Question:

A thick cylinder has inner diameter 20 mm and outer diameter 40 mm, subjected to internal pressure of 100 MPa. Tensile stress is positive and compressive stress negative. The sum of radial and hoop stresses (in MPa) at radius 15 mm is \(\underline{\hspace{2cm}}\) (round off to two decimal places).

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In thick cylinders, $\sigma_r + \sigma_\theta$ is constant and equals $2A$, independent of radius.
Updated On: Jan 13, 2026
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Correct Answer: 66 - 67

Solution and Explanation

Inner radius: \[ r_i = 10\ \text{mm}, r_o = 20\ \text{mm}. \] Lame's constants: \[ A = \frac{p r_i^2}{r_o^2 - r_i^2} = \frac{100 \times 100}{400 - 100} = 33.333\ \text{MPa}. \] \[ B = \frac{p r_i^2 r_o^2}{r_o^2 - r_i^2} = \frac{100 \times 100 \times 400}{300} = 13333.33\ \text{MPa} \cdot \text{mm}^2. \] Radial stress: \[ \sigma_r(r) = A - \frac{B}{r^2}. \] Hoop stress: \[ \sigma_\theta(r) = A + \frac{B}{r^2}. \] Sum at \( r = 15\ \text{mm} \): \[ \sigma_r + \sigma_\theta = (A - B/r^2) + (A + B/r^2) = 2A = 66.666 \ \text{MPa}. \] Rounded to two decimal places: \[ 66.67\ \text{MPa}. \]

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