14 three moment equation
DESCRIPTION
Three Moment Equations for analysisTRANSCRIPT
Department of Civil EngineeringUniversity of Engineering and Technology, Taxila, Pakistan
Three Moment Equation
Theory of Structure - I
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Lecture Outlines
Introduction
Proof of Three Moment Equation
Example
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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 RML MC MC MR
P1 P2 P3 P4
WL WR
LL LR
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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 LRCL1CR1
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 LRCL2CR2
R’
ML /EIL
MC /EIL
MC /EIR
MR /EIR
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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 RRLL CCCC
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L
LC
L
LL
L
LL
LLL
CLL
L
L
LL
L
L
LLL
EILM
EILM
EIxA
LLEIMLL
EIM
Lx
EIA
LCC
36
32
21
31
211)(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
R
RR
RRR
CRR
R
R
RR
R
R
RRR
EILM
EILM
EIxA
LLEIMLL
EIM
Lx
EIA
LCC
36
32
21
31
211)(1
21
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Equating
and simplifying yields
)(2121 RRLL CCCC
RR
RR
LL
LL
R
RR
R
R
L
LC
L
LL
LIxA
LIxA
ILM
IL
ILM
ILM 662
General Equation
(1)
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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.
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L C C RC R
LL
KLLLKRLR
PL PR
w
R
RR
L
LLRR
R
RRLL
L
LL
R
RR
R
R
L
LC
L
LL
ILw
ILwkk
ILPkk
ILP
ILM
IL
ILM
ILM
442
333
23
2
If the areas and centroidal distances for their M/EI diagrams are substituted in to 3-Moment equation,
(2)
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Special Case: If the moment of inertia is constant for the
entire span, IL = IR.
44
233
3232 RRLLRRRRLLLLRRRLCLL
LwLwkkLPkkLPLMLLMLM
(3)
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Example: Determine the reactions at the supports for
the beam shown. The moment of inertia of span AB is one half that of span BC.
15k3 k/ft
I0.5 I
25 ft 15 ft 5 ft
A CB
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ML = 0 LL = 25ft IL = 0.5I PL = 0 wL = 3k/ft kL = 0
MC = MB
LR = 20ft IR = I PR = 15k wR = 0 kR = 0.25
MR = 0
Data
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Substituting the values in equation 2,
ftkM
IIIIM
B
B
.5.177
05.0*425*325.025.020*150020
5.02520
33
2
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For span AB:
kVV
F
kA
A
M
AF
BL
BL
y
y
y
B
xx
6.440754.30
;0
4.30
0)5.12(755.177)25(
;0
0;0
75 k
A B
12.5’ 12.5’
VBL
177.5k.ftAy
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For span BC:
kVV
F
kC
C
M
BR
BR
y
y
y
B
6.1201538.2
;0
38.2
0)15(155.177)20(
;0
15 k
B C
15 ft 5 ft
VBR
177.5k.ft Cy
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A free body diagram of the differential segment of the beam that passes over roller at B is shown in figure.
kB
B
F
y
y
y
2.57
06.126.44
0
By
44.6 k 12.6 k
177.5k.ft 177.5k.ft
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Practice Problems:
Chapter 9
Example 9-11 to 9-13 and Exercise
Structural Analysis by R C Hibbeler
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