# Calculating Steel Universal I-Beam for Simply Supported Beam

• qems
In summary, the beam needs to be at least 203x133 in order to support the weight it will be subjected to.f

#### qems

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 teh 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:
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 teh 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).