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DRAFT –1
Chapter 3 Truss Element
y
Example 2
1
(1)
2
4000
lb
Given:
20”
Elements 1 and 2: Aluminum
50”
Element 3: steel
(3)
A
(1)
= 1.5in
2
(2)
A
(2)
= 1.0in
2
40”
A
(3)
= 1.0in
2
3
4
x
Required:
Find stresses and displacements using hand calculations.
30”
Solution
Calculate the stiffness constants:
in
lb
L
AE
K
in
lb
L
AE
K
in
lb
L
AE
K
5
6
3
5
6
2
5
6
1
10
0
.
6
50
10
10
30
10
5
.
2
40
10
10
1
10
5
.
7
20
10
10
5
.
1
×
=
×
×
=
=
×
=
×
×
=
=
×
=
×
×
=
=
Calculate the Element matrix equations.
Element (1)
u
2
u
4
(1)
u
1
u
3
1
2
Denoting the Spring constant for element (1) by k
1
, and the stiffness matrix by K
(1)
, the
stiffness matrix in global coordinates is given as,
ME 273 Lecture Notes © by R. B. Agarwal
3-16
DRAFT –1
Chapter 3 Truss Element
1
2
3
4
2
2
1
[K
g
]
(1)
= K
1
For element (1),
θ
= 0
0,
therefore
=1, c
2
= 1
= 0, s
2
=0, and cs = 0
1
2 3 4
1
[k
(1)
] = k
1
u
4
lement (2)
2
u
3
or this element,
θ
= 90
0
, Therefore,
(2)
= cos
θ
= 0, c
2
= 0
θ
= 90
0
3
u
he stiffness matrix is,
u
6
5
6 3 4
2
2
5
[k
g
]
(2)
= k
2
c cs -c -cs
cs s
2
-cs -s
2
2
c
2
-cs c
2
cs
3
-cs -s
2
cs s
2
4
1 0 -1 0
c
s
0 0 0 0
2
1 0 1 0
3
0 0 0 0
4
E
F
c
s = sin
θ
= 1, s
2
= 1
5
c cs -c -cs
cs= 0
T
cs s
2
-cs -s
2
6
c
2
-cs c
2
cs
3
-cs -s
2
cs s
2
4
ME 273 Lecture Notes © by R. B. Agarwal
3-17
DRAFT –1
Chapter 3 Truss Element
5 6 3 4
5
[k
g
]
(2)
= k
2
lement 3
u
4
2
u
3
= cos(126.9
0
) = -0.6, c
2
= .36
= sin(126.9
0
, s
2
= .64
7
u
8
7
[k
g
]
(3)
= k
3
0 0 0 0
0 1 0 -1
6
0 0 0 0
3
0 -1 0 1
4
E
For element (3),
θ
= 126.9
0
.
c
(3)
4
θ
= 126.9
0
s
u
cs = -0.48
7
8 3 4
.36 -.48 -.36 .48
-.48 .64 .48 -.64
8
-.36 .48 .36 -.48
3
.48 -.64 -.48 .64
4
ssembling the global Matrix
mbly described earlier, the assembled matrix is,
[K
g
]
=
A
Following the procedure for asse
1
1
1
1
3
3
3
3
2
3
2
3
2
2
3
3
3
3
3
3
0
0
0
0
0
1
0
0
0
0
0
0
0
0
2
0
.36
.48
0
0
.36
.48
3
0
0
.48
.64
0
.48
.64
4
0
0
0
0
0
0
0
0
5
0
0
0
0
0
0
6
0
0
.36
.48
0
0
.36
.48
7
0
0
.48
.64
0
0
.48
.64
8
K
K
K
K
K
K
K
K
K
K
K
K
K
K
K
K
K
K
K
K
K
K
−
−
+
−
−
+
−
−
−
−
−
−
−
3
3
3
3
0
K
K
ME 273 Lecture Notes © by R. B. Agarwal
3-18
DRAFT –1
Chapter 3 Truss Element
The boundary conditions are:
1
= u
2
= u
5
= u
6
= u
7
= u
8
= 0
e will suppress the corresponding rows and columns. The reduced matrix is a 2 x2,
u
W
given below,
1
3
3
3
2
3
1
3
3
3
3
2
3
4
.36
.48
3
[
]
0.48
.64
4
The final equation is
.36
.48
4000
0.48
.64
8000
g
K
K
K
K
K
K
K
K
K
K
u
K
K
K
u
+
−
=
−
+
+
−
−
=
−
+
ubstituting values for k
1
, k
2
, and k
3
, we get
S
3
5
4
3
4
5
1
1
1
5
2
2
2
5
3
3
31
9.66
2.88
4000
10
2.88
6.34
8000
0.0000438
0.012414
(7.5 10 )( 0.0000438)
214
1.5
(2.5 10 )( 0.012414)
3015
1.0
(6 10 )( 0.0000
u
u
u
in
u
in
P
K u
psi
A
A
P
K u
psi
A
A
K u
P
A
A
σ
σ
σ
−
−
=
−
= −
= −
∆
×
−
=
=
=
= −
∆
×
−
=
=
=
=
∆
×
−
=
=
=
0
0
438cos 53.1
.012414sin 53.1 )
6119
1.0
psi
+
=
ME 273 Lecture Notes © by R. B. Agarwal
3-19