## DESIGN FORMULAS FOR COLUMNS

## DESIGN FORMULAS FOR COLUMNS

(OP)

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

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

## RE: DESIGN FORMULAS FOR COLUMNS

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.

## RE: DESIGN FORMULAS FOR COLUMNS

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

## RE: DESIGN FORMULAS FOR COLUMNS