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Module 4
Analysis of Statically Indeterminate
Structures by the Direct Stiffness Method
Version 2 CE IIT, Kharagpur
Lesson 30
The Direct Stiffness Method: Plane Frames
Version 2 CE IIT, Kharagpur
Instructional Objectives After reading this chapter the student will be able to 1. Derive plane frame member stiffness matrix in local co-ordinate system. 2. Transform plane frame member stiffness matrix from local to global co-
ordinate system. 3. Assemble member stiffness matrices to obtain the global stiffness matrix of
the plane frame. 4. Write the global load-displacement relation for the plane frame. 5. Impose boundary conditions on the load-displacement relation. 6. Analyse plane frames by the direct stiffness matrix method. 30.1 Introduction In the case of plane frame, all the members lie in the same plane and are interconnected by rigid joints. The internal stress resultants at a cross-section of a plane frame member consist of bending moment, shear force and an axial force. The significant deformations in the plane frame are only flexural and axial. In this lesson, the analysis of plane frame by direct stiffness matrix method is discussed. Initially, the stiffness matrix of the plane frame member is derived in its local co-ordinate axes and then it is transformed to global co-ordinate system. In the case of plane frames, members are oriented in different directions and hence before forming the global stiffness matrix it is necessary to refer all the member stiffness matrices to the same set of axes. This is achieved by transformation of forces and displacements to global co-ordinate system. 30.2 Member Stiffness Matrix Consider a member of a plane frame as shown in Fig. 30.1a in the member co-ordinate system ' . The global orthogonal set of axes '' zyx xyz is also shown in the figure. The frame lies in the xy plane. The member is assumed to have uniform flexural rigidity EI and uniform axial rigidity EA for sake of simplicity. The axial deformation of member will be considered in the analysis. The possible displacements at each node of the member are: translation in - and - direction and rotation about - axis.
'x 'y'z
Version 2 CE IIT, Kharagpur
Thus the frame members have six (6) degrees of freedom and are shown in Fig.30.1a. The forces acting on the member at end j and are shown in Fig. 30.1b. The relation between axial displacement and axial forces is derived in chapter 24. Similarly the relation between shear force, bending moment with translation along axis and rotation about axis are given in lesson 27. Combining them, we could write the load-displacement relation in the local co-ordinate axes for the plane frame as shown in Fig 30.1a, b as,
k
'y 'z
Version 2 CE IIT, Kharagpur
=
6
5
4
3
2
1
22
2323
22
2323
6
5
4
3
2
1
'
'
'
'
'
'
460
260
6120
6120
0000
260
460
6120
6120
0000
'
'
'
'
'
'
u
u
u
u
u
u
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LAE
LAE
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LEI
LAE
LAE
q
q
q
q
q
q
zzzz
zzzz
zzzz
zzzz
(30.1a)
This may be succinctly written as
{ } [ ]{ }''' ukq = (30.1b) where is the member stiffness matrix. The member stiffness matrix can also be generated by giving unit displacement along each possible displacement degree of freedom one at a time and calculating resulting restraint actions.
[ ]'k
30.3 Transformation from local to global co-ordinate system 30.3.1 Displacement transformation matrix In plane frame the members are oriented in different directions and hence it is necessary to transform stiffness matrix of individual members from local to global co-ordinate system before formulating the global stiffness matrix by assembly. In Fig. 30.2a the plane frame member is shown in local coordinate axes zyx and in Fig. 30.2b, the plane frame is shown in global coordinate axes xyz . Two ends of the plane frame member are identified by j and . Let and
be respectively displacements of ends k 321 ',',' uuu
654 ',',' uuu j and of the member in local coordinate system ' . Similarly and respectively are displacements of ends
k'' zyx 321 ,, uuu 654 ,, uuuj and k of the member in global co-ordinate system.
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Let be the angle by which the member is inclined to global x -axis. From Fig.30.2a and b, one could relate to as, 321 ',',' uuu 321 ,, uuu
sincos' 211 uuu += (30.2a)
cossin' 212 uuu += (30.2b)
33' uu = (30.2c) This may be written as,
Version 2 CE IIT, Kharagpur
=
6
5
4
3
2
1
6
5
4
3
2
1
100000
0000
0000
000100
0000
0000
'
'
'
'
'
'
u
u
u
u
u
u
lm
ml
lm
ml
u
u
u
u
u
u
(30.3a)
Where, cos=l and sin=m . This may be written in compact form as,
{ } [ ]{ }uTu =' (30.3b) In the above equation, [ is defined as the displacement transformation matrix and it transforms the six global displacement components to six displacement components in local co-ordinate axes. Again, if the coordinate of node
]T
j is and coordinate of node are ( 11, yx ) k ( )22 , yx , then,
Lxxl 12cos == and
Lyym 12sin == .
Where ( ) ( )212212 yyxxL += (30.4)
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30.3.2 Force displacement matrix
Let and be respectively the forces in member at nodes 321 ',',' qqq 654 ',',' qqq j and as shown in Fig. 30.3a in local coordinate system. and
are the forces in members at node k 321 ,, ppp
654 ,, ppp j and respectively as shown in Fig. 30.3b in the global coordinate system. Now from Fig 30.3a and b,
k
sin'cos' 211 qqp = (30.5a)
cos'sin' 212 qqp += (30.5b)
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33 'qp = (30.5c) Thus the forces in global coordinate system can be related to forces in local coordinate system by
=
6
5
4
3
2
1
6
5
4
3
2
1
'
'
'
'
'
'
100000
0000
0000
000100
0000
0000
q
q
q
q
q
q
lm
ml
lm
ml
p
p
p
p
p
p
(30.6a)
Where, cos=l and sin=m . This may be compactly written as,
{ } [ ] { }'qTp T= (30.6b) 30.3.3 Member global stiffness matrix From equation (30.1b), we have
{ } [ ]{ }''' ukq = Substituting the above value of { }'q in equation (30.6b) results in,
{ } [ ] [ ]{ }'' ukTp T= (30.7) Making use of equation (30.3b), the above equation may be written as
{ } [ ] [ ][ ]{ }uTkTp T '= (30.8) or
{ } [ ]{ }ukp = (30.9) The equation (30.9) represents the member load-displacement relation in global coordinate system. The global member stiffness matrix [ ]k is given by,
[ ] [ ] [ ][ ]TkTk T '= (30.10)
Version 2 CE IIT, Kharagpur
After transformation, the assembly of member stiffness matrices is carried out in a similar procedure as discussed for truss. Finally the global load-displacement equation is written as in the case of continuous beam. Few numerical problems are solved by direct stiffness method to illustrate the procedure discussed. Example 30.1 Analyze the rigid frame shown in Fig 30.4a by direct stiffness matrix method. Assume and . The flexural rigidity 441033.1;200 mIGPaE ZZ
== 204.0 mA =EI and axial rigidity EA are the same for both the beams.
Solution: The plane frame is divided in to two beam elements as shown in Fig. 30.4b. The numbering of joints and members are also shown in Fig. 30.3b. Each node has three degrees of freedom. Degrees of freedom at all nodes are also shown in the figure. Also the local degrees of freedom of beam element are shown in the figure as inset.
Version 2 CE IIT, Kharagpur
Formulate the element stiffness matrix in local co-ordinate system and then transform it to global co-ordinate system. The origin of the global co-ordinate system is taken at node 1. Here the element stiffness matrix in global co-ordinates is only given. Member 1: == 90;6 mL node points 1-2; 0=l and 1=m .
Version 2 CE IIT, Kharagpur
[ ] [ ] [ ]1 'Tk T k T =
3 3 3 3
6 6
3 3 31
3 3 3 3
6 6
3 3 3
1 2 3 4 5 6
1.48 10 0 4.44 10 1.48 10 0 4.44 100 1.333 10 0 0 1.333 10 0
4.44 10 0 17.78 10 4.44 10 0 8.88 101.48 10 0 4.44 10 1.48 10 0 4.44 10
0 1.333 10 0 0 1.333 10 04.44 10 0 8.88 10 4.44 10 0 17.
k
=
3
12345678 10
3
(1) Member 2: == 0;4 mL ; node points 2-3 ; 1=l and 0=m . [ ] [ ] [ ][ ]
9
8
7
6
5
4
1066.26101001088.810100
1010105010101050
00100.200100.2
1088.8101001066.2610100
1010105010101050
00100.200100.2987654
'
3333
3333
66
3333
3333
66
2
=
= TkTk T
(2) The assembled global stiffness matrix [ ]K is of the order 99 . Carrying out assembly in the usual manner, we get,
Version 2 CE IIT, Kharagpur
[ ]
=
66.2610033.13100000
10501050000
002000002000000
33.1310044.441044.488.8044.4
1050103.1338003.13330
00200044.405.200144.4048.1
00088.8044.477.17044.4
00003.1333003.13330
00044.4048.144.4048.1
K (3)
Version 2 CE IIT, Kharagpur
Version 2 CE IIT, Kharagpur
The load vector corresponding to unconstrained degrees of freedom is (vide 30.4d),
{ }
=
=
6
24
12
6
5
4
p
p
p
pk (4)
In the given frame constraint degrees of freedom are . Eliminating rows and columns corresponding to constrained degrees of freedom from global stiffness matrix and writing load-displacement relationship for only unconstrained degree of freedom,
987321 ,,,,, uuuuuu
Version 2 CE IIT, Kharagpur
=
6
5
4
3
44.441044.4
103.13380
44.405.2001
10
6
24
12
u
u
u
(5)
Solving we get,
=
3-
5-
-6
6
5
4
100.13-
101.695-
106.28
u
u
u
(6)
6 5
4 56.28 10 m., 1.695 10u u = =
Let be the support reactions along degrees of freedom
respectively (vide Fig. 30.4e). Support reactions are calculated by 987321 ,,,,, RRRRRR
9,8,7,3,2,1
+
=
6
5
4
3
9
8
7
3
2
1
9
8
7
3
2
1
33.13100
1050
002000
88.8044.4
03.1333
44.4048.1
10
654
u
u
u
p
p
p
p
p
p
R
R
R
R
R
R
F
F
F
F
F
F
=
+
=
92.25
40.25
57.12
85.16
59.22
42.11
92.1
40.1
57.12
14.1
59.22
57.0
24
24
0
18
0
12
9
8
7
3
2
1
R
R
R
R
R
R
(7)
Version 2 CE IIT, Kharagpur
Example 30.2 Analyse the rigid frame shown in Fig 30.5a by direct stiffness matrix method. Assume and . The flexural rigidity
5 4200 GPa ; 1.33 10 mZZE I= = 201.0 mA =
EI and axial rigidity EA are the same for all beams.
Solution: The plane frame is divided in to three beam elements as shown in Fig. 30.5b. The numbering of joints and members are also shown in Fig. 30.5b. The possible degrees of freedom at nodes are also shown in the figure. The origin of the global co- ordinate system is taken at A (node 1).
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Now formulate the element stiffness matrix in local co-ordinate system and then transform it to global co-ordinate system. In the present case three degrees of freedom are considered at each node.
Member 1: == 90;4 mL ; node points 1-2 ; 2 1 0x xlL
= = and
2 1 1y ymL
= = .
The following terms are common for all elements.
5 22
65 10 kN/m; 9.998 10 kNAE EIL L
= =
2 3
3
12 44.999 10 kN/m; 2.666 10 kN.mEI EIL L
= =
Version 2 CE IIT, Kharagpur
[ ] [ ] [ ][ ]
6
5
4
3
2
1
1066.201011033.10101
0105001050
10101050.010101050.0
1033.101011066.20101
0105001050
10101050.010101050.0654321
'
3333
55
3333
3333
55
3333
1
=
= TkTk T
(1) Member 2: == 0;4 mL node points 2-3; 1=l and 0=m . [ ] [ ] [ ][ ]
9
8
7
6
5
4
10666.210101033.11010
101105.00101105.00
00100.500100.5
1033.1101010666.21010
101105.00101105.00
00100.500100.5987654
'
3333
3333
66
3333
3333
65
2
=
= TkTk T
(2)
Member 3: == 270;4 mL ; node points 3-4 ; 2 1 0x xlL
= = and
2 1 1y ymL
= = .
Version 2 CE IIT, Kharagpur
[ ] [ ] [ ][ ]
12
11
10
9
8
7
1066.201011033.10101
0105001050
10101050.010101050.0
1033.101011066.20101
0105001050
10101050.010101050.0121110987
'
3333
55
3333
3333
55
3333
3
=
= TkTk T
(3) The assembled global stiffness matrix [ ]K is of the order 1212 . Carrying out assembly in the usual manner, we get,
[ ]
=
66.20133.100.1000000
0500005000000000
105.00.105.0000000
33.100.133.50.10.133.10.10000
050000.15.50000.15.00000
0.105.00.105.50000500000
00033.10.1033.50.10.133.100.1
0000.150.000.15.500005000
000005000.105.5000.1050.0
00000033.100.166.200.1
0000000500005000
0000000.1050.00.1050.0
103K
(4)
Version 2 CE IIT, Kharagpur
Version 2 CE IIT, Kharagpur
The load vector corresponding to unconstrained degrees of freedom is,
{ }
=
=
24
24
0
24
24
10
9
8
7
6
5
4
p
p
p
p
p
p
pk (5)
In the given frame, constraint (known) degrees of freedom are
. Eliminating rows and columns corresponding to constrained degrees of freedom from global stiffness matrix and writing load displacement relationship,
121110321 ,,,,, uuuuuu
=
9
8
7
6
5
4
3
33.51133.110
15.500015.00
105.50000500
33.10.1033.50.10.1
0.15.000.15.5000
005000.105.500
10
24
24
0
24
24
10
u
u
u
u
u
u
(6)
Solving we get,
=
3-
5-
2-
3-
5-
-2
9
8
7
6
5
4
103.85
105.65-
101.43
108.14-
103.84-
1043.1
u
u
u
u
u
u
(7)
Version 2 CE IIT, Kharagpur
Let be the support reactions along degrees of freedom respectively. Support reactions are calculated by
121110321 ,,,,, RRRRRR12,11,10,3,2,1
+
=
9
8
7
6
5
4
3
12
11
10
3
2
1
12
11
10
3
2
1
33.100.1000
05000000
0.105.0000
00033.100.1
00005000
0000.105.0
10
987654
u
u
u
u
u
u
p
p
p
p
p
p
R
R
R
R
R
R
F
F
F
F
F
F
=
+
=
42.19
28.28
99.10
43.3
71.19
99.0
42.19
28.28
99.10
43.3
71.19
99.0
0
0
0
0
0
0
12
11
10
3
2
1
R
R
R
R
R
R
(8)
Summary In this lesson, the analysis of plane frame by the direct stiffness matrix method is discussed. Initially, the stiffness matrix of the plane frame member is derived in its local co-ordinate axes and then it is transformed to global co-ordinate system. In the case of plane frames, members are oriented in different directions and hence before forming the global stiffness matrix it is necessary to refer all the member stiffness matrices to the same set of axes. This is achieved by transformation of forces and displacements to global co-ordinate system. In the end, a few problems are solved to illustrate the methodology.
Version 2 CE IIT, Kharagpur
Analysis of Statically Indeterminate Structures by the Direct Stiffness MethodThe Direct Stiffness Method: Plane FramesInstructional Objectives30.1 Introduction30.2 Member Stiffness Matrix30.3 Transformation from local to global co-ordinate system30.3.1 Displacement transformation matrix30.3.2 Force displacement matrix30.3.3 Member global stiffness matrix
Summary