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Need help sizing a steel I-beam

Wheelin66bronco

Wheelin66bronco

Pure Blood
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Star Valley, Wyoming
I found a few threads by arse_sidewards and the like, and a calculator posted by larboc
engineersviewpoint.blogspot.com

Beam Total Maximum Uniform Load & Reaction: ONLINE CALCULATOR

Designing a simple beam shear connection is easy and quite simple, of course. All that an engineer needs to do is to determine ...
engineersviewpoint.blogspot.com engineersviewpoint.blogspot.com
But I'm not smart enough to put in the info to get what I need.
Some assistance please.
It will span across shop, I want to set a I-beam for a rolling hoist.
It will be supported by vertical I-beams on the ends for which I poured extra deep spots into the floor to support it.
I'd like to see what the beam size would need to be for three different weight limits, and I can decide based on size, cost, and weight to get set via tractor bucket etc.
30' span
13' end legs(depending on beam width)
2,000, 4,000, 6,000, 8,000 capacity at center
Is square or tapered flange better?
Am I missing anything?
Example of rolling trolly:
www.mcmaster.com

McMaster-Carr

McMaster-Carr is the complete source for your plant with over 595,000 products. 98% of products ordered ship from stock and deliver same or next day.
www.mcmaster.com www.mcmaster.com
 
My goal is to be able to lift a steel shed vertical to back a trailer under.
I'd love 12" or less, but I'm guessing that's going to be out of the question.
I'm going to mount it all the way touching the ceiling.
 
if the beam is just supported under each end.

WL'3 / 48EI

deflection = weight at center x L x L x L divided by (48 x modulus of elasticity x moment of inertia)

deflection = 1000# x 360 x 360 x 360 divided by 48 x 29000000 x I)
deflection = 46,656,000,000 divided by 1,392,000,000 x I
deflection = 33.5 divided by I

so a beam with a moment of inertia of 100 would bend .335 inches at the center with a 1000# weight

you need a big beam for that span
 
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That calculator is for uniform load. What you are describing will be a dynamic point load.
 
First post on the site but have been lurking since the old place. The simplest way to size a beam is to divide the maximum moment by the allowable stress to find the section modulus.
S=Mmax/Allstress.

To find the max moment, Mmax, for a centered point load on a simply supported beam use the equation
Mmax = (Load*Length)/4.
Your live load is 8,000lbs, +10% impact + dead load. For now we'll use an even 10,000 lbs. Length in inches multiply 30ft*12
900,000inlbs = (30ft*12*10000lbs) / 4

The max stress for steel is 60,000 psi. For hoists and cranes a safety factor of 5 is used. The allowable stress then becomes:
12,000 psi = 60,000psi / 5.

The section modulus S=Mmax/Allstress becomes
75in^3 = 900,000inlbs / 12,000psi

Using the AISC book the W12" x 58# beam has a section modulus of 78, its greater than the 75 calculated which is the smallest beam that would work without looking at deflection.

Max deflection is limited to the (Length of span in inches / 600) in most cases. For 30 ft allowable deflection is .6 inches.
In the case of the above beam using the above equation from WLDWUN. A W12" x 58# beam has a moment of inertia of 475in^4

Maxdef = WL^3 / 48EI
Maxdef = (10,000lb*(30ft*12)^3) / (48*29,000,000*475in^4)
Maxdef = .75 > .60 need to find a beam with a larger moment of intertia

If we stay with a 12 inch beam a W12x96# has a moment of inertia of 833in^4 and if we calculate that and increase the dead load to 12000 for the heavier beam the deflection is around .482 an inch in the center of the span. The actual deflection will most likely be less as the total increase in beam weight is not a point load at the center.

If you can live with losing 4 inches in headroom and want to save in steel weight a W16" x 57# beam would have a similar deflection around .47 and section modulus of 92.2.

It's always going to be a give and take between headroom, beam weight, and flange width.

The beam flange itself also needs to be checked for deflection depending on how many wheels are being used on the trolley.
 
I think you need Beamboy or whatever the hell it is. Look for my thread on trailer ramps, there was links in there. I used a version of it when I sized the beam for my shop years ago. You can have a decent amount of deflection without risking the trolley wanting to move, there's enough friction at high loading.
 
Since im an ironworker and not an engeneer I do everything at home by "that looks about good enough" and im still alive


Id go minimum w18x86.
Or maybe w24x 55

No way would I want to span 30' with a 12" beam and lift that much weight regularly


Whats above it and what strenth does it have? Maybe you lag bolt the top flang of a 12" beam to the rafters every 24"
 
GUDA said:
First post on the site but have been lurking since the old place. The simplest way to size a beam is to divide the maximum moment by the allowable stress to find the section modulus.
S=Mmax/Allstress.

To find the max moment, Mmax, for a centered point load on a simply supported beam use the equation
Mmax = (Load*Length)/4.
Your live load is 8,000lbs, +10% impact + dead load. For now we'll use an even 10,000 lbs. Length in inches multiply 30ft*12
900,000inlbs = (30ft*12*10000lbs) / 4

The max stress for steel is 60,000 psi. For hoists and cranes a safety factor of 5 is used. The allowable stress then becomes:
12,000 psi = 60,000psi / 5.

The section modulus S=Mmax/Allstress becomes
75in^3 = 900,000inlbs / 12,000psi

Using the AISC book the W12" x 58# beam has a section modulus of 78, its greater than the 75 calculated which is the smallest beam that would work without looking at deflection.

Max deflection is limited to the (Length of span in inches / 600) in most cases. For 30 ft allowable deflection is .6 inches.
In the case of the above beam using the above equation from WLDWUN. A W12" x 58# beam has a moment of inertia of 475in^4

Maxdef = WL^3 / 48EI
Maxdef = (10,000lb*(30ft*12)^3) / (48*29,000,000*475in^4)
Maxdef = .75 > .60 need to find a beam with a larger moment of intertia

If we stay with a 12 inch beam a W12x96# has a moment of inertia of 833in^4 and if we calculate that and increase the dead load to 12000 for the heavier beam the deflection is around .482 an inch in the center of the span. The actual deflection will most likely be less as the total increase in beam weight is not a point load at the center.

If you can live with losing 4 inches in headroom and want to save in steel weight a W16" x 57# beam would have a similar deflection around .47 and section modulus of 92.2.

It's always going to be a give and take between headroom, beam weight, and flange width.

The beam flange itself also needs to be checked for deflection depending on how many wheels are being used on the trolley.
Click to expand...


Just trying to follow along, does that take into consideration the 3k lb beam weight?
 
2big bronco said:
Just trying to follow along, does that take into consideration the 3k lb beam weight?
Click to expand...
beam weight is a distributed weight and he recommended a larger beam due to deflection. He briefly points out that the (relatively small) distributed load won't affect deflection significantly.
 
To keep the beam size small, can you do a moveable A frame to support the beam and reduce the span? Think A frame on a beam trolley, but underside of trolley is a plate. Floor side has adjustment bolts to tighten it up to beam.
 
GUDA said:
First post on the site but have been lurking since the old place. The simplest way to size a beam is to divide the maximum moment by the allowable stress to find the section modulus.
S=Mmax/Allstress.

To find the max moment, Mmax, for a centered point load on a simply supported beam use the equation
Mmax = (Load*Length)/4.
Your live load is 8,000lbs, +10% impact + dead load. For now we'll use an even 10,000 lbs. Length in inches multiply 30ft*12
900,000inlbs = (30ft*12*10000lbs) / 4

The max stress for steel is 60,000 psi. For hoists and cranes a safety factor of 5 is used. The allowable stress then becomes:
12,000 psi = 60,000psi / 5.

The section modulus S=Mmax/Allstress becomes
75in^3 = 900,000inlbs / 12,000psi

Using the AISC book the W12" x 58# beam has a section modulus of 78, its greater than the 75 calculated which is the smallest beam that would work without looking at deflection.

Max deflection is limited to the (Length of span in inches / 600) in most cases. For 30 ft allowable deflection is .6 inches.
In the case of the above beam using the above equation from WLDWUN. A W12" x 58# beam has a moment of inertia of 475in^4

Maxdef = WL^3 / 48EI
Maxdef = (10,000lb*(30ft*12)^3) / (48*29,000,000*475in^4)
Maxdef = .75 > .60 need to find a beam with a larger moment of intertia

If we stay with a 12 inch beam a W12x96# has a moment of inertia of 833in^4 and if we calculate that and increase the dead load to 12000 for the heavier beam the deflection is around .482 an inch in the center of the span. The actual deflection will most likely be less as the total increase in beam weight is not a point load at the center.

If you can live with losing 4 inches in headroom and want to save in steel weight a W16" x 57# beam would have a similar deflection around .47 and section modulus of 92.2.

It's always going to be a give and take between headroom, beam weight, and flange width.

The beam flange itself also needs to be checked for deflection depending on how many wheels are being used on the trolley.
Click to expand...
Maybe the best first post ever written on irate. Well done.

Bonus question: Could he save beam weight by going to a rigid connection over the supports since he's not going to be reducing to shear requirements anyway?
 
Dethmachinefab said:
To keep the beam size small, can you do a moveable A frame to support the beam and reduce the span? Think A frame on a beam trolley, but underside of trolley is a plate. Floor side has adjustment bolts to tighten it up to beam.
Click to expand...
Do you have a picture of what you are talking about? I'm a bit confused with the description but it sounds interesting.
 
2big bronco said:
Since im an ironworker and not an engeneer I do everything at home by "that looks about good enough" and im still alive


Id go minimum w18x86.
Or maybe w24x 55

No way would I want to span 30' with a 12" beam and lift that much weight regularly


Whats above it and what strenth does it have? Maybe you lag bolt the top flang of a 12" beam to the rafters every 24"
Click to expand...
w24x68 8000lb point load at 15'
1665606789650.png

1665606823099.png
 
Dethmachinefab said:
Just an easy to move supporting member. Stow it up against the wall, rolls on beam flange like a trolley until its locked into place where needed.
Click to expand...
perfect idea.
use only when needed for heavier loads, can be used anywhere in the 30 foot span.
if used right in the middle, would reduce the deflection to 12.5% of what it would be without it
even if setting it to make a 20 foot span deflection goes down to 30%
 
WLDWUN said:
perfect idea.
use only when needed for heavier loads, can be used anywhere in the 30 foot span.
if used right in the middle, would reduce the deflection to 12.5% of what it would be without it
even if setting it to make a 20 foot span deflection goes down to 30%
Click to expand...
I have cripples made for mine. Let me pick the 4000lbs+ lathe off my trailer without it going badly. :homer:
 
The style that has one end fixed and the other on a sliding A frame is awesome too. Gets you a arc of reach at near unlimited capacity by shortening the beam or a light duty cantilever arm over the A frame.
 
junkytj said:
Maybe the best first post ever written on irate. Well done.

Bonus question: Could he save beam weight by going to a rigid connection over the supports since he's not going to be reducing to shear requirements anyway?
Click to expand...
Thanks. He probably could, I doubt it would be much of a savings. The calculations for rigid connections becomes a pain. They really aren't 100% rigid as the columns would flex as the beam deflects in turn moving the whole connection.

I attached the hand written calcs to clear it up some. I was off on my 10,000 lb approximate earlier. Using the W12 x 58# from earlier the overall weight is 1740 lbs. The overall force (beam+load+impact) on the beam would be 10,540 lbs. When used in the bending moment equation the section modulus required becomes 79.05 which is greater than the section modulus of a W12" x 58# which is 78. Jumping up to a W12 x 65# the section modulus becomes 87.5. Double check by recalculating for the larger W12 x 65# which is 1950lb overall weight, 10,750 overall force on the beam. The calculated section modulus is 80.625, less than what the beam can safely support at a 5:1 safety factor.

The moment of inertia "I" for the beam can be plugged into the deflection equation to get the deflection you require. A list that includes both the section modulus and moment of inertia for W-Beams can be found here. W-Beams - American Wide Flange Beams

If needed you can go back and double check that the beam selected has a greater section modulus than the calculated "S" due to forces.

If the dead load or live load changes "F" can be adjusted and the equations recalculated.

As was mentioned earlier the wheel loading impact on the flange thickness will need to be checked as well.
 

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2big bronco said:
Since im an ironworker and not an engeneer I do everything at home by "that looks about good enough" and im still alive


Id go minimum w18x86.
Or maybe w24x 55

No way would I want to span 30' with a 12" beam and lift that much weight regularly


Whats above it and what strenth does it have? Maybe you lag bolt the top flang of a 12" beam to the rafters every 24"
Click to expand...
Judging by all the suggestions so far, I think I need to drop it down to maaaybe 4k capacity. No way I'm lifting an w18 or 24 into place.
Above is trusses, but it runs parallel with them, and right in-between two of them so lagging up is not an option.
I'd like to go 10-12" beam if at all possible.
PXL_20221013_011651172~2.jpg
PXL_20221013_011640473~2.jpg
 
If it was me

I would fins one of the hoist company's, figure out what they are using for a similar span/weight and go from there

but I am not a rocket scientist with all this fancy math using action words
 
Another thing to consider is DOL, or duration of load. Your high concentrated load will only exist for a very short amount of time. If on the other hand, you were to suspend a significant load for a long period of time you would need a much more rigid beam. Ponder the usage of said beam, and the frequency thereof. As mentioned before, you’ve got a good starting point for the size of a new beam. The length is the issue. Long span beams usually require pretty significant size.
 
Ryan283 said:
Another thing to consider is DOL, or duration of load. Your high concentrated load will only exist for a very short amount of time. If on the other hand, you were to suspend a significant load for a long period of time you would need a much more rigid beam. Ponder the usage of said beam, and the frequency thereof. As mentioned before, you’ve got a good starting point for the size of a new beam. The length is the issue. Long span beams usually require pretty significant size.
Click to expand...
My 20t overhead crane with a 35’ span is only 2 24” x 104ish # per ft beams.
 
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