Question 5.8: A round cold-drawn 1018 steel rod has an 0.2 percent yield s...

C_{S} = 5.90/78.4 = 0.0753, and

σ =\frac {P}{A} =\frac {4P}{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:

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

From Eq. (5–42),

\overline {n} =\frac {1 + \sqrt {1 − \left (1 − z^{2}C^{2}_{S}\right )  \left ( 1 − z^{2}C^{2}_{σ})\right)}}{1 − z^{2}C^{2}_{S}}               (5–42)

\overline {n} =\frac {1 ± \sqrt {1 − \left [1 − (3.09)^{2}(0.0753^{2} )\right ]  \left [ 1 − (3.09)^{2}(0.082^{2})\right]}}{1 − (3.09)^{2}(0.0753^{2})}

The diameter is found deterministically:

d= \sqrt{\frac {4\overline {P}}{π \overline{Sy}/ \overline {n}}}=\sqrt {\frac {4(50 000)}{π(78 400)/1.416}}=1.072  in

\overline{Sy}= N(78.4, 5.90) kpsi, P = N(50, 4.1) kip, and d = 1.072 in. Then

A =\frac {πd^{2}}{4} =\frac {π(1.072^{2})}{4} = 0.9026  in^{2}

\overline {σ} =\frac {\overline{P}}{A} =\frac {(50 000)}{0.9026} = 55 400  psi

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

\hat {σ}_{σ} = C_{σ} \overline {σ} = 0.082(55 400) = 4540  psi

\hat {σ}_{S} = 5.90  kpsi

From Eq. (5–40)

z =\frac {m − μ_{m}}{\hat {σ}_{m}} =\frac {0 − μ_{m}}{\hat {σ}_{m}} = − \frac {μ_{m}}{\hat {σ}_{m}} =− \frac {μ_{S} − μ_{σ}}{(\hat {σ}^{2}_{S} + \hat {σ}^{2}_{σ})^{1/2}}               (5–40)

z = −\frac {78.4 − 55.4}{(5.90^2 + 4.54^2)^{1/2}}= −3.09

From Appendix Table A–10, R = (−3.09) = 0.999.

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