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A hollow steel tube with an inside diameter of 100 mm must carry a load of 380 kN. Calculate the outside diameter of the tube if the stress is limited to 110 MN/m2.

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2 parts
a

15.61 MPa\text{15.61 MPa}

b

1.17×10−41.17 \times 10^{-4}

Step-by-step solution

5 steps

Here's how to solve question 4:

Part a· 3 steps

The stress in the wire.

  1. Calculate the force exerted by the mass.

    The force (FF) is the weight of the mass, where g=9.81 m/s2g = 9.81 \text{ m/s}^2.

    F=m×gF = m \times g F=20 kg×9.81 m/s2F = 20 \text{ kg} \times 9.81 \text{ m/s}^2 F=196.2 NF = 196.2 \text{ N}

  2. Calculate the cross-sectional area of the wire.

    The diameter (DD) is 4 mm4 \text{ mm}, which is 4×10−3 m4 \times 10^{-3} \text{ m}. The area (AA) of a circular cross-section is given by π(D2)2\pi \left(\frac{D}{2}\right)^2. Use π=3.142\pi = 3.142.

    A=π(D2)2A = \pi \left(\frac{D}{2}\right)^2 A=3.142×(4×10−3 m2)2A = 3.142 \times \left(\frac{4 \times 10^{-3} \text{ m}}{2}\right)^2 A=3.142×(2×10−3 m)2A = 3.142 \times (2 \times 10^{-3} \text{ m})^2 A=3.142×(4×10−6 m2)A = 3.142 \times (4 \times 10^{-6} \text{ m}^2) A=1.2568×10−5 m2A = 1.2568 \times 10^{-5} \text{ m}^2

  3. Calculate the stress in the wire.

    Stress (σ\sigma) is force per unit area.

    σ=FA\sigma = \frac{F}{A} σ=196.2 N1.2568×10−5 m2\sigma = \frac{196.2 \text{ N}}{1.2568 \times 10^{-5} \text{ m}^2} σ=15611075.75 Pa\sigma = 15611075.75 \text{ Pa} σ=15.61×106 Pa\sigma = 15.61 \times 10^6 \text{ Pa} σ=15.61 MPa\sigma = \boxed{\text{15.61 MPa}}

Part b

1.17×10−41.17 \times 10^{-4}

The strain in the wire.

  1. Convert the change in length to meters.

    The stretch (ΔL\Delta L) is 0.35 mm0.35 \text{ mm}, which is 0.35×10−3 m0.35 \times 10^{-3} \text{ m}. The original length (LL) is 3 m3 \text{ m}.

  2. Calculate the strain.

    Strain (ϵ\epsilon) is the ratio of the change in length to the original length.

    ϵ=ΔLL\epsilon = \frac{\Delta L}{L} ϵ=0.35×10−3 m3 m\epsilon = \frac{0.35 \times 10^{-3} \text{ m}}{3 \text{ m}} ϵ=0.00011666...\epsilon = 0.00011666... ϵ=1.17×10−4\epsilon = \boxed{1.17 \times 10^{-4}}
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