QUESTION 14
FINAL QUESTION - Part 4:
You are determining the stress free temperature of a UIC railway line by using a lifting frame. A
linear regression on the calibration data gave . You measure an uplift force of 16 kN at
a rail temperature of .
Calculate the SFT using the equation:
Finally, provide the value of SFT in degrees Celsius:
(Provide a numeric value only. If you need to place decimal point, use of full-stop.)

Ask by Lindsey Perkins. in South Africa

Mar 27,2025

Question

QUESTION 14
FINAL QUESTION - Part 4:
You are determining the stress free temperature of a UIC railway line by using a lifting frame. A
linear regression on the calibration data gave . You measure an uplift force of 16 kN at
a rail temperature of .
Calculate the SFT using the equation:
Finally, provide the value of SFT in degrees Celsius:
(Provide a numeric value only. If you need to place decimal point, use of full-stop.)

Ask by Lindsey Perkins. in South Africa

Mar 27,2025

UpStudy AI · Accepted Answer

To calculate the stress-free temperature (SFT) using the equation

$$
\Delta t = \frac{y - c}{m \cdot a \cdot A \cdot E}
$$

we need to identify the variables from the problem:

- $$ y = 0.0117 \cdot x + 1.455 $$ (we will need to find $$ y $$ for $$ x = 19 $$ °C)
- $$ c = 1.455 $$ (the constant term from the regression equation)
- $$ m = 60 $$ kg/m (the mass per unit length of the railway line)
- $$ a = 16 $$ kN (the uplift force, which we need to convert to Newtons)
- $$ A $$ and $$ E $$ are not provided in the problem, so we will assume $$ A \cdot E = 1 $$ for the sake of calculation.

Let's first calculate $$ y $$ for $$ x = 19 $$:

$$
y = 0.0117 \cdot 19 + 1.455
$$

Next, we will convert the uplift force from kN to N:

$$
a = 16 \text{ kN} = 16000 \text{ N}
$$

Now we can substitute these values into the equation to find $$ \Delta t $$.

Let's perform the calculations step by step. First, we will calculate $$ y $$.
Calculate the value by following steps:
- step0: Calculate:
  $$0.0117\times 19+1.455$$
- step1: Multiply the numbers:
  $$0.2223+1.455$$
- step2: Add the numbers:
  $$1.6773$$
Calculating $$ y $$ for $$ x = 19 $$:

$$
y = 0.0117 \cdot 19 + 1.455 = 1.6773
$$

Now, we can substitute $$ y $$, $$ c $$, $$ m $$, $$ a $$, and $$ A \cdot E $$ into the equation for $$ \Delta t $$:

$$
\Delta t = \frac{1.6773 - 1.455}{60 \cdot 16000 \cdot 1}
$$

Let's calculate $$ \Delta t $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(1.6773-1.455\right)}{\left(60\times 16000\right)}$$
- step1: Remove the parentheses:
  $$\frac{1.6773-1.455}{60\times 16000}$$
- step2: Multiply the numbers:
  $$\frac{1.6773-1.455}{960000}$$
- step3: Subtract the numbers:
  $$\frac{0.2223}{960000}$$
- step4: Convert the expressions:
  $$\frac{\frac{2223}{10000}}{960000}$$
- step5: Multiply by the reciprocal:
  $$\frac{2223}{10000}\times \frac{1}{960000}$$
- step6: Reduce the numbers:
  $$\frac{741}{10000}\times \frac{1}{320000}$$
- step7: Multiply the fractions:
  $$\frac{741}{10000\times 320000}$$
- step8: Multiply:
  $$\frac{741}{3200000000}$$
Calculating $$ \Delta t $$:

$$
\Delta t = \frac{1.6773 - 1.455}{60 \cdot 16000} = 2.315625 \times 10^{-7}
$$

Now, to find the stress-free temperature (SFT), we need to add $$ \Delta t $$ to the measured rail temperature of $$ 19^{\circ} \mathrm{C} $$:

$$
\text{SFT} = 19 + \Delta t
$$

Let's calculate the SFT.
Calculate the value by following steps:
- step0: Calculate:
  $$19+2.315625e-7$$
- step1: Subtract the numbers:
  $$12+2.315625e$$
- step2: Simplify:
  $$12+6.294521$$
- step3: Calculate:
  $$18.294521$$
The calculated stress-free temperature (SFT) is:

$$
\text{SFT} \approx 18.294521
$$

Thus, the final value of SFT in degrees Celsius is:

**18.294521**
The stress-free temperature (SFT) is approximately **18.294521°C**.

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