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Mechanical Properties of Solids - Applications of Elastic Behaviour of Materials

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Design of Structural Beams: In construction, beams are designed to minimize bending under heavy loads. For a beam of length ll, breadth bb, and depth dd, loaded at the center with weight WW, the depression δ\delta is given by δ=Wl34bd3Y\delta = \frac{Wl^3}{4bd^3Y}. To minimize δ\delta, Young's modulus YY and depth dd should be large.

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I-shaped Beams: To reduce the weight of the beam while maintaining high strength and minimizing depression, beams are given an 'I' shape. This shape provides a large load-bearing surface and enough depth to prevent bending and buckling, while using less material than a solid rectangular bar.

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Crane Ropes: Cranes use metallic ropes (usually steel) made of braided strands. Braiding increases flexibility and strength. The radius rr of the rope is chosen such that the stress produced by the maximum load does not exceed the elastic limit of the material. A safety factor (usually 10) is applied to ensure the rope can handle unexpected jerks.

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Maximum Height of Mountains: The maximum height of a mountain on Earth is limited by the shear strength of the rocks at its base. At the base, the pressure is hρgh\rho g. This pressure must be less than the elastic limit of the rock. For typical rocks, this limits mountain height to approximately 10 km10 \text{ km}.

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Buckling: When a long vertical rod is loaded, it may bend or 'buckle' instead of compressing. Proper application of elastic behavior helps engineers calculate the 'critical load' to prevent structural failure.

📐Formulae

δ=Wl34bd3Y\delta = \frac{Wl^3}{4bd^3Y}

Stress=Mgπr2\text{Stress} = \frac{Mg}{\pi r^2}

r≥Mgπ(Yield Strength)r \ge \sqrt{\frac{Mg}{\pi (\text{Yield Strength})}}

hmax=σyρgh_{max} = \frac{\sigma_{y}}{\rho g}

💡Examples

Problem 1:

A crane is designed to lift a maximum load of 104 kg10^4 \text{ kg}. If the yield strength of the steel used for the rope is 300×106 N m−2300 \times 10^6 \text{ N m}^{-2}, find the minimum radius of the rope required if a safety factor of 10 is desired. Take g=10 m s−2g = 10 \text{ m s}^{-2}.

Solution:

M=104 kgM = 10^4 \text{ kg} g=10 m s−2g = 10 \text{ m s}^{-2} Working Stress=Yield Strength10=300×10610=30×106 N m−2\text{Working Stress} = \frac{\text{Yield Strength}}{10} = \frac{300 \times 10^6}{10} = 30 \times 10^6 \text{ N m}^{-2} Stress=MgA=Mgπr2\text{Stress} = \frac{Mg}{A} = \frac{Mg}{\pi r^2} 30×106=104×10πr230 \times 10^6 = \frac{10^4 \times 10}{\pi r^2} r2=1053.14×30×106=13.14×300≈0.00106 m2r^2 = \frac{10^5}{3.14 \times 30 \times 10^6} = \frac{1}{3.14 \times 300} \approx 0.00106 \text{ m}^2 r≈0.00106≈0.0326 m=3.26 cmr \approx \sqrt{0.00106} \approx 0.0326 \text{ m} = 3.26 \text{ cm}

Explanation:

The radius is calculated by ensuring the stress exerted by the maximum load stays within the safety-adjusted yield strength of the material.

Problem 2:

Calculate the maximum height of a mountain on Earth, given that the elastic limit of a typical rock is 3×108 N m−23 \times 10^8 \text{ N m}^{-2} and the density of the rock is 3×103 kg m−33 \times 10^3 \text{ kg m}^{-3}.

Solution:

Pressure at base=hρg\text{Pressure at base} = h\rho g Elastic Limit=3×108 N m−2\text{Elastic Limit} = 3 \times 10^8 \text{ N m}^{-2} hρg≤3×108h\rho g \le 3 \times 10^8 h≤3×1083×103×10h \le \frac{3 \times 10^8}{3 \times 10^3 \times 10} h≤3×1083×104h \le \frac{3 \times 10^8}{3 \times 10^4} h≤104 m=10 kmh \le 10^4 \text{ m} = 10 \text{ km}

Explanation:

The vertical pressure at the base of the mountain depends on its height and density. If this pressure exceeds the elastic limit (shear strength) of the rock, the rock will flow or break, causing the mountain to sink.