14 three moment equation
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14 three moment equationTRANSCRIPT
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
EI
LM
EI
LM
EI
xA
LLEI
MLL
EI
M
Lx
EI
A
LCC
36
3
2
2
1
3
1
2
11)(
121
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
EI
LM
EI
LM
EI
xA
LLEI
MLL
EI
M
Lx
EI
A
LCC
36
3
2
2
1
3
1
2
11)(
121
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Equating
and simplifying yields
)(2121 RRLL CCCC
RR
RR
LL
LL
R
RR
R
R
L
LC
L
LL
LI
xA
LI
xA
I
LM
I
L
I
LM
I
LM 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
I
Lw
I
Lwkk
I
LPkk
I
LP
I
LM
I
L
I
LM
I
LM
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*4
25*325.025.0
20*1500
20
5.0
2520
33
2
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For span AB:
kV
V
F
kA
A
M
AF
BL
BL
y
y
y
B
xx
6.44
0754.30
;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:
kV
V
F
kC
C
M
BR
BR
y
y
y
B
6.12
01538.2
;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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