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PhysicsCh 8 Mechanical Properties of Solids

Mechanical Properties of Solids: Class 11 Physics · Chapter 8

Physics Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published · Updated

  • 80 min
  • 23 practice questions
  • 26 MCQs

What you will learn in Mechanical Properties of Solids

  • Distinguish elastic from plastic behaviour and explain how forces between atoms give rise to stress
  • Define tensile, compressive, shearing and hydraulic stress and the matching longitudinal, shearing and volume strains
  • State Hooke's law and read every region of a stress–strain curve, from the proportional limit to fracture
  • Define Young's modulus, shear modulus, bulk modulus, compressibility and Poisson's ratio, and use them to calculate deformations
  • Derive the elastic potential energy of a stretched wire and its energy density ½ × stress × strain
  • Apply elastic properties to design: rope thickness, beam sag and the I-section, pillars and the height of mountains

Original companion text aligned to NCERT Physics Class 11, Chapter 8, pp. 167–179. Not affiliated with or endorsed by NCERT.

8.1 Introduction

In Chapter 6 we treated a wheel or a rod as a rigid body, whose shape never changes however hard we push. That was a useful model, not the truth. Stand on a steel bench and it sags by a fraction of a millimetre; pluck a sitar string and it lengthens; press your thumb into an eraser and it dents. Every real solid changes its size or shape when forces act on it, even when the change is far too small to see.

What happens when the forces are taken away? Here solids fall into two broad families. A steel spring, a rubber band or a diving board springs back to its original size and shape: it shows elasticity (प्रत्यास्थता), and the deformation it suffered was elastic. A ball of wet clay keeps whatever shape you squeeze it into. Such materials show plasticity (सुघट्यता); kneaded atta dough and modelling clay come close to being perfectly plastic (Fig. 8.1).

3/3The spring returns to its first length (elastic); the dough stays stretched (plastic).

Fig. 8.1Elastic and plastic behaviour: the spring recovers, the dough does not.

Two objects lie side by side: a steel spring fixed to a wall and a roll of atta dough on a board, both the same starting length, marked by a dashed line. In the first step both are at rest. In the second step both are pulled to the right and are clearly longer. In the third step the pull is removed: the spring has returned to the dashed line, labelled springs back, elastic, while the dough stays long, labelled stays long, plastic.

Definition
Elasticity

The tendency of a deformed body to return to its original size and shape when the forces deforming it are taken away.

ExampleA stretched rubber band shrinking back when you let go.

Engineers depend on this distinction. Before the steel for the Chenab rail bridge in Jammu and Kashmir, whose deck stands 359 m above the riverbed, could be ordered, its designers had to know how much each grade of steel stretches, bends and twists under load, and at what load it would stop springing back. The same knowledge decides how thick a lift cable must be, why a railway rail has a narrow waist and wide flanges, and why a glass tumbler shatters when dropped while a steel one only dents. This chapter builds the ideas needed to answer such questions.

Quick check
1 of 2
A potter presses a thumb into a lump of wet clay and lifts it away. The dent stays. Which describes the clay?
Show answer

Answer (C) plastic, because the deformation remained

A plastic material keeps the deformation after the force is removed. Every solid deforms under a force, so deforming alone does not make it elastic: an elastic body would have sprung back.

8.2 Stress and Strain

Hang a bucket of water from a steel wire. The wire lengthens a little and then stops: the bucket is in equilibrium. Something inside the wire must now be pulling up on the bucket with a force equal to its weight. That something is the set of forces between neighbouring atoms. Stretching drags the atoms slightly further apart than their natural spacing, and the bonds between them pull back like tiny stretched springs (Fig. 8.2). Squeeze a solid instead, and the atoms are pushed closer than they like, so the bonds push back.

2/3Pulled apart, the bonds pull back: a restoring force opposes the stretch.

Fig. 8.2A spring model of a solid: bonds resist any change in the spacing of atoms.

Three rows of five atoms are joined by zigzag springs horizontally and straight bonds vertically. In the first step the atoms sit at their natural spacing and there is no net force. In the second step outward arrows pull the ends apart, the spacing is larger and green arrows labelled restoring point inwards: the bonds pull back. In the third step inward arrows squeeze the block, the spacing is smaller and the restoring arrows point outwards: the bonds push back.

These internal forces are restoring forces. In equilibrium, the restoring force across any cross-section balances the external force on it. What matters to the material is not the force alone but how widely it is spread: 100 N hanging from a cotton thread is very different from 100 N on a steel bar. So we define stress (प्रतिबल) as the restoring force per unit area of the surface across which it acts:

stress=FA
(8.1)
where
F
the restoring force, equal in size to the applied force in equilibrium
A
the area of the surface across which it acts

Stress has the unit of pressure, the pascal: 1 Pa=1 N m−2. Its dimensional formula is [M L−1 T−2]. Engineering stresses run to millions or even hundreds of millions of pascals, so the megapascal (1 MPa=106 Pa) and gigapascal (109 Pa) are handy units.

Depending on how the forces are applied, a solid is stretched or squashed, sheared, or squeezed from every side. Each case has its own stress and its own measure of deformation, called strain (विकृति): the change in some dimension divided by its original value. Fig. 8.3 shows all of them on the same block.

1/4Equal and opposite pulls at right angles to the end faces: the length increases.

Fig. 8.3Three kinds of stress: longitudinal (tensile or compressive), shearing and hydraulic.NCERT Fig. 8.1 · p. 168

A square block with a dashed outline of its original shape. In the first step forces pull its top and bottom outward and it becomes taller and slightly narrower: tensile stress, length grows. In the second step forces push inward and it becomes shorter and wider: compressive stress, length shrinks. In the third step its base sits on a floor and a force along the top face slides it sideways into a slanted shape: shear, shape changes. In the fourth step the block sits in water with arrows pressing on all four sides and it shrinks evenly: hydraulic stress, volume shrinks.

Pulling or pushing along a length. Grip a rod by its ends and pull. The forces act along its axis, at right angles to the end faces, and the stress inside is tensile. Turn the forces round so that they push the ends together and the stress becomes compressive. Either way it acts along the length, so the two are grouped as longitudinal stress. The length changes from L to L±ΔL, and the deformation is measured by the

longitudinal strain=ΔLL
(8.2)

Sliding one face past another. Now let the two forces act along two opposite faces instead of across them, like your palm pushing the top of an eraser whose base is glued to the desk (Fig. 8.4). The restoring force per unit area of that face is a shearing stress, also called a tangential stress. The top face moves sideways by Δx relative to the bottom face, a distance L below it, so every vertical edge tilts through an angle θ. The shearing strain is

3/3Shearing strain =Δx/L=tan⁡θ, close to θ for small angles.

Fig. 8.4Shearing strain: the top face shifts by Δx relative to a face a distance L away.

A rectangular eraser stands on a desk with its base glued down. In the first step it is at rest. In the second step a force F pushes its top face to the right and the desk holds the base with an equal force to the left, so the block leans into a parallelogram. In the third step the original vertical edge is shown dashed, the top face's sideways shift is marked delta x, the height between the faces is marked L, and the small angle theta between the dashed edge and the slanted edge is marked at the bottom.

shearing strain=ΔxL=tan⁡θ
(8.3)

In practice θ is tiny, so we can use the small-angle approximation, with θ in radians:

shearing strain≈θ
(8.4)

How good is this? At θ=5∘=0.0873 rad we get tan⁡θ=0.0875: the two differ by less than 0.3 %. Shear strains in loaded metal parts are usually thousands of times smaller still.

Squeezing from all sides. Drop a steel ball to the bottom of a deep lake. The water presses on every bit of its surface, always at right angles and equally hard everywhere, so the ball becomes very slightly smaller but stays perfectly round. The restoring force per unit area, the hydraulic stress, simply equals the extra pressure p of the water, and the ball regains its size when it is brought back up. The strain here is the

volume strain=ΔVV
(8.5)

Each strain is a ratio of two lengths or two volumes (or, for shear, an angle in radians), so strain is a pure number with no unit and no dimensions.

Quick check
1 of 3
A nylon fishing line of cross-sectional area 0.50 mm2 holds a fish of weight 40 N. What is the tensile stress in the line, in MPa?
MPa
Show answer

Answer 80 MPa

σ=F/A=40/(0.50×10−6) Pa=8.0×107 Pa=80 MPa. A common slip is to forget that 1 mm2=10−6 m2.

8.3 Hooke's Law

Load a wire, a spring or a rod gently and measure how far it deforms. Double the load and the deformation doubles; triple it and it triples (Fig. 8.5). As long as the deformation stays small, stress and strain rise and fall together in a fixed ratio, a rule known as Hooke's law.

LawHooke's law

As long as a deformation is small, the stress in a body and the strain it produces keep a constant ratio.

stress=k×strain
(8.6)
where
k
the modulus of elasticity of the material

3/3The points lie on a straight line through the origin: Hooke's law.

Fig. 8.5Equal additions of load give equal additions of extension.

On the left a spring hangs from a bracket next to a vertical scale, with a pointer at its lower end. In the first step it holds one mass and the matching point is plotted on a load against extension graph on the right. In the second step it holds two masses, stretches twice as far and a second point appears at twice the height and twice the distance. In the third step three masses give a third point, and a dashed straight line through the origin passes through all three, labelled straight line.

The constant k is the modulus of elasticity (प्रत्यास्थता गुणांक). Because strain has no unit, k has the unit of stress, the pascal. A large modulus means a stiff material, one that needs a large stress for even a small strain. Each kind of stress has its own modulus; we meet them in Section 8.5.

Nothing deeper forces Hooke's law to be true: it is empirical, a pattern found by measurement. Metals follow it well, but only up to a certain stress, while rubber and many body tissues curve away from a straight line almost from the start. The next section shows where the law gives out.

Quick check
1 of 3
Within its proportional region, a wire stretches by 0.40 mm under a load of 50 N. What load would stretch it by 1.0 mm?
Show answer

Answer (D) 125 N

Load is proportional to extension: 50×1.0/0.40=125 N. Choosing 100 N treats the extension as if it had doubled; it has grown 2.5 times.

8.4 Stress–Strain Curve

How does a material behave all the way from a gentle tug to breaking point? Clamp a sample, say a wire, in a testing machine and stretch it slowly while the machine logs two numbers: the pulling force and the extension. This is a tensile test. Turning each pair into a stress (force ÷ original area) and a strain (extension ÷ original length) and plotting them gives the material's stress–strain curve, a fingerprint of how it deforms. Squeezing or shearing the sample instead gives other curves of the same kind.

Fig. 8.6 sketches the curve for a ductile metal such as mild steel. The strains in its first part are tiny, well under 1 %, so the drawing stretches that part out to make it visible. Follow the curve from the origin O:

5/5The sample breaks at F.

Fig. 8.6Stress against strain for a ductile metal, from the first load to fracture (the elastic part is drawn much wider than to scale).NCERT Fig. 8.2 · p. 169

A stress against strain graph. In the first step a straight line rises steeply from the origin O to point P, labelled Hooke's law holds. In the second step the line bends slightly as it continues to the yield point Y, labelled still elastic; a dashed level marks the yield strength sigma y. In the third step the curve rises gently to a point K, and a dashed line drawn from K parallel to OP reaches the strain axis to the right of O; the gap from O is labelled permanent set. In the fourth step the curve reaches its highest point U, at the ultimate strength sigma u. In the fifth step the curve falls to the fracture point F, marked with a cross.

  • O to P: the proportional region. A straight line: Hooke's law holds. Take the load off anywhere along it and the sample is exactly as long as before. P marks the proportional limit.
  • P to Y: elastic but not proportional. The graph bends away from the straight line, but no harm is done yet: unload and the sample retraces its path to O. The end of this stretch, Y, is the yield point or elastic limit, and the stress there is the material's yield strength σy.
  • Beyond Y: plastic flow. The curve flattens: layers of atoms slip past one another, and the sample stretches a lot for little extra stress. Let the load off at K and the sample comes back along a line parallel to OP, not along the curve, ending up longer than it started. That leftover strain is a permanent set, and the deformation is plastic.
  • U: the ultimate tensile strength. The curve peaks at the ultimate tensile strength σu. From here on the sample thins at one spot, forming a neck, and that spot keeps stretching under a falling force.
  • F: fracture. The sample breaks at the narrowest part of the neck.

The gap between U and F tells you how a material fails. If F lies far beyond U, the material can be drawn into wires or bent into shape before it breaks: it is ductile (तन्य), like copper, gold or mild steel. If U and F almost coincide, the material breaks soon after it stops being elastic, with little warning: it is brittle (भंगुर), like glass, cast iron or a china cup (Fig. 8.7).

1/3Copper: U and F are far apart, so it stretches a lot before it breaks.

Fig. 8.7Ductile, brittle and elastomeric behaviour (strain scales differ between the curves).NCERT Fig. 8.3 · p. 170

Each step shows one curve on stress and strain axes. In the first step, copper: a steep straight start, a long gently rising curve to its highest point U, then a fall to the fracture point F far to the right; a dashed bracket between U and F is labelled far apart. In the second step, cast iron: a nearly straight line that ends abruptly at a cross labelled U and F together. In the third step, rubber: a curve that rises slowly over a huge strain and then turns steeply upward before it breaks, labelled huge strain, no permanent set. The strain axis is not to the same scale in the three steps.

Rubber plays by different rules. Pull a rubber band to three or four times its length and it still returns to its first length, yet its graph curves almost from the first moment, and it never passes through a plastic stage before it snaps. A material whose elastic stretch is huge but far from proportional is an elastomer. Your body has some: the wall of the aorta, the large artery leaving the heart, balloons out at every heartbeat. Its curve rises gently and then sharply, so the artery gives easily at first and then resists.

Table 8.2Typical strengths of some materials (real samples vary with composition and treatment)NCERT Table 8.1 · p. 170
MaterialYield strength σy (MPa)Ultimate strength σu (MPa)Behaviour
Structural (mild) steel250400–550ductile
Aluminium alloy 6061275310ductile
Copper (annealed)70220very ductile
Grey cast iron—150–300brittle
Soda-lime glass—about 50brittle
Human bone (compact)—about 130fairly brittle
InteractiveTensile test lab

A testing machine holds a sample between two grips beside its stress–strain graph. You choose mild steel, copper, cast iron or rubber and set the wire's diameter and length. A slider stretches the sample step by step (Play test does it automatically) and the point traces the material's curve, with readouts of force, extension, stress and the starting slope. Release load unloads the sample: in the elastic region it returns to zero, beyond the yield point it follows a straight line back and keeps a permanent set. Stretch far enough and the sample necks and breaks. A toggle zooms the graph onto the small-strain region.

Fig. 8.8Choose a material, then drag the slider or press Play test to stretch the sample. Release load shows whether it comes back.
Quick check
1 of 3
A steel wire is loaded beyond its yield point and the load is then removed. What happens?
Show answer

Answer (A) it keeps some extra length, a permanent set

Beyond the yield point the deformation is partly plastic. Unloading follows a line parallel to the elastic line, so some strain remains. Returning along the same curve happens only below the elastic limit.

8.5 Elastic Moduli

Designers work hard to keep structures in the straight-line part of the curve, below the proportional limit, where deformations are small and fully reversible. There the ratio of stress to strain is constant. This modulus of elasticity is a property of the material (it varies a little with temperature), not of the size or shape of the sample. Because there are different kinds of stress and strain, there are several moduli; Table 8.5 collects them.

8.5.1 Young's Modulus

Stretch a short steel rod, or squeeze it with a stress of the same size, and its length changes by the same amount either way. So a single constant serves for both, the Young's modulus Y (यंग गुणांक): longitudinal stress σ divided by the longitudinal strain ε it produces.

Y=σε
(8.7)

For a rod of length L and cross-sectional area A that a force F stretches by ΔL, (8.1) and (8.2) give

Y=F/AΔL/L=FLA ΔL
(8.8)

Its unit is the pascal, like that of stress. Turned round, the same equation predicts the extension of any wire, ΔL=FL/(AY): a long, thin wire of a material with a small modulus stretches most.

Table 8.3Elastic constants of some materials at room temperature (typical values)NCERT Table 8.1 · p. 170
MaterialY (GPa)G (GPa)B (GPa)Poisson's ratio
Steel200791600.29
Aluminium7026760.33
Copper120451400.34
Glass7029400.22
Lead165.6460.44
Tungsten4111613100.28
Rubber0.01–0.1—about 2about 0.5

Why are crane hooks, rails and building frames made of steel? Hang the same 100 N load from wires 1 m long and 1 mm2 in cross-section: a steel wire stretches by 0.50 mm, a copper one by 0.83 mm and an aluminium one by 1.4 mm. Steel's modulus is among the largest of common materials, so it deforms least. Glass is about as stiff as aluminium; concrete, bone and wood are softer, at roughly 10–30 GPa; rubber's modulus is thousands of times smaller.

Example 8.1NCERT Example 8.1 · p. 171

Stretching a footbridge hanger

A footbridge over a busy road hangs from vertical steel rods. One rod is 2.5 m long and 16 mm in diameter, and it carries a tension of 30 kN. Taking Y=2.0×1011 Pa for steel, find (a) the stress in the rod, (b) the strain and (c) the extension. (d) The steel's yield strength is 250 MPa. By what factor is the stress below it? Ignore the rod's own weight.

Try it yourself first. Then reveal the solution one step at a time.

Example 8.2NCERT Example 8.2 · p. 171

Two wires in a row

A garden lamp of mass 6.0 kg hangs from a steel wire 2.0 m long, joined end to end with an aluminium wire 1.5 m long; both are 1.0 mm in diameter. Take Ys=2.0×1011 Pa for steel, Ya=7.0×1010 Pa for aluminium and g=9.8 m/s2. (a) Find the extension of each wire and the total drop of the lamp. (b) What lamp mass would make the total extension 3.0 mm?

Try it yourself first. Then reveal the solution one step at a time.

Example 8.3NCERT Example 8.3 · p. 171

How much does a porter's shin bone shorten?

A railway porter carries a 40 kg suitcase on his head while standing on both feet. Each shin bone (tibia) is about 38 cm long, with an effective cross-sectional area of 3.2 cm2. Taking Y=1.6×1010 Pa for bone in compression and g=9.8 m/s2, find (a) the extra compression of each shin bone due to the suitcase, (b) the strain and (c) the compression when he lifts one foot to step forward.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
Two wires of the same material carry the same load. Wire Q is twice as long as wire P and has twice its diameter. How does the extension of Q compare with that of P?
Show answer

Answer (D) half as much

ΔL=FL/(AY) with A∝d2: doubling L doubles the extension, doubling d divides it by 4, so Q stretches 2/4=1/2 as much. 'Twice' forgets that the area depends on the diameter squared.

8.5.2 Shear Modulus

For shearing, the stress is the tangential force per unit area of the face along which it acts, and the strain is the angle of tilt. Their ratio is the shear modulus G (अपरूपण गुणांक), also called the modulus of rigidity:

G=σsΔx/L=FLA Δx
(8.10)

By (8.4) the strain is simply θ, so

G=FA θ
(8.11)
σs=G θ

The unit of G is again the pascal. The G column of Table 8.3 shows that, for a given material, the shear modulus is smaller than Young's modulus, typically between a third and two-fifths of it: it is easier to make layers of atoms slide over each other than to pull them apart. Only solids have a shear modulus. A liquid has no shape of its own, so even the gentlest sideways push makes it flow instead of settling at a fixed angle.

Shear turns up all over engineering: in a bolt that stops two plates sliding past each other, in a twisted drive shaft, and in the wire of a coil spring, which is twisted rather than stretched as the spring extends.

Example 8.4NCERT Example 8.4 · p. 172

A bridge on rubber pads

Many bridges rest on thick rubber (elastomeric) bearing pads that let the deck expand and contract with temperature. One pad measures 50 cm×40 cm in plan and 6.0 cm in height, with its base fixed to the pier. On a hot afternoon the expanding deck drags the top of the pad sideways with a force of 18 kN. The rubber's shear modulus is 0.90 MPa. Find (a) the shearing stress, (b) the shearing strain and (c) the sideways shift of the top of the pad.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 2
Why does a liquid have no shear modulus?
Show answer

Answer (A) it flows under any sideways force instead of reaching a fixed strain

A shear modulus needs a fixed strain for a given shearing stress. A liquid keeps flowing as long as the sideways force acts. Its molecules do attract each other, which is why liquids hold together.

8.5.3 Bulk Modulus

A body in a fluid whose pressure rises by p is squeezed equally from all sides, as in Section 8.2: the hydraulic stress is p and the volume strain is ΔV/V. Their ratio defines the bulk modulus B (आयतन गुणांक):

B=−pΔV/V
(8.12)

Squeezing always shrinks a stable body, so ΔV and p have opposite signs, and the minus sign makes B a positive number. Its reciprocal, the compressibility k (संपीड्यता), answers the opposite question: by what fraction does the volume shrink for each pascal of extra pressure?

k=1B=−1Δp ΔVV
(8.13)

Unlike Y and G, the bulk modulus applies to liquids and gases as well as to solids. A fluid has no shape of its own, but it does have a volume, and squeezing it reduces that volume. Table 8.4 and Fig. 8.9 compare the three states of matter.

Table 8.4Bulk modulus and compressibility in the three states of matterNCERT Table 8.3 · p. 173
SubstanceB (Pa)k=1/B (Pa−1)
Steel1.6×10116.3×10−12
Mercury2.85×10103.5×10−11
Water2.2×1094.5×10−10
Air at constant temperature, 1 atm1.0×1051.0×10−5

2/2One extra atmosphere shrinks steel by 0.00006 %, water by 0.005 % and air (at constant temperature) by 50 %.

Fig. 8.9Adding one atmosphere of pressure: steel and water barely change, air halves its volume.

Three identical cylinders with pistons stand side by side, filled with a steel block, water and air drawn as scattered molecules. In the first step there are no weights. In the second step an equal weight is placed on each piston, giving the same extra pressure. The steel and water pistons do not visibly move; the air piston drops to half its height and the molecules crowd into the lower half. The volume changes are labelled 0.00006 percent for steel, 0.005 percent for water and 50 percent for air.

For the same extra pressure, water shrinks about 70 times as much as steel, and air at constant temperature over 20 000 times as much as water. Why such a range? Go back to the spring model of Fig. 8.2. In a solid every atom is already pressed against stiff bonds on all sides, so moving the atoms even slightly closer takes an enormous pressure. A liquid's molecules also touch, but they are held more loosely. A gas is mostly empty space, and squeezing it only crowds the molecules into less of that space; this is also why a gas's compressibility depends strongly on its pressure and temperature.

Table 8.5Which modulus for which deformationNCERT Table 8.4 · p. 173
DeformationWhat changesStrainModulusApplies to
Stretch or squash along a lengthlengthΔL/LYoung's, Y=FL/(A ΔL)solids
Slide one face past anothershape onlyθ≈Δx/Lshear, G=F/(Aθ)solids
Squeeze equally from all sidesvolume onlyΔV/Vbulk, B=−p/(ΔV/V)solids, liquids and gases
Example 8.5NCERT Example 8.5 · p. 174

Squeezed at the bottom of the sea

A research submersible dives to a depth of 6000 m carrying a solid steel test ball. Take the density of seawater as 1030 kg/m3, g=9.8 m/s2, B=1.6×1011 Pa for steel and B=2.3×109 Pa for seawater. Find the fractional decrease in volume, compared with the surface, of (a) the steel ball and (b) the seawater at that depth.

Try it yourself first. Then reveal the solution one step at a time.

Looking ahead: in Chapter 14 you will find that these moduli also set the speed of sound. A stiff, light medium carries sound fast, which is why sound travels about four times faster in water than in air, and faster still in steel.

Quick check
1 of 3
Why is there a minus sign in B=−p/(ΔV/V)?
Show answer

Answer (B) so that B is positive, since a rise in pressure lowers the volume

An increase in pressure (p>0) gives a decrease in volume (ΔV<0), so the ratio p/(ΔV/V) is negative; the minus sign makes B positive.

8.5.4 Poisson's Ratio

Stretch a broad rubber band and watch its width: it narrows as it lengthens (Fig. 8.10). A steel wire does the same, far less visibly. The strain along the force, the longitudinal strain, comes with a strain at right angles to it, the lateral strain. If a wire of diameter d and length L lengthens by ΔL while its diameter shrinks by Δd, the lateral strain is Δd/d. Below the elastic limit the sideways strain keeps step with the lengthwise one, so their ratio is a fixed property of the material, named after the French mathematician Siméon Denis Poisson (1781–1840): Poisson's ratio (पॉयसन अनुपात):

2/2Length grows by ΔL while the diameter shrinks by Δd.

Fig. 8.10A stretched rod lengthens and narrows; the ratio of the two strains is Poisson's ratio.

In the first step a rubber rod lies horizontally with its length L marked below and its diameter d marked across its middle. In the second step forces pull its ends outward: the rod is longer, marked L plus delta L, and thinner, marked d minus delta d, and a dashed outline shows its original size.

ν=Δd/dΔL/L=ΔdΔL⋅Ld
where
ν
Poisson's ratio

No units survive the division, so Poisson's ratio is just a number. For most metals it lies between about 0.25 and 0.35: close to 0.29 for steel and 0.33 for aluminium (Table 8.3). For rubber it is almost 0.5, the value at which the volume stays the same as the material stretches. The lesson is general: a force in one direction produces strains in other directions too, so a single modulus cannot describe everything a solid does. A material that behaves the same in every direction needs two independent constants, for example Y and Poisson's ratio, to describe its elastic behaviour fully.

Example

A copper wire gets thinner

A copper wire 3.0 m long and 2.0 mm in diameter is stretched elastically by 1.5 mm. Poisson's ratio for copper is 0.34. By how much does its diameter decrease?

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 2
A steel rod with Poisson's ratio 0.30 is stretched so that its length increases by 0.10 %. By what percentage does its diameter decrease?
%
Show answer

Answer 0.03 %

Lateral strain =0.30×0.10 %=0.030 %. A common slip is to divide by 0.30 instead of multiplying.

8.5.5 Elastic Potential Energy in a Stretched Wire

Pulling a wire longer means dragging its atoms apart against their bonds, and that takes work. Below the elastic limit none of it is wasted: the wire holds it as elastic potential energy and pays it back as it shortens. That is how a bow launches an arrow and a gulel (catapult) flings a pebble. How much energy does a wire hold?

DerivationEnergy stored in a stretched wire

Let a wire of natural length L and cross-sectional area A be stretched slowly, so that it is always in equilibrium. When its extension is x, (8.8) gives the force needed to hold it:

F(x)=YAL x

The force grows in proportion to the extension, like a spring of constant YA/L. Stretching it by a further small amount dx takes work dW=F dx, a thin strip under the force–extension line (Fig. 8.11). Adding the strips from x=0 to the final extension l:

W=∫0lYAL x dx=12 YAL l2

This is the area of the triangle under the line, 12Fl, where F=YAl/L is the final force. In terms of the strain l/L and the volume AL, using Y (l/L)=σ:

W=12 Y(lL)2(AL)=12 σ ε (AL)

The work is stored as elastic potential energy, U=W. Dividing by the volume AL gives the energy stored per unit volume, the energy density:

u=12 σ ε
(8.14)

3/3All the strips together make the triangle: W=12Fl.

Fig. 8.11The work done in stretching a wire is the area under its force–extension line.

A graph of force against extension. In the first step a straight line rises from the origin to a final force F at extension l, labelled F equals YA over L times l. In the second step a thin vertical strip under the line is highlighted, labelled dW equals F dl. In the third step the whole triangle under the line is shaded and labelled W equals one half F l.

Since σ=Yε, the energy density can also be written as u=12Yε2 or u=σ2/2Y. At a given stress, a material with a smaller Young's modulus stores more energy per unit volume. Good spring materials combine a high yield strength with a moderate modulus, so they can store a lot of energy without being permanently bent.

Example

Energy in a stretched steel wire

A steel wire 2.0 m long with a cross-section of 1.0 mm2 is stretched by 2.0 mm (Y=2.0×1011 Pa). Find (a) the final force, (b) the energy stored and (c) the energy per unit volume.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 2
The extension of a wire, kept within the elastic limit, is doubled. The elastic energy stored becomes
Show answer

Answer (D) four times as large

U=12(YA/L) l2 grows as the square of the extension. 'Twice' forgets that the force doubles too.

8.6 Applications of Elastic Behaviour of Materials

Every bridge, building, crane and aircraft is designed with the ideas of this chapter. The designer must know how much each part will deform under the loads it will meet, and must make sure no part is ever stressed beyond its elastic limit, with a generous margin for surprises. Four examples show the reasoning.

Lifting ropes. A crane hook or a lift cage hangs from a steel rope. The rope must never be stretched into its plastic region, or it would stay permanently longer and weaker. So the stress Mg/A caused by the heaviest load M must stay below the yield strength σy, which sets a minimum cross-sectional area:

A≥Mgσy
(8.15)
Example

How thick a crane rope?

A tower crane at a building site in Pune is rated to lift 4.0 tonnes (1 t=1000 kg). Its rope is steel with a yield strength of 250 MPa. (a) Find the least cross-sectional area, and the radius, of a solid rope that would just stay elastic. (b) For safety, the rope is designed for 5 times the rated load. What radius does that need?

Try it yourself first. Then reveal the solution one step at a time.

2/249 thin wires in 7 strands: the same steel area, far more flexible.

Fig. 8.12A wire rope carries the load of a solid rod of the same steel area, but it can bend.

In the first step the cross-section of a solid round steel rod is shown, labelled one solid rod, strong but stiff. In the second step the rod fades and beside it appears the cross-section of a rope made of seven strands, each of seven round wires, inside a dashed outline, labelled 7 strands of 7 wires, same steel, bends easily.

Beams. A beam carries loads across a gap: a floor joist, a bridge girder, a plank laid over a drain. Its job is to bend as little as possible. For a bar of rectangular cross-section with breadth b and depth d, resting on supports a distance l apart and loaded by a weight W at the middle, the sag at the centre is

δ=Wl34bd3Y
(8.16)
where
δ
the sag (depression) at the centre
b
the breadth, the horizontal width of the cross-section
d
the depth, the vertical height of the cross-section
l
the span between the supports

Deriving this needs more calculus than we have yet, but we can read it. The sag grows rapidly with span, as l3, so long spans need extra support. A stiffer material (larger Y) helps. Most striking is the role of shape: the sag falls as 1/b but as 1/d3. Doubling the breadth halves the sag; doubling the depth cuts it to one-eighth. So beams are made deep rather than wide (Fig. 8.13).

1/3Wide and thin: the plank sags a lot.

Fig. 8.13Sag depends on 1/(bd3): depth matters far more than breadth.NCERT Fig. 8.7 · p. 175

A beam rests on two triangular supports with a weight W at its centre and its cross-section drawn at the right. In the first step the beam is a plank laid flat, wide and thin, and it sags visibly; a dashed outline shows its unloaded position and an arrow marks the sag delta. In the second step the same plank stands on its edge, tall and narrow, and barely sags, labelled 25 times less sag. In the third step an I-shaped steel beam, with wide top and bottom flanges joined by a thin web, carries the load with almost no sag, labelled deep but light.

Why do steel girders have an I-shaped cross-section? When a beam sags, its top layers are squeezed and its bottom layers stretched, while the layers near the middle are hardly strained at all, so metal far from the middle does most of the work. An I-section puts most of its steel into two wide flanges, top and bottom, joined by a thin web: deep and stiff, yet light and cheap. The wide flanges also keep it from rolling over and folding sideways, the failure called buckling that threatens a tall, thin plank whenever the load is a little off-centre, as with vehicles crossing a bridge. Railway rails use the same idea in a modified form.

Example

A plank over a drain

During the monsoon, a wooden plank 20 cm wide and 4.0 cm thick is laid across a drain, with supports 2.0 m apart, for people to walk over. Take Y=1.0×1010 Pa for the wood. (a) How much does it sag when a person weighing 600 N stands at the middle? (b) What would the sag be if the same plank were set on its edge?

Try it yourself first. Then reveal the solution one step at a time.

Pillars. Columns and pillars carry loads in compression, and their ends matter. A pillar with rounded ends touches what it supports over a small area, so the stress there is large, and the pillar can tilt and buckle easily. A pillar whose ends spread into a broad base and a broad top shares the load over a larger area and is held firmly in line, so it can carry much more. The wide bases and bracketed tops of the stone pillars in many old buildings do exactly this job.

How high can a mountain be? Elasticity even limits the height of mountains. Picture a vertical column of rock from the summit of a mountain of height h down to its base (Fig. 8.14). Its weight, spread over its cross-section, presses on the rock at the base with a stress of about hρg, where ρ is the density of the rock. The base is not squeezed equally from all sides, since nothing presses on the slopes from outside, so this stress has a large shearing part. If it exceeds the stress at which hot, deep rock begins to flow, the base slowly squeezes out sideways and the mountain sinks until the stress falls back below that limit.

2/3If hρg exceeds the rock's flow stress, the base squeezes out and the mountain sinks.

Fig. 8.14The base of a mountain must stay below the stress at which rock flows.

In the first step a mountain is drawn with a thin vertical column of rock from its summit to its base and its height h marked; the stress at the base is about h rho g. In the second step a taller dashed outline rises above the mountain, labelled too tall, base rock flows, with arrows at the base pointing outward. In the third step two mountains stand side by side: a smaller one on Earth labelled about 9 km and a much taller one on Mars labelled about 25 km.

Example

The tallest possible mountain

Take the density of crustal rock as 2.7×103 kg/m3 and the stress at which it begins to flow as roughly 2.5×108 Pa. Estimate the greatest height of a mountain (a) on the Earth (g=9.8 m/s2) and (b) on Mars (g=3.7 m/s2), assuming similar rock.

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 3
The depth of a rectangular beam is doubled while its breadth, span, material and load stay the same. Its sag becomes
Show answer

Answer (C) one-eighth as large

δ∝1/d3, so doubling d divides the sag by 23=8. 'Half' would be right for doubling the breadth.
Recap

Mechanical Properties of Solids: key points and summary

  1. Real solids deform under forces. An elastic body regains its size and shape when the forces are removed; a plastic body stays deformed (Section 8.1).
  2. Stress, the restoring force on each unit of area (F/A), is measured in pascals; strain, a change in dimension divided by the original dimension, has no unit (Section 8.2).
  3. Tensile or compressive stress gives longitudinal strain ΔL/L; shearing stress gives shearing strain Δx/L=tan⁡θ≈θ; hydraulic stress gives volume strain ΔV/V (Section 8.2).
  4. Hooke's law: while deformations are small, stress is proportional to strain, and the constant is the modulus of elasticity (Law).
  5. A metal's stress–strain curve shows a proportional limit, a yield point (elastic limit), plastic flow with a permanent set, an ultimate tensile strength and fracture. Ductile materials break far beyond the ultimate strength, brittle ones close to it (Section 8.4).
  6. Elastomers such as rubber stretch elastically to very large strains but do not obey Hooke's law (Section 8.4).
  7. Young's modulus Y=FL/(A ΔL); metals, steel above all, have large Y, and in physics a larger modulus means 'more elastic' (Section 8.5.1).
  8. Shear modulus G=F/(Aθ), typically a third to two-fifths of Y; only solids have Y and G (Section 8.5.2).
  9. Bulk modulus B=−p/(ΔV/V) applies to all states of matter; compressibility k=1/B; gases are over a million times as compressible as solids (Section 8.5.3).
  10. Poisson's ratio is lateral strain divided by longitudinal strain: about 0.3 for metals and close to 0.5 for rubber (Section 8.5.4).
  11. A stretched wire stores elastic energy 12Fl; per unit volume u=12σε (Section 8.5.5).
  12. Design uses these ideas: rope area A≥Mg/σy with a safety factor, beam sag δ=Wl3/(4bd3Y) and the I-section, pillars with spread ends, and a greatest mountain height of about 10 km on the Earth (Section 8.6).
Practice · 23 questions

Practice questions with answers (NCERT “Exercises”)

  1. NumericalEasyNCERT Exercise 8.1

    Wire P is 2.4 m long with a cross-section of 2.0×10−6 m2; wire Q is 3.6 m long with a cross-section of 1.5×10−6 m2. Under the same load they stretch by exactly the same amount. Find the ratio YP/YQ of their Young's moduli.

    Show solution
    FLPAPYP=FLQAQYQ
    YPYQ=LP/APLQ/AQ=1.2×1062.4×106

    =0.50: wire P's material is half as stiff as wire Q's.

  2. NumericalMediumNCERT Exercise 8.2

    A tensile test on a metal wire gives the readings in the table. Plot them if you like.

    Table 8.6
    Strain (×10−3)Stress (MPa)
    00
    0.535
    1.070
    1.5105
    2.0140
    2.5158
    3.0166
    3.5170
    1. Find the Young's modulus of the material, in pascals.

      Pa
      Show solution

      Up to a strain of 2.0×10−3 the stress rises by 35 MPa for every 0.5×10−3 of strain, so

      Y=7.0×107 Pa1.0×10−3=7.0×1010 Pa
    2. Estimate the yield strength of the material.

      Show solution

      Answerabout 1.5×108 Pa

      The straight line would give 175 MPa at a strain of 2.5×10−3, but the reading is only 158 MPa: the material has left its straight-line region between 140 and 158 MPa. Its yield strength is roughly 1.5×108 Pa.

  3. Short answerEasyNCERT Exercise 8.3

    The stress–strain curves of two materials, P and Q, share the same axes and scales.

    Fig. 8.15Stress–strain curves of materials P and Q.NCERT Fig. 8.10 · p. 178

    Two curves start at the origin. Curve P, dashed, rises steeply in a straight line and ends with a cross at a moderate stress after only a small strain. Curve Q rises less steeply, bends over, keeps rising slowly to a higher greatest stress and then dips before ending with a cross at a large strain.

    1. Which material has the greater Young's modulus?

      Show answer

      Answer (A) P

      P's straight part is steeper, so a given strain needs more stress.
    2. Which is the stronger material, able to bear the greater stress before breaking?

      Show answer

      Answer (B) Q

      Q reaches a higher greatest stress (ultimate tensile strength) than P ever does. Stiffness and strength are different properties.
    3. Which material is brittle?

      Show answer

      Answer (A) P

      P breaks right after its straight part, with almost no plastic stretching; Q stretches a long way before it breaks.
  4. Short answerMediumNCERT Exercise 8.4

    Say whether each statement is true or false, and give a reason.

    1. A rubber band is more elastic than a steel wire of the same size, in the physicist's sense of the word.

      Show answer

      Answer False

      False. In physics 'more elastic' means a larger modulus. For the same stress steel deforms far less than rubber, so steel is the more elastic.
    2. When a coil spring is stretched, its stiffness is decided mainly by the Young's modulus of its wire, because the wire gets longer.

      Show answer

      Answer False

      False. The total length of wire hardly changes. Each turn of the coil is twisted, so the wire is in shear, and the spring's stiffness depends mainly on the shear modulus.
  5. NumericalMediumNCERT Exercise 8.5

    A hanging planter uses two wires, each 0.80 mm in diameter. A steel wire 1.8 m long hangs from a hook and holds a 2.0 kg pot at its lower end; from the bottom of that pot a copper wire 0.80 m long holds a second pot of 3.0 kg. Take Y=2.0×1011 Pa for steel, Y=1.2×1011 Pa for copper and g=9.8 m/s2.

    1. Find the extension of the steel wire, in mm.

      mm
      Show solution

      Area: A=π(4.0×10−4)2=5.03×10−7 m2. The steel wire holds both pots: T=(5.0)(9.8)=49 N.

      ΔLs=TLsAYs=8.8×10−4 m
    2. Find the extension of the copper wire, in mm.

      mm
      Show solution

      The copper wire holds only the lower pot: T=(3.0)(9.8)=29.4 N.

      ΔLc=TLcAYc=3.9×10−4 m
  6. NumericalMediumNCERT Exercise 8.6

    An anti-vibration mount is a rubber cube of edge 5.0 cm. One face is bonded to a vertical wall, and a 4.0 kg instrument is fixed to the opposite face, so its weight acts along that face. The shear modulus of the rubber is 0.40 MPa. How far does the outer face sag, in mm? (g=9.8 m/s2)

    mm
    Show solution

    The weight F=39.2 N acts along a face of area A=2.5×10−3 m2, and the faces are L=0.050 m apart.

    Δx=FLAG

    =1.96×10−3 m≈2.0 mm, a shearing strain of about 0.039.

  7. NumericalMediumNCERT Exercise 8.7

    Six identical hollow steel columns hold up an overhead water tank of total mass 3.0×105 kg. Each column has an outer radius of 40 cm and an inner radius of 32 cm. If the load is shared equally, find the compressive strain in each column. (Y=2.0×1011 Pa, g=9.8 m/s2)

    Show solution

    Load per column: F=(3.0×105)(9.8)/6=4.9×105 N.

    A=π(0.16−0.1024) m2=0.181 m2
    ε=FAY=1.35×10−5
  8. NumericalEasyNCERT Exercise 8.8

    An aluminium busbar with a 12.0 mm×25.0 mm rectangular cross-section is pulled with a force of 21.0 kN, well within its elastic limit. Find the strain. (Y=7.0×1010 Pa)

    Show solution
    A=(0.0120)(0.0250)=3.00×10−4 m2
    σ=2.10×1043.00×10−4=7.0×107 Pa
    ε=σY=1.0×10−3
  9. NumericalEasyNCERT Exercise 8.9

    The haul cable of a mountain ropeway has a radius of 2.0 cm. For safety, the stress in it must never exceed 1.5×108 Pa. What is the greatest tension it may carry, in newtons?

    N
    Show solution
    Fmax⁡=σmax⁡ πr2

    =(1.5×108)π(0.020)2 N=1.9×105 N, the weight of about 19 tonnes.

  10. NumericalHardNCERT Exercise 8.10

    A uniform signboard of mass 24 kg hangs level from three vertical wires, each 1.5 m long, attached at its two ends and its middle. The two end wires are brass (Y=1.0×1011 Pa) and the wire in the middle is steel (Y=2.0×1011 Pa). Each wire is to carry the same tension.

    1. What is the tension in each wire, in newtons?

      N
      Show solution

      Three equal tensions support the weight: T=(24)(9.8)/3=78.4 N.

    2. Find the ratio of the diameter of a brass wire to that of the steel wire.

      Show solution

      With equal tension and equal length, ΔL=TL/(AY) is the same for all three only if AY is the same:

      AbYb=AsYs
      dbds=YsYb=2≈1.41
  11. NumericalHardNCERT Exercise 8.11

    A 2.0 kg metal ball is fastened to a steel wire whose natural length is 0.80 m and cross-section 2.0 mm2, and swung round in a vertical circle. At the lowest point it moves at 2.0 revolutions per second. Ignore the wire's mass and take r=0.80 m, Y=2.0×1011 Pa and g=9.8 m/s2.

    1. Find the tension in the wire at the lowest point, in newtons.

      N
      Show solution
      T=mg+mω2r

      With ω=2π(2.0)=12.6 rad/s:

      T=(2.0)(9.8+126.3) N=272 N
    2. Find the extension of the wire there, in mm.

      mm
      Show solution
      ΔL=TLAY

      =5.4×10−4 m≈0.54 mm. The stress, 1.4×108 Pa, is safely below steel's yield strength.

  12. NumericalMediumNCERT Exercise 8.12

    In a laboratory test, 2.000 L of hydraulic oil in a strong cylinder has the pressure on it raised by 200 atm, and its volume falls by 26.0 mL. Take 1 atm=1.013×105 Pa.

    1. Find the bulk modulus of the oil, in pascals.

      Pa
      Show solution
      p=(200)(1.013×105)=2.03×107 Pa
      ∣ΔV∣V=26.02000=0.0130
      B=p∣ΔV∣/V=1.56×109 Pa
    2. Compare it with the bulk modulus of air at constant temperature, about 1.0×105 Pa, and say, in everyday language, what makes the gap so huge.

      Show solution

      Answerabout 1.6×104 times larger

      Boil/Bair≈1.6×104: the oil is about 16 000 times as hard to compress. Its molecules are already in contact, so squeezing them closer means pushing against strong repulsive forces. The molecules of air are, on average, about ten molecular diameters apart and barely interact, so the gas gives way easily.

  13. NumericalMediumNCERT Exercise 8.13

    Deep in the ocean the pressure is 400 atm above atmospheric. Seawater has a density of 1.025×103 kg/m3 at the surface and a bulk modulus of 2.3×109 Pa. Find its density at that depth, in kg/m³. (1 atm=1.013×105 Pa)

    kg/m³
    Show solution

    A fixed mass keeps its mass while its volume falls to V(1−p/B), so ρ′=ρ/(1−p/B).

    pB=4.05×1072.3×109=0.0176
    ρ′=10251−0.0176=1.043×103 kg/m3

    The water there is about 1.8 % denser than at the surface.

  14. NumericalEasyNCERT Exercise 8.14

    A glass marble lies in a pressure vessel in which the hydraulic pressure is raised by 60 atm. Using B=4.0×1010 Pa for glass, find the fractional decrease in its volume. (1 atm=1.013×105 Pa)

    Show solution
    ∣ΔV∣V=pB

    =(6.08×106)/(4.0×1010)=1.5×10−4, a decrease of about 0.015 %.

  15. NumericalMediumNCERT Exercise 8.15

    A solid aluminium sphere of radius 5.0 cm is placed under a hydraulic pressure of 1.2×107 Pa. By how much does its volume decrease, in cm³? (B=7.6×1010 Pa)

    cm³
    Show solution
    V=43π(0.050)3=5.24×10−4 m3
    ∣ΔV∣=pVB=8.3×10−8 m3

    That is about 0.083 cm3.

  16. NumericalEasyNCERT Exercise 8.16

    By how much must the pressure on 5.0 L of mercury be increased to compress it by 0.050 %? Give the answer in atmospheres. (B=2.85×1010 Pa, 1 atm=1.013×105 Pa)

    atm
    Show solution
    p=B ∣ΔV∣V=(2.85×1010)(5.0×10−4)

    =1.43×107 Pa≈141 atm. The volume of 5.0 L does not matter.

  17. Quick answerMedium

    A metal wire is pulled through a die so that its length doubles while its volume stays the same. Under the same load, its extension becomes

    Show answer

    Answer (C) 4 times as large

    At constant volume, doubling the length halves the area, and ΔL=FL/(AY) grows by 2×2=4.
  18. Quick answerMedium

    Assertion (A): Steel is more elastic than rubber. Reason (R): For the same stress, the strain in steel is much smaller than in rubber.

    Show answer

    Answer (A) A and R are true, and R explains A

    Both are true. A larger modulus (less strain for the same stress) is exactly what 'more elastic' means in physics, so R explains A.
  19. Quick answerMedium

    A steel wire (Y=2.0×1011 Pa) is under a tensile stress of 1.0×108 Pa. Find the elastic energy stored per unit volume, in J/m³.

    J/m³
    Show answer

    Answer 25000 J/m³

    u=σ2/2Y=(1.0×1016)/(4.0×1011)=2.5×104 J/m3.
  20. Quick answerMedium

    A material keeps exactly the same volume when it is stretched slightly. Its Poisson's ratio is

    Show answer

    Answer (C) 0.5

    ΔV/V=ΔL/L+2 Δd/d=ε(1−2ν), which is zero when ν=0.5. Rubber comes close to this.
  21. Quick answerHard

    Assertion (A): The bulk modulus of a gas kept at constant temperature increases as its pressure increases. Reason (R): For a gas at constant temperature pV is constant, so its bulk modulus equals its pressure.

    Show answer

    Answer (A) A and R are true, and R explains A

    From pV= constant, V dp+p dV=0, so B=−V dp/dV=p. A gas at higher pressure is harder to compress, and R explains A.
  22. Quick answerEasy

    Match each situation with the elastic constant that mainly decides it.

    Tap an item on the left, then its match on the right.

    Show answer
    Answer
    • a lift cable stretching : Young's modulus
    • a drive shaft twisting : shear modulus
    • a ball sinking in deep sea : bulk modulus
    • a stretched wire thinning : Poisson's ratio
    Stretching is longitudinal strain (Y), twisting is shear (G), all-round pressure changes the volume (B), and sideways thinning is lateral strain (Poisson's ratio).
  23. Quick answerMedium

    Wires A and B are of the same material and length, and the diameter of A is twice that of B. The same force stretches each. What is the ratio UA/UB of the elastic energies stored?

    Show answer

    Answer (D) 1/4

    For the same force, U=12F ΔL=F2L/(2AY)∝1/A; A has four times the area, so it stores a quarter of the energy.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbookPractice question here
NCERT Exercise 8.1Practice Q1
NCERT Exercise 8.2Practice Q2
NCERT Exercise 8.3Practice Q3
NCERT Exercise 8.4Practice Q4
NCERT Exercise 8.5Practice Q5
NCERT Exercise 8.6Practice Q6
NCERT Exercise 8.7Practice Q7
NCERT Exercise 8.8Practice Q8
NCERT Exercise 8.9Practice Q9
NCERT Exercise 8.10Practice Q10
NCERT Exercise 8.11Practice Q11
NCERT Exercise 8.12Practice Q12
NCERT Exercise 8.13Practice Q13
NCERT Exercise 8.14Practice Q14
NCERT Exercise 8.15Practice Q15
NCERT Exercise 8.16Practice Q16

Quick check: MCQs on Mechanical Properties of Solids

  1. 8.1 Introduction2 questions
  2. 8.2 Stress and Strain3 questions
  3. 8.3 Hooke's Law3 questions
  4. 8.4 Stress–Strain Curve3 questions
  5. 8.5.1 Young's Modulus3 questions
  6. 8.5.2 Shear Modulus2 questions
  7. 8.5.3 Bulk Modulus3 questions
  8. 8.5.4 Poisson's Ratio2 questions
  9. 8.5.5 Elastic Potential Energy in a Stretched Wire2 questions
  10. 8.6 Applications of Elastic Behaviour of Materials3 questions
Glossary · 15

Key terms (English–Hindi)

bulk modulusआयतन गुणांकaayatan gunank
The ratio of hydraulic stress to volume strain, B=−p/(ΔV/V); defined for solids, liquids and gases.First seen in the chapter →
compressibilityसंपीड्यताsampeedyata
The reciprocal of the bulk modulus: the fractional decrease in volume per unit increase in pressure.First seen in the chapter →
ductileतन्यtanya
Able to stretch a long way plastically before breaking, like copper; a brittle material, like glass, breaks soon after its elastic limit.First seen in the chapter →
elastic limitप्रत्यास्थता सीमाpratyasthata seema
The greatest stress a material can bear and still recover fully; the stress at this yield point is the yield strength.First seen in the chapter →
elasticityप्रत्यास्थताpratyasthata
The property of a body that makes it return to its original size and shape when the deforming forces are removed.First seen in the chapter →
elastomer
A material, such as rubber, that stretches elastically to very large strains but does not obey Hooke's law.First seen in the chapter →
modulus of elasticityप्रत्यास्थता गुणांकpratyasthata gunank
The ratio of stress to strain where Hooke's law holds; a property of the material.First seen in the chapter →
permanent set
The strain that remains when a body stressed beyond its elastic limit is unloaded.First seen in the chapter →
plasticityसुघट्यताsughatyata
The property of a material that keeps the new shape it is given after the deforming force is removed, as clay does.First seen in the chapter →
Poisson's ratioपॉयसन अनुपातPoisson anupat
Lateral strain divided by longitudinal strain in a stretched body; a pure number, about 0.3 for metals.First seen in the chapter →
shear modulusअपरूपण गुणांकaparupan gunank
The ratio of shearing stress to shearing strain, G=F/(Aθ).First seen in the chapter →
strainविकृतिvikriti
The fractional change in a dimension of a body (length, angle of shear or volume); it has no unit.First seen in the chapter →
stressप्रतिबलpratibal
The restoring force per unit area inside a deformed body, F/A, measured in pascals.First seen in the chapter →
ultimate tensile strength
The greatest stress a material reaches in a tensile test, after which it necks and breaks.First seen in the chapter →
Young's modulusयंग गुणांकYoung gunank
The ratio of longitudinal stress to longitudinal strain, Y=FL/(A ΔL).First seen in the chapter →

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