7. Rod has a mass which follows a normal distribution M~N(40,1.52) g. Determine the distributions of
reactions on the rod.
B
85 mm
220 mm
45°
A
Solution
45°
Ay
M
45°
 M  0;
 F  0;
 F  0;
A
NB
Ax
N B (0.22)  M (9.81)  (0.22  0.085) / 2  cos 45  0;
x
Ax  N B sin45  0;
y
Ay  M (9.81)  N B cos 45  0.
Since the rod mass follows the normal distribution M~N(0.04,0.00152) kg, we have
N 
B
N
B
 M (9.81)(0.305 / 2)cos 45
 0.19 N
0.22
 (9.81)(0.305 / 2)cos 45
 M
 7.2  103
0.22
 A   N sin 45  0.19sin 45 =0.14 N
x
B
 A   N sin 45  7.2 103 sin 45 =5.1 103
x
B
After known the distributions of N B and Ax , we have
 A  M (9.81)   N cos 45  0.26 N
y
A 
y
B

NB
cos 45
  
2
(9.81)   1.6  102
2
M
Thus, the support distributions respectively as, NB ~ N(0.19, (7.2×10-3)2) N, Ax ~ N(0.14, (5.1×10-3)2) N
and Ay ~ N(0.26, (1.6×10-2)2) N.
Ans.