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Lecture No. : 3
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Remember :F1
1 2
k1 k2
F3
3
k3
F2
d1 d2 d3
k11F1
F2 = k21
F = K D
F3 k31
k12
k22
k32
k13
k23
k33
d1
d2
d3
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k11F1
F2 = k21
F3 k31
k12
k22
k32
k13
k23
k33
d1
d2d3
First column in Stiffness matrix
d1 =1 d2 =0 d3 =0
1 2
k1 k2
3
k3
d1=1
k11
k21
k31
=
F1
F2
F3
Remember :
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k11F1
F2 = k21
F3 k31
k12
k22
k32
k13
k23
k33
d1
d2
d3
Second column in Stiffness matrix
d1 =0 d2 =1 d3 =0
k12
k22
k32
=
F1
F2
F3
1 2
k1 k2
3
k3
d2=1
Remember :
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k11F1
F2 = k21
F3 k31
k12
k22
k32
k13
k23
k33
d1
d2
d3
Third column in Stiffness matrix
d1 =0 d2 =0 d3 =1
k13
k23
k33
=
F1
F2
F3
1 2
k1 k2
3
k3
d3=1
Remember :
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a
b
D
q
D1= D cos q
D1 D2
D2= D sin q
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D
P
P
AE
LPD=
AE
L
P D=
Relation between force andDisplacement in truss element
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Degrees of freedom (DOF)
d1
d2
d1
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AB
C 4 m
3
Example 1:
Construct the stiffness matrix for the shown
truss where EA is constant for all members
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Modeling
A
B
C
4
3F1d1
F2 d2
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F = K D
k11F1
F2=
k21
k12
k22
d1
d2
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First column in Stiffness matrix
d1 =1 d2 =0k11F1
F2
=k21
k12
k22
d1
d2
A
B
C
4
3F1d
1
F2
d2
k11F1
F2= k21
k12
k22
1
0
k11 F1
F2
=k21
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First column in Stiffness matrix
d1 =1 d2 =0k11F1
F2
=k21
k12
k22
d1
d2
A
B
C
4
3d1
1
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A
B
C
4
3
1A
1
q
A cos qEL
AE
LAE
L
A cos qEL
AB
C 4 m3
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A
B
C
4
3
1A
1
q
A x 0.8E
5
AE
4AE
4
A x 0.8E
5
AB
C 4 m3
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4
3q
0.16 EA
0.25 EA 0.25 EA
0.16 EA
AE
4AE
4
A x 0.8E
5
A x 0.8E5
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A
B
C
4
30.25 EA
0.16 EA
q
0.16 EA
0.25 EA 0.25 EA
0.16 EA
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A
B
C
4
30.25 EA
0.16 EA
q 0.16 EA cos q
0.16 EA sin q
cos q=0.8sin q=0.6
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A
B
C
4
30.25 EA
0.16 EA cos q
0.16 EA sin q
0.16 EA x 0.8
0.16 EA x 0.6
0.128 EA
0.096 EA
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A
B
C
4
30.25 EA
0.128 EA
0.096 EA
0.378 EA
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A
B
C
4
3
0.096 EA
0.378 EA
A
B
C
F1
F2
F1 = 0.378 EAF2 = 0.096 EA
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A
B
C
4
3
0.096 EA
0.378 EA
F1 = 0.378 EAF2 = 0.096 EA
k11 F1
F2
=k21
k11 =k21
0.378 EA
0.096 EA
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second column in Stiffness matrix
d1 =0 d2 =1k11F1
F2
=k21
k12
k22
d1
d2
A
B
C
4
3F1d
1
F2
d2
k11F1
F2
= k21
k12
k22
0
1
k12 F1
F2
=k22
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second column in Stiffness matrix
d1 =0 d2 =1k11F1
F2
=k21
k12
k22
d1
d2
A
B
C
4
3
d2
1
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A
B
C
4
3
1
A
1
a
A cos aEL
0
A cos aEL
0
cos a = 0.6L = 5
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A
B
C
4
3
1
A
1
a
A x 0.6E5
0
A x 0.6E
5
0
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4
3
0.12 EA
0 0
0.12 EA
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A
B
C
4
30
0.12 EA
q
0.12 EA
0 0
0.12 EA
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A
B
C
4
30
0.12 EA
q 0.12 EA cos q
0.12 EA sin q
cos q=0.8sin q=0.6
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A
B
C
4
30
0.12 EA cos q
0.12 EA sin q
0.12 EA x 0.8
0.12 EA x 0.6
0.096 EA
0.072 EA
cos q=0.8sin q=0.6
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A
B
C
4
3
0.096 EA
0.072 EA
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A
B
C
4
3
0.072 EA
A
B
C
F1
F2
F1 = 0.096 EAF2 = 0.072 EA
0.096 EA
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A
B
C
4
3 k12 F1F2=k22k12 =k22
0.096 EA
0.072 EA
F1 = 0.096 EAF2 = 0.072 EA
0.072 EA
0.096 EA
S
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Summary
k11F1
F2=
k21
F = K D
k12
k22
d1
d2
k11 =k21
0.378 EA
0.096 EA
k12 =k22
0.096 EA
0.072 EA
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k11 =k21
0.378 EA
0.096 EA
k12 =k22
0.096 EA
0.072 EA
K =
0.378 EA
0.096 EA
0.096 EA
0.072 EA
K =0.378
0.096
0.096
0.072
EA
E l 2
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AB
C 4 m
3
Example 2:
Calculate the joint displacements for the
shown truss due to the shown force whereEA = 105 kN for all members
12 kN
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k11F1
F2=
k21
F = K D
k12
k22
d1
d2
From the previous example
K =0.378
0.096
0.096
0.072
EA
F th i l
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From the previous example
K =
0.378
0.096
0.096
0.072
EA
EA = 105 kN
K = 0.378
0.096
0.096
0.072
105 = 378
96
96
72
100
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AB
C 4 m
312 kN
A
B
C
F1
F2
F1
F2=
0
-12
Force vector
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=378
96
96
72
100d1
d2
0
-12
d1
d2= 100
378
96
96
72
0
-12
1
-1
d1
d2 =105
4
-5.33
-5.33
21
0
-12
1
=
0.0006
-0.0025=
0. 6
-2.5m mm
E l 3
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AB
C 4 m
3
Example 3:
Calculate the joint displacements for the
shown truss due to the shown force whereE = 2000 kN/cm2, ABA= 100 cm
2, ACA= 200 cm2
12 kN
M d li
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Modeling
A
B
C
4
3F1d1
F2 d2
EABA= 2x105 kN
EACA= 4x105
kN
E = 2000 kN/cm2
, ABA= 100 cm2
, ACA= 200 cm2
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k11F1
F2=
k21
F = K D
k12
k22
d1
d2
First column in Stiffness matrix k k d
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First column in Stiffness matrix
d1 =1 d2 =0k11F1
F2
=k21
k12
k22
d1
d2
A
B
C
4
3F1d1
F2 d2
k11F1
F2
=k21
k12
k22
1
0
k11 F1
F2
=k21
First column in Stiffness matrix k k d
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First column in Stiffness matrix
d1 =1 d2 =0k11F1
F2
=k21
k12
k22
d1
d2
A
B
C
4
3d1
1
1
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A
B
C
4
3
1A
1
q
A cos qEL
AE
LAE
L
A cos qEL
EABA= 2x105 kN
EACA= 4x105 kN
LBA= 400 cmLCA= 500 cm
cos q=0.8
1
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A
B
C
4
3
1A
1
q
4x105x0.8
500
2x105
4002x105
400
4x105x0.8
500
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4
3q
640
500 500
640
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A
B
C
4
3500
640
q
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A
B
C
4
3500
640
q 640 cos q
640 sin q
cos q=0.8sin q=0.6
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A
B
C
4
3500
512
384
1012
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A
B
C
4
3
384
1012
A
B
C
F1
F2
F1 = 1012F2 = 384
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A
B
C
4
3
384
1012
F1 = 1012F2 = 384
k11 F1
F2
=k21
k11 =k21
1012
384
second column in Stiffness matrix k11F k12 d
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second column in Stiffness matrix
d1 =0 d2 =1k11F1
F2
=k21
k12
k22
d1
d2
A
B
C
4
3F1d1
F2 d2
k11F1
F2
=k21
k12
k22
0
1
k12 F1
F2
=k22
second column in Stiffness matrix k11F k12 d1
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second column in Stiffness matrix
d1 =0 d2 =1k11F1
F2
=k21
k12
k22
d1
d2
A
B
C
4
3
d2
1
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A
B
C
4
3
1
A
1
a
A cos aEL
0
A cos aEL
0
EABA= 2x105 kN
EACA= 4x105 kN
LBA= 400 cmLCA= 500 cm
cos a=0.6
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A
B
C
4
3
1
A
1
a
4x105x0.65
0 0
4x105x0.6
5
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4
3
480
0 0
480
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A
B
C
4
30
480
q
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A
B
C
4
30
480
q 480 cos q
480 sin q
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A
B
C
4
3
384
288
F 38
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A
B
C
4
3
288
A
B
C
F1
F2
F1 = 384F2 = 288
384
F 384
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A
B
C
4
3 k12 F1F2=k22k
12 =k22
384
288
F1 = 384F2 = 288
288
384
Summary
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K =1012
384
384
288
k11 =
k21
1012
384
k12 =k22
384
288
F1
F2=
0
-12
=d1
d2
0
-12
1012
384
384
288
Summary
d0 1012 384
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=d1
d2
0
-12
d1
d2 =103
2
-2.67
-2.67
7.03
0
-12
1
=
0.0320
-0.0843 =
0.320
-0.843cm mm
1012
384
384
288
d1
d2=
0
-12
-1
1012
384
384
288
Calculation of internal force in truss member
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Calculation of internal force in truss member
a
b
q
DbxDby
DaxDa
y
a
b
q
DbxDby
DaxDay
-
-
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a
b
q
Dbx
DbyDax
Day-
-D= ( )cos q + ( )sin q
bxD
ax- D
byD
ay-
AEL
N D=( )cos q+ ( )sin qbx Dax- Dby Day-
AE
L
N= [ ]
Example 4:
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AB
C 4 m
3
Example 4:
Calculate the internal forces for the shown
truss due to the shown force whereE = 2000 kN/cm2, ABA= 100 cm
2, ACA= 200 cm2
12 kN
EABA= 2x105 kN EACA= 4x10
5 kN
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EABA= 2x10 kN EACA= 4x10 kN
D= ( )cos q + ( )sin qbx Dax- Dby Day-
d1
d2=
0.0320
-0.0843 cm
D = ( )cos 0+ ( )sin 0Ax DBx- DAy DBy-BA = 0.032 cm
+ ( )sin 37= ( )cos 37Ax DCx- DAy DCy-CA = -0.025 cmNBA = 2 x 10
5 x 0.032 / 400 = 16 kN
AE
L
ND
=
NCA = 4 x 105 x -0.025 / 500 = - 20 kN
From example 3:
Tension
Compression
Check of resultsAB
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Check of results
C 4 m
312 kN
For equilibrium of
joint A :
FCA
FBA qA
12 kNS Fy = 0S Fx = 0FCA sin q - 12 = 0 FCA = 12 / sin q = 20 kN
FCA cos q - FBA = 0 FBA = 20 x cos q = 16 kN
For large trusses
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For large trusses
a
b
q
F1d1
F2 d2
F3d3
F4 d4
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a
b
q
F1d1=1
F2
F3
F4
d = cos q
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a
b
q
F1d1=1
F2
F3
F4
d = cos q
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a
b
q
F1d1=1
F2
F3
F4
EA/L cos q
EA/L cos q
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a
b
q
F1d1=1
F2
F3
F4
EA/L cos q
EA/L cos q
= EA/L cos2q
= EA/L cos qsin q
= - EA/L cos2q
= - EA/L cos qsin q
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a
b
q
F1d1=1
F2
F3
F4
= EA/L cos2q
= EA/L cos qsin q
= - EA/L cos2q
= - EA/L cos qsin q
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a
b
q
F1
F2
F3
F4
d = sin qd2=1
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a
b
q
F1
F2
F3
F4
d = sin qd2=1
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a
b
q
F1
F2
F3
F4
EA/L sin q
EA/L sin q
d2=1
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a
b
qF1
F2
F3
F4
EA/L sin q
EA/L sin q
= EA/L sin2q
= EA/L sin qcos q
= - EA/L sin qcos q
= - EA/L sin2
q
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a
b
qF1
F2
F3
F4
= EA/L sin2q
= EA/L sin qcos q
= - EA/L sin qcos q
= - EA/L sin2
q
d2=1
Example 5:
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A B
O
5
p
Calculate the internal forces for the shown
truss due to the shown forces whereEA = 4x105 kN for all members
200 kN
100 kN
C D E
150 13560 45
Modeling
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A B
O
5
C D E
150 13560 45
F1d1
d2F2k11F1
F2
=k21
k12
k22
d1
d2
Modeling
First column in Stiffness matrix
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a
b
q
F1d1=1
F2
F3
F4
= EA/L cos2q= EA/L cos qsin q
= - EA/L cos2
q
= - EA/L cos qsin q
k11=
k21
d1 =1 d2 =0
S EA/L cos2qS EA/L cos qsin q
Second column in Stiffness matrix
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a
b
qF1
F2
F3
F4
= EA/L sin2q
= EA/L sin qcos q
= - EA/L sin qcos q
= - EA/L sin2q
d2=1
d1 =0 d2 =1
k12=
k22 S EA/L sin2qS EA/L cos qsin q
A B C D E
150135
60 45
L= 500 / sin
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O
5 m150
135
12719851039S
282.9282.9282.9.707.70756670745OE
300519.6173.2.866.569357760OD
080001080050090OC
-282.9282.9282.9.707-.707566707135OB
-173100300.5-.8664001000150OA
member L sin qcos qEAL
EA/Lcos2q
EA/Lsin2q
EA/Lsinqcosq
L= 500 / sin qEA = 4x105 kN
EA/L2
EA/L EA/Lsin
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12719851039S
cos2q sin2q sinqcosq
k11=
k21
S EA/L cos2qS EA/L cos qsin q
k12=
k22 S EA/L sin2qS EA/L cos qsin q
k11
K= k21
k12
k22
1039
= 127
127
1985
Force vector
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A B
O
5200 kN
100 kN
C D E
150 13560 45
F1
F2=
100
-200=F
F = K D
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F = K Dk11F1
F2=
k21
k12
k22
d1
d2
100
-200
=
1039
127
127
1985
d1
d2
d1
d2
1039
127
127
1985=
-1100
-200
=
0. 109
-0. 108
Internal forces in truss elements
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( )cos q+ ( )sin qbx Dax- Dby Day-AE
LN=
a
b
q
Dbx
Dby
DaxDay
[ ]
AFor member OA
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A
O
150= 400
EA
L
= 0.5sin q= -.866cos q= 150oq
( )cos q+ ( )sin qAx Dox- DAy Doy-AE
L
NoA= [ ]
Dox = 0.109Doy= -0.108
- 0.109 cos 150+ 0.108sin 150400NoA= [ ]59.4 kNNoA= Tension
150
330
Another way for member AO
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A
O
330
= 400EA
L
= -0.5sin q=.866cos q= 330oq
( )cos q+ ( )sin qox DAx- Doy DAy-AE
L
NAo= [ ]
Dox = 0.109Doy= -0.108
0.109 cos 330 - 0.108sin 330400NAo= [ ]59.4 kNNAo= Tension
A B C D E
150135
60 45
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O
5 m150
-0.4.707.70756645OE
27.8660.569360OD
86.41080090OC
86.8.707-.707566135OB
59.40.5-.866400150OA
member sin qcos qEAL
N
- 0.019 cos q + 0.0108sin qAEL
Noi = [ ]
Example 6:
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C
B
D
4 m
4
Calculate the joint displacements for theshown truss due to the shown forces where
the axial stiffness = 400 kN/cm for all members
100 kN
A
50 kN
Modeling
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d1d2
C
B
D
A
d3d4
d5
D C
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-0.50.50.5.707-.707400566135BD
0.50.50.5.707.70740056645AC
010-10400400270DA
0010-1400400180CD
0101040040090BC
001014004000AB
member L sin qcos qEAL
cos2q sin2q sinqcosq
BA
F = K D
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k11F1
F2
=
k21
F3 k31
k12
k22
k32
k13
k23
k33
F4
F5
k41
k51
k42
k52
k43
k53
k14
k24
k34
k15
k25
k35
k44
k54
k45
k55
d1
d2
d3
d4
d5
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a
b
q
F1
d1=1
F2
F3
F4
= EA/L cos2q= EA/L cos qsin q
= - EA/L cos2q
= - EA/L cos qsin q
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a
b
q F1
F2
F3
F4
= EA/L sin2q
= EA/L sin qcos q
= - EA/L sin qcos q
= - EA/L sin2q
d2=1
d1d2
d3d4First column in Stiffness matrix
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C
B
D
A d5
-0.50.50.5BD
0.50.50.5AC
010DA
001CD010BC
001AB
member cos2q sin2q sinqcosq
d1 =1k11
k21
k31
k41
k51
[S cos2q]DA,DB,DC
=EA
L
[S cosq sinq]DA,DB,DC[-cos2q]DC[-cosq sinq]DC[-cos2q]DB
=
1.5
-.5
-1
0
-.5
EA
L
d1d2
d3d4Second column in Stiffness matrix
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C
B
D
A d5
-0.50.50.5BD
0.50.50.5AC
010DA
001CD010BC
001AB
member cos2q sin2q sinqcosq
d2 =1k12
k22
k32
k42
k52
[S sin2q]DA,DB,DC=
EA
L
[S cosq sinq]DA,DB,DC
[-sin2q]DC
[-cosq sinq]DC =
-.5
1.5
0
0
0.5
EA
L
[-cosq sinq]DB
d1d2
d3d4Third column in Stiffness matrix
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C
B
D
A d5
-0.50.50.5BD
0.50.50.5AC
010DA
001CD010BC
001AB
member cos2q sin2q sinqcosq
d3 =1k13
k23
k33
k43
k53
[S cos2q]CA,CB,CD=EA
L
[S cosq sinq]CA,CB,CD
[-cos2q]DC[-cosq sinq]DC
[-cos2q]CB
=
-1
0
1.5
0.5
0
EA
L
d1d2
d3d4Fourth column in Stiffness matrix
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C
B
D
A d5
-0.50.50.5BD
0.50.50.5AC
010DA
001CD010BC
001AB
member cos2q sin2q sinqcosq
d4 =1k14
k24
k34
k44
k54
[S sin2q]CA,CB,CD=
EA
L[S cosq sinq]CA,CB,CD[-sin2q]CD
[-cosq sinq]CD
=
0
0
0.5
1.5
0
EA
L
[-cosq sinq]CB
d1d2
d3d4Fifth column in Stiffness matrix
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C
B
D
A d5
-0.50.50.5BD
0.50.50.5AC
010DA
001CD010BC
001AB
member cos2q sin2q sinqcosq
d5 =1k15
k25
k35
k45
k55 [S cos2q]BA,BC,BD
=EA
L
[-cos2q]BD[-cosq sinq]BD
=
-.5
0.5
0
0
1.5
EA
L[-cos2q]BC[-cosq sinq]BC
k11 1.5 k12 -.5 k13 -1 k14 0 k15 -.5
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k21
k31
k41
k51
=
-.5
-1
0
-.5
EA
L
k22
k32
k42
k52
=
1.5
0
0
0.5
EA
L
k23
k33
k43
k53
=
0
1.5
0.5
0
EA
L
k24
k34
k44
k54
=
0
0.5
1.5
0
EA
L
k25
k35
k45
k55
=
0.5
0
0
1.5
EA
L
k11 1.5 k12 -.5 k13 -1 k14 0 k15 -.5
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k21
k31
k41
k51
=
-.5
-1
0
-.5
EA
L
k22
k32
k42
k52
=
1.5
0
0
0.5
EA
L
k23
k33
k43
k53
=
0
1.5
0.5
0
EA
L
k24
k34
k44
k54
=
0
0.5
1.5
0
EA
L
k25
k35
k45
k55
=
0.5
0
0
1.5
EA
L
1.5
=
-.5
K -1
-.5
1.5
0
-1
0
1.5
0
-.5
0
0.5
0.5
0
0
0
0.5
-.5
0.5
0
1.5
0
0
1.5
EA
L
d1d2
d3d4Force vector
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C
B
D
A d5
C
B
D
100 kN
A
50 kN
F1
F2
=F3
F4
F5
50
-100
0
0
0
F = K D
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=
d1
d2
d3
d4
d5
50
-100
0
0
0
1.5
-.5
-1
-.5
1.5
0
-1
0
1.5
0
-.5
0
0.5
0.5
0
0
0
0.5
-.5
0.5
0
1.5
0
0
1.5
EA
L
400
D = K-1 F
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=
d1
d2
d3
d4
d5
50
-100
0
0
0
1.5
-.5
-1
-.5
1.5
0
-1
0
1.5
0
-.5
0
0.5
0.5
0
0
0
0.5
-.5
0.5
0
1.5
0
0
1.5
-1
1
400
D = K-1 F
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=
d1
d2
d3
d4
d5
50
-100
0
0
0
2
.5
1.5
.5
.875
.375
1.5
.375
1.875
-.5
.5
-.125
-.125
-.625
.375
-.5
-.125
-.625
.5
-.125
.375
.875
-.125
-.125
.875
1
400
D = K-1 F
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=
d1
d2
d3
d4
d5
0.1250
-0.15625
-0.09375
-0.03125
-0.09375
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Summary
ModelingF d
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A
B
C 4
3F1d1
F2 d2
F = K Dk11F1
F2=
k21
k12
k22
d1
d2
First column in Stiffness matrixk11F1
F2
=k21
k12
k22
d1
d2
Summary
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d1 =1 d2 =0
A
B
C
4
3d1
1
second column in Stiffness matrix
k11F1
F2
=k21
k12
k22
d1
d2
Summary
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second column in Stiffness matrix
d1 =0
d2 =1
F2 k21 k22 2
A
B
C
d2
1
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a
b
q
F1d
1=1
F2
F3
F4
= EA/L cos2q= EA/L cos qsin q
= - EA/L cos2q
= - EA/L cos qsin q
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a
b
q F1
F2
F3
F4
= EA/L sin2q
= EA/L sinqcos
q
= - EA/L sin qcos q
= - EA/L sin2q
d2=1
For internal forces in truss elementsSummary
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( )cos q+ ( )sin qbx Dax- Dby Day-AE
L
N=
a
b
q
Dbx
Dby
Dax
Day
[ ]
For internal forces in truss elements
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Questions