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A load of 40 kN causes a compressive stress of 16 MPa in a square brass bar - NSC Technical Sciences - Question 5 - 2022 - Paper 1

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A load of 40 kN causes a compressive stress of 16 MPa in a square brass bar. Young's modulus for brass is 90 GPa. The original length of the bar is 300 mm. 5.1 Dete... show full transcript

Worked Solution & Example Answer:A load of 40 kN causes a compressive stress of 16 MPa in a square brass bar - NSC Technical Sciences - Question 5 - 2022 - Paper 1

Step 1

5.1.1 The length, x, of one side of the square brass bar. Give your answer in millimetres.

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Answer

To find the length of one side, we start from the formula for stress:

σ=FAσ = \frac{F}{A}

Where:

  • σ = stress = 16 , \text{MPa} = 16 \times 10^6 , \text{Pa}
  • F = load = 40 , \text{kN} = 40 \times 10^3 , \text{N}
  • A = area = x^2 (where x is one side of the square bar)

Substituting the values, we have:

16×106=40×103x216 \times 10^6 = \frac{40 \times 10^3}{x^2}

Rearranging gives:

x2=40×10316×106=0.0025 m2x^2 = \frac{40 \times 10^3}{16 \times 10^6} = 0.0025 \, \text{m}^2

Thus, taking the square root:

x=0.05 m=50 mmx = 0.05 \, \text{m} = 50 \, \text{mm}

Step 2

5.1.2 The strain caused by the load

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Answer

The strain (ε) can be calculated using:

ε=σEε = \frac{σ}{E}

Where:

  • E = Young's modulus = 90 , \text{GPa} = 90 \times 10^9 , \text{Pa}

Substituting the known values:

ε=16×10690×109=1.78×10−4ε = \frac{16 \times 10^6}{90 \times 10^9} = 1.78 \times 10^{-4}

Step 3

5.1.3 The change in length, in millimetres, caused by the load

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Answer

The change in length (ΔL) can be calculated with:

ΔL=L×εΔL = L \times ε

Where L is the original length of the bar:

  • L = 300 , \text{mm} = 0.3 , \text{m}

Thus, substituting values:

ΔL=0.3×1.78×10−4=0.0000534 m=0.0534 mmΔL = 0.3 \times 1.78 \times 10^{-4} = 0.0000534 \, \text{m} = 0.0534 \, \text{mm}

Step 4

5.2.1 The fluid pressure in the hydraulic system when in equilibrium

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Answer

In equilibrium, the pressure (P) can be calculated using:

P=FAP = \frac{F}{A}

For Piston B:

  • F = 20 , \text{kN} = 20 \times 10^3 , \text{N}
  • The area (A) for Piston B (diameter = 200 mm):
A=π(d2)2=π(0.22)2=π4×(0.2)2=3.142×10−2 m2A = \pi \left( \frac{d}{2} \right)^2 = \pi \left( \frac{0.2}{2} \right)^2 = \frac{\pi}{4} \times (0.2)^2 = 3.142 \times 10^{-2} \, \text{m}^2

Then:

P=20×1033.142×10−2=636.54×103 PaP = \frac{20 \times 10^3}{3.142 \times 10^{-2}} = 636.54 \times 10^3 \, \text{Pa}

Step 5

5.2.2 The force, F, that must be exerted onto Piston A

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Answer

To find the force F, we can use the pressure calculated earlier:

P=FAAP = \frac{F}{A_A}

Where A_A is the area for Piston A:

  • Diameter = 50 mm = 0.05 m:
AA=π(0.052)2=3.142×10−3 m2A_A = \pi \left( \frac{0.05}{2} \right)^2 = 3.142 \times 10^{-3} \, \text{m}^2

Now substituting:

  • Using P = 636.54 \times 10^3 , \text{Pa}:
F=P×AA=636.54×103×3.142×10−3=2000 N=2 kNF = P \times A_A = 636.54 \times 10^3 \times 3.142 \times 10^{-3} = 2000 \, \text{N} = 2 \, \text{kN}

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