14 three moment equation

14 three moment equation

1. Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan
Three Moment
Equation
Theory of Structure - I

2. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
2
Lecture Outlines
Introduction
Proof of Three Moment Equation
Example

3. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
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. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
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
M
L
M
C
M
C
M
R
P
1
P
2
P
3
P
4
W
L W
R
L
L
L
R

5. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
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’
X
L X
R
L
L
L
R
C
L
1
C
R
1
R’
A
L
/EI
L
A
R
/EI
R

6. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
6
Conjugate Beam (internal
moments)
Using the principle of superposition, the M /
EI diagram for the internal moments is
shown.
L’ L
L
L
R
C
L
2
C
R
2
R’
M
L
/EI
L
M
C
/EI
L
M
C
/EI
R
M
R
/EI
R

7. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
7
In particular A
L
/EI
L
and A
R
/EI
R
represent the
total area under their representative M / EI
diagrams; and x
L
and x
R
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 +-=+

8. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
8
() ()
L
LC
L
LL
L
LL
LL
L
C
LL
L
L
L
L
L
L
L
LL
EI
LM
EI
LM
EI
xA
LL
EI
M
LL
EI
M
L
x
EI
A
L
CC
36
3
2
2
1
3
1
2
11
)(
1
21
++=
ú
û
ù
ê
ë
é
÷
ø
ö
ç
è
æ
÷
÷
ø
ö
ç
ç
è
æ
+÷
ø
ö
ç
è
æ
÷
÷
ø
ö
ç
ç
è
æ
+=+
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
RR
R
C
RR
R
R
R
R
R
R
R
RR
EI
LM
EI
LM
EI
xA
LL
EI
M
LL
EI
M
L
x
EI
A
L
CC
36
3
2
2
1
3
1
2
11
)(
1
21
++=
ú
û
ù
ê
ë
é
÷
ø
ö
ç
è
æ
÷
÷
ø
ö
ç
ç
è
æ
+÷
ø
ö
ç
è
æ
÷
÷
ø
ö
ç
ç
è
æ
+=+

9. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
9
Equating
and simplifying yields
)(
2121
RRLL
CCCC +-=+
åå --=+
÷
÷
ø
ö
ç
ç
è
æ
++
RR
RR
LL
LL
R
RR
R
R
L
L
C
L
LL
LI
xA
LI
xA
I
LM
I
L
I
L
M
I
LM 66
2
General Equation
(1)

10. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
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. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
11
L C C RC R
L
L
K
L
L
L
K
R
L
R
P
L P
R
w
( ) ( )
R
RR
L
LL
RR
R
RR
LL
L
LL
R
RR
R
R
L
L
C
L
LL
I
Lw
I
Lw
kk
I
LP
kk
I
LP
I
LM
I
L
I
L
M
I
LM
44
2
33
3
2
3
2
------=+
÷
÷
ø
ö
ç
ç
è
æ
++ åå
If the areas and centroidal distances for their
M/EI diagrams are substituted in to 3-Moment
equation,
(2)

12. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
12
Special Case:
If the moment of inertia is constant for the
entire span, I
L
= I
R
.
( ) ( ) ( )
44
2
33
3232 RRLL
RRRRLLLLRRRLCLL
LwLw
kkLPkkLPLMLLMLM ------=+++ åå
(3)

13. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
13
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.
15k
3 k/ft
I0.5 I
25 ft 15 ft 5 ft
A C
B

14. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
14
M
L
= 0
L
L
= 25ft
I
L
= 0.5I
P
L
= 0
w
L
= 3k/ft
k
L
= 0
M
C
= M
B
L
R
= 20ft
I
R
= I
P
R
= 15k
w
R
= 0
k
R
= 0.25
M
R
= 0
Data

15. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
15
Substituting the values in equation 2,
( )
ftkM
IIII
M
B
B
.5.177
0
5.0*4
25*3
25.025.0
20*15
00
20
5.0
25
20
3
3
2
-=
----=+÷
ø
ö
ç
è
æ
++ å

16. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
16
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
=
=+-
=
=
=+--
=
==
å
å
å
75 k
A B
12.5’ 12.5’
V
BL
177.5k.ft
A
y

17. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
17
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
=
=+-
=
=
=++
=
å
å
15 k
B C
15 ft 5 ft
V
BR
177.5k.ft
C
y

18. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
18
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
=
=--
=å
B
y
44.6 k 12.6 k
177.5k.ft 177.5k.ft

19. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
19
Practice Problems:
Chapter 9
Example 9-11 to 9-13 and Exercise
Structural Analysis by R C Hibbeler

20. Department of
Civil
Engineering
University of
Engineering
and
Technology,
Taxila,
Pakistan
20