Question 16.2: Determine the longest unsupported length L for which  the S1...

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