Physics Help Forum Maximum strain of steel?

 Equilibrium and Elasticity Equilibrium and Elasticity Physics Help Forum

 May 25th 2012, 01:06 PM #1 Junior Member   Join Date: May 2012 Posts: 1 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=E*strain 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
 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.
 May 29th 2012, 12:32 PM #3 Physics Team     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.

 Tags maximum, space elevator, steel, strain

 Thread Tools Display Modes Linear Mode

 Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post MMM Kinematics and Dynamics 1 Mar 2nd 2015 04:07 AM jijilulu Kinematics and Dynamics 8 Feb 11th 2014 11:19 AM botee Equilibrium and Elasticity 4 Feb 1st 2013 10:11 PM mola Advanced Optics 1 May 24th 2009 10:34 PM werehk General Physics 1 Jul 1st 2008 04:21 AM