Problem about tensile stress during thermal contraction

[issacnewton]

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 $\Delta x_s$
from its reduced length L_s. Similarly, let copper wire be stretched by amount
$\Delta x_c$ from its reduced length L_c. Let L = 2 m be the original length of each wire. Since

$$\mbox{Young's Modulus}\, =\frac{\mbox{Tensile Stress}}{\mbox{Tensile Strain}}$$

we have

$$\frac{F}{A}= Y_s \,\frac{\Delta x_s}{L_s} = Y_c \,\frac{\Delta x_c}{L_c}$$

where Y_s and Y_c are Young's Moduli for the steel and copper.
They are

$$Y_s=20 \times 10^{10} N/m^2$$

$$Y_c= 11 \times 10^{10} N/m^2$$

so from above equation, we get , rearranging


$$\Delta x_s = \left(\frac{Y_c}{Y_s}\right)\left(\frac{L_s}{L_c}\right) \Delta x_c$$

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

$$\Delta x_s + \Delta x_c= \Delta L_s + \Delta L_c$$

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

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

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

where $\alpha_s \cdots \alpha_c$ are Linear thermal coefficients of the steel and the copper respectively.

So, we have two equations in variables, $\Delta x_s$ and $\Delta x_c$

we can solve for one of them now.


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

The values for $\alpha_s \cdots \alpha_c$ are

$$\alpha_s = 11 \times 10^{-6} (^{\circ}C)^{-1}$$
$$\alpha_c=17 \times 10^{-6} (^{\circ}C)^{-1}$$

solving we get, for various quantities described above,

$$\Delta x_c= 7.2555 \times 10^{-4} m$$

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

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

$$\Delta x_s =\Delta L_s + \Delta L_c -\Delta x_c$$

$$\Delta x_s = 3.97 \times 10^{-4} m$$

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

$$k= \Delta L_s - \Delta x_s=\Delta x_c - \Delta L_c = 0.43 \times 10^{-4} m$$

Now the tension , F is given by

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

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 $\Delta T = 20^{\circ}\, C$, as positive.

 

[issacnewton]

any inputs ... anybody ?

 

[tiny-tim]

hi IssacNewton!

$$Y_s=20 \times 10^{10} N/m^2$$
…
$$\Delta x_s = 3.97 \times 10^{-4} m$$
…
Now the tension , F is given by

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

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 $\Delta T = 20^{\circ}\, C$, as positive.

i do get about 125 N

 

[issacnewton]

Oh no...

math mistakes...:tongue2:

 

[rwisz]

But in this case, the thermal contraction would actually cause the wire to expand. So, the value of \Delta T should be negative, i.e. -20^{\circ} \, C . This would result in a negative value for \Delta x_c and a positive value for \Delta x_s, which would give a tension of 125 N. So, the issue was with the sign of \Delta T.

Additionally, it is important to note that the values of Young's Moduli for steel and copper may vary depending on the specific composition and properties of the materials being used. These values are just estimates and may not be exact for the materials in the problem. This could also contribute to a slight difference in the final tension value. It is always important to double check the values used in calculations to ensure accuracy.