# Beam calculations

## Main Question or Discussion Point

<<moderator note, moved from a homework forum, hence the template.>>

1. Homework Statement

I'm trying to work out what 2.29 is in an equation an engineer has used for what i believe is supposed to be the second moment of area. See photo for report.

## Homework Equations

See screenshot photo

## The Attempt at a Solution

See photo. I tried to work back to determine his units but they don't seem to tie in. 2.29 * 8.2 * 4.675^3 = 1919

But he's using kN/m and m^3 to give an answer of cm^4. Surely this doesn't work?

Also why would he use the length of the beam and load without a safety factor to determine the second moment of area.

I see he uses 2.29 in both calcs so i presume it relates to, or could be the second moment if area but it would need something to cancel out the newtons and every calc I've done for the second moment of area for an i beam doesn't give any 2.29

#### Attachments

• 24.9 KB Views: 248
• 52.1 KB Views: 286
• 13.4 KB Views: 320
Last edited by a moderator:

Related General Engineering News on Phys.org
Chestermiller
Mentor
Can you please show a clean diagram of the beam with supports and loading, together with a diagram showing the dimensions of the cross section? Also showing support constraint conditions. It is really impossible to make sense out of the diagrams you have provided (except for the first).

Hi Chestermiller and thank you for reading my post.
I don't have any diagrams myself as I was more just trying to view the calculations to understand how they are done. I have a photo of the finished beams in place if that may help to visualize and I have also done a diagram for the cross section of the 254 x 102 x 25 universal beam.

From viewing his bending moment calculation he has taken a UDL simply supported beam at a span of 4.675m

I have also added a couple close ups of the calculations he has used. He has used the sum of the calculations from the total amount of dead and imposed loads in kN/m at 11.4 kN/m and 8.2 kN/m for inner and outer beam respectively.

Hope this helps.

#### Attachments

• 13.4 KB Views: 297
• 17 KB Views: 209
• 234.2 KB Views: 267
• 207.9 KB Views: 239
Chestermiller
Mentor
All I'm asking for is a sketch, with the dimensions on it (including the cross section).

Sorry, would this be sufficient?

Many thanks
Jonathan

#### Attachments

• 34.1 KB Views: 237
• 27.1 KB Views: 239
Chestermiller
Mentor
Sorry, would this be sufficient?

Many thanks
Jonathan
I'm really having a lot of trouble understanding the details of this. What is the "dead load" supposed to be, and where is it applied to the beam (if at all)? What does S.F. stand for? Have you tried setting up this calculation on your own (from scratch) to see how your results compare? What did you calculate for the moment of inertial about the neutral axis (I will do the calculation and compare)? I would be interested in seeing all your own calculations for this problem.

Chestermiller
Mentor
For the moment of inertial about the neutral axis, I get 3252 cm^4

For the moment of inertial about the neutral axis, I get 3252 cm^4
I get also 3252.38 cm^4. Which doesn't tie in to any of his calculations and I have no idea why he would use the span of the beam and the sum of the loads per kN/m to calculate this(His Ixx calculations). His calculations to me do noy make sense but I'm no expert in the structural field.

As i said these are not my calculations. I'm just trying to decipher them and understand how you would calculate a beam size required, for a span of a gap with a specific load as the only experience I have is calculating shear forces and bending monents within a beam from academic study. Also using the bending equation B/I = σ/y = E/R. But wasn't sure how i could actually use all of this to determine a required beams dimensions so wanted to study the report that an engineer has do e recently for my parents house, but cannot understand his workings.

The dead load I presume is the ststic load, the weight of the roof, wall, floor and beam.

The imposed would be the live load, i.e accounting for people etc.

S.f is the safety factor. Sorry I wasn't clear on that.

Many thanks
Jonathan

Last edited:
Chestermiller
Mentor
I think we can make sense out of this, but it is difficult to understand his diagrams. You have provided a diagram of the distributed load, but I would like to see where the dead loads fit into the picture. Probably what he has done is derived an equation for the maximum tensile stress (at the outside of the bend) and compared that with the ultimate strength of the material in conjunction with a safety factor. I think you are very close to having this all deciphered.

Last edited: