How Do You Calculate the Mass of a Steel I-Beam?

  • Thread starter Thread starter CollegeStudent
  • Start date Start date
  • Tags Tags
    Density Mass
Click For Summary
SUMMARY

The mass of a steel I-beam measuring 6.44 m in length, 32.2 cm in height, and 14.0 cm in width, with a thickness of 2.00 cm, can be calculated using the density of steel at 7.56 x 103 kg/m3. The cross-sectional area is determined by calculating the area of the horizontal and vertical sections, resulting in 112.4 cm² or 0.01124 m². The final mass can be computed by multiplying the cross-sectional area by the length and the density of the steel.

PREREQUISITES
  • Understanding of geometric calculations for cross-sectional areas
  • Familiarity with units of measurement conversion (cm² to m²)
  • Knowledge of density and mass calculations
  • Basic principles of structural engineering and I-beam design
NEXT STEPS
  • Learn how to calculate the mass of different structural shapes using density
  • Study the properties and applications of steel I-beams in construction
  • Explore unit conversion techniques for area and volume measurements
  • Investigate the role of cross-sectional area in structural integrity and load-bearing capacity
USEFUL FOR

Engineering students, structural engineers, and anyone involved in construction or materials science will benefit from this discussion on calculating the mass of a steel I-beam.

CollegeStudent
Messages
109
Reaction score
0

Homework Statement


A steel I beam is 6.44 m long, 32.2 cm high and 14.0 cm wide. The top bottom and sides are all 2.00 cm thick. If the density of the steel is 7.56 x 10^3 kg/m^3, find the mass of the beam.


Homework Equations


I'm not sure which equations to use actually.


The Attempt at a Solution


So what I was thinking was

The 2 horizontal parts are 14.0cm * 2.00cm = 28 *2 = 56 cm²
And the 1 vertical part is (32.2cm-4cm) *2cm = 56.4 cm²

So that is 112.4 cm²

Converting this to m² would be .01124m²

After that I'm kind of lost, and I don't even know if THAT'S correct. The thing that's throwing me off is the fact they say there's a length, width, height, and a thickness. Whereas I was taught that thickness and height are the same thing. I don't know how to approach this.

Could someone explain what the difference between height and thickness here...and if I'm on the right track

Thanks everyone
 
Physics news on Phys.org
Welcome to PF CollegeStudent!

The reason why there are four dimensions given instead of three is that an I-beam looks like this (like a capital letter I) in cross section):

Code:
              w
-------------------------   
                            t
--------        --------
        |      |
        |      |
        |   t  |  h
        |      |
        |      |
--------        --------
                            t
-------------------------

And "l" is the length of the beam, i.e the distance that this extends "out of your screen." Does that make sense?

EDIT: I attached a screen shot in case that ASCII diagram looks messed up in other browsers.
 

Attachments

  • i_beam.png
    i_beam.png
    719 bytes · Views: 468
One thing that is ambiguous is whether h is the height from the bottom surface of the bottom horizontal section to the top surface of the top horizontal section (i.e. does it include the two thicknesses of those sections), or is it JUST the height of the vertical section, excluding the thicknesses of the horzontal parts. Whatever, just assume something, state explicitly what you assumed, and do the calculation.
 
Your calculation of the cross sectional area looks OK.

Finish the calculation and determine the mass of the beam.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 44 ·
2
Replies
44
Views
5K
  • · Replies 11 ·
Replies
11
Views
4K