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7/31/2019 Advanced Structure
1/20
University of Technology
Building & Construction Dept.
Structural Engineering Branch
Fourth Stage Structural Eng. Branch
Dr. Alaa K. Abdul Karim
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Matrix Displacement (Stiffness) Method
:nIntroductio
The matrix displacement method, which is also known as the stiffness
method, is a method of analysis that uses the stiffness properties of the elements of
a structure to form a set of simultaneous equation relating displacements of the
structure to loads acting on the structure. The governing matrix equation is of the
form :
= .KF
Where F is a column matrix of external loads, K is the stiffness matrix of the
structure, and is a column matrix of external displacements. For a given set of
external loads, the corresponding displacements are obtained by solving the set of
simultaneous equations. Force in the elements of the structure are obtained through
the use of the calculated displacements and the element stiffness properties. In
analyzing a structure by the matrix displacement method, it is advantageous to
consider the structure in a broad fashion. A structure is considered to be an
assembly of structural elements connected at a finite number of points referred to
as nodal point and loaded only at these points.
The following basic conditions are satisfied at each node :
1. The equations of equilibrium.2. The compatibility of displacements.3. The force - displacement relationship.
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In skeletal structures, it is assumed that the element of the structure is straight and
prismatic thus, the position of the nodes is located at places where abrupt changes
in geometry, loading or material properties occur. Examples of these natural
subdivisions are shown in Fig. below.
One Dimensional Elements :
A one dimensional element may be represented by a straight line whose ends,
such as 1 and 2, are nodal points. These elements are referenced in a coordinate
system Known as local or element coordinate system. In this system , x- axis is
defined by element axis which is a line joining the two nodes of the elements. One
- dimensional elements are used when the geometry, material properties, and
dependent variables such as displacements can all be expressed in terms of one
independent space coordinate which is measured along the element axis.
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Pin-Jointed Bar Element:
The pin-jointed bar or truss element shown in Fig. below is the simplest
structural element and is assumed to be pin connected at both the ends. The bar
element is also assumed to have constant cross-sectional area (A) and modulus of
elasticity (E) over its length (L), external loads are applied at the nodes and effect
of self-weight is neglected. Thus for a plane structure, this element has four
degrees of freedom, two at each node.
Beam Element:
The beam element shown in Fig. below, is also known as frame element. For a
plane structure, this element has six degrees of freedom, three at each node, ( axial
and in-plane transverse displacements, and in-plane rotation.
Global axis
YLocal axis
Y'
F1
F2
F4
F3
X'
X
1
2
DOF= 4
Y
F1
F2
F5
F4
X1
2
DOF= 6
F3 F6
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Stiffness Matrix of Elastic Spring
A typical elastic spring is shown in Fig. below. Forces exist at nodes 1 and 2.
Assume that node 2 is fixed and node 1 is displaced by 1 due to forceF1,1 (case A).
Appling Equilibrium of forces ( Fx = 0 )F1,1= - F2,1 = k1
Similarly assume that node 1 is fixed and node 2 is displaced by 2 due
to force F2,2 (case B).
Appling Equilibrium of forces ( Fx = 0 )F2,2= - F1,2 = k2
By using the principle of superposition for case A and B (adding up
algebraically the two cases A &B.) we obtain :
Node 1 Node 2
F1
, 1
F2 ,
2K
Node 1 Node 2
K1
F1,1 F2,1
Node 1 Node 2
K 2
F1,2 F2,2
Node 1 Node 2
K 2
F1 F2
1
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F1= F1,1 + F 1,2
F 2= F 2,1 + F 2,2
Sub. The value of F1,1 , F 1,2 , F 2,1 and F 2,2
F 1= k1 - k2
F 2= - k1 + k2
In matrix notation, these equations may be re- written as :
=
2
1
2
1
.KK
kK
F
F
{ } [ ]{ }= .KF
Spring Assemblage
Actual structures generally consist of basic structural componentsstringers, beams, thin plate, etc. properly fastened into an assemblage.
For example, a truss is usually considered as an assemblage of axial
force members, whereas a building frame consists of an assemblage of
beams and columns. Consequently, it is vitally important to be able to
form the total structure stiffness matrices for the separate components.
Consider the assemblage of springs A and B shown:
Node 1 Node 2
Ka2
F1
F21
Node 3
Kb3
F3
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The first step is to assemble the stiffness matrix for each element :
For spring a :
=
2
1
2
1.
aa
aa
KK
kK
F
F
For spring b :
=
3
2
3
2.
bb
bb
kK
kK
F
F
using the principle of superposition and applying the rule of matrix
addition, we obtain :
+
=
3
2
1
3
2
1
.
0
0
bb
bbaa
aa
KK
KKKK
KK
F
F
F
Example 1:Analyze the linear spring system shown in Fig. below which has two
linear springs connected a series with spring stiffness Ka and Kb .
Node 1 Node 2
KaF1
F2
Node 3
KbF3
Node 1 Node 2
Ka
Node 3
KbP
Element 1 Element 2
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For element 1 : For element 2 :
=
2
1
2
1.
aa
aa
KK
kK
F
F
=
3
2
3
2.
bb
bb
kK
kK
F
F
overall stiffness matrix for all system (assemblage of system).
+
=
3
2
1
3
2
1
.
0
0
bb
bbaa
aa
KK
KKKK
KK
F
F
F
1 = 0F2 = 0
F3 = P
+=
3
2.
0
bb
bba
KK
kKK
P
solve the two equations to obtain the values of2 and 3
2 = P/Ka 3 = P[(Ka+Kb) / KaKb]
to find reaction F1= - Ka2= - Ka(P/Ka) = - P
finally the internal forces in the elements may be obtain using the
stiffness matrix for each element.
For element 1 :
=
aaa
aa
KpKK
kK
F
F
/
0.
2
1
F1 = - Ka * p/Ka = - p
F2 = Ka * p/Ka = pKa
P P
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Example 2:The axially loaded bar system shown in Fig. below is formed from
three different segments. Applied loads of 100 and 150 kN are acted at
points 2 &3. Compute the displacements at these two nodes. Then
determine the reactions at the fixed ends of the system and draw the
axial force diagram. Use E= 30 kN/mm2
.
For element A :
=
2
1
2
1.
aa
aa
KK
kK
F
Fwhere : K=AE/L
=
2
1
2
1.
11
11
L
AE
F
F
mmkNL
EAK
a
aa
a/120
1000
30*10*40 2===
=
2
1
2
1.
120120
120120
F
F
For element B :
=
3
2
3
2.
bb
bb
KK
kK
F
F
=
3
2
3
2.
9090
9090
F
F
For element C :
=
4
3
4
3.
cc
cc
KK
kK
F
F
1
1000 mm
A=40*102
mm2Element 1
150 kN
2 3
4
A=60*102
mm2
2000 mm
A=40*102
mm2
100 kN
A B C
2000 mm
mmkNL
EAK
b
bbb
/902000
30*10*60 2===
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=
4
3
4
3.
6060
6060
F
F
overall matrix
=
=
=
=
+
+
=
=
=
=
=
0
?
?
0
.
606000
606090900
09090120120
00120120
?
150
100
?
4
3
2
1
4
3
2
1
F
F
F
F
=
3
2.
15090
90210
150
100
The solution of this pair of simultaneous equations is :
mm
=
731.1
218.1
3
2
Now, we can find the reactions at nodes 1 &4 (F1 and F4 ).
=
=
=
=
=
0
731.1
218.1
0
.606000
00120120
4
3
2
1
4
1
F
F
kNF
F
=
86.103
16.146
4
1
The internal forces in the elements are:
For element A:
kNF
FA
=
=
16.146
16.146
218.1
0.
120120
120120
2
1
mmkNL
EAK
c
cc
c/60
2000
30*10*40 2===
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146.1646.16
100
For element B:
kNF
FB
=
=
16.46
16.46
731.1
218.1.
9090
9090
3
2
For element C:
kNF
FC
=
=
84.103
84.103
0
731.1.
6060
6060
4
3
146.16 46.16 103.84103.84
150
Tension
compression
146.16
46.16
0
-103.84
Axial Force diagram
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Example 3:Analyze the linear spring system shown in Fig. below ,determine the
displacements of nodes 2 and 3 , also find the reactions at nodes 1 and 4 .
where : K2 =K4 = K and K3 =2K K1= K5 = K/2 .
First we put the forces and displacements at the positive direction in
each node .
The element stiffness matrix for each element can be written as :
For element 1 : For element 2 :
=
3
1
11
11
3
1.
KK
kK
F
F
=
2
1
22
22
2
1.
kK
kK
F
F
1
K1
K2F2
2
3 4K3F1K5
F3
F4
K4
1
K1
K2
2P
2
3 4K3
K4
K5
P
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For element 3 : For element 4 :
=
3
2
33
33
3
2.
KK
kK
F
F
=
4
2
44
44
4
2.
kK
kK
F
F
For element 5 :
=
4
3
55
55
4
3.
KK
kK
F
F
overall stiffness matrix for all system (assemblage of system).
+
++
++
+
=
4
3
2
1
5454
553131
434322
1221
4
3
2
1
.
0
0
KKKK
KKKKKK
KKKKKK
KKKK
F
F
F
F
For this system, nodes 1 and 4 are fixed (1 = 4=0 ). Also contain twounknown displacements (2 and 3 ) and unknown reaction forces (F1,F4).
F2 = -p
F3 = 2P
++
++=
3
2
5313
3432.
2 KKKK
KKKK
P
P
=
3
2.
32
24
2 KK
KK
P
P
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-P = 4K2 2K32P = -2K2 + 3K3
solve the above equations to get the value of 2 and 3
and
The ratio of61
5.12*
83
2== P
KKP
Now To find the reactions (F1 and F4 )
8
7
2
5.1
81
P
K
P
KK
P
KF =
=
244 = KF
884
P
K
PKF =
=
K
P
82 =
K
P
2
5.13 =
31221 = KKF
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Pin-Jointed Plane Bar Element:
The pin-jointed bar or truss element shown in Fig. below is the simplest type of
truss element. The bar element is assumed to have constant cross-sectional area (A)
and modulus of elasticity (E) over its length (L), also assumed to be homogeneous
and isotropic.
The stiffness matrix based on local coordinate system for this type
of elements as follows:
[ ]
=
0000
0101
0000
0101
L
AEK
e
For element :
{ } [ ] { }eee KF =
Global axis
Y
Local axis
Y'
F1'
F2'
F4'
F3'
X'
X
1
2
DOF= 4
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In other terms, We rewrite the equation as:
=
4
3
2
1
4
3
2
1
0000
0101
0000
0101
L
AE
F
F
F
F
For transformation of the stiffness matrix based on global coordinate
system, referring to the above Fig. we will seen that :
sincos 433 FFF +=
cossin 434 FFF +=
similar equations hold at node 1. if we put = cos and = sin ,the above set of four equations may be written as :
=
4
3
2
1
4
3
2
1
00
00
00
00
F
F
F
F
F
F
F
F
OR
[ ]
=
00
00
00
00
T
{ } [ ]{ }FTF =
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also :
{ } [ ]{ }= T
{ } [ ] { }= 1T
[ ] [ ] [ ] [ ]TKTK = 1
the stiffness matrix based on global coordinate system for bar
(truss) element :
[ ]
=
22
22
22
22
scsscs
csccsc
scsscs
csccsc
L
AEK
e
where C =cos and S = sin
Example 4:
Using the stiffness displacement method to find displacements at node
1 and hence solve for all the internal member forces and support
reactions.
Given:
E = 200 kN/mm2
A = 6400 mm2
45o
60o
40 kN
3 m
1
2 3 4
1
23
F1
F2
F3
F4
F5
F6
F7
F8
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ELE. NODE O
COS SIN C.S C2
S2
L MM AE/L
1 1-2 135 -0.707 0.707 -0.5 0.5 0.5 4242 301.74
2 1-3 90 0 1 0 0 1 3000 426.67
3 1-4 30 0.866 0.5 0.433 0.75 0.25 6000 213.33
For Ele. 1
=
4
3
2
1
4
3
2
1
.
87.15087.15087.15087.150
87.15087.15087.15087.150
87.15087.15087.15087.150
87.15087.15087.15087.150
F
F
F
F
For Ele. 2
=
6
5
2
1
6
5
2
1
.
67.426067.4260
0000
67.426067.42600000
F
F
FF
For Ele.3
=
8
7
2
1
8
7
2
1
.
33.5337.9233.5337.92
37.9216037.92160
33.5337.9233.5337.9237.9216037.92160
F
F
FF
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Overall stiffness matrix
=
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
33.5337.92000033.5337.92
37.92160000037.92160
0067.42600067.4260
00000000
000087.15087.15087.15087.150
000087.15087.15087.15087.15033.5337.9267.426087.15087.15087.63048.58
37.921600087.15087.15048.5887.310
F
F
F
F
F
FF
F
3 = 4 =5 = 6 =7 =8 = 0F1= 0 and F2 = -40
=
2
1
87.63048.58
48.5887.310
40
0
21 48.5887.3100 = 21 87.63048.5840 +=
solve (1) and (2) to get :
1= - 0.012142= - 0.06453
To find the reactions :
=
06453.0
01214.0
33.5337.92
37.92160
67.4260
00
87.15087.150
87.15087.150
8
7
6
5
4
3
F
F
F
F
F
F
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F3 = -7.903 kN
F4 = 27.53 kN
F5 = 0
F6 = 27.53F7 = 7.903
To find the internal forces, first we obtain the local displacement due to
global displacement for each element.
For element one
{ } [ ] { }eee T =
=
=
=
=
=
0
0
0645.0
01214.0
.
707.0707.000
707.0707.000
00707.0707.0
00707.0707.0
4
3
2
1
4
3
2
1
0
0053.0
037.0
4
3
2
1
=
==
=
{ } [ ] { }eee KF =
=
=
=
=
=
0
0
053.0
037.0
.
0000
074.301074.301
0000
074.301074.301
4
3
2
1
4
3
2
1
F
F
F
F
Tension member
Similarly for elements 2 &3 .
0
164.11
0
164.11
3
2
1
=
=
=
=
F
F
F
F
F1/
1
2
X/
Y/
F3/
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