85
[T*] =
0 0 0
0 0 0
x y z
x y z
C C C
C C C
L
E
E
0 0 0
x y z
C C C
L
2
2
2
u
v
w
3.46
E = 30 106 psi
Reduce the given figure by symmetry.
A(1) = A(2) = A(3) = 10 in.2
(Reducing given A(3) by half)
86
Data for reduced truss
Element
°
C
C2
S2
CS
1
36.9°
0.8
0.6
0.64
0.36
0.48
2
0°
1.0
0
1.0
0
0
3
90°
0
1.0
0
1.0
0
AE
22
22
C CS C CS
CS S CS S
1
3
3
0.64 0.48 0.64 0.48
0.36 0.36 0.48 0.36
u
v
Finally
1 1 3 3
1
1
3
3
1.6 1.2 1.6 1.2
1.6 1.2 1.6 1.2
1.2 0.9 1.2 0.9
u v u v
u
u
v
[k(2)] = (10 in.2)
6
2
30 10 lbf
in.
1
8ft
1 ft
12in.
1.0 0 1.0 0
0 0 0 0
1.0 0 1.0 0
0 0 0 0
and
1 1 2 2
1
3.125 0 3.125 0
u v u v
u
2
in.
6ft
12in.
0000
0 1.0 0 1.0
2 2 3 3
2
0 0 0 0
u v u v
u
4.725 0 1.2
0 0.01 4.167
1.2 0 5.067
(3) =
5
3
4.167 10 lbf
in.
= 1000 psi

(3) = 1000 psi (T)
3.47
Using symmetry
AE
1
2
3
4
4
0.0756 0 0.0192 0 0
0 0.0811 0.0667 0.0192 0.0144
0.0192 0.0667 0.0811 0 0
0 0.0192 0 0.0756 0.0192
0 0.0144 0 0.0192 0.0811
u
v
v
u
v
=
2
0
0
0
P
P









Using the simultaneous equation solver, we obtain
P
AE
P
AE
v3 = – 433
P
AE
, u4 = 50
P
AE
P
E
P
AE
P
E
4
u
v
E
P
2
0
405
u
v
AE
AE
1
1
0.8 0.4 – 0.8 – 0.4
0.2 – 0.4 – 0.2
u
v
4.47
2
2
0.8 0.4
Symmetry 0.2
u
v
4.47
AE
2
2
0.8 0.4 0.8 0.4
0.2 0.4 0.2
u
v
4
3
1
3
2
00
Symmetry
u
v
Boundary conditions
v1 = u4 = u3 = 0
1 2 2 3 4
0.8 0.4
0.8 1
1
4.47 8 4.47 4.47
0.8 0.4
2.4 0.4 0.4
4.47 4.47 4.47 4.47 4.47
0.4 0.2 0.2
0.4 0.6
4.47 4.47 4.47 4.47 4.47
0.4 0.2 0.5
0.2 0.5
4.47 4.47 4.47 4 4
0.2 0.5
0.4 0.2 0.5
4.47 4.47 4 4.47 4
00
0
0
u u v v v
u
2
2
3
4
u
v
v
v
F2y = – 20 kN F3y = – 10 kN
1
2
3
x
x
F
F
21
21 1
8
5 5 5 5 5 5
21
611
2
5 5 5 5 5 5 5 5 5 5
1 1 1
3
12
00
u
u
1
2
4
u
u
v
=
3
4
2
2.285 10
8.9381 10
1.0276 10







(1) =
E
L
[– C – S C S]
1
1
2
2
u
v
u
v
(1) =
9
210 10
25
2 1 2 1
5 5 5 5
3
4
3
2.285 10
0
8.9381 10
9.553 10









(1) = – 67.08 106
2
N
mm
(2) =
9
210 10
8
[– 1 0 1 0]
3
2
2.285 10
0
0
1.0276 10









(2) = 59.99 106
2
N
mm
= 60.0 MPa (T)
2
N
mm
9
3
2
N
mm
9
210 10
2 1 2 1
2
0
1.0276 10



= 19.99 106
2
N
mm
= 20.00 MPa (T)
3.49
6
2 30 10
30
0 0 0 0
0 1 0 1
Element (2)
= 0°
[k(2)] =
6
2 30 10
30
(1) (2)
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
Element (3)
= 135°
6
2 30 10
(2) (3)
0.5 0.5 0.5 0.5
2 2 2 2
0.5 0.5 0.5 0.5
2 2 2 2
0.5 0.5 0.5 0.5
2 2 2 2
00
00
0 1 1
2 0 2 0 0 0
0 2 0 0 0 2
2 0 2.707 0.707 0.707 0.707
97
{
} = [C {d}
[C =
E
L
[– C – S C S]
(1) =
6
30 10
30
[0 – 1 0 1]
1
1
3
3
0
0
0
0
u
v
u
v
(1) = 0
(2) =
6
30 10
30
[– 1 0 1 0]
1
1
2
2
0
0
0
0.00283
u
v
u
v
(2) = 0
(3) =
6
30 10
30 2
[0.707 – 0.707 – 0.707 0.707]
0
– 0.00283
0
0
3.50
98
0 0 0 0
0 1 0 1
0 0 0 0
0 1 0 1
[k(2)] = 2 106
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
[k(3)] = 1.414 106
0.5 0.5 0.5 0.5
0.5 0.5 0.5 0.5
0.5 0.5 0.5 0.5
0.5 0.5 0.5 0.5
Assembling the stiffness matrices
[K] = 106
0.707 0.707 0.707 0.707 0 0
0.707 2.707 0.707 0.707 0 2
0.707 0.707 2.707 0.707 2.0 0
0.707 0.707 0.707 0.707 0 0
0 0 2.0 0 2.0 0
0 2.0 0 0 0 2.0
0.707 0.707 0 0 0 0
0.707 0.707 0 0 0 0
0 0 1 0 0 0
Applying the boundary conditions
v1 = u2 = u3 = v3 = 0
1
0
x
F
1
0
1,000,000 0
u



3–2 = 0
3.51
[k(1)] = 2 106
(1) (2)
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
1
3
T
6 5 6
56
66
2.5 10 8.66 10 1.414 10
8.66 10 1.5 10 0
1.414 10 0 1.134 10
6 5 6
56
66
66
5 5 6
5 6 5
2.5 10 8.66 10 2 10
8.66 10 1.5 10 0
1.414 10 0 1.155 10
1.414 10 0 2.38 10
5 10 8.66 10 1.5 10
8.66 10 1.5 10 8.66 10
55
56
5 5 5
6 5 6
56
66
0 5 10 8.66 10
0 8.66 10 1.5 10
4.483 10 2.588 10 4.483 10
1.673 10 9.659 10 1.673 10
8.66 10 1 10 0
1.5 10 0 3 10
2
AE u
1x
L
0.004
–30.4
0.006
–38.4
0.008
–41.6
0.010
–40.6
p min = – 41.67 lbin.
3.53
du =
x d x du
U =
0
x
xx
vd
dv
dv = A(x) dx A(x) = A0
1x
L
1
L
3.54
(1)
1
(1)
2
x
x
f
f
1500
3000
2
1
1
1