Moment
Distribution Method
Moment
‐
Distribution
Method
Structural Analysis
By
Aslam Kassimali
Theory of Structures‐II
M Shahid Mehmood
Department of Civil Engineering
Swedish College of Engineering & Technology, Wah Cantt
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Moment Distribution Method: Determining Member End Moments in a Frame, Study notes of Acting
The Moment Distribution Method (MDM) used to determine the end moments of members in a frame. It includes the calculation of distribution factors, fixed end moments, and the balancing of moments at all joints. The process continues until the unbalanced moments are negligibly small.
What you will learn
- What is the process for balancing moments at all joints in the Moment Distribution Method (MDM)?
- What are fixed end moments and how are they calculated using the Moment Distribution Method (MDM)?
- What is the Moment Distribution Method (MDM) and how is it used to determine member end moments in a frame?
- How are distribution factors calculated in the Moment Distribution Method (MDM)?
- How are unbalanced moments calculated using the Moment Distribution Method (MDM)?
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Moment
Distribution Method
Moment
Distribution
Method
Structural Analysis
By
Aslam Kassimali
Theory of Structures
‐II
M Shahid Mehmood
Department of Civil Engineering
Swedish College
of Engineering & Technology,
Wah Cantt
Moment
Distribution Method
Classical method.
Used for Beams and Frames.
Developed by Hardy Cross in 1924.
Used by Engineers for analysis of small structures.
It does not involve the solution of many simultaneous equations.
Definitions and TerminologySign Convention •
Counterclockwise member end moments are considered positive.
Clockwise
moments
on
joints
are
considered
positive
Clockwise
moments
on
joints
are
considered
positive
Member Stiffness •
Consider a prismatic beam AB, which is hinged at end A and fixed
d
at end B.
A
B
L
EI = constant
Member Stiffness
If we apply a moment M at the end A, the beam rotates by anangle
at the hinged end A and develops a moment M
BA
at the
g
g
p
BA
fixed end B, as shown.
M
BA
= carryover moment
A
B
θ
M = applied moment
M
BA
carryover moment
The relationship between the applied moment M and the rotation
L
EI = constant
The relationship between the applied moment M and the rotation θ
can be established using the slope
deflection equation.
Member Stiffness
when the modulus of elasticity for all the members of a structure isthe same (constant), it is usually convenient to work with the
(^
y
relative bending stiffness of members in the analysis.“
The relative bending stiffness, K, of a member is obtained by
dividing its bending stiffness,
, by 4E.”
K
I L
K E
K
Now suppose that the far end B of the beam is hinged as shown.
A
B
L
EI = constant
Member Stiffness
The relationship between the applied moment M and the rotation θ
of the end A of the beam can now be determined by using the
y
g
modified slope
‐deflection equation.
By substituting M
rh
= M,
r^
, and
= FEM
rh
= FEM
hr
= 0 into
MSDE, we obtain
^ θ
EI L
M θ
M = applied moment
A
B
L
EI = constant
Member Stiffness
R l ti
hi
b/
li d
d
t
M
d
th
t ti
R
elationship b/w applied end moment M and the rotation
fixed is
member
of
end
far if
θ
EI L
hinged is
member
of
end
far if
θ
L EI L
M
Bending stiffness of a member
fixed is
member
of
end
far if
EI L
l^
b
d
ff
f
b
hinged is
member
of
end
far if
L EI L
K
Relative bending stiffness of a member
fixed is
member
of
end
far if
I L
(9)^10
hinged is
member
of
end
far if
I L
L
K
Carryover Moment
Let us consider again the hinged
‐fixed beam of Figure.
A
B
θ
M = applied moment
M
BA
= carryover moment
L
EI = constant
When a moment M is applied at the hinged end A of the beam, amoment M
BA
develops at the fixed end B.
The moment MBA
is termed the
carryover moment
Carryover Moment
When the far end of the beam is hinged as shown, the carryovermoment M
BA
is zero.
BA
θ
M = applied moment
A
B
L
EI = constant
θ
M
applied moment
fixed is
member
of
end
far if
M
M
hinged is
member
of
end
far if
M
BA
Carryover Factor (COF)
The
ratio
of
the
carryover
moment
to
the
applied
moment
(M
BA
/M) is called the carryover factor of the member.”
(^
BA
)^
y
f^
f
It represents the fraction of the applied moment M that is carriedover to the far end of the member. By dividing Eq. 12 by M, we canexpress the carryover factor (COF) as
fixed is
member
of
end
far if
COF
hinged is
member
of
end
far if
Suppose that a moment M is applied to the joint B, causing it torotate by an angle
as shown in figure below
rotate by an angle
as shown in figure below.
M
= applied moment
D
B
A
θ
θ
θθ
E = constant
L
, I 2
2
C
L
, I
1
L
, I
3
To determine what fraction of applied moment is resisted by eachof the three members AB, BC, and BD, we draw free
body diagrams
,^
,^
,^
y
g
of joint B and of the three members AB, BC, and BD.
By considering the moment equilibrium of the free body of joint B(∑
M = 0) we write
M
B
= 0), we write
BD
BC
BA
M
M
M
M
M
BD
BC
BA (
)^
BD
BC
BA
M
M
M
M
A
B
B
B
D
M
M
BA
M
BA M
BC
M
BD
M
BD
B
M
BC
BD
M
BC
B
17
C
Substitution of Eq. 15 through Eq. 17 into the equilibrium equationEq 14 yieldsEq. 14 yields
3
3
2
2
1
1
^ θ
EI L
EI L
EI L
M
in which
represents the sum of the bending stiffnesses of all
(^
)^
(^
)^
∑ − = + + − =
B
BD
BC
BA
K
K
K
K
∑
B K
the members connected to joint B.“
Th
t ti
l^
tiff
f^
j i t
i^
d fi
d
th
t
Th
e rotational stiffness of a joint is defined as the moment
required to cause a unit rotation of the joint.” From Eq. 18, we can see that the rotational stiffness of a joint isequal to the sum of the bending stiffnesses of all the members
i idl
t d t
th
j i t
rigidly connected to the joint.
19
The negative sign in Eq. 18 appears because of the sign convention.To express member end moments in terms of the applied momentM, we first rewrite Eq. 18 in terms of the relative bending
,^
q
g
stiffnesses of members as
(^
)^
(^
)
∑ − = + + − =
K E K K K E M
(^
)^
(^
)
∑
∑
B
B
BD
BC
BA K
E
M
K E K K K E
M θ
By substituting Eq. 19 into Eqs. 15 through 17, we obtain
K
M
K
K
M
B BA
BA
∑
20
M
K
K
M
B BC
BC
∑