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 L_{1}, L_{2}, and L_{3}, 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 L_{1}′ L_{1}, L_{2}, and L_{2}′, respectively, as shown in Figure 4.2. In this beam consider consecutive spans AB and BC carrying udl of intensity w_{1} and w_{2}, respectively, as shown in Figure 4.2. Say support moments at A, B, and C are M_{A}, M_{B}, M_{C}, 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 XX is
Support moments M_{A}, M_{B}, M_{C} 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 C_{1} is a constant of integration.
At support B, where x = 0, slope (say),
so,
or constant of integration, C_{1} = +EIi_{B}
Now
Integrating Eq. (4.2), we get
where C_{2} is another constant of integration.
At support B, x = 0, deflection y = 0,
so,
Constant
Finally
At the support A, x = L_{1}, 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 i′_{B} = − i′_{B}
∴ 2M_{B} (L_{1} + L_{2}) + M_{A}L_{1} + M_{C}L_{2}
This is a wellknown “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
Moreover, the beam is symmetrically loaded about its centre; therefore, M_{B} = M_{C} and reactions R_{A} = R_{D} and R_{B} = R_{C}. Using Clapeyron’s theorem of three moments for spans BA and BC
But M_{A} = 0, M_{B} = B_{C},
so,
Figure 4.4 Example 4.1
Taking moments at support B of the beam, only on left side of B
or
Reaction,
Reaction, R_{A} = 0.4wL = Reaction, R_{D}
Total load on beam = 4wL
= R_{A} + R_{B} + R_{C} + R_{D}
But R_{B} = R_{C}, due to symmetry, therefore,
2R_{A} + 2 R_{B} = 3wL
or +2 × 0.4wL + 2R_{B} = 3wL
Reaction, R_{B} = 1.1wL = Reaction, R_{C}
Bending moment at centre of each span
Figure 4.5 shows the BM diagram for the continuous beam with four points of contraflexure P_{1}, P_{2}, P_{3}, and P_{4}, respectively. Bending moment at centre of span AB and DC
= 0.125wL^{2} − 0.05wL^{2}
= +0.075wL^{2}
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
M_{B} = M_{C} = −0.1 wL^{2}
Reaction R^{B} = R_{C} = 1.1 wL
Take a section XX at a distance of x from B, bending moment
Integrating Eq. (4.7), we get
where C_{1} 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 C_{2} is another constant of integration.
Constant C_{2} = 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. [R_{A} = 9.4 kN, R_{B} = 30.5, R_{C} = 8.1 kN, M_{B} = –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 L_{1} and L_{2}, 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 XX, bending moment at the section will be
= 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 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
(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 i_{B} = –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 a′_{1} and a′_{2} are positive as per convention and support moments M_{A}, M_{B}, M_{C} 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)
Using the equation of three moments for spans AB and CB
or M_{A} + 4M_{B} + MC = −132.5 (i)
But M_{A} = 0
So, 4M_{B} + M_{C} = −132.5 (ii)
Using the equation of three moments for spans BC and DC
But
15M_{C} = −197.5
M_{C} = −13.166 kNm
M_{B} = −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 L_{1} and L_{2} of a continuous beam. Supports A, B, and C are not at one level. Support B is below support A by δ_{1} and below support C by δ_{2} 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 XX, 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 i_{B} = –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, i_{B} + –i′_{B} = 0
Adding Eqs (4.16) and (4.17)
or M_{A} L_{1} + 2M_{B} (L_{1} + L_{2}) + M_{C}L_{2}
From this equation M_{A}, M_{B}, and M_{C} 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 kNm^{2}.
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 M_{A} = M_{C} = 0
Using the equation derived in Article 3.3
Total load on beam 

= 
6 × 2 + 6 = 18kN 
Moreover 
M_{B} 
= 
R_{A} × 6 − 6 × 2 × 3 

−1.95 
= 
6R_{A} − 36 
Reaction, 
R_{A} 
= 
+ 5.675 kN 
also 
M_{B} 
= 
4R_{C} − 6 × 2 

−1.95 
= 
4R_{C} −12 
Reaction, 
R_{C} 
= 
+2.5125 kN 
Reaction, 
R_{B} 
= 
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/mm_{2}, I = 30,000 cm_{4}.
Ans. [M_{A} = M_{C} = 0, M_{B} = –84.43 kNm, R_{A} = 41.51 kN, R_{B} = 135.46 kN, R_{C} = 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 a′_{1} and a′_{2} are plotted considering spans AB and BC supported independently as shown in Figure 4.14. At any section if:
M′_{x} = BM due load on span, considering span independently
M″_{x} = 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 M_{A} 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
Putting the value of M_{A} in Eq. (iii)
Solving Eqs (v) and (vi), we get
Moment, M_{B} = −20.625 kNm
Putting the value of M_{B} in Eq. (vi)
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

M_{C} 
= 
4R_{D} − 60 × 2 = −32.334 
Reaction, 
R_{D} 
= 
21.9 kN 

M_{B} 
= 
8R_{D} + 4R_{C} − 6 × 60 − 60 × 2 = −20.625 

8R_{D} + 4R_{C} 
= 
459.375 

8 × 21.9 + 4R_{C} 
= 
459.375 
Reaction, 
R_{C} 
= 
71.0 kN 
Moments about A
Putting the values and after solution.
Reaction, 
R_{B} 
= 
41.1 kN 
Total load on beam 

= 
60 + 4 × 15 + 60 = 180 kN 
Reaction 
R_{A} 
= 
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. [M_{A} –21.43 kNm, M_{B} = –32.14 kNm, M_{C} = 0, R_{A} = 17.86, R_{B} = 48.57 kN, R_{C} = 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 I_{1}, but moment of inertia of section of beam BC is I_{2}. If I_{1} = 2I_{2}, determine support moments and support reactions.
Solution: Let us consider a′_{1} and a′_{2} 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 M_{A} = M_{C} = 0,L_{1} = 6m, L_{2} = 4 m
I_{1 }= 2I_{2}, putting these values
or 14M_{B} = −324 −120 = −444
Moment
Taking M_{A} = 0, M_{B} = –31.71 kNm, M_{C} = 0, support moment diagram is drawn. Resultant BM diagram with positive and negative areas is shown in Figure 4.17(b).

M_{B} 
= 
6R_{A} −12 × 6 × 3 = −31.71 
Reaction, 
R_{A} 
= 
30.71 kN 
Moreover 
M_{B} 
= 
4R_{C} − 20 × 2 = −31.71 
Reaction, 
R_{C} 
= 
2.07 kN 
Reaction, R_{B} = 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
Reactions
Taking moments about A,
10 × 2 + 1 + 20 × 2 × 3 
= 
4R′_{B} 
R′_{B} 
= 
35kN 
R′_{A} 
= 
20 + 40 − 35 = 25 kN 
Span AB
Origin at A,
Putting the values of R′_{A} w_{1}, w_{2}, etc.
Now using Clapeyron’s theorem of three moments, note that M_{A} = M_{C} = 0
= −250 − 514.28
M_{B} = −34.74 kNm

MB 
= 
4RA − 2 × 10 × 3 − 2 × 20 × 1 = −34.74 
Reaction, 
R_{A} 
= 
16.325 kN 
Moreover 
M_{B} 
= 
7R_{C} − 40 × 5 = −34.74 
Reaction, 
R_{C} 
= 
23.60 kN 
Reaction, 
R_{B} 
= 
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 kNm^{2}.
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
Maximum BM
Span CD
Maximum BM
Using the theorem of three moments with a sinking of support for spans AB and CB, note that M_{A} = 0
Putting the values
Using the equation of three moments for spans BC and DC.
Now δ_{3} = 0
Support C is 6 mm higher than support B
or δ_{3} = −0.0006m
Note that M_{D} = 0
or
15M_{B} = −298.5
M_{B} = −19.9 kNm
Support moment,
From Eq. (iii)
Support moment,
Resultant BM diagram for the beam is shown in Figure 4.20(b). Shaded areas are positive BM.
Support Reactions
Moments about point B
Reaction,
Moments about C
Reaction,
Moreover
Total load on beam
Reaction,
R_{B} = 160 − R_{A} − R_{C} − R_{D}
= 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: M_{B}, support moment at B (where AB is cantilever)
Figure 4.21 Problem 4.4
Span BC
Maximum bending moment under load,
Span DC
Maximum bending moment of centre
In this problem, moment, M_{D} = 0, since end D is simply supported.
Using Clapeyron’s theorem of three moments for spans BC and DC,
Putting the values
But
−40 × 6 + 20M_{C} + 960 = 0
M_{C} = − 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
4R_{D} − 20 × 4 × 2 = M_{C} = −6
R_{D} = 31 kNM
Moreover
19R_{D} + 6R_{C} − 20 × 4 (10 − 2) − 60 × 2 
= 
M_{B} 
10 × 31 6R_{C} − 640 − 120 
= 
−40 
310 + 6R_{C} − 760 
= 
−40 
6R_{C} 
= 
410 
Reaction, R_{C} = 68.333 kN
Reaction,

R_{B} 
= 
Total load − R_{C} − R_{D} 


= 
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,
Bending moment at D, M_{D} = 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, 
R_{B} = +1.53125 W 


Reaction 
R_{D} = +0.53125 W 
Total load on beam 
= 2 W 
Reaction, 
R_{C} = 2 W − 1.53125 − 0.53125 

= −0.0625 W 
Figure 4.23 shows the SF diagram of beam, in which BB′ = 1.53125 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
AD′DD″C is the bending moment diagram, where
Figure 4.24 Problem 4.6
Consider imaginary span A′A of zero length, and using Clapeyron’s theorem of three moments for spans A′A AB, we get
Spans AB and BC
Consider imaginary span CC′ of zero length.
Span BC and CC′
So,
From Eq. (i),
From Eq. (iii),
Putting the values of M_{A} and M_{C} in Eq. (ii)
4 (−12 − 0.5M_{B}) + 16M_{B} + 4 (1.25 − 0.25M_{B})
= −116
or
−48 − 2M_{B} + 16M_{B} + 5 − M_{B}
= −116
13M_{B} = −116 + 48 − 5 = −73
Support moment

M_{B} 
= 
− 5.615 kNm 

M_{A} 
= 
−12 − 0.5M_{B} =−12 + 05 × 5.615 


= 
−9.192 kNm 

M_{C} 
= 
1.25 − 0.25M_{B} 


= 
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. P_{1}, P_{2}, P_{3}, P_{4}, and P_{5}. 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 midpoint C, to the same level as of ends. A concentrated load W is applied at the midpoint of AC and another concentrated load kW is applied at midpoint 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 M_{A} = M_{B} = 0, simply supported ends
Taking moments about C
Take a section XX, at a distance of x from A as shown.





Integrating Eq. (i)
But at , at A as given in the problem
So, 0 = 0 − omitted + C_{1}
Constant C_{1} = 0
Integrating Eq. (ii), we get
So, 0 = 0 − omitted term + C_{2}
Constant C_{2} = 0
But at midpoint C, x = L, y = 0, putting the value






or 
1 + k = 4 

or 
k = 3 
(iv) 
Moment,
Taking moments about C,
Moments about C
Slope at C
Putting R_{A} 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 cm^{4} and from BC it is 12,000 cm^{4}. If E = 210 kN/mm^{2}, 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, M_{C} = 0, because end C is a simple support.
Span AAB
Imaginary span AA′ and AB equation of three moments.
Taking I_{1} common throughout
(because level of A is higher than level of B by 1 mm)
Now using the theorem of three moments for spans AB and BC, and noting that I_{1} is different than I_{2}, equation can be modified as
Multiplying throughout by I_{1},
But I_{1} = 1.5I_{2}, putting this value, we get
6M_{A} + 30M_{B} + 9M_{C} = −540 −810 + 37.8 + 37.8
6M_{A} + 30M_{B} + 9M_{C} = −1274.4 kNm
But M_{C} = 0
So, 
6M_{A} + 30M_{B} = −1274.4 kNm 
(ii) 

6M_{A} + 3M_{B} = −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
 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?
 0.125wL^{2}
 0.10wL^{2}
 0.08wL^{2}
 None of these
 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?
 20 kNm
 30 kNm
 40 kNm
 None of these
 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?
 30 kN
 40 kN
 50 kN
 None of these
 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
 –90 kNm
 –60 kNm
 –45 kNm
 None of these
 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?
 4
 3
 2
 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 cm^{4}.
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. [M_{A} –360 kNm, M_{B} = –1,230 kNm, M_{C} = –480 kNm, R_{A} = 377.5, R_{B} = 869.4, R_{C} = 673.1 kN].
Figure 4.27 Exercise 4.3
4.4 A continuous beam ABC, having two spans, AB = L_{1} and BC=L_{2}, 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} m^{4} (Figure 4.30).
Ans. [M_{A} = + 6 kNm, M_{B} = –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. [M_{A} = – 41.25 kNm, M_{B} = –7.5 kNm, M_{C} = –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. [M_{A} = 0, M_{B} = –7 kNm, M_{C} = –12 kNm; M_{A} = M_{B} = M_{D} = 0, M_{C} = –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 I_{1} and that for span BC is I_{2}. If I_{1} = 0.5I_{2}, determine supports moments and reactions.
Ans. [M_{B} = –81 kNm, R_{A} = 31.5, R_{B} = 144, R_{C} = 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 cm^{4} and from B to D is 12,000 cm^{4}. If E = 210 kN/mm^{2}, determine support moments and support reactions.
Ans. [M_{A} = –28.83 kNm, M_{B} = –13.92 kNm, M_{C} = –20 kNm, R_{A} = 27.98 kN, R_{B} = 40.5 kN, R_{C} = 41.52 kN].