Question 5.9: Rework Ex. 5–8 with lognormally distributed stress and stren...

C_{S} = 5.90/78.4 = 0.0753, and  C_{σ} = C_{P} = 4.1/50 = 0.082. Then

σ =\frac {P}{A} =\frac {4P}{πd^{2}}

C_{n} =\sqrt { \frac {C^{2}_{S} + C^{2}_{σ}}{1 + C^{2}_{σ}}} =\sqrt { \frac {0.0753^{2} + 0.082^{2}}{1 + 0.082^{2}}} =0.1110

From Table A–10, z = −3.09. From Eq. (5–45),

μ_{n} = \overline {n} = exp \left [−z \sqrt {ln \left (1 + C^{2}_{n}\right)} + ln \sqrt {1 + C^{2}_{n}} \right] \dot{=} exp \left [C_{n} \left (− z +\frac {C_{n}}{2} \right) \right]                     (5-45)

n = exp \left[ −(−3.09) \sqrt {ln(1 + 0.111^{2})}+ln \sqrt {1 + 0.111^{2}} \right]=1.416

d= \sqrt{\frac {4\overline {P}}{π \overline{Sy}/ \overline {n}}}=\sqrt {\frac {4(50 000)}{π(78 400)/1.416}}=1.0723  in

Check

S_{y} = LN(78.4, 5.90), P = LN(50, 4.1) kip. Then

A =\frac {πd^{2}}{4} =\frac {π(1.0723^{2})}{4} = 0.9031  in^{2}

\overline {σ} =\frac {\overline{P}}{A} =\frac {(50 000)}{0.9031} = 55 365  psi

C_{P} = C_{σ} =\frac {4.1}{50} = 0.082

\hat {σ}_{σ} = C_{σ} \overline {σ} = 0.082(55 367) = 4540  psi

From Eq. (5–43),

z = − \frac {μ_{ln S} − μ_{ln σ}}{(\hat{σ}^{2}_{ln S} + \hat {σ}^{2}_{ln σ})^{1/2}}=-\frac {ln \left(\frac {μ_{S}}{μ_{σ}} \sqrt {\frac {1 + C^{2}_{σ}}{1 + C^{2}_{S}}}\right)}{\sqrt {ln\left [ \left(1 + C^{2}_{S}\right) \left (1 + C^{2}_{σ}\right)\right]}}                   (5-43)

z =-\frac {ln \left(\frac {78.4}{55.365} \sqrt {\frac {1 + 0.082^{2}}{1 + 0.0753^{2}}}\right)}{\sqrt {ln\left [ \left(1 + 0.0753^{2}\right) \left (1 + 0.082^{2}\right)\right]}}= −3.1343

Appendix Table A–10 gives R = 0.99950.

Table A–10
Cumulative Distribution Function of Normal (Gaussian) Distribution