How to Calculate Safety Factor for Beam Columns?

[derail]

Beam held up by two columns

I apologize in advance if you think this should really go in the homework help area, I am not sure because this isn't a homework problem, but rather my struggle to understand and carry out concepts. If it needs to be re-posted, or moved, I understand. :)

In a building a beam is held up by two columns, and only two columns. I need to show that the columns are strong enough to hold up the beam. I thought the best way to do this would be to find the safety factor, and by showing that this value was high enough I would know that it would work.

Safety factor can by shown by, Ultimate load/actual load


Or, Design load/ expected load


I know the Tensile yield strength is 345 MPa (50 ksi) and tensile ultimate strength 450 MPa (65 ksi) of the column.

The cross sectional area of the column I found to be, 8.109 in^2, which is also 0.20596 meters. It is an I beam that I calculated the cross sectional area for.

How do I find out the force that the two columns can hold? That way I can use the safety factor with the load method. Or is that the wrong way to do it?



Keep in mind, I am trying to find the safety factor, so I can't assume its anything to help me find other data.

 

[Simon Bridge]

You want something like this?
http://www.efunda.com/formulae/solid_mechanics/columns/calc_column_critical_load.cfm

 

[derail]

That looks very helpful but I am not sure, that website incorporates column buckling which I wasn't going to include as a factor. Then again, column buckling could be a major factor in this situation.

What is really hinges on then is, would it be possible for a column to not be strong enough to hold the beam but not buckle in the true sense? Or perhaps I don't understand the concept of column buckling.

 

[SteamKing]

What is really hinges on then is, would it be possible for a column to not be strong enough to hold the beam but not buckle in the true sense? Or perhaps I don't understand the concept of column buckling.

That's why you must check the columns to see if buckling will be a problem. In other words, will the critical load in the column where buckling is imminent produce an axial stress in the column which is less than the elastic limit of the material?

The thing to realize about buckling, it is a phenomenon which is related to the stability of the column under load. Long, skinny columns can buckle with relatively little load applied to them, while short, squat columns will not exhibit any instability until the material starts to fail for other reasons.

This article illustrates some of the key concepts to learn about how to analyze buckling in columns:

http://www.ah-engr.com/som/10_buckling/text_10-1.htm

 

[derail]

How does the situation change when there are two columns instead of just one?

Should I try and see if the columns will hold up outside of the buckling situation?

 

[Bobbywhy]

derail, It may be useful to discover how others have already dealt with and solved your issues. We may, with some humility, look backwards to the Romans two thousand years ago, keeping in mind the huge influence Roman archeticture has had on today's. See, for example:

http://www.ancient.eu/Roman_Architecture/
http://en.wikipedia.org/wiki/List_of_Roman_victory_columns
http://www.oneonta.edu/faculty/farberas/arth/arth200/politics/roman_architecture.html

Cheers, Bobbywhy

 

[derail]

I am not designing columns from scratch, these columns are actually I beams. I am just trying to prove that they are strong enough to hold the weight of the beam and roof. So I am trying to find the force they can hold. I thought I could do this with the safety factor. How does the fact that there are two columns effect the column buckling equation?

 

[derail]

I am trying the method and equation at http://www.efunda.com/formulae/solid_mechanics/columns/calc_column_critical_load.cfm

How do I calculate the Young's Modulus? This requires stress which requires force. What force would be used?

 

[AlephZero]

How do I calculate the Young's Modulus?

You don't calculate it. It is a property of the material. If you don't know it, you look it up in a table of material data.

For steel at room temperature, it is about 30 x 10⁶ psi, or 200 GPa.

 

[Simon Bridge]

I am not designing columns from scratch, these columns are actually I beams. I am just trying to prove that they are strong enough to hold the weight of the beam and roof. So I am trying to find the force they can hold. I thought I could do this with the safety factor.

Why not check your proposed construction against existing regulations for your jurisdiction?

 

[SteamKing]

I am not designing columns from scratch, these columns are actually I beams. I am just trying to prove that they are strong enough to hold the weight of the beam and roof. So I am trying to find the force they can hold. I thought I could do this with the safety factor.

Most columns in buildings and other structures are constructed from readily available material, like I-beams. The equation for column buckling and a good set of steel tables will allow you to check buckling for your design, but I emphasize, only if you know how.

How does the fact that there are two columns effect the column buckling equation?

The fact that you asked this question and other similar basic ones strongly suggests that you obtain the help of a competent structural engineer. You shouldn't be checking the design of a structure if you have no experience or training doing so.

 

[berkeman]

Most columns in buildings and other structures are constructed from readily available material, like I-beams. The equation for column buckling and a good set of steel tables will allow you to check buckling for your design, but I emphasize, only if you know how.



The fact that you asked this question and other similar basic ones strongly suggests that you obtain the help of a competent structural engineer. You shouldn't be checking the design of a structure if you have no experience or training doing so.

I agree. Thread is closed.