1-2. The two aluminum rods support the vertical force of P
stress for the aluminum is Sa
N(30,42 )kN . The allowable tensile
N(150, 202 )Pa . Determine the diameters of the two rods to
make sure that the probabilities of failure of them both are less than 104 . Assume that P and Sa
are independent.
C
60°
A
B
P
Fig. 1.2.1
Solution
Internal Loadings: The force developed in each rod can be determined by using the method of
joints. Consider the equilibrium of joint A as shown in Fig. 1.2.2,
y
FAC
   Fy  0 ; FAC  sin 60  P  0 ;
  Fx  0 ;
FAC  1.155P
FAB
60°
x
 FAC  cos 60  FAB  0 ; FAB  0.5FAC =0.577P
P
For rod AB, the probability of failure p f is
Fig. 1.2.2
F
0.577 P




p f  Pr( S AB  Sa )  Pr(Y  Sa  S AB  0)  Pr Y  Sa  AB  0   Pr Y  Sa 
 0  (1)
A
A




Since P ~ N(30,42 )kN , Sa ~ N(150,2 )MPa , and P and Sa are independent, Y also follows a
normal distribution, Y ~ N (Y ,  Y2 ) .



0.577 
 0.577 
4
Y  Sa  
  P  150    d 2  3  10
 A 
AB 


 4 
(2)
2


2

0.577 
2
 0.577  2
2
 Y   S2a  


20

4

 P
2   
 A 
  d AB 
 4 
(3)
Equation (1) can be written as
 Y  Y  Y 
  Y 
4
p f  PrY  0   Pr

  
  10    3.719 
Y 
 Y
 Y 
Therefore,
 Y
Y
 3.719
(4)
From (1), (2) and (3), we can get d AB  8.33mm .
Ans.
Similarly, For rod AC, the probability of failure p f is
F
1.155P




p f  Pr( S AC  Sa )  Pr(Y  Sa  S AC  0)  Pr Y  Sa  AC  0   Pr Y  Sa 
 0  (5)
A
A




we can get d AC  11.786mm .
Ans.