformly distributed loads). If there are n supports for a

continuous beam and if the two ends are simply supported, there will be

(n – 2) intermediate supports and (n – 2) equations will be formed so as

to determine the support moments at intermediate supports.

But i′B = − i′B

∴ 2MB (L1 + L2) + MAL1 + MCL2

Example A continuous beam ABCD is carried over three equal spans of length L each. It carries a udl of intensity w through its length as shown in the figure. Determine the (i) support moments, (ii) support reactions, (iii) deflection at centre of span BC, if EI is the flexural rigidity of the beam. Draw the BM diagram of the continuous beam.

The beam is simply supported at ends A and D, so bending moments

MA = MD = 0

Moreover, the beam is symmetrically loaded about its centre; therefore, MB = MC and reactions RA = RD and RB = RC. Using Clapeyron’s theorem of

three moments for spans AB and BC

MAL + 2MB (L + L) + MCL (from page 11)

But MA = 0, MB = MC,

so,

Taking moments at support B of the beam, only on left side of B

or

Reaction,

Reaction, RA = 0.4wL = Reaction, RD

Total load on beam = 3wL

= RA + RB + RC + RD

But RB = RC, due to symmetry, therefore,

2RA + 2 RB = 3wL or 2 × 0.4wL + 2RB = 3wL

Reaction, RB = 1.1wL = Reaction, RC

Bending moment at centre of each span

The figure above shows the BM diagram for the continuous beam with four points of contraflexure P1, P2, P3, and P4, respectively. Bending moment at centre of span AB and CD

= 0.125wL2 − 0.05wL2

= +0.075wL2

Deflection at the Centre of Span BC

Considering span BC only, with known values of moments

MB = MC = −0.1 wL2

Reaction RB = RC = 1.1 wL

If we take a section XX at a distance of x from B, the bending moment

Integrating this equation, we get

where C1 is constant of integration.

At the centre of middle span, slope is zero due to symmetric loading, so

Constant,

Now,

Integrating this equation we get

where C2 is another constant of integration.

At x = 0, y = 0, therefore, EI × 0 = −0 + 0 − 0 − 0 + C2 Constant C2 = 0

Finally,

Deflection at centre,

We should note that in this problem we have taken uniform section throughout. The section of the beam can change from span to span or within the span, depending upon the requirement.

Exercise 5.1 A continuous beam ABC is carried over two spans AB and BC, in which AB = 6 m and BC = 4 m. Span AB carries a udl of 4 kN/m and span BC carries a udl of 6 kN/m, as shown in the figure. Calculate the support reactions and support moments.

THEOREM OF THREE MOMENTS—ANY TYPE OF LOADING

So far we have considered only uniformly distributed loading over the spans of a

continuous beam. But a beam can be subjected to any type of loading and the

Clapeyron’s theorem can be modified for any type of loading on a continuous

beam. Let us consider two spans AB and BC of spans lengths L1 and L2,

respectively, of a continuous beam. Spans carry the combination of loads as

shown in Figure 4.

Figure 4 Loads on a continuous beam

THEOREM OF THREE MOMENTS—ANY TYPE OF LOADING

Considering the beam of spans AB and BC, independently as simply supported,

the bending moment diagrams of two spans can be drawn. Then support moment

diagrams will be as shown by AA′B′C′C. Consider span AB, origin at A, x positive

towards right, section X-X, bending moment at the section will be

M = M′x + M″x

M= BM considering span as simply supported

+ BM due to support moments

Figure 4 Loads on a continuous beam

or

Multiplying both the sides by x dx and integrating

or

where a′1 = Area of BM diagram of span AB, considering this as simply

supported,

a″1 = Area of BM diagram of span AB due to support moments,

= Distance of centroid of a′1 from end A, and

= Distance of centroid of area a″1 (due to support moments from A).

Moreover, from A

(from the figure, lined area))

a′ BM diagram support moments From , we can write

Similarly considering the span CB, taking origin at C and x positive towards left, another equation can be written:

But slope at iB = –i′B because in portion AB we have taken x positive towards right and in portion CB we have taken x positive towards left. Adding these last two equations:

or

Area a′1 and a′2 are positive as per convention and support moments MA, MB, MC are negative.

Figure 4 Support moments diagram

Example A continuous beam ABCD, 12 m long supported over spans AB = BC = CD = 4 m each, carries a udl of 20 kN/m run over span AB, a concentrated load of 40 kN at a distance of 1 m from point B on span BC, and a load of 30 kN at the centre of the span CD. Determine support moments and draw the BM diagram for the continuous beam.

The figure shows a continuous beam ABCD, with equal spans. AB carries a udl of 20 kN/m. Span BC carries a concentrated load of 40 kN at E, 1 m from B. Span CD carries a point load of 30 kN at F, at centre of CD.

Let us construct a′ diagrams (BM diagrams considering each span independently as S.S.)

Span AB Maximum BM

For parabola AG′B, area

Span BC Maximum BM

Triangle BE′C, area

(about B)

(about C)

Span CD, Maximum BM

Using the equation of three moments for spans AB and CB

4MA + 16MB + 4MC = −320 − 210 = −530 or MA + 4MB + MC = −132.5 (i)

But MA = 0

So, 4MB + MC = −132.5 (ii)

Using the equation of three moments for spans BC and DC

But

MD = 0

4MB + 16MC = −150 − 180 = −330 4MB + 16MC = −330 (iii)

From Equations (ii) and (iii)

15MC = −197.5

MC = −13.166 kNm MB = −29.8335 kNm

Taking

BB′ = −29.8335 kNm CC′ = −13.166 kNm

Drawing lines AB′, B′C′, C′D, the diagram AB′C′DA, is the BM diagram due to support moments. There are four points of contraflexure in the BM diagram for continuous beam when a′ diagrams are superimposed over a″ diagram, the BM diagram due to support moments. Positive and negative areas of the BM diagram are also marked.

SUPPORTS NOT AT SAME LEVEL There are two spans AB and BC of lengths L1 and L2 of a continuous beam. Supports A, B, and C are not at same one level. Support B is below support A by δ1 and below support C by δ2 as shown in Figure 5 (a). These level differences are very small in comparison to span length, say of the order of 0.1% of span length. For spans AB and BC, bending moment diagrams are plotted considering the spans independently as shown by the shaded diagrams. Say

Figure 5 Continuous beam

MA

MB

L1

MC

MB

L2

Span AB: Consider a section X-X, at a distance of x from A

or

or

and .

Similarly consider span CB, origin at C, x positive towards left

It may be noted that iB = –i′B slope, because for span AB, we have taken x

positive towards right but for span CB, we have taken x positive towards left.

So, iB + –i′B = 0

Adding and

or MA L1 + 2MB (L1 + L2) + MCL2

From this equation MA, MB, and MC can be determined, after determining

Example A continuous beam ABC, 10 m long, spans AB = 6 m, span BC = 4 m, carries a udl of 2 kN/m over AB and a concentrated load of 6 kN at centre of BC. Support B is below the level of A by 10 mm and below the level of C by 5 mm. Determine support moments and support reactions, if EI = 6,000 kNm2.

Figure (a) shows the loads on the beam. Let us draw BM diagrams considering the spans as independently supported.

Span AB

area

Span CB

area

Support moments MA = MC = 0

Using the equation derived in Fixed Beams, Support Moments:

Total load on beam = 6 × 2 + 6 = 18kN

Moreover MB = RA × 6 − 6 × 2 × 3

−1.95 = 6RA − 36

Reaction, RA = + 5.675 kN

also MB = 4RC − 6 × 2

−1.95 = 4RC −12

Reaction, RC = +2.5125 kN

Reaction, RB = 18 − 5.675 − 2.5125 = 9.8125 kN.

CONTINUOUS BEAM WITH FIXED END For a continuous beam with a fixed end, the equations for support moments can be derived considering the slope and deflection at fixed end to be zero. Figure 6 (a) shows two consecutive spans AB and BC of a continuous beam. End A of the beam is fixed. Bending moment diagrams a′1 and a′2 are plotted considering spans AB and BC supported independently as shown in Figure 6. At any section if: M′x = BM due load on span, considering span independently M″x = BM due to support moments, then

Figure 6 Support moments

Considering origin as B and x positive towards left:

But at fixed end:

or

where MA is the fixing couple at fixed end A.

This relationship can also be obtained by considering an imaginary span A′A of zero length and bending moment at A′, M′A = 0 using Clapeyron’s theorem for two spans A′A and AB.

or

If the other end of the continuous beam is also fixed, a similar equation can be made by considering an imaginary span to the right of the other fixed end, and then applying the theorem of three moments.

Example A continuous beam ABCD 14 m long rests on supports A, B, C, and D all at the same level. AB = 6 m, BC = 4 m, CD = 4 m. Support A is a fixed support. It carries two concentrated loads of 60 kN each, at a distance of 2 m from end A and end D as shown in the figure. There is a udl of 15 kN/m over span BC. Find the moments and reactions at the supports. The figure shows the continuous beam ABCD, with fixed end at A. Let us first draw BM diagrams considering each span to be simply supported.

Span AB Maximum BM,

Origin at A,

Origin at B,

Span BC

Span CD

Considering imaginary span AA′ of zero length and using Clapeyron’s theorem for two spans A′A and AB,

or

Spans AB and BC

Spans BC and CD

But 4MB + 16MC = −600 (iv)

Putting the value of MA in equation (iii)

MB + 4MC = −150 From eq. (iv) (vi) Solving eqs. (v) and (vi), we get 16 MB = −330

Moment, MB = −20.625 kNm, Putting the value of MB in eq. (vi) 4 MC = −150 + 20.625 Moments,

The figure shows shaded diagrams are a′1, a′2, and a′3 BM diagrams. Bending moment diagram due to support moments is superimposed on these diagrams to get resultant BM at any section.

MC = 4RD − 60 × 2 = −32.334

Reaction, RD = 21.9 kN

MB = 8RD + 4RC − 6 × 60 − 60 × 2 = −20.625

8RD + 4RC = 459.375

8 × 21.9 + 4RC = 459.375

Reaction, RC = 71.0 kN

Support Reactions

Moments about A 14RD + 10RC + 6RB −12 × 60 − 4 × 15 × 8 − 60 × 2 = MA = −56.354 kNm

Reaction, RB = 41.1 kN

Total load on beam = 60 + 4 × 15 + 60 = 180 kN

Reaction RA = 180 − 41.1 − 71.0 − 21.9

= 46 kN

Putting the values in the solution:

Example A continuous beam ABC, fixed at end A, supported over spans AB = BC = 6 m each. There is a udl of 10 kN/m over AB and a concentrated load of 40 kN at centre of BC as shown in the figure. While the supports A and C remain at the same level, the level of support B is 1 mm below due to sinking. Moment of inertia of beam from A to B is 18,000 cm4 and from BC it is 12,000 cm4. If E = 210 kN/mm2, determine support moments and draw BM diagram. Let us first draw a diagram for both spans area,

a′2 is a triangle with

Moment, MC = 0, because end C is a simple support.

Span AA´B Imaginary span AA′ and AB equation of three moments.

Taking I1 common throughout

(because level of A is higher than level of B by 1 mm)

12MA + 6MB = − 577.8 (i) Now using the theorem of three moments for spans AB and BC, and noting that I1 is different than I2, equation can be modified as

Multiplying throughout by I1,

But I1 = 1.5I2, putting this value, we get

6MA + 30MB + 9MC = −540 −810 + 37.8 + 37.8 6MA + 30MB + 9MC = −1274.4 kNm But MC = 0

So, 6MA + 30MB = −1274.4 kNm (ii)

6MA + 3MB = −288.9 kNm From eq (i) (iii)

From these two equations Support moments,

Superimposing diagram over this, we get resultant bending

moment diagram. Shaded parts show positive BM. There are three

points of contraflexure as shown.