How Does End Fixation Affect Slenderness Ratio and Axial Load in Steel Columns?

[beegcheef]

Hi, I have the following question to complete and i was wondering if anyone can help..

Compare the slenderness ratio and maximum safe axial load that a universal column of section 152 x 152 x 37 can carry if it is made from grade 43 steel to BS5950; 1985 and 6.5m in height when its ends are fixed as follows:

a) both ends pin-jointed, and
b) one end direction-fixed and other end pin-jointed.

i know the effective length for both ends pin-jointed is l=L and one end direction_fixed, other pin-jointed is l=0.85L. I have to use that with the formula:

S.R = l/k where S.R is slenderness ratio, l= effective length and k = least radius of gyration.

I also have these formulae:

I=k²A where I = second moment of area, k= radius of gyration and A = cross-sectional area.

F=σA where F= safe axial load, σ= allowable stress and A=cross sectional area

I also got a table from BS5950:Part 1:1985 that shows the British Standards recommendations for allowable slendreness ratio

If anyone could help i would appreciate it greatly

thanx in advance

 

[FredGarvin]

Your question makes little sense.

Compare the slenderness ratio and maximum safe axial load that a universal column of section 152 x 152 x 37 can carry if it is made from grade 43 steel to BS5950; 1985

It's obvious that you want to compare the buckling loads of a column to a standard...

and 6.5m in height when its ends are fixed as follows

...but what does the rest of the question mean? Can you write the question dowwn exactly as it is written?

 

[beegcheef]

Hi Fred, thanks for replying to my post. however, that is exactly how the question is written word for word.

Compare the slenderness ratio and maximum safe axial load that a universal column of section 152 x 152 x 37 can carry if it is made from grade 43 steel to BS5950; 1985 and 6.5m in height when its ends are fixed as follows:

a) both ends pin-jointed, and
b) one end direction-fixed and other end pin-jointed.

thanx

 

[FredGarvin]

The only thing I am wondering is in the dimensions of the column. You give 152x152x37. Is the 37 a wall thickness since you give the height as 6.5m? I am assuming so.

For some reason I read that problem about 20 times and it didn't click. Now it is. Perhaps I needed more coffee.

So are you wondering where to start? Basically, you are looking at a change in restraints for two scenarios. Since the cross section doesn't change, the radius of gyration will not change. So, you have 2 boundary conditions. That means you should calculate the maximum safe load for 2 situations (4 if the BS spec also specifies a different material than the grade 43). Since I don't have access to the BS spec you referenced, I would assume that it spells out some minimums in the safe load. Take your 2 results and then compare them to the spec.

It would be a lot easier to help if you had some specific questions.

 

[beegcheef]

thanx for your reply... i was sort of stuck on the dimensions aswell. I have done the following calculations and they do fall in the table i got but if i could run it past you could you tell me if it makes sense. here it is:

I=bd³/12 = 37 x 152³/12 = 11.93 x 106mm4(sorry i don't have a "ten to the power of 6"symbol)

A=152 x 37 = 5624 mm²

K=√I/A = √11930000/5624 = 46mm


for a) both ends pin-jointed

l=L =1x6.5 = 6.5m = 6.5 x 10³ mm

S.R=l/K = 6.5 x 10³/46 = 141.3

for b) one end direction-fixed and other end pin-jointed

l=L =0.85 x 6.5 = 5.5m = 5.5 x 10³ mm

S.R=l/K = 5.5 x 10³/46 = 119.5

on the table i got for slenderness ratio of 140 the allowable stress for flange thickness 17 - 40 mm is 83

for slenderness ratio of 120 the allowable stress for flange thickness 17 - 40 mm is 107

does this sound like I am on the right track here? thanks for your help

 

[FredGarvin]

Reexamine your calculation for the area. If it is a square hollow column, you'll need to subtract the area of the smaller area from the area of the larger.

 

[beegcheef]

Ok.. I've got some figures from the british standards universal column table. it isn't hollow but is an "H" column. so...

for serial size 152 x 152 x 37:

Cross sec. area (A) = 47.1cm = 471mm²
Radius of gyration (k) = 38.7mm

so slenderness ratio for a):

S.R = l/k = 6.5 x 10³/38.7 = 168
for flange thickness of 11.5 the allowable stress is 62 Nmm-²

and for b):

S.R = l/k = 5.5 x 10³/38.7 = 142
for flange thickness of 11.5 the allowable stress is 83 Nmm-²

then the maximum safe axial load for a):

F=ơA = 62 x 471 = 29202N = 29KN

and for b):

F=ơA = 83 x 471 = 39093N = 39KN


does this sound like it to you? once again thanks very much

 

[beegcheef]

Ok.. I've got some figures from the british standards universal column table. it isn't hollow but is an "H" column. so...

for serial size 152 x 152 x 37:

Cross sec. area (A) = 47.1cm = 471mm²
Radius of gyration (k) = 38.7mm

so slenderness ratio for a):

S.R = l/k = 6.5 x 10³/38.7 = 168
for flange thickness of 11.5 the allowable stress is 62 Nmm-²

and for b):

S.R = l/k = 5.5 x 10³/38.7 = 142
for flange thickness of 11.5 the allowable stress is 83 Nmm-²

then the maximum safe axial load for a):

F=ơA = 62 x 471 = 29202N = 29KN

and for b):

F=ơA = 83 x 471 = 39093N = 39KN


does this sound like it to you? once again thanks very much

 

[FredGarvin]

I would believe those numbers. It looks like you have the right approach. I do have to ask where did your flange thickness of 11.5mm came from?

Now ask yourself if the two end allowable load results make sense to you.

 

[beegcheef]

I got the 11.5mm flange thickness from the same british standards table for that particular column.

do you mean by the two end results as in the 29N and the 39N or do each of the results make sense with how they are fixed (both pin-jointed, etc...)?