Department of Civil Engineering

University of Engineering and Technology, Taxila, Pakistan

Three Moment

Equation

Theory of Structure - I

2

Lecture Outlines

Introduction

Proof of Three Moment Equation

Examples

3

Introduction

Developed by French Engineer Clapeyron in 1857.

This equation relates the internal moments in a continuous beam at three points of support to the loads acting between the supports.

By successive application of this equation to each span of the beam, one obtains a set of equations that may be solved simultaneously for the unknown internal moments at the support.

4

Proof: Real Beam

A general form of three moment equation can

be developed by considering the span of a

continuous beam.

L C R

ML MC MC MR

P1 P2 P3 P4

WL WR

LL LR

5

Conjugate Beam (applied

loads)

The formulation will be based on the

conjugate-beam method.

Since the “real beam” is continuous over the

supports, the conjugate-beam has hinges at

L, C and R.

L’

XL XR

LL LR CL1 CR1

R’

AL /EIL AR /EIR

6

Conjugate Beam (internal

moments)

Using the principle of superposition, the M/EI

diagram for the internal moments is shown.

L’ LL LR CL2 CR2

R’

ML /EIL

MC /EIL

MC /EIR

MR /EIR

7

In particular AL/EIL and AR/EIR represent the

total area under their representative M/EI

diagrams; and xL and xR locate their

centroids.

Since the slope of real beam is continuous

over the center support, we require the shear

forces for the conjugate beam.

)(2121

RRLLCCCC

8

L

LC

L

LL

LL

LL

LL

L

C

LL

L

L

L

L

L

L

L

LL

EI

LM

EI

LM

LEI

xA

LLEI

MLL

EI

M

Lx

EI

A

LCC

36

3

2

2

1

3

1

2

11)(

1

21

Summing moments about point L’ for left

span, we have

Summing moments about point R’ for the

right span yields

R

RC

R

RR

RR

RR

RR

R

C

RR

R

R

R

R

R

R

R

RR

EI

LM

EI

LM

LEI

xA

LLEI

MLL

EI

M

Lx

EI

A

LCC

36

3

2

2

1

3

1

2

11)(

1

21

9

Equating

and simplifying yields

)(2121

RRLLCCCC

RR

RR

LL

LL

R

RR

R

R

L

L

C

L

LL

LI

xA

LI

xA

I

LM

I

L

I

LM

I

LM 662

General Equation

10

Eq. Modification for point load

and uniformly distributed load

Summation signs have been added to the

terms on the right so that M/EI diagrams for

each type of applied load can be treated

separately.

In practice the most common types of

loadings encountered are concentrated and

uniform distributed loads.

11

L C C R C R

LL

KLLL KRLR

PL PR

w

R

RR

L

LL

RR

R

RR

LL

L

LL

R

RR

R

R

L

L

C

L

LL

I

Lw

I

Lwkk

I

LPkk

I

LP

I

LM

I

L

I

LM

I

LM

442

33

3

2

3

2

If the areas and centroidal distances for their

M/EI diagrams are substituted in to 3-Moment

equation,

12

Special Case:

If the moment of inertia is constant for the

entire span, IL = IR.

44

2

33

3232 RRLL

RRRRLLLLRRRLCLL

LwLwkkLPkkLPLMLLMLM

Three-Moment Theorem

Any number of spans

Symmetric or non-symmetric

Procedure:

1) Draw a free body diagram of the

first two spans.

2) Label the spans L1 and L2 and the

supports (or free end) A, B and C

as show.

3) Use the Three-Moment equation to

solve for each unknown moment,

either as a value or as an

equation.

4) Move one span further and repeat

the procedure.

13

Three-Moment Theorem

Procedure:

5) In a 3 span beam, the mid-moment from

step 3 above (B), could now be solved

using the two equations from step 4 and

3 together, by writing 2 equations with 2

unknowns.

6) Repeat as needed, always moving one

span to the right and writing a new set of

moment equations.

7) Solve 2 simultaneous equations for 3

spans, or 3 equations for more than 3

spans, to get the interior moments.

8) Once all interior moments are known,

solve for reactions using free body

diagrams of individual spans.

9) Draw shear and moment diagrams as

usual. This will also serve as a check

for the moment values. 14

Three-Moment Theorem SUMMARY:

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