Civil Engineering Reference
In-Depth Information
Figure 5-14
Force Density Example
(Drawing by the author)
The force density can be selected as any convenient value. The value can be the
same for all elements or it can vary. In meshes with no applied load, 1.0 is often
chosen for the initial value. In this example, with an applied load, the initial value
will be chosen to be representative of the force density anticipated.
q
i
= F
i
/L
i
= 10.0/2000 = 0.005 for all I
(5.5.2-1)
Find:
C (X
C
,Y
C
)
F
i
For all bars
Solution:
F
= 0 For node C
(ref 5.1.3-1)
=> F
1
(X
A
-X
C
)/L
1
+ F
2
(X
B
-X
C
)/L
2
+ F
3
(X
D
-X
C
)/L
3
= 0
F
= 0 For node C
(ref 5.1.3-2)
=> F
1
(Y
A
-Y
C
)/L
1
+ F
2
(Y
B
-Y
C
)/L
2
+ F
3
(Y
D
-Y
C
)/L
3
- P
C
= 0
Note: The equations in the X-direction are uncoupled from the equations in the Y-
direction.
Let F
i
/L
i
= g
i
where g
i
is the Force Density
(5.5.2-2)
Rearranging:
-(g
1
+g
2
+g
3
)X
C
+ g
1
X
A
+ g
2
X
B
+ g
3
X
D
- P
CX
= 0
-(g
1
+g
2
+g
3
)Y
C
+ g
1
Y
A
+ g
2
Y
B
+ g
3
Y
D
- P
CY
= 0
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