Transcript Moment Distribution

ERT 348 Controlled Environment Design 1
Moment Distribution
Ms Siti Kamariah Binti Md Sa’at
School of Bioprocess Engineering,
UniMAP
[email protected]
Introduction
Such structures are indeterminate, i.e. there are
more unknown variables than can be solved
using only the three equations of equilibrium.
This section deals with continuous beams and
propped cantilevers.
An American engineer, Professor Hardy Cross,
developed a very simple, elegant and practical
method of analysis for such structures called
Moment Distribution.
Indeterminate structures
Bending (Rotational) Stiffness
A fundamental relationship which exists in the
elastic behaviour of structures and structural
elements is that between an applied force
system and the displacements which are
induced by that system,
Force = Stiffness x Displacement
P
=kxδ
Stifness, k = P/δ
when δ = 1.0 (i.e. unit displacement) the
stiffness is: ‘the force necessary to maintain a
UNIT displacement, all other displacements
being equal to zero.’
Bending (Rotational) Stiffness
The displacement can be a shear displacement,
an axial displacement, a bending (rotational)
displacement or a torsional displacement, each
in turn producing the shear, axial, bending or
torsional stiffness.
When considering beam elements in continuous
structures using the moment distribution method
of analysis, the bending stiffness is the principal
characteristic which influences behaviour.
Example:
The force (MA) necessary to maintain this
displacement can be shown (e.g. Using
McCaulay’s Method) to be equal to (4EI)/L.
If the bending stiffness of the beam is equal to
(Force/1.0), therefore k = (4EI)/L.
This is known as the absolute bending stiffness
of the element.
General Principles & Definition
Member stiffness factor
K 
4 EI
L
Joint stiffness factor
 The total stiffness factor of joint A is
K T   K  4000  5000  1000  10000
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd
General Principles & Definition
Member relative stiffness factor
 Quite often a continuous beam or a frame will be
made from the same material
 E will therefore be constant
KR 
I
L
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd
Carry-Over Moment
When the beam element deforms due to the
applied rotation at end A, an additional moment
(MB) is also transferred by the element to the
remote end if it has zero slope (i.e. is fixed).
 The moment MB is known as the carry-over
moment.
Carry-Over Moment
Carry-over (CO) factor
M
A
 4 EI 

 A ;
 L 
M
B
 2 EI 

 A
 L 
 Solving for  and equating these eqn,
M B  0 .5 M A
 The moment M at the pin induces a moment of
M’ = 0.5M at the wall
 In the case of a beam with the far end fixed, the
CO factor is +0.5
Carry-Over Moment
Therefore be stated that ‘if a moment is applied
to one end of a beam then a moment of the
same sense and equal to half of its value will be
transferred to the remote end provided that it is
fixed.’
If the remote end is ‘pinned’, then the beam is
less stiff and there is no carry-over moment.
Pinned End
‘the stiffness of a pin-ended
beam is equal to ¾ × the stiffness of a
fixed-end beam.’
Free Bending Moment
When a beam is free to rotate at both ends, no
bending moment can develop at the supports,
then the bending moment diagram resulting from
the applied loads on the beam is known as the
Free Bending Moment Diagram.
Fixed Bending Moment
 When a beam is fixed at the ends (encastre) such that it
cannot rotate, i.e. zero slope at the supports, then
bending moments are induced at the supports and are
called Fixed-End Moments.
 The bending moment diagram associated only with the
fixed-end moments is called the Fixed Bending Moment
Diagram.
Fixed Bending Moment
+
Example calculation 1:
1
2
1
2
Propped Cantilever
The fixed-end moment for propped cantilevers
(i.e. one end fixed and the other end simply
supported) can be derived from the standard
values given for encastre beams as follows.
Consider the propped cantilever, which supports
a uniformly distributed load as indicated.
Propped Cantilever
 The structure can be considered to be the superposition
of an encastre beam with the addition of an equal and
opposite moment to MB applied at B to ensure that the
final moment at this support is equal to zero.
Example calculation 2:
Solutions:
Example: Solution
Example: Solution
Distribution Factor
Distribution Factor (DF)
 That fraction of the total resisting moment supplied
by the member is called the distribution factor (DF)
DF i 
Mi
DF 
K
M
K

K i
  Ki
Consider 1 case
Stiffness of span BA = KBA = (I1/L1)
Stiffness of span BC = KBC = (I2/L2)
 Total stiffness at the support = ∑K = KBA + KBC
 K BA 

K 
 The moment absorbed by beam BA M 1  M applied x 
 The moment absorbed by beam AB M 2  M applied
 K AB 
x

K 
Example Calculation 1:
4 EI
K 
L
6
K BA 
4 E (120 )( 10 )
K BC 
4 E ( 240 )( 10 )
 4 E ( 40 )( 10 ) mm
6
4
/m
3
6
DF BA 
DF BC 
 4 E ( 60 )( 10 ) mm
6
4
4 E ( 40 )
4 E ( 40 )  4 E ( 60 )
4 E ( 60 )
4 E ( 40 )  4 E ( 60 )
 0 .4
 0 .6
4
/m
DF 
K
K
Solution:
Note that the above results could also have been
obtained if the relative stiffness factor is used
( FEM ) BC  
( FEM ) CB 
wL
2
  8000 kNm
12
wL
12
2
 8000 kNm
Solution:
As a result, portions of this moment are
distributed in spans BC and BA in accordance
with the DFs of these spans at the joint
Moment in BA is 0.4(8000) = 3200Nm
Moment in BC is 0.6(8000) = 4800Nm
These moment must be carried over since
moments are developed at the far ends of the
span
Solution
Example Calculation 2 (Example 12.2)
Determine the internal moment at each support of the
beam. The moment of inertia of each span is indicated.
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd
Solution
A moment does not get distributed in the overhanging
span AB
So the distribution factor (DF)BA =0
Span BC is based on 4EI/L since the pin rocker is not at
the far end of the beam
6
K BC 
4 E ( 300 )( 10 )
K CD 
4 E ( 240 )( 10 )
6
 300 (10 ) E
4
6
3
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd
6
 320 (10 ) E
Solution
DF BC  1  ( DF ) BA  1  0  1
DF CB 
300 E
300 E  320 E
DF CD  0 . 516 ;
Due to overhang,
 0 . 484
DF DC  0
( FEM ) BA  2000 N ( 2 m )  4000 Nm
( FEM ) BC  
( FEM ) CB 
wL
2
  2000 Nm
12
wL
12
2
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd
 2000 Nm
Solution
 The overhanging span requires the internal moment
to the left of B to be +4000Nm.
 Balancing at joint B requires an internal moment of
–4000Nm to the right of B.
 -2000Nm is added to BC in order to satisfy this
condition.
 The distribution & CO operations proceed in the
usual manner.
 Since the internal moments are known, the moment
diagram for the beam can be constructed.
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd
Solution
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd
Stiffness-Factor Modifications
Member pin supported at far end
 As shown the applied moment M rotates end A by
an amt 
 To determine , the shear in the conjugate beam at
A’ must be determined
M
B'
0
V 'A   
V 'A (L) 
3L
EI
 M 
1M  2 

L L   0
2  EI   3 
3 EI
L

Stiffness-Factor Modifications
Member pin supported at far end (cont’d)
 The stiffness factor in the beam is
K 
3 EI
L
 The CO factor is zero, since the pin at B does not
support a moment
 By comparison, if the far end was fixed supported,
the stiffness factor would have to be modified by ¾
to model the case of having the far end pin
supported
Stiffness-Factor Modifications
Symmetric beam & loading
 The bending-moment diagram for the beam will
also be symmetric
 To develop the appropriate stiffness-factor
modification consider the beam
 Due to symmetry, the internal
moment at B & C are equal
 Assuming this value to
be M, the conjugate
beam for span BC is shown
Stiffness-Factor Modifications
Symmetric beam & loading (cont’d)
 M C'  0
V 'B   
K
2 EI
L
M  L
- V 'B ( L )  
L   0
 EI   2 
ML
2 EI
 M 
2 EI

L

 Moments for only half the beam can be distributed
provided the stiffness factor for the center span is
computed
Stiffness-Factor Modifications
Symmetric beam with asymmetric loading
 Consider the beam as shown
 The conjugate beam for its center span BC is
shown
 Due to its asymmetric loading, the internal moment
at B is equal but opposite to that at C
Stiffness-Factor Modifications
Symmetric beam with asymmetric loading
 Assuming this value to be M, the slope  at each
end is determined as follows:
 M C'  0
 1  M  L  5 L
- V 'B ( L )   
 
2
EI
 
 2  6
V 'B   
K
6 EI
L
ML
6 EI
 M 
6 EI
L

 1  M  L  L 
 
    0
 2  EI   2   6 
Example Calculation 3 (Example 12.4)
Determine the internal moments at the supports of the beam shown
below. The moment of inertia of the two spans is shown in the figure.
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd
Solution
 The beam is roller supported at its far end C.
 The stiffness of span BC will be computed on the
basis of K = 3EI/L
 We have:
K
AB 
K BC 
4 EI
6

4 E (120 )( 10 )
L
3
3 EI
3 E ( 240 )( 10 )
L
6
 160 (10 ) E
6

4
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd
6
 180 (10 ) E
Solution
DF
AB

DF BA 
DF BC 
DF CB 
160 E
  160 E
0
160 E
160 E  180 E
180 E
160 E  180 E
180 E
 0 . 4706
 0 . 5294
1
180 E
( FEM ) BC  
wL
8
2

 6000 ( 4 )
8
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd
2
  12000 Nm
Solution
The forgoing data are entered into table as shown.
The moment distribution is carried out.
By comparison, the method considerably simplifies the
distribution.
The beam’s end shears & moment diagrams are shown.
Chapter 12: Displacement Method of Analysis: Moment Distribution
Structural Analysis 7th Edition
© 2009 Pearson Education South Asia Pte Ltd