Page 277 - Shigley's Mechanical Engineering Design

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252 Mechanical Engineering Design
Check S y = N(78.4, 5.90) kpsi, P = N(50, 4.1) kip, and d = 1.072 in. Then
2
πd 2 π(1.072 ) 2
A = = = 0.9026 in
4 4
P ¯ (50 000)
¯ σ = = = 55 400 psi
A 0.9026
4.1
C P = C σ = = 0.082
50
ˆ σ σ = C σ ¯σ = 0.082(55 400) = 4540 psi
ˆ σ S = 5.90 kpsi
From Eq. (5–40)
78.4 − 55.4
z =− =−3.09
2
2 1/2
(5.90 + 4.54 )
From Appendix Table A–10, R =
(−3.09) = 0.999.

EXAMPLE 5–9 Rework Ex. 5–8 with lognormally distributed stress and strength.
Solution C S = 5.90/78.4 = 0.0753, and C σ = C P = 4.1/50 = 0.082. Then
P 4P
= =
A πd 2

2 2
2
C + C σ 0.0753 + 0.082 2
S
C n = 2 = 2 = 0.1110
1 + C 1 + 0.082
σ
From Table A–10, z =−3.09. From Eq. (5–45),

2
¯ n = exp −(−3.09) ln(1 + 0.111 ) + ln 1 + 0.111 2 = 1.416

4P ¯ 4(50 000)
d = = = 1.0723 in
πS y /¯n π(78 400)/1.416
¯
Check S y = LN(78.4, 5.90), P = LN(50, 4.1) kip. Then
2
πd 2 π(1.0723 )
A = = = 0.9031
4 4
P ¯ 50 000
¯ σ = = = 55 365 psi
A 0.9031
4.1
C σ = C P = = 0.082
50
ˆ σ σ = C σ μ σ = 0.082(55 367) = 4540 psi
From Eq. (5–43), ⎛ ⎞
78.4 1 + 0.082 2
ln ⎝ ⎠
55.365 1 + 0.0753 2
=−3.1343
z =−
2
ln[(1 + 0.0753 )(1 + 0.082 )]
2
Appendix Table A–10 gives R = 0.99950.