60
Chapter 2
requirements.
Load = (~b, ~p) = (6000,90)
Tensile ultimate 4130 steel = ~’, }F = (156,000;4300) psi
1
Pfait, re -- 1000 R = 0.999 so t = -3.0912(~ -3)
The cross-sectional area A = ~r
2
The standard deviation ~i = (OA/Or)dr 2rc?~r
We are given from manufacturing
0.015
~r = -t- 2.-- ~ ? for 99% of the samples zr = 4-2.576
~r = 5.83 x 10 -3 ~ ~ 0.005?
The applied stress is
(]}, ~e) (6000,
(6, ~) -- (~4, SA) (~.~2,
~, 6000
from Eq, (2.22)
with
~e = 90 lb
Figure 2.7 A tension sample.