Solved Examples

How to Use This Problem Bank

Ten sections of theory, and now the part that actually earns marks. What follows is 49 worked problems, arranged easy first and hard last, covering everything from "which modulus is this?" up to the multi-step arrangements that decide ranks.

Work them with a pen and paper. Cover the solution, try it, then compare — including the check at the end of each one, because the check is where the marks usually leak away.

The four questions to ask before you write anything

Nearly every mistake in this chapter is made in the first ten seconds, before any arithmetic starts. Ask these four, in this order:

1. Which deformation is it? Is the body getting longer or shorter (longitudinal), is its shape changing while its volume stays put (shearing), or is it being squeezed from every side at once (hydraulic)? That single answer picks the modulus: $Y$, $G$ or $B$.
2. Which area do I divide by? For longitudinal stress, $A$ is the face the force is perpendicular to — the cross-section. For shearing stress, $A$ is the face the force acts along. Getting this backwards is the single most expensive slip in the chapter.
3. Which length is on the bottom? The strain is always the change divided by the original dimension. For shear, the length underneath is the distance between the two faces, measured perpendicular to the sliding.
4. Is the load constant along the body? If the body's own weight matters, or the bar tapers, or it is spinning, the tension is different at every cross-section and you cannot use one $\frac{FL}{AY}$ for the whole thing.

Key Point — the master formulas, all in one place: $$Y = \frac{F/A}{\Delta L/L} = \frac{FL}{A\,\Delta L}, \qquad G = \frac{F/A}{\Delta x/L} = \frac{F}{A\theta}, \qquad B = -\frac{\Delta p}{\Delta V/V}$$ $$\sigma = -\frac{\Delta d/d}{\Delta L/L}, \qquad U = \frac{1}{2}F\,\Delta L, \qquad u = \frac{1}{2}\times\text{stress}\times\text{strain}$$ The minus sign in $B$ is there so that $B$ comes out positive — pressure up means volume down. The minus sign in $\sigma$ is there for the same reason: pulling a rod longer makes it thinner, so the two strains have opposite signs and the ratio would otherwise be negative.

Notation

$\sigma$ on its own always means Poisson's ratio, never stress. Stress is written as $\frac{F}{A}$, or named in words, or given a subscript: $\sigma_L$ for longitudinal, $\sigma_s$ for shearing, $\sigma_h$ for hydraulic and $\sigma_y$ for a yield strength. The shear modulus is $G$ and the bulk modulus is $B$. Elsewhere $\sigma$ often denotes stress, with $\nu$ or $\mu$ for Poisson's ratio, so check the convention in whatever else you read.

The material data used below

Unless a problem states its own numbers, every solution here uses this table. Each problem also restates the constants it uses inside its own solution, so you never have to scroll back.

Material	$Y$ (Pa)	$G$ (Pa)	$B$ (Pa)	$\rho$ (kg/m$^3$)
Steel	$2.0 \times 10^{11}$	$0.84 \times 10^{11}$	$1.6 \times 10^{11}$	7800
Copper	$1.2 \times 10^{11}$	$0.42 \times 10^{11}$	$1.4 \times 10^{11}$	8900
Brass	$0.91 \times 10^{11}$	$0.36 \times 10^{11}$	$0.61 \times 10^{11}$	8500
Aluminium	$0.70 \times 10^{11}$	$0.25 \times 10^{11}$	$0.72 \times 10^{11}$	2700
Glass	$0.65 \times 10^{11}$	$0.23 \times 10^{11}$	$0.37 \times 10^{11}$	2500
Bone (compression)	$9.4 \times 10^{9}$	—	—	1900
Water	—	0	$2.2 \times 10^{9}$	1000

Typical values for the chapter's worked problems. Real samples vary by several per cent, and answers below are quoted to two or three significant figures for that reason.

On the value of $g$: every problem states whether it uses $9.8$ m/s$^2$ or $10$ m/s$^2$, and no problem mixes the two. Where a source of data is quoted in atmospheres, one atmosphere is taken as $1.013 \times 10^{5}$ Pa.

[Board Important] Every solution below writes the formula on its own line before any number goes into it. Do the same in the exam. A correct formula with an arithmetic slip still earns most of the marks; a bare number earns none.

Solved Examples

Part 1: Which Stress Is It, and Which Modulus?

Four warm-ups whose whole content is the decision in step 1 above. Get fast at these and the rest of the chapter becomes arithmetic.

Example 1: Sorting five everyday deformations

For each situation below, name the type of stress, the matching strain, and the modulus you would need to put a number on the deformation.

(a) A nylon tow-rope pulling a stalled car. (b) A brass coin lying on the bed of a deep lake. (c) A rubber pad glued between a washing machine and the floor, pushed sideways as the drum spins. (d) A granite pillar holding up a roof. (e) The blade of a screwdriver as you turn a stiff screw.

Solution:

1. Run the test each time. Longer or shorter along one axis? Longitudinal. Shape changed, volume kept? Shearing. Squeezed inward from every side? Hydraulic.

2. Work through them.

Situation	Stress	Strain	Modulus
(a) nylon tow-rope	longitudinal, tensile	$\frac{\Delta L}{L}$	$Y$
(b) coin on the lake bed	hydraulic	$\frac{\Delta V}{V}$	$B$
(c) rubber pad under the machine	shearing	$\theta$	$G$
(d) granite pillar	longitudinal, compressive	$\frac{\Delta L}{L}$	$Y$
(e) twisted screwdriver blade	shearing (torsion)	$\theta$	$G$

3. The two that catch people. (b) is not a "squashing" problem in the $Y$ sense — the water pushes on the coin from all directions at once, not on two opposite faces, so it is a volume problem and needs $B$. And (e) is not a $Y$ problem either: twisting a rod slides each thin layer over the next, which is shear applied around an axis.

Final Answer: (a) tensile, $Y$; (b) hydraulic, $B$; (c) shearing, $G$; (d) compressive, $Y$; (e) shearing, $G$.

Takeaway: Tensile and compressive are the same stress with opposite signs, and both use $Y$. Only "from every side at once" gets you $B$, and only "sliding one face across another" gets you $G$.

Example 2: Two claims to judge

State whether each of these is true or false, and give the reason.

(a) Young's modulus of rubber is greater than that of steel. (b) The stretching of a coiled spring is determined by its shear modulus.

Solution:

1. (a) is FALSE. For the same stress, rubber stretches enormously and steel barely moves. Since $$Y = \frac{\text{stress}}{\text{strain}}$$ a big strain for a given stress means a small $Y$. Steel sits near $2 \times 10^{11}$ Pa and rubber near $10^{6}$ Pa — a factor of about a hundred thousand the other way.

2. Why people get it wrong. In everyday speech "elastic" means "stretchy", so rubber sounds like it should win. In physics the modulus measures resistance to deformation, so the stiff material has the larger $Y$.

3. (b) is TRUE. Pulling the two ends of a helical spring apart does not stretch the wire lengthwise in any important way. What actually happens is that each little element of the coiled wire gets twisted about the wire's own axis. Twisting is shear, so the spring constant of a helical spring is set by $G$, not by $Y$.

4. The check that makes it obvious. Unwind the spring into a straight wire of the same length and hang the same load on it. It extends by a fraction of a millimetre. Wound into a coil, the same wire and the same load give centimetres. The extra compliance comes entirely from twisting, and twisting is governed by $G$.

Final Answer: (a) false — steel's $Y$ is about $10^{5}$ times rubber's; (b) true — a helical spring extends by torsion of its wire, which is a shear deformation.

Takeaway: "Stretches more" means a smaller modulus, not a larger one. And a spring is a shear device wearing a stretching disguise.

Example 3: One copper cube, one force, three different answers

A solid copper cube of edge 4.0 cm is acted on by a force of magnitude $8.0 \times 10^{3}$ N, applied three different ways: (a) as a pull perpendicular to two opposite faces, (b) as a tangential push along the top face with the bottom face fixed, and (c) as a hydraulic pressure of the same magnitude of force per unit area on every face. For copper take $Y = 1.2 \times 10^{11}$ Pa, $G = 0.42 \times 10^{11}$ Pa and $B = 1.4 \times 10^{11}$ Pa. Find the deformation in each case.

Solution:

1. The stress is the same number in all three cases. Each face has area $$A = (0.040)^{2} = 1.6 \times 10^{-3} \text{ m}^2$$ $$\frac{F}{A} = \frac{8.0 \times 10^{3}}{1.6 \times 10^{-3}} = 5.0 \times 10^{6} \text{ Pa}$$ That is the whole point of the problem: identical stress, three different responses.

2. (a) Longitudinal. $$\varepsilon = \frac{F/A}{Y} = \frac{5.0 \times 10^{6}}{1.2 \times 10^{11}} = 4.17 \times 10^{-5}$$ $$\Delta L = \varepsilon L = 4.17 \times 10^{-5} \times 0.040 = 1.67 \times 10^{-6} \text{ m}$$ about $1.7$ micrometres longer.

3. (b) Shearing. $$\theta = \frac{F/A}{G} = \frac{5.0 \times 10^{6}}{0.42 \times 10^{11}} = 1.19 \times 10^{-4} \text{ rad}$$ $$\Delta x = \theta L = 1.19 \times 10^{-4} \times 0.040 = 4.76 \times 10^{-6} \text{ m}$$ about $4.8$ micrometres sideways.

4. (c) Hydraulic. $$\frac{\Delta V}{V} = \frac{\Delta p}{B} = \frac{5.0 \times 10^{6}}{1.4 \times 10^{11}} = 3.57 \times 10^{-5}$$ $$\Delta V = 3.57 \times 10^{-5} \times (0.040)^{3} = 2.29 \times 10^{-9} \text{ m}^3$$ a contraction of about $2.3$ cubic millimetres.

5. Rank them. The shear displacement is the biggest, because $G$ is the smallest of the three moduli. That ordering — $G < Y < B$ for a typical metal — is worth carrying around.

Final Answer: (a) $\Delta L = 1.7 \times 10^{-6}$ m; (b) $\Delta x = 4.8 \times 10^{-6}$ m with $\theta = 1.19 \times 10^{-4}$ rad; (c) $\Delta V = 2.3 \times 10^{-9}$ m$^3$.

Takeaway: The stress does not know which modulus you are going to use — you do. Same force, same area, same $\frac{F}{A}$, and three completely different deformations, because the geometry of how the force is applied is what selects the modulus.

Example 4: Reading the question, four sentences at a time

For each one-line description, write down the formula you would reach for first.

(i) A steel rod is clamped between two immovable walls and then heated. (ii) A solid glass sphere is lowered to a depth of 500 m in the sea. (iii) A rubber block bonded to the floor has its top surface pushed 2.0 mm sideways. (iv) A wire is stretched so that its length rises by $0.1\%$, and its diameter is measured before and after.

Solution:

1. (i) The rod wants to expand by $\alpha L\,\Delta T$ and is not allowed to. The forced-back strain is $\alpha\,\Delta T$, so the stress is $$\frac{F}{A} = Y\alpha\,\Delta T$$ Longitudinal stress, so $Y$ — and notice neither $L$ nor $A$ survives.

2. (ii) Pressure on every side at once. Get the pressure from $\Delta p = \rho gh$, then $$\frac{\Delta V}{V} = -\frac{\Delta p}{B}$$

3. (iii) One face slides over another, shape changes, volume does not. $$G = \frac{F/A}{\Delta x/L} \qquad \text{with } A \text{ the bonded face and } L \text{ the block's height}$$

4. (iv) Two strains at right angles to each other, so this is Poisson's ratio: $$\sigma = -\frac{\Delta d/d}{\Delta L/L}$$ and if the volume is also wanted, $\frac{\Delta V}{V} = (1 - 2\sigma)\frac{\Delta L}{L}$.

Final Answer: (i) $\frac{F}{A} = Y\alpha\,\Delta T$; (ii) $\frac{\Delta V}{V} = -\frac{\Delta p}{B}$ with $\Delta p = \rho gh$; (iii) $G = \frac{F}{A\theta}$; (iv) $\sigma = -\frac{\Delta d/d}{\Delta L/L}$.

[NEET Important] In an objective paper you will not have time to derive anything. Train yourself to go from the words of the stem to the formula in one step, exactly as above — that is what the first four examples are for.

Takeaway: Identify the deformation before you touch a calculator. Three lines of correct identification beat two pages of arithmetic done with the wrong modulus.

Part 2: Elongation Under a Load

Now the workhorse. Everything here is one formula rearranged, but the rearrangements are where the marks are.

Key Point: $$\Delta L = \frac{FL}{AY} \qquad \text{and therefore} \qquad \Delta L \propto \frac{L}{r^{2}} \text{ for a given load and material}$$ Notice what is not in it: the shape of the ends, how the load got there, or the material's density (unless the wire's own weight is part of the load).

Example 5: A structural steel rod, start to finish

A steel rod of radius 8.0 mm and length 1.5 m is pulled along its length by a force of 45 kN. Taking $Y = 2.0 \times 10^{11}$ Pa, find (a) the stress, (b) the strain and (c) the elongation.

Solution:

1. Cross-sectional area. The force is perpendicular to the circular end faces, so $$A = \pi r^{2} = \pi\left(8.0 \times 10^{-3}\right)^{2} = 2.011 \times 10^{-4} \text{ m}^2$$

2. (a) Stress. $$\frac{F}{A} = \frac{45 \times 10^{3}}{2.011 \times 10^{-4}} = 2.238 \times 10^{8} \text{ Pa}$$

3. (b) Strain, straight from the definition of $Y$: $$\varepsilon = \frac{F/A}{Y} = \frac{2.238 \times 10^{8}}{2.0 \times 10^{11}} = 1.119 \times 10^{-3}$$ which is $0.112\%$.

4. (c) Elongation. $$\Delta L = \varepsilon L = 1.119 \times 10^{-3} \times 1.5 = 1.68 \times 10^{-3} \text{ m} = 1.68 \text{ mm}$$

5. Is the rod safe? A structural steel yields somewhere around $2.5 \times 10^{8}$ Pa. We are at $2.24 \times 10^{8}$ Pa — inside the elastic region, but only by about $10\%$. A real designer would not accept that margin, which is exactly what the next few problems are about.

Final Answer: stress $2.24 \times 10^{8}$ Pa, strain $1.12 \times 10^{-3}$, elongation $1.68$ mm.

Takeaway: Do it in the order stress $\to$ strain $\to$ elongation, always. The intermediate stress is the number you compare against the yield strength, and skipping straight to $\Delta L$ throws away the one check that tells you whether the answer means anything.

Example 6: A copper bar of rectangular section

A copper bar of rectangular cross-section 12.0 mm by 24.0 mm is pulled in tension by a force of 38 000 N, and the deformation stays elastic. Taking $Y_{\text{copper}} = 1.2 \times 10^{11}$ Pa, find the strain, and the extension of a 2.0 m length of it.

Solution:

1. The area is a rectangle, not a circle. No $\pi$ anywhere: $$A = 12.0 \times 10^{-3} \times 24.0 \times 10^{-3} = 2.88 \times 10^{-4} \text{ m}^2$$

2. Stress. $$\frac{F}{A} = \frac{38\,000}{2.88 \times 10^{-4}} = 1.319 \times 10^{8} \text{ Pa}$$

3. Strain. $$\varepsilon = \frac{1.319 \times 10^{8}}{1.2 \times 10^{11}} = 1.100 \times 10^{-3}$$ that is, $0.110\%$.

4. Extension of 2.0 m. $$\Delta L = 1.100 \times 10^{-3} \times 2.0 = 2.20 \times 10^{-3} \text{ m} = 2.20 \text{ mm}$$

5. A useful way to store the answer. A strain of $1.10 \times 10^{-3}$ means "$1.1$ mm of stretch for every metre of bar", whatever the length. Quoting the strain rather than the extension is what makes an answer portable.

Final Answer: strain $1.10 \times 10^{-3}$ (about $0.11\%$); a 2.0 m length extends by $2.20$ mm.

Takeaway: Strain is the answer that does not depend on how long the bar happens to be. Give the strain and the extension follows for any length.

Example 7: The most a chairlift cable may carry

A ski-lift is hung from a steel cable of radius 1.2 cm. The design rule is that the stress in the cable must never exceed $1.2 \times 10^{8}$ Pa. What is the largest load the cable may carry, in newtons and in kilograms? Take $g = 9.8$ m/s$^2$.

Solution:

1. Area of the cable. $$A = \pi r^{2} = \pi\left(1.2 \times 10^{-2}\right)^{2} = 4.524 \times 10^{-4} \text{ m}^2$$

2. Rearrange the definition of stress. The permitted stress and the area fix the permitted force: $$F_{\max} = \left(\frac{F}{A}\right)_{\max} \times A = 1.2 \times 10^{8} \times 4.524 \times 10^{-4}$$ $$F_{\max} = 5.43 \times 10^{4} \text{ N}$$

3. Convert to a mass. $$m_{\max} = \frac{F_{\max}}{g} = \frac{5.43 \times 10^{4}}{9.8} = 5.54 \times 10^{3} \text{ kg}$$ about $5.5$ tonnes.

4. Two things the answer does not depend on. The length of the cable does not appear — a 10 m cable and a 500 m cable of the same section break at the same load. Neither does $Y$: how far it stretches on the way is a different question from how much it can hold.

Final Answer: about $5.4 \times 10^{4}$ N, which is a load of roughly $5.5 \times 10^{3}$ kg.

Takeaway: A strength question needs an area and a limiting stress, and nothing else. If you found yourself reaching for $Y$ or for the length, you answered a stiffness question by mistake.

Example 8: Designing a wire backwards from the stretch you will allow

A steel wire 8.0 m long is to carry a load of 4.0 kN, and it must not stretch by more than 2.0 mm. Take $Y = 2.0 \times 10^{11}$ Pa. What is the smallest diameter that will do, and what stress does the wire then run at?

Solution:

1. Start from the elongation formula and make $A$ the subject. $$\Delta L = \frac{FL}{AY} \qquad \Longrightarrow \qquad A = \frac{FL}{Y\,\Delta L}$$

2. Substitute, with everything in SI. $$A = \frac{4.0 \times 10^{3} \times 8.0}{2.0 \times 10^{11} \times 2.0 \times 10^{-3}} = \frac{3.2 \times 10^{4}}{4.0 \times 10^{8}} = 8.0 \times 10^{-5} \text{ m}^2$$

3. Turn the area into a diameter. $$r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{8.0 \times 10^{-5}}{\pi}} = 5.046 \times 10^{-3} \text{ m}$$ $$d = 2r = 1.009 \times 10^{-2} \text{ m} \approx 10.1 \text{ mm}$$

4. The stress it then runs at. $$\frac{F}{A} = \frac{4.0 \times 10^{3}}{8.0 \times 10^{-5}} = 5.0 \times 10^{7} \text{ Pa}$$ comfortably inside steel's elastic region, so the design is limited by stiffness, not by strength.

5. Which way to round. A smaller wire stretches more, so you must round the diameter up, never down. A 10 mm wire would be very slightly too thin; specify 11 mm or 12 mm.

Final Answer: a minimum diameter of about $10.1$ mm, running at a stress of $5.0 \times 10^{7}$ Pa.

Takeaway: Real design questions run the formula backwards. And when you round a designed dimension, always round in the direction that makes the structure safer.

Example 9: A long rope stretching under nothing but its own weight

A uniform steel rope of length 250 m hangs vertically from a winch with nothing attached to its lower end. Steel has density $7800$ kg/m$^3$ and $Y = 2.0 \times 10^{11}$ Pa. Take $g = 9.8$ m/s$^2$. Find (a) the greatest stress anywhere in the rope, (b) how much the rope stretches, and (c) the longest such rope that could hang without snapping, if steel's breaking stress is $4.0 \times 10^{8}$ Pa.

Solution:

1. Where is the tension largest? At the top. A cross-section at a distance $x$ below the top carries everything hanging under it, a length $L - x$, so $$T(x) = \rho A g\,(L - x)$$ The bottom carries nothing; the top carries the whole rope.

2. (a) The maximum stress, at $x = 0$: $$\left(\frac{F}{A}\right)_{\max} = \frac{\rho ALg}{A} = \rho gL = 7800 \times 9.8 \times 250 = 1.911 \times 10^{7} \text{ Pa}$$ The area has cancelled, which is the whole reason a thicker rope does not help.

3. (b) The stretch. The tension varies along the rope, so add up the stretch of each little piece: $$\Delta L = \int_0^L \frac{T(x)\,dx}{AY} = \frac{\rho g}{Y}\int_0^L (L - x)\,dx = \frac{\rho g}{Y}\left[\frac{L^{2}}{2}\right]$$ $$\Delta L = \frac{\rho gL^{2}}{2Y} = \frac{7800 \times 9.8 \times (250)^{2}}{2 \times 2.0 \times 10^{11}}$$ $$\Delta L = \frac{4.7775 \times 10^{9}}{4.0 \times 10^{11}} = 1.194 \times 10^{-2} \text{ m} = 11.9 \text{ mm}$$

4. Read the factor of two. $\frac{\rho gL^{2}}{2Y}$ is exactly what you would get by hanging half the rope's weight from the end of a weightless rope. The average tension along the rope is half the maximum, and "average tension" is all a linear material cares about.

5. (c) The longest rope that holds. Set the top stress equal to the breaking stress: $$\rho gL_{\max} = 4.0 \times 10^{8} \qquad \Longrightarrow \qquad L_{\max} = \frac{4.0 \times 10^{8}}{7800 \times 9.8} = \frac{4.0 \times 10^{8}}{7.644 \times 10^{4}} = 5.23 \times 10^{3} \text{ m}$$ about $5.2$ km — and again independent of how thick you make it. The 250 m rope of parts (a) and (b) is running at about a twentieth of that limit, which is why it is nowhere near failing.

Final Answer: (a) $1.91 \times 10^{7}$ Pa at the top; (b) $11.9$ mm of stretch; (c) a maximum hanging length of about $5.2$ km.

Takeaway: Self-weight elongation is $\frac{\rho gL^{2}}{2Y}$, the same as putting half the weight at the free end — and the maximum hanging length $\frac{\sigma_{\text{break}}}{\rho g}$ does not care about the thickness at all, because both the weight and the area grow together.

Example 10: The rope and the cage together

The same 250 m steel rope, of cross-sectional area $4.0$ cm$^2$, now carries a mine cage of mass 1200 kg at its lower end. With $\rho = 7800$ kg/m$^3$, $Y = 2.0 \times 10^{11}$ Pa and $g = 9.8$ m/s$^2$, find the total elongation and the maximum stress.

Solution:

1. Superpose. The rope is a linear elastic body, so the extension caused by the cage and the extension caused by the rope's own weight simply add.

2. The cage's contribution. Its weight is carried in full by every cross-section: $$\Delta L_1 = \frac{MgL}{AY} = \frac{1200 \times 9.8 \times 250}{4.0 \times 10^{-4} \times 2.0 \times 10^{11}}$$ $$\Delta L_1 = \frac{2.94 \times 10^{6}}{8.0 \times 10^{7}} = 3.675 \times 10^{-2} \text{ m} = 36.8 \text{ mm}$$

3. The rope's own contribution, from the previous problem: $$\Delta L_2 = \frac{\rho gL^{2}}{2Y} = 1.194 \times 10^{-2} \text{ m} = 11.9 \text{ mm}$$

4. Total. $$\Delta L = 36.8 + 11.9 = 48.7 \text{ mm}$$ Almost five centimetres — which is why a mine cage does not stop where you expect it to.

5. Maximum stress, again at the top. The top section carries the cage plus the whole rope: $$F_{\text{top}} = Mg + \rho ALg = 1200 \times 9.8 + 7800 \times 4.0 \times 10^{-4} \times 250 \times 9.8$$ $$F_{\text{top}} = 11\,760 + 7644 = 19\,404 \text{ N}$$ $$\left(\frac{F}{A}\right)_{\max} = \frac{19\,404}{4.0 \times 10^{-4}} = 4.85 \times 10^{7} \text{ Pa}$$

6. A number worth noticing. The rope's own weight is $7644$ N against the cage's $11\,760$ N — it is nearly two-thirds of the payload. At 500 m it would exceed it. That is the real limit on how deep a single-rope hoist can go.

Final Answer: total elongation $48.7$ mm; maximum stress $4.85 \times 10^{7}$ Pa at the top of the rope.

Takeaway: Superpose, do not average. The load's own $\frac{MgL}{AY}$ plus the rope's own $\frac{\rho gL^{2}}{2Y}$ — two separate terms, added, with the factor of two belonging only to the second.

Example 11: Four hollow columns under a building

A structure of mass 60 000 kg rests on four identical hollow steel columns, and the load is shared equally. Each column has an inner radius of 25 cm and an outer radius of 50 cm. Taking $Y = 2.0 \times 10^{11}$ Pa and $g = 9.8$ m/s$^2$, find the compressive strain in each column.

Solution:

1. Load per column. $$F = \frac{Mg}{4} = \frac{60\,000 \times 9.8}{4} = 1.47 \times 10^{5} \text{ N}$$

2. Area of an annulus — subtract, do not use the mean radius. $$A = \pi\left(r_o^{2} - r_i^{2}\right) = \pi\left(0.50^{2} - 0.25^{2}\right) = \pi\left(0.25 - 0.0625\right)$$ $$A = \pi \times 0.1875 = 0.5890 \text{ m}^2$$

3. Stress. $$\frac{F}{A} = \frac{1.47 \times 10^{5}}{0.5890} = 2.496 \times 10^{5} \text{ Pa}$$

4. Strain. $$\varepsilon = \frac{F/A}{Y} = \frac{2.496 \times 10^{5}}{2.0 \times 10^{11}} = 1.25 \times 10^{-6}$$

5. Feel the size of it. A column 5 m tall shortens by $1.25 \times 10^{-6} \times 5 = 6.2$ micrometres — about a tenth the width of a human hair. Buildings do not visibly settle under load because steel columns are enormously oversized for the stress, and they are oversized because the real enemy is not crushing but buckling.

Final Answer: a compressive strain of about $1.25 \times 10^{-6}$ in each column.

Takeaway: For a hollow section, $A = \pi(r_o^{2} - r_i^{2})$. Every year somebody writes $\pi(r_o - r_i)^{2}$ instead, which here would give $0.196$ m$^2$ and an answer three times too big.

Part 3: Comparing Two Wires

The single most-asked question type in the chapter. Nothing here needs a calculator if you set it up as a ratio.

Key Point: For the same load, $$\Delta L \propto \frac{L}{r^{2}Y}$$ Long, thin and floppy stretches most. Write the ratio first and cancel everything that is common before any number goes in.

Example 12: Two wires that stretch by exactly the same amount

A steel wire of length 5.2 m and cross-sectional area $2.5 \times 10^{-5}$ m$^2$ stretches by the same amount as a copper wire of length 4.0 m and cross-sectional area $3.6 \times 10^{-5}$ m$^2$ when both carry the same load. What is the ratio of Young's modulus of steel to that of copper?

Solution:

1. Write the elongation for each. Same load $F$ in both: $$\Delta L_s = \frac{FL_s}{A_sY_s}, \qquad \Delta L_c = \frac{FL_c}{A_cY_c}$$

2. Set them equal — that is the whole content of the phrase "stretches by the same amount": $$\frac{FL_s}{A_sY_s} = \frac{FL_c}{A_cY_c}$$

3. $F$ cancels. Rearranging for the modulus ratio: $$\frac{Y_s}{Y_c} = \frac{L_s}{L_c} \times \frac{A_c}{A_s}$$

4. Substitute. $$\frac{Y_s}{Y_c} = \frac{5.2}{4.0} \times \frac{3.6 \times 10^{-5}}{2.5 \times 10^{-5}} = 1.30 \times 1.44 = 1.87$$

5. Sanity check the direction. The steel wire is longer and thinner, and yet it stretches only as much as the copper one. It must therefore be the stiffer material — $Y_s > Y_c$, and the ratio should come out above 1. It does. If you had got $0.53$, you inverted a fraction.

Final Answer: $\frac{Y_s}{Y_c} = 1.87$, so steel is about $1.9$ times as stiff as copper.

Takeaway: "Same load, same extension" is a ratio equation, not two separate calculations. You never need to know $F$, and you never need to know either modulus on its own.

Example 13: Longer, thinner and softer, all at once

Wire B is twice as long as wire A, has half the radius, and is made of a material whose Young's modulus is one third of A's. Both carry the same load. By what factor does B stretch more than A?

Solution:

1. Start from the proportionality. $$\Delta L \propto \frac{L}{r^{2}Y}$$

2. Take the ratio term by term. $$\frac{\Delta L_B}{\Delta L_A} = \frac{L_B}{L_A} \times \left(\frac{r_A}{r_B}\right)^{2} \times \frac{Y_A}{Y_B}$$

3. Substitute the three factors. $$= 2 \times \left(\frac{1}{1/2}\right)^{2} \times \frac{1}{1/3} = 2 \times 4 \times 3 = 24$$

4. Read where the 24 comes from. Doubling the length costs a factor of 2. Halving the radius costs a factor of 4, because the area went down by four. Thirding the modulus costs a factor of 3. Multiply.

5. The lesson in one line. Radius is the most powerful lever you have, because it enters squared. Halving a wire's thickness has the same effect on its stretch as making it four times longer.

Final Answer: wire B stretches $24$ times as much as wire A.

Takeaway: Radius enters as $r^{2}$, so thickness beats length and modulus for sheer leverage. Deal with the squared factor first and the rest is easy.

Example 14: Three wires under the same load — which stretches most?

Three wires each carry a load of 20 N. Wire P is steel, 2.0 m long, radius $0.40$ mm; wire Q is copper, 3.0 m long, radius $0.50$ mm; wire R is aluminium, 1.5 m long, radius $0.30$ mm. Take $Y_{\text{steel}} = 2.0 \times 10^{11}$ Pa, $Y_{\text{copper}} = 1.2 \times 10^{11}$ Pa and $Y_{\text{aluminium}} = 0.70 \times 10^{11}$ Pa. Rank them by extension.

Solution:

1. Areas first. $$A_P = \pi\left(0.40 \times 10^{-3}\right)^{2} = 5.027 \times 10^{-7} \text{ m}^2$$ $$A_Q = \pi\left(0.50 \times 10^{-3}\right)^{2} = 7.854 \times 10^{-7} \text{ m}^2$$ $$A_R = \pi\left(0.30 \times 10^{-3}\right)^{2} = 2.827 \times 10^{-7} \text{ m}^2$$

2. Now $\Delta L = \frac{FL}{AY}$ for each. $$\Delta L_P = \frac{20 \times 2.0}{5.027 \times 10^{-7} \times 2.0 \times 10^{11}} = 3.98 \times 10^{-4} \text{ m}$$ $$\Delta L_Q = \frac{20 \times 3.0}{7.854 \times 10^{-7} \times 1.2 \times 10^{11}} = 6.37 \times 10^{-4} \text{ m}$$ $$\Delta L_R = \frac{20 \times 1.5}{2.827 \times 10^{-7} \times 0.70 \times 10^{11}} = 1.516 \times 10^{-3} \text{ m}$$

3. Rank. $$\Delta L_R > \Delta L_Q > \Delta L_P \qquad 1.52 \text{ mm} > 0.64 \text{ mm} > 0.40 \text{ mm}$$

4. Check the stresses too, because they decide safety. $$\left(\frac{F}{A}\right)_R = \frac{20}{2.827 \times 10^{-7}} = 7.07 \times 10^{7} \text{ Pa}$$ which is the largest of the three, and already approaching aluminium's yield. The thinnest wire is both the stretchiest and the closest to failure — two separate reasons, both traceable to the same small area.

Final Answer: aluminium stretches most ($1.52$ mm), then copper ($0.64$ mm), then steel ($0.40$ mm).

Takeaway: The shortest wire here stretches the most. Length is only one of three factors, and being thin and soft beat being short.

Example 15: Cut the wire in half, then use both halves

A wire of length $L$ extends by $1.2$ mm under a certain load. The wire is now cut into two equal halves. (a) One half alone is used to carry the same load — what is its extension? (b) Instead, the two halves are hung side by side and together carry the same load — now what?

Solution:

1. (a) One half alone. Same material, same area, half the length, same force: $$\Delta L \propto L \qquad \Longrightarrow \qquad \Delta L^{\,\prime} = \frac{1.2}{2} = 0.60 \text{ mm}$$

2. (b) Two halves side by side. This is a parallel arrangement. Both halves have the same length and the same area, so both stretch by the same amount, and each therefore carries half the load: $$F_{\text{each}} = \frac{F}{2}$$

3. Now apply the formula to one of them, with half the length and half the force: $$\Delta L^{\,\prime\prime} = \frac{(F/2)(L/2)}{AY} = \frac{1}{4}\times\frac{FL}{AY} = \frac{1.2}{4} = 0.30 \text{ mm}$$

4. Cross-check with the spring picture. A wire is a spring of stiffness $k = \frac{YA}{L}$. Halving $L$ doubles $k$; putting two such springs in parallel doubles it again. Four times the stiffness, so a quarter of the extension. Same answer.

5. And what if the halves are rejoined end to end? Then you are back to the original wire — a series pair of stiffness-$2k$ springs is a spring of stiffness $k$ — and the extension is $1.2$ mm again, as it must be.

Final Answer: (a) $0.60$ mm; (b) $0.30$ mm.

Takeaway: Parallel wires share the extension and divide the load; series wires share the tension and their extensions add. Or in one sentence: in parallel the stiffnesses add, in series the compliances add — exactly like springs, because a wire is a spring with $k = \frac{YA}{L}$.

Part 4: Wires Joined Together

Three arrangements, three different bookkeeping rules. Get the rule right and the arithmetic is the easy part.

Example 16: A copper wire and a steel wire joined end to end

A copper wire 1.8 m long and a steel wire 1.2 m long, both of diameter $2.0$ mm, are joined end to end and hung from a support. A load is attached to the free end and the total elongation of the pair is measured as $0.60$ mm. Find the load, and how the elongation splits between the two wires. Take $Y_{\text{copper}} = 1.2 \times 10^{11}$ Pa and $Y_{\text{steel}} = 2.0 \times 10^{11}$ Pa.

Solution:

1. The rule for series. Cut anywhere and you find the same tension, so each wire carries the full load $W$. The extensions add.

2. Common area. $$A = \pi\left(1.0 \times 10^{-3}\right)^{2} = 3.142 \times 10^{-6} \text{ m}^2$$

3. Write the total extension in terms of $W$. $$\Delta L_{\text{total}} = \frac{W}{A}\left(\frac{L_c}{Y_c} + \frac{L_s}{Y_s}\right)$$

4. Evaluate the bracket. $$\frac{L_c}{Y_c} = \frac{1.8}{1.2 \times 10^{11}} = 1.500 \times 10^{-11}, \qquad \frac{L_s}{Y_s} = \frac{1.2}{2.0 \times 10^{11}} = 0.600 \times 10^{-11}$$ $$\text{sum} = 2.100 \times 10^{-11} \text{ m}^3/\text{N}$$

5. Solve for $W$. $$W = \frac{A\,\Delta L_{\text{total}}}{2.100 \times 10^{-11}} = \frac{3.142 \times 10^{-6} \times 0.60 \times 10^{-3}}{2.100 \times 10^{-11}}$$ $$W = \frac{1.885 \times 10^{-9}}{2.100 \times 10^{-11}} = 89.8 \text{ N}$$

6. Split the elongation. The two pieces are in the ratio $1.500 : 0.600$, that is $5 : 2$: $$\Delta L_c = \frac{5}{7}\times 0.60 = 0.429 \text{ mm}, \qquad \Delta L_s = \frac{2}{7}\times 0.60 = 0.171 \text{ mm}$$ Copper is both longer and softer, so it does most of the stretching. Their sum is $0.60$ mm, as required.

Final Answer: the load is about $89.8$ N; the copper stretches $0.43$ mm and the steel $0.17$ mm.

Takeaway: In series, factor the common tension out and add the compliances $\frac{L}{AY}$. The softer, longer wire always takes the larger share of the extension.

Example 17: A load at each joint

A steel wire 1.2 m long hangs from a ceiling. From its lower end a mass of $6.0$ kg is hung, and from that same point a brass wire $0.8$ m long hangs, carrying a further $4.0$ kg at its lower end. Both wires have diameter $0.30$ cm. Find the elongation of each wire. Take $Y_{\text{steel}} = 2.0 \times 10^{11}$ Pa, $Y_{\text{brass}} = 0.91 \times 10^{11}$ Pa and $g = 9.8$ m/s$^2$.

Solution:

1. The rule here. The tension is not the same in both wires, because there is a load hanging between them. Make an imaginary cut in each wire and add up what hangs below it.

2. Cut the brass wire. Below the cut hangs only the lower mass: $$F_{\text{brass}} = 4.0 \times 9.8 = 39.2 \text{ N}$$

3. Cut the steel wire. Below that cut hangs the $6.0$ kg, the brass wire and the $4.0$ kg: $$F_{\text{steel}} = (6.0 + 4.0)\times 9.8 = 98.0 \text{ N}$$ (The wires' own weights are neglected, as the data implies.)

4. Common area. The radius is $0.15$ cm $= 1.5 \times 10^{-3}$ m: $$A = \pi\left(1.5 \times 10^{-3}\right)^{2} = 7.069 \times 10^{-6} \text{ m}^2$$

5. Elongation of the steel wire. $$\Delta L_s = \frac{98.0 \times 1.2}{7.069 \times 10^{-6} \times 2.0 \times 10^{11}} = \frac{117.6}{1.414 \times 10^{6}} = 8.32 \times 10^{-5} \text{ m}$$ that is, $0.0832$ mm.

6. Elongation of the brass wire. $$\Delta L_b = \frac{39.2 \times 0.8}{7.069 \times 10^{-6} \times 0.91 \times 10^{11}} = \frac{31.36}{6.433 \times 10^{5}} = 4.88 \times 10^{-5} \text{ m}$$ that is, $0.0488$ mm.

7. A check on the ratio. Brass carries $0.4$ of the steel's load over $\frac{2}{3}$ of the length with $0.455$ of the modulus, so its extension should be $0.4 \times \frac{2}{3} \div 0.455 = 0.586$ of the steel's. And $\frac{4.88}{8.32} = 0.586$. Consistent.

Final Answer: the steel wire elongates by $0.083$ mm, the brass wire by $0.049$ mm.

Takeaway: Cut below each joint and re-count what is hanging. A load applied part way down changes the tension above it but not below it, and that is the whole problem.

Example 18: A rigid bar carried by three wires, all pulling equally

A rigid bar of mass 20 kg hangs symmetrically from three vertical wires, each $1.8$ m long. The two outer wires are copper and the middle one is steel. What must the ratio of their diameters be if all three are to carry the same tension? Take $Y_{\text{copper}} = 1.2 \times 10^{11}$ Pa, $Y_{\text{steel}} = 2.0 \times 10^{11}$ Pa and $g = 9.8$ m/s$^2$.

Solution:

1. The tension in each, from the force balance. Equal tensions plus symmetry give $$3T = Mg \qquad \Longrightarrow \qquad T = \frac{20 \times 9.8}{3} = 65.3 \text{ N}$$

2. The condition the bar is rigid. Because the bar cannot bend or tilt, all three wires must extend by exactly the same amount. They all start at the same length, so all three have the same strain: $$\varepsilon_{\text{copper}} = \varepsilon_{\text{steel}}$$

3. Turn equal strain into a condition on the areas. Strain is $\frac{F/A}{Y}$, and the tensions are equal by requirement, so $$\frac{T}{A_cY_c} = \frac{T}{A_sY_s} \qquad \Longrightarrow \qquad A_cY_c = A_sY_s$$

4. Hence $$\frac{A_c}{A_s} = \frac{Y_s}{Y_c} \qquad \Longrightarrow \qquad \frac{d_c^{2}}{d_s^{2}} = \frac{Y_s}{Y_c}$$ $$\frac{d_c}{d_s} = \sqrt{\frac{Y_s}{Y_c}} = \sqrt{\frac{2.0 \times 10^{11}}{1.2 \times 10^{11}}} = \sqrt{1.667} = 1.29$$

5. Read it physically. Copper is the softer material, so a copper wire must be fatter — by about $29\%$ in diameter — to stretch as little as the steel one under the same pull. Make them the same thickness instead and the stiff steel wire would take the lion's share of the load.

Final Answer: $\frac{d_{\text{copper}}}{d_{\text{steel}}} = \sqrt{\frac{Y_s}{Y_c}} = 1.29$, with each wire carrying $65.3$ N.

Takeaway: A rigid bar is a statement about strains, not about forces. Write "all extensions equal" first, then combine it with the force balance — that pair of equations solves every problem of this family.

Example 19: Where must the load hang for the bar to stay level?

A light rigid bar $1.0$ m long hangs horizontally from two vertical wires attached at its two ends. Both wires are $1.5$ m long with cross-sectional area $1.0$ mm$^2$; the left one is steel and the right one is copper. A load of 100 N is to be hung from the bar. (a) Where must it be attached for the bar to stay horizontal? (b) What tension does each wire then carry, and (c) how far does the bar drop? Take $Y_{\text{steel}} = 2.0 \times 10^{11}$ Pa and $Y_{\text{copper}} = 1.2 \times 10^{11}$ Pa.

Solution:

1. The horizontal condition. If the bar stays horizontal, both wires must stretch by the same amount: $$\frac{T_sL}{AY_s} = \frac{T_cL}{AY_c} \qquad \Longrightarrow \qquad \frac{T_s}{T_c} = \frac{Y_s}{Y_c} = \frac{2.0}{1.2} = \frac{5}{3}$$ Same length, same area, so the stiffer wire simply takes proportionally more.

2. (b) Combine with the force balance $T_s + T_c = 100$: $$T_s = \frac{5}{8}\times 100 = 62.5 \text{ N}, \qquad T_c = \frac{3}{8}\times 100 = 37.5 \text{ N}$$

3. (a) Now the torque balance fixes the position. Take moments about the point where the steel wire meets the bar, and let the load hang a distance $x$ from that end: $$100x = T_c \times 1.0 = 37.5 \qquad \Longrightarrow \qquad x = 0.375 \text{ m}$$ So the load must hang $37.5$ cm from the steel end — nearer the stiff wire, which is what you would guess, because the stiff wire needs the bigger share.

4. (c) The drop. Use either wire; both must give the same number: $$\Delta L = \frac{T_sL}{AY_s} = \frac{62.5 \times 1.5}{1.0 \times 10^{-6} \times 2.0 \times 10^{11}} = \frac{93.75}{2.0 \times 10^{5}} = 4.69 \times 10^{-4} \text{ m}$$ Check on the copper side: $$\frac{37.5 \times 1.5}{1.0 \times 10^{-6} \times 1.2 \times 10^{11}} = \frac{56.25}{1.2 \times 10^{5}} = 4.69 \times 10^{-4} \text{ m}$$ The same $0.469$ mm, which confirms the bar really is level.

Final Answer: the load must hang $0.375$ m from the steel wire; $T_{\text{steel}} = 62.5$ N and $T_{\text{copper}} = 37.5$ N; the bar drops $0.469$ mm.

[JEE Tip] This is the three-equation template that JEE keeps coming back to: force balance, torque balance, and compatibility of extensions. Two of them are mechanics you already know; the third is the only new thing this chapter adds.

Takeaway: When a rigid body hangs on more than one wire, the geometry supplies the missing equation. Force balance alone is not enough — you always need the statement about how the extensions must be related.

Part 5: Things Being Squashed — Bones, Pillars and Columns

Compression is longitudinal stress with the sign flipped, so the formulas are identical. What changes is that the numbers now come from bodies and buildings.

Example 20: The human pyramid

In a circus act the whole of a balanced human pyramid rests on the legs of one performer lying on his back. The combined mass of everybody in the act, together with the tables and planks, is 350 kg. The performer at the bottom himself has a mass of 65 kg. Each of his thighbones is $0.45$ m long with an effective radius of $2.2$ cm. Taking $Y_{\text{bone}} = 9.4 \times 10^{9}$ Pa and $g = 9.8$ m/s$^2$, find how much each thighbone is compressed by the extra load.

Solution:

1. Work out what the legs actually carry. The performer's own body is not pressing down on his own thighbones through the plank — the question asks for the compression caused by the extra load above him: $$m_{\text{supported}} = 350 - 65 = 285 \text{ kg}$$

2. Turn it into a force, and split it between two legs. $$W = 285 \times 9.8 = 2793 \text{ N}$$ $$F_{\text{per bone}} = \frac{W}{2} = 1396.5 \text{ N}$$

3. Cross-sectional area of one femur. $$A = \pi r^{2} = \pi\left(2.2 \times 10^{-2}\right)^{2} = 1.521 \times 10^{-3} \text{ m}^2$$

4. The stress in the bone. $$\frac{F}{A} = \frac{1396.5}{1.521 \times 10^{-3}} = 9.18 \times 10^{5} \text{ Pa}$$

5. The compression. $$\Delta L = \frac{FL}{AY} = \frac{1396.5 \times 0.45}{1.521 \times 10^{-3} \times 9.4 \times 10^{9}}$$ $$\Delta L = \frac{628.4}{1.430 \times 10^{7}} = 4.40 \times 10^{-5} \text{ m}$$ about $0.044$ mm, or $4.4 \times 10^{-3}$ cm.

6. As a fraction. $$\varepsilon = \frac{\Delta L}{L} = \frac{4.40 \times 10^{-5}}{0.45} = 9.77 \times 10^{-5} \approx 0.0098\%$$

Final Answer: each thighbone shortens by about $4.4 \times 10^{-5}$ m, a strain of roughly $0.01\%$.

Takeaway: Subtract the bottom performer's own mass before you start. And notice how small the answer is: bone has a modest $Y$, but it is short and fat, and area beats modulus every time.

Example 21: What load would a thighbone actually fail at?

Bone will fail in compression at a stress of about $1.7 \times 10^{8}$ Pa. For the femur of the previous problem, of effective radius $2.2$ cm, find the force that would break it, and express it as a multiple of the body weight of a 65 kg person. Take $g = 9.8$ m/s$^2$.

Solution:

1. Use the same area. $$A = \pi\left(2.2 \times 10^{-2}\right)^{2} = 1.521 \times 10^{-3} \text{ m}^2$$

2. Breaking force. $$F_{\text{break}} = \sigma_{\text{break}} \times A = 1.7 \times 10^{8} \times 1.521 \times 10^{-3}$$ $$F_{\text{break}} = 2.585 \times 10^{5} \text{ N}$$ about 26 tonnes-weight on one bone.

3. As a multiple of body weight. A 65 kg person weighs $65 \times 9.8 = 637$ N, so $$\frac{F_{\text{break}}}{W_{\text{body}}} = \frac{2.585 \times 10^{5}}{637} = 406$$

4. So why do people break their femurs? Because real falls do not load the bone squarely along its axis. A sideways impact bends the bone, and in bending the stress concentrates on the outer surface and can be tens of times the average — and bone is far weaker in tension and in shear than it is in straight compression. Standing on it, the femur is absurdly over-engineered; twisting it is another matter.

5. Back to the pyramid. The pyramid produced $9.18 \times 10^{5}$ Pa in the bone. That is $$\frac{9.18 \times 10^{5}}{1.7 \times 10^{8}} = 0.0054$$ about half of one per cent of the failure stress. The act is safe by a factor of nearly 200 — as far as the thighbone is concerned.

Final Answer: about $2.6 \times 10^{5}$ N, roughly 406 times the person's own body weight.

Takeaway: Compare every stress you calculate against the material's limit. An answer of "$9.2 \times 10^{5}$ Pa" means nothing on its own; "half a per cent of the breaking stress" means everything.

Example 22: A concrete pillar under a water tank

A concrete pillar of square cross-section 40 cm by 40 cm and height $4.0$ m carries a water tank of total mass 90 tonnes. Concrete has $Y = 3.0 \times 10^{10}$ Pa and a crushing strength of about $2.0 \times 10^{7}$ Pa. Take $g = 9.8$ m/s$^2$. Find the stress, the strain and the shortening of the pillar, and say whether it is safe.

Solution:

1. The load. $$F = Mg = 90 \times 10^{3} \times 9.8 = 8.82 \times 10^{5} \text{ N}$$

2. The area. $$A = 0.40 \times 0.40 = 0.160 \text{ m}^2$$

3. The compressive stress. $$\frac{F}{A} = \frac{8.82 \times 10^{5}}{0.160} = 5.51 \times 10^{6} \text{ Pa}$$

4. Strain and shortening. $$\varepsilon = \frac{5.51 \times 10^{6}}{3.0 \times 10^{10}} = 1.84 \times 10^{-4}$$ $$\Delta L = \varepsilon L = 1.84 \times 10^{-4} \times 4.0 = 7.35 \times 10^{-4} \text{ m} = 0.735 \text{ mm}$$

5. Is it safe? $$\frac{5.51 \times 10^{6}}{2.0 \times 10^{7}} = 0.276$$ The pillar runs at about $28\%$ of its crushing strength — a safety factor of roughly $3.6$ against crushing. That is a normal design margin for a static load, though a real engineer would also check buckling and the strength of the foundation.

Final Answer: stress $5.51 \times 10^{6}$ Pa, strain $1.84 \times 10^{-4}$, shortening $0.735$ mm; safe, at about $28\%$ of the crushing strength.

Takeaway: The whole calculation is $\frac{F}{A}$, then divide by $Y$, then multiply by $L$. What makes it an engineering answer rather than a physics answer is the last line, where you compare it with the limit.

Example 23: A reinforced column — how does the load split?

A short column consists of a concrete block of square section 30 cm by 30 cm with four steel reinforcing rods embedded in it, of total cross-sectional area $20$ cm$^2$. The column carries an axial load of 400 kN. Steel and concrete are bonded so that they shorten together. Take $Y_{\text{steel}} = 2.0 \times 10^{11}$ Pa and $Y_{\text{concrete}} = 3.0 \times 10^{10}$ Pa. Find how the load divides, and the stress in each material.

Solution:

1. This is a parallel arrangement. Bonded together and the same length, so both materials must have the same strain $\varepsilon$. The forces they carry are what differ.

2. Areas. The steel occupies part of the section, so the concrete area is what is left: $$A_s = 20 \times 10^{-4} = 2.0 \times 10^{-3} \text{ m}^2$$ $$A_c = 0.30 \times 0.30 - 2.0 \times 10^{-3} = 0.090 - 0.002 = 0.088 \text{ m}^2$$

3. Write each force in terms of the common strain. Since $F = YA\varepsilon$, $$F_s + F_c = \left(Y_sA_s + Y_cA_c\right)\varepsilon = P$$

4. Evaluate the bracket. $$Y_sA_s = 2.0 \times 10^{11} \times 2.0 \times 10^{-3} = 4.0 \times 10^{8} \text{ N}$$ $$Y_cA_c = 3.0 \times 10^{10} \times 0.088 = 2.64 \times 10^{9} \text{ N}$$ $$\text{sum} = 3.04 \times 10^{9} \text{ N}$$

5. The common strain. $$\varepsilon = \frac{P}{3.04 \times 10^{9}} = \frac{400 \times 10^{3}}{3.04 \times 10^{9}} = 1.316 \times 10^{-4}$$

6. Back-substitute for the forces and stresses. $$F_s = 4.0 \times 10^{8} \times 1.316 \times 10^{-4} = 5.26 \times 10^{4} \text{ N}$$ $$F_c = 2.64 \times 10^{9} \times 1.316 \times 10^{-4} = 3.47 \times 10^{5} \text{ N}$$ Their sum is $4.00 \times 10^{5}$ N, as it must be. $$\left(\frac{F}{A}\right)_s = Y_s\varepsilon = 2.63 \times 10^{7} \text{ Pa}, \qquad \left(\frac{F}{A}\right)_c = Y_c\varepsilon = 3.95 \times 10^{6} \text{ Pa}$$

7. The point of reinforcing. The steel is only $2.2\%$ of the area but carries $13.2\%$ of the load, and it runs at almost seven times the stress of the concrete — exactly the ratio $\frac{Y_s}{Y_c}$. In a parallel arrangement, stress divides in the ratio of the moduli.

Final Answer: steel carries $52.6$ kN at $2.63 \times 10^{7}$ Pa; concrete carries $347$ kN at $3.95 \times 10^{6}$ Pa; the common strain is $1.32 \times 10^{-4}$.

Takeaway: Same strain, so stresses are in the ratio of the moduli, and forces in the ratio of $YA$. The stiff material always grabs more than its share of the load — which is the reason reinforcement works and also the reason it can fail first.

Part 6: Shear — Sliding, Not Stretching

The one place where using the wrong area is almost guaranteed unless you are deliberate about it.

Key Point: $$G = \frac{F/A}{\Delta x / L} = \frac{F}{A\theta}$$ Here $A$ is the face the force acts along — the face being dragged — and $L$ is the perpendicular distance between the dragged face and the fixed one. For small $\theta$, $\theta$ in radians equals $\frac{\Delta x}{L}$.

Example 24: A copper cube stuck to a wall

A solid copper cube of edge $8.0$ cm has one face cemented firmly to a vertical wall. A mass of 150 kg is hung from the opposite (outer) face. For copper $G = 0.42 \times 10^{11}$ Pa. Take $g = 9.8$ m/s$^2$. Find the vertical deflection of the loaded face.

Solution:

1. Identify the deformation. The hanging weight drags the outer face downwards while the cemented face cannot move. One face slides relative to the other, parallel to itself — that is shear.

2. The shearing force. $$F = mg = 150 \times 9.8 = 1470 \text{ N}$$

3. The right area. The force acts along the outer face, so $A$ is the area of that face: $$A = (0.080)^{2} = 6.4 \times 10^{-3} \text{ m}^2$$

4. Shearing stress. $$\frac{F}{A} = \frac{1470}{6.4 \times 10^{-3}} = 2.297 \times 10^{5} \text{ Pa}$$

5. Shearing strain. $$\theta = \frac{F/A}{G} = \frac{2.297 \times 10^{5}}{0.42 \times 10^{11}} = 5.47 \times 10^{-6} \text{ rad}$$

6. The deflection. The perpendicular distance between the fixed face and the loaded one is the cube's edge: $$\Delta x = \theta L = 5.47 \times 10^{-6} \times 0.080 = 4.38 \times 10^{-7} \text{ m}$$ about $0.44$ micrometres.

Final Answer: the loaded face drops by about $4.4 \times 10^{-7}$ m, with a shearing strain of $5.5 \times 10^{-6}$ rad.

Takeaway: In shear, both $A$ and $L$ mean something different from what they mean in tension. $A$ is the face the force runs along; $L$ is measured across the block, perpendicular to the sliding. Get those two right and the rest is one division.

Example 25: The force needed to move a block a stated amount

A brass block with a base $10$ cm by $10$ cm and a height of $8.0$ cm is bonded to a rigid bench. What tangential force applied to its top face will move that face sideways by $0.010$ mm? For brass, $G = 0.36 \times 10^{11}$ Pa.

Solution:

1. Work out the shearing strain first, because it needs no force at all: $$\theta = \frac{\Delta x}{L} = \frac{0.010 \times 10^{-3}}{0.080} = 1.25 \times 10^{-4} \text{ rad}$$ Note $L$ is the height, $8.0$ cm, not an edge of the loaded face.

2. Turn strain into stress with the modulus. $$\frac{F}{A} = G\theta = 0.36 \times 10^{11} \times 1.25 \times 10^{-4} = 4.50 \times 10^{6} \text{ Pa}$$

3. The loaded face is the top, so $$A = 0.10 \times 0.10 = 1.0 \times 10^{-2} \text{ m}^2$$

4. The force. $$F = \frac{F}{A}\times A = 4.50 \times 10^{6} \times 1.0 \times 10^{-2} = 4.5 \times 10^{4} \text{ N}$$ about $45$ kN, or four and a half tonnes-weight — to move the top of a brass block by one hundredth of a millimetre.

5. Why so enormous? Because $G$ for a metal is around $10^{10}$ Pa. Metals resist shear almost as fiercely as they resist stretching, which is exactly why you cannot deform a machine part with your hands.

Final Answer: about $4.5 \times 10^{4}$ N.

Takeaway: Go strain $\to$ stress $\to$ force when the displacement is given, and force $\to$ stress $\to$ strain when the force is given. Both directions are the same three quantities; only the order changes.

Example 26: A rubber pad in an engine mounting

A rubber block $12$ cm by $12$ cm and $3.0$ cm thick is bonded between two steel plates, the lower one fixed. A horizontal force of 480 N is applied to the upper plate. The rubber has $G = 8.0 \times 10^{5}$ Pa. Find the shearing stress, the angle of shear in degrees, and the sideways displacement of the upper plate.

Solution:

1. The bonded face area. $$A = 0.12 \times 0.12 = 1.44 \times 10^{-2} \text{ m}^2$$

2. Shearing stress. $$\frac{F}{A} = \frac{480}{1.44 \times 10^{-2}} = 3.33 \times 10^{4} \text{ Pa}$$

3. Shearing strain. $$\theta = \frac{F/A}{G} = \frac{3.33 \times 10^{4}}{8.0 \times 10^{5}} = 4.17 \times 10^{-2} \text{ rad}$$

4. In degrees. $$\theta = 4.17 \times 10^{-2} \times \frac{180}{\pi} = 2.39°$$

5. The displacement. $$\Delta x = \theta L = 4.17 \times 10^{-2} \times 0.030 = 1.25 \times 10^{-3} \text{ m} = 1.25 \text{ mm}$$

6. Compare with the brass block above. Almost a hundred times less force, and the displacement is a hundred times bigger — because $G$ for rubber is about $45\,000$ times smaller. That enormous compliance in shear is precisely what an engine mounting is for: it lets the engine move a millimetre or two instead of transmitting the vibration into the chassis.

7. A caution about the small-angle step. At $\theta = 0.0417$ rad, $\tan\theta = 0.04174$, so using $\theta$ instead of $\tan\theta$ costs about $0.06\%$. Fine here. For rubber sheared through $20°$ or more it would not be.

Final Answer: shearing stress $3.33 \times 10^{4}$ Pa, angle of shear $2.39°$, displacement $1.25$ mm.

Takeaway: A small $G$ is a design feature, not a defect. Anywhere you want motion to be absorbed rather than transmitted, you put in something with a shear modulus five orders of magnitude below the metal around it.

Example 27: Shearing a rivet

Two steel plates are joined by a single rivet of diameter $8.0$ mm, and the plates are pulled apart with a force of $3.0$ kN. (a) Find the shearing stress in the rivet. (b) If the plates themselves have a cross-sectional area of $60$ mm$^2$ at the joint, find the tensile stress in a plate and compare. (c) If the rivet material can stand a shearing stress of $1.5 \times 10^{8}$ Pa, how many such rivets are needed for a joint carrying 20 kN?

Solution:

1. How the rivet is loaded. The two plates try to slide past each other, and the rivet is cut across by that sliding. The relevant area is the rivet's circular cross-section, the plane on which the shearing happens: $$A_{\text{rivet}} = \pi r^{2} = \pi\left(4.0 \times 10^{-3}\right)^{2} = 5.027 \times 10^{-5} \text{ m}^2$$

2. (a) Shearing stress in the rivet. $$\frac{F}{A} = \frac{3.0 \times 10^{3}}{5.027 \times 10^{-5}} = 5.97 \times 10^{7} \text{ Pa}$$

3. (b) Tensile stress in a plate. $$\frac{F}{A} = \frac{3.0 \times 10^{3}}{60 \times 10^{-6}} = 5.0 \times 10^{7} \text{ Pa}$$ The two are comparable — which is exactly how a joint should be designed. A joint far weaker than the parts it joins is a wasted structure; a joint far stronger is wasted metal.

4. (c) How many rivets for 20 kN. Each rivet may carry $$F_{\text{allowed}} = 1.5 \times 10^{8} \times 5.027 \times 10^{-5} = 7.54 \times 10^{3} \text{ N}$$ $$n = \frac{20 \times 10^{3}}{7.54 \times 10^{3}} = 2.65$$

5. Round up, never down. You cannot fit two-thirds of a rivet, and rounding down would overload the joint. Three rivets are needed, and they will then run at $\frac{20\,000}{3 \times 5.027 \times 10^{-5}} = 1.33 \times 10^{8}$ Pa, safely under the limit.

Final Answer: (a) $5.97 \times 10^{7}$ Pa; (b) $5.0 \times 10^{7}$ Pa in the plate, comparable to the rivet's; (c) three rivets.

Takeaway: The plate is in tension, the rivet is in shear, and the two use different areas. The plate's area is its section perpendicular to the pull; the rivet's is its own circular section lying in the sliding plane. Confusing them is the classic error in every joint problem.