A 10cm long cylindrical rod of steel is elongated to become 12.1cm long by the a
ID: 1825686 • Letter: A
Question
A 10cm long cylindrical rod of steel is elongated to become 12.1cm long by the application of a force of 10kN.
1- What is the ration of initial diameter to final diameter, assuming no net volume change?
2- At this load, what proportion of the true stress is the applied stress?
3- For this moment of deformation, what is the true strain and engineering strain?
4- if the initial diameter is 5mm, by how much will the specimen contract when the 10kN load is removed if the Young's modulus for steel is 210GPa?
Could you please provide me the word solution?
Explanation / Answer
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1) intial volume = final volume
10*pi*r^2=12.1*pi*r2^2
(r2/r)^2=10/12.1
r2/r1=10/11
2)true stress / apllied atress =
(load/(intial surface area ))/(load/final surface area)
=(r2/r1)^2=100/121
3)strain = 2.1/10
=.21
eng strain =.21
4)(l/l)*210gpa=10/pi*r^2
l=10*10/3.14/(5*10^3)^2/210gpa
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