Question 16.1: A 2-m-long pin-ended column of square cross section is to be...

(a) For the 100-kN Load. Using the given factor of safety, we make

P_{cr} = 2.5(100  kN) = 250  kN               L = 2  m                E = 13  GPa

in Euler’s formula (16.11) and solve for I. We have

P_{cr} =\frac{π^{2}EI}{L^{2}}                         (16.11)

I =\frac{P_{cr}L^{2}}{π^{2}E} = \frac{(250 × 10^{3}  N) (2  m)^{2}}{π^{2} (13 × 10^{9}  Pa)} = 7.794 × 10^{-6}  m^{4}

Recalling that, for a square of side a, we have I = a^{4}/12, we write

\frac{a^{4}}{12} = 7.794 × 10^{-6}  m^{4}                 a = 98.3   mm ≈ 100  mm

We check the value of the normal stress in the column:

σ = \frac{P}{A} = \frac{100  kN}{(0.100  m)^{2}} = 10  MPa

Since σ is smaller than the allowable stress, a 100 × 100-mm cross section is acceptable.

(b) For the 200-kN Load. Solving again Eq. (16.11) for I, but making now P_{cr} = 2.5(200) = 500  kN, we have

I = 15.588 × 10^{-6}  m^{4}

\frac{a^{4}}{12} = 15.588 × 10^{-6}                     a = 116.95  mm

The value of the normal stress is

σ = \frac{P}{A} =\frac{200  kN}{(0.11695  m)^{2}} = 14.62  MPa

Since this value is larger than the allowable stress, the dimension obtained is not acceptable, and we must select the cross section on the basis of its resistance to compression. We write

A =\frac{P}{σ_{all}} = \frac{200  kN}{12  MPa} = 16.67 × 10^{-3}  m^{2}

a^{2} = 16.67 × 10^{-3}  m^{2}                      a = 129.1  mm

A 130 × 130-mm cross section is acceptable.