Answer
$n_{d}$ = 1.43 Ans
Work Step by Step
Problem 1-9
Given:
Max. load = 1.10 P
Min. area = $(0.95)^{2}$ A
Min. strength = 0.85 S
Analysis:
We will use the design factor from eq (1-1)
$n_{d}$=$\frac{loss-of-function-parameter}{maximum
-allowable-parameter}$
$n_{d}$=$\frac{1.10}{(0.85)(0.95^{2})}$
$n_{d}$ = 1.43 Ans
