Physics by Lamhi: Not Your Boring Physics

CLASS 11 · CHAPTER 8 · MATTER

Mechanical Properties of Solids

No real solid is perfectly rigid. Every one of them stretches, compresses, or shears a little under load, and exactly how much, and whether it ever fully bounces back, is the entire subject of this chapter.

01 · See it

Watch it happen

Elastic — releases back to zero strain

0
Stress (relative)
0.0%
Strain

Drag the strain slider up slowly. Past the elastic limit, the curve you’re tracing bends away from the straight line, that’s the material still stretching, but no longer proportionally. Now pull the slider back down: the point doesn’t retrace the curve it came up on. It drops straight down a parallel line instead, and stops short of zero, the slider physically won’t go any lower than that either. That gap is permanent deformation, and it’s there even at zero load.

02 · Derive it

Where the formula comes from

When a force FF acts normal to a cross-section of area AA, the body develops an internal restoring force per unit area called stress, and responds with a fractional change in length called strain:

stress=FA,strain=ΔLL\text{stress} = \dfrac{F}{A}, \qquad \text{strain} = \dfrac{\Delta L}{L}

For small deformations these are proportional, Hooke’s law, and the ratio that connects them is a property of the material alone, its Young’s modulus YY:

Young's modulus
Y=stressstrain=F/AΔL/L=FLA ΔL\textcolor{#e08a1e}{Y} = \dfrac{\text{stress}}{\text{strain}} = \dfrac{F/A}{\Delta L/L} = \dfrac{FL}{A\,\Delta L}

The same ratio idea applies to the other two ways a solid can deform: shearing stress divided by shear strain θ\theta gives the shear modulus GG, and hydraulic pressure divided by fractional volume change gives the bulk modulus BB:

G=F/Aθ,B=−pΔV/VG = \dfrac{F/A}{\theta}, \qquad B = -\dfrac{p}{\Delta V/V}

all three are just “how much restoring stress per unit of deformation,” applied to stretching, shearing, and uniform squeezing respectively.

03 · Break it

Where the shortcut stops working

“Elastic” in everyday speech means stretchy. In physics it means something much narrower: returns completely to its original shape once the load is removed, regardless of how much or how little it deformed to begin with. A steel rod that stretches a fraction of a millimetre and springs back is just as elastic, in this exact sense, as a rubber band that stretches several centimetres and springs back. Steel simply has a much larger Young’s modulus, it’s stiffer, which is a statement about how much stress it takes to deform it, not about whether it counts as elastic.

A second trap sits right next to this one: the straight-line region of the stress-strain graph is where Hooke’s law holds, stress proportional to strain, but the material stays elastic a little past that too, all the way to the yield point. Proportional and elastic are not the same claim; the graph is proportional only on O to A, but still fully recoverable out to B.

Push past the yield point, though, and unloading no longer retraces the loading curve. The simulation above shows exactly this: pull the strain past the bend and release, and the material unloads along a new straight line, parallel to the original elastic slope, but landing at a nonzero strain when the stress reaches zero. That leftover strain is permanent, no amount of waiting undoes it. This is precisely why a bent paper clip stays bent.

04 · Master it

Apply it under exam conditions

Q1. A steel wire of length 2 m and cross-sectional area 1×10⁻⁶ m² is stretched by 0.5 mm under a load. Given Y = 2×10¹¹ N/m² for steel, find the applied load.

F=YA ΔLL=(2×1011)(1×10−6)(0.5×10−3)2≈50 NF = \dfrac{Y A\,\Delta L}{L} = \dfrac{(2\times10^{11})(1\times10^{-6})(0.5\times10^{-3})}{2} \approx \textcolor{#e08a1e}{50\text{ N}}

Q2. Two wires of the same material and length, but with radii in the ratio 1:2, are stretched by the same force. Find the ratio of their elongations.

Since ΔL=FL/(AY)\Delta L = FL/(AY) and A∝r2A \propto r^2, elongation scales as 1/r21/r^2 for fixed F,L,YF, L, Y:

ΔL1ΔL2=r22r12=41=4:1\dfrac{\Delta L_1}{\Delta L_2} = \dfrac{r_2^2}{r_1^2} = \dfrac{4}{1} = \textcolor{#e08a1e}{4:1}
05 · FAQs

Quick answers

Does a higher Young's modulus mean a material is 'more elastic'?+

No, it means the material is stiffer, it deforms less for a given stress. Elasticity is about whether a material returns to its original shape at all; the modulus only measures how much it resists deforming in the first place. Steel has a huge Young's modulus and is still elastic within its limit; so is rubber, with a tiny one.

Is a material still elastic in the region between the linear part and the yield point?+

Yes. Hooke's law (stress proportional to strain) only describes the initial straight-line region. Between there and the yield point, the material is still elastic, it fully recovers when unloaded, but stress and strain are no longer proportional. 'Elastic' and 'obeys Hooke's law' are not the same claim.

Why does removing the load not undo the deformation past the yield point?+

Past the yield point, some of the deformation becomes permanent (plastic). On unloading, the material still behaves elastically, but along a different, parallel line, one that doesn't return to zero strain. The gap left behind is the permanent set.

What's actually different between stress and pressure? They have the same units.+

Pressure is an external force applied per unit area. Stress is the internal restoring force per unit area that the body develops in response, equal and opposite to what's deforming it. Same dimensional formula, different physical role.

Why do beams and railway tracks use an I-shaped cross-section instead of a solid rectangle?+

Bending sag depends much more strongly on depth than on breadth (it falls off as depth cubed). An I-beam puts most of its material as far from the centre as possible, top and bottom, maximising effective depth for a given weight, instead of wasting material near the middle where it barely helps.