Notices

structural steel strength

Old 07-20-2020, 11:30 AM
  #1  
Senior MemberCaptains Club Member
Thread Starter
 
Join Date: Jul 2006
Location: S.E. Michigan
Posts: 1,145
Likes: 0
Received 278 Likes on 142 Posts
Default structural steel strength

First, I don't know the proper terminology, so be gentle.

How much weight could a 3ft long 1x2 solid steel bar fixed at one end support with the weight at the unsupported end?
Deflection?

I have some of this material and want to bolt it to my front end loader as a fork. I will not be picking up pallets.
Just chunks of tree trunks/ small boulders

Thanks.


Would 2 pieces support 1000# ? 1"x2" hot rolled mill grade steel. 3/4" bolts 14 " apart.
In my mind this is way stronger than what I need, but it is what I have.

Last edited by InTheNet; 07-20-2020 at 11:53 AM.
Old 07-20-2020, 11:35 AM
  #2  
Senior MemberCaptains Club Member
 
Garett's Avatar
 
Join Date: Feb 2003
Posts: 24,745
Likes: 0
Received 1,137 Likes on 640 Posts
Default

What kind of metal are we talking here, cold roll, chrome molly, etc?

What size is 1x2? inches or are you talking feet?
Old 07-20-2020, 11:37 AM
  #3  
Senior Member
 
Join Date: Aug 2009
Location: N E Florida
Posts: 1,685
Likes: 0
Received 292 Likes on 179 Posts
Default

The problem will be with the attachment point strength.
Old 07-20-2020, 11:39 AM
  #4  
Senior Member
 
Join Date: Jun 2007
Location: Spring Hill, FL
Posts: 7,742
Received 897 Likes on 460 Posts
Default

Originally Posted by Stim View Post
The problem will be with the attachment point strength.
Exactly. When you say bolt on, that is the weak point in my mind.
Old 07-20-2020, 11:46 AM
  #5  
Senior MemberCaptains Club Member
Thread Starter
 
Join Date: Jul 2006
Location: S.E. Michigan
Posts: 1,145
Likes: 0
Received 278 Likes on 142 Posts
Default

1"x2" hot rolled steel mill grade

I would have 14 inches bolted to the bucket with a 3ft overhang Bolts near one end and 14 inches

3/4" bolts


Old 07-20-2020, 11:52 AM
  #6  
Senior MemberCaptains Club Member
 
Join Date: Dec 2003
Location: Northern Neck, VA
Posts: 6,385
Received 2,637 Likes on 1,305 Posts
Default

I don't think anyone can verify a random unseen untested piece a metal as to tensile or "strength"

It sounds like it would be fine from experience of field engineering stuff.

what the worse thing? you will drop an old tree stump?

Guessing the whole thing will tip over or your weld/bolt will break first.
Old 07-20-2020, 11:57 AM
  #7  
Admirals Club Admiral's Club Member
 
Join Date: Nov 2009
Location: Tarpon Springs, FL
Posts: 8,435
Received 4,006 Likes on 2,265 Posts
Default

Hopefully you're not drilling your forks to bolt this on.
Old 07-20-2020, 11:59 AM
  #8  
Senior MemberCaptains Club Member
Thread Starter
 
Join Date: Jul 2006
Location: S.E. Michigan
Posts: 1,145
Likes: 0
Received 278 Likes on 142 Posts
Default

[QUOTE=rocksandblues;13871962

what the worse thing? you will drop an old tree stump?

Guessing the whole thing will tip over or your weld/bolt will break first.[/QUOTE]

Just trying to ease my mind, but I pretty well will have someone hold my beer when I try. I will only be out 4 bolts if they fail.
Old 07-20-2020, 12:14 PM
  #9  
Senior Member
 
Join Date: Aug 2012
Posts: 209
Received 56 Likes on 44 Posts
Default

Generally cold rolled is stronger than hot rolled. I made a hay spear out of 1 1/2" round cold rolled. It is supported for 6" and then is 42" long. It has no problem picking up a 5'x5' round bale which weighs about 1000 lbs. It doesn't answer your question but it gives you a reference. As others have said attachment will be critical, use grade 8 bolts at a minimum. If you are attaching to the inside of bottom of the bucket make sure you have a piece of angle or something underneath to spread the load. Or better yet build a receiver that the fork goes in to and make it with large flanges on it and then bolt that to the bucket and just pin the fork to it.
Old 07-20-2020, 12:29 PM
  #10  
Admirals Club Admiral's Club Member
 
Join Date: Nov 2008
Location: Jupiter, FL
Posts: 2,557
Received 379 Likes on 235 Posts
Default

Call it a yield strength of 45 KSI to avoid plastic deformation. If rectangular, the 1x2" would be 2 in2 cross sectional area.

Tensile stress is load over area - so one bar could carry 45,000 lb / in2 x 2 in2 = 90,000 lbs. Put a 10X safety factor and you get 9,000 lbs.

In practice, you will induce bending stress so you would need to factor that into the equation but it sounds like you are lifting much lighter loads and supporting with 2 bars.

As others said, attachment method is likely the limiting load carrying location.
Old 07-20-2020, 12:44 PM
  #11  
Admirals Club Admiral's Club Member
 
Join Date: Feb 2012
Location: Beaufort, NC
Posts: 2,011
Received 1,100 Likes on 509 Posts
Default

Originally Posted by lobsdiver1 View Post
Call it a yield strength of 45 KSI to avoid plastic deformation. If rectangular, the 1x2" would be 2 in2 cross sectional area.

Tensile stress is load over area - so one bar could carry 45,000 lb / in2 x 2 in2 = 90,000 lbs. Put a 10X safety factor and you get 9,000 lbs.

In practice, you will induce bending stress so you would need to factor that into the equation but it sounds like you are lifting much lighter loads and supporting with 2 bars.

As others said, attachment method is likely the limiting load carrying location.
That formula would apply if he was hanging the load from the bar positioned vertically. This is a cantilevered beam, shear and bending moment applies, not tensile. Also would not assume more than 36ksi for yield of unknown steel.
Old 07-20-2020, 12:52 PM
  #12  
Admirals Club Admiral's Club Member
 
Join Date: Nov 2008
Location: Jupiter, FL
Posts: 2,557
Received 379 Likes on 235 Posts
Default

Originally Posted by lobsdiver1 View Post
Call it a yield strength of 45 KSI to avoid plastic deformation. If rectangular, the 1x2" would be 2 in2 cross sectional area.

Tensile stress is load over area - so one bar could carry 45,000 lb / in2 x 2 in2 = 90,000 lbs. Put a 10X safety factor and you get 9,000 lbs.

In practice, you will induce bending stress so you would need to factor that into the equation but it sounds like you are lifting much lighter loads and supporting with 2 bars.

As others said, attachment method is likely the limiting load carrying location.
Just realized you are loading the bar the opposite way so it is all bending stress, assume all the load is out at the end of the 3'. I = bh**3/12. Assume your h is the 1" so I = 2x (1 x 1 x 1 )/12 = 0.1667

Stress = (Force x distance) x c/I so Force x 36 in x 0.5 /0.1667 = F x 108 = 45,000, so F = 415 lbs

Call it 800 pounds with 2 but assuming all the load is out at the end of the 3'.
Old 07-20-2020, 01:00 PM
  #13  
Senior Member
 
Join Date: Oct 2009
Posts: 10,367
Received 1,707 Likes on 921 Posts
Default

Is this going to be a point load at the end or distributed along the entire length of the bar? If a point load 3 feet from the attachment point, a lot less weight than you think it will. And it will deflect quite a bit.
Old 07-20-2020, 01:13 PM
  #14  
Admirals Club Admiral's Club Member
 
Join Date: Feb 2012
Location: Beaufort, NC
Posts: 2,011
Received 1,100 Likes on 509 Posts
Default

Originally Posted by lobsdiver1 View Post
Just realized you are loading the bar the opposite way so it is all bending stress, assume all the load is out at the end of the 3'. I = bh**3/12. Assume your h is the 1" so I = 2x (1 x 1 x 1 )/12 = 0.1667

Stress = (Force x distance) x c/I so Force x 36 in x 0.5 /0.1667 = F x 108 = 45,000, so F = 415 lbs

Call it 800 pounds with 2 but assuming all the load is out at the end of the 3'.
Run it again with b=1 and h=2.
Old 07-20-2020, 01:39 PM
  #15  
Senior Member
 
Join Date: Mar 2009
Posts: 294
Received 156 Likes on 53 Posts
Default


All the cool kids use computers now a days. I'm just old fashioned.
Old 07-20-2020, 02:05 PM
  #16  
Senior Member
 
Join Date: Mar 2013
Location: coastal tx
Posts: 697
Likes: 0
Received 716 Likes on 443 Posts
Default

drilling holes in the bar to put bolts thru will weaken it too
ubolts or straps around it would be better
Old 07-20-2020, 02:33 PM
  #17  
Senior MemberCaptains Club MemberPLEDGER
 
Join Date: Aug 2001
Location: West Carolina
Posts: 22,017
Received 1,594 Likes on 862 Posts
Default

Originally Posted by 5tmorris View Post

All the cool kids use computers now a days. I'm just old fashioned.
Could you be a little more specific?
Old 07-20-2020, 02:36 PM
  #18  
Senior MemberCaptains Club Member
 
Join Date: Dec 2003
Location: Northern Neck, VA
Posts: 6,385
Received 2,637 Likes on 1,305 Posts
Default

Originally Posted by 5tmorris View Post

All the cool kids use computers now a days. I'm just old fashioned.

you all willing to put your stamp on it?
Old 07-20-2020, 02:38 PM
  #19  
Senior Member
 
Join Date: Mar 2009
Posts: 2,191
Likes: 0
Received 573 Likes on 366 Posts
Talking

Originally Posted by Shag View Post
Could you be a little more specific?
HaHa 5/16" deflection too general for you. What I love about this site.
Old 07-20-2020, 02:49 PM
  #20  
Admirals Club Admiral's Club Member
 
Join Date: Aug 2017
Posts: 326
Received 144 Likes on 97 Posts
Default

Originally Posted by 5tmorris View Post

All the cool kids use computers now a days. I'm just old fashioned.


I would think the height would be 1" and width 2". That's the only practical way to attach with 3/4" bolts. Otherwise the edge distance is 1/8". And OP is trying to drill a 2" deep hole. With my assumptions the steel forks would yield at a much lower load.

Thread Tools
Search this Thread

Contact Us - Archive - Advertising - Cookie Policy - Privacy Statement - Terms of Service - Do Not Sell My Personal Information -

Copyright © 2018 MH Sub I, LLC dba Internet Brands. All rights reserved. Use of this site indicates your consent to the Terms of Use.