m4l28.pdf Beams Continue 01

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Module 
4 
 
Analysis of Statically 
Indeterminate 
Structures by the Direct 
Stiffness Method 
Version 2 CE IIT, Kharagpur 
 
 
 
 
 
 
 
 
Lesson 
28 
The Direct Stiffness 
Method: Beams 
(Continued) 
 
Version 2 CE IIT, Kharagpur 
 
Instructional Objectives 
After reading this chapter the student will be able to 
1. Derive member stiffness matrix of a beam element. 
2. Assemble member stiffness matrices to obtain the global stiffness matrix for a 
beam. 
3. Write the global load-displacement relation for the beam. 
4. Impose boundary conditions on the load-displacement relation of the beam. 
5. Analyse continuous beams by the direct stiffness method. 
 
 
28.1 Introduction 
In the last lesson, the procedure to analyse beams by direct stiffness method has 
been discussed. No numerical problems are given in that lesson. In this lesson, 
few continuous beam problems are solved numerically by direct stiffness method. 
 
Example 28.1 
Analyse the continuous beam shown in Fig. 28.1a. Assume that the supports are 
unyielding. Also assume that EI is constant for all members. 
 
 
 
The numbering of joints and members are shown in Fig. 28.1b. The possible 
global degrees of freedom are shown in the figure. Numbers are put for the 
unconstrained degrees of freedom first and then that for constrained 
displacements. 
 
Version 2 CE IIT, Kharagpur 
 
 
 
The given continuous beam is divided into three beam elements Two degrees of 
freedom (one translation and one rotation) are considered at each end of the 
member. In the above figure, double headed arrows denote rotations and single 
headed arrow represents translations. In the given problem some displacements 
are zero, i.e., 0876543 ====== uuuuuu from support conditions. 
 
In the case of beams, it is not required to transform member stiffness matrix from 
local co-ordinate system to global co-ordinate system, as the two co-ordinate 
system are parallel to each other. 
 
 
 
First construct the member stiffness matrix for each member. This may be done 
from the fundamentals. However, one could use directly the equation (27.1) 
given in the previous lesson and reproduced below for the sake convenience. 
 
Version 2 CE IIT, Kharagpur 
 
[ ]
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−−−
−
−
=
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
L
EI
k
zzzz
zzzz
zzzz
zzzz
4626
612612
2646
612612
22
2323
22
2323
 (1) 
 
The degrees of freedom of a typical beam member are shown in Fig. 28.1c. 
Here equation (1) is used to generate element stiffness matrix. 
 
 
Member 1: , node points 1-2. mL 4=
 
 The member stiffness matrix for all the members are the same, as the length 
and flexural rigidity of all members is the same. 
 
[ ]
1
3
5
6
0.1375.05.0375.0
375.01875.0375.01875.0
5.0375.00.1375.0
375.01875.0375.01875.0
'
1356..
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−−−
−
−
= zzEIk
fodGlobal
 (2) 
 
On the member stiffness matrix, the corresponding global degrees of freedom 
are indicated to facilitate assembling. 
 
Member 2: , node points 2-3. mL 4=
 
[ ]
2
4
1
3
0.1375.05.0375.0
375.01875.0375.01875.0
5.0375.00.1375.0
375.01875.0375.01875.0
2413..
2
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−−−
−
−
= zzEIk
fodGlobal
 (3) 
 
Member 3: , node points 3-4. mL 4=
 
 
Version 2 CE IIT, Kharagpur 
 
[ ]
7
8
2
4
0.1375.05.0375.0
375.01875.0375.01875.0
5.0375.00.1375.0
375.01875.0375.01875.0
7824..
3
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−−−
−
−
= zzEIk
fodGlobal
 (4) 
 
The assembled global stiffness matrix of the continuous beam is of the 
order . The assembled global stiffness matrix may be written as, 88×
 
 
[ ]
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−−−
−
−
−
−−−
−−−
−
−
=
1875.0375.0001875.00375.00
375.00.100375.005.00
001875.0375.001875.00375.0
00375.00.10375.005.0
1875.0375.000375.01875.00375.0
001875.0375.01875.0375.0375.00
375.05.0000375.00.25.0
00375.05.0375.00.05.00.2
zzEIK
 (5) 
 
Now it is required to replace the given members loads by equivalent joint loads. 
The equivalent loads for the present case is shown in Fig. 28.1d. The 
displacement degrees of freedom are also shown in Fig. 28.1d. 
 
Version 2 CE IIT, Kharagpur 
 
 
 
Thus the global load vector corresponding to unconstrained degree of freedom 
is, 
 
{ } ⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧ −=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧=
33.2
5
2
1
p
p
pk (6) 
 
Writing the load displacement relation for the entire continuous beam, 
 
 
 
⎪⎪
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪⎪
⎪
⎨
⎧
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−−−
−
−
−
−−−
−−−
−
−
=
⎪⎪
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪⎪
⎪
⎨
⎧ −
8
7
6
5
4
3
2
1
8
7
6
5
4
3
18703750001870037500
3750010037500500
00187037500187003750
0037500103750050
18703750003750187003750
00187037501870375037500
37505000037500250
003750503750005002
332
5
u
u
u
u
u
u
u
u
....
....
....
....
.....
.....
.....
......
EI
p
p
p
p
p
p
.
zz
 
(7) 
 
where is the joint load vector, { }p { }u is displacement vector. 
 
Version 2 CE IIT, Kharagpur 
 
We know that 0876543 ====== uuuuuu . Thus solving for unknowns and 
, yields 
1u
2u
 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧ −
2
1
0.25.0
5.00.2
33.2
5
u
u
EI zz (8) 
 
 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧ −
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−
−=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
333.2
5
0.25.0
5.00.2
75.3
1
2
1
zzEIu
u
 (9) 
 
 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧−=
909.1
977.21
zzEI
 
Thus displacements are, 
 
zzzz EI
u
EI
u 909.1 and 977.2 21 =−= (10) 
 
The unknown joint loads are given by, 
 
 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧−
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−
=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
909.1
977.21
375.00
5.00
0375.0
05.0
0375.0
375.00
8
7
6
5
4
3
zz
zz EI
EI
p
p
p
p
p
p
 (11) 
 
 
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−
−
−=
715.0
955.0
116.1
488.1
116.1
715.0
 
 
Version 2 CE IIT, Kharagpur 
 
The actual reactions at the supports are calculated as, 
 
 
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−
−
−=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−
−
−+
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
−=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
+
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
284.3
715.1
116.1
489.1
116.10
716.5
715.0
955.0
116.1
488.1
116.1
715.0
4
67.2
0
0
9
5
8
7
6
5
4
3
8
7
6
5
4
3
8
7
6
5
4
3
p
p
p
p
p
p
p
p
p
p
p
p
R
R
R
R
R
R
F
F
F
F
F
F
 (12) 
 
 
Member end actions for element 1 
 
 
⎪⎪⎭
⎪⎪⎬⎪⎨=
⎫
⎪⎪⎩
⎪⎧
−
−
−
⎪⎭
⎪⎪⎬
⎫
⎪⎩
⎪⎪⎨
⎧
−
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
−
−−−
−
−
+
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎩
⎪⎪⎨
⎧
=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎩
⎪⎪⎨
⎧
977.2
116.1
488.1
116.1
977.2
0
0
1
0.1375.05.0375.0
375.01875.375.01875.0
5.0375.00.1375.0
375.01875.0375.01875.0
0
0
0
0
4
3
2
1
zz
zzEI
q
q
q
q
 
 
 ⎪⎪⎥⎢⎪⎪ 00 EI 
 
 
 
 
 
 (13) 
 
 
 
 
Member end actions for element 2 
 
 
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−
−−−
−
−
+=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
909.1
0
977.2
0
1
0.1375.05.0375.0
375.01875.0375.01875.0
5.0375.00.1375.0
375.01875.0375.01875.0
4
3
2
1
zz
zz EI
EI
q
q
q
q
 
 
 
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−
=
58.4
4.5
98.2
6.4
 (14) 
Version 2 CE IIT, Kharagpur 
 
Member end actions for element 3 
 
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−
−−−
−
−
+
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−
=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
0
0
909.1
0
1
0.1375.05.0375.0
375.01875.0375.01875.0
5.0375.00.1375.0
375.01875.0375.01875.0
67.2
0.4
67.2
0.4
4
3
2
1
zz
zz EI
EI
q
q
q
q
 
 
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−
=
72.1
28.3
58.4
72.4
 (15) 
 
Example 28.2 
Analyse the continuous beam shown in Fig. 28.2a. Assume that the supports are 
unyielding. Assume EI to be constant for all members. 
 
 
 
The numbering of joints and members are shown in Fig. 28.2b. The global 
degrees of freedom are also shown in the figure. 
 
The given continuous beam is divided into two beam elements. Two degrees of 
freedom (one translation and one rotation) are considered at each end of the 
member. In the above figure, double headed arrows denote rotations and single 
headed arrow represents translations. Also it is observed that displacements 
 from support conditions. 06543 ==== uuuu
First construct the member stiffness matrix for each member. 
 
 
Member 1: , node points 1-2. mL 4=
 
Version 2 CE IIT, Kharagpur 
 
The member stiffness matrix for all the members are the same, as the length and 
flexural rigidity of all members is the same. 
 
 (1) [ ]
1
3
5
6
0.1375.05.0375.0
375.01875.0375.01875.0
5.0375.00.1375.0
375.01875.0375.01875.0
'
1356..
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−−−
−
−
= zzEIk
fodGlobal
 
On the member stiffness matrix, the corresponding global degrees of freedom 
are indicated to facilitate assembling. 
 
Member 2: , node points 2-3. mL 4=
 
[ ]
2
4
1
3
0.1375.05.0375.0
375.01875.0375.01875.0
5.0375.00.1375.0
375.01875.0375.01875.0
2413..
2
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−−−
−
−
= zzEIk
fodGlobal
 (2) 
 
The assembled global stiffness matrix of the continuous beam is of order 66× . 
The assembled global stiffness matrix may be written as, 
 
 
[ ]
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−
−−−
−−−
−
−
=
1875.0375.001875.00375.0
375.00.10375.005.0
001875.01875.0375.0375.0
1875.0375.01875.0375.0375.00
00375.0375.00.15.0
375.05.0375.005.02
zzEIK (3) 
 
 
Now it is required to replace the given members loads by equivalent joint loads. 
The equivalent loads for the present case is shown in Fig. 28.2c. The 
displacement degrees of freedom are also shown in figure. 
Version 2 CE IIT, Kharagpur 
 
 
 
 
 
Thus the global load vector corresponding to unconstrained degree of freedom 
is, 
 
{ } ⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧=
67.6
0
2
1
p
p
pk (4) 
 
Writing the load displacement relation for the entire continuous beam, 
 
Version 2 CE IIT, Kharagpur 
 
 (5) 
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−
−
−−−
−−−
−
−
=
⎪⎪
⎪⎪
⎪
⎭
⎪⎪
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
6
5
4
3
2
1
6
5
4
3
1875.0375.001875.00375.0
375.00.10375.005.0
001875.01875.0375.0375.0
1875.0375.01875.0375.0375.00
00375.0375.00.15.0
375.05.0375.005.02
67.6
0
u
u
u
u
u
u
EI
p
p
p
p
zz
 
We know that 06543 ==== uuuu . Thus solving for unknowns and , yields 1u 2u
 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
2
1
0.15.0
5.00.2
67.6
0
u
u
EI zz (6) 
 
 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−
−=⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
67.6
0
0.25.0
5.00.1
75.1
1
2
1
zzEIu
u
 
 
 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧−=
627
90511
.
.
EI zz
 (7) 
 
Thus displacements are, 
 
zzzz EI
u
EI
u 62.7 and 905.1 21 =−= 
 
The unknown joint loads are given by, 
 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧−
⎥⎥
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎢⎢
⎣
⎡
−−=
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
62.7
905.11
0375.0
05.0
375.0375.0
375.00
6
5
4
3
zz
zz EI
EI
p
p
p
p
 
 
Version 2 CE IIT, Kharagpur 
 
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
−
−
−=
714.0
95.0
14.2
857.2
 (8) 
 
The actual support reactions are, 
 
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
−
=
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
−
−
−+
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
−
=
⎪⎪
⎪
⎭
⎪⎪
⎪
⎬
⎫
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
286.9
62.7
86.7
857.22
714.0
95.0
14.2
857.2
10
67.6
10
20
6
5
4
3
R
R
R
R
 (9) 
 
Member end actions for element 1 
 
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−
−−−
−
−
+
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−
=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
905.1
0
0
0
1
0.1375.05.0375.0
375.01875.0375.01875.0
5.0375.00.1375.0
375.01875.0375.01875.0
66.6
10
66.6
10
4
3
2
1
zz
zz EI
EI
q
q
q
q
 
 
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−
=
565.8
714.10
707.5
285.9
 (10) 
 
Member end actions for element 2 
 
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−
⎥⎥
⎥⎥
⎦
⎤
⎢⎢
⎢⎢
⎣
⎡
−
−−−
−
−
+
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
−
=
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
62.7
0
905.1
0
1
0.1375.05.0375.0
375.01875.0375.01875.0
5.0375.00.1375.0
375.01875.0375.01875.0
66.6
10
66.6
10
4
3
2
1
zz
zz EI
EI
q
q
q
q
 
 
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
=
0
856.7565.8
14.12
 
 
Version 2 CE IIT, Kharagpur 
 
Summary 
In the previous lesson the beam element stiffness matrix is derived from 
fundamentals. Assembling member stiffness matrices, the global stiffness matrix 
is generated. The procedure to impose boundary conditions on the load-
displacement relation is discussed. In this lesson, a few continuous beam 
problems are analysed by the direct stiffness method. 
 
Version 2 CE IIT, Kharagpur 
 
	Analysis of Statically Indeterminate Structures by the Direct Stiffness Method
	The Direct Stiffness Method: Beams (Continued)
	Instructional Objectives
	28.1 Introduction
	Summary