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Deformation of Solids: Question 9

Syllabus 6.1, 6.2

Structured AS 10 marks

A steel support strut in a model bridge has an original (unloaded) length of 0.800 m0.800\text{ m} and a uniform circular cross-section of diameter 6.0 mm6.0\text{ mm}. When a compressive force of 2.8 kN2.8\text{ kN} is applied along its length, the strut shortens by 0.40 mm0.40\text{ mm}. The strut obeys Hooke's law throughout this compression, with stress and strain related in the same way as for extension.

(a) Calculate the cross-sectional area of the strut. [2]

(b) Calculate the compressive stress in the strut when the 2.8 kN2.8\text{ kN} force is applied. [2]

(c) Calculate the compressive strain in the strut when the 2.8 kN2.8\text{ kN} force is applied. [2]

(d) Determine the Young modulus of the steel. [2]

(e) Calculate the elastic strain energy stored in the strut at this compression. [2]

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Worked solution

Part (a): Cross-sectional area

The strut has diameter d=6.0 mmd = 6.0\text{ mm}, so its radius is: r=d2=6.0×10−32=3.0×10−3 mr = \frac{d}{2} = \frac{6.0\times10^{-3}}{2} = 3.0\times10^{-3}\text{ m}

The cross-sectional area is: A=πr2=π×(3.0×10−3)2=π×9.0×10−6A = \pi r^2 = \pi \times (3.0\times10^{-3})^2 = \pi \times 9.0\times10^{-6} A=2.83×10−5 m2A = 2.83\times10^{-5}\text{ m}^2

Part (b): Compressive stress

First convert the force to newtons: F=2.8 kN=2800 NF = 2.8\text{ kN} = 2800\text{ N}

Stress is force per unit cross-sectional area: σ=FA=28002.83×10−5\sigma = \frac{F}{A} = \frac{2800}{2.83\times10^{-5}} σ=9.89×107 Pa\sigma = 9.89\times10^{7}\text{ Pa}

Part (c): Compressive strain

Convert the shortening to metres: x=0.40 mm=4.0×10−4 mx = 0.40\text{ mm} = 4.0\times10^{-4}\text{ m}

Strain is the (compressive) extension per unit original length: ε=xL=4.0×10−40.800\varepsilon = \frac{x}{L} = \frac{4.0\times10^{-4}}{0.800} ε=5.0×10−4\varepsilon = 5.0\times10^{-4}

(Strain is a ratio of two lengths, so it has no units.)

Part (d): Young modulus

The Young modulus is the ratio of stress to strain, within the limit of proportionality: E=σε=9.89×1075.0×10−4E = \frac{\sigma}{\varepsilon} = \frac{9.89\times10^{7}}{5.0\times10^{-4}} E=1.98×1011 PaE = 1.98\times10^{11}\text{ Pa}

As a check, using E=FLAxE=\dfrac{FL}{Ax} directly: E=2800×0.8002.83×10−5×4.0×10−4=22401.132×10−8=1.98×1011 PaE = \frac{2800\times0.800}{2.83\times10^{-5}\times4.0\times10^{-4}} = \frac{2240}{1.132\times10^{-8}} = 1.98\times10^{11}\text{ Pa} Both methods agree. This value is close to 200 GPa200\text{ GPa}, consistent with the accepted Young modulus of steel (typically about 190190–210 GPa210\text{ GPa}).

Part (e): Elastic strain energy

Since the strut obeys Hooke’s law throughout this compression, the elastic strain energy stored is the area under a force–extension graph up to this point, a triangle: Ep=12FxE_p = \tfrac12 F x

Substituting F=2800 NF = 2800\text{ N} and x=4.0×10−4 mx = 4.0\times10^{-4}\text{ m}: Ep=12×2800×4.0×10−4=12×1.12E_p = \tfrac12 \times 2800 \times 4.0\times10^{-4} = \tfrac12 \times 1.12 Ep=0.56 JE_p = 0.56\text{ J}

Final answers

  • (a) Cross-sectional area =2.83×10−5 m2= \boxed{2.83\times10^{-5}}\text{ m}^2
  • (b) Compressive stress =9.89×107 Pa= \boxed{9.89\times10^{7}}\text{ Pa}
  • (c) Compressive strain =5.0×10−4= \boxed{5.0\times10^{-4}} (no units)
  • (d) Young modulus =1.98×1011 Pa= \boxed{1.98\times10^{11}}\text{ Pa} (≈198 GPa\approx 198\text{ GPa}, consistent with steel)
  • (e) Elastic strain energy =0.56 J= \boxed{0.56}\text{ J}