Question 16.2: Determine the longest unsupported length L for which the S1...
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. 16.22). Assume σ_{Y} = 250 MPa and E = 200 GPa.
From App. C we find that, for an S100 × 11.5 shape,
A = 1460 mm^{2} r_{x} = 41.7 mm r_{y} = 14.6 mm
If the 60-kN load is to be safely supported, we must have
σ_{all} = \frac{P}{A} = \frac{60 × 10^{3} N}{1460 × 10^{-6} m^{2}} = 41.1 × 10^{6} Pa
We must compute the critical stress σ_{cr}. Assuming L/r is larger than the slenderness specified by Eq. (16.25), we use Eq. (16.24) with (16.23) and write
σ_{e} = \frac{π^{2}E}{(L/r)^{2}} (16.23)
σ_{cr} = 0.877σ_{e} (16.24)
\frac{L}{r} = 4.71 \sqrt{\frac{E}{σ_{Y}}} (16.25)
σ_{cr} = 0.877 σ_{e} = 0.877 \frac{π^{2}E}{(L/r)^{2}}
= 0.877 \frac{π^{2}(200 × 10^{9} Pa)}{(L/r)^{2}} = \frac{1.731 × 10^{12} Pa}{(L/r)^{2}}
Using this expression in Eq. (16.26) for σ_{all}, we write
σ_{all} = \frac{σ_{cr}}{1.67} (16.26)
σ_{all} = \frac{σ_{cr}}{1.67} = \frac{1.037 × 10^{12} Pa}{(L/r)^{2}}
Equating this expression to the required value of σ_{all}, we write
\frac{1.037 × 10^{12} Pa}{(L/r)^{2}} = 1.41 × 10^{6} Pa L/r = 158.8
The slenderness ratio from Eq. (16.25) is
\frac{L}{r} = 4.71 \sqrt{\frac{200 × 10^{9}}{250 × 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 × 10^{-3} m} = 158.8 L =2.32 m