Chapter 8: Mechanical Properties of Solids Case Study Questions | physics Class 11 CBSE

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Chapter 8: Mechanical Properties of Solids Case Study Questions | physics Class 11 CBSE | ByteNotes.in

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An engineering student is testing a structural steel rod in a laboratory. The rod has a radius of 10 mm and a length of 1.0 m. A tensile force of 100 kN is applied along its length. The student records the elongation and calculates the stress and strain. The Young's modulus of structural steel is given as 2.0 × 10¹¹ N m⁻². The student also notes that when the force is removed, the rod returns exactly to its original dimensions, confirming elastic behaviour. The experiment is repeated with increasing loads and a stress-strain graph is plotted. The linear region is observed up to a certain point, beyond which the proportionality breaks down. The student refers to Table 8.1 from the NCERT textbook to compare properties of different materials.

Question 1: Define stress and calculate the stress experienced by the steel rod when a 100 kN force is applied over its cross-sectional area.

Stress is defined as the restoring force developed per unit area inside a deformed body: stress = F/A.
Cross-sectional area A = πr² = π × (10⁻²)² = 3.14 × 10⁻⁴ m².
Stress = 100 × 10³ / 3.14 × 10⁻⁴ = 3.18 × 10⁸ N m⁻² (Pascal).

Question 2: Calculate the elongation of the steel rod and determine the strain. State the region of the stress-strain curve this corresponds to.

Elongation ΔL = (F/A × L)/Y = (3.18 × 10⁸ × 1.0) / (2.0 × 10¹¹) = 1.59 × 10⁻³ m = 1.59 mm.
Strain = ΔL/L = 1.59 × 10⁻³ / 1.0 = 1.59 × 10⁻³ (or 0.16%), which is dimensionless.
Since the rod returns to its original length on removal of force, this corresponds to the elastic (linear) region O–A of the stress-strain curve where Hooke's law is obeyed.

Question 3: If the same experiment is performed on an aluminium rod of identical dimensions (Y = 70 × 10⁹ N m⁻²), compare the elongation with that of steel under the same force. What does this comparison reveal about the relative elasticity of the two materials?

For aluminium: ΔL = (F × L)/(A × Y) = (10⁵ × 1.0)/(3.14 × 10⁻⁴ × 70 × 10⁹) = 4.55 × 10⁻³ m = 4.55 mm — approximately 2.86 times the elongation of steel (1.59 mm).
Steel elongates less than aluminium under the same force because its Young's modulus (200 GPa) is nearly three times larger than that of aluminium (70 GPa).
Scientifically, the material that stretches less for a given load is more elastic; therefore, steel is more elastic than aluminium, confirming why steel is preferred for heavy-duty structural applications.

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Case Set 2

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A factory quality-control team tests metal wires for a cable manufacturer. Two wires are loaded on a universal testing machine: the first is mild steel and the second is made of glass. As the tensile load is gradually increased, the stress-strain curves are plotted for both. The steel wire shows a long plateau between the yield point and ultimate tensile strength before fracturing, while the glass wire fractures almost immediately after reaching its maximum stress with virtually no plastic deformation. The team records the yield strength of mild steel as approximately 300 × 10⁶ N m⁻² and notes that glass has no well-defined yield point. These observations are critical for selecting materials for construction of a multi-storey building.

Question 1: Based on the stress-strain observations, classify mild steel and glass as ductile or brittle, and state the feature of the stress-strain curve that supports your answer.

Mild steel is ductile because the ultimate tensile strength point (D) and fracture point (E) are far apart on its stress-strain curve, indicating large plastic deformation before fracture.
Glass is brittle because points D and E are very close together on its curve, meaning fracture occurs almost immediately after reaching the ultimate tensile strength with negligible plastic deformation.
The key distinguishing feature is the separation between D (ultimate tensile strength) and E (fracture point) on the stress-strain graph.

Question 2: Explain the significance of the yield point on the stress-strain curve of mild steel, and state why structures must be designed to operate below this point.

The yield point (B) is the stress beyond which the material does not return to its original dimensions upon removal of load; the corresponding stress is the yield strength σ_y.
Beyond the yield point, plastic deformation begins — atoms rearrange irreversibly — and even at zero stress, a permanent set (residual strain) remains.
Engineering structures must be designed so that applied stresses remain below σ_y to ensure all deformations are elastic and fully recoverable, preventing permanent distortion or failure of the component.

Question 3: For a multi-storey building, the team must choose between mild steel and glass for load-bearing columns. Analyse which material is preferable using the concepts of yield strength, ductility, and the safety advantage offered by plastic deformation before fracture.

Mild steel is strongly preferred for load-bearing columns because its high yield strength (≈300 × 10⁶ N m⁻²) ensures it can withstand large compressive and tensile stresses before permanent deformation begins.
Steel's ductility means that if accidentally overloaded, it undergoes significant visible plastic deformation (necking, elongation) before fracturing, providing crucial warning time for evacuation or corrective action.
Glass, being brittle, would fracture suddenly and without warning upon exceeding its ultimate tensile strength — catastrophically dangerous in a structural member supporting thousands of tonnes of load.

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A civil engineering team is designing a horizontal beam of length 4 m, breadth 0.1 m, and depth 0.2 m, made of steel (Young's modulus 2 × 10¹¹ N m⁻²), to span a bridge opening and support a central load. The team uses the formula δ = Wl³/(4bd³Y) to calculate the central sag. They also consider replacing the rectangular beam with an I-shaped cross-section of the same length and material to reduce weight. A colleague suggests increasing the breadth instead of the depth, arguing it would be equally effective. The team refers to NCERT Chapter 8 on elastic behaviour of materials to resolve the disagreement and finalise the optimal cross-section for the bridge beam.

Question 1: State the formula for the sag of a centrally loaded beam and identify which geometric parameter most effectively reduces sag, giving the mathematical reason.

The sag formula is δ = Wl³/(4bd³Y), where W is the central load, l is span length, b is breadth, d is depth, and Y is Young's modulus.
Since δ ∝ d⁻³, doubling the depth reduces sag by a factor of 8; since δ ∝ b⁻¹, doubling the breadth reduces sag by only a factor of 2.
Therefore, increasing depth is far more effective than increasing breadth in reducing the sag of a beam under a given central load.

Question 2: Explain the problem of buckling in deep rectangular beams and why the I-shaped cross-section is the standard engineering solution.

A very deep rectangular beam, while effective against bending (large d reduces δ), becomes unstable under off-centre loads and buckles sideways — particularly problematic in bridges with moving, unpredictable traffic.
The I-shaped cross-section resolves this: wide flanges provide large effective depth (resisting bending) while the web provides lateral stability (preventing buckling), combining the benefits of depth and stability.
The I-section also removes material from the middle of the web where its bending-resistance contribution is minimal, reducing weight and cost without sacrificing structural strength.

Question 3: If the depth of the beam is doubled from 0.2 m to 0.4 m while all other parameters remain constant, calculate the factor by which the sag changes. Also explain why simply maximising depth is not always the practical solution in bridge design.

Since δ ∝ d⁻³, doubling d from 0.2 m to 0.4 m reduces sag by a factor of 2³ = 8; the new sag is one-eighth of the original.
However, simply maximising depth is impractical because a very deep narrow beam is susceptible to buckling when the traffic load is not perfectly centred — which is inevitable on a real bridge.
Additionally, a very deep beam increases the structural height of the bridge (clearance issues), adds to self-weight in unpredictable ways, and complicates construction; the I-section is the practical compromise balancing sag reduction, buckling resistance, and economy.

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A materials science student is studying the behaviour of rubber (an elastomer) versus a steel wire by plotting their stress-strain curves. The rubber sample stretches to nearly five times its original length and snaps back, while the steel wire shows a well-defined linear region followed by plastic deformation before fracture. The student notices that rubber's curve is non-linear from the very start and there is no well-defined yield point or plastic region. The student also reads about the elastic tissue of the aorta, which has a similar non-linear elastic curve and plays a vital role in cardiovascular function. The student wonders why rubber, despite being so stretchable, is not considered 'more elastic' than steel in physics.

Question 1: What are elastomers? State two properties that distinguish their stress-strain curve from that of a metal.

Elastomers are substances (such as rubber and the elastic tissue of the aorta) that can be stretched to cause very large strains and still return to their original shape upon removal of the force.
Their stress-strain curve is non-linear over most of the elastic region, meaning they do not obey Hooke's law — unlike metals, which have a clear linear (proportional) region.
Elastomers have no well-defined plastic region or yield point, whereas metals show a distinct yield point (B), plastic region (B–D), and fracture point (E) on their stress-strain curves.

Question 2: Explain why rubber is not considered more elastic than steel, despite stretching far more under the same force.

In physics, elasticity is measured by the modulus of elasticity (Young's modulus Y = stress/strain); the material that stretches less for a given load — i.e., has higher Y — is more elastic.
Steel has a very high Young's modulus (200 × 10⁹ N m⁻²) and resists deformation strongly, while rubber has an extremely low effective modulus and deforms greatly under small forces.
A material which stretches more for the same stress has a lower modulus and is therefore less elastic in the scientific sense; the everyday notion of 'stretchy = elastic' is a misnomer.

Question 3: Explain how the non-linear elastic behaviour of the aorta (an elastomer) is ideally suited to its cardiovascular function, and compare this with the elastic behaviour required of a steel structural beam.

The aorta must expand significantly under the high pressure of each heartbeat and recoil forcefully to sustain blood pressure between beats; its large elastic strain range (elastomer behaviour) allows this without permanent deformation.
Its non-linear stress-strain curve means initial stretching requires little force (accommodating large blood volume), while at higher strains the curve steepens (increasing stiffness), preventing over-expansion — a naturally optimised pressure-buffer.
A steel beam, by contrast, requires a high Young's modulus so that deformation (sag, elongation) under structural loads remains minimal and precisely predictable; large strains in a structural beam would be catastrophic, making steel's linear, low-strain elastic behaviour ideal for engineering structures.

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Case Set 5

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A submarine is designed to operate at a depth of 3000 m in the Indian Ocean. Engineers must calculate the effect of water pressure on the submarine's hull and internal components. The bulk modulus of water is 2.2 × 10⁹ N m⁻², and the bulk modulus of steel (hull material) is 160 × 10⁹ N m⁻². The density of seawater is 1000 kg m⁻³ and g = 10 m s⁻². Engineers also compare the compressibility of steel, water, and air to decide on materials for different parts of the submarine, referring to Table 8.3 of NCERT Chapter 8. The concept of hydraulic stress and volume strain is central to ensuring the structural integrity of the vessel.

Question 1: Define hydraulic stress and volume strain as they apply to the submarine hull at a depth of 3000 m. Calculate the hydraulic pressure at this depth.

Hydraulic stress is the restoring force per unit area developed inside a solid body when it is subjected to uniform fluid pressure on all sides; it equals the hydraulic pressure in magnitude.
Volume strain is the ratio of the change in volume (ΔV) to the original volume (V): volume strain = ΔV/V.
Hydraulic pressure at 3000 m: p = hρg = 3000 × 1000 × 10 = 3 × 10⁷ N m⁻².

Question 2: Calculate the fractional compression (ΔV/V) of water at 3000 m depth using its bulk modulus. What does this tell us about water's compressibility?

ΔV/V = p/B = (3 × 10⁷)/(2.2 × 10⁹) = 1.36 × 10⁻² = 1.36%.
Despite a pressure of 30 MPa (about 300 times atmospheric pressure), water is compressed by only 1.36%, demonstrating that water is nearly incompressible — its bulk modulus is very high relative to gases.
This near-incompressibility of water is why hydraulic systems can transmit force efficiently, and why submarine hulls experience high stress but minimal volume change in the surrounding water.

Question 3: Compare the compressibility of steel (bulk modulus 160 GPa), water (2.2 GPa), and air (1.0 × 10⁻⁴ GPa). Explain the molecular basis for this hierarchy and its implication for the submarine design.

Compressibility k = 1/B: steel (k = 1/160 GPa ≈ 6.25 × 10⁻¹² Pa⁻¹) is least compressible; water (k = 1/2.2 GPa ≈ 4.5 × 10⁻¹⁰ Pa⁻¹) is moderately compressible; air (k = 1/10⁻⁴ GPa = 10⁴ GPa⁻¹) is about a million times more compressible than steel.
Steel's tight inter-atomic coupling gives it enormous resistance to compression; water molecules are bound but less rigidly; gas molecules are poorly coupled and widely spaced, allowing easy compression.
For the submarine hull, steel's very high bulk modulus means the hull barely changes volume even under 3 × 10⁷ Pa pressure, ensuring structural integrity; air-filled compartments must be pressurised or reinforced to avoid collapse under the much higher external water pressure.

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Chapter 8: Mechanical Properties of Solids | Case Study Questions

Passage

Question 1: Define stress and calculate the stress experienced by the steel rod when a 100 kN force is applied over its cross-sectional area.

Question 2: Calculate the elongation of the steel rod and determine the strain. State the region of the stress-strain curve this corresponds to.

Question 3: If the same experiment is performed on an aluminium rod of identical dimensions (Y = 70 × 10⁹ N m⁻²), compare the elongation with that of steel under the same force. What does this comparison reveal about the relative elasticity of the two materials?

Passage

Question 1: Based on the stress-strain observations, classify mild steel and glass as ductile or brittle, and state the feature of the stress-strain curve that supports your answer.

Question 2: Explain the significance of the yield point on the stress-strain curve of mild steel, and state why structures must be designed to operate below this point.

Question 3: For a multi-storey building, the team must choose between mild steel and glass for load-bearing columns. Analyse which material is preferable using the concepts of yield strength, ductility, and the safety advantage offered by plastic deformation before fracture.

Passage

Question 1: State the formula for the sag of a centrally loaded beam and identify which geometric parameter most effectively reduces sag, giving the mathematical reason.

Question 2: Explain the problem of buckling in deep rectangular beams and why the I-shaped cross-section is the standard engineering solution.

Question 3: If the depth of the beam is doubled from 0.2 m to 0.4 m while all other parameters remain constant, calculate the factor by which the sag changes. Also explain why simply maximising depth is not always the practical solution in bridge design.

Passage

Question 1: What are elastomers? State two properties that distinguish their stress-strain curve from that of a metal.

Question 2: Explain why rubber is not considered more elastic than steel, despite stretching far more under the same force.

Question 3: Explain how the non-linear elastic behaviour of the aorta (an elastomer) is ideally suited to its cardiovascular function, and compare this with the elastic behaviour required of a steel structural beam.

Passage

Question 1: Define hydraulic stress and volume strain as they apply to the submarine hull at a depth of 3000 m. Calculate the hydraulic pressure at this depth.

Question 2: Calculate the fractional compression (ΔV/V) of water at 3000 m depth using its bulk modulus. What does this tell us about water's compressibility?

Question 3: Compare the compressibility of steel (bulk modulus 160 GPa), water (2.2 GPa), and air (1.0 × 10⁻⁴ GPa). Explain the molecular basis for this hierarchy and its implication for the submarine design.

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