Thermal Expansion

Question

Steel wire of radius 9 mm is cooled from 21° C to 0° C. How much tension would you have to apply to the wire to stretch it back to its original length? Young's modulus for steel is 2.0 1011 N/m2, and the linear coefficient of thermal expansion is 1.2 10-5 C−1.

Answer 1

Hi Alanna, 

 

Let's call the original length of the wire LW (for length warm) and the shorter length of the cooler wire LC (for length cold).  

 

In cooling from 21C to 0C, the wire shortens by a*(21C - 0C)*LW, where a is the coefficient of thermal expansion.  In other words, the new length (LC) is

LC = LW - a*(21C - 0C)*LW

LC = (1 - a*(21C - 0C))*LW

 

In order to stretch the wire back to its original length, you must elongate it by the amount that it shortened.  This induces a tensile strain equal to the change in length [a*(21C - 0C)*LW] divided by the unstressed length [LC].

tensile strain = [a*(21C - 0C)*LW]/LC

 

Using Hooke's law, (stress = E*strain):

tensile stress = E*[a*(21C - 0C)*LW]/LC

tensile stress = E*[a*(21C - 0C)*LW]/[(1 - a*(21C - 0C))*LW]

tensile stress = E*[a*21/(1 - a*21)]

 

The tension is tensile stress times cross-sectional area.

 

I hope that this helps you!  If you have any other questions, please feel free to contact me for tutoring.  I live in the South Bay, and I am also available for online tutoring.

 

Answer 2

Young's Modulus E  = stress / strain         E =  (F/A)/(?L/ L)        F = E A (?L/ L) = E x pi x r^2 α ?T

Area of cross section of wire = A = pi x r^2

Extension of wire ?L =  L α ?T

where L = original length of wire

α = linear coefficient of thermal expansion

?T = change in temperature

 

Substitute and simplify.