Go Back   Physics Help Forum > High School and Pre-University Physics Help > Equilibrium and Elasticity

Equilibrium and Elasticity Equilibrium and Elasticity Physics Help Forum

LinkBack Thread Tools Display Modes
Old May 25th 2012, 01:06 PM   #1
Junior Member
Join Date: May 2012
Posts: 1
Question Maximum strain of steel?

Alright, so I have this final project in physics which was due yesterday and i'm turning it in late because i have a problem with understanding one thing.
so the project it about space elevators and the teacher asked me to make calculations about what is the maximum height of strongest modern material (steel) as a rod, untill it tears itself down... I think of using the equation:

F/A for steel is 500x10^6
and E for steel is 200x10^6

i don't know how to calculate strain though, as i don't know initial length or change in length so could you please help me with this? with explanations
exibo177 is offline   Reply With Quote
Old May 28th 2012, 10:51 AM   #2
Senior Member
Join Date: Apr 2008
Location: HK
Posts: 886
It is correct to calculate strain in this way. But you must have the initial length first before you can get the maximum change in length. Perhaps it is measured by yourself in the project?
Good results were achieved and the new task is to become a good doctor.
werehk is offline   Reply With Quote
Old May 29th 2012, 12:32 PM   #3
Physics Team
ChipB's Avatar
Join Date: Jun 2010
Location: Naperville, IL USA
Posts: 2,269
Are you asking what's the longest length of a steel rod that can hang without breaking under its own weight? This doesn't involve strain at all - you just need to find the ultimate yield stress of the material, and set that equal to F/A, where F (force) is due to the weight of the material being hung. For a steel rod the weight is length x cross-sectional area x density x g. If we let L = length and r = density you get:

stress = F/A = LArg/A = Lrg

The yield stress of steel is about 400 MPa, and the density is about 7.8 g/cm^3. You can get similar data for other materials here: http://www.askmehelpdesk.com/islam/q...ik-573046.html . So:

stress = Lrg
400 x 10^6 N/m^2 = L (7.8 g/cm^3 x 1 Kg/1000g x 10^6 cm^3/m^3) x 9.8 m/s^2 = L x 76.4 x 10^3 N/m^3
Rearrange to solve for L:
L = 400 x 10^6/76.4 x 10^3 = 5.2 x 10^3 m.

Hence a rod that is greater than 5.2 Km in length would break under its own weight. Of course this assumes that the acceleration due to gravity 'g' is constant, which for a rod only 5 Km in length is a reasonable assumption. But if you had a really tall space elevator you would have to take into account the fact that 'g' gets smaller as you increase your disnace from the ground.

Last edited by ChipB; May 29th 2012 at 12:45 PM.
ChipB is offline   Reply With Quote

  Physics Help Forum > High School and Pre-University Physics Help > Equilibrium and Elasticity

maximum, space elevator, steel, strain

Thread Tools
Display Modes

Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
How heavy is 1kg steel and 1kg wood MMM Kinematics and Dynamics 1 Mar 2nd 2015 04:07 AM
Steel balls vibration jijilulu Kinematics and Dynamics 8 Feb 11th 2014 11:19 AM
Strain tensor botee Equilibrium and Elasticity 4 Feb 1st 2013 10:11 PM
What double slit maximum is closest to the 2nd single slit secondary maximum mola Advanced Optics 1 May 24th 2009 10:34 PM
Steel wire werehk General Physics 1 Jul 1st 2008 04:21 AM

Facebook Twitter Google+ RSS Feed