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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.
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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
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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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