Question 16.CA.2: Determine the longest unsupported length L for which the S10...
Determine the longest unsupported length L for which the S100 × 11.5 rolled-steel compression member AB can safely carry the centric load shown (Fig. 16.23). Assume \sigma_Y = 250 MPa and E = 200 GPa.
From Appendix D, for an S100 × 11.5 shape,
A = 1460 mm² r_x = 41.7 mm r_y = 14.6 mm
If the 60-kN load is to be safely supported,
\sigma_{\text {all }}=\frac{P}{A}=\frac{60 \times 10^3 N}{1460 \times 10^{-6} m^2}=41.1 \times 10^6 PaTo compute the critical stress \sigma_{cr}, we start by assuming that L/r is larger than the slenderness specified by Eq. (16.25). We then use Eq. (16.24) with Eq. (16.23) and write
\frac{L}{r}=4.71 \sqrt{\frac{E}{\sigma_Y}} (16.25)
\sigma_{cr} =0.877 \sigma_e (16.24)
\sigma_e =\frac{\pi^2 E}{(L / r)^2} (16.23)
\begin{aligned}\sigma_{cr} &=0.877 \sigma_e=0.877 \frac{\pi^2 E}{(L / r)^2} \\&=0.877 \frac{\pi^2\left(200 \times 10^9 Pa\right)}{(L / r)^2}=\frac{1.731 \times 10^{12} Pa}{(L / r)^2}\end{aligned}Using this expression in Eq. (16.26),
\sigma_{\text {all}}=\frac{\sigma_{cr}}{1.67}=\frac{1.037 \times 10^{12} Pa}{(L / r)^2}Equating this expression to the required value of \sigma_{\text {all}} gives
\frac{1.037 \times 10^{12} Pa}{(L / r)^2}=41.1 \times 10^6 Pa \quad L / r=158.8The slenderness ratio from Eq. (16.25) is
\frac{L}{r}=4.71 \sqrt{\frac{200 \times 10^9}{250 \times 10^6}}=133.2Our assumption that L/r is greater than this slenderness ratio is correct. Choosing the smaller of the two radii of gyration:
\frac{L}{r_y}=\frac{L}{14.6 \times 10^{-3} m}=158.8 \quad L=2.32 m