A hollow mild steel tube 6 m long has an external diameter of 150 mm. The bore diameter is 85 mm throughout the length of the tube. The tube is subjected to an axial tensile load of 65 tonnes and the extension was measured to be 3 mm. Determine the value of Youngs modulus for the tube.
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Answer:
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To answer your question, we need to use the formula for Young's modulus, which is the ratio of stress to strain for a material under axial loading.
$$E = \frac{\sigma}{\epsilon}$$
where $E$ is Young's modulus, $\sigma$ is stress, and $\epsilon$ is strain.
To find the stress, we need to use the formula for the cross-sectional area of a hollow tube, which is the difference between the areas of the outer and inner circles.
$$A = \pi \left( \frac{D^2 - d^2}{4} \right)$$
where $A$ is the cross-sectional area, $D$ is the outer diameter, and $d$ is the inner diameter.
Substituting the given values, we get:
$$A = \pi \left( \frac{(0.15)^2 - (0.085)^2}{4} \right)$$
$$A \approx 0.012 \text{ m}^2$$
The stress is then the force divided by the area:
$$\sigma = \frac{F}{A}$$
where $F$ is the force.
Substituting the given values, we get:
$$\sigma = \frac{65 \times 10^3}{0.012}$$
$$\sigma \approx 5.42 \times 10^6 \text{ N/m}^2$$
To find the strain, we need to use the formula for the change in length divided by the original length.
$$\epsilon = \frac{\Delta L}{L}$$
where $\epsilon$ is the strain, $\Delta L$ is the change in length, and $L$ is the original length.
Substituting the given values, we get:
$$\epsilon = \frac{0.003}{6}$$
$$\epsilon \approx 5 \times 10^{-4}$$
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Answer:
Already given before... If u remember