How to calculate if a vertical beam is stable under a particular load?

[buytree]

What all the parameters to be calculated to find a vertical standing beam subjected to axial loading stands without buckling or stands stable?

My HW problem has a theoretical question, a vertical beam is fixed on both ends. Loaded on the top with axial load. Show that with an example calculation that the beam can be stable below what load?

 

[PhanthomJay]

If you know the length and other properties of the beam, are you familiar with the Euler buckling equations for various end conditions?

 

[buytree]

Yes, do you say just to calculate the Elastic Buckling Strength using the formula Pc = ((PI^2)*E*I)/(KL)^2?? Where K is defined for four different supporting conditions.
After calculating how do you say that the calculated value states that the beam is stable under that load?

 

[berdan]

Hmm,maybe what you can do ,is actually write an equilibrium state for beam that has buckled.That means,that you looking for situation in which your beam can be under an equilibrium in buckled state.
Write an equilibrium equations for a cross section of a beam-you will get a moment equation.Substitute that moment with EIV``,divide by EI,you will get differential equation.
The solution of the equation will give you the load upon which you can have various equilibrium states,which can happen even if the beam isn't straight.

Well,thats actually the way you get Euler equation.Bellow that force there is no stable state,but only when the beam is straight.

 

[buytree]

An attempt to solve the problem, This is an alternate method from Eulers.
Vertical Beam details UC 152 x 152 x 37,
Length L = 4000 mm
Depth D = 161.8 mm
Width B = 154.4 mm
Flange Thk T = 11.5 mm
Web Thk t = 8 mm
Moment of Inertia Ix = 2210 x 10^4 mm^4
Moment of Inertia Iy = 706 x 10^4 mm^4
Radius of Gyration rx = 68.5 mm
Radius of Gyration ry = 38.7 mm
Yield Strength = 250 x 10^6 N/m^2

Load acting P = 2000 kgs (Total weight of the square block sitting over the cross section of the beam, includes self weight). The movement is restrained in all directions at the top and bottom of the beam. This is all my assumption.

Calculating the compressive stress,
Sigma = Load/CS Area = 2000/(0.1618 x 0.1544) = 80057.898 Kg/m^2

Allowable compressive stress = 0.6 x yield strength = 0.6 x 250 x 10^6 x .102 (to convert it into Kg/m^2) = 15300000 Kg/m^2.

80057.898 < 15300000 so it is stable.

Calculating the bending stress,
sigma bx = Bending Moment/(Moment of Inertia x /(Depth/2))
= (Load x Distance)/(Ix/(D/2))
= (2000 x .1618)/((2.210 x 10^-4)/(4/2)) (Substuting in meters, Kg)
= 2928506.79 Kg/m^2 ( I doubt if it is right)
sigma by = Bending Moment/(Moment of Inertia y /(Depth/2))
= (Load x Distance)/(Ix/(D/2))
= (2000 x .1544)/((.706 x 10^-4)/(4/2)) (Substuting in meters, Kg)
= 8747875.354 Kg/m^2 ( I doubt if it is right)

Allowable bending stress sigma bc
Slenderness ratio L/ry = 4000/38.7 = 103.36
For Yield Strength of 250 MPa and Slenderness ratio L/ry = 103.36 from table 5.1 from ISO 800 code book (which I got it from online, if needed I can attach a copy).
Allowable bending stress sigma bc = 77.312 MPa
= 77.312 x 10^6 /9.81
= 7880937.8 Kg/m^2
2928506.79<7880937.8 It is stable
8747875.354>7880937.8 It is not stable

So the size of the beam should be increased.
Is this a valid proof??

 

[nvn]

buytree: Can you convert everything to correct SI units (N, mm, MPa)?

 

[PhanthomJay]

This is getting quite complex. If you really want to get into buckling behavior, you could read a whole book or take a full semester course on column instabilty (and probably still not fully understand it).

buytree: Your method is unclear. I don't see a value that includes Young's Modulus (E) in your equation. Buckling depends upon beam stiffness ( a steel column with high E modulus buckles at a far greater load than a wood column of the same cross section, but with a much lower E modulus). It must enter into whatever equations are being used.

For long slender columns (l/r greater than 200 or so), Euler's formula (Pi^2EI/(kL)^2) yields the critical load up to which point the column is stable, and beyond which buckling occurs under slight lateral deformations. Your example seems to indicate an 'intermediate' column, for which the critical load due to inelastic behavior must be determined by empirical equations such as given in AISC/LRFD manuals.

 

[buytree]

nvn : I will attach an excel sheet with the correct SI units in an hour.

 

[nvn]

buytree: I cannot open proprietary file formats. Always use a portable format, such as plain text (.txt), .png, or .pdf.

Also, what is your column effective length factor, ke? Is your column pinned-pinned (ke = 1.0), fixed-fixed (ke = 0.50), or what? Another name for pinned-pinned is simply-supported.

 

[buytree]

nvn: I have done the calculation in SI units. Pl check belownvn : I have done the calculation in SI units.Vertical Beam details UC 152 x 152 x 37,
Length L = 4 m
Depth D = 0.1618 m
Width B = 0.1544 m
Flange Thk T = 0.0115 m
Web Thk t = 0.008 m
Moment of Inertia Ix = 2.210 x 10^-5 m^4
Moment of Inertia Iy = 0.706 x 10^-5 m^4
Radius of Gyration rx = 0.0685 m
Radius of Gyration ry = 0.0387 m
Yield Strength = 250 x 10^6 N/m^2

Load acting P = 19620 N (Total weight of the square block sitting over the cross section of the beam, includes self weight). The movement is restrained in all directions at the top and bottom of the beam. This is all my assumption.

Calculating the compressive stress,
Sigma = Load/CS Area = 19620/(0.1618 x 0.1544) = 785367.98 N/m^2

Allowable compressive stress = 0.6 x yield strength = 0.6 x 250 x 10^6 = 150,000,000 N/m^2.

785367.98 < 150,000,000 so it is stable. It can hold the load without getting collapsed.

Calculating the bending stress,
sigma bx = Bending Moment/(Moment of Inertia x /(Depth/2))
= (Load x Distance)/(Ix/(D/2))
= (19620 x .1618)/((2.210 x 10^-5)/(4/2)) (Substituting in meters, N)
= 71821628.96 N/m^2 ( I doubt if it is right)

sigma by = Bending Moment/(Moment of Inertia y /(Depth/2))
= (Load x Distance)/(Ix/(D/2))
= (19620 x .1544)/((0.706 x 10^-5)/(4/2)) (Substituting in meters, N)
= 858166572.2 N/m^2 ( I doubt if it is right)

Allowable bending stress sigma bc
Slenderness ratio L/ry = 4/0.0387 = 103.36
For Yield Strength of 250 MPa and Slenderness ratio L/ry = 103.36 from table 5.1 from ISO 800 code book (which I got it from online, if needed I can attach a copy).
Allowable bending stress sigma bc = 77.312 MPa
= 77.312 x 10^6
= 77312000 N/m^2
71821628.96<77312000 It is stable
858166572.2>77312000 It is not stable
According to the IS 800 code, the allowable compressive and bending stress should be less than the calculated compressive and bending stress. I believe my HW needs to prove how the beam stability can be checked and if it withstands the applied load. In the above example if I reduce the load then the beam should be alright. If anyone is okay with my calculation.

Thanks nvn. My Ke should be 0.5 because my beam is restrained from any movement. Give me some time I am doing the Eulers buckling calculation too.

 

[buytree]

This is getting quite complex. If you really want to get into buckling behavior, you could read a whole book or take a full semester course on column instabilty (and probably still not fully understand it).

buytree: Your method is unclear. I don't see a value that includes Young's Modulus (E) in your equation. Buckling depends upon beam stiffness ( a steel column with high E modulus buckles at a far greater load than a wood column of the same cross section, but with a much lower E modulus). It must enter into whatever equations are being used.

For long slender columns (l/r greater than 200 or so), Euler's formula (Pi^2EI/(kL)^2) yields the critical load up to which point the column is stable, and beyond which buckling occurs under slight lateral deformations. Your example seems to indicate an 'intermediate' column, for which the critical load due to inelastic behavior must be determined by empirical equations such as given in AISC/LRFD manuals.

Euler's formula

Critical load Pc = (Pi^2EI/(kL)^2) = (3.14^2 * 250 * 10^6 * 2.210 x 10^-5)/((0.5*4)^2) = 13632.39 N.

So the applied load should always be less than the critical load from euler's formula, then the beam will be stable. That makes sense a big time. Thanks PhanthomJay.

Since my load is 19620 N, the beam is not stable. My above calculation provides additional proof.

 

[buytree]

buytree: See question I added to post 9.

I answered you at the end of post 10, My Ke should be 0.5 because my beam is restrained from any movement, fixed. If you have any questions in any of my above calculation please ask me, I would like to learn it. It is going to be very useful in future. Thanks again.

 

[nvn]

buytree: (1) Is your axial load (P = 19 620 N) a coaxial load, applied exactly on the centroid of your I-beam? Or is your applied load offset from the I-beam centroidal axis?

(2) Is this question a school assignment?

(3) The I-beam cross-sectional area is A = 4710 mm^2, not 161.8*154.4 mm^2.

(4) Slenderness ratio is ke*L/ry, not L/ry.

 

[buytree]

Answer to nvn:
1. The load is actually a tank of 19620N is sitting on top of the vertical standing beam. I assume the tank's load is applied over the cross section of the beam. (not on the centroid or at any offset).
2. It is a HW problem.
3. Yes you are right, A=4710 mm^2.
4. Yes Ke = 0.5, so my slenderness ratio should be multiplied by 0.5. It is my mistake.

Can you check my above calculation pl!

 

[nvn]

buytree: (1) Your answer in post 14 indicates your applied load resultant is concentric with your column. Therefore, there is no primary bending moment, nor primary bending stress, applied to your column. Therefore, delete your sigma_bx and sigma_by bending stress calculations.

(3) Recompute your axial compressive stress, using the correct I-beam cross-sectional area.

(4) Correct your slenderness ratio, and redo your calculations. (You can use mm and MPa.)