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Calculate the stress on the supports of a petrol tank and check for resistance conditions.

Physics

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Answer

2.30 MPa\text{2.30 MPa}

Step-by-step solution

7 steps

To check the condition of resistance, we need to calculate the stress on the supports and compare it to an allowable stress for the material. Since no allowable stress is provided, we will calculate the stress and state that further verification requires this value.

  1. Calculate the volume of petrol in the tank.

    The internal dimensions are length L=8 mL = 8 \text{ m}, width W=5 mW = 5 \text{ m}, and height H=2 mH = 2 \text{ m}.

    V=L×W×H=8 m×5 m×2 m=80 m3V = L \times W \times H = 8 \text{ m} \times 5 \text{ m} \times 2 \text{ m} = 80 \text{ m}^3
  2. Calculate the mass of the petrol.

    The density of petrol is ρ=8.8 kg/m3\rho = 8.8 \text{ kg/m}^3.

    mpetrol=ρ×V=8.8 kg/m3×80 m3=704 kgm_{\text{petrol}} = \rho \times V = 8.8 \text{ kg/m}^3 \times 80 \text{ m}^3 = 704 \text{ kg}
  3. Calculate the weight of the petrol.

    Using the acceleration due to gravity g=9.81 m/s2g = 9.81 \text{ m/s}^2.

    Wpetrol=mpetrol×g=704 kg×9.81 m/s2=6906.24 NW_{\text{petrol}} = m_{\text{petrol}} \times g = 704 \text{ kg} \times 9.81 \text{ m/s}^2 = 6906.24 \text{ N}
  4. Calculate the total weight supported by the four supports.

    The weight of the tank is Wtank=500 NW_{\text{tank}} = 500 \text{ N}.

    Wtotal=Wtank+Wpetrol=500 N+6906.24 N=7406.24 NW_{\text{total}} = W_{\text{tank}} + W_{\text{petrol}} = 500 \text{ N} + 6906.24 \text{ N} = 7406.24 \text{ N}
  5. Calculate the total cross-sectional area of the four supports.

    Each support has a diameter D=32 mm=0.032 mD = 32 \text{ mm} = 0.032 \text{ m}.
    The radius of one support is R=D/2=0.032 m/2=0.016 mR = D/2 = 0.032 \text{ m} / 2 = 0.016 \text{ m}.
    The cross-sectional area of one support is:

    Aone support=πR2=π(0.016 m)2=π×0.000256 m2≈0.00080425 m2A_{\text{one support}} = \pi R^2 = \pi (0.016 \text{ m})^2 = \pi \times 0.000256 \text{ m}^2 \approx 0.00080425 \text{ m}^2

    Since there are four supports, the total cross-sectional area is:

    Atotal=4×Aone support=4×0.00080425 m2≈0.003217 m2A_{\text{total}} = 4 \times A_{\text{one support}} = 4 \times 0.00080425 \text{ m}^2 \approx 0.003217 \text{ m}^2
  6. Calculate the compressive stress on the supports.

    Stress σ\sigma is defined as force per unit area.

    σ=WtotalAtotal=7406.24 N0.003217 m2≈2302390 Pa\sigma = \frac{W_{\text{total}}}{A_{\text{total}}} = \frac{7406.24 \text{ N}}{0.003217 \text{ m}^2} \approx 2302390 \text{ Pa}

    Converting to megapascals (MPa):

    σ≈2.30 MPa\sigma \approx 2.30 \text{ MPa}
  7. Check the condition of resistance.

    The condition of resistance is verified if the calculated stress (σ\sigma) is less than or equal to the allowable stress (σallowable\sigma_{\text{allowable}}) for the material of the supports.

    σ≤σallowable\sigma \le \sigma_{\text{allowable}}

    Since the problem does not provide an allowable stress value for the support material, we cannot definitively verify the condition of resistance. We have calculated the stress on the supports.

    The stress on the supports is approximately 2.30 MPa\boxed{\text{2.30 MPa}}.
    To verify the condition of resistance, this calculated stress must be compared to the allowable stress of the support material, which is not provided in the problem statement.

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