01-21-2010, 12:12 PM
Columns are structural elements that sustain load in compression/tension and in flexion. As a result, the structural strength of column is comprised of two components.
Lets look at the classical equation for the capacity of column expressed as the bearing capacity or allowable stress. This is expressed as:-
σ = p/A (compression) OR σ = p/A (tension)
This equation could be rewritten as:-
σ = p/A OR σ = p/A
Where:-
σ = stress
p = applied vertical load
A = cross section area of the column
M = applied moment
y = the distance of the most stressed fiber from the neutral axis
I = moment of inerter of the column section
Z = section modulus = I/y
From this equation, it is obvious that the area of the given section will be same despite the orientation of the column. What can change with the orientation of the column are:-
y ( the distance of the most stressed fiber from the neutral axis)
I ( moment of inerter of the column section).
So if the column is so oriented that the longer side is in the same direction as the span of the primary (principal) beam, the moment of inerter will be greater as such the capacity of the column will be greater in this direction than it will be when the shorter side is aligned with the span of the primary (principal) beam, in which case the moment of inerter will be minimal.
Lets consider a 300mm x 600mm concrete column. When oriented such that the longer side (b = 300mm, d = 600mm, y = 600/2 = 300) is aligned with the direction of span of the principal beam, the moment of inerter I = = (300 x 600^3)/12 = 5.4 10^9
This implies that the section modulus Z = 5.4 10^9/300 = 1.8 x 10^7. For the regular case in which we have a compressive force and moment, stress experienced σ = p/A
= p/A = p/A +
If we reverse the orientation of the column i.e. with the shorter side aligned in the direction of the span of the principal beam, (b = 600mm, d = 300mm, y = 300/2 = 150mm), then Z = ( 600 x 300^3)/(12 x 150) = 9 x 10^6. This implies that the stress on the section is σ = p/A
= p/A = p/A +
Since in either of the two orientations, A = area is same, vertical load (P) is the same and the moment (M) is the same, then the ratio of the stress experienced by the section when oriented with the longer side in alignment to the direction of the span of the primary (principal) beam and that in case the orientation is reversed is given as:- (p/A + )/( p/A + ). From this, it is obvious that the orientation of the column could determine the carrying capacity of a given column.
To boil down further this discussion, assume a vertical load p = 100 kN and moment M = 10KNM
Again A = 600 x 300 = 180000
For the first case (b =300mm, d = 600mm), σ = p/A
= + = 0.56 + 0.02 = 0.58N/mm^2
For the second case (b =600mm, d = 300mm), σ = p/A
= + = 0.56+1.11=1.67N/mm^2. Then the ratio of:-
= = 0.35
This is to say that the first case will experience a stress that is only about a third of that experienced in the second
This confirm that the orientation of the column goes to a great length in determining the load bearing capacity of a column as such should be chosen with care. The orientation will also influence the choice of the modification factor which is to be employed towards the determination of the effective length of the column that could in turn, be used in the classification of the column as short, intermediate or long slender column which will also influence the options for its design. How do you chose the initial sizes of your column? Well this is a bit a difficult question to answer. A general guide is to take a look at the loads (vertical load, shear force and the moment), the geometric form of the column, concrete grade and the its effective length. If you are using a soft ware, the software will help you in making your choice. But if you are making a hand calculation, first calculate the critical load as to ascertain that the a trial size that you have chosen will be able to withstand the load, then calculate the stress and compare it with the design or allowable stress. If yes, terminate the design or move to the next phase, if no chose a new trial section. Note that for the sake of economy, the resultant stress that the column is to sustain should be very close the upper limit as allowed in the relevant code.
Regards
Teddy
Lets look at the classical equation for the capacity of column expressed as the bearing capacity or allowable stress. This is expressed as:-
σ = p/A (compression) OR σ = p/A (tension)
This equation could be rewritten as:-
σ = p/A OR σ = p/A
Where:-
σ = stress
p = applied vertical load
A = cross section area of the column
M = applied moment
y = the distance of the most stressed fiber from the neutral axis
I = moment of inerter of the column section
Z = section modulus = I/y
From this equation, it is obvious that the area of the given section will be same despite the orientation of the column. What can change with the orientation of the column are:-
y ( the distance of the most stressed fiber from the neutral axis)
I ( moment of inerter of the column section).
So if the column is so oriented that the longer side is in the same direction as the span of the primary (principal) beam, the moment of inerter will be greater as such the capacity of the column will be greater in this direction than it will be when the shorter side is aligned with the span of the primary (principal) beam, in which case the moment of inerter will be minimal.
Lets consider a 300mm x 600mm concrete column. When oriented such that the longer side (b = 300mm, d = 600mm, y = 600/2 = 300) is aligned with the direction of span of the principal beam, the moment of inerter I = = (300 x 600^3)/12 = 5.4 10^9
This implies that the section modulus Z = 5.4 10^9/300 = 1.8 x 10^7. For the regular case in which we have a compressive force and moment, stress experienced σ = p/A
= p/A = p/A +
If we reverse the orientation of the column i.e. with the shorter side aligned in the direction of the span of the principal beam, (b = 600mm, d = 300mm, y = 300/2 = 150mm), then Z = ( 600 x 300^3)/(12 x 150) = 9 x 10^6. This implies that the stress on the section is σ = p/A
= p/A = p/A +
Since in either of the two orientations, A = area is same, vertical load (P) is the same and the moment (M) is the same, then the ratio of the stress experienced by the section when oriented with the longer side in alignment to the direction of the span of the primary (principal) beam and that in case the orientation is reversed is given as:- (p/A + )/( p/A + ). From this, it is obvious that the orientation of the column could determine the carrying capacity of a given column.
To boil down further this discussion, assume a vertical load p = 100 kN and moment M = 10KNM
Again A = 600 x 300 = 180000
For the first case (b =300mm, d = 600mm), σ = p/A
= + = 0.56 + 0.02 = 0.58N/mm^2
For the second case (b =600mm, d = 300mm), σ = p/A
= + = 0.56+1.11=1.67N/mm^2. Then the ratio of:-
= = 0.35
This is to say that the first case will experience a stress that is only about a third of that experienced in the second
This confirm that the orientation of the column goes to a great length in determining the load bearing capacity of a column as such should be chosen with care. The orientation will also influence the choice of the modification factor which is to be employed towards the determination of the effective length of the column that could in turn, be used in the classification of the column as short, intermediate or long slender column which will also influence the options for its design. How do you chose the initial sizes of your column? Well this is a bit a difficult question to answer. A general guide is to take a look at the loads (vertical load, shear force and the moment), the geometric form of the column, concrete grade and the its effective length. If you are using a soft ware, the software will help you in making your choice. But if you are making a hand calculation, first calculate the critical load as to ascertain that the a trial size that you have chosen will be able to withstand the load, then calculate the stress and compare it with the design or allowable stress. If yes, terminate the design or move to the next phase, if no chose a new trial section. Note that for the sake of economy, the resultant stress that the column is to sustain should be very close the upper limit as allowed in the relevant code.
Regards
Teddy