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! Simple Frames! Frame-Member Stiffness Matrix! Displacement and Force Transformation Matrices! Frame-Member Global Stiffness Matrix
! Special Frames! Frame-Member Global Stiffness Matrix
FRAME ANALYSIS USING THE STIFFNESSMETHOD
2
Simple Frames
3
Frame-Member Stiffness Matrix
0 0 00 - AE/LAE/L
4EI/L - 6EI/L2 2EI/L6EI/L2 00
6EI/L2 - 12EI/L3 6EI/L212EI/L3 00
0
2EI/L
-6EI/L2
0
- 6EI/L2
12EI/L3
0
-6EI/L2
4EI/L
0
6EI/L2
-12EI/L3
AE/L
0
0
-AE/L
0
0
m
x´y´
i
j
3´
5´
6´
2´
4´
1´
3´ 6´2´ 4´1´ 5´
[k´]
4´5´6´
1´2´3´
6EI/L24EI/L
6EI/L24EI/L
AE/L
AE/L
6EI/L2
6EI/L2
12EI/L3
12EI/L3
2EI/L
d 1́ = 1
AE/L
AE/L
d 2́ = 1
12EI/L3
12EI/L3
d 3́ = 1
2EI/L
6EI/L2
6EI/L2
6EI/L26EI/L2AE/L
2EI/L
6EI/L2
6EI/L2d 6́
= 1
6EI/L2
6EI/L22EI/L
6EI/L2
12EI/L3
12EI/L3
4EI/L4EI/L12EI/L3
12EI/L3
6EI/L2
6EI/L2AE/L
6EI/L2d 4́ = 1
d 5́ = 1AE/L
AE/L
4
m
i
j
m
i
j
x´y´
x
y
Displacement and Force Transformation Matrices
12
3
45
6
θyθx
4´
5´
6´
1´
2´
3´
5
λx
q4 = q4´ cos θx - q5´ cos θy
q5 = q4´ cos θy + q5´ cos θx
q6 = q6´
λy
i
jθy
θx
y´
x´
4´5´
6´
1´
2´
3´
θy
θx
m
i
j
x
y
12
3
45
6
Force Transformation
Lxx ij
x
−=λ
Lyy ij
y
−=λ
−=
6'
5'
4'
6
5
4
10000
qqq
λλλλ
qqq
xy
yx
−
−
=
6'
5'
4'
3'
2'
1'
6
5
4
3
2
1
1000000000000000010000000000
qqqqqq
λλλλ
λλλλ
qqqqqq
xy
yx
xy
yx
[ ] [ ] [ ]'qTq T=
6
[q] = [T]T[q´]
= [T]T ( [k´][d´] + [q´F] )
= [T]T [k´][d´] + [T]T [q´F]
[q] = [T]T [k´][T][d] + [T]T [q´F] = [k][d] + [qF]
Therefore, [k] = [T]T [k´][T]
[qF] = [T]T [q´F]
[q] = [T]T[q´]
[d´] = [T][d]
[k] = [T]T [k´][T]
7
[q] = [T]T[q´]= [T]T ( [k´][d´] + [q´F] ) = [T]T[k´][d´] + [T]T[q´F] = [T]T [k´][T][d] + [T]T [q´F]
Frame Member Global Stiffness Matrix
[k] [qF][ k ] = [ T ]T[ k´ ][T] =
Ui
Vi
Mi
Uj
Vj
Mj
Vj Mj
- λiy6EIL2
λix6EIL2
2EIL
λjy6EIL2
- λjx6EIL2
4EIL
Ui Vi Mi
- λiy6EIL2
λix6EIL2
4EIL
λjy6EIL2
λjx6EIL2
-
2EIL
Uj
AEL
- λixλiy)(12EIL3
AEL
λiy2 +
12EIL3
λix2 )(
λix6EIL2
λix6EIL2
AEL
λiyλjx -
12EIL3
λixλjy)-(
AEL
λixλjx +
12EIL3
λiyλjy)-(
λjy6EIL2
AEL
λjx2 +
12EIL3
λjy2 )(
λjy6EIL2
λjx6EIL2
-
AEL
- λjxλjy)(12EIL3
- λjx6EIL2
AEL
λixλjy -
12EIL3
λiyλjx)-(
AEL
- λixλiy)(12EIL3
- λiy6EIL2
- λiy6EIL2
AEL
λixλjx +
12EIL3
λiy λjy)-(
)(AEL
λix2 +
12EIL3
λiy2
AEL
λiyλjy +
12EIL3
λix λjx )-(
AEL
λiyλjy +
12EIL3
λixλjx)-(
12EIL3
λjx2 )
AEL
- ( λixλjy- λiyλjx )12EIL3
AEL
λiyλjx -
12EIL3
λixλjy)-(
AEL
- λjxλjy)(12EIL3
AEL λjy
2 + (
8
5 kN
6 m
6 m
AB
C
Example 1
For the frame shown, use the stiffness method to:(a) Determine the deflection and rotation at B.(b) Determine all the reactions at supports.(c) Draw the quantitative shear and bending moment diagrams.E = 200 GPa, I = 60(106) mm4, A = 600 mm2
9
5 kN
6 m
6 m
AB
C kN/m666.667(6m)
)m10)(60mkN1012(20012
3
462
6
3 =××
=
−
LEI
kN/m200006m
)mkN10)(200m10(600 2
626
=××
=
−
LAE
kN2000(6m)
)m10)(60mkN106(2006
2
462
6
2 =××
=
−
LEI
mkN80006m
)m10)(60mkN104(2004
462
6
•=××
=
−
LEI
mkN40006m
)m10)(60mkN102(2002
462
6
•=××
=
−
LEI
Global :
AB
C
1
2
78 9
4
6 5
12 3
10
Global :
AB
C
1
2
78 9
4
6 5
12 3
A B14´
5´ 6´
1´
2´ 3´
Local :
5´
2´
1´ 3´
4´
6´
2
Using Transformation Matrix:
� Member Stiffness Matrix
[ ]
−−−−
−−−
−
=
LEILEILEILEILEILEILEILEI
LAELAELEILEILEILEILEILEILEILEI
LAEAE/L
/4/60/2/60/6/120/6/120
00/00//2/60/4/60/6/120/6/120
00/00
k'
22
2323
22
2323
Mi
Vj
Mj
Vi
Nj
Ni
θi ∆j θj∆i δjδi
11
A B14´
5´ 6´
1´
2´ 3´
Local :
[q] = [q´]
-> [k]1 = [k´]1
Stiffness Matrix: Member 1
Global:
AB
C
1
2
78 9
4
6 5
12 3
4 6 5 1 2 34
6
5
1
2
3
20000
0
0
-20000
0
0
0
666.667
2000
0
-666.667
2000
0
2000
8000
0
-2000
4000
-20000
0
0
20000
0
0
0
-666.667
-2000
0
666.667
-2000
0
2000
4000
0
-2000
8000
[k]1 =
12
Local:
5´
2´
1´ 3´
4´
6´
2
[q]2 = [ T ]T[ q´]2
q1´
q3´
q2´
q4´q5´
q6´
q1
q3
q2
q7
q8q9
[T]T
Stiffness Matrix: Member 2
=
123789
4´0000
0-1
5´000100
6´000001
1´0
0-1
000
2´100000
3´001000
90o
λjx = cos (-90o) = 0λjy = sin (-90o) = -1
λix = cos (-90o) = 0λiy = sin (-90o) = -1
Global:
AB
C
1
2
78 9
4
6 5
12 3
13
[k]2 = [ T ]T[ k´ ]2[ T ]
1´ 2´ 3´ 4´ 5´ 6´1´
2´
3´
4´
5´
6´
20000
0
0
-20000
0
0
0
666.667
2000
0
-666.667
2000
0
2000
8000
0
-2000
4000
-20000
0
0
20000
0
0
0
-666.667
-2000
0
666.667
-2000
0
2000
4000
0
-2000
8000
[k´]2 =
1 2 3 7 8 9
666.667 20001
2
3
7
8
9
2000
0
0
-666.667
-666.667
0
0
2000
2000
20000 0
0
0
0
-20000
-20000
0
0
8000 -2000
-2000
0
0
4000
4000
666.667 0
0
-2000
-2000
20000 0
0 8000
[k]2 =
14
[k]14 6 5 1 2 3
4
6
5
1
2
3
20000
0
0
-20000
0
0
0666.667
20000
-666.667
2000
02000
80000
-2000
4000
-200000
020000
0
0
0-666.667
-20000
666.667
-2000
02000
40000
-2000
8000
1 2 3 7 8 9666.667 20001
2
3
7
8
9
2000
00
666.667
-666.667
0
0
2000
2000
20000 0
0
0
0
-20000
-20000
0
0
8000 -2000-2000
0
0
4000
4000
666.667 0
0
-2000
2000
20000 0
0 8000
[k]2
Global Stiffness Matrix:
20000
0
-20000
0
0
0
8000
0
-2000
4000
0
-2000
0
4000
-20000
0
0
20666.667
-2000
2000
-2000
16000
20666.667
0
2000
[K]
4
5
1
2
3
4 5 1 2 3
Global:
AB
C
1
2
78 9
4
6 5
12 3
15
AB
C
Q4 = 0Q5 = 0
Q1 = 5Q2 = 0
Q3 = 0
D4
D5
D1
D2
D3
+
0 0
0 0
0
D4
D5
D1
D2
D3
=
0.01316 m
0.01316 m9.199(10-4) rad
-9.355(10-5) m
-1.887(10-3) rad
5 kN
6 m
6 m
AB
C
1
2
Global:
1
2
78 9
4
6 5
12 3
5 kN
=
4
5
1
2
3
4 5 1 2 3-20000 0 0
20666.667 00
2000
2000
20666.667 -2000
-2000 16000
0-2000
4000
08000 0 -2000 4000
-200000
20000
0
0
[Q] = [K][D] + [QF]
16
D4 = 0.01316
D5 = 9.199(10-4)
D6 = 0
D1 = 0.01316
D2 = -9.355(10-5)
D3 = -1.887(10-3)
15 kN
6 m
6 m
AB
C
2
q4
q5
q6
q1
q2
q3
0
0
-1.87
0
1.87
-11.22
Member 1
A B11
2 3
4
6 5
A B1
1.87 kN 11.22 kN�m1.87 kN
4 6 5 1 2 3
4
6
5
1
2
3
20000
0
0
-20000
0
0
0
666.667
2000
0
-666.667
2000
0
2000
8000
0
-2000
4000
-20000
0
0
20000
0
0
0
-666.667
-2000
0
666.667
-2000
0
2000
4000
0
-2000
8000
[q]1 = [k]1[d]1 + [qF]1
17
5 kN
1.87 kN 11.22 kN�m
5 kN1.87 kN
18.77 kN�m
Member 2
q1
q3
q2
q7
q8
q9
D1 = 0.01316
D3 = -1.887(10-3)
D2 = -9.355(10-5)
D7 = 0
D8 = 0
D9 = 0
5
11.22
-1.87
-5
1.87
18.77
1 2 3 7 8 9
666.667 20001
2
3
7
8
9
2000
0
0
-666.667
-666.667
0
0
2000
2000
20000 0
0
0
0
-20000
-20000
0
0
8000 -2000
-2000
0
0
4000
4000
666.667 0
0
-2000
-2000
20000 0
0 8000
[q]2 = [k]2[d]2 + [qF]2
15 kN
6 m
6 m
AB
C
2 1
2 3
78 9
2 2
18
AB
C
Bending moment diagram
A B1
1.87 kN 11.22 kN�m1.87 kN5 kN
6 m
6 m
AB
C
1.87 kN
18.77 kN�m5 kN
1.87 kN
AB
C
Shear diagram
5
5
+
-1.87-
+
18.77
11.22
-11.22
-
5 kN
1.87 kN 11.22 kN�m
5 kN
1.87 kN
18.77 kN�m
2
19
Deflected shape
AB
C
AB
C
Bending moment diagram +
18.77
11.22
-11.22
-
D3=-0.00189 rad
D3=-0.00189 rad
D1=13.16 mmD4=13.16 mm
D4
D5
D1
D2
D3
=
0.01316 m
0.01316 m9.199(10-4) rad
-9.355(10-5) m
-1.887(10-3) rad
Global :
AB
C
1
2
78 9
4
6 5
12 3
D5=0.00092 rad
20
4.5 m
6 m 6 m
3 kN/m
Example 2
For the beam shown, use the stiffness method to:(a) Determine the deflection and rotation at B(b) Determine all the reactions at supports(c) Draw the quantitative shear and bending moment diagrams.E = 200 GPa, I = 60(106) mm4, A = 600 mm2 for each member.
A
B C
21
Global 2
1
1
2 3
4
56
7
8 9
2 7
8 9
1
2 3
[FEM] 9 kN�m
9 kN
9 kN�m
9 kN
3 kN/m
2
Members1
2 3
4
56 1
4.5 m
6 m 6 m
3 kN/m
22
4.5 m
6 m 6 m
3 kN/m
7.5 m
λx = cos θx = 6/7.5 = 0.8
Member 1:
θx
θy 1
2 3
4
56 1
λy = cos θy = 4.5/7.5 = 0.6
mkNm
mkNmL
AE
/160005.7
)/10200)(10600( 2626
=
××=
−
mkNm
mmkNLEI
/33.341)5.7(
)1060)(/10200(12123
4626
3
=
××=
−
kNm
mmkNLEI
1280)5.7(
)1060)(/10200(662
4626
2
=
××=
−
mkNm
mmkNLEI
•=
××=
−
64005.7
)1060)(/10200(44 4626
mkNm
mmkNLEI
•=
××=
−
32005.7
)1060)(/10200(22 4626
23
[ km ] = [ T ]T[ k´ ][T ] =
λx = cos θx
Member m:
θx
θy Uj
VjMj
Ui
Vi
Mim
λy = cos θy
Ui
Vi
Mi
Uj
Vj
Mj
Vj Mj
- λiy6EIL2
λix6EIL2
2EIL
λjy6EIL2
- λjx6EIL2
4EIL
Ui Vi Mi
- λiy6EIL2
λix6EIL2
4EIL
λjy6EIL2
λjx6EIL2
-
2EIL
Uj
AEL
- λixλiy)(12EIL3
AEL
λiy2 +
12EIL3
λix2 )(
λix6EIL2
λix6EIL2
AEL
λiyλjx -
12EIL3
λixλjy)-(
AEL
λixλjx +
12EIL3
λiyλjy)-(
λjy6EIL2
AEL
λjx2 +
12EIL3
λjy2 )(
λjy6EIL2
λjx6EIL2
-
AEL
- λjxλjy)(12EIL3
- λjx6EIL2
AEL
λixλjy -
12EIL3
λiyλjx)-(
AEL
- λixλiy)(12EIL3
- λiy6EIL2
- λiy6EIL2
AEL
λixλjx +
12EIL3
λiy λjy)-(
)(AEL
λix2 +
12EIL3
λiy2
AEL
λiyλjy +
12EIL3
λix λjx )-(
AEL
λiyλjy +
12EIL3
λixλjx)-(
12EIL3
λjx2 )
AEL
- ( λixλjy- λiyλjx )12EIL3
AEL
λiyλjx -
12EIL3
λixλjy)-(
AEL
- λjxλjy)(12EIL3
AEL λjy
2 + (
24
λx = cos θx = 6/7.5 = 0.8
Member 1:
θx
θy 1
2 3
4
56 1
λy = cos θy = 4.5/7.5 = 0.6
4.5 m
6 m 6 m
3 kN/m
7.5 m
=
4
5
6
1
2
3
4 1
-10362.879
-7516.162
768
10362.879
768
7516.162
2
-7516.162
-5978.451
-1024
7516.162
5978.451
-1024
3
-768
1024
3200
768
-1024
6400
10362.879
-768
7516.162
-10362.879
-7516.162
-768
5
7516.162
5978.451
1024
-7516.162
-5978.451
1024
6
-768
1024
6400
768
-1024
3200
[k1]
25
λx = cos 0o = 1.0, λy = cos 90o = 0
Member 2:
2 7
8 9
1
2 3
4.5 m
6 m 6 m
3 kN/m
7.5 mmkN
mmkNm
LAE
/200006
)/10200)(10600( 2626
=
××=
−
mkNm
mmkNLEI
/667.666)6(
)1060)(/10200(12123
4626
3
=
××=
−
kNm
mmkNLEI
2000)6(
)1060)(/10200(662
4626
2
=
××=
−
mkNm
mmkNLEI
•=
××=
−
80006
)1060)(/10200(44 4626
mkNm
mmkNLEI
•=
××=
−
40006
)1060)(/10200(22 4626
26
[k2] =0
2EI/L
-6EI/L2
- 6EI/L2
6EI/L2
0
6EI/L2
-12EI/L3
0
0
0
0
-AE/L
0
0
0
6EI/L2
0
- 6EI/L2
- 12EI/L3
0
0
6EI/L2
2EI/L
0
-6EI/L2
0 - AE/L
0
0
4EI/L
12EI/L3
4EI/L
12EI/L3
AE/L
AE/L
3
8
9
2
7
1
3 92 71 8
[k2] =0
4000
-2000
- 2000
2000
0
2000
-666.667
0
0
0
0
-20000
0
0
0
2000
0
- 2000
- 666.667
0
0
2000
4000
0
-2000
0 - 20000
0
0
8000
666.667
8000
666.667
20000
20000
3
8
9
2
7
1
3 92 71 8
2 7
8 9
1
2 3
27
=
4
5
6
1
2
3
4 1
-10362.879
-7516.162
768
10362.879
768
7516.162
2
-7516.162
-5978.451
-1024
7516.162
5978.451
-1024
3
-768
1024
3200
768
-1024
6400
10362.879
-768
7516.162
-10362.879
-7516.162
-768
5
7516.162
5978.451
1024
-7516.162
-5978.451
1024
6
-768
1024
6400
768
-1024
3200
[k1]
[k2] =0
4000
-2000
- 2000
2000
0
2000
-666.667
0
0
0
0
-20000
0
0
0
2000
0
- 2000
- 666.667
0
0
2000
4000
0
-2000
0 - 20000
0
0
8000
666.667
8000
666.667
20000
20000
3
8
9
2
7
1
3 92 71 8
28
D1
D3
D2 =4.575(10-4) m
-5.278(10-4) rad
-1.794(10-3) m
Global:
1
21
2 3
4
5 6
7
8 9
4.5 m
6 m 6 m
3 kN/m
7.5 m9 kN
9 kN�m
9 kN
0
0
0
Q1
Q3
Q2
D1
D3
D2
0
9
9+1
2
3
1
30362.9
768
7516.16
2
7516.16
6645.12
976
3
768
976
14400=
9 kN�m
9 kN
29
λx = cos θx = 6/7.5 = 0.8
Member 1:
θx
θy 1
2 3
4
56 1
λy = cos θy = 4.5/7.5 = 0.6
0
0
0
D1 = 4.575(10-4)
D2 = -1.794(10-3)
D3 = -5.278(10-4)
0
0
0
0
0
0
+ =
q4
q6
q5
q1
q2q3
4
5
6
1
2
3
5 1 2 34 6
k1
1
11.37 kN
11.37 kN
0.09 kN
1.19 kN�m
0.09 kN
0.50 kN�m 1
9.15 kN6.75 kN 1.19 kN�m
9.15 kN
6.75 kN
0.50 kN�m
9.15
0.50
6.75
-9.15
-6.75
-1.19
=
30
Member 2:
2 7
8 9
1
2 3
0
9
9
0
9
-9
+=
q1
q3
q2
q7
q8q9
D1 = 4.575(10-4)
D2 = -1.794(10-3)
D3 = -5.278(10-4)
0
0
0
1
2
3
7
8
9
2 7 8 91 3
k2
9.15
1.19
6.75
-9.15
11.25
-14.70
=
9 kN�m
9 kN
9 kN�m
9 kN
3 kN/m
2[FEM]
3 kN/m
29.15 kN
6.75 kN
1.19 kN�m
9.15 kN11.25 kN
14.70 kN�m
31
3 kN/m
29.15 kN
6.75 kN
1.19 kN�m
9.15 kN11.25 kN
14.70 kN�m
1
9.15 kN6.75 kN 1.19 kN�m
9.15 kN
6.75 kN
0.50 kN�m
3 kN/m
All Reactions
9.15 kN11.25 kN
14.70 kN�m
9.15 kN
6.75 kN
0.50 kN�m
32
-1.19
Shear diagram (kN)
Deflected shape
-0.09
-0.09
6.75
3 kN/m
29.15 kN
6.75 kN
1.19 kN�m
9.15 kN11.25 kN
14.70 kN�m
1
11.37 kN
11.37 kN
0.09 kN
1.19 kN�m
0.09 kN0.50 kN�m
D3 =-5.278(10-4) rad
D2 = -1.79 mm
D1 = 0.46 mm
D1
D3
D2 =4.575(10-4) m
-5.278(10-4) rad
-1.794(10-3) m -11.25
-
+
Bending-moment diagram (kN�m)
-14.70
0.5
33
Example 3
For the beam shown, use the stiffness method to:(a) Determine the deflection and rotation at B.(b) Determine all the reactions at supports.(c) Draw the quantitative shear and bending moment diagrams.E = 200 GPa, I = 60(106) mm4, A = 600 mm2 for each member.
4.5 m
6 m 3 m
10 kN
A
BC15 kN
20 kN�m
3 kN/m
3 m
34
[FEM] 13 kN/m 11.25 kN
wL/2 = 11.25 kN
wL2/12 = 14.06 kN�m
14.06 kN�m
Global
21
2 7
8 9
1
2 3
4.5 m
6 m 3 m
10 kN
A
BC15 kN
20 kN�m
3 kN/m
3 m
7.5 kN�m
5 kN
7.5 kN�m
5 kN
2
10 kN
λx = cos θx = 6/7.5 = 0.8
1
Members1
2 3
4
56
1
2 3
4
56
7
8 9
θy = 53.13
θx =36.87
λy = cos θy = 4.5/7.5 = 0.6
6.75 kN
9 kN
11.25(0.6) = 6.75 kN
11.25(0.8) = 9 kN
35
Global:
1
21
2 3
4
5 6
7
8 9
Q1 = 15
Q3 = 20
Q2 = 0
D1
D3
D2
-6.75
-14.06 + 7.5
9 + 5+1
2
3
1
30362.9
768
7516.16
2
7516.16
6645.2
976
3
768
976
14400=
[FEM]13 kN/m 11.25 kN
11.25 kN
14.06 kN�m
14.06 kN�m7.5 kN�m
5 kN
7.5 kN�m
5 kN
2
10 kN
6.75 kN
9 kN
11.25(0.6) = 6.75 kN
9 kN
10 kN
A
BC15 kN
20 kN�m
3 kN/m
14.06 kN�m7.5 kN�m
6.75 kN
5 kN9 kN
15 kN20 kN�m
36
D1
D3
D2 =1.751(10-3) m
2.049(10-3) rad
-4.388(10-3) m
1
21
2 3
4
5 6
7
8 9
37
Member 1:
0
0
0
D1 = 1.751(10-3)
D2 = -4.388(10-3)
D3 = 2.049(10-3)
-6.75
14.06
9
-6.75
9
-14.06
+ =
q4
q6
q5
q1
q2q3
4
5
6
1
2
3
5 1 2 34 6
k1
6.51
26.46
24.17
-20.01
-6.17
4.89
=
[FEM]
13 kN/m 11.25 kN
11.25 kN
14.06 kN�m
14.06 kN�m
6.75 kN
9 kN
11.25(0.6) = 6.75 kN
9 kN
θx
θy 1
2 3
4
56 1
λx = cos 36.87o = 0.8, λy = cos 53.13o = 0.6
14.06 kN�m
14.06 kN�m
6.75 kN
9 kN
11.25(0.6) = 6.75 kN
9 kN
38
3 kN/m
7.5 m1
19.71 kN
19.71 kN
7.07 kN
4.89 kN�m
15.43 kN
26.46 kN�m
=
q4
q6
q5
q1
q2q3
6.51
26.46
24.17
-20.01
-6.17
4.89
θx = 36.87o
θy = 53.13o 1
2 3
4
56 1
3 kN/m
7.5 m
20.01 kN
4.89 kN�m
6.51 kN
24.17 kN
26.46 kN�m
6.17 kN
39
Member 2:
2 7
8 9
1
2 3
0
7.5
5
0
5
-7.5
+=
q1
q3
q2
q7
q8q9
D1 = 1.751(10-3)
D2 = -4.388(10-3)
D3 = 2.049(10-3)
0
0
0
1
2
3
7
8
9
2 7 8 91 3
k2
35.02
15.12
6.17
-35.02
3.83
-8.08
=
35.02 kN
6.17 kN
15.12 kN�m
35.02 kN3.83 kN
8.08 kN�m
[FEM]
7.5 kN�m
5 kN
7.5 kN�m
5 kN
2
10 kN 2
10 kN
7.5 kN�m
5 kN
7.5 kN�m
5 kN
40
3 kN/m
7.5 m
20.01 kN
4.89 kN�m
6.51 kN
24.17 kN
26.46 kN�m
6.17 kN
35.02 kN
6.17 kN
15.12 kN�m
35.02 kN3.83 kN
8.08 kN�m
2
10 kN
10 kN
A
BC15 kN
20 kN�m
3 kN/m
6 m 3 m3 m
6.51 kN
24.17 kN
26.46 kN�m
35.02 kN3.83 kN
8.08 kN�m
41
Shear diagram (kN)
Bending-moment diagram (kN�m)
3 kN/m
7.5 m1
19.71 kN
19.71 kN
7.07 kN
4.89 kN�m
15.43 kN
26.46 kN�m
35.02 kN
6.17 kN
15.12 kN�m
35.02 kN3.83 kN
8.08 kN�m
2
10 kN
15.43D1
D3
D2 =1.751(10-3) m
2.049(10-3) rad
-4.388(10-3) m
Deflected shape
D3 = 2.05(10-3) rad
D2 = -4.39 mm
D1 =1.75 mm
-26.46
4.89
-15.12 -8.08
-7.07
6.17
-3.83
42
Special Frames
43
i j
θjθi 4 *
5 * 6*
1*
2*3 *
λix = cos θi
λiy = sin θi
λjx = cos θj
λjy = sin θj
[ q* ] = [ T ]T[ q´ ]
1
Stiffness matrix
i j
4´
5´ 6´
1´
2´3´
[ T ]T
−
−
=
6'
5'
4'
3'
2'
1'
*6
5*
*4
*3
2*
*1
1000000000000000010000000000
qqqqqq
λλλλ
λλλλ
qqqqqq
jxjy
jyjx
ixiy
iyix1*2*3*4*5*6*
1´ 2´ 3´ 4´ 5´ 6´
44
[ ]
−
−
=
1000000000000000010000000000
jxjy
jyjx
ixiy
iyix
λλλλ
λλλλ
T
1´2´3´4´5´6´
1* 2* 3* 4* 5* 6*
� Member Stiffness Matrix
1´2´3´4´5´6´
1´ 2´ 3´ 4´ 5´ 6´
[ ]
−−−−
−−−
−
=
LEILEILEILEILEILEILEILEI
LAELAELEILEILEILEILEILEILEILEI
LAEAE/L
/4/60/2/60/6/120/6/120
00/00//2/60/4/60/6/120/6/120
00/00
k'
22
2323
22
2323
45
[ k ] = [ T ]T[ k´ ][T] =
Ui
Vi
Mi
Uj
Vj
Mj
Vj Mj
- λiy6EIL2
λix6EIL2
2EIL
λjy6EIL2
- λjx6EIL2
4EIL
Ui Vi Mi
- λiy6EIL2
λix6EIL2
4EIL
λjy6EIL2
λjx6EIL2
-
2EIL
Uj
AEL
- λixλiy)(12EIL3
AEL
λiy2 +
12EIL3
λix2 )(
λix6EIL2
λix6EIL2
AEL
λiyλjx -
12EIL3
λixλjy)-(
AEL
λixλjx +
12EIL3
λiyλjy)-(
λjy6EIL2
AEL
λjx2 +
12EIL3
λjy2 )(
λjy6EIL2
λjx6EIL2
-
AEL
- λjxλjy)(12EIL3
- λjx6EIL2
AEL
λixλjy -
12EIL3
λiyλjx)-(
AEL
- λixλiy)(12EIL3
- λiy6EIL2
- λiy6EIL2
AEL
λixλjx +
12EIL3
λiy λjy)-(
)(AEL
λix2 +
12EIL3
λiy2
AEL
λiyλjy +
12EIL3
λix λjx )-(
AEL
λiyλjy +
12EIL3
λixλjx)-(
12EIL3
λjx2 )
AEL
- ( λixλjy- λiyλjx )12EIL3
AEL
λiyλjx -
12EIL3
λixλjy)-(
AEL
- λjxλjy)(12EIL3
AEL λjy
2 + (
46
40 kN
4 m 4 m
3 m
7.416 m
22.02 o
20 kN200 kN�m
6 kN/m
Example 4
For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative bending moment diagrams and qualitativedeflected shape.Take I = 200(106) mm4 , A = 6(103) mm2, and E = 200 GPa for all members.Include axial deformation in the stiffness matrix.
47
1Local
12
40 kN
4 m 4 m
3 m
7.416 m
22.02 o
20 kN200 kN�m
6 kN/m
4
5 6
7
8 922.02 o
Global
2
1´
2´ 3´
4´
5´ 6´
1´
2´3´
4´
5´ 6´
1 *
2* 3 *
48
mkNm
mkNmL
AE /101508
)/10200)(006.0( 3262
×=×
=
mkNm
mmkNLEI
•×=×
= 3426
10208
)0002.0)(/10200(44
mkNLEI
•×= 310102
kNm
mmkNLEI 3
2
426
2 1075.3)8(
)0002.0)(/10200(66×=
×=
mkNm
mmkNLEI /109375.0
)8()0002.0)(/10200(1212 3
3
426
3 ×=×
=
[ ]
−−−−
−−−
−
=
LEILEILEILEILEILEILEILEI
LAELAELEILEILEILEILEILEILEILEI
LAEAE/L
/4/60/2/60/6/120/6/120
00/00//2/60/4/60/6/120/6/120
00/00
k'
22
2323
22
2323
Mi
Vj
Mj
Vi
Nj
Ni
θi ∆j θj∆i δjδi
49
2
1´
2´ 3´
4´
5´ 6´
1´
2´3´
1 4´
5´ 6´
Local
8 m
8 m
03750
0
- 3750- 937.5
0
0375010000
0-3750
0 - 150000
00
0
10000
-3750
- 3750
3750
0
3750
-937.5
0
0
00
-150000
0
0
[ k´]1 = [ k´]2 =20000
937.5
20000
937.5
150000
150000
θi ∆j θj∆i δjδi
Mi
Vj
Mj
Vi
Nj
Ni
50
θi =22.02o
Member 1:
1
1 *
2 * 3 *
4
5 6
Global1´
2´3´
1 4´
5´ 6´Local
θj = 0o
λjx = cos (0o) = 1,λjy = sin (0o) = 0
λix = cos (22.02o) = 0.927,λiy = sin (22.02o) = 0.375
40 kN40 kN�m40 kN�m
20 kN20 kN
1
[FEM]
q1´
q3´
q2´
q4´
q5´
q6´
q*1
q*3
q*2
q4
q5q6
[ T ]1T
00
0
00
0
00
0
000
000
000
1 0
1
01
00
0 0
0.9270.375
- 0.375 0.927
10 0
00
[ q* ] = [ T ]T[ q´ ]
=
1´ 4´ 5´ 6´2´ 3´
1*
2*
3*
4*
5*
6*
51
θi =22.02o1
1 *
2 * 3 *
4
5 6
Global1´
2´3´
1 4´
5´ 6´Local
θj = 0o
[ k* ]1 = [ T ]1T[ k´ ]1[ T ]1
[ k* ]1 = 103
1*2*3*456
1*
129.04651.811-1.406
-139.0580.351-1.406
2*
51.811
21.892
3.476-56.240
-0.8693.476
3*
-1.4063.47620.000.00-3.7510.00
4
-139.058-56.240
01500
0
5
0.351-0.869-3.75
00.938-3.75
6
-1.4063.47610.00
0-3.75
20
52
θi =22.02o1
1 *
2 * 3 *
4
5 6
Globalθj = 0o
40 kN40 kN�m40 kN�m
20 kN20 kN
1
[FEM]
[ qF* ] = [ T ]T[ qF´]
[ qF*] = [ T ]1T
0
40
20
0
20
-40
40 kN40 kN�m
20 kN
1
40 kN�m
18.547.5
-7.5018.5440020-40
=
1*
2*
3*
4
5
6
53
Member 2
q1´
q3´
q2´
q4´
q5´
q6´
q4
q6
q5
q7
q8q9
[ T ]2T
0.927-0.375
0.3750.927
1
00
0
00
0
00
0
000
000
000
1
0.927-0.375
0.3750.927
00
0 0
0 0
00
[ q ] = [ T ]T[ q´ ]
=
1´ 4´ 5´ 6´2´ 3´
456
78
9
6 kN/m2
24 kN
24 kN
32 kN�m
32 kN�m
[ q´F]
2
1´
2´ 3´
4´
5´ 6´
[ q´]
λix = λjx = cos (-22.02o) = 0.927,λiy = λjy = sin (-22.02o) = -0.375
2
45 6
[ q ]7
8 922.02o
22.02o
54
2
1´
2´ 3´
4´
5´ 6´
[ q´ ]
[ k ]2 = [ T ]2T[ k´]2[ T ]2
2
45 6
[ q ]7
8 922.02o
22.02o
[ k ]2 = 103
456
78
9
4
129.046-51.811
1.406-129.046
51.8111.4056
5
-51.81121.8923.476
51.811-21.892
3.476
6
1.4063.476
20-1.406-3.476
10
7
-129.04651.811-1.406
129.046-51.811-1.406
8
51.811-21.892-3.476
-51.81121.892-3.476
9
1.4063.476
10-1.406-3.476
20
55
6 kN/m2
24 kN
24 kN
32 kN�m
32 kN�m
[ q´F ]
[ qF* ] = [ T ]T[ qF´ ]
[ qF ] = [ T ]2T
0
32
24
0
24
-32
8.99822.249328.99822.249-32
=
4
5
6
7
8
9
6 kN/m2
32 kN�m
32 kN�m
[ qF ]
22.249 kN
8.998kN
22.249 kN
8.998 kN
2
45 6
[ q ]7
8 922.02o
22.02o
56
[ k* ]1 = 103
1*2*3*456
1*129.04651.811-1.406
-139.0580.351-1.406
2*51.811
21.892
3.476-56.240-0.8693.476
3*-1.4063.47620.000.00-3.7510.00
4-139.058-56.240
01500
0
50.351-0.869-3.75
00.938-3.75
6-1.4063.47610.00
0-3.75
20
Global Stiffness:
12
1 *
2* 3 *
4
5 6
7
8 9
2*
7
8 9
[ k ]2 = 103
456
78
9
4
129.046-51.811
1.406-129.046
51.8111.4056
5
-51.81121.8923.476
51.811-21.892
3.476
6
1.4063.476
20-1.406-3.476
10
7
-129.04651.811-1.406
129.046-51.811-1.406
8
51.811-21.892-3.476
-51.81121.892-3.476
9
1.4063.476
10-1.406-3.476
20
57
Global: 40 kN 20 kN200 kN�m
6 kN/m 12
1 *
2 * 3 *
4
5 6
7
8 9
6 kN/m2
32 kN�m
32 kN�m
[ qF ]
22.249 kN
8.998 kN
22.249 kN
8.998 kN
40 kN40 kN�m
20 kN
1
40 kN�m
18.54 kN7.5
[ q*F ]
2 *
7
8 9
[ Q ] = [ K ][ D ] + [ QF ]
40 kN�m
7.5
40 kN�m 32 kN�m
20 kN 22.249 kN
8.998 kN
20 kN200 kN�m
D1*
D4
D3*
D5
D6
= 0.0
= 0.0
= 0.0
= -20
= -200
Q1*
Q4
Q3*
Q5
Q6
= 103
1*
3*
4
5
6
1*
129.046
-1.406
-139.058
0.352
-1.406
3*
-1.406
20
0
-3.75
10
4
-139.058
0
279.046
-51.811
1.406
5
0.351
-3.75
-51.811
22.829
-0.274
6-1.406
10
1.406
-0.274
40
+ 8.998
-7.5
40
20 +22.249
-40 + 32
58
Global: 40 kN 20 kN200 kN�m
6 kN/m 12
1 *
2 * 3 *
4
5 6
7
8 9
2 *
7
8 9
D1*
D4
D3*
D5
D6
=
-0.0205 m
-0.0112 rad
-0.0191 m
-0.0476 m
-0.0024 rad
59
1
1 *
2 * 3 *
4
5 6
[ q ]
40 kN 40 kN�m
20 kN
1
40 kN�m
18.547.5
[ qF* ]Member 1: [ q*] = [ k*][ d*] + [ qF*]
q1*
q3*
q2*
q4
q5q6
D1*=-0.0205
D3*=-0.0112D2*= 0.0
D4= -0.0191D5= -0.0476D6=-0.0024
-7.50
40
18.54
0
20
-40
+
q1*
q3*
q2*
q4
q5
q6
0
022.63 kN
-8.49 kN19.02 kN7.87 kN�m
=
40 kN
22.63 kN
8.49 kN
19.02 kN
7.87 kN�m
= 103
1*2*3*456
1*
129.04651.811-1.406
-139.0580.351-1.406
2*
51.811
21.892
3.476-56.240
-0.8693.476
3*
-1.4063.47620.000.00-3.7510.00
4
-139.058-56.240
01500
0
5
0.351-0.869-3.75
00.938-3.75
6
-1.4063.47610.00
0-3.75
20
60
Member 2 : [ q ] = [ k ][ d ] + [ qF ]
q4
q6
q5
q7
q8q9
8.998
3222.249
8.99822.249
-32
+
2
4
5 6
[ q ] 7
8 96 kN/m
2
32 kN�m
32 kN�m
[ qF ]
22.249 kN
8.998 kN
22.249 kN
8.998 kN
D4= -0.0191
D6=-0.0024D5= -0.0476
D7 = 0D8 = 0D9 = 0
q4
q6
q5
q7
q8
q9
8.49 kN
-207.87 kN�m-39.02 kN
9.51 kN83.51 kN-248.04 kN�m
= 8.49 kN
39.02 kN
207.87 kN�m
9.51 kN83.51 kN
248.04 kN�m
6 kN/m
=103
456
78
9
4
129.046-51.811
1.406-129.046
51.8111.4056
5
-51.81121.8923.476
51.811-21.892
3.476
6
1.4063.476
20-1.406-3.476
10
7
-129.04651.811-1.406
129.046-51.811-1.406
8
51.811-21.892-3.476
-51.81121.892-3.476
9
1.4063.476
10-1.406-3.476
20
61
40 kN
22.63 kN
8.49 kN
19.02 kN
7.87 kN�m
8.49 kN
39.02 kN
207.87 kN�m
9.51 kN
83.51 kN
248.04 kN�m6 kN/m
22.5 kN81 kN
22.5 kN
33 kN
8.49 kN
20.98 kN
Bending-moment diagram (kN�m)
-
+
207.87
-248.04
83.56
+7.87
Deflected shape
D3*=-0.0112 rad
D1*=- 20.5 mm
D6=-0.0024 rad
D1*
D4
D3*
D5
D6
=
-0.0205 m
-0.0112 rad
-0.0191 m
-0.0476 m
-0.0024 radD4=-19.1 mmD5=-47.6 mm
62
3 m
80 kN�m20 kN/m
20o
50 kN
4 m 2 m
A B
C
Example 5
For the beam shown:(a) Use the stiffness method to determine all the reactions at supports.(b) Draw the quantitative free-body diagram of member.(c) Draw the quantitative bending moment diagramsand qualitative deflected shape.Take I = 400(106) mm4 , A = 60(103) mm2, and E = 200 GPa for all members.
63
Global
1
2 3
4
5 6
7*9*
8*
1
2 2
1´2´
3´
4´5´
6´
3 m
80 kN�m20 kN/m
20o
50 kN
4 m 2 m
A B
C
Local
1´
2´ 3´
4´
5´ 6´1
76.31o
64
3 m
80 kN�m20 kN/m
20o
50 kN
4 m 2 m
A B
C
1
2 3
4
5 61
Member 1:
kN/m103000m4
)kN/m10)(200m10(60
3
2623
×=
××=
−
LAE
mkN1080m4
)m10)(400kN/m104(2004
3
4626
•×=
××=
−
LEI
mkN1040m4
)m10)(400kN/m102(2002
3
4626
•×=
××=
−
LEI
kN1030m)(4
)m10)(400kN/m106(2006
3
2
4626
2
×=
××=
−
LEI
kN/m1015m)(4
)m10)(400kN/m1012(20012
3
3
4626
3
×=
××=
−
LEI
65
d1
d3
d2
d4
d5
d6
q1
q3
q2
q4
q5q6
0
26.67
40
040
-26.67
+= 103
123
45
6
1 4 5 6
030
40
0-30
80
3000
0
0
-30000
0
2
015
30
0-15
30
3
030
80
0-30
40
-30000
0
3000
0
0
0-15
-30
015
-30
3 m
80 kN�m20 kN/m
20o
50 kN
4 m 2 m
A B
C
26.67 kN�m26.67 kN�m
40 kN40 kN
1
20 kN/m
Member 1: [ q ] = [ k ][ d ] + [ qF ]
1
2 3
4
5 61
56.31o
76.31o
66
Member 2:
2
1´2´
3´
4´5´
6´
45 6
7*9*
8*
2
i
j
kN/m103324m3.61
)kN/m10)(200m10(60
3
2623
×=
××=
−
LAE
mkN1088.64m3.61
)m10)(400kN/m104(2004
3
4626
•×=
××=
−
LEI
kN1036.83m)(3.61
)m10)(400kN/m106(2006
3
2
4626
2
×=
××=
−
LEI
kN/m1020.41m)(3.61
)m10)(400kN/m1012(20012
3
3
4626
3
×=
××=
−
LEI
mkN1044.32m3.61
)m10)(400kN/m102(2002
3
4626
•×=
××=
−
LEI
[ k´ ]2 = 103
1´2´3´4´5´6´
1´ 4´ 5´ 6´3324
00
-332400
2´0
20.4136.83
0-20.4136.83
3´0
36.8388.64
0-36.8344.32
-332400
3324
00
0-20.41-36.83
020.41
-36.83
036.8344.32
0-36.8388.64
67
45 6
7*9*
8*
2
i
j 76.31oλjx = cos (-76.31o) = 0.24,λjy = sin (-76.31o) = -0.97
[ q* ] = [ T ]T[ q´ ]
q1´
q3´
q2´
q4´q5´q6´
q4
q6
q5
q7*q8*q9*
=
4567*8*9*
1´ 4´000
0.24
0-0.97
5´000
0.970.24
0
6´000001
0.55
0-0.83
000
2´0.830.55
0000
3´001000
λix = cos (-56.31o) = 0.55,λiy = sin (-56.31o) = -0.83
56.31o
2
i
j
1´2´
3´
4´5´
6´
[ k* ]2 = [ T ]T[ k´]2[ T ] = 103
4567*8*9*
4 7* 8* 9*5 6-452.884643.585-35.786205.452
-35.786-759.668
1787.474-2689.923
-8.717-759.6683139.053
-8.717
30.64620.43144.321
-35.786-8.71788.643
1036.923
30.646-1524.780
-452.8841787.474
30.646
-1524.7802307.582
20.431643.585
-2689.92320.431
30.64620.43188.643
-35.786-8.71744.321
68
[k]1 = 103
123456
1 4 5 60
30400
-3080
3000
00
-300000
20
1530
0-1530
30
30800
-3040
-300000
3000
00
0-15-300
15-30
1
2 3
4
5 6
7*9*
8*
1
2
[k*]2 = 103
4567*8*9*
4 7* 8* 9*5 6-452.884643.585-35.786205.452
-35.786-759.668
1787.474-2689.923
-8.717-759.6683139.053
-8.717
30.64620.43144.321
-35.786-8.71788.643
1036.923
30.646-1524.780
-452.8841787.474
30.646
-1524.7802307.582
20.431643.585
-2689.92320.431
30.64620.43188.643
-35.786-8.71744.321
69
Global:
= -50= 0= -80= 0= 0
Q4
Q6
Q5
Q7*
Q9*
D4
D6
D5
D7*
D9*
0
-26.6740
00
+
D4
D6
D5
D7*
D9*
-2.199x10-5 m
-2.840x10-4 rad-3.095x10-4 m
0.979x10-3 m6.161x10-4 rad
=
1
2 3
4
5 6
7*9*
8*
1
23 m
80 kN�m20 kN/m
20o
50 kN
4 m 2 m
A B
C 26.67 kN�m
26.67 kN�m
40 kN40 kN
1
20 kN/m
80 kN�m50 kN
= 103
45679
4 7* 9*5 6
205.452
-452.884643.585-35.786
-35.786
30.646-9.56944.321
-35.78688.643
-452.884
4036.923
30.646-1524.780
30.646
-1524.780232.582
-9.569643.58520.431
30.646-9.569
168.643-35.78644.321
70
q1
q3
q2
q4
q5q6
0
26.67
40
040
-26.67
+= 103
123
45
6
1 4 5 6
030
40
0-30
80
3000
0
0
-30000
0
2
015
30
0-15
30
3
030
80
0-30
40
-30000
0
3000
0
0
0-15
-30
015
-30
Member 1:
q1
q3
q2
q4
q5q6
65.97 kN
24.59 kN�m
36.12 kN
-65.97 kN
43.88 kN-40.11 kN�m
=
0
0
0
D4
D5
D6
40.11 kN�m24.59 kN�m
43.88 kN36.12 kN
1
20 kN/m
65.97 kN65.97 kN
1
2 3
4
5 61
71
q4
q6
q5
q7*
q8*q9*
Member 2:
q4
q6
q5
q7*
q8*q9*
15.97 kN
-39.89 kN�m
-43.88 kN
0 kN
46.69 kN0 kN�m
=
= 103
456
7*8*
9*
4 7* 8* 9*5 6-455.21651.27
-35.73
210.67
-35.73
-769.1
1769.34-2678.93
-8.84
-769.13128.82
-8.84
30.5720.26
44.32
-35.73-8.84
88.64
1036.923
30.57-1508.14
-455.211769.34
30.57
-1508.142296.15
20.26
651.27-2678.93
20.26
30.5720.26
88.64
-35.73-8.84
44.32
D4
D6
D5
D*7
0
D*9
15.97 kN43.88 kN 39.89 kN�m
46.69 kN
2
45 6
7*9*
8*
2
i
j
72
Bending-moment diagram (kN�m)
+
40.11 kN�m24.59 kN�m
43.88 kN36.12 kN
1
20 kN/m
65.97 kN65.97 kN
15.97 kN43.88 kN 39.89 kN�m
46.69 kN
2
-24.59-
-40.11
- +
39.89
D4
D6
D5
D7*D9*
-2.199x10-5 m
-2.840x10-4 rad-3.095x10-4 m
0.979x10-3 m6.161x10-4 rad
=
Deflected shape
D4=-2.2x10-5 m
D5=-3.1x10-4 m
D7*=0.979x10-3 mD6 = -2.84x10-4 rad
D9*=6.161x10-4 rad
1
2 3
4
5 6
7*9*
8*
1
2
![1 CST ELEMENT STIFFNESS MATRIX Strain energy –Element Stiffness Matrix: –Different from the truss and beam elements, transformation matrix [T] is not](https://img.dokumen.tips/doc/110x75/56649d6f5503460f94a518a4/1-cst-element-stiffness-matrix-strain-energy-element-stiffness-matrix-different.jpg)
