Calculating Steel Universal I-Beam for Simply Supported Beam

[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

 

[stewartcs]

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).

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

 

[oswaler]

Hi Rob,

Thank you for reaching out to us. I can definitely provide some guidance on calculating the size of a Steel Universal I-Beam for your simply supported beam.

First, let's start with the basics. A simply supported beam is a beam that is supported at two points, and the load is distributed evenly along the length of the beam. In your case, the load is 1.5 tonnes (or 3349 lbs) and the length of the beam is 14.5ft.

To calculate the required size of the beam, we will use some basic structural engineering principles. The first step is to determine the maximum bending moment (BM) that the beam will experience. This can be calculated using the formula BM = wl^2/8, where w is the load per unit length (in your case, 3349 lbs/14.5ft = 231 lbs/ft) and l is the length of the beam.

BM = (231 lbs/ft)(14.5ft)^2/8 = 5574.94 ft-lbs

Next, we need to calculate the section modulus (S) of the beam, which is a measure of the beam's resistance to bending. This can be calculated using the formula S = I/c, where I is the moment of inertia of the cross-section of the beam and c is the distance from the neutral axis to the outermost fiber of the beam.

To determine the moment of inertia, we need to know the cross-sectional shape of the beam. Let's assume that you are using a standard S-shaped beam, also known as an American Standard I-Beam. The moment of inertia for this shape can be found in a table or calculated using a formula. For a 203 x 133 beam, the moment of inertia is 416.1 in^4.

Now, we need to determine the distance from the neutral axis to the outermost fiber of the beam. This is known as the section modulus (Z) and can be found in a table or calculated using the formula Z = (bd^2)/6, where b is the width of the beam and d is the depth of the beam.

For a 203 x 133 beam, the section modulus is 133.5 in^3.

Finally, we can calculate the required depth (D) of the beam using the formula BM = (S)(E)(D), where E is the modulus of