Page 276 - Shigley's Mechanical Engineering Design

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Failures Resulting from Static Loading 251
Failure will occur when the stress is greater than the strength, when ¯n < 1, or when
y < 0.

2 2
ln μ n − ln 1 + C ln μ n / 1 + C
0 − μ y μ y n n
z = =− =− ˙ =− (5–44)
ln 1 + C n ln 1 + C n
ˆ σ y σ y 2 2
Solving for μ n gives

. C n

μ n =¯n = exp −z ln 1 + C n 2 + ln 1 + C 2 n = exp C n − z + (5–45)

2
Equations (5–42) and (5–45) are remarkable for several reasons:
• They relate design factor ¯n to the reliability goal (through z) and the coefficients of
variation of strength and stress.
• They are not functions of the means of stress and strength.
• They estimate the design factor necessary to achieve the reliability goal before
decisions involving means are made. The C S depends slightly on the particular
material. The C σ has the coefficient of variation (COV) of the load, and that is gen-
erally given.

EXAMPLE 5–8 A round cold-drawn 1018 steel rod has an 0.2 percent yield strength S y = N(78.4, 5.90)
kpsi and is to be subjected to a static axial load of P = N(50, 4.1) kip. What value of
the design factor ¯n corresponds to a reliability of 0.999 against yielding (z =−3.09)?
Determine the corresponding diameter of the rod.
Solution C S = 5.90/78.4 = 0.0753, and

P 4P
= =
A πd 2
Since the COV of the diameter is an order of magnitude less than the COV of the load
or strength, the diameter is treated deterministically:

4.1
C σ = C P = = 0.082
50
From Eq. (5–42),

2
2
2
1 1 [1 ( 3.09) (0.0753 )][1 ( 3.09) (0.082 )]
2
n 1.416
2
2
1 ( 3.09) (0.0753 )
The diameter is found deterministically:
Answer
4P ¯ 4(50 000)
d = = = 1.072 in
πS y /¯n π(78 400)/1.416
¯