Hi all,
I have a query regarding orientation of columns in buildings how does it influence the design. Does orientation of column is essential?. And how do you guys initially decide the column sizes?
Hi,
Is it steel column OR RCC?
For steel: (General idea only)
(1) Sq. HSS columns  does not matter
(2) Wide Flange Col:  Interior  Web parallel to Joists
 exteriao  Web perpendicular to exterior wall.
(3) Also based on length & width of building. Wind & EQ force also to be consider in which direction they act.
(11112009, 07:49 PM)king12 Wrote: Hi all,
I have a query regarding orientation of columns in buildings how does it influence the design. Does orientation of column is essential?. And how do you guys initially decide the column sizes?
Ya my intention was for RCC columns
(11112009, 10:52 PM)tgbyhn1234 Wrote: Hi,
Is it steel column OR RCC?
For steel: (General idea only)
(1) Sq. HSS columns  does not matter
(2) Wide Flange Col:  Interior  Web parallel to Joists
 exteriao  Web perpendicular to exterior wall.
(3) Also based on length & width of building. Wind & EQ force also to be consider in which direction they act.
(11112009, 07:49 PM)king12 Wrote: Hi all,
I have a query regarding orientation of columns in buildings how does it influence the design. Does orientation of column is essential?. And how do you guys initially decide the column sizes?
Dear King12,
Yes! The orientation is essential for multistore buildings or 1 store warehouse, for example.
Imagine that you have a rectangular (projection) building, as this very simple example:
Calculate the wind forces in X an Y direction. You agree that the wind forces in Y (up) direction is greater than X forces? Now imagine the building stability if you rotate all columns in 90 degree! Probably you'll experience instability in Y direction.
Regards
Dell Brett
Dear Dell Brett,
Thanks for your valuable advice. Is it that 50 % of the colums should be oriented in Xdir and remaining 50% in Y dir?, if possible for a multi storeyed structure
hi
for very special cases that could be done for simple and symetric buildings for ex.,,, but what is essencial for a good structural response for dynamic loading like wind an eartquake, because that what´s we are talking , you sould provide equal stiffness and resistance in both "directions" ( cotes are used because in some structures it's nor easy do define two ortogonal directions). This can be achieved by providing a structural sistem where the center os mass of the floors are coincident ( well that is almosts impossibele to the simplifications in the whole process versus the as built building ) with the center of rigidity. This avoids torsional problems and you have a structural scheme that is more simple to understand and wich response is more predictable. Remenber the structures behave like their were build and not like they apear in fancy structural software. So 1) predesign for gravity+live loads with or not considering in this early stage lateral loads . 2 ) minimum of 3 bays for direction ti achieve minumum redundancie 3) if bays length are similar in both directions as a rule of thumb think only in terms of section enertia to be similar in bothe directions. 4 ) built your model in ETABS or SAP or other and refin the design to achive the closest possibel center of mass to center of rigidity.
This explanaton was very simplifyed but i hope it could contibute for something.
best regards
the software posted links sould be used for evaluation and testing purposes, if you like it buy it !! Always read the disclaimer and act acoordly
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
I am perplexed to find out that the equations that I wrote for the column design appeared only in parts. The main parts are missing for the reasons that I cannot explain (may be due to the browser that I am using).
I would like to know if I am permitted to repost.
Regards
Teddy
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.
Let’s 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 + My/I (compression) OR σ = p/A  My/I (tension)
This equation could be rewritten as:
σ = p/A + M/Z OR σ = p/A  M/Z
Where:
σ = stress
p = applied axial 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 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 beam, in which case the moment of inerter will be minimal.
Let’s consider a 300mm x 600mm reinforced 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 beam, the moment of inerter I = bd^3/12 = (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 allowable σ = p/A + M/Z = p/A + M/1.8 x 10^7 = p/A + 5.56 x10^8M
If we reverse the orientation of the column i.e. with the shorter side aligned in the direction of the span of the 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 + M/Z
= p/A + M/ 9 x 10^6 = p/A + 1.11 x10^7M
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 span of the major principle beam and that in case the orientation is reversed is given as: (p/A + 5.56 x10^8M )/( p/A +1.11 x10^7M ). 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 + M/Z
= (100 x 10^3/180000) + (10 x 10^6/5.4 x 10^9)
= 0.56 + 0.02 = 0.58N/mm^2
For the second case (b =600mm, d = 300mm), σ = p/A + M/Z
= (100 x 10^3/180000) + (10 x 10^6/9 x 10^6) = 0.56+1.11=1.67N/mm^2
(First case (b = 300 and d = 600)) / (Second case (b = 600 and d = 300))= 0.35
This is to say that the first case will experience a stress that is only about a third of that experienced by the second case or that the first case has about three times the load carrying ability as the second: which implies 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 and 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 trial section 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 section has the capacity to withstand the load, terminate the design or move to the next phase, if not 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 to the upper limit as allowed in the relevant code.
Regards
Teddy
