14 three moment equation

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

2. Lecture Outlines
 Introduction
 Proof of Three Moment Equation
 Example
Department of 2

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.
Department of 3

4. Proof: Real Beam
A general form of three moment equation can
be developed by considering the span of a
continuous beam.
P1 P2 P3 P4
WL WR
ML MC MC MR
L C R
LL LR
Department of 4

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.
AL /EIL AR /EIR
L’ LL CL 1 CR 1 LR R’
XL XR
Department of 5

6. Conjugate Beam (internal
moments)
 Usingthe principle of superposition, the M /
EI diagram for the internal moments is
shown.
MC /EIL
MR /EIR
ML /EIL MC /EIR
L’ LL CL2 CR2 LR R’
Department of 6

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.
C L1 + C L2 = −(C R1 + C R2 )
Department of 7

8.  Summing moments about point L’ for left
span, we have
1 AL 1 1  M L  1  1  M  2 
C L1 + C L2 = ( xL ) +   ( LL ) LL  +  C ( LL ) LL 
LL EI L LL 2  EI L
 



 3  2  EI L

 3 
AL xL M L LL M C LL
= + +
EI L 6 EI L 3EI L
 Summing moments about point R’ for the
right span yields
1 AR 1 1  M R  1  1  M  2 
C R1 + C R2 = ( xR ) +   ( LR ) LR  +  C ( LR )  LR 
LR EI R LR 2  EI R
 



 3  2  EI R

 3 
AR xR M R LR M C LR
= + +
EI R 6 EI R 3EI R
Department of 8

9. General Equation
 Equating
C L1 + C L2 = −(C R1 + C R2 )
 and simplifying yields
M L LL  LL LR  M R LR 6 AL xL 6 AR xR
+ 2M C  +  +
I  = −∑ −∑
IL  L IR  IR I L LL I R LR
(1)
Department of 9

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.
Department of 10

11. PL PR
w
L KLLL C C KRLR R C R
LL
 If the areas and centroidal distances for their
M/EI diagrams are substituted in to 3-Moment
equation,
 LL LR  M R LR
( ) ( )
2 2 3 3
M L LL PL LL PR LR wL LL wR LR
+ 2M C  +  + = −∑ kL − kL − ∑
3 3
I  kR − kR − −
IL  L IR  IR IL IR 4I L 4I R
(2)
Department of 11

12. Special Case:
 If the moment of inertia is constant for the
entire span, IL = IR.
( ) ( )
3 3
wL LL wR LR
M L LL + 2M C ( LL + LR ) + M R LR = − ∑ PL LL k L − k L − ∑ PR LR k R − k R
2 3 2 3
− −
4 4
(3)
Department of 12

13. Example:
 Determinethe reactions at the supports for
the beam shown. The moment of inertia of
span AB is one half that of span BC.
3 k/ft
15k
A C
0.5 I B I
25 ft 15 ft 5 ft
Department of 13

14. Data
M =0 M = MB M =0
L C R
L = 25ft L = 20ft
L R
I = 0.5I I =I
L R
P =0 P = 15k
L R
w = 3k/ft w =0
L R
k =0 k = 0.25
L R
Department of 14

15.  Substituting the values in equation 2,
 25 20 
0 + 2M B  + +0 = 0−∑
15 * 20 2
(
0.25 − 0.253 − )
3 * 253
−0
 0.5I I  I 4 * 0.5I
M B = −177.5k . ft
Department of 15

16.  For span AB:
∑ F = 0; A
x x =0
75 k
∑ M = 0;B
− Ay (25) − 177.5 + 75(12.5) = 0
VBL
Ay = 30.4k
A B
∑F y = 0;
Ay
12.5’ 12.5’ 177.5k.ft
30.4 − 75 + VBL = 0
VBL = 44.6k
Department of 16

17.  For span BC:
∑M B = 0;
15 k
C y (20) + 177.5 + 15(15) = 0
C y = 2.38k VBR
∑F y = 0; B
15 ft 5 ft
C
177.5k.ft
2.38 − 15 + VBR = 0 Cy
VBR = 12.6k
Department of 17

18. A free body diagram of the differential
segment of the beam that passes over roller
at B is shown in figure.
∑F y =0 177.5k.ft 177.5k.ft
B y − 44.6 − 12.6 = 0
B y = 57.2k 44.6 k 12.6 k
By
Department of 18

19. Practice Problems:
 Chapter 9
 Example 9-11 to 9-13 and Exercise
 Structural Analysis by R C Hibbeler
Department of 19

20. Department of 20