Matakuliah
Tahun
: S0725 – Analisa Struktur
: 2009
Analysis of Indeterminate Structure
Session 05-22
Contents
•3 Equations Method
•Flexibility Method
•Slope Deflection Method
•Cross Method/ Moment Distribution Method
•Matrix Analysis
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3 Equation Method
This method will be used for
analizing the indeterminate
structure ( support reaction,
internal loads )
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3 Equation Method
Degree of indeterminacy 
P1
i=r–3
i=7–3 =4
q1
P2
q2
A
E
B
C
D
Gambar 1.1 . Struktur statis tak tentu
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3 Equation Method
Principles :
jokl
jokr
celah
P1
q1
P2
jokr
jokl
k-1
k+1
ll
k
lk
Gambar 1.2.a. Putran sudut akibat beban luar
Pada bagian konstruksi diantara 2 perletakan yang
berdekatan diberikan kebebasan untuk berputar sudut satu
sama lain. Sebagai akibatnya akan timbul CELAH pada
balok di tempat tumpuan ( perletakan ) sebagai akibat dari
adanya beban luar. ( Gambar 1.2.a )
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3 Equation Method
Princiles :
Mk
Mk-1
j'kl
Mk+1
j'kr
k-1
k+1
ll
k
lk
Gambar 1.2.b. Momen lentur negatif
Pada hakekatnya balok ini adalah menerus , utuh dan tidak boleh ada
celah , maka harus ada MOMEN pada tumpuan / perletakan antara
yang berfungsi mengembalikan celah tadi menjadi utuh kembali. Sebagai
gaya statis kelebihan harus dipilih berupa momen-momen pada
perletakan antara, umumnya momen – momen yang bekerja adalah
MOMEN NEGATIF. ( Gambar 1.2.b )
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3 Equation Method
Compatibility eq :
Sebagai suatu syarat kompatibilitas , joint ‘k’ merupakan
rotasi yang kompatibel, sehingga persamaan
kompatibilitasnya menjadi
jokl + jokr
=
j’kl + j’kr
j’  diambil nilai harga mutlaknya.
Sehingga harga Mk yang positif berarti bekerja sebagai
Dimana : jo dan
momen lentur negatif , dimana ini berarti bahwa kita
harus MERUBAH
didapatkan.
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tanda
gaya
statis
kelebihan
yang
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3 Equation Method
PERSYARATAN KOMPATIBILITAS :
Jika tidak ingin merubah tanda tersebut maka harus
dimasukkan anggapan – anggapan bahwa momen peralihan
merupakan momen lentur positif, sehingga persamaan
kompatibilitas menjadi :
(j
o
o
kl + j kr ) + ( j’kl + j’kr ) = 0
Sehingga hasil-hasil momen peralihan sudah langsung
berikut tandanya menyatakan MOMEN LENTUR
SEBENARNYA , jika hasilnya negatif , berarti bekerja
sebagai momen lentur negatif.
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3 Equation Method
Karena
nilai j’kl tergantung pada Mk-1 dan Mk
nilai j’kr tergantung pada Mk dan Mk+1
m a k a ...
dari persamaan kompatibilitas di atas akan selalu
didapatkan maksimum sebanyak 3 momen tumpuan
antara yang terlibat dalam persamaan kompatibilitas ,
sehingga metoda ini disebut juga 
PERSAMAAN TIGA MOMEN
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3 Equation Method
Dengan mennjau nilai putaran sudut akibat momen –
momen peralihan persamaan kompatibilitas dapat dituliskan
sebagai 
j’kl
j’kr
j’kl
l
( lr / 3EI )
l
(lr / 6EI )
= Mk ( l / 3EI ) + Mk-1 ( l / 6EI )
= Mk
+
j’kr
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l
+ Mk+1
l
l
+
l
= Mk-1 ( l / 6EI )+Mk {( l / 3EI )+( r / 3EI )}+ Mk+1 ( r / 6EI )
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3 Equation Method
Sehingga persamaan 3 momen pada tumpuan k ...
Mk-1 ( l / 6EI )+Mk {( l / 3EI )+( r / 3EI )}+ Mk+1 ( r / 6EI ) + ( jokl + jokr ) = 0
l
l
l
l
Agar persamaan ini dapat digunakan secara efisien dan
tepat maka diperlukan rumus – rumus dari berbagai type
beban , baik dari beban luar maupun momen peralihan
( momen pada tumpuan antara ).
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3 Equation Method
C a t a t a n ...
Jika pada perletakan ujung – ujung adalah
SENDI & ROL, maka i jumlah perletakan
antara.
Sedangkan jika perletakan ujung adalah
JEPIT, maka perletakan jepit ini
diperlakukan sebagai perletakan antara ,
dengan MENGUBAH menjadi SENDI
dan Momen Jepit sebagai gaya
13
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3 Equation Method
M
Gambar 1.3. Struktur dengan perletakan Jepit yang diubah  Sendi +
Momen jepit
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Method Analysis
While analyzing indeterminate
structures, it is necessary to
satisfy (force) equilibrium,
(displacement) compatibility
and
force-displacement
relationships
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Method Analysis
(a) Force equilibrium is satisfied when the reactive
forces hold the structure in stable equilibrium, as the
structure is subjected to external loads
(b) Displacement compatibility is satisfied when the
various segments of the structure fit together
without intentional breaks, or overlaps
(c) Force-displacement requirements depend on the
manner the material of the structure responds to the
applied loads, which can be linear/nonlinear/viscous
and elastic/inelastic; for our study the behavior is
assumed to be linear and elastic
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Method Analysis
(a) Force equilibrium is satisfied
when the reactive forces hold the
structure in stable equilibrium, as
the structure is subjected to
external loads
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Method Analysis
(b)
Displacement
compatibility is satisfied when the
various segments of the structure
fit together without intentional
breaks, or overlaps
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Method Analysis
(c)
Force-displacement
requirements depend on the manner
the material of the structure
responds to the applied loads,
which can be linear/nonlinear/viscous
and elastic/inelastic; for our study the
behavior is assumed to be linear and
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Method Analysis
Two methods are available to analyze
indeterminate structures, depending on whether
we satisfy force equilibrium or displacement
compatibility conditions –
They are:
Force method and
Displacement Method
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Method Analysis
Force
Method
satisfies
displacement compatibility and forcedisplacement relationships; it treats the
forces as unknowns –
Two methods which we will be studying
are Method of Consistent Deformation
and (Iterative Method of) Moment
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Distribution
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Method Analysis
Displacement
Method
satisfies force equilibrium and forcedisplacement relationships; it treats the
displacements as unknowns –
Two available methods are Slope
Deflection
Method
and
Stiffness
(Matrix) method
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Method Analysis
Method
Unknowns
Equations used for
Forced
Forces
Compatibility and force Flexibility
displacement
Displacem
Method
Displaceme Equilibrium and force
displacement
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Coefficient of the
unknowns
Stiffness
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Flexibility Method
Two methods are available to analyze indeterminate structures,
depending on whether we satisfy force equilibrium or displacement
compatibility conditions - They are: Force method and
Displacement Method
• Force Method satisfies displacement compatibility and forcedisplacement relationships; it treats the forces as unknowns - Two
methods which we will be studying are Method of Consistent
Deformation and (Iterative Method of) Moment Distribution
• Displacement Method satisfies force equilibrium and forcedisplacement relationships; it treats the displacements as unknowns
- Two available methods are Slope Deflection Method and
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Stiffness (Matrix) method
Flexibility Method
FORCED METHOD
This method is also known as flexibility
method or compatibility method.
In this method the degree of static
indeterminacy of the structure is determined
and the redundants are identified.
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VTU Programme
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Flexibility Method
FORCED METHOD
A coordinate is assigned to each
redundant.
Thus,P1, P2 - - - - - -Pn are the
redundants
at the coordinates 1,2, - - - - - n.If all the
redundants are removed , the resulting
structure known as released structure, is
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VTU Programme
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Flexibility Method
FORCED METHOD
This released structure is also known as basic
determinate structure. From the principle of
super position the net displacement at any
point in statically indeterminate structure is some
of the displacements in the basic structure due
to the applied loads and the redundants.
This is known as the compatibility condition and
may be expressed by the equation;
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Flexibility Method
FORCED METHOD
∆1 = ∆1L + ∆1R
∆2 = ∆2L + ∆2R
|
|
|
|
|
|
∆n = ∆nL + ∆nR
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Where ∆1 - - - - ∆n = Displ. At Co-ord.at 1,2 - -n
∆1L ---- ∆nL = Displ.At Co-ord.at 1,2 - - - - -n
Due to aplied loads
∆1R ----∆nR = Displ.At Co-ord.at 1,2 - - - - -n
Due to Redudants
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Flexibility Method
FORCED METHOD
The above equations may be return as [∆] = [∆L] + [∆R] - - - - (1)
∆1 = ∆1L + δ11 P1 + δ12 P2 + - - - - - δ1nPn
∆2 = ∆2L + δ21 P1 + δ22 P2 + - - - - - δ2nPn
|
|
|
|
|
|
|
|
|
|
- - - - - (2)
∆n = ∆nL + δn1 P1 + δn2 P2 + - - - - - δnnPn
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Flexibility Method
FORCED METHOD
∆ = [∆ L] + [δ] [P]
- - - - - - (3)
[P]= [δ]-1 {[∆] – [∆ L]}
- - - - - - (4)
If the net displacements at the redundants are zero then
∆1, ∆2 - - - - ∆n =0,
Then  [P]
= - [δ] -1 [∆ L]
- - - - - -(5)
The redundants P1,P2, - - - - - Pn are Thus determined
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Slope Deflection
• Slope deflection equations
The slope deflection equations express the member end
moments in terms of rotations angles. The slope
deflection equations of member ab of flexural rigidity EIab
and length Lab are:
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Slope Deflection
• Slope deflection equations
where θa, θb are the slope angles of ends a
and b respectively, Δ is the relative lateral
displacement of ends a and b. The absence
of cross-sectional area of the member in these
equations implies that the slope deflection
method neglects the effect of shear and
axial deformations.
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Slope Deflection
• Slope deflection equations
The slope deflection equations can also be written using the
stiffness factor .
and the chord rotation
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Slope Deflection
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Slope Deflection
• Slope deflection equations
When a simple beam of length Lab and
flexural rigidity EIab is loaded at each end
with clockwise moments Mab and Mba,
member end rotations occur in the same
direction. These rotation angles can be
calculated using the unit dummy force
method or the moment-area theorem
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Slope Deflection
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Slope Deflection
• Slope deflection equations
Joint equilibrium conditions imply that each joint with a degree of
freedom should have no unbalanced moments i.e. be in equilibrium.
Therefore,
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Slope Deflection Method
FORCED METHOD
∆ = [∆ L] + [δ] [P]
- - - - - - (3)
[P]= [δ]-1 {[∆] – [∆ L]}
- - - - - - (4)
If the net displacements at the redundants are zero then
∆1, ∆2 - - - - ∆n =0,
Then  [P]
= - [δ] -1 [∆ L]
- - - - - -(5)
The redundants P1,P2, - - - - - Pn are Thus determined
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Slope Deflection Method
Consider portion AB of a
continuous beam, shown
below,
subjected
to
a
distributed load w(x) per unit
length
and
a
support
settlement of  at B; EI of the
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Slope Deflection Method
A
A
B

=
B
(i) Due to externally
applied loads
(ii) Due to rotation A
at support A
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
 = rigid body motion
= /L
FEMBA
FEMAB
+
A
MAB=(4EIA)/L
MBA=(2EIA)/L
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Slope Deflection Method
+
B
A
(iii) Due to rotation
B at support B
B
MAB=(2EIB)/L
MBA=(4EIB)/L
L
+
MAB=(-6EI)/L2
A
(iv) Due to differential
settlement of  (between A
and B)
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MBA=(-6EI)/L2
B

L
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Slope Deflection Method
M
 (FEM)
 M
 M 
 M 
 (FEM)

AB
AB
AB
AB
AB
AB

2EI θ B

L
M

BA
L
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
A
L
6EI Δ
2EI
3Δ

(2θ

θ

)  (FEM)
2
B
A
AB
L
L
L
 (FEM)
4EI θ B
4EI θ
 M
 M 
 M 
 (FEM)

BA
BA
BA
BA
BA
2EI θ
A
L
6EI Δ
2EI
3Δ

(θ

2θ

)  (FEM)
2
B
A
BA
L
L
L
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Slope Deflection Method
A
=

B
(FEM)BA
(FEM)AB
+
(FEM)BA
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(FEM)BA/2
43
Slope Deflection Method
+
A
MAB=(3EIA)/L
+

MAB=(3EI)/L2
 = PL3/(3EI),
M = PL = (3EI/L3)(L) = 3EI/L2
MAB = [(FEM)AB - (FEM)BA/2]+(3EIA)/L -(3EI)/L2
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Modified FEM at end A
Slope Deflection Method
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Slope Deflection Method
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Slope Deflection Method
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Slope Deflection Method
This method is at the core of the moment
distribution method, and is also very powerful.
Consider a beam of length L, subjected to end
moments (clockwise positive), and downward
transverse loads either distributed or concentrated
The end slopes are θA θB.
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Slope Deflection Method
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Slope Deflection Method
M AB
M BA
2EI

( 2 A  B )  M FA .......... ...(1)
L
2EI

( A  2B )  M FB .......... ...( 2)
L
MFA , MFB
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are the numerical values of the
fixed-end moments, e.g. wL2/8,
PL/8, Pab2/L2, etc…
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Slope Deflection Method
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Slope Deflection Method
Here,
 A  C  0, and M BA  M BC  0
2EI
PL
( 2 B ) 
L
8
2EI

( 2 B )
L
M BA 
M BC
8EI
PL
PL2

B  
 B 
and
L
8
64EI
PL
M BC 
 62.5 kNm
16
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Moment Distribution Method
(Method developed by Prof.
Hardy Cross in 1932)
The method solves for the joint
moments in continuous beams
and
rigid frames by successive
approximation.
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Moment Distribution Method
Thus the Moment Distribution
Method (also known as the
Cross Method) became the
preferred calculation
technique for reinforced
concrete structures.
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Moment Distribution Method
The description of the moment distribution method by
Hardy Cross is a little masterpiece.
He wrote: "Moment Distribution.
The method of moment distribution is this:
• Imagine all joints in the structure held so that
they cannot rotate and compute the moments at
the ends of the members for this condition;
•at each joint distribute the unbalanced fixed-end
moment among the connecting members in
proportion to the constant for each member defined as
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"stiffness";
55
Moment Distribution Method
•multiply the moment distributed to each
member at a joint by the carry-over factor at
the end of the member and set this product at
the other end of the member;
•distribute these moments just "carried
over";
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Moment Distribution Method
•repeat the process until the moments to
be carried over are small enough to be
neglected; and
•add all moments - fixed-end
moments, distributed moments, moments
carried over - at each end of each member to
obtain the true moment at the end." [Cross
1949:2]
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Moment Distribution Method
1. Restrain all possible displacements.
2. Calculate Distribution Factors:
The distribution factor DFi of a member connected to
any joint J is
where S is the rotational stiffness , and is given by
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Moment Distribution Method
3. Determine carry-over factors
The carry-over factor to a fixed end is always 0.5,
otherwise it is 0.0.
4. Calculate Fixed End Moments.
These could be due to in-span loads, temperature
variation and/or
relative displacement between the ends of a
member.
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Moment Distribution Method
5. Do distribution cycles for all joints simultaneously
Each cycle consists of two steps:
1. Distribution of out of balance moments Mo,
2.Calculation of the carry over moment at the far end of each member.
The procedure is stopped when, at all joints, the out of balance moment is a
negligible value. In this case, the joints should be balanced and no carry-over
moments are calculated.
6. Calculate the final moment at either end of each member.
This is the sum of all moments (including FEM) computed during the distribution
cycles.
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Moment Distribution Method
Consider the continuous beam ABCD,
subjected to the given loads,
as shown in Figure below. Assume that
only rotation of joints occur
at B, C and D, and that no support
displacements occur at B, C and
D. Due to the applied loads in spans AB,
BC and CD, rotations occur at B, C and
D.
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Moment Distribution Method
PROBLEMS
15 kN/m
10 kN/m
3m
A
I
8m
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B
I
6m
C
D
I
8m
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Moment Distribution Method
Step I
The joints B, C and D are locked in position before any load is
applied on the beam ABCD; then given loads are applied on the
beam. Since the joints of beam ABCD are locked in position, beams
AB, BC and CD acts as individual and separate fixed beams,
subjected to the applied loads; these loads develop fixed end
moments.
15 kN/m
-80 kN.m
-80 kN.m -112.5kN.m
3m
A
8m
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B
150 kN
B
6m
112.5 kN.m -53.33 kN.m
C
10 kN/m
53.33 kN.m
C
D
8m
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Moment Distribution Method
Step I
The joints B, C and D are locked in
position before any load is applied on
the beam ABCD; then given loads are
applied on the beam. Since the joints
of beam ABCD are locked in position,
beams AB, BC and CD acts as
individual and separate fixed beams,
subjected to the applied loads; these64
loads develop fixed end moments.
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Moment Distribution Method
15 kN/m
-80 kN.m
-80 kN.m -112.5kN.m
3m
A
8m
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B
150 kN
B
6m
112.5 kN.m -53.33 kN.m
C
10 kN/m
53.33 kN.m
C
D
8m
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Moment Distribution Method
In beam AB
Fixed end moment at A = -wl2/12 = - (15)(8)(8)/12 = - 80 kN.m
Fixed end moment at B = +wl2/12 = +(15)(8)(8)/12 = + 80 kN.m
In beam BC
Fixed end moment at B = - (Pab2)/l2 = - (150)(3)(3)2/62 = 112.5 kN.m
Fixed end moment at C = + (Pab2)/l2 = + (150)(3)(3)2/62 = +
112.5
kN.m
In
beam
AB
Fixed end moment at C = -wl2/12 = - (10)(8)(8)/12 = - 53.33 kN.m
Fixed end moment at D = +wl2/12 = +(10)(8)(8)/12 = + 53.33kN.m
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Moment Distribution Method
Step II
Since the joints B, C and D were fixed
artificially (to compute the the fixed-end
moments), now the joints B, C and D are
released and allowed to rotate. Due to the joint
release, the joints rotate maintaining the
continuous nature of the beam. Due to the joint
release, the fixed end moments on either side
of joints B, C and D act in the opposite direction
now, and cause a net unbalanced moment to67
occur at the joint
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Moment Distribution Method
150 kN
15 kN/m
10 kN/m
3m
A
B
I
8m
Released moments
I
6m
-80.0
+112.5
Net unbalanced moment
+32.5
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C
I
D
8m
-112.5
+53.33
-59.17
-53.33
-53.33
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Moment Distribution Method
Step III
These unbalanced moments act at the joints and
modify the joint moments at B, C and D, according to
their relative stiffnesses at the respective joints. The
joint moments are distributed to either side of the joint
B, C or D, according to their relative stiffnesses. These
distributed moments also modify the moments at the
opposite side of the beam span, viz., at joint A in span
AB, at joints B and C in span BC and at joints C and D
in span CD. This modification is dependent on the
carry-over factor (which is equal to 0.5 in this case);
when this carry over is made, the joints on opposite69
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Moment Distribution Method
Step IV
The carry-over moment becomes the
unbalanced moment at the joints to
which they are carried over. Steps 3
and 4 are repeated till the carry-over or
distributed moment becomes small.
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Moment Distribution Method
Step V
Sum up all the moments at
each of the joint to obtain the
joint moments.
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Moment Distribution Method
Stiffness and Carry-over Factors
Stiffness = Resistance offered by member to a unit displacement or
rotation at a point, for given support constraint conditions
MB
MA
A
A
B
A
RA
A clockwise moment MA is
applied at A to produce a +ve
bending in beam AB. Find A
and MB.
RB
L
E, I – Member properties
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Moment Distribution Method
Using method of consistent deformations
MA
A
B
A
L
M A L2
A  
2 EI
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fAA
B
L
A
1
f AA
L3

3EI
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Moment Distribution Method
Applying the principle of consistent deformation,
 A  R A f AA  0  R A  
M A L R A L2 M A L
A 


EI 2EI 4EI
3M A
M A 
2L

M 4EI
4EI
 A ; hence k  A 
L
A
L
Stiffness factor = k = 4EI/L
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74
Moment Distribution Method
Considering moment MB,
MB + MA + RAL = 0
MB = MA/2= (1/2)MA
Carry - over Factor = 1/2
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75
Moment Distribution Method
Distribution Factor
Distribution factor is the ratio
according to which an externally
applied unbalanced moment M at
a joint is apportioned to the various
members mating at the joint
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76
Moment Distribution Method
Distribution Factor
moment M
A
I1
L1
B
I3
L3
I2
L2
C
M
A
B
MBA
D
MBC
C
MBD
At joint B
M - MBA-MBC-MBD = 0
D
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77
Moment Distribution Method
Distribution Factor
i.e.,
 4 E1 I1   4 E2 I 2   4 E3 I 3 
 B
  
  
 
L
L
L
1
2
3
 
 


 K BA  K BC  K BD  B
 B 
M
M

K BA  K BC  K BD   K
 K
M BA  K BA B   BA
K

Similarly
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M = MBA + MBC + MBD

 M  ( D.F ) BA M


 K 
M BC   BC  M  ( D.F ) BC M
K 


 K 
M BD   BD  M  ( D.F ) BD M
K 


78
Moment Distribution Method
Modified Stiffness Factor
The stiffness factor changes when the far end of the beam is simplysupported.
MA
A
A
B
L
RA
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RB
79
Moment Distribution Method
Modified Stiffness Factor
As per earlier equations for deformation, given in Mechanics of Solids
text-books.
M AL
3EI
M A 3EI  3  4 EI 
K AB 

  

A
L  4  L 
3
 ( K AB ) fixed
4
A 
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80
Moment Distribution Method
Solution of that problems above
Fixed end moments
M AB
M BC
M CD
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wl 2
(15)(8) 2
  M BA  

 80 kN.m
12
12
wl
(150)(6)
  M CB  

 112.5 kN.m
8
8
wl 2
(10)(8) 2
  M DC  

 53.333 kN.m
12
12
81
Moment Distribution Method
Solution of that problems above
Stiffness Factors (Unmodified Stiffness)
K
AB
K BC
K CD
K DC
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4EI
( 4)(EI )
K


 0.5EI
BA
L
8
4EI
( 4)(EI )
 K CB 

 0.667EI
L
6
 4EI  4

 EI  0.5EI

 8  8
4EI

 0.5EI
8
82
Moment Distribution Method
Solution of that problems above
Distribution Factors
DFAB 
DF

DF

BA
BC
DFCB 
DFCD 
DF
DC
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K BA
K
K
wall
K
0.5EI
 0.0
0.5   (wall stiffness )

0.5EI
 0.4284
0.5EI  0.667EI

0.667EI
 0.5716
0.5EI  0.667EI

0.667EI
 0.5716
0.667EI  0.500EI

0.500EI
 0.4284
0.667EI  0.500EI
BA
BA
K BA  K BC
K BC
K BA  K BC
K CB
K

K
CD
K
CB
CD
K CB  K CD
K
 DC  1.00
K DC
83
Moment Distribution Method
Solution of that problems above
Moment Distribution Table
Joint
A
Member
AB
Distribution Factors
Computed end moments
0
-80
B
BA
C
BC
CB
D
CD
0.4284 0.5716 0.5716 0.4284
80
-112.5
112.5
DC
1
-53.33
53.33
-33.82 -25.35
-53.33
9.289
-26.67
-12.35
9.662
9.935 7.446
12.35
4.968
4.831 6.175
3.723
-2.84
-6.129 -4.715
-3.723
-3.146
-1.42 -1.862
-2.358
1.798
1.876 1.406
2.358
0.938
0.9 1.179
0.703
-0.402
-0.536
-1.187 -0.891
-0.703
99.985
-99.99
Cycle 1
Distribution
Carry-over moments
13.923 18.577
6.962
-16.91
Cycle 2
Distribution
Carry-over moments
7.244
3.622
Cycle 3
Distribution
Carry-over moments
-2.128
-1.064
Cycle 4
Distribution
Carry-over moments
1.348
0.674
Cycle 5
Distribution
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Summed up
moments
-69.81
96.613
-96.61
0
84
Moment Distribution Method
Solution of that problems above
Computation of Shear
Forces
15 kN/m
10 kN/m
150 kN
B
C
A
D
I
8m
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I
3m
I
3m
8m
85
Moment Distribution Method
Solution of that problems above
Computation of Shear
Forces
Simply-supported
60
60
75
75
40
8.726
-8.726
16.665
-16.67
12.079
-12.5
12.498
-16.1
16.102
0
56.228
63.772
75.563
74.437
53.077
40
reaction
End reaction
due to left hand FEM
-12.08
End reaction
due to right hand FEM
Summed-up
moments
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0
27.923
86
Moment Distribution Method
Solution of that problems above
Shear Force and Bending Moment Diagrams
52.077
75.563
2.792 m
56.23
27.923
74.437
3.74 m
63.77
S. F. D.
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87
Moment Distribution Method
Solution of that problems above
Shear Force and Bending Moment Diagrams
Mmax=+38.985 kN.m
Max=+ 35.59 kN.m
126.704
31.693
35.08
-69.806
3.74 m
48.307
84.92
-99.985
Bina Nusantara University
98.297
2.792 m
-96.613
B. M. D
88
Moment Distribution Method
Solution of that problems above
Simply-supported bending moments at center of span
Mcenter in AB = (15)(8)2/8 = +120 kN.m
Mcenter in BC = (150)(6)/4 = +225 kN.m
Mcenter in AB = (10)(8)2/8 = +80 kN.m
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89
Moment Distribution Method
MOMENT DISTRIBUTION METHOD FOR
NONPRISMATIC MEMBER
The section will discuss moment
distribution method to analyze
beams and frames composed of
non prismatic members.
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90
Moment Distribution Method
MOMENT DISTRIBUTION METHOD FOR
NONPRISMATIC MEMBER
First the procedure to obtain the necessary carryover factors, stiffness factors and fixed-end
moments will be outlined. Then the use of values
given in design tables will be illustrated. Finally
the analysis of statically indeterminate structures
using the moment distribution method will be
outlined
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91
Moment Distribution Method
Stiffness and Carry-over Factors
Use moment-area method to find the stiffness and
carry-over factors of the non-prismatic beam.
MA
PA
MB
A
B

PA  ( K A ) AB  A
A
M A  K AB A
M B  C AB M A
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CAB= Carry-over factor of moment MA from A to B
92
Moment Distribution Method
Stiffness and Carry-over Factors
A (= 1.0)
MA
B (= 1.0)
MB
A
B
MA(KA)
MB=CABMA
=CABKA
(a)
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A
B
MA=CBAMB
=CBAKB
MB(KB)
(b)
93
Moment Distribution Method
Use of Betti-Maxwell’s reciprocal theorem
requires that the work done by loads in case
(a) acting through displacements in case (b) is
equal to work done by loads in case (b) acting
through displacements in case (a)
M A (0)  M B (1)  M A (1.0)  M B (0.0)
C AB K A  C BA K B
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94
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95
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96
Moment Distribution Method
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97
Moment Distribution Method
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98
Moment Distribution Method
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99
Moment Distribution Method
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100
Moment Distribution Method
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101
Moment Distribution Method
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102
Moment Distribution Method
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103
Moment Distribution Method
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104
Moment Distribution Method
Stiffness-Factor Modification
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105
Moment Distribution Method
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106
Moment Distribution Method
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107
Moment Distribution Method
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108
Moment Distribution Method
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109
Moment Distribution Method
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110
Moment Distribution Method
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111
Moment Distribution Method
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112
Moment Distribution Method
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113
Moment Distribution Method
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114
Moment Distribution Method
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115
Moment Distribution Method
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116
Moment Distribution Method
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117
Moment Distribution Method
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118
Moment Distribution Method
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119
Moment Distribution Method
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120
Moment Distribution Method
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121
Moment Distribution Method
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122
Moment Distribution Method
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123
Moment Distribution Method
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124
Moment Distribution Method
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125
Moment Distribution Method
Symmetric Beam and Loading
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126
Moment Distribution Method
Symmetric Beam with Antisymmetric Loading
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127
Moment Distribution Method
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128
Moment Distribution Method
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129
Moment Distribution Method
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130
Moment Distribution Method
Moment Distribution for frames:
No sidesway
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131
Moment Distribution Method
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132
Moment Distribution Method
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133
Moment Distribution Method
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134
Moment Distribution Method
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135
Moment Distribution Method
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136
Moment Distribution Method
Moment Distribution for frames: sidesway
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137
Moment Distribution Method
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138
Moment Distribution Method
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139
Moment Distribution Method
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140
Moment Distribution Method
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141
Moment Distribution Method
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142
Moment Distribution Method
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143
Moment Distribution Method
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144
Moment Distribution Method
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145
Moment Distribution Method
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146
Moment Distribution Method
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147
Matrix Analysis
FLEXIBILITY AND STIFFNESS MATRICES : SINGLE CO-ORD.
D= δ x P
D=PL3

/3EI = δ x P
δ=L3
/3EI
δ=Flexibility Coeff.
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P=K x D
P=K x PL3 / 3EI
K=3EI / L3
K=Stiffness Coeff.
VTU Programme
148
Matrix Analysis
FLEXIBILITY AND STIFFNESS MATRICES : SINGLE CO-ORD.
D= ML / EI
M=K x D =K x ML/EI
D= δ x M=ML/EI
 K=EI / L
 δ=L / EI
δ=Flexibility Coeff.
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K=Stiffness Coeff.
δ X K= 1
VTU Programme
149
Matrix Analysis
FLEXIBILITY AND STIFFNESS MATRICES : TWO COORDINATE SYSTEM
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VTU Programme
150
Matrix Analysis
FLEXIBILITY AND STIFFNESS MATRICES : TWO COORDINATE SYSTEM
Unit Force At Co-ord.(1)
Unit Force At Co-ord.(2)
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δ11=L3 / 3EI
δ21=L2 / 2EI
δ12=δ21 =L2 / 2EI
δ22=L / EI
VTU Programme
151
Matrix Analysis
STIFFNESS MATRIX
K11=12EI / L3
K21= – 6EI / L2
Unit Displacement at (1)
Forces at Co-ord.(1) & (2)
K12= – 6EI / L2
Unit Displacement at (2)
P1=K11D1+K12D
2
P2=K21D1+K22D
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2

P1
P2
K22=4EI / L
K11 K12 D1
=
K21 K22 D2
Forces at Co-ord.(1) & (2)
12EI / L3– 6EI / L2
K = – 6EI / L24EI / L
VTU Programme
152
Matrix Analysis
Develop the Flexibility and stiffness matrices for frame
ABCD with reference to Coordinates shown
The Flexibility matrix can be developed by
applying unit force successively at
coordinates (1),(2) &(3) and evaluating the
displacements at all the coordinates
δij =displacement at I due to unit load at j
δij = ∫ mi mj / EI x ds
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VTU Programme
153
Matrix Analysis
Unit Load at (1)
Unit Load at (3)
Unit Load at (2)
Portion
DC
CB
BA
I
I
4I
4I
Origin
D
C
B
Limits
0-5
0 - 10
0 - 10
m1
x
5
5-x
m2
0
x
10
m3
-1
-1
VTU Programme
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-1
154
Matrix Analysis
δ11 = ∫m1.m1dx / EI = 125 / EI
δ21 = δ12 = ∫m1.m2dx / EI = 125 / 2EI
δ31 = δ13 = ∫m1.m3dx / EI = -25 / EI
δ22 = ∫m2.m2dx / EI = 1000 / 3EI
δ23 = δ32 = ∫m2.m3dx / EI = -75 / 2EI
δ33 = ∫m3.m3dx / EI = 10 / EI
 δ = 1 / 6EI
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750
375
-150
375
2000
-225
-150
-225
60
VTU Programme
155
Matrix Analysis
INVERSING THE FLEXIBILITY MATRIX [ δ ]
THE STIFENESS MATRIX [ K ] CAN BE OBTAINED
K = EI
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0.0174
0.0028
0.0541
0.0028
0.0056
0.0282
0.0541
0.0282
0.3412
VTU Programme
156