Question

Steel train rails are laid in 12.0-m-long segments placed end to end. The rails are laid on a winter day when their temperature is −9.0∘C. (a) How much space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 33.0∘C? (b) If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 33.0∘C?

Answer 1

Answer:

a) Space = $6.05 x 10^{-3} m = 0.605 cm$

b) Stress= $-100.8 x 10^{6} Pa$

Explanation:

1) Data Given

$L = 12 m , T_i = -9 C \degree, T_f = 33 C \degree$

2) Calculate the space using Linear thermal expansion formula

We need to use Linear thermal expansion formula since the space created would be a change on 1 dimension, the increase of the temperature will increase the length of the steel.  The formula is given by:

$\Delta L = L_i \alpha_{steel} \Delta T$

We have everything except the $\alpha_{steel}$ , so we look for this on a book and we find that $\alpha_{steel} = 1.2 x 10^{-5} C^{-1}$, so we can replace.

$\Delta L = 12 m (1.2 x 10^{-5} C^{-1}) (33 C \degree -(-9 C \degree)) = 12 m (1.2 x 10^{-5} C^{-1}) 42 C \degree =6.048 x 10^{-3} m = 0.6048cm$

3) Calculate the stress of the steel

The Stress is the ratio of applied force F to a cross section area - defined as

$\sigma = \frac{F_n}{A}$

Since we don't have the force and the Area, we need to look for another way to find the stress.

For this we can use the concept called Young's Modulus, defined as : "the mechanical property that measures the stiffness of a solid material", and the formula for this is given by:

$Y =\frac{F L}{A \Delta L}$ (1)

Solving $\frac{F}{A}$ from the previous formula we have this:

$\frac{F}{A}$  = (Y  Δ L)/L  (2)

From the Linear thermal expansion formula we can solve like this

$\frac{\Delta L}{L}$ =  α  ΔT  (3)

And replacing equation (3) into equation (2) we have:

$\frac{F}{A}$  = Y α ΔT (4)

We have that the Young's Modulus for the steel is 20x10^{10} Pa, so replacing into equation (4)

$\frac{F}{A}$ = $20x10^{10}$ Pa (1.2x10^-5 C^-1) (42C) = $100.8 *10^{6}$ Pa  

That represent the absolute value for the Stress, the sign on this case would be negative since there is a compression.