Mass And Density Question

1. Jan 30, 2013

CollegeStudent

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
I'm not sure which equations to use actually.

3. 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
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 30, 2013

cepheid

Staff Emeritus
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 (Text):

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.

Attached Files:

• i_beam.png
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3. Jan 30, 2013

cepheid

Staff Emeritus
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.

4. Jan 30, 2013

SteamKing

Staff Emeritus
Your calculation of the cross sectional area looks OK.

Finish the calculation and determine the mass of the beam.