15.8. Continuous Beams - 15.9. Bending Moment Diagram for Continuous Beams

Oct 25, 2026

Continuous beams are defined as beams supported on more than two supports, subjected to external loading such as uniformly distributed load. The deflection shape leads to hogging moments (negative) over intermediate supports and sagging moments (positive) at midspan. Since end moments differ between simply supported and fixed continuous beams, intermediate support moments are determined using Clapeyron’s three-moment equation. With span lengths and areas of bending-moment diagrams, the resultant bending moment diagram and shear-force diagrams can be constructed for given problems.

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15.8. CONTINUOUS BEAMS

Definition and moment behavior

15.9. BENDING MOMENT DIAGRAM FOR CONTINUOUS BEAMS

Clapeyron’s equation of three moments

Derivation and application

Worked problem (15.9)

■0000000.000000000 # 15.8.CONTINUOUS BEAMS x₁=Distance of C.G.of the B.M.diagram due to vertical loads on BC from B Continuous beam is a beam which is supported on more than two supports.Fig.15.16shows such a beam,which is subjected to some external loading(here a uniformly distributedload).The deflection curve for the beam is shown by dotted line.The deflection curve is havingconvexity upwards over the intermediate supports,and concavity upwards over the mid of thespan.Hence there will be hogging moments (L.e,negative)over the intermediate supports andsagging moments(i.e.,positive)over the mid of the span.The end supports of a simply sup-ported continuous beam will not be subjected to any bending moment.But the end support offixed continuous beam will be subjected to fixing moments.If the moments over the interme-diate supports are known,then the B.M.diagram can be drawn. x₂=Distance of C.G.of the B.M.diagram due to vertical loads on CD from D. The equation(15.12)is known as theequation of three moments or Clapeyron's equation. 15.9.1Derivation of Clapeyron's Equation of three Moments.Fig.15.17 showsthe length BCD(two consecutive spans)of a continuous beam which is shown in Fig.15.16.LetMg,Mcand Mpare the support moments at B,C and D respectively. (b) Fig.15.16 The Fig.15.16 shows a simply supported continuous beam.In this figure the end sup-ports atA and E will not be subjected to any bending moment.Hence in this case Ma=Mg=0.Fig.15.16(a)shows a continuous beam with fixedends atA andE.Here the end supportsat A and E will be subjected to fixing moments.Hence MAand Mg willnot be zero. (c) Fig.15.16(a) (d) # 15.9.BENDING MOMENT DIAGRAM FOR CONTINUOUS BEAMS In Art.15.8 it is mentioned that if the moments over the intermediate supports of acontinuous beam are known,then the B.M,diagram can be drawn easily.The moments overthe interrmediate supports are determined by using Clapeyron's theorem of three momenswhich states that: Resultant B.M.Diagram Fig.15.17 Let L₁=Length of span BCL2=Length of span CD IfBC andCD are any two consecutive span ofa continuous beam subjected to an externalB,C and D are given by,loading,the a₁=Area of B.M.diagram due to vertical loads on span BCa₂=Area of B.M.diagram due to vertical loads on span CDa₁'=Area of B.M.diagram due to support moments Mgand Mca₂'=Area of B.M.diagram due to support moments Mcand MDx₁=Distance of C.G.of B.M.diagram due to vertical loads on BCx₂=Distance of C.G.of B.M.diagram due to vertical loads on CD₁'=Distance of C.G.of B.M.diagram due to support moments on BC₂'=Distance of C.G.of B.M.diagram due to support moments on CD. ..(15.12) where L₁=Length of span BCL₂=Length of span CD a²=Area of B.M.diagram due to vertical loads on span BCa₂=Area of B.M.diagram due to vertical loads on span CD Fig.15.17(b)and(c)shows the B.M.diagrams due to vertical loads and due to supportsmoments respectively. (i)Consider the span BC Let M=B.M.due to vertieal loads at a distance x from B(sagging)M'=B.M.due to support moments at a distance x from B(hogging) ∴Net B.M.at a distance x from B is given by, Multiplying by x to both sides,we get Integrating from zero to L₁,we get ..(i)(∵M.dx=Area of B.M.diagram of length dx.And x.M,dx=Moment of area of B.M.diagram of length dx about B. or Substituting the limits in L.H.S.of equation(i),we have =a₁x1-α₁'x₁'EI(L1θc-yc)-(0-yg)]=a1ˣ1-a₁'x₁'or But deflection at B and C are zero.Hence yB=0 and yc=0.Hence above equationbecomes as ..(ii) [EIL1.θc=a₁x₁-a₁'x₁' Buta₁'=Area of B.M.diagram due to supports moments=Area of trapezium BCKJ=(Mg+Mc)×L₁ x₁′=Distance of C.G.of area BCKJ from Band Substituting the values of a₁and x₁′in equation(ii),we get ...(iii) or (ii)Consider the span CD Similarly considering the span CD and taking D as origin and x positive to the left,itcan be shown that [In the above case the slope at C(i.e,θg)will have opposite sign than that given byequation(iii

FAQs

What characterizes a continuous beam and how do its moments vary along the span?

A continuous beam is supported on more than two supports. The deflection curve is convex upward over intermediate supports (hogging/negative moments) and concave upward near midspan (sagging/positive moments).

How are support end moments different for simply supported versus fixed continuous beams?

In a simply supported continuous beam, end supports are not subjected to bending moments. In a fixed continuous beam, end supports are subjected to fixing moments, so the end moments are not zero.

What is Clapeyron’s equation of three moments used for in continuous beams?

It is used to determine the moments at intermediate supports. Once intermediate moments are known, the bending moment diagram for the continuous beam can be drawn.

Oct 25, 2026

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