7. The cable supports two independently and normally distributed forces F1 ~ N(15, 22) Nand F2 ~ N(10,
12) N shown below. If the maximum tension of the cable is 20 N, determine the reliability of the system.
A
45°
D
45°
C
B
30°
F2
F1
Solution
TAB
y
TCD
y
TBC
x
x
TBC
F2
F1
Connector B:
F
x
 0; TBC cos30  TAB cos 45
F
y
 0; TBC sin 30  TAB sin 45  F1
Solve the above equations, we have
TAB  0.897 F1
TBC  0.732F1
Connector C:
F
x
 0; TBC cos30  TCD cos 45  F2 cos 45
Solve it, we have
TCD  0.897 F1  F2
Therefore, we can obtain the distributions of all forces
TAB ~ N (13.45,1.792 ) N , TBC ~ N (10.98,1.4642 ) N and TCD ~ N (3.46, 2.052 ) N
The reliability can be determined by if the largest force is smaller than the allowable tension, thus we
have:
R  P(Y  0) , where Y  FAB  20 .
Finally, we have R  99.99%
ANS.