# Physics Elastic limit

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#### Maverick56

##### Full Member
10+ Year Member
A walkway is supported by pairs of hollow carbon steel pipes, each having an outer diameter of 2.19 cm and walls that are 2.55 mm thick. Each pair of pipes supports a section of walkway weighing 5060 N.
If the elastic limit for carbon steel is 4.1 x 10^8 Pa, how many people can safely stand on a single section of the walkway? Assume an average of 664 N/person.

what do you do with the thickness of the steel in this question?

do you find the inner surface area and subtract it from the outer surface area?

if that is the case, then I think the equation would be like this:

(5060 + 440p) = (4.1*10^8)*[pi*2.19*10^2 - (pi*2.19*10^2 - pi*2.55*10^-3)]

this is just my guess right now. I am just trying dimensional analysis and inferring the relationship between thickness and strength

edit: looks like person below me got it right. I shouldn't try to answer questions when tired

Last edited:
Yield strength = Force/Area

Area of steel is pi*(.0219^2 - .01835^2)*2 = .000897 m^2

F = Yield Strength * Area = .000897*4.1*10^8 = 367770 N

Force = 664* # of People + 5060N

#people = (367770-5060)/664 = 546 people

Edit : 2 steel beams, so area calculation is doubled.

I don't believe this question would be on the MCAT.