Mechanical Properties of Solids

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XI, Chapter 8.

Which is more elastic: a rubber band or a steel wire?

Almost everyone says rubber, because it stretches so much. Physics says steel, and the reason reveals what "elastic" really means. Hold on to that puzzle. By the end of this chapter you will not only know the answer, you will be able to compute how much a bridge cable stretches, how deep a beam must be, and why mountains stop growing at about 10 km.

Solids that deform: elasticity and plasticity

Mechanics so far treated bodies as rigid, unchanging in shape and size. Real solids are not. Push, pull or twist any solid and its atoms are displaced slightly from their equilibrium spacing. The forces between atoms behave like tiny springs, and they pull the atoms back. That inner response is why a deformed body can recover.

Elasticity is the property of regaining the original shape and size when the deforming force is removed. Plasticity is the opposite tendency: the body stays deformed, as putty or wet clay does. No material is perfectly elastic over all loads. Every solid is elastic up to some limit and plastic beyond it.

Stress and strain

When an external deforming force acts on a body, an equal and opposite internal restoring force develops. Stress measures how intensely this restoring force acts, per unit area:

$$\text{Stress} = \dfrac{F}{A}$$

Its SI unit is the newton per square metre, N m⁻², called the pascal (Pa), and its dimensions are $[M\,L^{-1}\,T^{-2}]$, the same as pressure. Stress is not a vector: although force has a direction, stress cannot be assigned a single direction, because it depends on the plane you cut through the body. The effect of the stress is strain, the fractional change in size or shape. Strain is a ratio of two like quantities, so it is dimensionless and has no unit.

There are three ways to deform a solid, each with its own stress and strain.

A wire stretched by a load is the standard picture of longitudinal stress. A book pushed sideways on a table shows shear. A sphere deep in water shows hydraulic stress. In a wire hung from a ceiling with a weight F at the bottom, the ceiling pulls up with F and the weight pulls down with F, but the tension at any section is F, not 2F. The stress is $F/A$.

Hooke's law

Robert Hooke found in 1676 that for small deformations, stress is proportional to strain:

$$\text{Stress} = k \times \text{strain}$$

The constant k is the modulus of elasticity. It has the units of stress (Pa) and depends only on the material and the type of deformation. Hooke's law holds only up to a certain limit. Beyond it, the response stops being proportional.

The stress-strain curve

Stretch a metal wire gradually, plotting stress against strain, and the curve has distinct regions.

If the gap between D and E is small, the material breaks soon after reaching its maximum stress: it is brittle (glass, cast iron). If the gap is large, it can be drawn out a long way first: it is ductile (copper, steel), which is why such metals can be pulled into wires.

Rubber and the aorta. The stress-strain curve of an elastomer is very different. It can be stretched to many times its length and still recover (a huge elastic region), but the curve is non-linear, so Hooke's law fails over most of the range, and there is no clearly defined plastic region.

Pause and predict: a wire is loaded until the point C on the plastic part of the curve and then unloaded slowly. Will the strain return to zero?

Check your answer: no. It follows a path roughly parallel to the straight OA part and ends at a non-zero strain, the permanent set. The wire is now longer than it started.

Beyond the textbook. For the exam: rubber is elastic over a large strain but does not obey Hooke's law. In reality: the loading and unloading curves of rubber do not coincide. The two curves enclose a loop, and the area of the loop equals the energy lost as heat in each stretch-and-release cycle. This is called hysteresis. It is why a rubber tyre warms up as it rolls and why rubber pads are used to absorb vibrations and shocks. Why it matters: "elastic" and "obeys Hooke's law" are different statements, and the curve of rubber shows how.

Young's modulus: stretching and squashing

For a rod or wire of original length L and area A, pulled by a force F that extends it by ΔL:

$$Y = \dfrac{F/A}{\Delta L/L} = \dfrac{FL}{A\,\Delta L}$$

Y is a property of the material, not of the specimen. A thick wire and a thin wire of the same steel have the same Y, though the thick one needs far more force for the same extension. Rearranging gives the working form $\Delta L = FL/(AY)$. Notice that a wire is really a spring: $F = (YA/L)\,\Delta L$, with spring constant $k = YA/L$.

Back to the puzzle. To stretch a steel wire and an aluminium wire of identical size (cross-section 0.1 cm²) by the same 0.1%, you need a force of 2000 N for steel but only 690 N for aluminium. A material is more elastic when it gives a smaller strain for the same stress, i.e. when its modulus is larger. Steel's Y is thousands of times that of rubber. The rubber band stretches more, but it is the steel that resists deformation and springs back more strongly. Elastic means "resists deformation and recovers", not "stretches easily".

Worked example: a loaded steel wire. A steel wire 3.0 m long with a cross-section of 1.0 mm² carries a load of 200 N. Find the extension and the energy stored. Take Y = 2.0 × 10¹¹ Pa.

Common wrong steps. Leaving the area in mm² (a million-fold error), and using $F\,\Delta L = 0.6$ J for the energy. The force grew from 0 to 200 N as the wire stretched, so the average force is 100 N and the energy is half of $F\,\Delta L$.

Worked example: comparing two wires. Wire B is made of the same material as wire A, but is twice as long and has twice the radius. The same force pulls both. How do their extensions compare?

Common wrong step. Making $\Delta L$ depend on r instead of $r^2$. The extension depends on the cross-sectional area, which is proportional to the square of the radius.

Shear modulus: sliding layers

Apply a tangential force F to the top face of area A of a block, and the block's shape tilts by an angle θ. The shearing strain is $\Delta x/L = \tan \theta \approx \theta$, and the shear modulus (modulus of rigidity) is

$$G = \dfrac{F/A}{\theta} = \dfrac{FL}{A\,\Delta x}$$

For most metals G is roughly $Y/3$. Lead has a low G ($5.6 \times 10^9$ Pa), steel has $84 \times 10^9$ Pa and tungsten $150 \times 10^9$ Pa.

Worked example: a sheared block. A lead cube of edge 10 cm is pushed along its top face with a tangential force of $2.8 \times 10^4$ N. Take $G = 5.6 \times 10^9$ Pa. How far does the top face slide?

Common wrong step. Using Y instead of G. Shear is not a stretch, so Young's modulus does not apply.

Bulk modulus: squeezing from all sides

Under hydraulic pressure p, a body's volume falls by ΔV. The bulk modulus is the pressure divided by the volume strain:

$$B = -\dfrac{p}{\Delta V/V}$$

The minus sign is needed because an increase in pressure causes a decrease in volume, so ΔV is negative, and B comes out positive. Its reciprocal, the compressibility $k = 1/B$, has units Pa⁻¹.

Worked example: water under pressure. The pressure on a sample of water rises by $1.1 \times 10^7$ Pa (about 110 atmospheres). Find the fractional change in volume. $B(\text{water}) = 2.2 \times 10^9$ Pa.

Common wrong step. Dropping the sign, or quoting a positive change in volume. Pressure squeezes, so the volume decreases.

Beyond the textbook. For the exam: air at STP has B of about $1.0 \times 10^5$ Pa, tiny compared with solids and liquids. In reality: a gas has no single fixed bulk modulus. For an ideal gas held at constant temperature, B equals its pressure (so about $10^5$ Pa at one atmosphere), and for a gas compressed quickly (adiabatically) it is 1.4 times the pressure for air. This is why sound, which compresses air quickly, travels at a speed set by the adiabatic value.

Which modulus for which material?

Poisson's ratio and elastic energy

When a wire is stretched, it gets thinner. Poisson's ratio is the ratio of lateral strain to longitudinal strain, $(\Delta d/d)/(\Delta L/L)$. It is dimensionless, with values of about 0.28 to 0.30 for steel and about 0.33 for aluminium alloys.

Stretching a wire takes work against the restoring forces, and this work is stored as elastic potential energy. Since the force rises uniformly from zero to F during the stretch, the energy is the area of a triangle:

$$U = \dfrac{1}{2}F\,\Delta L, \qquad u = \dfrac{1}{2} \times \text{stress} \times \text{strain} = \dfrac{1}{2}Y(\text{strain})^2$$

For the steel wire above, the stress is $200/10^{-6} = 2 \times 10^8$ Pa and the strain is $10^{-3}$, so $u = 1/2 \times 2 \times 10^8 \times 10^{-3} = 10^5$ J m⁻³. The volume is $AL = 3 \times 10^{-6}$ m³, so $U = 10^5 \times 3 \times 10^{-6} = 0.3$ J, the same as $1/2\,F\Delta L$.

Engineering with elasticity

Crane ropes. A rope must never be stressed to its yield strength, so designers use a large safety factor.

Worked example: sizing a crane rope. A crane must lift loads up to 10 tonnes ($10^4$ kg). The steel has a yield strength of $3 \times 10^8$ Pa. Find the smallest radius if the rope were loaded to the yield point, and the radius if a safety factor of 10 is used. Take g = 9.8 m s⁻².

Common wrong step. Multiplying the radius by 10 for a safety factor of 10. Area, not radius, controls the stress, and area goes as $r^2$.

Beams. A beam of length l, breadth b and depth d, resting on two supports with a load W at the centre, sags at the middle by

$$\delta = \dfrac{W l^3}{4 b d^3 Y}$$

Because of $d^3$, a deeper beam is far stiffer: doubling the depth cuts the sag to one eighth, while doubling the breadth only halves it. But a very deep, thin beam bends sideways (buckles). The compromise is the I-section, with most material in the top and bottom flanges and a thin web between them. Similarly, a pillar with flared, distributed ends carries a larger load than one with rounded ends.

Mountains. At the base of a mountain of height h, the pressure from the weight of rock is $h\rho g$. If it exceeds the elastic limit of the rock, the rock flows. With a critical stress of about $3 \times 10^8$ Pa and density $3 \times 10^3$ kg m⁻³, $h = 3 \times 10^8/(3 \times 10^3 \times 10) = 10^4$ m, so mountains on Earth are limited to about 10 km.

Beyond the textbook. For the exam: the height of a mountain is limited by the elastic limit of rock, $h = (\text{critical stress})/(\rho g)$, about 10 km on Earth. In reality: h goes inversely with g, and Mars has a surface gravity about 38% of the Earth's. Its tallest volcano, Olympus Mons, rises roughly 22 km, more than twice the height of anything on Earth. Why it matters: the same elasticity formula predicts different mountain limits on different planets.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Gravitation supplies g and the weight hρg that limits mountain height.
• Next: Mechanical Properties of Fluids continues with the bulk modulus and pressure, and with the property solids have and fluids lack: resistance to shear.
• Earlier idea reused: Work, Energy and Power introduced the spring energy $1/2\,kx^2$, which is the same as the strain energy of a stretched wire.
• Later idea: Thermal Properties of Matter shows how heated, constrained solids develop stress.