Determine the allowable compressive stress for a steel column?

Question

Determine the allowable compressive stress for a steel column 254x254 by 198kg/m. The column is 7m long and is fixed at both ends.

Answer 1

The column section is 254 mmx 254 mm or 10" x 10" . Weight in kg/m = 198 kg .
for solid square section wt/m would be 0.254 x0.254 x 7.85 = 0.5064506 ton/m = 506 kg/m .
So, it may be a fabricated hollow square section and is not commercially produced .
Thickness of walls( t )will be 0.198 / (0. 254x 4x 7.85 ) m or 24.8257 mm say 1".
Section properties/slenderness ratio are primary requirement to deal with columns and is to be calculated first or to be obtained from chart.
In absence of that one can only proceed with certain assumptions. Here are those applicable :-

For a carbon steel compressive member (l/r<30) ,
allowable direct compressive stress = 17,000 psi or 1190 kg/sq.cm.

For a carbon seel short column ( l/r>30 but <120),
allowable compressive stress = 17,000 - 0.4875 ( l/r) psi. ( ASCI ) )

For a carbon steel long columns( l/r >120 ),
allowable compressive stress < critical stress
( P/ A } = { n X3.142^2 xE I / L^2 } { Euler }
You may find the answer from above information.


Guess /apprximation :
column section radii of gyration < (9 / 12 )^ 0.5 " or 0.866 " or 22 mm
column length = 7 m or 275.52"
slenderness ratio l/r > 275.52/ 0.866 = 318 >120
the column seems to be a long column ,
L= 7 M. = 700 cm.
n = 4 , both ends fixed., axial loading considered.
section = 254 mm x 254 mm x25.4 mm , hollow box - section (guessed)
E = 2 X 10^6 kg/cm^2
find I , r , area A from chart for hollow square section with stiffeners and put these values in Euler's formula to find allowable compressive stress for the column .
Do not forget to deduct column self weight to arrive at the safe load figure.
Fill the hollow with concrete , to strenghten the column . It will be further economic and will carry much higher load than calculated above.

Answer 2

Allowable Compressive Stress

Answer 3

You will have to calculate the L/r ratio of the steel column to determine the allowable buckling strength.

Answer 4

this is far more complicated than you've allowed for in your question. the axial capcity for such a long column will be governed by buckling in the member which needs careful consideration

Answer 5

what is the equivalent American shape? I could give you an answer then.

254 is 10 inches?
198 kg/m =133 plf?

Our steel book has no such shape.

Answer 6

400 joules

Answer 7

depends which steel code you are using