Calculating Steel Universal I-Beam for Simply Supported Beam

  • Thread starter Thread starter qems
  • Start date Start date
  • Tags Tags
    Beam
Click For Summary
SUMMARY

This discussion focuses on calculating the appropriate size of a Steel Universal I-Beam to replace an old wooden beam for a simply supported structure. The user, Rob, determined a total weight of 2233 lbs, factoring in a safety load of 3349 lbs. He attempted calculations using bending moment (BM = wl^2/8) and deflection (D = 5/384 x wl^3/EI) formulas but found a 203 x 133 beam insufficient. The response emphasizes the importance of considering maximum stress and allowable stress in beam selection.

PREREQUISITES
  • Understanding of structural engineering principles
  • Familiarity with Steel Universal I-Beam specifications
  • Knowledge of bending moment and deflection formulas
  • Awareness of local building codes and safety factors
NEXT STEPS
  • Research allowable stress for Steel Universal I-Beams
  • Learn about the effects of live load on beam design
  • Explore advanced beam calculation methods using software tools
  • Review local building codes regarding beam specifications and safety factors
USEFUL FOR

Structural engineers, architects, and construction professionals involved in beam design and load calculations will benefit from this discussion.

qems
Messages
1
Reaction score
0
Hi, great forum

My apologies for my first post being a question, but I am trying to calculate the size of a Steel Universal I-Beam that I will need to replace an old wooden beam.

It is a simply supported beam that needs to hold up a uniformly distributed load of around 1.5tonnes, over a length of 14.5ft. The beam is supported at either end by an upright column.

I have calculated this weight using the following:
Surface Area of Slate & Wood Roof = 103 sq ft @ 16lbs / sq ft
= 1624 lbs

Surface area of Wood Panal and Partially Glazed wall = 100sq ft @ 6lbs / sq ft =609lbs

TOTAL Weight = 2233 lbs

Safety Factor = 2233*1.5
= 3349lbs

(I also included an assumed live load of approximately 40lb sq ft, but did not show this in the calculatiosn above.)

I have tried using BM = wl^2/8 and D= 5/384 x wl^3/EI , but I am not having much luck as I am getting figures that say a 203 x133 will be sufficient but this seems light.

Any help would be greatly appreciated. If you could show me the calculations it would be fantastic so I can see where I am going wrong.

Rob
 
Last edited:
Engineering news on Phys.org
qems said:
Hi, great forum

My apologies for my first post being a question, but I am trying to calculate the size of a Steel Universal I-Beam that I will need to replace an old wooden beam.

It is a simply supported beam that needs to hold up a uniformly distributed load of around 1.5tonnes, over a length of 14.5ft. The beam is supported at either end by an upright column.

I have calculated this weight using the following:
Surface Area of Slate & Wood Roof = 103 sq ft @ 16lbs / sq ft
= 1624 lbs

Surface area of Wood Panal and Partially Glazed wall = 100sq ft @ 6lbs / sq ft =609lbs

TOTAL Weight = 2233 lbs

Safety Factor = 2233*1.5
= 3349lbs

(I also included an assumed live load of approximately 40lb sq ft, but did not show this in the calculatiosn above.)

I have tried using BM = wl^2/8 and D= 5/384 x wl^3/EI , but I am not having much luck as I am getting figures that say a 203 x133 will be sufficient but this seems light.

Any help would be greatly appreciated. If you could show me the calculations it would be fantastic so I can see where I am going wrong.

Rob

Hi gems,

You should be considering the max stress that the beam will be subjected to and select the appropriate beam based on it's allowable stress (bending, shear, etc).

This should help you get started: http://www.efunda.com/formulae/solid_mechanics/beams/casestudy_display.cfm?case=simple_uniformload

Also, check your local building codes for the proper live load if you haven't already, plus any other applicable considerations.

I would also suggest using the factor of safety with respect to the actual and allowable stress and not just simply adding "extra" load to the structure.

Hope this helps.

CS
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
4K
Replies
3
Views
2K
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K