Question 10.02: Determine the longest unsupported length L for which the S10...

From Appendix C we find that, for an \mathrm{S} 100 \times 11.5 shape,

A=1460 \mathrm{~mm}^{2} \quad r_{x}=41.7 \mathrm{~mm} \quad r_{y}=14.6 \mathrm{~mm}

If the 60-\mathrm{kN} load is to be safely supported, we must have

\sigma_{\text {all }}=\frac{P}{A}=\frac{60 \times 10^{3} \mathrm{~N}}{1460 \times 10^{-6} \mathrm{~m}^{2}}=41.1 \times 10^{6} \mathrm{~Pa}

We must compute the critical stress \sigma_{\text {cr }}. Assuming L / r is larger than the slenderness specified by Eq. (10.41),

{\frac{L}{r}}=4.71{\sqrt{\frac{E}{\sigma_{Y}}}}    (10.41)

we use Eq. (10.40)

\sigma_{\mathrm{cr}}=0.877\sigma_{e}       (10.40)

with (10.39)

\sigma_{e}={\frac{\pi^{2}E}{\left(L/r\right)^{2}}}      (10.39)

and write

\begin{aligned} \sigma_{\mathrm{cr}} & =0.877 \sigma_{e}=0.877 \frac{\pi^{2} E}{(L / r)^{2}} \\ & =0.877 \frac{\pi^{2}\left(200 \times 10^{9} \mathrm{~Pa}\right)}{(L / r)^{2}}=\frac{1.731 \times 10^{12} \mathrm{~Pa}}{(L / r)^{2}} \end{aligned}

Using this expression in Eq. (10.42) for \sigma_{\text {all, }}, we write

\sigma_{\text {all }}=\frac{\sigma_{\mathrm{cr}}}{1.67}=\frac{1.037 \times 10^{12} \mathrm{~Pa}}{(L / r)^{2}}

Equating this expression to the required value of \sigma_{\text {all }}, we write

\frac{1.037 \times 10^{12} \mathrm{~Pa}}{(L / r)^{2}}=1.41 \times 10^{6} \mathrm{~Pa} \quad L / r=158.8

The slenderness ratio from Eq. (10.41) is

\frac{L}{r}=4.71 \sqrt{\frac{200 \times 10^{9}}{250 \times 10^{6}}}=133.2

Our assumption that L / r is greater than this slenderness ratio was correct.

Choosing the smaller of the two radii of gyration, we have

\frac{L}{r_{y}}=\frac{L}{14.6 \times 10^{-3} \mathrm{~m}}=158.8 \quad L=2.32 \mathrm{~m}