Beam design 3

[ChadInColo]

Here is the deal. I need a beam to hold (2) 12.5 kip loads, 35 feet apart, with supports 5 feet outside each load. Force diagram is, x=0 at left, up force 12.5 kip at x=0, 12.5 kip down at x=5, 12.5 kip down at x=40, 12.5 kip up at x=45. The 12.5 kip loads are including a factor of safety of 2 built in (estimated load is actually about 6 kip at each point)

It has been too long since I did bending moment calculations, so I want some back up on what size I-beam I need. Thanks for any help.

 

[desertfox]

Hi ChadInColo

To work out your max stress first calculate your max bending
moment on the beam, this occurs at the centre and as a value
of 68.75kip/ft.
If your basing your design on stress then use the stress for bending formula:-


stress= My/I

where M=bending moment

I = second moment of area of the beam which you
can get from tables

y = distance from the neutral axis of the beam to the
extreme surface for symmetrical shapes this is
half the beam depth.

If you select a beam from the standards then you can get y,I
and you know M so you can calculate your stress and compare it with your safe working stress.

From here then it is easy to calculate the deflection of the beam if you need to, any strength of materials book will give you the formula.

hope this helps

regards desertfox

 

[boo1]

Desertfox assumed fixed end condition, with pinned end condition the bending is 81.24 kip/ft assuming L/360 defection, and 45ft ubraced length: W14x74 works for these loads.

Is minor axis bending to be considered?

 

[desertfox]

Hi boo1

I assumed a simply supported beam and used the loads given

to find the maximum bending moment ie:-


at the beam midpoint B.M.=(12.5x22.5)-(12.5x17.5)kipsft


However the error I did make was I used 17ft instead of 17.5 ft so the actual value of bending moment should have been 62.5 kips ft.


regards desertfox

 

[PEinc]

Isn't it amazing that different engineers get different answers for a basic structural calculation?

Refer to AISC, 9th Edition, page 2-298, Diagram 9.

For this symmetrical, simple span: loads = 2 ea. @ 12.5 kips; reactions = 12.5 kips; max. shear = 12.5 kips; maximum bending moment = 5' x 12.5 kips = 62.5 ft-kips (or kip-ft).

IF max. deflection = L/360 = 1.5 inches, Moment of Inertia (I) must be 618 in.4. Use a W14x61 (I = 640 in.4, Sx = 99.2 in.3)

This is a long beam with no lateral support of the compression flange. Therefore, the allowable bending stress will be much lower than Fb = 0.66 x Fy.

Bending stress = fb = 62.5 x 12 / 99.2 = 7.56 ksi.

Remember, in the US, new steel is commonly available today with Fy = 50 ksi. However, deflection is a function of the moment of inertia, not yield strength of steel. Therefore, if deflection is critical, check required section modulus based on the unbraced length of the beam AND check the deflection of the beam you choose to use. You must meet both bending and deflection requirements. To be thorough, you should also check beam shear and web crippling/yielding.

 

[Lutfi]

I like to offer the following comments to all of the above:

1. W14 for 45 foot span is too shallow
2. You need to add the bending moment due to the beam weight (M = w * l^2 / 8 )
3. I would calculate the bending moment based on the real load, not the safety factor. AISC has its own safety factor built in the allowable stresses. If you use a SF of 2 in your loads, and compound those SF built in AISC allowable, then you are over sizing the beam!
4. I would not use an allowable stress of 0.6FY for such span without proper correction for the unbraced length
5. You must real define the compression flange un-braced length. We can assume that it is 35 feet,; but I am not sure.
6. The boundary conditions for the beam ends are unknown. I was amazed how some engineers are quick to make assumption that they are simply supported (most cases this is true; but not always).
7. You need to keep in mind economy when selecting the final beam size. Some beams are more available than another and that could impact cost and delivery schedule.
8. It has been my 20 years of experience that shear never controls the design (Please do not mistake this by not checking it).
9. Max bending moment is simply the area under the shear curve.

 

[PEinc]

Lutfi makes excellent points.

There is a safety factor already in the allowable stresses. However, we do not know why ChadinColo's beam must be designed for twice the load. It may not be for an "extra" safety factor. It could be for future loading. It may be for uncertainty of the design live load.

Lutfi is correct that the weight of the beam should be included but, in this case, it is relatively minor. 60 or 70 plf dead load adds another 0.3 inches of deflection and an additional 1.87 ksi bending stress - still not much , especially if it is 50 ksi steel. But, Lutfi is correct.

If you add the weight of the beam, the 81.24 kip-ft moment indicated by boo1 is correct.

Shear usually does not control unless the beams is short but heavily loaded - which we do not have in this case.

We don't know what the beam is being used for or if it is for a temporary or permanent application. If ChadinColo wants to add in the beam's dead load, he can use AISC's Manual of Steel Construction, 9th Edition, Diagram 1 on page 2-296 and add the results to those for the live load from Diagram 9.

 

[trainguy]

This beam seems really long, too shallow, and has good potential for some kind of compression flange instability. M/S is nowhere close to the approach required here. The allowable stress based on compression flange instability is much less than yield for such a slender beam.

Draw it to scale, show it to a senior structural engineer, and see what he/she says...

tg

 

[ChadInColo]

Sorry I didn't give more information, the beam will be temporary, it will be used to lift an old 40x35 barn enough to replace a couple sections of sill plate and foundation. I added in the fos of 2 because the weight is a complete estimate. I estimated bd-ft of lumber, used oak's denisity (it's not, but I don't know what it is, it is very old and hard, and I didn't add up 50 years of dust, tin on the roof...) I am planning on using 2 beams, but could use 3 if it turns out to be cheaper. The ends of the beam will be set on cribbing, I would guess that is approximated by a simple support on the ends.

If that changes anyone's answer, let me know. And yes, I know that this is dangerous, make sure everything is secure...and what everyone has said on beam size is approximately what I was thinking, but like I said originally, it has been a while since I have done any of these calculations, and was just looking for verification for my peace of mind. Thanks for all the help, this is a great site.

Chad