1. Example(3):
Determine the maximum stress in the
2200 lb
reinforced weld of the bracket plate in fig.
2 in 2 in 4.5 in
(7.6) . Assume that the load varies from zero to
the maximum value
Solution:
The throat area of each weld is:
A= 0.707×0.25× (6 – 0.5) ×1.2 = 1.16
6 in
2
in
For each weld, r1=2 in, and the total polar O
second moment of area is:
l2 5.5 2 2
J = 2A ×
12 + r1 2 = 2 x1.16 ×
12 + 2 = 15.1 in ¼ in weld
The torque is
T = 2200 × 6.5 = 14300 lb-in Fig. (7.6)
The distance r in equation (7.4) is
r = 2.75 2 + 2 2 = 3.35 in
Hence, the maximum torsional stress, by equation (7.4), is
14300 × 3.35
s= = 3172.5 psi
15.1
This stress is resolved into a vertical component
3172.5 × 2
sv = = 1894 psi
3.35
And a horizontal component
3172.5 × 2.75
sh = = 2604.3 psi
3.35
The direct stress has only a vertical component,
F 2200
s vd = = = 948.3 psi
2A 2 ×1.16
The total vertical stress is
svt = 1894 +948.3 = 2842.3 psi
Therefore the resultant stress is
s = 2842.3 2 + 2604.3 2 = 3855 psi
With a concentration factor of k’ = 2.7 from table (7.2) the significant stress is
smax = 3855 × 2.7 = 10408.5 psi