Why is a crane rope made of many thin wires instead of one thick rod?

The cross-sectional area is fixed by the requirement that stress stay below the yield strength: A ≥ Mg/σy. For a 10-tonne crane that gives a radius near 3 cm once a safety factor of about ten is applied. A single solid wire of that radius would behave as a rigid rod and would not coil over pulleys. NCERT therefore notes that ropes are made of many thin wires braided like pigtails — this gives the same total area for strength while keeping the rope flexible and easy to manufacture.

What is the beam-depression formula and what does each symbol mean?

For a bar of length l, breadth b and depth d, supported near its ends and loaded at the centre by a load W, the sag at the centre is δ = Wl³ / (4bd³Y), where Y is Young's modulus. The depression grows with the cube of the span length, falls inversely with breadth b, falls with the cube of depth d, and falls inversely with Y. Depth and span dominate because they enter as cubes.

Why is the cross-section of a bridge girder shaped like the letter I?

Because δ ∝ 1/d³, increasing depth is far more effective than increasing breadth at resisting sag. But a tall thin bar tends to buckle sideways if the load shifts, which is unavoidable with moving traffic. The I-section is the compromise: the deep web supplies depth to resist bending while the wide flanges provide a large load-bearing surface and resist buckling. The shape gives strength without the weight and cost of a solid block.

Why can a mountain on Earth not be much taller than about 10 km?

A mountain's base is not under uniform compression, so the rock experiences a shear component roughly equal to hρg, the stress from the weight of material above. When hρg exceeds the critical shearing stress at which rock flows (about 30 × 10⁷ N m⁻²), the base yields. Equating hρg to that limit with ρ ≈ 3 × 10³ kg m⁻³ and g ≈ 10 m s⁻² gives h ≈ 10 km — more than the height of Mt. Everest.

Why does a pillar with distributed (spread) ends support more load than one with rounded ends?

A pillar with rounded ends concentrates the contact and is freer to bow sideways under load, so it buckles at a lower load. Distributing the ends — flaring the top and base — restrains that sideways bowing and spreads the contact stress, so the column carries more before failing. This is why classical columns and bridge piers are widened at their ends.

Which elastic property limits a crane rope and which limits a mountain?

A crane rope is limited by its tensile yield strength (elastic limit under longitudinal stress): the rope must not be stretched past the point of permanent set. A mountain is limited by the critical shearing stress of rock, because the base of a mountain is under shear rather than uniform bulk compression. Different loads call out different failure thresholds — tension for the rope, shear for the rock.

Applications of Elastic Behaviour — NEET Physics

Why elastic behaviour drives design

Every load-bearing structure is a managed deformation. A beam under traffic sags; a crane rope under a hanging load stretches; a column under a roof compresses. The engineer's task is to keep that deformation small and, above all, elastic — the material must spring back, not take a permanent set. NCERT §8.6 frames the whole subject around this constraint: design so that the working stress stays inside the elastic region of the stress–strain curve, well below the yield point.

Two material constants do almost all the work. Young's modulus \(Y\) sets how stiff a material is in tension or compression — large \(Y\) means small strain for a given stress. The yield strength \(\sigma_y\) sets the ceiling — the stress beyond which deformation becomes permanent. The applications below are simply these two ideas applied to ropes, beams, columns and rock.

Crane and elevator ropes

A crane lifts a load \(W = Mg\) on a rope of cross-sectional area \(A\). The longitudinal stress in the rope is \(W/A\). To keep the rope from stretching past its elastic limit, this stress must stay below the yield strength \(\sigma_y\). That single inequality fixes the minimum thickness.

\[ A \ge \dfrac{W}{\sigma_y} = \dfrac{Mg}{\sigma_y} \tag{NCERT 8.15} \]

NCERT 8.6 worked estimate

Size the steel rope of a crane with a lifting capacity of 10 tonnes (\(M = 10^4\) kg). Mild steel has yield strength \(\sigma_y \approx 300 \times 10^6~\text{N m}^{-2}\). Take \(g = 9.8~\text{m s}^{-2}\).

Minimum area. \(A \ge \dfrac{Mg}{\sigma_y} = \dfrac{10^4 \times 9.8}{300 \times 10^6} = 3.3 \times 10^{-4}~\text{m}^2\), which corresponds to a radius of about 1 cm for a circular cross-section.

Apply the safety margin. A factor of about ten on the load is standard, so NCERT recommends a thicker rope of radius about 3 cm.

The design twist. A single solid wire of radius 3 cm would be, in NCERT's words, practically a rigid rod — it could not wind round the pulleys. So the rope is made of many thin wires braided together, like pigtails, giving the same total area for strength while keeping flexibility and easy manufacture.

Figure 1. The required cross-sectional area is set by \(A \ge Mg/\sigma_y\). Realising that area as one solid rod gives a rigid bar; braiding many thin wires gives the same area with the flexibility a rope needs.

Bending of a loaded beam

A bridge deck or a floor joist is a beam — supported near its ends and loaded across its span. The beam must not bend too much, and it must not break. NCERT analyses the canonical case: a bar of length \(l\), breadth \(b\) and depth \(d\), supported near its ends and carrying a central load \(W\). The downward sag of the centre, called the depression \(\delta\), is

\[ \delta = \dfrac{W\,l^{3}}{4\,b\,d^{3}\,Y} \tag{NCERT 8.16} \]

This compact formula carries the entire design logic of beams. Read off the dependence on each variable and the engineering choices follow directly.

∝ l³ Span length

Sag grows with the cube of the span. Doubling the unsupported length increases depression eightfold — keep spans short.

∝ 1/d³ Depth

Sag falls with the cube of depth. Doubling the depth cuts depression to one-eighth — the most powerful lever.

∝ 1/b Breadth

Sag falls only linearly with breadth. Widening a beam helps far less than deepening it.

∝ 1/Y Young's modulus

A stiffer material sags less. Steel (large \(Y\)) is preferred over wood or concrete for the same span.

Figure 2. A beam supported near its ends and loaded at the centre sags by \(\delta = Wl^3/(4bd^3Y)\). The dashed line is the unloaded position; the depression \(\delta\) is measured at midspan.

Why beams are I-shaped

The depression formula points to deep, slender beams. But there is a competing failure mode. A tall thin bar, unless the load sits exactly over its centreline, tends to bow sideways and collapse — a failure NCERT calls buckling. With moving traffic on a bridge, the load never stays exactly placed, so a plain deep rectangular bar is unsafe.

The resolution is the I-section. Concentrate the material in two wide flanges at top and bottom, joined by a thin vertical web. The flanges sit at maximum depth, where they do the most to resist bending, and they present a broad surface that resists sideways buckling. The web maintains the depth between them while using little material. The result, in NCERT's phrasing, is a section that "provides a large load-bearing surface and enough depth to prevent bending," cutting weight and cost without sacrificing strength.

Figure 3. Three cross-sections for the same load. The solid rectangle wastes material near the neutral axis; the thin deep bar resists sag but buckles; the I-section places wide flanges at maximum depth and links them with a slim web — strength with minimum weight.

Foundation check

The \(Y\) in the depression formula is the same Young's modulus from the elastic moduli chapter — revise how \(Y\) is measured before relying on it here.

Pillars and columns

Columns and pillars carry compressive loads in buildings and bridges. Their failure mode is again buckling — a long column under compression can suddenly bow out and collapse well before the material is crushed. The end conditions matter a great deal. NCERT contrasts two shapes: a pillar with rounded ends supports a smaller load than one with a distributed (flared) shape at the ends.

A rounded end is nearly free to pivot, so the column can bow sideways easily and buckles at a low load. Distributing the ends — broadening the top and base — restrains that sideways motion and spreads the contact stress over a larger area, allowing the column to carry more before failing. This is why classical stone columns, and the piers of bridges, are widened where they meet the load and the ground.

Figure 4. End conditions decide a column's load capacity. Rounded ends pivot freely and buckle early; distributed ends restrain sideways bowing and carry a larger load before failure.

Maximum height of a mountain

The same elastic reasoning sets a ceiling on geology. A mountain of height \(h\) presses on its base with a force per unit area \(h\rho g\), where \(\rho\) is the rock density and \(g\) is gravity. Crucially, the base is not under uniform compression — the sides of the mountain are free, so there is a shear component, approximately \(h\rho g\) itself. If this shear stress exceeds the critical value at which rock flows, the base gives way and the mountain cannot grow taller.

NCERT 8.6 estimate

Estimate the maximum height of a mountain on Earth, given a critical (elastic-limit) shearing stress for rock of \(30 \times 10^7~\text{N m}^{-2}\), density \(\rho = 3 \times 10^3~\text{kg m}^{-3}\), and \(g = 10~\text{m s}^{-2}\).

Set the shear stress to its limit. The shear stress at the base is about \(h\rho g\). Equate it to the critical shearing stress: \(h\rho g = 30 \times 10^7~\text{N m}^{-2}\).

Solve for \(h\). \(h = \dfrac{30 \times 10^7}{(3 \times 10^3)(10)} = 10^4~\text{m} = 10~\text{km}\), which is more than the height of Mt. Everest.

Which modulus governs which failure

The recurring NEET test across this subtopic is whether you can pair an application with the right elastic property. A single table fixes it.

Application	Governing relation	Limiting property
Crane / elevator rope thickness	`A ≥ Mg/σy`	Tensile yield strength \(\sigma_y\)
Sag of a loaded beam	`δ = Wl³/4bd³Y`	Young's modulus \(Y\)
I-section vs solid bar	maximise depth, resist buckling	Young's modulus \(Y\) (stiffness)
Column / pillar capacity	end conditions vs buckling	Young's modulus \(Y\) (stiffness)
Maximum mountain height	`hρg = critical shear stress`	Critical shearing stress of rock
Stretching of a coil spring	only shape changes, no Δlength	Shear modulus \(G\)

Quick recap

Applications of elastic behaviour in one breath

Rope thickness: keep stress below yield strength, \(A \ge Mg/\sigma_y\); a 10-tonne crane needs \(r \approx 3\) cm after a safety factor of ten — braided thin wires for flexibility.
Beam sag: \(\delta = Wl^3/(4bd^3Y)\). Depression \(\propto l^3,\ \propto 1/d^3,\ \propto 1/b,\ \propto 1/Y\).
Depth and span dominate (cubes); breadth and \(Y\) are first-power levers.
I-section: flanges at maximum depth resist bending and buckling; slim web saves weight and cost.
Pillars: distributed (flared) ends carry more than rounded ends because they resist sideways buckling.
Mountain height \(\approx 10\) km: set \(h\rho g\) equal to the critical shearing stress of rock, not bulk compression.

Applications of Elastic Behaviour