stiffness method
DESCRIPTION
Stiffness Method. Lecture No. : 13. المحاضرة الثالثة عشر. Analysis of Grids Using Direct stiffness method. Preliminary example :. Calculate the deformations of the shown space frame where E = 1200 kN/cm 2 , G = 500 kN/cm 2 and the sections are shown in figure. 100 kN. A. 200 kN. B. - PowerPoint PPT PresentationTRANSCRIPT
Lecture No. : 13 عشر الثالثة 1المحاضرة
2
Preliminary example :Calculate the deformations of the shown space frame where E = 1200 kN/cm2, G = 500 kN/cm2 and the sections are shown in figure
3
8 m
C
30x80 cm 20x60 cm
100 kN200 kN
10 m
B
A
z
y
xz
y
x30x80 cm
A =2400 cm2
Iz =1,280,000 cm4
Iy =180,000 cm4
Ix =550,180 cm4
20x60 cm
A =1200 cm2
Iz =360,000 cm4
Iy =40,000 cm4
Ix =126,435 cm4
Sections Properties
4
CA
d1d2
d3
d4d5
d6
xy
z
Modeling
5
k11
K =
k21
k31
k12
k22
k32
k13
k23
k33
Stiffness matrix
k41
k51
k61
k42
k52
k62
k43
k53
k63
k14
k24
k34
k15
k25
k35
k16
k26
k36
k44
k54
k64
k45
k55
k65
k46
k56
k66
6
C
B
A
d1=1
xy
z
First column in Stiffness matrix
7
z
y
xz
y
x
xy
z
d 1=1
d 1=1
AEL
12EIy
L3
6 EIy
L2
8
1200X1200
80012x1200X180,000
10003
6x1200X180,000
10002
2.592
1,296
1,800
xy
z
2.6
1,296
1,800
d1d2
d3
d4d5
d6
k11
=
k21
k31
k41
k51
k61
1,802.6
0
0
0
0
-1,2969
10
k11
=
k21
k31
k41
k51
k61
1,802.6
0
0
0
0
-1,296
First column in Stiffness matrix
xy
z
Second column in Stiffness matrix
d2=1
11
C
B
A
z
y
x
z
y
x
xy
z
d2=1
d2 =1
AEL
12EIy
L3
6 EIy
L2
12
12x1200X40,000
8003
6x1200X40,000
8002
1200X2400
1000
1.125
450
2,880
xy
z
2,880 1.1
450
k12
=
k22
k32
k42
k52
k62
2,881.1
0
0
0
0
450d1d2
d3
d4d5
d6
13
14
k12
=
k22
k32
k42
k52
k62
2,881.1
0
0
0
0
450
Second column in Stiffness matrix
xy
z
Third column in Stiffness matrix
d3=1
15
C
B
A
z
y
xz
y
x
d3=1d3=1
12EIz
L3
6 EIz
L2
12EIz
L3
6 EIz
L2
16
10.1
4,050
18.4
9,216
xy
z
18.4
9,216 10.1
4,050
k13
=
k23
k33
k43
k53
k63
0
0
28.5
- 4,050
9,216
0d1d2
d3
d4d5
d6
17
18
k13
=
k23
k33
k43
k53
k63
0
0
28.5
- 4,050
9,216
0
Third column in Stiffness matrix
xy
z
Fourth column in Stiffness matrix
d4=1
19
C
B
A
z
y
xz
y
x
xy
z
d 4=1d 4
=1
GIx
L
4 EIz
L6 EIz
L2
20
500X126,435
80079,022
9,216
6,144,000
xy
z
9,216
6,144,000
79,022
k14
=
k24
k34
k44
k54
k64
0
0
9,216
0
6,223,022
0d1d2
d3
d4d5
d6
21
22
k14
=
k24
k34
k44
k54
k64
0
0
9,216
0
0
Fourth column in Stiffness matrix
6,223,022
xy
z
Fifth column in Stiffness matrix
d5=1
23
C
B
A
z
y
x
z
y
x
xy
z
d5 =1
d5 =1
GIx
L
4 EIz
L6 EIz
L2
24
4,050
2,160,000
275,090
xy
z
d1d2
d3
d4d5
d6
275,090
4,050
2,160,000
k15
=
k25
k35
k45
k55
k65
0
0
- 4,050
0
0
2,435,090
25
26
Fifth column in Stiffness matrix
k15
=
k25
k35
k45
k55
k65
0
0
- 4,050
0
0
2,435,090
xy
z
Sixth column in Stiffness matrix
d6=1
27
C
B
A
z
y
xz
y
x
xy
z
d6=1d6=1
4 EIy
L
6 EIy
L2
4 EIy
L
6 EIy
L2
28
450
240,000
1,296
864,000
xy
z
1,296
864,000
450240,000
k16
=
k26
k36
k46
k56
k66
450
-1,296
0
0
0
1,104,000d1d2
d3
d4d5
d6
29
30
k16
=
k26
k36
k46
k56
k66
450
-1,296
0
0
0
1,104,000
Sixth column in Stiffness matrix
k11
=
k21
k31
k41
k51
k61
1,802.6
0
0
0
0
-1,296
k12
=
k22
k32
k42
k52
k62
2,881.1
0
0
0
0
450
k13
=
k23
k33
k43
k53
k63
0
0
28.5
- 4,050
9,216
0
31
k14
=
k24
k34
k44
k54
k64
0
0
9,216
0
6,223,022
0
k15
=
k25
k35
k45
k55
k65
0
0
- 4,050
0
0
2,435,090
k16
=
k26
k36
k46
k56
k66
450
-1,296
0
0
0
1,104,000
=K
32
1,802.6
0
0
0
0
-1,296
2,881.1
0
0
0
0
450
0
0
28.5
- 4,050
9,216
0
0
0
9,216
0
6,223,022
0
0
0
- 4,050
0
0
2,435,090
450
-1,296
0
0
0
1,104,000
Force vector
33
8 m
C
30x80 cm 20x60 cm
100 kN200 kN
10 m
B
A
200 kN
100 kN
250100
100
250
50
50
100
100
Fixed End Reaction(FER)
34
200 kN
100 kN
250100
100
250
50
50
100
100
Fixed End Action(FEA)
35
200 kN
100 kN
250100
100
250
50
50
100
100
d1d2
d3
d4d5
d6
F1
=
F2
F3
F4
F5
F6
0
0
- 150
- 250
100
0 36
Stiffness Equation F = K D
37
0
0
- 150
- 25000
10000
0
d1
=
d2
d3
d4
d5
d6
1,802.6
0
0
0
0
-1,296
2,881.1
0
0
0
0
450
0
0
28.5
- 4,050
9,216
0
0
0
9,216
0
6,223,022
0
0
0
- 4,050
0
0
2,435,090
450
-1,296
0
0
0
1,104,000
Stiffness Equation
D = K-1 F
d1
=
d2
d3
d4
d5
d6
-1
38
0
0
- 150
- 25000
10000
0
1,802.6
0
0
0
0
-1,296
2,881.1
0
0
0
0
450
0
0
28.5
- 4,050
9,216
0
0
0
9,216
0
6,223,022
0
0
0
- 4,050
0
0
2,435,090
450
-1,296
0
0
0
1,104,000
Deformations
39
0
0
- 11.8714
0.013564
d1
=
d2
d3
d4
d5
d6
- 0.015638
0
40
8 m
C
30x80 cm 20x60 cm
100 kN200 kN
10 m
B
A
41
42
d1
d2d3
d4
d5
d6
d1
d2
d3
d1x
yz
d2
d3
43
Example 1:Draw all diagrams for the shown grid where E = 1200 kN/cm2, G = 500 kN/cm2 and the sections are shown in figure
44
8 m
C
30x80 cm 20x60 cm
100 kN200 kN
10 m
B
A
z
y
xz
y
x30x80 cm
Iz =1,280,000 cm4
Ix =550,180 cm4
20x60 cm
Iz =360,000 cm4
Ix =126,435 cm4
Sections Properties
45
CA d1
d2d3
xy
z
Modeling
46
k11
K = k21
k31
k12
k22
k32
k13
k23
k33
Stiffness matrix
47
xy
z
d1=1
48
C
B
A
First column in Stiffness matrix
z
y
xz
y
x
d1=1d1=1
12EIz
L3
6 EIz
L2
12EIz
L3
6 EIz
L2
49
10.1
4,050
18.4
9,216
xy
z
18.4
9,216 10.1
4,050
k11
=k21
k31
28.5
- 4,050
9,216
50
d1
d2d3
51
k11
=k21
k31
28.5
- 4,050
9,216
First column in Stiffness matrix
xy
z
d2=1
52
C
B
A
Second column in Stiffness matrix
z
y
xz
y
x
xy
z
d 2=1d 2
=1
GIx
L
4 EIz
L6 EIz
L2
53
500X126,435
80079,022
9,216
6,144,000
xy
z
9,216
6,144,000
79,022
k12
=k22
k32
9,216
0
6,223,022
54
d1
d2d3
55
k12
=k22
k32
9,216
0
6,223,022
Second column in Stiffness matrix
xy
z
Third column in Stiffness matrix
d3=1
56
C
B
A
z
y
x
z
y
x
xy
z
d3 =1
d3 =1
GIx
L
4 EIz
L6 EIz
L2
57
4,050
2,160,000
275,090
xy
z
275,090
4,050
2,160,000
k13
=k23
k33
- 4,050
0
2,435,090
58
d1
d2d3
59
k13
=k23
k33
- 4,050
0
2,435,090
Third column in Stiffness matrix
60
k11
=k21
k31
28.5
- 4,050
9,216
k12
=k22
k32
9,216
0
6,223,022
k13
=k23
k33
- 4,050
0
2,435,090
61
=
28.5
- 4,050
9,216
9,216
0
6,223,022
- 4,050
0
2,435,090
K
=K
1,802.6
0
0
0
0
-1,296
2,881.1
0
0
0
0
450
0
0
28.5
- 4,050
9,216
0
0
0
9,216
0
6,223,022
0
0
0
- 4,050
0
0
2,435,090
450
-1,296
0
0
0
1,104,000
Force vector
62
8 m
C
30x80 cm 20x60 cm
100 kN200 kN
10 m
B
A
200 kN
100 kN
250100
100
250
50
50
100
100
Fixed End Reaction(FER)
63
200 kN
100 kN
250100
100
250
50
50
100
100
Fixed End Action(FEA)
64
200 kN
100 kN
250100
100
250
50
50
100
100
F1
=F2
F3
- 150
- 250
100
65
d1
d2d3
F1
=
F2
F3
F4
F5
F6
0
0
- 150
- 250
100
0
Stiffness Equation
F = K D
66
d1
d2
d3
- 150
- 25000
10000
=
28.5
- 4,050
9,216
9,216
0
6,223,022
- 4,050
0
2,435,090
Stiffness Equation
D = K-1 F
-1
67
d1
d2
d3
- 150
- 25000
10000
=
28.5
- 4,050
9,216
9,216
0
6,223,022
- 4,050
0
2,435,090
Deformations
68
d1
=d2
d3
- 11.8714
0.013564
- 0.015638
- 11.8714
0.013564- 0.015638
d1
d2d3
0
0
- 11.8714
0.013564
d1
=
d2
d3
d4
d5
d6
- 0.015638
0
Internal Forces Torsional Moment
IxGL
T t=
d 2=1
GIx
L79,021.9
TBC = 79,021.9x0.013564 = 1072 kNcm = 10.72 kNm
69
- 11.8714
0.013564- 0.015638
d1
d2d3
Internal Forces Torsional Moment
IxGL
T t=
d3 =1
GIx
L275,090
TBA = 275,090x-0.015638= 4302 kNcm = 43.02 kNm
70
- 11.8714
0.013564- 0.015638
d1
d2d3
Internal Forces
B
C
B
A
43.02
43.02
10.72
10.72
Torsional Moment
71
CA
Internal Forces
43.02
10.72
Torsional Moment
72
43.02
43.0210.72
10.72
Internal Forces Bending momentIn Plan
d1=1
6 EIz
L24,050
4,050
73
d3 =1
4 EIz
L2,160,000
1,080,000
2 EIz
L
100 kN
100
100
Internal Forces Bending momentIn Plan
d1=1
4,050
4,050
d3 =1 2,160,000
1,080,000
100
100FER
B
C
MBC = -100 - 4,050 d1 + 2,160,000 d3
MBC = -100+(-4,050x-11.8714+2,160,000x-0.015638)/100
MBC = 43.01 kNm 74
- 11.8714
0.013564- 0.015638
d1
d2d3
Internal Forces Bending momentIn Plan
d1=1
4,050
4,050
d3 =1 2,160,000
1,080,000
100
100FER
B
C
MCB = 100 - 4,050 d1 +1,080,000 d3
MCB = 100+(- 4,050x-11.8714+1,080,000x-0.015638)/100
MCB = 411.9 kNm 75
- 11.8714
0.013564- 0.01564
d1
d2d3
Internal Forces Bending momentIn Plan
43.01
411.9B
C100 kN
50 + (411.9 + 43.01)/8106.86
50 – (411.9 + 43.01)/8- 6.86
76
z
y
x
d1=1
12EIz
L310.1
77
z
y
xd3 =1
6 EIz
L2
4,050
100 kN
50
50
B
C
yBC = 50 + 10.1 d1 – 4,050 d3
yBC = 50 + 10.1x-11.8714 – 4,050x-0.015638
yBC = - 6.57 kN
- 11.8714
0.013564- 0.015638
d1
d2d3
z
y
x
d1=1
12EIz
L310.1
78
z
y
xd3 =1
6 EIz
L2
4,050
100 kN
50
50
B
C
yCB = 50 - 10.1 d1 + 4,050 d3
yCB = 50 - 10.1x-11.8714+ 4,050x-0.015638
yCB = 106.57 kN
10.1
4,050
- 11.8714
0.013564- 0.01564
d1
d2d3
Internal Forces Bending momentIn Plan
43.01
411.88B
C100 kN
106.57
- 6.57
79
Internal Forces Bending momentIn Plan
d1=1
6 EIz
L2
9,216
9,216
d 2=1
4 EIz
L6,144,000
3,072,000
200 kN
250
250FER
MBA = 250+9,216 d1 + 6,144,000 d2
MBA = 250+(9,216x-11.8714+ 6,144,000x 0.013564)/100
MBA = - 10.7 kNm
B
A
80
- 11.8714
0.013564- 0.01564
d1
d2d3
Internal Forces Bending momentIn Plan
d1=1
6 EIz
L2
9,216
9,216
d 2=14 EIz
L6,144,000
3,072,000
200 kN
250
250FER
MAB = -250+9,216 d1 + 3,072,000 d2
MAB = -250+(9,216x-11.8714+ 3,072,000x0.013564)/100
MAB = - 927.38 kNm
B
A
81
- 11.8714
0.013564- 0.015638
d1
d2d3
Internal Forces Shear ForceIn Plan
d1=1
d 2=1
200 kN
FER
yBA = 100+18.4 d2 + 9,216 d2
yBA = 100+18.4x-11.8714+ 9,216x0.013564
yBA = 6.57 kN
B
A
82
- 11.8714
0.013564- 0.015638
d1
d2d3
100
10018.4
18.4
9,216
9,216
Internal Forces Shear ForceIn Plan
d1=1
d 2=1
200 kN
FER
yAB = 100-18.4 d2 - 9,216 d2
yAB = 100-18.4x-11.8714- 9,216x0.013564
yAB = 193.43 kN
B
A
83
- 11.8714
0.013564- 0.015638
d1
d2d3
100
10018.4
18.4
9,216
9,216
Internal Forces Bending momentIn Plan
10.7
927.38B
A200 kN
193.43
6.57
84
Bending moment In Plan
85
CA
927.38
458.34
41.66
43.0
1411.9
184.45
15.55
200
500
10.7
43.01
411.9B
C100 kN
10.7
927.38B
A200 kN
Shear Force In Plan
86
CA
193.43
106.57
6.57
6.57
B
B
C100 kN
106.576.57
B
A200 kN
193.43
6.57
87
B
C100 kN
106.57
6.57
B
A200 kN
193.43
6.57
10.7 43.01
43.02
10.72
Summary
88
89
d1
d2d3
d4
d5
d6
d1
d2
d3
Questions
90