9.1 The mechanical workshop need a hydraulic press - NSC Mechanical Technology Fitting and Machining - Question 9 - 2017 - Paper 1

To calculate the pressure exerted on Piston A, we employ the formula:

$P_A = \frac{F_A}{A_A}$

Where:

• $F_A = 550 \text{ N}$
• $A_A$ can be calculated from the diameter found in step 9.1.1.

Thus:

$P_A = \frac{550}{A_A}$

We find $P_A$ to be approximately $108,268 \text{ kPa}.$

To find the force exerted on Piston B, we use:

$F_B = P_B \times A_B$

Given that pressure is the same on both pistons, we apply:

$F_B = 108,268 \times 10^3 \times A_B$

After calculations: $F_B = 2.76 \text{ kN}$.

The bush material is subjected to compressive stress.

The stress ($\sigma$) in the material can be calculated using:

$\sigma = \frac{F}{A}$

Where:

• $F = 23 \text{ kN} = 23000 \text{ N}$
• A can be computed from the cross-sectional area:

$A = A_{outer} - A_{inner} = \pi \left(\left(0.04 \text{ m}\right)^2 - \left(0.03 \text{ m}\right)^2\right)$

This results in: $$\sigma = 4181.18 \text{ MPa}.$

Given the gear ratio and teeth:

• Motor gear: 30 teeth
• Driven gear: 80 teeth

Using the gear ratio:

$\text{Gear ratio} = \frac{T_{driven}}{T_{motor}} = \frac{80}{30} = 2.67$

If the driven gear rotates at 90 rpm then:

$$\text{Motor frequency} = \frac{90}{2.67} \approx 480 \text{ rpm}.$

1. No slip occurs, ensuring consistent power transfer.
2. More accurate torque and speed control, resulting in improved operation.