9475 2475 2475 0
9475 2475 0
5975 3500
5975
2
2
3
4
u
v
v
v
0
– 50
0
– 50
Solving simultaneously
u3 = 0.135 10–2 m
34
x
f
= – 18.93 kN
3.33 (a)
67
94
(210 10 ) (5.0 10 )
(1) (2)
0.5 0.5 0.5 0.5
1111
1111
Element 1–3 ;
= 180°
C2 = 1.0, CS = 0, S2 = 0
[k1–3] =
94
(210 10 ) (5 10 )
10
(1) (3)
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
[k1–3] = 105 105
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
N/m
Element 1–4 ;
= 270°
C2 = 0, CS = 0, S2 = 1.0
(1) (4)
0 0 0 0
Boundary conditions are
The final matrix (assembled)
1
3
1
0
100 10
x
y
F
F
= 105
210 105
105 125
1
1
u
v
0 = 210 u1 – 105 v1 v1 = 2 u1
3
1–2 =
9
210 10
5m
[0707 – 0707 – 0707 0707]
–3
–3
6.897 10
14.0 10
0
0
1–2 = 210.9 MPa (T)
1–3 =
9
210 10
10
[1.0 0 –1.0 0]
3
3
6.897 10
14.0 10
0
0
1–3 = – 144.8 MPa (C)
(b)
A = 5 10–4 m2, E = 210 109
2
N
m
44 N
m
[k(2)] = 2.1 107
11
3
1
44
33
44 N
m
uv
11
00
uv
8
AE
1 0 1 0
0 0 0 0
1 0 1 0
0 0 0 0
0 0 0 0
0 1 0 1
0 0 0 0
4
0 1 0 1
AE
[k2–4] =
0.80 0.40 0.80 0.40
0.40 0.20 0.40 0.20
0.80 0.40 0.80 0.40
8.94
0.40 0.20 0.40 0.20
AE
0 0 0 0
0 1 0 1
0 0 0 0
4
0 1 0 1
AE
1 0 1 0
0 0 0 0
1 0 1 0
8
0 0 0 0
AE
Boundary conditions
u1 = v1 = u3 = v3 = u2 = 0, v2 = – 0.05 in.
Applying the boundary conditions and superimposing the [k]s
AE
2
4
4
0.02
0.272 0.0447 0.0223
0.0447 0.214 0.0447
0.0223 0.0447 0.272
v
u
v
=
0
0
0
3
30 10
2
0
0.02
u
v
(5) =
3
30 10
0
0.02
AE
(1) (6)
0.75 0.433
1
u
4
3
76
[ ] [ ]
2
0.48 0 0.64
[ ] [ ]
Element 1–3
0.64 0 0.48
0.48 0 0.64
[ ] [ ]
[ ] [ ]
Element 1–4
000
000
[ ] [ ]
[ ] [ ]
Element 1–5
0.356 0.444 0.178
0.178 0.222 0.0889
[k] = 31305
[ ] [ ]
[ ] [ ]
Applying the boundary conditions where all deflections at node 2, 3, 4 and 5 are zero.
The global equations are
0
0
64905 13899 5572
5572 6950 33023
1
1
1
u
w
0 = 64905 u1 – 13899 v1 – 5572 w1 (1)
– 10 = – 13899 u1 + 69906 v1 + 6950 w1 (2)
0 = – 5572 u1 + 6950 v1 + 33023 w1 (3)
From (1) and (3)
From (2) and (3)
From (4) and (5), we get
77
1–2 = 42 106 [0.8 0 – 0.6 – 0.8 0 0.6]
5
4
5
3.0183 10
1.5171 10
2.6837 10
0
0
0
2
m
kN
1000
1–3 = 42 106 [0.8 0 – 0.6 – 0.8 0 0.6]
5
4
5
3.0183 10
1.5171 10
2.6837 10
0
0
0
2
m
kN
1000
1–4 = 52500 103 [0 1 0 0 –1 0]
5
4
5
3.0183 10
1.5171 10
2.68374 10
0
0
0
1–4 = – 7965
2
kN
m
(C)
kN
7965
2
m
Force equilibrium at node 1
x direction
y direction
5
z direction
3.41
L1–2 =
2 2 2
(12 0) ( 3 0) ( 4 0)
L1–2 = 13 m
2 2 2
L1–4 =
2 2 2
(14 12) (6 3) (0 4)
L1–4 = 10.05 m
12 34
Element
j i j i j i
x y z
i j i j i j
x x y y z z
C C C
L L L
2 2 2
2
Element
1 2 0.852 0.213 0.284 0.053 0.071 0.095
x x y x z y y z
C C C C C C C C C
80
1.203 10
(1)
12 3 4 12 3 4
13 13 13 13 13 13
E
L
3
4
4
2.766 10
1.024 10
0
0
0
(1) = 41.0 MPa (T)
(2) =
(2)
E
L
[0 0 1 0 0 –1]
3
4
4
2.766 10
1.024 10
1.203 10
0
0
0
(2) = 8.42 MPa (T)
(3) =
(3)
E
L
2 9 4 2 9 4
10.05 10.05 10.05 10.05 10.05 10.05
3
4
4
2.766 10
1.024 10
1.203 10
0
0
0
(3) = – 11.58 MPa (C)
3.42
Element 1–5
15
L
108
51
yy
0 ( 36)
6
30 10
1–5 =
15
E
L
[– Cx – Cy – Cz Cx Cy Cz]
1
1
1
5
5
5
u
v
w
u
v
w
6
30 10
0
0
0
41
xx
36
148.4
u4 = – 1.1305 w4
148.4
AE
– 4000 =
148.4
AE
[1.529 (– 1.1305 w4) + 2.2357 w4]
w4 =
6
1171501.87
6 30 0
–
1
1–4 =
14
E
L
[– Cx – Cy – Cz Cx Cy Cz]
1
1
1
4
4
4
u
v
w
u
v
w
1–4 =
6
30 10
148.4
[– 0.2426 0 – 0.9704 0.2426 0 0.9704]
0
0
0
0.00863
0
– 0.00683
3.44 Derive Equation (3.7.21)
=
2x
f
A
2x
AE
L
1
u
u
=
E
L
[–1 1]
1
u
u
Now in 3–D
{d
= [T*]d}
where by Equation (3.7.7)