Date post: | 06-Nov-2015 |
Category: |
Documents |
Upload: | a-bashir-asalai |
View: | 26 times |
Download: | 1 times |
0 | P a g e
12/6/2015
Analysis Of Frame By Direct Stiffness Method | Abdelghani Asalai
UNIVERSITY OF TRIPOLI
CE 609 ASSIGNMENT II
Lecturer: Dr. Ramadan Murad
1 | P a g e
FRAME ANALYSIS USING THE DIRECT STIFFNESS
METHOD:
For the frame shown, use the stiffness method to:
(a) Determine the deflection and rotation at B & C.
(b) Determine all the reactions at supports.
E=A=I=Constant =1
100 kN.m
100 kN
4m
4m
4m
E = I = A =1
2 | P a g e
Transformation of Local Matrix
3
4
1
A
B
1
2
6
5
=90.0
Local
42
B C
7
6 9
5
8
Local
=0.0
3
C
D11
7
9
12
8
10
=90.0
Local
3 | P a g e
3
4
4m
4m
4m1
2
3
A
B C
D11
1
7
2
69
12
5
8
10
4 | P a g e
=
1 2 3 4 5 6
1
2
3
4
5
6[
0 0
0 0
012
36
20
12
26
2
06
24
0
6
22
0 0
0 0
0 12
36
20
12
36
2
06
22
0
6
24
]
1 =
1 2 3 4 5 6
1
2
3
4
5
6[ 1
40 0
1
40 0
03
16
3
80
3
16
3
8
03
81 0
3
8
1
2
1
40 0
1
40 0
0 3
163
80
3
163
8
03
8
1
20
3
81 ]
5 | P a g e
=
[ 0 0 0 0
0 0 0 0
0 0 1 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0 0 1]
Where
C=Cos ()
S= Sin ()
Transformation matrix for member No.1 , where angle = 900 is
1 =
[ 0 1 0 0 0 0
1 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 1 0
0 0 0 1 0 0
0 0 0 0 0 1]
6 | P a g e
Frame member global stiffness matrix
1 =
1 2 3 4 5 6
1
2
3
4
5
6[ 3
160
3
83
160
3
8
01
40 0
1
40
3
80 1
3
80
1
2
3
160
3
8
3
160
3
8
0 1
40 0
1
40
3
80
1
2
3
80 1 ]
Local Matrix for member 2:
2 =
4 5 6 7 8 9
4
5
6
7
8
9[ 1
40 0
1
40 0
03
16
3
80
3
16
3
8
03
81 0
3
8
1
2
1
40 0
1
40 0
0 3
163
80
3
163
8
03
8
1
20
3
81 ]
7 | P a g e
Transformation matrix for member No.2 , where angle = 00 is
2 =
[ 1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1]
Frame member global stiffness matrix
2 =
4 5 6 7 8 9
4
5
6
7
8
9[ 1
40 0
1
40 0
03
16
3
80
3
16
3
8
03
81 0
3
8
1
2
1
40 0
1
40 0
0 3
163
80
3
163
8
03
8
1
20
3
81 ]
8 | P a g e
Local Matrix for member 3:
3 =
7 8 9 10 11 12
7
8
9
10
11
12[ 1
40 0
1
40 0
03
16
3
80
3
16
3
8
03
81 0
3
8
1
2
1
40 0
1
40 0
0 3
163
80
3
163
8
03
8
1
20
3
81 ]
Transformation matrix for member No.3 , where angle = 900 is
3 =
[ 0 1 0 0 0 0
1 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 1 0
0 0 0 1 0 0
0 0 0 0 0 1]
9 | P a g e
Frame member global stiffness matrix
3 =
7 8 9 10 11 12
7
8
9
10
11
12[ 3
160
3
83
160
3
8
01
40 0
1
40
3
80 1
3
80
1
2
3
160
3
8
3
160
3
8
0 1
40 0
1
40
3
80
1
2
3
80 1 ]
10 | P a g e
Global stiffness matrix for the whole frame could be established by gathering the global matrixs for each member into one large matrix taking
into account the nodes in the joints.
[
]
[
]
[
]
11 | P a g e
The global Matrix for the whole frame in its final form is as follows:
=
1
2
3
4
5
6
7
8
9
10
11
12
1 2 3 4 5 6 7 8 9 10 11 12
[ 3
160
3
83
160
3
80 0 0 0 0 0
01
40 0 1/4 0 0 0 0 0 0 0
3/8 0 13
80
1
20 0 0 0 0 0
3
160
3
8
7
160
3
81
40 0 0 0 0
0 1
40 0
7
16
3
80
3
16
3
80 0 0
3
80
1
2
3
8
3
82 0
3
8
1
20 0 0
0 0 0 1
40 0
7
160
3
83
160
3
8
0 0 0 0 3
163
80
7
163
80
1
40
0 0 0 03
8
1
23
83
82
3
80
1
2
0 0 0 0 0 0 3
160
3
8
3
160
3
8
0 0 0 0 0 0 0 1
40 0
1
40
0 0 0 0 0 0 3
80
1
2
3
80 1 ]
{} = []{}
occurs in the unconstrained nodes , where in node 1 there is rotational displacement and in node 2 and 3
there is vertical, horizontal and rotational displacement , but in node 4 there is no displacement because joint
is totally constrained .
{
10010000000 }
[ 1
3
80
1
20 0 0
3
8
7
160
3
81
40 0
0 07
16
3
80
3
16
3
81
2
3
8
3
82 0
3
8
1
2
0 1
40 0
7
160
3
8
0 0 3
163
80
7
163
8
0 03
8
1
23
83
82 ] 1
=
{
566.57 1493.9448.52187.331083.5648.52268.2 }
{} = {} + []{}
12 | P a g e
We obtain the reactions acting on the support as follows
{
12101112}
=
{
00000}
+
[ 3
83
160
3
80 0 0
0 0 1/4 0 0 0 0
0 0 0 0 3
160
3
8
0 0 0 0 0 1
40
0 0 0 0 3
80
1
2]
{
566.57 1493.9448.52187.331083.5648.52268.2 }
=
{
2.612.13102.612.13272.24}
100kN.m
4m
4m
4m
A
B
C
D102.6kN
2.6kN
12.3kN
272.24kN.m
12.3kN
100kN