Problem about tensile stress during thermal contraction

  • #1
issacnewton
983
22
Hi

I am trying to solve this problem. I will present my solution though I couldn't get the tension right.

At 40 Celsius , there is no tension in the wire. When the wires are cooled , they contract and the tension increases. But the tensile stress, which is force per unit area, will remain the same for both the wires, even if they are different in constitution.
I am neglecting the increase in radius, hence the area of the wires, because its negligible.

When the tensions develop , let the steel wire be stretched by amount [itex]\Delta x_s[/itex]
from its reduced length Ls. Similarly, let copper wire be stretched by amount
[itex]\Delta x_c[/itex] from its reduced length Lc. Let L = 2 m be the original length of each wire. Since

[tex]\mbox{Young's Modulus}\, =\frac{\mbox{Tensile Stress}}{\mbox{Tensile Strain}}[/tex]

we have

[tex]\frac{F}{A}= Y_s \,\frac{\Delta x_s}{L_s} = Y_c \,\frac{\Delta x_c}{L_c}[/tex]

where Ys and Yc are Young's Moduli for the steel and copper.
They are

[tex]Y_s=20 \times 10^{10} N/m^2 [/tex]

[tex]Y_c= 11 \times 10^{10} N/m^2 [/tex]

so from above equation, we get , rearranging


[tex] \Delta x_s = \left(\frac{Y_c}{Y_s}\right)\left(\frac{L_s}{L_c}\right) \Delta x_c [/tex]

Since amount of stretching must be equal to the thermal contraction, we have

[tex] \Delta x_s + \Delta x_c= \Delta L_s + \Delta L_c [/tex]

where [itex] \Delta L_s \cdots \Delta L_c [/itex] are change in the length of the steel and copper rods due to cooling. Also note that

[tex] L_s = L\left[1-\alpha_s (\Delta T) \right] [/tex]

[tex] L_c= L\left[1-\alpha_c(\Delta T) \right] [/tex]

where [itex]\alpha_s \cdots \alpha_c [/itex] are Linear thermal coefficients of the steel and the copper respectively.

So, we have two equations in variables, [itex]\Delta x_s[/itex] and [itex]\Delta x_c[/itex]

we can solve for one of them now.


[tex] \Delta x_c=\left[1+\frac{Y_c}{Y_s}\frac{(1-\alpha_s \Delta T)}{(1-\alpha_c \Delta T)} \right]^{-1} L (\alpha_s +\alpha_c) \Delta T [/tex]

The values for [itex]\alpha_s \cdots \alpha_c [/itex] are

[tex]\alpha_s = 11 \times 10^{-6} (^{\circ}C)^{-1} [/tex]
[tex]\alpha_c=17 \times 10^{-6} (^{\circ}C)^{-1} [/tex]

solving we get, for various quantities described above,

[tex]\Delta x_c= 7.2555 \times 10^{-4} m [/tex]

[tex]\Delta L_c=\alpha_c L(\Delta T) =(17 \times 10^{-6})(2)(20)=6.8 \times 10^{-4} m[/tex]

[tex]\Delta L_s=\alpha_s L(\Delta T) =(11 \times 10^{-6})(2)(20)=4.4 \times 10^{-4} m[/tex]

[tex]\Delta x_s =\Delta L_s + \Delta L_c -\Delta x_c [/tex]

[tex] \Delta x_s = 3.97 \times 10^{-4} m [/tex]

and the junction shifts by the distance of , say k,

[tex]k= \Delta L_s - \Delta x_s=\Delta x_c - \Delta L_c = 0.43 \times 10^{-4} m [/tex]

Now the tension , F is given by

[tex]F= Y_s(A)\frac{\Delta x_s}{L(1-\alpha_s \Delta T)} = 119.093 \, N [/tex]

I got the value of k right, but the book's answer for F is 125 N. Is there anything wrong with
the calculations ?

I had taken the value of [itex]\Delta T = 20^{\circ}\, C [/itex], as positive.
 

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Answers and Replies

  • #2
issacnewton
983
22
any inputs ... anybody ?
 
  • #3
tiny-tim
Science Advisor
Homework Helper
25,838
255
hi issacnewton! :smile:
[tex]Y_s=20 \times 10^{10} N/m^2 [/tex]

[tex] \Delta x_s = 3.97 \times 10^{-4} m [/tex]

Now the tension , F is given by

[tex]F= Y_s(A)\frac{\Delta x_s}{L(1-\alpha_s \Delta T)} = 119.093 \, N [/tex]

I got the value of k right, but the book's answer for F is 125 N. Is there anything wrong with
the calculations ?

I had taken the value of [itex]\Delta T = 20^{\circ}\, C [/itex], as positive.

i do get about 125 N :confused:
 
  • #4
issacnewton
983
22
Oh no..........:cry:

math mistakes....:tongue2:
 

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