130
Chapter 2
Divide top and bottom by s and set N= ~/~ for structural factor of
safety
t2 _ (N- 1)
2
(2.202)
(CvsN) 2 q- C2vs
from Eq. (2.194) C~s = (0.0266433) 2 = 0.000709
If N= 1 Cvs= +0.1395 Eq. (2.188) then
( CvsN) 2 = 0.01946
In this case dropping Cv2~ causes in the denominator the following
percent error for N= 1
1 (C~sN) 2 + (C~s)2. x 100 = 3.52%
Now drop C~. in Eq. (2.202) and take the square root
N-1
t -
(2.203)
CvsN
with C~s= :1:0.1395 Eq. (2.188)
Now
t = :t:7.168 N - 1
(2.204)
N
If a safety factor of 3 is desired
t = -~ (+7.168)
t = -t-4.7787
Table 2.3
ef 10~
6
Now when t= 4-7.168 in Eq. (2.199) substituting in Eq. (2.204)
-t- 7.168 = 4-7.168-+ N = 4-[U- 1]
for + sign
N=N-1
N-1
N