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Presentation on Matrix analysis of trusses
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1
! 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
2
2-Dimension Trusses
3
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
4
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´
5
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
6
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)
7
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
8
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
9
[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
10
+
−
−=
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
11
+
−−
−−=
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
12
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
13
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.
14
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
15
-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
16
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
17
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
18
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 :
19
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
20
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
21
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
22
∆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+
23
[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=
24
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
25
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
26
∆ 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.
27
∆ 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
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
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
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
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
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
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
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
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
36
Special Trusses (Inclined roller supports)
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
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
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
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
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
][*][
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
][*][
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*
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*
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*
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
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
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
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:
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
][*][
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
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
][*][
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
][*][
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
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
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*
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
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
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
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.
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*
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
][][
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*
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*
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*
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*
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
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
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
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
71
Space-Truss Analysis
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´]:
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
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
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
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.
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
78
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
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]
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
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
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
83
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
