Chapter 4. Continuous Beams – Experimental Stress Analysis

4

Continuous Beams

4.1 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. In this chapter, we will only consider the effect of static, concentrated, and distributed loads for the analysis of reactions and support moments. Figure 4.1 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. A 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. 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 Clapeyron’s theorem of three moments. Civil engineers are generally entrusted with the job of analyzing support moments, support reactions, and bending moments in a continuous beam for the optimum design of beam sections.

Figure 4.1 Continuous beam

4.2 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 L1L1, L2, and L2′, respectively, as shown in Figure 4.2. In this beam consider consecutive spans AB and BC carrying udl of intensity w1 and w2, respectively, as shown in Figure 4.2. Say 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. Take two spans BA and BC independently and draw BM diagram of each considering simple supports at ends. Maximum bending moment of BA will occur at its centre and be equal to similarly maximum bending moment at centre of span BC will be as shown in Figure 4.3.

Figure 4.2 Continuous beam

Figure 4.3 BM diagrams over two supports

4.2.1 Span BA (Independently)

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

Support moments MA, MB, MC are shown in diagram. Bending moment at section due to support moments

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

or

Integrating Eq. (4.1), we get

where C1 is a constant of integration.

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

so,

 

EIiB = 0–0+0+0+C1

or constant of integration,           C1 = +EIiB

Now

Integrating Eq. (4.2), 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 consider span BC, origin at B, x positive towards right and proceeding in the same manner as before, we can write the equation

Note that slope i′B = –i′B because in portion BA, x is taken positive towards left and in portion BC, x is taken positive towards right.

Adding Eqs (4.4) and (4.5), we get

But         iB = − iB

∴         2MB (L1 + L2) + MAL1 + MCL2

This is a well-known Clapeyron’s theorem of three moments” (for support moments for three consecutive supports for a continuous beam carrying uniformly distributed loads). If there are n supports for a continuous beam, 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.

Example 4.1 A continuous beam ABCD is carried over three equal spans of length L each. It carries a udl of intensity w through its centre length as shown in Figure 4.4. Determine (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 continuous beam.

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 BA and BC

 

MAL + 2MB (L + L) + MCL

But                 MA = 0, MB = BC,

so,

Figure 4.4 Example 4.1

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 = 4wL

                               = 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

Figure 4.5 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 DC

= 0.125wL2 − 0.05wL2

= +0.075wL2

Figure 4.5 BM diagram

Figure 4.6 Example 4.1

4.2.2 Deflection at Centre of Span BC

Consider span BC only, with known values of moments

            MB = MC = −0.1 wL2

Reaction   RB = RC = 1.1 wL

Take a section XX at a distance of x from B, bending moment

Integrating Eq. (4.7), 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 Eq. (4.9), 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,

Please 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. Reader may find out deflection in the centre of span AB or CD.

Exercise 4.1 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 Figure 4.7. Calculate support reactions and support moments.

Figure 4.7 Exercise 4.1

Ans. [RA = 9.4 kN, RB = 30.5, RC = 8.1 kN, MB = –15.6 kNm].

4.3 THEOREM OF THREE MOMENTS—ANY TYPE OF LOADING

In the last chapter we have considered only uniformly distributed load 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 4.8(a). 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′CB. 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

 

= BM considering span as simply supported

   + BM due to support moments

Figure 4.8 Support moments diagrams

or

Multiplying both the sides by x dx and integrating

or

where a1 = Area of BM diagram of span AB, considering this as simply supported,
         a1 = Area of BM diagram of span AB due to support moments,
            = Distance of centroid of a1 from end A, and
           = Distance of centroid of area a1 (due to support moments from A).

 

Moreover, from A

(as shown in Figure 4.9)

Figure 4.9 a′ BM diagram support moments

From Eq. (4.12), we can write

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

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 the Eqs (4.13) and (4.14)

or

Area a1 and a2 are positive as per convention and support moments MA, MB, MC are negative.

Example 4.2 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 (Figure 4.10).

Solution: Figure 4.10 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 diagram considering each span independently as S.S.)

Span AB Maximum BM

Parabola AG′B

Figure 4.10 Example 4.2

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 Eqs (ii) and (iii)

15MC = −197.5

MC = −13.166 kNm

MB = −29.8335 kNm

Taking

BB′ = −29.8335 kNm

CC′ = −13.166 kNm

Draw lines AB′, B′C′, C′D, the diagram AB′C′DA, is the BM diagram due to support moments. Reader can note that 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.

Figure 4.11 Exercise 4.2

Exercise 4.2 A continuous beam ABCD, 16 m long with spans AB = 6 m, BC = CD = 5 m each, carries transverse loads as shown in Figure 4.11. Determine support moments and draw BM diagram of the beam.

Ans. [MA = MD = 0, MB = –12.23 kNm, MC = –9.44 kNm].

4.4 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 one level. Support B is below support A by δ1 and below support C by δ2 as shown in Figure 4.12(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 4.12 Support moments

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

or

or

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 Eqs (4.16) and (4.17)

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

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

Example 4.3 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.

Solution: Figure 4.13 (a) shows loads on the beam. Let us draw BM diagrams considering the spans as independently supported.

Span AB

Figure 4.13 Example 4.3

area

Span CB

area

Support moments            MA = MC = 0

Using the equation derived in Article 3.3

 

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.

 

Exercise 4.3 A girder 15 m long is supported at ends and has an intermediate support at 10 m from one end. It carries a concentrated load of 120 kN at the middle of 5 m span and a uniformly distributed load of 10 kN/m run over a span of 10 m. The central support is 1 cm lower than the end supports. Calculate support moments and support reactions. Given E = 200 kN/mm2, I = 30,000 cm4.

Ans. [MA = MC = 0, MB = –84.43 kNm, RA = 41.51 kN, RB = 135.46 kN, RC = 43.03 kN].

4.5 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 4.14(a) shows two consecutive spans AB and BC of a continuous beam. End A of the beam is fixed. Bending moment diagrams a1 and a2 are plotted considering spans AB and BC supported independently as shown in Figure 4.14. At any section if:

 

Mx = BM due load on span, considering span independently

Mx = BM due to support moments, then

Considering original B and x positive towards right:

Figure 4.14 Support moments

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 4.4 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 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 Figure 4.15. There is a udl of 15 kN/m over span BC. Find the moments and reactions at the supports.

Solution: Figure 4.15(a) shows continuous beam ABCD, with fixed end at A. Let us first draw BM diagrams considering each span to be simply supported.

Figure 4.15 Example 4.4

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 Eq. (iii)

MB + 4MC = −150                  From Eq. (iv)         (vi)

 

Solving Eqs (v) and (vi), we get

 

16MB = −330

 

Moment,         MB = −20.625 kNm

Putting the value of MB in Eq. (vi)

 

4MC = −150 + 20.625

Moment,

Figure 4.15(b) 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.

 

Support Reactions

 

 

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

 

Moments about A

 

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

 

Putting the values and after solution.

 

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

Exercise 4.4 A continuous beam ABC 10 m long is fixed at end A, simply supported at B and C, on spans AB = BC = 5 m. It carries 40 kN load at centre of each span, as shown in Figure 4.16.

Figure 4.16 Exercise 4.4

Determine support moments and support reactions.

Ans. [MA –21.43 kNm, MB = –32.14 kNm, MC = 0, RA = 17.86, RB = 48.57 kN, RC = 13.57 kN].

Problem 4.1 A continuous beam ABC 10 m long is supported on two spans AB = 6 m, BC = 4 m. Span AB carries a udl of 12 kN/m while span BC carries a concentrated load of 20 kN at its centre. Moment of inertia of section of beam AB is I1, but moment of inertia of section of beam BC is I2. If I1 = 2I2, determine support moments and support reactions.

Solution: Let us consider a1 and a2 diagrams.

Span AB

Moment of inertia of the beam in two spans is different; the theorem of three moments can be modified as follows:

Figure 4.17 Problem 4.1

But            MA = MC = 0,L1 = 6m, L2 = 4 m

 

I1 = 2I2, putting these values

or         14MB = −324 −120 = −444

Moment

Taking MA = 0, MB = –31.71 kNm, MC = 0, support moment diagram is drawn. Resultant BM diagram with positive and negative areas is shown in Figure 4.17(b).

 

Support reactions

 

 

MB

=

6RA −12 × 6 × 3 = −31.71

Reaction,

RA

=

30.71 kN

Moreover

MB

=

4RC − 20 × 2 = −31.71

Reaction,

RC

=

2.07 kN

 

Reaction,            RB = 12 × 6 + 20 − 30.71 − 2.07 = 59.22 kN

Problem 4.2 A continuous beam ABC of length 11 m is loaded as shown in Figure 4.18. Determine support moment B and reactions at supports.

Solution: Let us first calculate and for the two spans AB and BC.

Span BC

Maximum BM,

Figure 4.18 Problem 4.2

Figure 4.19 Problem 4.2

Span AB

Reactions

Taking moments about A,

 

10 × 2 + 1 + 20 × 2 × 3

=

4RB

RB

=

35kN

RA

=

20 + 40 − 35 = 25 kN

 

Span AB

Origin at A,

Putting the values of R′A w1, w2, etc.

Now using Clapeyron’s theorem of three moments, note that MA = MC = 0

      = −250 − 514.28

MB = −34.74 kNm

 

Support reactions

 

 

MB

=

4RA − 2 × 10 × 3 − 2 × 20 × 1 = −34.74

Reaction,

RA

=

16.325 kN

Moreover

MB

=

7RC − 40 × 5 = −34.74

Reaction,

RC

=

23.60 kN

Reaction,

RB

=

10 × 2 + 20 × 2 + 40 − RA − RC = 100 − 16.325 − 23.60

 

 

=

60.075 kN

Problem 4.3 A beam ABCD, 12 m long supported over three spans, AB = BC = CD = 4 m each. On AB there is a central point load of 40 kN, on BC there is a uniformly distributed load of 20 kN/m run throughout and on CD there is a central point load of 40 kN. The level of the support B is 6 mm below the levels of A, C, and D supports. Determine support moments and support reactions. Draw the BM diagram of continuous beam. EI = 6,000 kNm2.

Solution: Figure 4.20(a) shows the continuous beam ABCD with lengths and loads on each span.

Span AB

Figure 4.20 Problem 4.3

Span BC

Maximum BM

Span CD

Maximum BM

Using the theorem of three moments with a sinking of support for spans AB and CB, note that MA = 0

Putting the values

Using the equation of three moments for spans BC and DC.

Now                δ3 = 0

 

Support C is 6 mm higher than support B

or                δ3 = −0.0006m

Note that                MD = 0

or

From Eqs (i) and (iii)

 

15MB = −298.5

MB = −19.9 kNm

 

Support moment,

From Eq. (iii)

 

4MC = −153.5 + 19.9
= −133.6

 

Support moment,

 

MC = −33.4 kNm

 

Resultant BM diagram for the beam is shown in Figure 4.20(b). Shaded areas are positive BM.

Support Reactions

Moments about point B

 

4RA − 40 × 2 = MB = −19.9
RA = 15.025 kN

Reaction,

Moments about C

 

4RD − 40 × 2 = MC = −33.4
RD = 11.65 kN

Reaction,

Moreover

Total load on beam

 

40 + 40 + 20 × 4 = 160 kN

 

Reaction,

 

RB = 160 − RARCRD

= 160 −15.025 − 71.725 −11.65

= 61.6 kN

 

Problem 4.4 A continuous beam ABCD, cantilevered at end A, is supported on two spans BC and CD, where AB = 2 m, BC = 6 m, CD = 4 m. On span BC, a point load of 60 kN acts at E, at a distance of 2 m from B. On span CD, there is a udl of 20 kN/m at end A, a point load of 20 kN as shown in Figure 4.21(a). Determine support moments and draw BM diagram of continuous beam. Determine support reactions also.

Solution: MB, support moment at B (where AB is cantilever)

 

= −20 × 2 =−40 kNm

Figure 4.21 Problem 4.4

Span BC

Maximum bending moment under load,

Span DC

Maximum bending moment of centre

In this problem, moment, MD = 0, since end D is simply supported.

 

MB = –40 kNm (already calculated)

Using Clapeyron’s theorem of three moments for spans BC and DC,

Putting the values

But

          −40 × 6 + 20MC + 960 = 0

                MC = − 36kNm

In Figure 4.21(b), diagram AB′C′DCBA shows the support moment diagram, which is superimposed on a′1 and a′2 diagrams to get resultant BM diagram as shown. Positive BM areas are shaded.

Support Reaction

Taking moments about C

         4RD − 20 × 4 × 2 = MC = −6

               RD = 31 kNM

Moreover

 

19RD + 6RC − 20 × 4 (10 − 2) − 60 × 2

=

MB

10 × 31 6RC − 640 − 120

=

−40

310 + 6RC − 760

=

−40

6RC

=

410

Reaction,           RC = 68.333 kN

Reaction,

 

 

RB

=

Total load − RCRD

 

 

=

20 + 60 + 20 × 4 − 31 −68.333

 

 

=

160 − 99.333

 

 

=

60.667 kN

Problem 4.5     A continuous beam ABCD, 5 m long, is cantilevered at end A, span BC = CD = 2 m. There is no load on span BC, and on span CD there is a concentrated load W. At end A there is a concentrated load W. All supports are at same level. Determine support moments and draw BM and SF diagrams for continuous beam.

Solution: Bending moment,

 

MB = –W × 1 = –W (BBFigure 4.22)

Bending moment at D, MD = 0, simply supported end.

There is no load on span BC, so

Span CD

Maximum BM under load,

Figure 4.22 Problem 4.5

Using Clapeyron’s theorem of three moments for spans BC and CD, we get

AB′C′D is the support moment diagram of beam. Superimposing this over BM diagram considering the spans independently simply supported, AB′PA is (–ve) BM diagram and PC′DE′PC′ is the (+) BM diagram.

Support Reaction

Taking moments about C,

Reaction,

RB = +1.53125 W


Moreover,

Reaction

RD = +0.53125 W

Total load on beam

= 2 W

Reaction,

RC = 2 W − 1.53125 − 0.53125

 

= −0.0625 W

Figure 4.23 SF diagram

Figure 4.23 shows the SF diagram of beam, in which BB′ = 1.53125 W↑,

 

C″C′ = 0.0625 W ↓, DD′ = 0.53125 W

Problem 4.6     A two span continuous beam ABC, AB = 4 m, BC = 4 m, fixed at both the ends, carries a udl over AB, of 6 kN/m run and a moment 20 kNm (CCW) at centre of BC as shown in Figure 4.24(a). Determine support moments at A, B, and C. Draw the resultant BM diagram.

Solution: Let us consider spans AB and BC independently as simply supported and calculate a

Span AB

Span BC

ADDDC is the bending moment diagram, where

 

DD′ = +10 kNm, DD″ = -kNm

Figure 4.24 Problem 4.6

Consider imaginary span AA of zero length, and using Clapeyron’s theorem of three moments for spans AA AB, we get

Spans AB and BC

Consider imaginary span CC′ of zero length.

Span BC and CC′

So,

From Eq. (i),

 

MA = − 12 − 0.5MB           (iv)

From Eq. (iii),

 

MC = 1.25 − 0.25 MB           (v)

Putting the values of MA and MC in Eq. (ii)

          4 (−12 − 0.5MB) + 16MB + 4 (1.25 − 0.25MB)

                   = −116

or

         −48 − 2MB + 16MB + 5 − MB

                   = −116

         13MB = −116 + 48 − 5 = −73

Support moment

 

 

MB

=

− 5.615 kNm

 

MA

=

−12 − 0.5MB =−12 + 05 × 5.615

 

 

=

−9.192 kNm

 

MC

=

1.25 − 0.25MB

 

 

=

1.25 − 0.25 (−5.615)

 

 

=

1.25 + 02.5 (−5.615)

 

 

=

1.25 + 1.404 = +2.654 kNm

TakingAA′, =-9.192 kNm, BB′ =-5.615 kNm, CC′ =+2.654 kNm, draw AA′B′C′CA diagram on support moment diagram.

Superimposing the a′1 and a′1 diagrams of spans AB and BC, we get resultant bending moment diagram as shown in Figure 4.24 (b). The reader can notice that there are five points of contraflexure, i.e. P1, P2, P3, P4, and P5. Shaded parts of the BM diagram show positive bending moments.

Problem 4.7     A beam AB of length 2L is simply supported at ends. It is propped at the mid-point C, to the same level as of ends. A concentrated load W is applied at the mid-point of AC and another concentrated load kW is applied at mid-point of CB. For what value of k slope at the end A will be zero? Determine slope at C. EI is the flexural rigidity of the beam.

Solution: Figure 4.25(a) shows the loading diagram of the beam. Let us calculate a for spans AB and BC.

Span AB

Figure 4.25 Problem 4.7

Span BC

Using Clapeyron’s theorem of three moments for spans AC, CB, knowing that MA = MB = 0, simply supported ends

Taking moments about C

Take a section X-X, at a distance of x from A as shown.

 


BM at section

 


or


(i)

Integrating Eq. (i)

But at , at A as given in the problem

So,                    0 = 0 − omitted + C1

Constant           C1 = 0

Integrating Eq. (ii), we get

So,                    0 = 0 − omitted term + C2

Constant           C2 = 0

But at mid-point C, x = L, y = 0, putting the value


But

 


or

 

or

1 + k = 4

 

or

k = 3

(iv)

Moment,

Taking moments about C,

Moments about C

Slope at C

Putting RA in Eq. (ii)

Slope at C,

Problem 4.8     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 Figure 4.26. 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.

Solution: Let us first draw a diagram for both spans

area,

a′2 is a triangle with

Figure 4.26 Problem 4.8

Note that moment, MC = 0, because end C is a simple support.

Span AAB

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 moment,

Superimposing diagram over this, we get resultant bending moment diagram. Shaded parts show positive BM. There are three points of contraflexure as shown.

MULTIPLE CHOICE QUESTIONS
  1. A continuous beam ABCD carried over three supports with equal spans L each. There is a udl of intensity w throughout the length of the beam. What is support moment at middle two supports?
    1. 0.125wL2
    2. 0.10wL2
    3. 0.08wL2
    4. None of these
  2. A continuous beam 8 m long supported over two spans 4 m each carries a udl of 10 kN/m over its entire length. What is the support moment at central support?
    1. 20 kNm
    2. 30 kNm
    3. 40 kNm
    4. None of these
  3. A continuous beam 8 m long, supported over two spans 4 m each carries a udl of 10 kN/m. If the reaction at one end is 15 kN, what is the reaction at central support?
    1. 30 kN
    2. 40 kN
    3. 50 kN
    4. None of these
  4. A continuous beam 12 m long, supported over two spans 6 m each, carries concentrated loads of 40 kN each at the centre of each span. Bending moment at central support is
    1. –90 kNm
    2. –60 kNm
    3. –45 kNm
    4. None of these
  5. A continuous beam ABC of length 2L is supported on two equal spans AB = BC = L. A concentrated load W is applied at centre of AB. Another concentrated load P is applied at the centre of BC. If slope at end A is zero, what is the ratio of P/W?
    1. 4
    2. 3
    3. 2
    4. None of these

Answers

 

1. (b)

2. (a)

3. (c)

4. (c)

5. (b)

EXERCISES

 

4.1 A continuous beam of length 2L, supported over two equal spans, carries a udl of w intensity throughout its length. Determine support moments and reactions.

Ans.

4.2 A continuous beam ABCD 13 m long simply supported with spans AB = 4 m, BC = 5 m, CD = 4 m, carries a udl of 12 kN/m run over AB and CD and 16 kN/m over BC. Determine support reactions and support moments. E = 210 GPa, I = 4,000 cm4.

Ans. [16.5, 71.5, 71.5, 16.5 kN, 0, –30.09, –30.09, 0 kNm].

4.3 A continuous beam of length 36 m is supported over A, B, and C as shown in Figure 4.27. Span AB is 12 m long and span BC is 16 m long. The overhang is equal on both the sides. The beam carries a udl of 4.5 kN/m from D to B, 16 m and 6.0 kN/m from B to E, 20 m. Draw SF and BM diagrams of beam.

Ans. [MA –360 kNm, MB = –1,230 kNm, MC = –480 kNm, RA = 377.5, RB = 869.4, RC = 673.1 kN].

Figure 4.27 Exercise 4.3

4.4 A continuous beam ABC, having two spans, AB = L1 and BC=L2, carries a udl of intensity w throughout its entire length. The beam simply rests on end supports. If the support B sinks by δ below the level of supports A and C, show that reaction at B is

Figure 4.28 Exercise 4.5

4.5 For the uniform beam shown in Figure 4.28, determine support moments and support reactions. Draw BM diagram for the continuous beam ABC.

Ans.

4.6 A continuous beam ABC, hinged at end B, roller supported at A and C, is subjected to a bending couple M at C as shown in Figure 4.29. Determine support reactions. Draw BM diagram of beam.

Ans.

Figure 4.29 Exercise 4.6

Figure 4.30 Exercise 4.7

Figure 4.31 Exercise 4.8

4.7 Draw the BM diagram of a continuous beam ABC of uniform section throughout, if prop at B sink by 2 mm below the common level of supports A and C. Given E = 200 GPa, I = 8 × 10–6 m4 (Figure 4.30).

Ans. [MA = + 6 kNm, MB = –5.2 kNm].

4.8 A beam ABCD 12 m long is supported on A, B, C, and D with equal spans of 4 m each. It is sub jected to transverse loads; 40 kN at centre of AB and udl of 10 kN/m over span CD. Determine support moments and support reaction (Figure 4.31).

Ans. [0, –13.33, –6.66, 0 kNm; 16.667, 25.0, 20.0, 18.333 kN].

4.9 A continuous beam ABC, 12 m long, is hinged at its centre as shown in Figure 4.32. Draw the BM diagram of the beam if it carries a point load of 60 kN at the centre of AB and is subjected to a bending couple of 60 kNm at the centre of BC.

Ans. [MA = – 41.25 kNm, MB = –7.5 kNm, MC = –3.75 kNm].

Figure 4.32 Exercise 4.9

4.10 A continuous beam ABCD, 12 m long, is carried over spans AB = BC = 4 m, CD = 4 m. It carries a udl of 10 kN/m over BC, and a point load 10 kN at centre of CD. Determine support moments. What are support moments if end A is roller supported and is not hinged as shown in Figure 4.33?

Ans. [MA = 0, MB = –7 kNm, MC = –12 kNm; MA = MB = MD = 0, MC = –13.75 kNm].

Figure 4.33 Exercise 4.10

4.11 A continuous beam ABC 12 m long is supported on two spans AB = BC = 6 m each. Span AB carries a udl of 15 kN/m run and span BC carries a udl of 24 kN/m. Moment of inertia of beam AB is I1 and that for span BC is I2. If I1 = 0.5I2, determine supports moments and reactions.

Ans. [MB = –81 kNm, RA = 31.5, RB = 144, RC = 58.5 kN].

4.12 A continuous beam ABCD, 10 m long, fixed at end A, supported over spans AB and BC; AB = 5 m, BC = 4 m with overhang CD = 1 m. There is a udl of 10 kN/m over span AB, a concentrated load of 40 kN at the centre of BC and a concentrated load of 20 kN at free end D. While the supports A and C remain at the same level, the support B sinks by 1 mm. The moment of inertia of beam from A to B is 18,000 cm4 and from B to D is 12,000 cm4. If E = 210 kN/mm2, determine support moments and support reactions.

Ans. [MA = –28.83 kNm, MB = –13.92 kNm, MC = –20 kNm, RA = 27.98 kN, RB = 40.5 kN, RC = 41.52 kN].