Download - 05 Matrix Truss
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! Fundamentals of the Stiffness Method! Member Local Stiffness Matrix! Displacement and Force Transformation Matrices! Member Global Stiffness Matrix! Application of the Stiffness Method for Truss
Analysis! Trusses Having Inclined Supports, Thermal Changes
and Fabrication Errors! Space-Truss Analysis
TRUSSES ANALYSIS
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2-Dimension Trusses
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Fundamentals of the Stiffness Method
� Node and Member Identification
� Global and Member Coordinates
� Degrees of Freedom
12
3 4
2
1
3
(x1, y1)
(x3, y3)
(x2, y2)
(x4, y4)
1
23
4
56
78
34
56
78
x
y
�Known degrees of freedom D3, D4, D5, D6, D7 and D8� Unknown degrees of freedom D1 and D2
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AE/LAE/L
x djAE/L x d´j
AE/L
x diAE/LAE/L x d´i
Member Local Stiffness Matrix
x´y´
i
j
q´i
q´j
AE/L
AE/L
jii dL
AEdL
AEq ''' −=
AE/LAE/Ld´ i = 1
d´ j = 1
x di
x dj
jij dL
AEdL
AEq ''' +−=
−
−=
j
i
j
i
dd
LAE
''
1111
''
−
−=
1111
]'[L
AEk
[q´] = [k´][d´] ----------(1)
q´j
q´i
x´y´
x´y´
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x´y´
m
i
j
(xi,yi)
(xj,yj)x
y
Displacement and Force Transformation Matrices
θy
θx
22 )()(cos
ijij
ijijxx
yyxx
xxL
xx
−+−
−=
−== θλ
22 )()(cos
ijij
ijijyy
yyxx
yyL
yy
−+−
−=
−== θλ
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x
y
Global
m
i
jdjx
djy
dix
diy
djx
djyx´y´
m
i
j
Local
d´i
d´j
dix
diy
d´j
d´i
� Displacement Transformation Matrices
yiyxixi ddd θθ coscos' +=
=
jy
jx
iy
ix
yx
yx
j
i
dddd
dd
λλλλ00
00''
θy
θx
yjyxjxj ddd θθ coscos' +=
=
yx
yxTλλ
λλ00
00][
λx λy
[d´] = [T][d] ----------(2)
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x
y
θy
θx
m
i
j
x
y
Global
x´y´
m
i
j
Local
q´i
q´j
� Force Transformation Matrices
θy
θx
xiix qq θcos'=
yiiy qq θcos'=
xjjx qq θcos'=
yjjy qq θcos'=
=
j
i
y
x
y
x
jy
jx
iy
ix
qqqq
''
00
00
λλ
λλ
=
y
x
y
x
TT
where
λλ
λλ
00
00
][
λx
λy
[q] = [T]T[q´] ----------(3)
qjx
qjy
qix
qiy
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Member Global Stiffness Matrix
[ q ] = [ T ]T ([k´][d´] + [q´F] ) = [ T ]T [ k´ ][T][d] + [ T ]T [q´F] = [k][d] + [qF]
[ k ] [ k ] = [ T ]T[ k´ ][T]
[qF] = [ T ]T [q´F]
[q] = [T]T[q´] ----------(3)
Substitute ( [q´] = [k´][d´] + [q´F] ) into Eq. 3, yields the result,
λyλx
λxλx
−λyλx
−λxλx
λyλy
λxλy
−λyλy
−λxλy
λyλx
λxλx
−λyλx
−λxλx
λyλy
λxλy
−λyλy
−λxλy
V U VU
[ k ] = AEL
VUV
U
00λyλx
λyλx00-11
1-1
AEL[ k ] =
00
λy
λx
λy
λx
00
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[Qa] = [K][D] + [QF]
[Qk] = [K11][Du] + [K12][Dk] + [QF]
Reaction Boundary Condition
Unknown DisplacementJoint Load
Equilibrium Equation:
Partitioned Form:
Application of the Stiffness Method for Truss Analysis
[Du] = (([Qk] - [QF]) - [K12][Dk])[Ku] -1
Qk
Qu
Du
Dk
=K12
K22
K11
K21
+QF
k
QFu
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+
−
−=
jF
iF
j
i
j
i
dd
LAE
''
''
1111
''
jy
jx
iy
ix
yx
yx
dddd
λλλλ00
00
+
−
−=
jF
iF
jy
jx
iy
ix
yx
yx
j
i
DDDD
LAE
''
0000
1111
''
λλλλ
Member Forces
x
y
θy
θx
x´y´
m
i
j
q´i
q´j
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+
−−
−−=
jF
iF
jy
jx
iy
ix
yxyx
yxyx
j
i
DDDD
LAE
''
''
λλλλλλλλ
Dyi
Dxi
Dyj
Dxj
q´j = AEL −λx −λy λx λy qj´
F+
x
y
θy
θx
x´y´
m
i
j
Member Forces
q´i
q´j
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Member Forces
Dyi
Dxi
Dyj
Dxj
qm = AEL −λx −λy λx λy qj´F+
x
y
θy
θx
x´y´
m
i
j
Member Forces
qm
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3 m
3 m
4 m 4 m
80 kN
50 kN
5 m
5 m5 m
Example 1
For the truss shown, use the stiffness method to:(a) Determine the deflections of the loaded joint.(b) Determine the end forces of each member and reactions at supports.Assume EA to be the same for each member.
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Ljyy
Lixx ijij
ij
�)(�)(� −+
−=λ
cosθx = λx cosθy = λy
1
2
3
434
56
78
1
2
(0,0)(-4,-3)
(-4,3)
(4,-3)
31
23 m
3 m
4 m 4 m
80 kN
50 kN
5 m
5 m5 m
Member
#1
#2
λx λy
#3
λyλx
λxλx
λyλy
λxλy
λyλx
λxλx
λyλy
λxλy
−λyλx
−λxλx
−λyλy
−λxλy
−λyλx
−λxλx
−λyλy
−λxλy
Vi Uj VjUi
[ k ]m = AEL
Vi
Uj
Vj
Ui
31
2
-4/5 = -0.8
-4/5 = -0.8
4/5 = 0.8
-3/5 = -0.6
3/5 = 0.6
-3/5 = -0.6
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-0.48
1.08-0.48
1.92-0.48
0.64
0.36
-0.481
2
1 2
[K] = AE5
1
2
3
434
56
78
1
2
31
2
0.48
0.64
0.36
0.48-0.48
-0.64
-0.36
-0.48
-0.48
-0.64
-0.36
-0.48[ k ]1 =
2 3 41
AE5
23
4
1
[ k ]2 =
2 5 61
AE5
25
6
1
-0.48
0.64
0.36
-0.480.48
-0.64
-0.36
0.48
0.48
-0.64
-0.36
0.48
[ k ]3 =
2 7 81
AE5
27
8
1
-0.48
0.64
0.36
-0.480.48
-0.64
-0.36
0.48
0.48
-0.64
-0.36
0.48
31
2
34
56
78
Member
#1
#2
#3
λx
-0.8
- 0.8
0.8
λy
-0.6
0.6
-0.6
λx2 λx λy λy
2
0.64 0.48 0.36
0.64 -0.48 0.36
0.64 -0.48 0.36
0.48
0.64
0.36
0.48
-0.48
0.64
0.36
-0.48
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34
56
78
1
2
31
2
Global
Q1 = -50
Q2 = -80
D1
D2
D1
D2=
-250.65/AE
-481.77/AE
3 m
3 m
4 m 4 m
80 kN
50 kN
5 m
5 m5 m
1
2
1
-0.48
2
-0.48
1.08
1.92
= AE5
80 kN
50 kN
0
0+1
2
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1
2
3
434
56
78
1
2
31
2
Local
= -97.9 kN (C)
[q´F]1 = AE5 0.8 0.6 -0.8 -0.6 D2=
D1=
D4=D3=
-481.77/AE-250.65/AE
0.00.0
= +17.7 kN (T)
D2=D1=
D6=D5=
-481.77/AE-250.65/AE
0.00.0
= -17.7 kN (C)
80 kN
50 kN 36.87o
97.9 kN
17.7 kN
17.7 kN
#1 -0.8 -0.6#2 -0.8 0.6#3 0.8 -0.6
[q´F]2 = AE5 0.8 -0.6 -0.8 0.6
[q´F]3 = AE5 -0.8 +0.6 +0.8 -0.6 D2=
D1=
D8=D7=
-481.77/AE-250.65/AE
0.00.0
Member
#1#2#3
λx λy
[ ] [ ]F
yj
xj
yi
xi
xxyxmF q
DDDD
LAEq ']'[ +
−−= λλλλ
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80 kN
50 kN 36.89o
97.9 kN
17.7 kN
17.7 kN
80 kN
50 kN
Member
#1#2#3
λx
-0.8-0.80.8
λy
-0.60.6-0.6
1
2
3
434
56
78
1
2
31
2
97.9(0.8)=78.32 kN
97.9(0.6)=58.74 kN
17.7(0.8)=14.16 kN
17.7(0.6)=10.62 kN
17.7(0.8)=14.16 kN
17.7(0.6)=10.62 kN
ΣFx ´ = 0:+ 17.7 + 17.7 +50cos 36.89 - 97.9cos73.78 - 80cos53.11 = 0, O.K
Check :
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Example 2
For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the deflections of the loaded joint.The support B settles downward 2.5 mm. Temperature in member BDincrease 20 oC. Take α = 12x10-6 /oC, AE = 8(103) kN.
A
BC
D
8 kN
4 kN
4 m
3 m+20o C
∆B = 2.5 mm
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λyλx
λxλx
λyλy
λxλy
λyλx
λxλx
λyλy
λxλy
−λyλx
−λxλx
−λyλy
−λxλy
−λyλx
−λxλx
−λyλy
−λxλy
∆B = 2.5 mm
A
BC
D
8 kN
4 kN
4 m
3 m+20o C
Vi Uj VjUi
[ k ]m = AEL
Vi
Uj
Vj
Ui
12
3 4
(0,0)
(-4,-3)
(-4,0)
(0,-3)
2
1
3
1
23
4
56
78
Member
#1
#2
λx λy
#3
-4/4 = -1
-4/5 = -0.8
0
0
-3/5 = -0.6
-3/3 = -1
Ljyy
Lixx ijij
ij
�)(�)(� −+
−=λ
cosθx = λx cosθy = λy
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0.096
0.128
0.072
0.096
0
0
0.333
0
0
0.25
0
0
0
0.25
0
00
-0.25
0
0
0
-0.25
0
0[k]1 = 8x103
2 3 41
234
1
[k]3 = 8x103
2 7 81
278
1
0
0
0.333
00
0
-0.333
0
0
0
-0.333
0
Member
#1
#2
#3
λx
-1
- 0.8
0
λy
0
-0.6
-1
λx2/L λx λy/L λy
2/L
0.25 0 0
0.128 0.096 0.072
0 0 0.333
0.096
0.128
0.072
0.096
-0.096
-0.128
-0.072
-0.096-0.096
-0.128
-0.072
-0.096
[k]2 = 8x103
2 5 61
256
1
12
3 4
(0,0)
(-4,-3)
(-4,0)
(0,-3)
2
1
3
1
23
4
56
78
1
23
4
56
78
[K] = 8x103
21
21
0.096
0.378
0.405
0.096
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∆B = 2.5 mm
1.536 kN
1.152 kN
2+20oC1.536 kN
1.152 kN
2+20oC
1.92 kΝ =α(∆T1)AE = (12x10-6)(20)(8x103)
1.92 kN
1.536 kN
1.152 kN
1.536 kN
1.152 kN
12
34
(0,0)
(-4,-3)
(-4,0)
(0,-3)
2
1
3
1
23
4
56
78
∆B = 2.5 mm
+20o C
-1.536
1.536-1.152
1.152
+
1
2
6
5
q1
q2
q1
q5
q2
q6
-1.536
-1.152+1
2
0
-2.5x10-3
5
6-0.096
-0.128
-0.072
-0.096+ 8x103
5 6
= 8x103
21
21
0.096
0.128
0.072
0.096 d1
d2
d1
d5 = 0d2
d6 = -2.5x10-3
0.096
0.128
0.072
0.096
= 8x103
2 5 61
256
1
0.096
0.128
0.072
0.096
-0.096
-0.128
-0.072
-0.096-0.096
-0.128
-0.072
-0.096
[q] = [k]m[d] + [qF]Member 2:
q1
q2
-1.536
-1.152+1
2= 8x103
21
0.096
0.128
0.072
0.096 d1
d2
1.92
1.44+
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[Q] = [K][D] + [QF]
Global:
A
BC
D
8 kN
4 kN
4 m
3 m+20o C
∆B = 2.5 mm
12
3 4
(0,0)
(-4,-3)
(-4,0)
(0,-3)
2
1
3
1
23
4
56
78
Q1 = -4
Q2 = -8= 8x103
21
21
0.096
0.378
0.405
0.096 D1
D2
-1.536
-1.152+1.92
1.44+
D1
D2
-0.8514x10-3 m
-2.356x10-3 m=
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Member
#1
#2
#3
λx λy
Local12
3 4
2
1
3
1
23
4
56
78
[q´F]1 = 8x103 1.0 0.0 -1.0 0.04
= -1.70 kN (C)
D2=D1=
D4=D3=
0.00.0
-0.8514x10-3
-2.356x10-3
-1.92+[q´F]2 = 8x103 0.8 0.6 -0.8 -0.65
= -2.87 kN (C)
D2=D1=
D6=D5=
-0.00250.0
-0.8514x10-3
-2.356x10-3
= -6.28 kN (C)
D2=D1=
D8=D7=
0.00.0
-0.8514x10-3
-2.356x10-3[q´F]3 = 8x103 0.0 1.0 0.0 -1.0
3
2+20oC
1.92 kN
1.92 kN
-1 0
- 0.8 -0.6
0 -1
[ ] [ ]F
yj
xj
yi
xi
xxyxmF q
DDDD
LAEq ']'[ +
−−= λλλλ
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4 kN
8 kN
6.28 kN
1.70 kN
2.87 kN
12
3 4
2
1
3
1
23
4
56
78
Member
#1
#2
#3
cosθx
-1
- 0.8
0
cosθy
0
-0.6
-1
[q´]m
-1.70
-2.87
-6.28
4 kN
8 kN
2
1
3
1.70 kN
6.28 kN
2.87(0.8) = 2.30 kN
2.87(0.6) = 1.72 kN
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∆ AD = + 3 mm
AB
C
3 m
D
8 kN
4 kN
4 m 4 m∆ = - 4 m
m
Example 3
For the truss shown, use the stiffness method to:(a) Determine the end forces of each member and reactions at supports.(b) Determine the displacement of the loaded joint.Take AE = 8(103) kN.
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∆ AD = + 3 mm
AB
C
D
8 kN
4 kN
3 m
4 m 4 m
∆ = - 4 mm
21
4
3
5
1
2
34
56
78
1
2 3 4
(0,0)
(-4,-3) (4,-3)(0,-3)
λyλx
λxλx
λyλy
λxλy
λyλx
λxλx
λyλy
λxλy
−λyλx
−λxλx
−λyλy
−λxλy
−λyλx
−λxλx
−λyλy
−λxλy
Vi Uj VjUi
[ k ]m = AEL
Vi
Uj
Vj
Ui
Member
#1
#2
λx λy
#3
#4
#5
-4/5 =-0.8
0
4/5 = 0.8
4/4 = 1
4/4 = 1
-3/5 = -0.6
-3/3 = -1
-3/5 = -0.6
0
0
Ljyy
Lixx ijij
ij
�)(�)(� −+
−=λ
cosθx = λx cosθy = λy
![Page 28: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/28.jpg)
28
[k]1 = 8x103
2 3 41
234
1
0.096
0.128
0.072
0.096
-0.096
-0.128
-0.072
-0.096-0.096
-0.128
-0.072
-0.096
0.096
0.128
0.072
0.096
[k]2 = 8x103
2 5 61
256
1
0
0
0.333
00
0
-0.333
0
0
0
0.333
0
0
0
-0.333
0
[k]3 = 8x103
2 7 81
278
1
-0.096
0.128
0.072
-0.0960.096
-0.128
-0.072
0.096
0.096
-0.128
-0.072
0.096-0.096
0.128
0.072
-0.096
Member
#1
#2
λy
#3
λx
-0.8
0
0.8
-0.6
-1
-0.6
λx2/L λy
2/Lλxλy/L
0.128 0.0720.096
0 0.3330
0.128 0.072-0.096
21
4
3
5
1
2
34
56
78
1
2 3 4
(0,0)
(-4,-3) (4,-3)(0,-3)
5 m
3 m 5 m
4 m 4 m
![Page 29: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/29.jpg)
29
21
4
3
5
1
2
34
56
78
1
2 3 4
(0,0)
(-4,-3) (4,-3)(0,-3)
5 m
3 m 5 m
4 m 4 m
Member
#4
#5
λyλx
1
1
0
0
λx2/L λy
2/Lλxλy/L
0
0.25
0
00
-0.25
0
0
0
0.25
0
0
0
-0.25
0
0
0
0.25
0
00
-0.25
0
0
0
0.25
0
0
0
-0.25
0
0[k]5= 8x103
6 7 85
678
5
0.25 00
0.25 00
Global Stiffness Matrix
[K] = 8x103
2 5 71
257
1
34 6
8
[k]4= 8x103
4 5 63
456
3
![Page 30: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/30.jpg)
30
Global Stiffness Matrix
0.2560.0 0.477
0.0 0.00.0
0.0 0.0-0.128 0.096
-0.1280.096
0.50-0.25
-0.250.378
[K] = 8x103
2 5 71
257
1
[k]1 = 8x103
2 3 41
234
1
0.096
0.128
0.072
0.096
-0.096
-0.128
-0.072
-0.096-0.096
-0.128
-0.072
-0.096
0.096
0.128
0.072
0.096
[k]2 = 8x103
2 5 61
256
1
0
0
0.333
00
0
-0.333
0
0
0
0.333
0
0
0
-0.333
0
[k]3 = 8x103
2 7 81
278
1
-0.096
0.128
0.072
-0.0960.096
-0.128
-0.072
0.096
0.096
-0.128
-0.072
0.096-0.096
0.128
0.072
-0.096
0
0.250
-0.25
0
00
0
0
0.25
0
-0.250
0
0
0[k]4= 8x103
4 5 63
456
3
0
0.250
-0.25
0
00
0
0
0.25
0
-0.250
0
0
0[k]5= 8x103
6 7 85
678
5
![Page 31: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/31.jpg)
31
∆AE/L = 10.67 kN
∆ AD = + 3 mm1
3.84 kN
2.88 kN
3.84 kN
2.88 kN
∆ AD = + 3 mm
AB
C
D
8 kN
4 kN
3 m
4 m 4 m
∆ = - 4 mm
1
2
21
4
3
534
56
78
Global Fixed end forces
0.00.0
257
1∆AE/L = 4.8 kN
4.8 kN
∆AE/L = 10.67 kN
10.67 kN∆ = -4 m
m
2 Fixed End3.84 kN
2.88 kN-3.84-2.88 + 10.67 = 7.79
![Page 32: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/32.jpg)
32
∆ AD = + 3 mm
AB
C
D
8 kN
4 kN
3 m
4 m 4 m
∆ = -4 mm
Global:
1
2
21
4
3
534
56
78
[Q] = [K][D] + [QF]
Q1 = 4
Q5 = 0Q2 = -8
Q7 = 0
= 8x103
2 5 71
257
1
-0.250.500.00.0
0.378-0.250.096
-0.128
-0.1280.00.0
0.256
0.0960.0
0.4770.0 D1
D5
D2
D7
-3.84
0.07.79
0.0
+
D1
D5
D2
D7
6.4426x10-3 m
2.6144x10-3 m-5.1902x10-3 m
5.2288x10-3 m
=
8 kN
4 kN
![Page 33: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/33.jpg)
33
D1
D5
D2
D7
6.4426x10-3 m
2.6144x10-3 m-5.1902x10-3 m
5.2288x10-3 m
=
-4.8+D2
D1
00
= -1.54 kN (C)
1
2
21
4
3
534
56
78
Member forces
= -3.17 kN (C)
10.67+
[q´F]1 = 8x103 0.8 0.6 -0.8 -0.65
[q´F]2 = 8x103 0.0 1.0 0.0 -1.03
D2
D1
0D5
Member
#1
#2
λiyλix
-0.8 -0.6
0 -1
1
4.8 kN
4.8 kN∆ AD
= + 3 mm
10.67 kN
10.67 kN
2
[ ] [ ]F
yj
xj
yi
xi
xxyxmF q
DDDD
LAEq ']'[ +
−−= λλλλ
![Page 34: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/34.jpg)
34
D1
D5
D2
D7
6.4426x10-3 m
2.6144x10-3 m-5.1902x10-3 m
5.2288x10-3 m
=
1
2
21
4
3
534
56
78
00
0D5
[q´F]4 = 8x103 -1.0 0.0 1.0 0.04
= 5.23 kN (T)
0D5
0D7
[q´F]5 = 8x103 -1.0 0.0 1.0 0.04
= 5.23 kN (T)
D2
D1
0D7
[q´F]3 = 8x103 -0.8 0.6 0.8 -0.65
= -6.54 kN (C)
Member λx λy
#3#4
#5
0.8 -0.61 0
1 0
[ ] [ ]F
yj
xj
yi
xi
xxyxmF q
DDDD
LAEq ']'[ +
−−= λλλλ
![Page 35: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/35.jpg)
35
1
2
21
4
3
534
56
78
-0.8
0
0.81
1
-0.6
-1
-0.60
0
Member
#1
#2
λx
#3#4
#5
λy [q´]
4 kN8 kN
3.17 kN
3.17 kN
6.54 kN
6.54 kN
1.54 kN
1.54 kN
5.23 kN5.23 kN
4 kN8 kN
0.92 kN
4 kN
3.17 kN 3.92 kN
-1.54
-3.17
-6.545.23
5.23
![Page 36: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/36.jpg)
36
Special Trusses (Inclined roller supports)
![Page 37: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/37.jpg)
37
λix = cos θi
λiy = sin θi
λjx = cos θj
λjy = sin θj
[ q* ] = [ T ]T[ q´ ] 1
2
3*
4*
56
78
3
2
4
5
1
x *
y *
θi
x
y
θj
1
i
j
q´i
q´j
q*3
q1
q*4
q2
q´iq´j
[T]T
00λiyλix
λjyλjx00 [ T ] = [[ T ]T]T =
=00
λiy
λix
λjy
λjx
00
1
i
j1
2
3*
4*
Transformation Matrices
![Page 38: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/38.jpg)
38
[ k ] = [ T ]T[ k´ ][T]
λjyλjx
λjxλjx
−λiyλjx
−λixλjx
λjyλjy
λjxλjy
−λiyλjy
−λixλjy
λiyλix
λixλix
−λjyλix
−λjxλix
λiyλiy
λixλiy
−λjyλiy
−λjxλiy
Vi Uj VjUi
[ k ]m = AEL
Vi
Uj
Vj
Ui
00λiyλix
λjyλjx00-11
1-1
AEL[ k ]m =
00
λiy
λix
λjy
λjx
00
![Page 39: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/39.jpg)
39
Example 5
For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the displacement of the loaded joint. AE is constant.
3 m
4 m
45o
30 kN
![Page 40: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/40.jpg)
40
3 m
4 m45o
2
1
3
1
2
5
63*
4*
Member 1:
15
63*
4*
[q*]θi = 0,λix = cos 0 = 1,λiy = sin 0 = 0
θij = -45o = 135o,λix = cos (-45o) = 0.707,λiy = sin(- 45o) = -0.707
45o
[q*] = [T*]T[q´] + [T*]T[q´F]
q5
q3*
q6
q4*
q´iq´j
[T*]T
=0001
-0.7070.707
00
q´i q´j
i j1
![Page 41: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/41.jpg)
41
15
63*
4*
[q*]θi = 0 ;λix = cos 0 = 1,λiy = sin 0 = 0
θij = -45o = 135o,λix = cos (-45o) = 0.707,λiy = sin(- 45o) = -0.707
45o
0-0.707
00.707
00
10
AE4
1-1
-11[k*]1 =
0001
-0.7070.707
00
Member 1:
[k*]1 = AE 63*4*
53*
-0.1250.125
0-0.1768
4*
0.125-0.125
00.1768
5
0.1768-0.1768
00.25
6
0000
q´i q´j
i j1
3 m
4 m45o
2
1
3
1
2
5
63*
4*
[ ]TL
AETk T
−
−=
1111
][*][
![Page 42: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/42.jpg)
42
Member 2:q´i
q´j
i
j
22
1
2
3*4*
θi = -90o = 270o,λix = cos(-90o) = 0,λiy = sin(-90o) = -1
90o+45o
=135o
90o
θj = -135o = 215o ,λix = cos (-135o) = -0.707,λiy = sin(- 135o) = -0.707
0-0.707
0-0.707
-10
00
AE3
1-1
-11[k*]2 =
00-10
-0.707-0.707
00
[k*]2 = AE
2 4*1
23*4*
1
0.16670.1667-0.2357
0
0.16670.1667-0.2357
0
0000
-0.2357-0.23570.3333
03*
3 m
4 m45o
2
1
3
1
2
5
63*
4*
[ ]TL
AETk T
−
−=
1111
][*][
![Page 43: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/43.jpg)
43
[ ]TL
AETk T
−
−=
1111
][][
36.87o
Member 3:
θi = θj = 36.87o ;λix = λjx = cos (36.87o) = 0.8,λiy = λjy = sin(36.87o) = 0.6
00.6
00.8
0.60
0.80
AE5
1-1
-11[k]3 =
00
0.60.8
0.60.800
3
1
2
5
63
q´i
q´j
i
j36.87o
[k]3 = AE
6 25
612
5
0.0960.128-0.096-0.128
0.0720.096-0.072-0.096
-0.096-0.1280.0960.128
-0.072-0.0960.0720.096
1
3 m
4 m45o
2
1
3
1
2
5
63*
4*
![Page 44: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/44.jpg)
44
3 m
4 m45o
2
1
3
1
2
5
63*
4*
[k*]1 = AE 63*4*
53*
-0.1250.125
0-0.1768
4*
0.125-0.125
00.1768
5
0.1768-0.1768
00.25
6
0000
5
6 4*
Global Stiffness:
[k*]2 = AE
2 4*1
23*4*
1
0.16670.1667-0.2357
0
0.16670.1667-0.2357
0
0000
-0.2357-0.23570.3333
03*
[k]3 = AE
6 25
612
5
0.0960.128-0.096-0.128
0.0720.096-0.072-0.096
-0.096-0.1280.0960.128
-0.072-0.0960.0720.096
1[K] = AE 2
3*
11
00.0960.128
2
-0.23570.40530.096
0.2917-0.2357
03*
![Page 45: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/45.jpg)
45
Global :
[Q] = [K][D] + [QF]
Q1 = 30
Q3*= 0
Q2 = 0D1
D3*
D2 = AE 23*
11
00.0960.128
2
-0.23570.40530.096
0.2917-0.2357
03*
D1
D3*
D2 =352.5
-127.3-157.5AE
1
3 m
4 m45o
30 kN
3 m
4 m45o
2
1
3
1
2
5
63*
4*
5
6 4*
![Page 46: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/46.jpg)
46
[q´F]1 = AE 00
0D3*
-1 0 0.707 -0.7074
= -22.50 kN, (C)
D2
D1
0D3*
[q´F]2 = AE 0 1 -0.707 -0.7073
= -22.50 kN, (C)
00
D2
D1
[q´F]3 = AE -0.8 -0.6 0.8 0.65
= 37.50 kN, (T)
D1
D3*
D2 =352.5
-127.3-157.5AE
1
Member Forces :
Member
#1
λiyλix λjx λjy
#2
#3
1 0 0.707 -0.707
0 -1 -0.707 -0.707
0.8 0.6 0.8 0.6
3 m
4 m45o
2
1
3
1
2
5
63*
4*
5
6 4*
[ ] [ ]F
yj
xj
yi
xi
xxyxmF q
DDDD
LAEq ']'[ +
−−= λλλλ
![Page 47: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/47.jpg)
47
36.87o 45o45o
3 m
4 m45o
2
1
3
1
2
5
63*
4*
Member
Member Force (kN)
[q´]2[q´]1 [q´]3
-22.50 -22.50 37.50
22.50 kN
22.50 kN
37.50 kN
22.50 kN
7.50 kN31.82 kN
Reactions :
3 m
4 m45o
30 kN
![Page 48: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/48.jpg)
48
Example 6
For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the displacement of the loaded joint. AE is constant.
3 m
4 m
45o
30 kN
4 m
![Page 49: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/49.jpg)
49
15
63*
4*
[q*]θi = 0o,λix = cos 0o = 1,λiy = sin 0o = 0
θij = -45o
λix = cos (-45o) = 0.707,λiy = sin(- 45o) = -0.707
45o
[q*] = [T*]T[q´] + [T*]T[q´F]
q5
q3*
q6
q4*
q´iq´j
[T*]T
=0001
-0.7070.707
00
q´i q´j
i j1
3 m
4 m 4 m
2
1
3
4
5
12
5
6 3*4*
78Member 1:
![Page 50: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/50.jpg)
50
3 m
4 m 4 m
15
63*
4*
[q*]θi = 0o,λix = cos 0o = 1,λiy = sin 0o = 0
θij = -45o = 315o,λix = cos (-45o) = 0.707,λiy = sin(- 45o) = -0.707
45o
0-0.707
00.707
00
10
AE4
1-1
-11[k*]1 =
0001
-0.7070.707
00
[k*]1 = AE 63*4*
53*
-0.1250.125
0-0.1768
4*
0.125-0.125
00.1768
5
0.1768-0.1768
00.25
6
0000
q´i q´j
i j1
2
1
3
4
5
12
5
6 3*4*
78Member 1:
[ ]TL
AETk T
−
−=
1111
][*][
![Page 51: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/51.jpg)
51
q´i
q´j
i
j
22
1
2
3*4*
θi = -90o,λix = cos(-90o) = 0,λiy = sin(-90o) = -1
90o
θj = -135o = 215o,λix = cos (-135o) = -0.707,λiy = sin(- 135o) = -0.707
0-0.707
0-0.707
-10
00
AE3
1-1
-11[k*]2 =
00-10
-0.707-0.707
00
[k*]2 = AE
2 4*1
23*4*
1
0.16670.1667-0.2357
0
0.16670.1667-0.2357
0
0000
-0.2357-0.23570.3333
03*
3 m
4 m 4 m
2
1
3
4
5
12
5
6 3*4*
78Member 2:
[ ]TL
AETk T
−
−=
1111
][*][
90o+45o
=135o
![Page 52: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/52.jpg)
52
36.87o
Member 3:
θi = θj = 36.87o ;λix = λjx = cos (36.87o) = 0.8,λiy = λjy = sin(36.87o) = 0.6
00.6
00.8
0.60
0.80
AE5
1-1
-11[k]3 =
00
0.60.8
0.60.800
3
1
2
5
63
q´i
q´j
i
j36.87o
[k]3 = AE
6 25
612
5
0.0960.128-0.096-0.128
0.0720.096-0.072-0.096
-0.096-0.1280.0960.128
-0.072-0.0960.0720.096
1
3 m
4 m 4 m
2
1
3
4
5
12
5
6 3*4*
78
[ ]TL
AETk T
−
−=
1111
][*][
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53
3 m
4 m 4 m
θi = θij = 0o; λix = λjx = cos 0o = 1, λiy = λjy = sin 0o = 0
00
01
00
10
AE4
1-1
-11[k]1 =
0001
0100
q´i q´j
i j1
2
1
3
4
5
12
5
6 3*4*
78Member 4:
47
8
[q]1
2
[k]4 = AE 278
17
00.25
0-0.25
8
0000
1
0-0.25
00.25
2
0000
[ ]TL
AETk T
−
−=
1111
][*][
![Page 54: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/54.jpg)
54
Member 5:
θi = - 8.13o;λix = cos (- 8.13o) = 0.9899,λiy = sin(- 8.13o) = -0.1414
00.6
00.8
-0.14140
0.98990
AE5
1-1
-11
5
q´i
q´j
i
j36.87o
3 m
4 m 4 m
2
1
3
4
5
12
5
6 3*4*
78
53*4*
7
8
[k*]5 =00
-0.14140.9899
0.60.800
[k*]5 = AE
4* 83*
4*78
3*
0.0960.128
0.02263-0.1584
0.0720.096
0.01697-0.1188
-0.1188-0.1584-0.0280.196
0.016970.022630.004-0.028
7
θ j = 36.87o ;
λ jx = cos (36.87o ) = 0.8,
λ jy = sin
(36.87o ) = 0.6
[ ]TL
AETk T
−
−=
1111
][*][
8.13o
![Page 55: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/55.jpg)
550 -0.2357-0.2357
0
[k*]1 = AE 63*4*
53*
-0.1250.125
0-0.1768
4*
0.125-0.125
00.1768
5
0.1768-0.1768
00.25
6
0000
[k*]3 = AE
6 25
612
5
0.0960.128-0.096-0.128
0.0720.096-0.072-0.096
-0.096-0.1280.0960.128
-0.072-0.0960.0720.096
1
3 m
4 m 4 m
2
1
3
4
5
Global Stiffness:1
2
5
6 3*4*
78
5
6 4*
78
[k*]2 = AE
2 4*1
23*4*
1
0.16670.1667-0.2357
0
0.16670.1667-0.2357
0
0000
-0.2357-0.23570.3333
03*
[k]4 = AE 278
17
00.25
0-0.25
8
0000
1
0-0.25
00.25
2
0000
[k*]5 = AE
4* 83*
4*78
3*
0.0960.128
0.02263-0.1584
0.0720.096
0.01697-0.1188
-0.1188-0.1584-0.0280.196
0.016970.022630.004-0.028
7
[K] = AE 23*
11 2
0.0960.378
0.40530.096
3*
0.4877
![Page 56: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/56.jpg)
56
Global :30 kN
4 m 4 m
3 m2
1
3
4
5
12
5
6 3*4*
78
5
6 4*
78
[Q] = [K][D] + [QF]
Q1 = 30
Q3*= 0
Q2 = 0D1
D3*
D2
D1
D3*
D2 =86.612
-13.791-28.535AE
1
= AE 23*
11
00.0960.378
2
-0.23570.40530.096
0.4877-0.2357
03*
![Page 57: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/57.jpg)
57
3 m
4 m 4 m
2
1
3
4
5
12
5
6 3*4*
78
5
6 4*
78
D1
D3*
D2 =86.612
-13.791-28.535AE
1
[q´F]1 = AE 00
0D3*
-1 0 0.707 -0.7074
= -2.44 kN, (C)
D2
D1
0D3*
[q´F]2 = AE 0 1 -0.707 -0.7073
= -6.26 kN, (C)
00
D2
D1
[q´F]3 = AE -0.8 -0.6 0.8 0.65
= 10.43 kN, (T)
Member Forces :
Member
#1
λiyλix λjx λjy
#2
#3
1 0 0.707 -0.707
0 -1 -0.707 -0.707
0.8 0.6 0.8 0.6
[ ] [ ]F
yj
xj
yi
xi
xxyxmF q
DDDD
LAEq ']'[ +
−−= λλλλ
![Page 58: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/58.jpg)
58
Member
#4
λiyλix λjx λjy
#5
3 m
4 m 4 m
2
1
3
4
5
12
5
6 3*4*
78
5
6 4*
78
D1
D3*
D2 =86.612
-13.791-28.535AE
1
D2
D1
00
[q´F]4 = AE -1 0 1 04
= -21.65 kN, (C)
0D3*
00
[q´F]5 = AE -0.9899 0.141 0.8 0.65
= 2.73 kN, (T)1 0 1 0
0.9899 -0.141 0.8 0.6
Member Forces :
[ ] [ ]F
yj
xj
yi
xi
xxyxmF q
DDDD
LAEq ']'[ +
−−= λλλλ
![Page 59: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/59.jpg)
59
36.87o 45o45o
81.87o
36.87o
Member
Member Force (kN)
[q´]1 [q´]2 [q´]3 [q´]4 [q´]5
-2.44 -6.26 10.43 -21.65 2.73
Reactions :30 kN
4 m 4 m
3 m2
1
3
4
5
12
5
6 3*4*
78
5
6 4*
78
6.26 kN10.43 kN
21.65 kN
2.73 kN
2.44 kN 2.44 kN5.90 kN
6.26 kN
19.47 kN
1.64 kN
6.54 kN
![Page 60: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/60.jpg)
60
AB
C
3 m
D
8 kN
4 kN
4 m36.87o
4 m
λjyλjx
λjxλjx
−λiyλjx
−λixλjx
λjyλjy
λjxλjy
−λiyλjy
−λixλjy
λiyλix
λixλix
−λjyλix
−λjxλix
λiyλiy
λixλiy
−λjyλiy
−λjxλiy
Vi Uj VjUi
[ k *]m = AEL
Vi
Uj
Vj
Ui
Example 7
For the truss shown, use the stiffness method to:(b) Determine the end forces of each member and reactions at supports.(a) Determine the displacement of the loaded joint.Take AE = 8(103) kN.
![Page 61: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/61.jpg)
61
Member 1:
λix = cos 73.74o = 0.28,λiy = sin 73.74o = 0.96
[q*] = [T*]T[q´] + [T*]T[q´F]
λjx = cos 36.87o = 0.8,λjy = sin 36.87o = 0.6
AB
C
3 m
D
8 kN
4 kN
4 m
36.87o
4 m
x *
y *
73.74o
x
y
36.87o
1
i
j1
2
3*
4*1
i
j
q´i
q´j
q3*
q1
q4*
q2
q´iq´j
=00
0.960.28
0.60.8
00
[T*]T
3
2
4
5
1
1
2
56
78
3*
4*
![Page 62: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/62.jpg)
62
[k]1 = 8x103
4* 1 23*
4*
12
3*
0.0960.128
-0.1536-0.0448
0.0720.096
-0.1152-0.0336
-0.0336-0.04480.053760.01568
-0.1152-0.15360.184320.05376
000.960.280.60.800
[k]1 =00
0.960.28
0.60.8
00
8x103
51
-1-11
AB
C
3 m
D
8 kN
4 kN
4 m
36.87o
4 m
3
2
4
5
1
1
2
56
78
3*
4*
[ ]TL
AETk T
−
−=
1111
][][
![Page 63: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/63.jpg)
63
x *
y*
36.87x
yMember 2:
λix = cos 36.87o = 0.8,λiy = sin 36.87o = 0.6
[q*] = [T*]T[q´] + [T*]T[q´F]
AB
C
3 m
D
8 kN
4 kN
4 m
36.87o
4 m
λjx = cos 0o = 1,λjy = sin 0o = 0
q3*
q5
q4*
q6
q´iq´j
[T*]T
[k] = [TT] AEL
[T]1-1
-11
q´i q´j
i j2
=00
0.60.8
0100
i j2 5
6
3*
4*
[k]2 = 8x103
4* 5 63*
4*
56
3*
00.25
-0.15-0.2
0000
0-0.20.120.16
0-0.150.090.12
3
2
4
5
1
1
2
56
78
3*
4*
![Page 64: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/64.jpg)
64
x
y
270o 1
2
784
Member 3:
AB
C
3 m
D
8 kN
4 kN
4 m
36.87o
4 m
λx = cos 323.13o = 0.8,λy = sin 323.13o = -0.6
Member 4:
[k]4 = 8x103
2 7 81
278
1
-0.0960.1280.096
-0.128
0.072-0.096-0.0720.096
0.096-0.128-0.0960.128
-0.0720.0960.072
-0.096
λx = cos 270o = 0, λy = sin 270o = -1
[k]3 = 8x103
0000
0.3330
-0.3330
00
00
2 5 61
256
10.333
0
-0.3330
x
y
323.13o3
1
2
56
3
2
4
5
1
1
2
56
78
3*
4*
![Page 65: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/65.jpg)
65
x
yMember 5:
AB
C
3 m
D
8 kN
4 kN
4 m
36.87o
4 m
λx = cos 0o = 1,λy = sin 0o = 0
00.25
0-0.25
0000
00.25
0-0.25
00
00
[k]5 = 8x103
6 7 85
678
5
5 78
56
3
2
4
5
1
1
2
56
78
3*
4*
![Page 66: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/66.jpg)
660.17568
-0.2-0.20.5
00
0 0
-0.0336-0.0448
-0.0448 -0.03360.0
0.2560.4740.0
[k]1 = 8x103
4* 1 23*
4*
12
3*
0.0960.128
-0.1536-0.0448
0.0720.096
-0.1152-0.0336
-0.0336-0.04480.053760.01568
-0.1152-0.15360.184320.05376
[k]4 = 8x103
2 7 81
278
1
-0.0960.1280.096
-0.128
0.072-0.096-0.0720.096
0.096-0.128-0.0960.128
-0.0720.0960.072
-0.096
00.25
0-0.25
0000
00.25
0-0.25
00
00
[k]5 = 8x103
6 7 85
678
5
[k]2 = 8x103
4* 5 63*
4*
56
3*
00.25
-0.15-0.2
0000
0-0.20.120.16
0-0.150.090.12
[k]3 = 8x103
0000
0.3330
-0.3330
00
00
2 5 61
256
10.333
0
-0.3330
[K] = 8x103
2 3* 51
23*
5
1
3
2
4
5
1
12
56
78
3*
4* 6
784*
![Page 67: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/67.jpg)
67
3
2
4
5
1
2
3*
4*
56
78A
BC
3 m
D
8 kN
4 kN
4 m
36.87o
4 m
1
Global: [Q] = [K][D] + [QF]
D1
D3*
D2
D5
1.988x10-3 m
1.996x10-4 m-2.0824x10-3 m
7.984x10-5 m
=
Q1 = 4
Q3*= 0
Q2 = -8
Q5 = 0
D1
D3*
D2
D5
0.17568-0.2
-0.20.5
00
0 0
-0.0336-0.0448
-0.0448 -0.03360.0
0.2560.4740.0
= 8x103
2 3* 51
23*
5
1
8 kN
4 kN
![Page 68: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/68.jpg)
68
AB
C
3 m
D
8 kN
4 kN
4 m
36.87o
4 m
3
2
4
5
1
2
3*
4*
56
78
1
Member forces
D1
D3*
D2
D5
1.988x10-3 m
1.996x10-4 m-2.0824x10-3 m
7.984x10-5 m
=
0D3*
D2
D1
[q´F]1 = 8x103 -0.28 -0.96 0.8 0.65
= 0.46 kN, (T)
0D3*
0D5
[q´F]2 = 8x103 -0.8 -0.6 1 04
= -0.16 kN, (C)
Member
#1
#2
λiyλix λjx λjy
0.28 0.96 0.8 0.6
0.8 0.6 1 0
[ ] [ ]F
yj
xj
yi
xi
xxyxmF q
DDDD
LAEq ']'[ +
−−= λλλλ
![Page 69: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/69.jpg)
69
D1
D*3
D2
D5
1.988x10-3 m
1.996x10-4 m-2.0824x10-3 m
7.984x10-5 m
=
3
2
4
5
1
2
3*
4*
56
78
1
[q´F]3 = 8x103 D2
D1
0D5
0 1 0 -13
= -5.55 kN
D2
D1
00
[q´F]4 = 8x103 -0.8 0.6 0.8 - 0.65
= -4.54 kN
0D5
00
[q´F]5 = 8x103 -1 0 1 04
= -0.16 kN
Member
#3
λiyλix λjx λjy
#4
#5
0 -1 0 -1
0.8 -0.6 0.8 -0.6
1 0 1 0
[ ] [ ]F
yj
xj
yi
xi
xxyxmF q
DDDD
LAEq ']'[ +
−−= λλλλ
![Page 70: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/70.jpg)
70
y*
x *
36.87o
36.87o 36.87o
AB
C
3 m
D
8 kN
4 kN
4 m
36.87o
4 m
3
2
4
5
1
2
3*
4*
56
78
1
0.46 -0.16 -5.55 -4.54 -0.16
0.46 kN
5.55 kN0.36 kN
4.54 kN
3.79 kN
2.72 kN
Member
Member Force (kN)
[q]2[q]1 [q]3 [q]4 [q]5
0.16 kN 0.16 kN
5.55 kN
![Page 71: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/71.jpg)
71
Space-Truss Analysis
![Page 72: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/72.jpg)
72
+
−
−=
Fj
Fi
j
i
j
i
dd
LEA
''
''
1111
''
[ ] [ ]F
jz
jy
jx
iz
iy
ix
q
dddddd
TL
EA '1111
+
−
−=
[ ]
=
zyx
zyxT
where
λλλλλλ000
000,
[q´] = [k´][d´] + [q´F]
= [k´][T][d] + [q´F]
Member Local Stiffness [k´]:
![Page 73: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/73.jpg)
73
Member Global Stiffness [km]:
[km]= [T]T[k´] [T]
[ ]
−
−
=zyx
zyx
z
y
x
z
y
x
m LEAk
λλλλλλ
λλλ
λλλ
000000
1111
000
000
![Page 74: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/74.jpg)
74
Global equilibrium matrix:
[Q] = [K][D] + [QF]
QI
QII
Du
Dk
KI,II
KII,II
KI,I
KII,I
Reaction Support Boundary Condition
Unknown DisplacementJoint Load
QFI
QFII
= +
Fixed End Forces
![Page 75: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/75.jpg)
75
q´j
q´i
i
j
m
i
j
m
=EAL
λxλx
λzλx
λyλx
λxλy
λyλy
λzλy
λxλz
λyλz
λzλz
λxλx
λzλx
λyλx
λxλy
λyλy
λzλy
λxλz
λyλz
λzλz
−λxλx
−λzλx
−λyλx
−λxλy
−λyλy
−λzλy
−λxλz
−λyλz
−λzλz
−λxλx
−λzλx
−λyλx
−λxλy
−λyλy
−λzλy
−λxλz
−λyλz
−λzλz
diy
dix
djx
diz
djy
djz
qiy
qix
qjx
qiz
qjy
qjz
qjy
qjx
qjzqiy
qix
qiz
+
−
−=
Fj
Fi
j
i
j
i
dd
LEA
''
''
1111
''
[ ] [ ] [ ]F
jz
jy
jx
iz
iy
ix
zyxzyxmj q
dddddd
LEAq '' +
−−−= λλλλλλ
![Page 76: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/76.jpg)
76
x
z
y
4 m4 m
3 m3 m
10 m
80 kN60 kN
O
Example 6
For the truss shown, use the stiffness method to:(b) Determine the end forces of each member.(a) Determine the deflections of the loaded joint.Take E = 200 GPa, A = 1000 mm2.
![Page 77: 05 Matrix Truss](https://reader034.vdocuments.site/reader034/viewer/2022042501/55cf9be4550346d033a7c3e7/html5/thumbnails/77.jpg)
77
λ1 = (-4/11.18)i + (3/11.18)j + (-10/11.18)k
= -0.3578 i + 0.2683 j - 0.8944 k
λ2 = (+4/11.18)i + (3/11.18)j + (-10/11.18)k= +0.3578 i + 0.2683 j - 0.8944 k
λ3 = (+4/11.18)i + (-3/11.18)j + (-10/11.18)k
= +0.3578 i - 0.2683 j - 0.8944 k
= -0.3578 i - 0.2683 j - 0.8944 k
λ4 = (-4/11.18)i + (-3/11.18)j + (-10/11.18)k
Member
#1
#2
#3
λx λy λz
#4
-0.3578 +0.2683 -0.8944
+0.3578 +0.2683 -0.8944
+0.3578 -0.2683 -0.8944
-0.3578 -0.2683 -0.8944
λm = λxi + λyj + λzk
12
34
12
3
(0, 0, 10)
(-4, 3, 0)
(4, 3, 0)
(4, -3, 0)
(-4, -3, 0)4
1
2
53
x
z
y
4 m4 m
3 m3 m
10 m
80 kN60 kN
O
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Member
#1#2
#3
λx λy λz
#4
-0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944
+0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944
Member Stiffness Matrix [k]6x6
[k]m =[k12]3x3
[k22]3x3
[k11]3x3
[k21]3x3
2 31
23
1
+0.80-0.240
+0.320
+0.320-0.096
+0.128
-0.240+0.072
-0.096
[k11]1 =AEL
2 31
23
1
+0.80-0.240
-0.320
-0.320+0.096
+0.128
-0.240+0.072
+0.096
[k11]2 =AEL
2 31
23
1
+0.80+0.240
-0.320
-0.320-0.096
+0.128
+0.240+0.072
-0.096
[k11]3 =AEL
2 31
23
1
+0.80+0.240
+0.320
+0.320+0.096
+0.128
+0.240+0.072
+0.096
[k11]4 =AEL
2 31
23
1
3.20.0
0.0
0.00.0
0.512
0.00.288
0.0
[KI,I] =AEL
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79
12
34
12
3
(0, 0, 10)
(-4, 3, 0)
(4, 3, 0)
(4, -3, 0)
(-4, -3, 0)4
1
2
53
x
z
y
4 m4 m
3 m3 m
10 m
80 kN60 kN
O
0.0-80
60
D3
D2
D1
0.00.0
0.0+
2 31
23
1
3.20.0
0.0
0.00.0
0.512
0.00.288
0.0
=AEL
[Q] = [K][D] + [QF]
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80
Global equilibrium matrix:
[Q] = [K][D] + [QF]
QI
QII
Du
Dk
KI,II
KII,II
KI,I
KII,I
Reaction Support Boundary Condition
Unknown DisplacementJoint Load
QFI
QFII
= +
Fixed End Forces
(AE/L) = (1x10-3)(200x106)/(11.18) = 17.89x103 kN
0.0-80
60
D3
D2
D1
0.00.0
0.0+
2 31
23
1
3.20.0
0.0
0.00.0
0.512
0.00.288
0.0
=AEL
0.0 mm-15.53 mm
6.551 mm=
D3
D2
D1
= LAE
0.0-277.8
+117.2
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81
Member forces:
q´F+
Member
#1#2
#3
λx λy λz
#4
-0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944
+0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944
dyi
dxi
dxj
dzi
dyj
dzj
[q´j]m = AEL −λx −λy −λz λx λy λz
D3
D2
D1
[ 0 ]
= +116.5 kN (T)
+0.3578 -0.2683 +0.8944[q´j]1 =AEL
0.0
117.2
-277.8L
AE
x
z
y
4 m4 m
3 m3 m
10 m
80 kN60 kN
O
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82
Member
#1#2
#3
λx λy λz
#4
-0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944
+0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944
= +32.61 kN (T)-0.3578 -0.2683 +0.8944[q´j]2 =AEL
0.0
117.2
-277.8L
AE
= -116.5 kN (T)-0.3578 +0.2683 +0.8944[q´j]3 =AEL
0.0
117.2
-277.8L
AE
= -32.61 kN (T)+0.3578 +0.2683 +0.8944[q´j]4 =AEL
0.0
117.2
-277.8L
AE
x
z
y
4 m4 m
3 m3 m
10 m
80 kN60 kN
O
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5
32.6 kN116.5 kN
116.5 kN
32.6 kN
Member
#1#2
#3
λx λy λz
#4
-0.3578 +0.2683 -0.8944+0.3578 +0.2683 -0.8944
+0.3578 -0.2683 -0.8944-0.3578 -0.2683 -0.8944
[q´j]m
116.5 32.6
-32.6-116.5
12
34
80 kN60 kN
R5x = (-32.6)(-0.3578) = 11.66 kN
R5y = (-32.6)(-0.2683) = 8.75 kN
R5z = (-32.6)(-0.8944) = 29.16 kN