What is the thickness needed for a blank, unstayed, full-hemispherical head with a radius of 700 mm under a pressure of 4000 kPa?

What is the thickness needed for a blank, unstayed, full-hemispherical head with a radius of 700 mm under a pressure of 4000 kPa?

• A. 13.8 mm
• B. 14.0 mm
• C. 14.8 mm
• D. 15.0 mm

Answer: (C)

To determine the thickness required for a blank, unstayed, full-hemispherical head under a specific internal pressure, several factors must be considered, including the internal pressure, the radius of the head, and the material's allowable stress. The formula used for calculating the thickness of a hemispherical head subjected to internal pressure is derived from the principles of pressure vessel design. Specifically, the thickness \( t \) necessary can be expressed as: \[ t = \frac{P \cdot R}{2 \cdot S} \] Where: - \( P \) is the internal pressure (in Pascals), - \( R \) is the radius of the head (in meters), - \( S \) is the allowable stress of the material (in Pascals). In this case, substituting the values provided: - The radius \( R \) is 700 mm (or 0.7 m), - The internal pressure \( P \) is 4000 kPa (or 4,000,000 Pa). For a typical material used in such applications, you would have a standard allowable stress value. Assuming a value for the allowable stress typically used in engineering calculations, substituting these quantities into the formula would yield a

To determine the thickness required for a blank, unstayed, full-hemispherical head under a specific internal pressure, several factors must be considered, including the internal pressure, the radius of the head, and the material's allowable stress.

The formula used for calculating the thickness of a hemispherical head subjected to internal pressure is derived from the principles of pressure vessel design. Specifically, the thickness ( t ) necessary can be expressed as:

[ t = \frac{P \cdot R}{2 \cdot S} ]

Where:

• ( P ) is the internal pressure (in Pascals),

• ( R ) is the radius of the head (in meters),

• ( S ) is the allowable stress of the material (in Pascals).

In this case, substituting the values provided:

• The radius ( R ) is 700 mm (or 0.7 m),

• The internal pressure ( P ) is 4000 kPa (or 4,000,000 Pa).

For a typical material used in such applications, you would have a standard allowable stress value. Assuming a value for the allowable stress typically used in engineering calculations, substituting these quantities into the formula would yield a