9.1 The mechanical workshop need a hydraulic press. The diameter of Piston B is 180 mm and moves up by 12 mm. The force applied on Piston A is 550 N. Piston A moves 60 mm downwards. Use the specifications shown in FIGURE 9.1 and calculate the following: (Hint: $V_A = V_B$) 9.1.1 The diameter of Piston A 9.1.2 The pressure exerted on Piston A 9.1.3 The force exerted on Piston B 9.2 Name the type of stress that the bush material is subjected to. 9.2.2 Calculate the stress in the material. Indicate the answer in MPa.

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

Question 9

9.1 The mechanical workshop need a hydraulic press. The diameter of Piston B is 180 mm and moves up by 12 mm. The force applied on Piston A is 550 N. Piston A moves ... show full transcript

Worked Solution & Example Answer:9.1 The mechanical workshop need a hydraulic press - NSC Mechanical Technology Welding and Metalwork - Question 9 - 2017 - Paper 1

Step 1

9.1.1 The diameter of Piston A

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Answer

To calculate the diameter of Piston A, we know that the volume of both pistons must be equal: $V_A = V_B$ .

We can use the formula for the volume of a cylinder:

$V = A \times L$

where $A$ is the cross-sectional area and $L$ is the stroke length.

The area $A_A$ for Piston A is given by:

$A_A = \pi \frac{D_A^2}{4}$

Given that the diameter of Piston B, $D_B$ , is 180 mm:

Convert to meters: $D_B = 0.18$ m.
Stroke length for Piston B is 12 mm = 0.012 m.

Calculate the volume for Piston B:

$V_B = A_B \times L_B = \pi \frac{(0.18)^2}{4} \times 0.012$

Solving gives us: $V_B = 5.08 \times 10^{-3} m^3$

Now set $V_B = V_A$ and substitute: $A_A \times 0.060 = 5.08 \times 10^{-3}$

Solving for $A_A$ gives: $A_A = 0.0847 m^2$

Now find $D_A$ :

ightarrow D_A = 0.32 ext{ m or } 320 ext{ mm}$$.

Step 2

9.1.2 The pressure exerted on Piston A

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Answer

The pressure exerted on Piston A can be calculated using:

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

Where:

$F_A = 550 N$ (force applied)
$A_A = 0.0847 m^2$ (area calculated above)

Substituting values gives us:

$P_A = \frac{550}{0.0847} = 6497.2 Pa = 6497.2 kPa$ .

Step 3

9.1.3 The force exerted on Piston B

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Answer

Using Pascal's principle, pressure at Piston A is equal to the pressure at Piston B ( $P_A = P_B$ ). Thus, we can write:

$P_B = \frac{F_B}{A_B}$

Where

We already processed $P_A = 6497.2 Pa$
Area $A_B = \pi \frac{(0.18)^2}{4} = 0.02545 m^2$

Setting the pressures equal gives:

$6497.2 = \frac{F_B}{0.02545}$

Solving for $F_B$ gives: $F_B ext{(Force)} = 6497.2 \times 0.02545 = 165.52 N$ .

Step 4

9.2 Name the type of stress that the bush material is subjected to.

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Answer

The type of stress that the bush material is subjected to is compressive stress.

Step 5

9.2.2 Calculate the stress in the material. Indicate the answer in MPa.

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Answer

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

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

$F = F_B$ , the force calculated earlier
$A$ is the cross-sectional area of the bush.

Assuming the area of the bush section is similar to that of Piston A, we will first ensure the area units are in square meters, convert the force to Newtons to calculate:

$\sigma = \frac{165.52 \: N}{A_{bush}}$

If $A_{bush}$ is similar as before (0.0847 m^2), we will proceed:
Thus,

ightarrow 1958.6 Pa \rightarrow 1.96 MPa$$.

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

Worked Solution & Example Answer:9.1 The mechanical workshop need a hydraulic press - NSC Mechanical Technology Welding and Metalwork - Question 9 - 2017 - Paper 1

9.1.1 The diameter of Piston A

9.1.2 The pressure exerted on Piston A

9.1.3 The force exerted on Piston B

9.2 Name the type of stress that the bush material is subjected to.

9.2.2 Calculate the stress in the material. Indicate the answer in MPa.

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