design formulas for columns
 

design formulas for columns
design formulas for columns
we have some new calculation standards in europe for
steel and i want ask you all about column calculation.
if we have a pipe as follows:
circular pipe

outer diameter d= 101.6 mm or l= 4 in.
wall thickness s=3.175 mm or s= 1/8 = 0.125 in.
column length l= 6096 mm or l= 20 ft
the following is with length in mm and force in n
modulus of elasticity = 210000 n/mm2
material s355 yield stress 355 n/mm2
effective length factor k=1 euler 2 with pinned ends
moment of inertia i = 1190067 mm4
area a = 982 mm2
radius of gyration r= 34.82 mm
slenderness ratio k*l/r = 175
the force type is static weight .
buckling stress with old euler theory is for this
stress = 3.14*3.14*210000 / (175*175 ) = 67.7 n/mm2
we all know that euler theory givs too high values for
small slenderness ratios.
i have learn in the school that if we have high values for
slenderness ratios then we use higher factors of safety
for allowable column stress calculations when we use
the old euler theory. for this case the safety factor
is in older books recommended to be about n=5 so
the allowable stress is then = 67.7 / 5 = 13.5 n/mm2
many newer theories gives higher values , why ?
at the same time aisc givs the upper limit for slenderness ratio to
200 and for eurocode 3 is this limit 250 for steel structures.
i do not use results for some structure !! this is only a "test" to see differencies
for calculation methods.
what is acceptable maximum compression
stress for this column ?
best regards
markku lavi
lamek oy
find a job or post a job opening
is this the same column that was discussed under the heading "column force"?
for a kl/r value of 175, the canadian code permits a resistant stress of 53.7 mpa for regular hss sections and 57.2 mpa for heat relieved hss sections (fy=350 mpa).
the allowable stress is a little involved and is determined by the following method:
lambda = (k*l/r)*sqrt(fy/(pi*pi*e))
phi = 0.9
e = 200,000 mpa
for non-stress relieved sections the value is determined below
for the above calculated lambda value
0 <= lambda <= 0.15
cr = phi*a*fy
0.15 < lambda <= 1.0
cr = phi*a*fy*(1.035-0.202*lambda-0.222*lambda*lambda)
1.0 < lambda <= 2.0
cr = phi*a*fy*(-0.111+0.636/lambda+0.087/(*lambda*lambda))
2.0 < lambda <= 3.6
cr = phi*a*fy*(0.009+0.877/(*lambda*lambda))
3.6 < lambda
cr = phi*a*fy*/(*lambda*lambda)) = phi*a*1970000/((kl/r)*(kl/r)) euler buckling formula (slightly modified)
for stress relieved sections, the cr is determined from values of lambda differing slightly from those above. i don't usually use the stress relieved sections because the cost generally outweighs the slight additional strength gain.
the resistant stress must be greater than or equal to the factored load applied. our load factors are 1.25 for dead load and 1.5 for live load.
research on columns indicates that for very slender columns, the load capacity is closely mirrored by the euler buckling strength.
for short or 'stub' columns, the strength is closely mirrored by the fy, ie., they 'squash'.
there is an intermediate range, covered by the above,where the buckling or failure strength (mixed mode of failure) is affected by the internal stresses.
the resistance of the column, (for non-stress relieved sections, by the canadian steel code is 53.7 mpa and not 13.5 mpa. our factor of safety is approximately 1.3 or 1.4 depending on the ll and dl (further increased by the phi factor). in the years i've designed structures, i've not encountered a factor of 5 (except, maybe for masonry <g>) and i'm not sure where it comes from either for working stress or for limit states.
to dik
it is not the same column, but it is a better
example (dimensions and form ).
we have the following values:
1 older theory gives 29.4 mpa
2 omega-method(old german) 27 mpa
3 eurocode3 load factor 1.2 58.2 mpa
( near euler??? )
regards
markku lavi

the new eurocode is close to the canadian value for heat relieved hss... 57.2 mpa and is likely a good representation of the actual strength.