ВУЗ: Не указан
Категория: Не указан
Дисциплина: Не указана
Добавлен: 09.06.2020
Просмотров: 203
Скачиваний: 3
DRAFT –1
Chapter 3 Truss Element
u
2
Element 2
θ
= 90
0
c = cos 90
0
= 0, c
2
= 0
1
u
1
s = sin 90
0
= cos 0
0
= 1, s
2
= 1
cs = 0
(2)
EA/L = 69 x 200 x (1/150) = 92 kN/mm
2
u
3
3
4
1 2
u
4
0
0
0
0
3
0
1
0 -1
4
[k
g
]
(2)
= (92) 0
0
0
0
1
0
-1
0
1
2
Element 3
u
6
θ
= 30
0
c = cos 30
0
= 0.866, c
2
= 0.75
3
u
5
s = cos 60
0
= .5, s
2
= 0.25
cs = 0.433
u
4
(3)
300 mm
EA/L = 207 x 100 x (1/300) = 69 kN/mm
θ
= 30
0
2
u
3
3
4
5
6
3
.75
.433 -.75
-.433
4
-.433 .25
-.433 -.25 (69)
[k
g
]
(3)
=
5
-.75 -.433 .75 .433
6
-.433 -.25 .433 .25
ME 273 Lecture Notes © by R. B. Agarwal
3-11
DRAFT –1
Chapter 3 Truss Element
Assembling the stiffness matrices
Since there are 6 deflections (or DOF), u
1
through u
6
, the matrix is 6 x 6. Now, we will
place the individual matrix element from the element stiffness matrices into the global
matrix according to their position of row and column members.
Element [1]
1
2 3 4 5 6
1
5 3 . 1
5 3 . 1
2
3
4
5
5 3 . 1
5 3 . 1
6
−
−
The blank spaces in the matrix have a zero value.
Element [2]
1 2 3 4 5 6
1
2
92
-92
3
4 -92
92
5
6
ME 273 Lecture Notes © by R. B. Agarwal
3-12
DRAFT –1
Chapter 3 Truss Element
Element [3]
1 2 3 4 5 6
1
2
3
51.7
29.9
-51.7
-29.9
4
29.9
17.2
-29.9
-17.2
5
-51.7
-29.9
51.7
29.9
6
-29.9
-17.2
29.9
17.2
Assembling all the terms for elements [1] , [2] and [3], we get the complete matrix
equation of the structure.
−
=
−
−
−
−
−
−
−
−
−
−
−
−
)
(
4
.
0
)
(
0
)
(
0
)
(
0
)
(
0
)
(
0
2
.
17
9
.
29
2
.
17
9
.
29
0
0
9
.
29
8
.
104
9
.
29
7
.
51
0
1
.
53
2
.
17
9
.
29
2
.
109
9
.
29
92
0
9
.
29
7
.
51
9
.
29
7
.
51
0
0
0
0
92
0
92
0
0
1
.
53
0
0
0
1
.
53
6
5
4
3
2
1
6
5
4
3
2
1
6
5
4
3
2
1
F
F
F
F
F
F
u
u
u
u
u
u
Boundary conditions
x and y directions, where as, node 2 is fixed in x-direction only
1
= u
2
= u
3
= 0.
herefore, all the columns and rows containing these elements should be set to zero. The
Node 1 is fixed in both
and free to move in the y-direction. Thus,
u
T
reduced matrix is:
−
=
−
−
4
.
0
0
2
.
17
9
.
29
2
.
17
9
.
29
8
.
104
9
.
29
6
5
4
u
u
−
−
0
2
.
17
9
.
29
2
.
109
u
ME 273 Lecture Notes © by R. B. Agarwal
3-13
DRAFT –1
Chapter 3 Truss Element
riting the matrix equation into algebraic linear equations, we get,
29.9u
4
- 29.9u
5
- 17.2u
6
= 0
4
lving, we get u
4
= -0.0043
ress, Strain and deflections
lement (1)
ote that u
1
, u2, u3, etc. are not coordinates, they
lement (2)
W
-29.9u
4
+ 104u
5
+ 29.9u
6
= 0
-17.2u
4
+ 29.9u
5
+ 17.2u
6
= -0.
so
u
5
= 0.0131
u
6
= -0.0502
S
E
N
are actual displacements.
5
5
5
2
0.0131
5.02 10
260
69 5.02 10
0.00347
,
0.00347 200 0.693
L
mm
L
mm
kN
E
mm
Reaction R
A
kN
σ
σ
−
−
∆
∈=
=
=
×
= ∈=
×
×
=
=
=
×
=
0.0131
L u
∆ =
=
E
4
5
5
2
3
0.0043
2.87 10
150
69 2.87 10
1.9803
(1.9803 10 )(200) 0.396
L
L
kN
E
mm
R
A
k
σ
σ
−
−
−
∆
∈=
=
=
×
= ∈=
×
×
=
=
=
×
=
0.0043
L u
N
∆ =
=
lement (3)
t (3) is at an angle 30
0
, the change in the length is found by adding the
E
Since elemen
displacement components of nodes 2 and 3 along the element (at 30
0
). Thus,
ME 273 Lecture Notes © by R. B. Agarwal
3-14
DRAFT –1
Chapter 3 Truss Element
0
0
0
5
6
4
0
0
5
5
2
cos 30
sin 30
cos 30
0.0131cos 30
0.0502sin 30
0.0043cos 30
0.0116 (
)
0.0116
3.87 10
300
207
3.87 10
0.0080
,
0.0080 100
0.800
L u
u
u
L
element is compressed
L
L
kN
E
mm
Axial force R
A
kN
σ
σ
−
−
∆ =
+
−
∆ =
−
+
= −
∆
−
∈=
=
= −
×
= ∈=
× −
×
= −
=
= −
×
= −
0
Factor of Safety
Factor of safety ‘
n’
is the ratio of yield strength to the actual stress found in the part.
0.0375
(1)
10.8
0.00347
0.0375
(2)
18.9
0.00198
0.0586
(3)
7.325
0.0080
y
y
y
S
Element
n
S
Element
n
S
Element
n
σ
σ
σ
=
=
=
=
=
=
=
=
=
The lowest factor of safety is found in element (3), and therefore, the steel bar is the most
likely to fail before the aluminum bar does.
Final Notes
- The example presented gives an insight into how the element analysis works. The
example problem is too simple to need a computer based solution; however, it
gives the insight into the actual FEA procedure. In a commercial FEA package,
solution of a typical problem generates a very large stiffness matrix, which will
require a computer assisted solution.
- In an FEA software, the node and element numbers will have variable subscripts
so that they will be compatible with a computer-solution
- Direct or equilibrium method is the earliest FEA method.
ME 273 Lecture Notes © by R. B. Agarwal
3-15