Physics by Lamhi: Not Your Boring Physics

CLASS 11 · CHAPTER 8 · MATTER

DETAILED NOTES

Mechanical Properties of Solids

Every solid gives a little under load, and the question this chapter answers is exactly how much, and what happens if you push past the point where it gives back. Stress, strain, the three moduli of elasticity, and the single most misunderstood idea in the chapter, the permanent set, are covered section by section with original worked examples. For the interactive stress-strain simulation, see the concept page.

8.1 Stress and Strain

Push, pull, or twist any real solid and it deforms, if only by an amount too small to see. Inside the body, the atoms resist being moved from their equilibrium spacing and develop an internal restoring force that opposes the deformation and tries to bring the body back to its original shape. Stress is simply how that restoring force is reported per unit area:

stress=restoring forcearea\text{stress} = \dfrac{\text{restoring force}}{\text{area}}

Since it is a force divided by an area, stress carries the same units as pressure, N/m2\text{N/m}^2 or pascal (Pa), and the same dimensional formula, [ML−1T−2][ML^{-1}T^{-2}], even though stress and pressure are conceptually different (pressure is always normal and usually external; stress can act in any direction and is the body’s own internal response).

Strain is the resulting fractional deformation, change relative to the original size, and since it is a ratio of two lengths, areas, or volumes, it is a pure number with no units and no dimensions. Depending on how the body is loaded, stress and strain come in three matching pairs:

8.1.1 Three kinds of stress and strain

  • Longitudinal (tensile or compressive) stress and strain: a force applied along the length of a rod or wire, stretching or compressing it. Strain here is ΔL/L\Delta L/L, the fractional change in length.
  • Shearing stress and strain: a tangential force applied parallel to a face, while the opposite face is held fixed, so the body’s shape distorts (a cube leans into a slanted box) without any real change in volume. Strain here is the angle of shear θ≈Δx/L\theta \approx \Delta x/L, the ratio of the sideways displacement to the perpendicular height.
  • Volume (hydraulic) stress and strain: a uniform force per unit area pressing inward from every direction at once, as a fluid does on a submerged body. The stress here is just the pressure change ΔP\Delta P, and the strain is the fractional change in volume, ΔV/V\Delta V/V.

8.2 Hooke's Law and the Stress-Strain Curve

For small deformations, stress and strain turn out to be directly proportional to each other, a relationship known as Hooke’s law:

Within the elastic limit, stress is directly proportional to strain.

stress=k×strain\text{stress} = k \times \text{strain}

where the constant kk is called the modulus of elasticity of the material, and which of the three moduli (sections 8.3–8.5) it refers to depends on which kind of stress and strain is being related. Hooke’s law is not universal, it is the statement that a stress-strain graph starts out as a straight line through the origin, and only an experiment (or the concept page’s simulation) shows how far that line actually extends before the material stops obeying it.

8.2.1 The elastic region: O to A

Load a wire gradually and plot the stress you apply against the strain it develops. From the unloaded state O up to a point A, the graph is a straight line through the origin, exactly as Hooke’s law predicts, and the slope of this line is the material’s Young’s modulus (§8.3). Point A is called the elastic limit (sometimes the yield point): load the wire to anywhere on OA and release it, and the stress-strain point retraces the same straight line exactly back down to O. Nothing is left behind, the wire returns to precisely its original length. This full reversibility, not just the straight line, is what the word “elastic” means here.

8.2.2 Beyond the elastic limit: yielding, plastic flow, and fracture

Push past A and the curve bends over, strain now grows much faster than stress for the same extra load, a behaviour called plastic flow. The curve climbs, with decreasing slope, to a peak at point B, the ultimate tensile strength, the largest stress the material can bear. Beyond B, something counterintuitive happens: the wire keeps stretching, often thinning visibly at one point (necking), while the stress it can actually sustain drops, until it snaps at the fracture point, E. A material that stretches a long way between A and E (a large plastic region) is called ductile (most metals); one that fractures soon after, or even at, the elastic limit, with almost no plastic region, is brittle (glass, cast iron, ceramics).

8.2.3 Unloading after yielding: the permanent set

Here is the detail that trips almost everyone up the first time: release the load from a point on OA and the wire returns to zero strain. Release it from a point past A, say somewhere between A and B, and it does not retrace the loading curve back down. It follows a new straight line instead, parallel to the original OA (same slope, same Young’s modulus, the material hasn’t changed, only its starting point has), down to zero stress at some strain that is greater than zero.

permanent set=εmax reached−stress at that pointY\text{permanent set} = \varepsilon_{\text{max reached}} - \dfrac{\text{stress at that point}}{Y}

That leftover strain is the permanent set: with absolutely no load left on it, the wire is now longer than it started, for good. Crucially, this depends on the highest strain the wire has ever reached, not on where you currently have it, pulling it back down part-way and re-stretching it later will not erase that history; the wire “remembers” the worst stretch it has survived. This is precisely the behaviour the interactive stress-strain simulation lets you feel directly: drag the strain slider past the elastic limit, release it, and watch the unloading line peel away from the loading curve instead of folding back onto it. The common misconception it is built to break is the assumption that a stretched wire always “springs back” to its original length, true only while you stay inside OA.

8.3 Young's Modulus

Young’s modulus, YY, is the modulus of elasticity for longitudinal stress and strain, the one that governs ordinary stretching and compression along a length:

Young's modulus
Y=longitudinal stresslongitudinal strain=F/AΔL/L\textcolor{#e08a1e}{Y} = \dfrac{\text{longitudinal stress}}{\text{longitudinal strain}} = \dfrac{F/A}{\Delta L/L}

Rearranged, this gives the single most useful formula in the chapter, the elongation a wire develops under a known load:

ΔL=FLAY\Delta L = \dfrac{FL}{AY}

which says exactly what intuition suggests: a longer or thinner wire stretches more for the same force, and a stiffer material (larger YY) stretches less.

8.3.1 Typical values and what they mean

Young’s modulus spans an enormous range. Steel is around 2×1011 Pa2\times10^{11}\text{ Pa}, one of the stiffest common engineering materials, which is exactly why it is the default choice for anything that must not visibly sag or stretch, bridge cables, rails, structural beams. Rubber, by contrast, sits around 105–106 Pa10^{5}\text{–}10^{6}\text{ Pa}, five to six orders of magnitude smaller, which is precisely why it is chosen when large, easily reversible deformation is the entire point, tyres, gaskets, elastic bands. A large YY means a material is rigid (resists stretching); it says nothing by itself about how strong or how brittle the material is, those are governed by the elastic limit and ultimate strength from §8.2, not by the slope OA.

Worked example

Elongation of a loaded steel wire

A steel wire of length 2.0 m and cross-sectional area 1.0×10−6 m21.0\times10^{-6}\text{ m}^2 (1.0 mm²) hangs a load that pulls on it with a force of 100 N. Taking Ysteel=2×1011 PaY_{\text{steel}} = 2\times10^{11}\text{ Pa}, find the elongation.

ΔL=FLAY=100×2.0(1.0×10−6)(2×1011)=2002×105=1.0×10−3 m=1.0 mm\Delta L = \dfrac{FL}{AY} = \dfrac{100\times2.0}{(1.0\times10^{-6})(2\times10^{11})} = \dfrac{200}{2\times10^{5}} = \textcolor{#e08a1e}{1.0\times10^{-3}\text{ m} = 1.0\text{ mm}}

A full metre of steel wire, under a load of only 100 N (about 10 kgf), stretches by just one millimetre, a direct illustration of how stiff YsteelY_{\text{steel}} really is.

8.4 Shear Modulus (Rigidity Modulus)

The shear modulus (or modulus of rigidity), GG, plays the same role as YY but for shearing stress and strain, a tangential force dragging one face of a body sideways relative to the opposite, fixed face:

Shear modulus
G=shearing stressshearing strain=F/AΔx/L=F/Aθ\textcolor{#e08a1e}{G} = \dfrac{\text{shearing stress}}{\text{shearing strain}} = \dfrac{F/A}{\Delta x/L} = \dfrac{F/A}{\theta}

where θ\theta is the (small) angle of shear. Because shearing a body sideways is, loosely, an “easier” deformation than stretching every bond along its length, GG is consistently smaller than YY for the same material, typically by a factor of two to three. The two are not independent: for an isotropic solid they are linked through Poisson’s ratio σ\sigma (§8.5) by Y=2G(1+σ)Y = 2G(1+\sigma), so once you know any two of YY, GG, and σ\sigma for a material, the third is fixed.

Worked example

Shear modulus of a rubber block

A rubber block is glued to a fixed surface at its base and has a square top face of area 25 cm², 2.0 cm above the base. A tangential force of 10 N applied to the top face displaces it sideways by 0.50 mm. Find the shear modulus of the rubber.

Shearing stress and strain first:

stress=FA=1025×10−4=4.0×103 Pa\text{stress} = \dfrac{F}{A} = \dfrac{10}{25\times10^{-4}} = 4.0\times10^{3}\text{ Pa}strain=ΔxL=0.50×10−32.0×10−2=0.025\text{strain} = \dfrac{\Delta x}{L} = \dfrac{0.50\times10^{-3}}{2.0\times10^{-2}} = 0.025G=4.0×1030.025=1.6×105 PaG = \dfrac{4.0\times10^{3}}{0.025} = \textcolor{#e08a1e}{1.6\times10^{5}\text{ Pa}}

a value squarely in the range expected for a soft rubber, three to four orders of magnitude below any metal’s shear modulus.

8.5 Bulk Modulus and Poisson's Ratio

The bulk modulus, BB, relates volume stress (a pressure change applied uniformly from all sides) to volume strain (the resulting fractional change in volume):

Bulk modulus
B=−ΔPΔV/V\textcolor{#e08a1e}{B} = -\dfrac{\Delta P}{\Delta V/V}

The minus sign is there purely to keep BB positive: an increase in pressure (ΔP>0\Delta P > 0) always produces a decrease in volume (ΔV<0\Delta V < 0), so the ratio on its own would be negative. Its reciprocal, k=1/Bk = 1/B, is the compressibility, how easy a substance is to squeeze. Gases have tiny BB (hugely compressible), liquids and solids have large, comparable BB (water is actually stiffer, bulk-modulus-wise, than many people expect, which is why water hammer and hydraulic systems work at all).

Poisson’s ratio, σ\sigma, captures a side effect of stretching: pull a wire longer and it also gets very slightly thinner. It is defined as the ratio of the (negative) lateral strain to the longitudinal strain that caused it:

σ=−lateral strainlongitudinal strain\sigma = -\dfrac{\text{lateral strain}}{\text{longitudinal strain}}

For real isotropic solids σ\sigma typically falls between about 0.2 and 0.4 (around 0.3 for most metals), and theory restricts it to the range −1<σ<0.5-1 < \sigma < 0.5, with 0.5 describing a material (like rubber, very nearly) that keeps its volume exactly constant while stretching.

Worked example

Bulk modulus of a compressed steel block

A steel block is subjected to a pressure increase of 2.0×107 Pa2.0\times10^{7}\text{ Pa}, which reduces its volume by 0.010%. Find the bulk modulus of the steel.

Volume strain is 0.010% expressed as a fraction:

ΔVV=0.010100=1.0×10−4\dfrac{\Delta V}{V} = \dfrac{0.010}{100} = 1.0\times10^{-4}B=ΔPΔV/V=2.0×1071.0×10−4=2.0×1011 PaB = \dfrac{\Delta P}{\Delta V/V} = \dfrac{2.0\times10^{7}}{1.0\times10^{-4}} = \textcolor{#e08a1e}{2.0\times10^{11}\text{ Pa}}

consistent with the accepted bulk modulus of steel, and, notice, of the same order as its Young’s modulus, solids genuinely resist being squashed almost as strongly as they resist being stretched.

8.6 Elastic Potential Energy in a Stretched Wire, and Applications

Stretching a wire does work on it, and as long as the wire stays within its elastic limit, that work is stored as recoverable elastic potential energy, released again the moment the load comes off. Since the force grows linearly with extension inside Hooke’s law, the work done is the area of the triangle under the stress-strain line, the average force times the total extension:

Elastic potential energy
U=12F ΔL=12×stress×strain×volume\textcolor{#e08a1e}{U} = \dfrac{1}{2}F\,\Delta L = \dfrac{1}{2}\times\text{stress}\times\text{strain}\times\text{volume}

The second form is worth noticing: 12×stress×strain\tfrac12 \times \text{stress}\times\text{strain} is the elastic energy stored per unit volume, so it applies equally to a stretched wire, a twisted shaft, or a compressed block, whatever kind of stress and strain is in play.

8.6.1 Why this matters in design

A crane cable or a bridge girder is never operated anywhere near its ultimate tensile strength (point B in §8.2), it is kept well inside the elastic region, usually at a stress that is some small fraction of the elastic limit itself, a factor of safety. The reason is exactly the permanent-set behaviour from §8.2.3: a cable loaded past its elastic limit even once comes away permanently longer and weaker, and repeating that even a little will eventually cause fatigue failure, so safe engineering design is bounded by the elastic limit, not by how strong the material is at the point of breaking.

It is also why metal, not wood, is the default choice for load-bearing beams in bridges and buildings: for a given cross-section, metal combines a far larger Young’s modulus (so it sags less under load) with a far higher elastic limit (so it can be loaded much harder before any permanent bending sets in), and modern beams are shaped (as a hollow box or an I-section rather than a solid bar) to concentrate material away from the centre, where bending stresses are smallest, and toward the top and bottom edges, where they are largest, giving much greater resistance to bending for very little extra weight.

Worked example

Energy stored in the stretched steel wire

Using the same wire as the §8.3 example (stretched by ΔL=1.0×10−3 m\Delta L = 1.0\times10^{-3}\text{ m} under a load of 100 N), find the elastic potential energy stored in it.

U=12F ΔL=12×100×(1.0×10−3)=5.0×10−2 JU = \dfrac{1}{2}F\,\Delta L = \dfrac{1}{2}\times100\times(1.0\times10^{-3}) = \textcolor{#e08a1e}{5.0\times10^{-2}\text{ J}}

A modest 0.05 J, it takes very little energy to produce an elastic deformation that is itself very small, which is also why a snapped, highly-stressed wire can release its stored energy violently and suddenly.

— From the NCERT exercises

Two of the chapter’s own exercise questions, with original worked solutions. (g=9.8 m/s2g = 9.8\text{ m/s}^2 throughout, as the chapter’s exercises specify.)

NCERT Exercise 8.1

A steel wire of length 4.7 m and cross-sectional area 3.0×10⁻⁵ m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area 4.0×10⁻⁵ m² under the same applied load. What is the ratio of the Young's modulus of steel to that of copper?

Solution

Both wires carry the same force FF and stretch by the same ΔL\Delta L, so from Y=FL/(AΔL)Y = FL/(A\Delta L), the ratio of the two moduli depends only on each wire’s length and area:

YsteelYcopper=Lsteel/AsteelLcopper/Acopper=Lsteel AcopperAsteel Lcopper\dfrac{Y_{\text{steel}}}{Y_{\text{copper}}} = \dfrac{L_{\text{steel}}/A_{\text{steel}}}{L_{\text{copper}}/A_{\text{copper}}} = \dfrac{L_{\text{steel}}\,A_{\text{copper}}}{A_{\text{steel}}\,L_{\text{copper}}}=4.7×(4.0×10−5)(3.0×10−5)×3.5=1.88×10−41.05×10−4≈1.79= \dfrac{4.7\times(4.0\times10^{-5})}{(3.0\times10^{-5})\times3.5} = \dfrac{1.88\times10^{-4}}{1.05\times10^{-4}} \approx \textcolor{#e08a1e}{1.79}

Steel’s Young’s modulus is about 1.79 times copper’s, consistent with steel being the noticeably stiffer of the two.

NCERT Exercise 8.2

Four identical hollow cylindrical steel columns support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load is distributed equally, calculate the compressional strain of each column. Young's modulus of steel is 2.0×10¹¹ Pa.

Solution

The total weight is shared equally among the four columns:

W=mg=50000×9.8=4.9×105 NW = mg = 50000\times9.8 = 4.9\times10^{5}\text{ N}Feach column=4.9×1054=1.225×105 NF_{\text{each column}} = \dfrac{4.9\times10^{5}}{4} = 1.225\times10^{5}\text{ N}

Each column is a hollow cylinder, so its load-bearing area is an annulus:

A=π(Rout2−Rin2)=π(0.602−0.302)=π(0.27)≈0.848 m2A = \pi(R_{\text{out}}^2 - R_{\text{in}}^2) = \pi(0.60^2 - 0.30^2) = \pi(0.27) \approx 0.848\text{ m}^2

The compressional (longitudinal) strain follows directly from Young’s modulus:

strain=F/AY=1.225×105/0.8482.0×1011≈1.444×1052.0×1011≈7.2×10−7\text{strain} = \dfrac{F/A}{Y} = \dfrac{1.225\times10^{5}/0.848}{2.0\times10^{11}} \approx \dfrac{1.444\times10^{5}}{2.0\times10^{11}} \approx \textcolor{#e08a1e}{7.2\times10^{-7}}

An almost unmeasurably small strain, which is exactly the point, well-designed structural columns operate so far inside the elastic region that their compression is imperceptible.

Want the interactive version instead? Try the stress-strain simulation → or take a practice paper →