Question

A circular cross-section shaft AB is subjected to loads as shown in the figure (neglecting bearing friction). The shaft is made of a material with [σ] = 500 MPa.

• Draw the internal force diagrams in shaft AB.

• Determine the diameters of the shaft sections according to the strength condition.

Answer 1

Phosphorus is primarily extracted from ores containing calcium phosphate, commonly found in minerals like apatite. To determine the minimum mass of such an ore required to obtain 1.00 kg of phosphorus, follow these steps:

1. Calculate the mass of calcium phosphate needed:

• Ca: 40.08 g/mol × 3 = 120.24 g/mol

• P: 30.97 g/mol × 2 = 61.94 g/mol

• O: 16.00 g/mol × 8 = 128.00 g/mol Total molar mass = 310.18 g/mol Mass fraction of phosphorus in Ca₃(PO₄)₂:$\frac{61.94}{310.18} \approx 0.1997$Therefore:$\text{Mass of calcium phosphate} = \frac{1.00\ \text{kg}}{0.1997} \approx 5.01\ \text{kg}$

• Assuming the ore contains 55.6% calcium phosphate by mass, the mass of calcium phosphate required is:$\text{Mass of calcium phosphate} = \frac{\text{Mass of phosphorus}}{\text{Mass fraction of phosphorus in calcium phosphate}}$The chemical formula for calcium phosphate is Ca₃(PO₄)₂. Molar mass of Ca₃(PO₄)₂:

2. Calculate the mass of the ore required:

• Since the ore contains 55.6% calcium phosphate by mass:$\text{Mass of ore} = \frac{\text{Mass of calcium phosphate}}{\text{Mass fraction of calcium phosphate in ore}}$$\text{Mass of ore} = \frac{5.01\ \text{kg}}{0.556} \approx 9.01\ \text{kg}$

Therefore, to obtain 1.00 kg of phosphorus, a minimum of approximately 9.01 kg of the ore must be processed.