N (2, 0.252 ) kN is applied to the bell crank. Assuming that the
1-6. A vertical force of F
allowable normal stress at the 10-mm diameter rod CD is Sa1 ~ N(80,2 )MPa , and the allowable
shear stress at the 6-mm diameter pin B, subjected to double shear is Sa 2 ~ N(60,2 )MPa . If
F , Sa1 and S a 2 are independent, determine the probabilities of failure of rod CD and pin B.
D
C
F
B
45o
A
400 mm
Fig. 1.6
Solution
Referring to the free-body diagram of the bell crank shown in Fig. 1.6.1.
FCD
F
By
Bx
Fig. 1.6.1
M B  0;
F (0.4)  FCD (0.26sin 45 )  0;
FCD  2.1757 F ;
Fx  0;
 Bx  FCD  0;
Bx  2.1757 F ;
  Fy  0;
By  F  0;
By  F ;
Thus, the force acting on pin B is FB  Bx2  By2  2.395F .
Since pin B is in double shear, we have VB 
The cross-section area of rod CD is ACD 
plane of pin B is AB 
 (0.0062 )
4
FB
 1.198F .
2
 (0.012 )
4
 78.540 106 m2 , and the area of the shear
 28.274 106 m2 . We obtain the normal stress S1 at rod CD
and shear stress S 2 at pin B as follows
S1 
FCD
2.1757 F

 0.0277 106 F ,
6
ACD 78.540 10
Set Y1  Sa1  S1 , then Y1
S2 
VB
1.198F

 0.0153 106 F .
6
AB 78.540 10
N (Y 1 ,  Y21 )MPa , where
Y 1 =S  S1  S  0.0277 106 F  80 106  55.4 106  24.6MPa
a1
a1
 Y 1   S2   S21  (5 106 )2  (0.0277 106  0.25 103 )2  8.541MPa .
a1
Thus, the probability of failure of rod CD is
p f  Pr(Y1  Sa1  S1  0)  Pr(
Y1  Y 1
Y1
Similarly, set Y2  Sa 2  S2 , then Y2

Y 1
Y1
)  (
Y 1
 Y1
)  (2.88)  2 103 .
N (Y 2 ,  Y22 )MPa , where
Y 2 =S  S 2  S  0.0153 106 F  60 106  30.6 106  29.4MPa
a2
a2
 Y 2   S2   S22  (6 106 )2  (0.0153 106  0.25 103 ) 2  7.116MPa .
a2
Thus, the probability of pin B is
p f  Pr(Y2  Sa 2  S2  0)  Pr(
Y2  Y 2
Y 2

Y 2
Y 2
)  (
Y 2
Y 2
)  (4.132)  1.8 105 .
Ans.