Question 10.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. 10.29). Assume \sigma_Y = 250 MPa and E = 200 GPa.
From Appendix E, for an S100 × 11.5 shape,
A=1460 mm ^2 \quad r_x=41.7 mm \quad 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 Pa
To compute the critical stress \sigma_{ cr } , we start by assuming that L/r is larger than the slenderness specified by Eq. (10.41). We then use Eq. (10.40) with Eq. (10.39) and write
\sigma_e=\frac{\pi^2 E}{(L / r)^2} (10.39)
\sigma_{ cr }=0.877 \sigma_e (10.40)
\frac{L}{r}=4.71 \sqrt{\frac{E}{\sigma_Y}} (10.41)
\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. (10.42),
\sigma_{ all }=\frac{\sigma_{ cr }}{1.67} (10.42)
=\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=158.8
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 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