Mechanical Properties of Solids
Why a steel wire hardly stretches while a rubber band stretches a lot, how to put a number on "stiffness", and how engineers use it to size ropes, beams and bridges. We start from atoms and springs and work up to exam-level problems.
Elasticity, plasticity and rigid bodies
Stretch a rubber band and let go: it snaps back. Squeeze a ball of wet clay and let go: it stays squashed. Push hard on a steel table and nothing you can see happens. These three everyday results are the whole chapter in miniature.
Every solid changes its shape or size at least a little when you apply a force. The force that causes the change is the deforming force. As soon as the body deforms, it develops an internal force that tries to undo the change. This is the restoring force. In equilibrium, the restoring force is equal and opposite to the applied force.
Elasticity
The property of a body to regain its original size and shape when the deforming force is removed. Examples: a steel wire within its limit, a spring, a rubber band.
Plasticity
The property of a body not to regain its original shape after the force is removed, so it stays permanently deformed. Examples: putty, wet clay, chewing gum, a metal bent too far.
Why solids are elastic: the atom-and-spring model
In a solid, every atom sits at an equilibrium position fixed by the forces from its neighbours. You can picture each pair of neighbouring atoms as joined by a tiny spring. Pull the solid and the atoms move apart, so the springs (interatomic forces) pull them back. Push and the atoms come closer, so the springs push them apart. The restoring force of a solid is just the sum of these interatomic forces.
If you pull too hard, the atoms slide past their neighbours into new equilibrium positions. After you let go they stay in the new places. That is plastic deformation. The largest stress the body can take and still recover completely is called its elastic limit.
Atoms joined by springs
Pull, then press Release: the stretched springs pull the atoms back. Now choose Allow past the limit, drag F beyond the black mark, and release again. The lattice settles at a longer length. That left-over stretch is the permanent set.
Three ideal cases
| Ideal body | Meaning | Closest real example |
|---|---|---|
| Perfectly elastic | Regains its shape completely, however often it is loaded | Quartz fibre (very nearly) |
| Perfectly plastic | Shows no tendency at all to regain its shape | Putty, wet clay, mud |
| Perfectly rigid | No change in shape or size for any force. Then \(Y = G = B = \infty\) | None. Every real body deforms a little. |
"Which is more elastic, rubber or steel?" Steel. In physics, "more elastic" means it takes a larger stress to produce the same strain, so the elastic modulus is larger. Rubber stretches easily because its modulus is tiny. \(Y_\text{steel} \approx 2\times10^{11}\) Pa is about \(10^{4}\)–\(10^{5}\) times the \(Y\) of rubber.
- "Young's modulus of a perfectly rigid body is …" Answer: infinite.
- Assertion–reason on "steel is more elastic than rubber".
- "The restoring force in a solid is due to …" Answer: interatomic (intermolecular) forces.
A glass marble and a rubber ball of the same size are dropped from the same height on a hard floor. The glass marble bounces higher. A student concludes "glass is more elastic than rubber". Is the conclusion right, and what does "more elastic" really compare?
Show solution
- "More elastic" compares the elastic modulus: how much stress is needed per unit strain. \(Y_\text{glass} \approx 6.5\times10^{10}\) Pa, while rubber is only about \(10^{6}\)–\(10^{7}\) Pa.
- So the conclusion "glass is more elastic than rubber" is correct. Glass deforms far less for the same stress.
- The bounce is only a rough guide, because it also depends on energy lost inside the material. The reliable test is always the modulus.
Yes: glass (and steel) are more elastic than rubber because their modulus is much larger.
A body is "perfectly rigid". What are its Young's modulus, shear modulus and bulk modulus? Why does no real body behave like this?
Show solution
- Perfectly rigid means zero strain for any stress.
- Each modulus = stress ÷ strain = (finite) ÷ 0 → \(\infty\). So \(Y = G = B = \infty\).
- Real atoms are held by finite "springs" (interatomic forces), so any force moves them a little. A perfectly rigid body is only an idealisation.
All three moduli are infinite; this is an ideal limit.
Try it
Stress and strain
Hang the same 10 kg from a thin wire and from a thick rod of the same metal. The thin wire stretches much more, because the same force is shared by fewer atoms. So force alone does not tell us how hard the material is being "worked". We need force per unit area. In the same way, a 1 mm stretch means a lot for a 10 cm wire and very little for a 10 m cable. So we compare the fractional change.
Stress is the restoring force per unit area that develops inside a deformed body. In equilibrium it equals the applied force per unit area: \(\text{stress} = F/A\).
Strain is the fractional change in a dimension: change ÷ original value. It is a ratio of two like quantities, so it has no unit and no dimensions.
Symbols and assumptions: σ = tensile or compressive stress (Pa); F = normal force (N); A = loaded cross-sectional area (m²).
Symbols and assumptions: Change in size and original size must use the same units. For a wire, longitudinal strain is ΔL/L, with L its original length and ΔL its extension.
Three kinds of stress, three kinds of strain
| Stress | How the force acts | What changes | Strain |
|---|---|---|---|
| Longitudinal: tensile (pull) or compressive (push) | Normal (perpendicular) to the cross-section, along the length | Length | Longitudinal strain \(= \Delta L/L\) |
| Shearing (tangential) | Parallel to the face, with the opposite face held | Shape (volume stays the same) | Shear strain \(= \Delta x/L = \tan\theta \approx \theta\) |
| Hydraulic (volume) | Equal normal force on every face, e.g. a body inside a fluid. Equals the pressure \(p\). | Volume (shape stays the same) | Volume strain \(= \Delta V/V\) |
One block, four kinds of stress
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Stress and pressure share the unit (Pa) and dimensions, but they are different ideas. Pressure is an external normal force per unit area pressing on a surface, and in a fluid at rest it acts equally in all directions. Stress is the internal restoring force per unit area. It can be normal (tensile or compressive) or tangential (shear), and its value depends on the direction of the surface you consider. Only hydraulic stress equals the pressure of the surrounding fluid.
Which area? For tensile/compressive stress, use the cross-section perpendicular to the force. For shear stress, use the face parallel to the force (the face the force slides). In shear strain \(\Delta x/L\), \(L\) is the distance between the fixed face and the moving face.
A wire 3 m long and 1 mm in diameter carries a load of 5 kg and stretches by 1.2 mm. Find the stress and the strain. \((g = 9.8\ \text{m s}^{-2})\)
Show solution
- \(F = 5\times9.8 = 49\ \text{N}\). Radius \(r = 0.5\ \text{mm} = 5\times10^{-4}\ \text{m}\), so \(A = \pi r^2 = 7.85\times10^{-7}\ \text{m}^2\).
- Stress \(= F/A = 49/(7.85\times10^{-7}) = 6.24\times10^{7}\ \text{N m}^{-2}\).
- Strain \(= \Delta L/L = 1.2\times10^{-3}/3 = 4.0\times10^{-4}\) (no unit).
- Bonus: \(Y = \text{stress}/\text{strain} = 1.56\times10^{11}\) Pa.
Stress ≈ 6.2 × 10⁷ N m⁻²; strain = 4 × 10⁻⁴.
Identify the stress in each case. (a) Legs of a table holding a heavy load. (b) A book pushed sideways along its top cover while the bottom cover is held. (c) A steel ball lying at the bottom of a deep lake. (d) The cable of a lift.
Show solution
- (a) Compressive (longitudinal): the legs are squeezed along their length.
- (b) Shearing: the force is parallel to the cover, the pages slide over each other and the shape changes.
- (c) Hydraulic: the water pushes equally on every part of the surface, so the volume decreases slightly and the shape stays the same.
- (d) Tensile: the cable is pulled at both ends.
Compressive · shear · hydraulic · tensile.
Try it
Hooke's law and the stress–strain curve
Hang 1 kg on a spring and it stretches 2 cm; hang 2 kg and it stretches 4 cm. For small deformations, the response is simply proportional. Robert Hooke stated this in 1676, and it holds for a wire or a rod as well as for a spring.
Symbols and assumptions: k here means an elastic modulus (Pa), not a spring constant (N/m). Stress and strain must be the corresponding pair, such as longitudinal stress and longitudinal strain.
What happens if we keep pulling?
Clamp a metal wire in a testing machine, increase the load slowly, and plot stress against strain. A ductile metal such as copper or mild steel gives the curve below. Learn the letters. NCERT and coaching tests use them directly.
- O → A, proportional region. A straight line; Hooke's law holds. Its slope is Young's modulus \(Y\). A is the proportional limit.
- A → B, elastic but not linear. Stress is no longer proportional to strain, yet the body still returns fully to O when unloaded.
- B, yield point (elastic limit). The stress here is the yield strength \(\sigma_y\). Beyond B, the material yields: strain grows quickly for little extra stress.
- B → D, plastic region. Unload at any point C here, and the wire comes back along a line parallel to OA. It does not reach O, so a strain is left: the permanent set.
- D, ultimate tensile strength \(\sigma_u\). The maximum stress the material can bear.
- D → E, necking. A neck forms; the wire stretches further even if the load is reduced, and breaks at the fracture point E.
Stress–strain curve explorer
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Shapes are schematic, not to scale. In a real metal, A and B occur at strains of only about 0.1–0.5%. Try this: stop inside O–B and press Unload (it goes back to O). Then stop beyond B and press Unload (it is left with a permanent set).
Ductile, brittle and elastomers
Ductile
Long plastic region: the ultimate point D and fracture point E are far apart. These materials can be drawn into wires or hammered into sheets. Examples: copper, aluminium, gold, mild steel.
Brittle
D and E are close together: the material breaks soon after the elastic limit, with almost no plastic flow. Examples: glass, cast iron, ceramics.
Elastomers
Huge elastic region (strain can be several hundred %), but the curve is not straight, so Hooke's law is not obeyed. There is no well-defined plastic region. Examples: rubber, the elastic tissue of the aorta (the large artery from the heart).
- Steeper initial slope → larger Young's modulus (more elastic, stiffer).
- Higher peak (D) → greater ultimate strength (stronger).
- Longer curve after B → more ductile. Breaks right after the straight part → brittle.
- On a strain–stress graph (strain on the y-axis), the slope is \(1/Y\), so the flatter line has the larger \(Y\).
- Proportional limit (A) ≠ elastic limit (B). Between them the body is elastic but non-linear.
- "Rubber obeys Hooke's law over a large range": false. Rubber is elastic over a large range, but not linear.
- Unloading from the plastic region is a straight line parallel to OA, not back along the curve.
From the graph above, find (a) Young's modulus and (b) the approximate yield strength of the material.
Show solution
- Pick a point on the straight part: stress \(=150\times10^{6}\ \text{N m}^{-2}\) at strain \(=0.002\).
- \(Y = \text{slope} = \dfrac{150\times10^{6}}{0.002} = 7.5\times10^{10}\ \text{N m}^{-2}\).
- The graph stops being straight and flattens at about \(300\times10^{6}\) N m⁻², so the yield strength \(\sigma_y \approx 3\times10^{8}\ \text{N m}^{-2}\).
(a) Y = 7.5 × 10¹⁰ N m⁻² (b) σy ≈ 3 × 10⁸ N m⁻²
Statement I: If a wire is loaded beyond its yield point and then unloaded, it returns along a line parallel to the initial straight portion. Statement II: Ultimate tensile strength is the stress at which the wire actually breaks. Which are correct?
Show solution
- Statement I is true: unloading from the plastic region is elastic recovery only, so it has the same slope \(Y\) as OA and leaves a permanent set.
- Statement II is false: ultimate tensile strength is the maximum stress (point D). The wire breaks later, at the fracture point E. For a ductile wire, the stress at E is lower than at D.
Statement I true, Statement II false.
Try it
Young's modulus
The modulus in Hooke's law gets a different name for each type of stress. For stretching or compressing along the length, it is Young's modulus \(Y\). It answers one question: how much stress does it take to produce a given fractional stretch? A big \(Y\) means a stiff material.
Symbols and assumptions: Y = Young’s modulus (Pa); F = axial force (N); L = original length (m); ΔL = extension (m); A = cross-sectional area (m²); r = wire radius (m); d = diameter = 2r; M = hanging mass (kg); g = gravitational acceleration (m/s²). The Mg form assumes a static hanging load and negligible wire weight.
- \(Y\) depends only on the material (and temperature), never on \(L\), \(A\) or the load. Cut a wire in half or double it, and \(Y\) is unchanged. What changes is the extension.
- \(\Delta L \propto L/d^{2}\): square the diameter! Doubling \(d\) cuts the extension to ¼.
- Diameter given? Halve it before using \(\pi r^{2}\). Convert mm² to m² (\(1\ \text{mm}^2 = 10^{-6}\ \text{m}^2\)).
- For most metals \(Y\) is about the same in tension and in compression.
Values you should recognise
| Material | Young's \(Y\) | Shear \(G\) | Bulk \(B\) |
|---|---|---|---|
| Steel | 200 | 84 | 160 |
| Iron | 190 | 70 | 100 |
| Copper | 110–120 | 42 | 140 |
| Brass | 91 | 36 | 61 |
| Aluminium | 70 | 25 | 72 |
| Glass | 65 | 23 | 37 |
| Wood | 13 | 10 | – |
| Bone | 9.4 | – | – |
| Water | – | 0 | 2.2 |
| Mercury | – | 0 | 25 |
| Air (at STP) | – | 0 | 1.0 × 10⁻⁴ |
All values in units of 10⁹ N m⁻² (GPa), e.g. steel \(Y = 2.0\times10^{11}\) Pa. Liquids and gases have no \(Y\) or \(G\); they have only a bulk modulus. Values from the NCERT tables (copper is listed as 1.1–1.2 × 10¹¹ Pa in different books).
Metals have large moduli (stiff). For most solids, \(G \approx Y/3\). Water is about 70× more compressible than steel, and air is about 20 000× more compressible than water.
Wire lab: stretch a real wire
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Set d = 0.50 mm and note ΔL. Now set d = 1.00 mm: ΔL drops to exactly ¼. Doubling L doubles ΔL. Changing the material changes ΔL as 1/Y. If the stress passes the yield strength, the wire turns red.
Measuring Y in the lab: Searle's method
- Two identical long wires, A and B, hang from a rigid support. A (reference) carries a millimetre main scale and a fixed weight. B (experimental) carries a vernier and a pan.
- Put a small initial load on B to straighten out kinks, and note the vernier reading.
- Add weights \(M\) one by one and note the extension \(\Delta l\) each time. Then remove them and check the readings come back (to stay within the elastic limit).
- Plot stress \((Mg/\pi r^{2})\) against strain \((\Delta l/L)\). The slope of the straight line is \(Y\); equivalently \(Y = \dfrac{Mg\,L}{\pi r^{2}\,\Delta l}\).
Why two wires? The reference wire cancels errors from the support yielding and from thermal expansion. Measure \(r\) at several points with a screw gauge, because \(r^{2}\) carries the biggest error.
A structural steel rod has radius 10 mm and length 1.0 m. A 100 kN force stretches it along its length. Find (a) the stress, (b) the elongation and (c) the strain. \(Y_\text{steel} = 2.0\times10^{11}\ \text{N m}^{-2}\).
Show solution
- \(A = \pi r^2 = \pi(10^{-2})^2 = 3.14\times10^{-4}\ \text{m}^2\).
- (a) Stress \(= \dfrac{F}{A} = \dfrac{10^{5}}{3.14\times10^{-4}} = 3.18\times10^{8}\ \text{N m}^{-2}\).
- (b) \(\Delta L = \dfrac{\text{stress}\times L}{Y} = \dfrac{3.18\times10^{8}\times1}{2\times10^{11}} = 1.59\times10^{-3}\ \text{m} = 1.59\ \text{mm}\).
- (c) Strain \(= \Delta L/L = 1.59\times10^{-3} = 0.16\%\).
3.18 × 10⁸ N m⁻²; 1.59 mm; 0.16%
A steel wire and a copper wire of the same length and the same diameter carry the same load. (a) Find the ratio of their extensions. (b) What diameter must the copper wire have, relative to the steel wire, so that both stretch equally? Take \(Y_s = 2.0\times10^{11}\) and \(Y_c = 1.1\times10^{11}\ \text{N m}^{-2}\).
Show solution
- (a) \(\Delta L = \dfrac{FL}{AY}\). With \(F, L, A\) equal, \(\Delta L \propto 1/Y\): \(\dfrac{\Delta L_s}{\Delta L_c} = \dfrac{Y_c}{Y_s} = \dfrac{1.1}{2.0} = 0.55\). The copper wire stretches about 1.8 times as much.
- (b) Equal extension needs \(A_cY_c = A_sY_s\), so \(\dfrac{d_c^{2}}{d_s^{2}} = \dfrac{Y_s}{Y_c} = 1.82\) and \(\dfrac{d_c}{d_s} = \sqrt{1.82} = 1.35\).
(a) ΔL_steel : ΔL_copper = 11 : 20 (b) copper wire ≈ 1.35 × thicker
A copper wire 2.2 m long and a steel wire 1.6 m long, both of diameter 3.0 mm, are joined end to end. When stretched by a load, the net elongation is 0.70 mm. Find the load. \((Y_c = 1.1\times10^{11},\ Y_s = 2.0\times10^{11}\ \text{N m}^{-2})\)
Show solution
- In series, both wires carry the same tension \(F\). Total \(\Delta L = \dfrac{F}{A}\left(\dfrac{L_c}{Y_c} + \dfrac{L_s}{Y_s}\right)\).
- \(A = \pi(1.5\times10^{-3})^2 = 7.07\times10^{-6}\ \text{m}^2\).
- \(\dfrac{L_c}{Y_c} + \dfrac{L_s}{Y_s} = \dfrac{2.2}{1.1\times10^{11}} + \dfrac{1.6}{2.0\times10^{11}} = 2.0\times10^{-11} + 0.8\times10^{-11} = 2.8\times10^{-11}\).
- \(F = \dfrac{\Delta L\cdot A}{2.8\times10^{-11}} = \dfrac{7\times10^{-4}\times7.07\times10^{-6}}{2.8\times10^{-11}} \approx 1.8\times10^{2}\ \text{N}\).
Load ≈ 180 N
A steel wire (length 1.5 m) hangs from the ceiling and holds a 4 kg mass. A brass wire (length 1.0 m) hangs from this mass and holds a 6 kg mass. Both wires have diameter 0.25 cm. Find the elongation of each. \((Y_s = 2.0\times10^{11},\ Y_b = 0.91\times10^{11}\ \text{N m}^{-2},\ g = 9.8)\)
Show solution
- Tension: the steel wire holds both masses, so \(F_s = (4+6)\times9.8 = 98\ \text{N}\). The brass wire holds only the lower one, so \(F_b = 6\times9.8 = 58.8\ \text{N}\).
- \(A = \pi(1.25\times10^{-3})^2 = 4.91\times10^{-6}\ \text{m}^2\).
- \(\Delta L_s = \dfrac{98\times1.5}{4.91\times10^{-6}\times2.0\times10^{11}} = 1.5\times10^{-4}\ \text{m}\).
- \(\Delta L_b = \dfrac{58.8\times1.0}{4.91\times10^{-6}\times0.91\times10^{11}} = 1.3\times10^{-4}\ \text{m}\).
Steel: 1.5 × 10⁻⁴ m; brass: 1.3 × 10⁻⁴ m
In a human pyramid, the total mass of the performers and props is 280 kg. All of it is supported by the legs of one performer (mass 60 kg) lying on his back. Each thighbone (femur) is 50 cm long with an effective radius of 2.0 cm. How much is each femur compressed? \((Y_\text{bone} = 9.4\times10^{9}\ \text{N m}^{-2})\)
Show solution
- Extra load on his legs = \(280-60 = 220\) kg (his own weight is not on his thighs). Each femur carries 110 kg: \(F = 110\times9.8 = 1078\ \text{N}\).
- \(A = \pi(0.02)^2 = 1.26\times10^{-3}\ \text{m}^2\).
- \(\Delta L = \dfrac{FL}{AY} = \dfrac{1078\times0.5}{1.26\times10^{-3}\times9.4\times10^{9}} = 4.6\times10^{-5}\ \text{m}\).
About 4.6 × 10⁻⁵ m (0.046 mm), tiny compared to 50 cm.
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Combined wires and a wire's own weight
In this section: F = axial load (N); ΔL = extension (m); Y = Young’s modulus (Pa); A = wire area (m²); L = original length (m); k = wire stiffness (N/m). The sliders use N/mm: 1 N/mm = 1000 N/m, so dividing N by N/mm gives extension in mm. Subscripts 1 and 2 identify the two wires.
A wire acts like a spring while it remains in its linear elastic range: \(F=k\Delta L\), with \(k=YA/L\). Young's modulus belongs to the material; stiffness belongs to the whole wire and also depends on its length and area. Two wires of the same metal need not have the same stiffness.
Draw how the wires are connected before choosing a formula. The useful question is whether they have the same tension or the same extension.
Series: one below the other
Neglecting wire weight, the same tension F acts in each segment. Extensions add: \(\Delta L=F/k_1+F/k_2\). Hence \(1/k_\text{eq}=1/k_1+1/k_2\). The softer segment stretches more.
Parallel: side by side
If a supported bar stays horizontal and both initially taut wires have equal extension, forces add: \(F=(k_1+k_2)\Delta L\). Thus \(k_\text{eq}=k_1+k_2\), and the force ratio is \(F_1/F_2=k_1/k_2\). A freely tilting bar may not give equal extensions.
Same wires, two connections
The drawing is schematic and extension is exaggerated. In parallel the bar is constrained to remain horizontal. The model neglects wire weight and assumes elastic behaviour throughout.
Equivalent Young's modulus of two joined wires
For two wires in series with the same area A and lengths L₁ and L₂, compare their total extension with one wire of length L₁ + L₂ and an effective modulus Y_eff. The force F is the same through both pieces:
\[\frac{F(L_1+L_2)}{AY_\text{eff}}=\frac{FL_1}{AY_1}+\frac{FL_2}{AY_2},\qquad Y_\text{eff}=\frac{L_1+L_2}{L_1/Y_1+L_2/Y_2}.\]
Y₁ and Y₂ are the individual moduli in Pa; L₁ and L₂ are lengths in m. For equal lengths, \(Y_\text{eff}=2Y_1Y_2/(Y_1+Y_2)\), the harmonic mean. The result is not the arithmetic mean: the softer material contributes more extension. If their areas differ, calculate each stiffness YA/L instead of using this same-area shortcut.
Why a hanging wire needs the factor ½
For a uniform hanging wire of mass M and length L, a point x above its free lower end supports only the wire below it. That mass is Mx/L, so \(T(x)=Mgx/L\): zero at the bottom and Mg at the top. The tension is not constant.
An element dx extends by \(d(\Delta L)=T(x)dx/(AY)\). Adding the extensions gives
Symbols and assumptions: M = wire mass (kg); m = additional end mass (kg); ρ = wire density (kg/m³); L = original length (m); x = distance above the free lower end (m); A = wire area (m²); Y = Young’s modulus (Pa); g = gravitational acceleration (m/s²); ΔL = extension (m).
The average tension determines total extension, but the maximum stress occurs at the top. For a wire hanging only under its own weight, top stress is ρgL. The ideal maximum length before reaching a specified breaking stress is \(L_\text{max}=\sigma_\text{break}/(\rho g)\), independent of area. A safe working limit is lower than the breaking limit.
Two wires have stiffnesses 200 N/mm and 100 N/mm. Compare their total extension under 200 N when connected in series and in parallel with equal extensions.
Show solution
- Series: \(\Delta L=200/200+200/100=3\) mm; each carries 200 N.
- Parallel: \(k=300\) N/mm, so \(\Delta L=200/300=0.667\) mm.
- Parallel loads: 133.3 N and 66.7 N. Their sum is 200 N; the stiffer wire carries twice as much.
3 mm in series; 0.667 mm in parallel
Check your understanding
Shear modulus (modulus of rigidity)
Put a thick book on the table and push its top cover sideways while the bottom stays put. The pages slide a little over one another and the book leans: its shape changes, but its volume does not. That is shear. The stiffness against this sliding is the shear modulus \(G\), also called the modulus of rigidity (some books use \(\eta\)).
Symbols and assumptions: G = shear modulus (Pa); F = tangential force (N); Δx = sideways displacement (m); A = loaded face area (m²); L = separation of the faces (m); θ = small shear angle (radians); Y = Young’s modulus (Pa).
Shearing a 10 cm cube
Rubber (G ≈ 6 × 10⁵ Pa, a typical value) leans visibly. For metals the real angle is only about 10⁻⁵ rad, so it is drawn 10 000× larger. Compare lead, aluminium and steel: θ ∝ 1/G.
- Fluids have \(G = 0\): they cannot resist a steady shear and just flow. So liquids and gases have only a bulk modulus.
- In \(\Delta x = L\theta\), \(L\) is the height between the two faces, not the side of the face that is pushed.
- The stress area is the face parallel to the force (top face), not the side face.
A square lead slab of side 50 cm and thickness 10 cm is subjected to a shearing force of \(9.0\times10^{4}\) N on its narrow face. The lower edge is riveted to the floor. How much does the upper edge move? \((G_\text{lead} = 5.6\times10^{9}\ \text{N m}^{-2})\)
Show solution
- The force acts along the narrow face: \(A = 0.50\times0.10 = 0.05\ \text{m}^2\).
- Shear stress \(= 9.0\times10^{4}/0.05 = 1.8\times10^{6}\ \text{N m}^{-2}\).
- The distance between the riveted edge and the pushed edge is \(L = 0.50\) m. \(\Delta x = \dfrac{\text{stress}\times L}{G} = \dfrac{1.8\times10^{6}\times0.5}{5.6\times10^{9}} = 1.6\times10^{-4}\ \text{m}\).
Δx ≈ 0.16 mm
One face of an aluminium cube of edge 10 cm is fixed to a vertical wall. A 100 kg mass hangs from the opposite face. Find the vertical deflection of that face. \((G_\text{Al} = 25\ \text{GPa},\ g = 9.8)\)
Show solution
- The weight \(F = 980\) N acts along (parallel to) the free face: \(A = 0.1\times0.1 = 0.01\ \text{m}^2\).
- \(\Delta x = \dfrac{FL}{AG} = \dfrac{980\times0.1}{0.01\times25\times10^{9}} = 3.9\times10^{-7}\ \text{m}\).
About 3.9 × 10⁻⁷ m (0.39 µm)
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Bulk modulus and compressibility
Take a steel ball deep into the sea. The water presses on it equally from every side, so it keeps its shape but becomes very slightly smaller. How strongly a material resists this all-round squeeze is measured by its bulk modulus \(B\). Its reciprocal, the compressibility \(k\), tells you how easily it is squeezed.
Symbols and assumptions: B = bulk modulus (Pa); p or Δp = added pressure (Pa); V = initial volume (m³); ΔV = signed volume change (m³), negative on compression. The finite-change expression assumes small compression and approximately constant B.
Symbols and assumptions: k = compressibility (Pa⁻¹), distinct from a spring constant; B = bulk modulus (Pa); Δp = pressure increase (Pa); V = initial volume (m³); ΔV = signed volume change (m³).
Symbols and assumptions: r = original sphere radius (m); Δr = signed radius change (m), negative on compression; p = added pressure (Pa); B = bulk modulus (Pa); V and ΔV = initial volume and its signed change. For a cube, a is its original side and Δa its change. The positive fractional decrease is p/(3B).
Squeeze it from all sides
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300 atm is roughly the pressure at the average depth of the Indian Ocean (3 km); 1100 atm is the bottom of the Mariana Trench (about 11 km). The bars on the right compare all four materials at the same pressure on a log scale.
- \(\Delta r/r = p/3B\), not \(p/B\). Fractional changes multiply by the power: volume ∝ \(r^{3}\).
- For a gas, the isothermal bulk modulus equals its pressure \(p\) (adiabatic: \(\gamma p\)). This is why air at 1 atm has \(B \approx 10^{5}\) Pa.
- A perfectly rigid body has \(B = \infty\) and compressibility 0.
The average depth of the Indian Ocean is about 3000 m. Find the fractional compression \(\Delta V/V\) of water at the bottom. \((B_\text{water} = 2.2\times10^{9}\ \text{N m}^{-2},\ \rho = 10^{3}\ \text{kg m}^{-3},\ g = 10\ \text{m s}^{-2})\)
Show solution
- Pressure at the bottom (above atmospheric): \(p = h\rho g = 3000\times10^{3}\times10 = 3\times10^{7}\ \text{N m}^{-2}\).
- \(\dfrac{\Delta V}{V} = \dfrac{p}{B} = \dfrac{3\times10^{7}}{2.2\times10^{9}} = 1.36\times10^{-2}\).
ΔV/V ≈ 1.36 × 10⁻², i.e. about 1.4%
A steel ball of volume 0.32 m³ is dropped into the Mariana Trench, where the water pressure is \(1.1\times10^{8}\) Pa. Find the change in its volume. \((B_\text{steel} = 1.6\times10^{11}\ \text{Pa})\)
Show solution
- \(\Delta V = \dfrac{pV}{B} = \dfrac{1.1\times10^{8}\times0.32}{1.6\times10^{11}} = 2.2\times10^{-4}\ \text{m}^3\) (a decrease).
ΔV ≈ 2.2 × 10⁻⁴ m³ decrease
100.0 litres of water under a pressure increase of 100.0 atm shrinks to 99.5 litres. Find \(B\) for water and compare it with air (\(B \approx 1.0\times10^{5}\) Pa). Take 1 atm \(= 1.013\times10^{5}\) Pa.
Show solution
- \(\Delta V/V = 0.5/100 = 5\times10^{-3}\); \(\Delta p = 100\times1.013\times10^{5} = 1.013\times10^{7}\) Pa.
- \(B = \dfrac{1.013\times10^{7}}{5\times10^{-3}} = 2.0\times10^{9}\ \text{Pa}\).
- \(B_\text{water}/B_\text{air} = 2.0\times10^{9}/1.0\times10^{5} = 2\times10^{4}\). Air is about 20 000 times more compressible.
B ≈ 2.0 × 10⁹ Pa, about 2 × 10⁴ times that of air
Try it
Poisson's ratio
Poisson's ratio appears in section 8.5.4 of the 2026–27 NCERT chapter. Learn its definition and distinguish signed lateral strain from the positive magnitude of contraction. Most familiar materials have a ratio between 0 and 0.5; the volume-change relation below assumes small strain.
Stretch a rubber band and watch it: it gets thinner as it gets longer. Most conventional materials do this. Materials with negative Poisson's ratio, called auxetic materials, can instead become wider when stretched. The longitudinal strain (along the force) comes with a lateral strain (across it) of opposite sign. Their ratio, within the elastic limit, is a constant of the material.
Symbols and assumptions: σ (or ν) = Poisson’s ratio; d = original diameter; Δd = signed diameter change; L = original length; ΔL = signed length change. The strains use matching length units. A contracting diameter has Δd < 0 while a stretched length has ΔL > 0.
Symbols and assumptions: V = original volume; ΔV = signed volume change; L = original length; ΔL = extension; σ = Poisson’s ratio. Both sides are dimensionless. This is the first-order small-strain result.
Longer and thinner
Try the rubber preset: the volume hardly changes (σ ≈ 0.5). With cork the rod gets longer without getting thinner, so its volume grows by the full strain. (Cork is why wine bottles can be corked easily.)
For an isotropic material \(Y = 2G(1+\sigma) = 3B(1-2\sigma)\), and eliminating \(\sigma\) gives \(\dfrac{9}{Y} = \dfrac{1}{B} + \dfrac{3}{G}\). Two quick uses: for the usual positive-ratio range \(0 \le \sigma \le 0.5\), \(G\) lies between \(Y/3\) and \(Y/2\) (hence "G ≈ Y/3"); and \(B \to \infty\) (incompressible) needs \(\sigma \to 0.5\).
A wire of diameter 2 mm and length 1 m is stretched by 1 mm. If \(\sigma = 0.3\), find the decrease in its diameter and the fractional change in volume.
Show solution
- Longitudinal strain \(= 10^{-3}/1 = 10^{-3}\).
- Lateral strain \(= \sigma\times10^{-3} = 3\times10^{-4}\), so \(\Delta d = 3\times10^{-4}\times2\ \text{mm} = 6\times10^{-4}\ \text{mm} = 0.6\ \mu\text{m}\).
- \(\Delta V/V = (1-2\times0.3)\times10^{-3} = 4\times10^{-4}\) (an increase).
Δd = 0.6 µm decrease; ΔV/V = +4 × 10⁻⁴
For a material, \(Y = 2.4\,G\). Find Poisson's ratio, and express \(B\) in terms of \(Y\).
Show solution
- \(Y = 2G(1+\sigma) \Rightarrow 2.4G = 2G(1+\sigma) \Rightarrow \sigma = 0.2\).
- \(Y = 3B(1-2\sigma) = 3B(0.6) = 1.8B \Rightarrow B = Y/1.8 \approx 0.56\,Y\).
σ = 0.2; B = Y/1.8
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Elastic potential energy of a stretched wire
This topic appears in section 8.5.5 of the 2026–27 NCERT chapter. It is the same idea as the energy of a spring, \(\tfrac12 kx^{2}\).
A stretched catapult rubber can launch a stone, so stretching stores energy. While a wire is being stretched, the restoring force grows from 0 up to its final value \(F\), in direct proportion to the extension (Hooke's law). The work done against it is the area under the force–extension line, which is a triangle.
Symbols and assumptions: U = stored elastic energy (J); F = final axial force (N); Δl = extension (m); Y = Young’s modulus (Pa); A = cross-sectional area (m²); L = original length (m); k = wire stiffness (N/m); x = extension in the spring notation. Loading is quasistatic and linear elastic.
Symbols and assumptions: u = elastic energy density (J/m³); U = total stored energy (J); Y = Young’s modulus (Pa); AL = initial wire volume (m³). Use dimensionless strain, not its percentage value.
Energy = area under the F–Δl line
Wire: L = 1 m, A = 1 mm², so k = YA/L. Doubling Δl makes U four times larger. Switch to vs work by gravity: a hung load does work Mg·Δl (the rectangle), but only half of it (the triangle) is stored.
A mass \(M\) hung on a wire stretches it by \(l\). Gravity does work \(Mgl\), but the wire stores only \(\tfrac12 Mgl\). If the mass is released suddenly, the extra half becomes kinetic energy. The mass overshoots (the maximum extension is \(2l\)), oscillates, and the energy is finally lost as heat. If you lower it slowly by hand, your hand absorbs that half.
A wire of length 2 m and cross-section 2 mm² \((Y = 2\times10^{11}\ \text{Pa})\) is stretched by 1 mm. Find the work done and the energy stored per unit volume.
Show solution
- \(k = \dfrac{YA}{L} = \dfrac{2\times10^{11}\times2\times10^{-6}}{2} = 2\times10^{5}\ \text{N m}^{-1}\).
- \(W = \tfrac12 k(\Delta l)^{2} = \tfrac12\times2\times10^{5}\times(10^{-3})^{2} = 0.1\ \text{J}\).
- Volume \(= AL = 4\times10^{-6}\ \text{m}^3\); \(u = 0.1/(4\times10^{-6}) = 2.5\times10^{4}\ \text{J m}^{-3}\).
- Check: strain \(= 5\times10^{-4}\), so \(\tfrac12 Y\varepsilon^{2} = \tfrac12\times2\times10^{11}\times2.5\times10^{-7} = 2.5\times10^{4}\) J m⁻³. ✓
W = 0.1 J; u = 2.5 × 10⁴ J m⁻³
Two wires of the same material, A and B, have lengths in the ratio 1 : 2 and radii in the ratio 2 : 1. They are stretched by the same force. Find the ratio of the elastic energies stored.
Show solution
- \(U = \tfrac12 F\Delta l\) and \(\Delta l = \dfrac{FL}{AY}\), so for the same \(F\): \(U = \dfrac{F^{2}L}{2AY} \propto \dfrac{L}{r^{2}}\).
- \(\dfrac{U_A}{U_B} = \dfrac{1}{2}\times\left(\dfrac{1}{2}\right)^{2} = \dfrac18\).
U_A : U_B = 1 : 8
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Thermal stress
Railway tracks have small gaps between rails, and bridges have expansion joints. Why? A rod that is heated wants to expand by \(\Delta L = L\alpha\Delta T\). If rigid supports stop it, the supports must squeeze it back by exactly that amount. The rod is then under a compressive stress, the thermal stress, without any visible change in length. If it is cooled instead, the stress is tensile.
Symbols and assumptions: L = original length (m); α = linear expansion coefficient (K⁻¹); ΔT = temperature increase (K or °C difference); Y = Young’s modulus (Pa); A = cross-sectional area (m²); F = force magnitude (N). Heating produces compression; the displayed stress is its magnitude. Supports are rigid and expansion is fully prevented.
- A longer rod expands more but also has more length to share the compression, so the force is the same. Do not put \(L\) into the answer.
- Doubling the area doubles the force but leaves the stress unchanged.
- Heating gives compressive stress; cooling a clamped rod gives tensile stress, which is how rails can crack in severe winters.
A brass rod of cross-section 1 cm² is clamped between rigid walls at 20 °C and heated to 120 °C. Find the thermal stress and the force on each wall. \((Y = 0.91\times10^{11}\ \text{Pa},\ \alpha = 1.8\times10^{-5}\ ^\circ\text{C}^{-1})\)
Show solution
- \(\Delta T = 100\ ^\circ\text{C}\); strain \(= \alpha\Delta T = 1.8\times10^{-3}\).
- Stress \(= Y\alpha\Delta T = 0.91\times10^{11}\times1.8\times10^{-3} = 1.64\times10^{8}\ \text{Pa}\).
- \(F = \text{stress}\times A = 1.64\times10^{8}\times10^{-4} = 1.64\times10^{4}\ \text{N}\). That is roughly the weight of a 1.7-tonne car, from just 100 °C of heating.
Stress ≈ 1.6 × 10⁸ Pa; F ≈ 1.6 × 10⁴ N
Try it
Applications: ropes, beams and mountains
1. How thick must a crane's rope be?
A crane must lift 10 tonnes \((10^{4}\ \text{kg})\). The steel rope must never be stressed beyond its yield strength, \(\sigma_y \approx 300\times10^{6}\ \text{N m}^{-2}\), or it would stretch permanently. So:
Symbols and assumptions: M = supported mass (kg); g = gravitational acceleration (m/s²); r = rope radius (m); σ_y = material yield stress (Pa). The numerical example uses M = 10⁴ kg and σ_y = 3 × 10⁸ Pa, before adding a safety factor.
2. Why beams are deep, and why girders are I-shaped
A beam of length \(l\), breadth \(b\) and depth \(d\), supported at its ends and loaded by \(W\) at the middle, sags at the centre by
Symbols and assumptions: δ = downward displacement at the middle (m); W = load force (N), not mass; l = distance between supports (m); b = breadth (m); d = vertical depth (m); Y = Young’s modulus (Pa). This is the small-deflection result for a rectangular beam supported at both ends and loaded centrally.
Beam bending lab
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Press Double the depth: δ falls to exactly 1/8. The I-section uses the same amount of metal as the b × d rectangle (same weight), arranged as two flanges and a thin web with twice the depth. It sags about 6.4 times less.
When a beam bends, its top layer is compressed and its bottom layer is stretched; the middle hardly does any work. An I-section removes metal from the middle and puts it in the top and bottom flanges. That gives a deep, stiff beam that is light and cheaper. The flanges also give a wide load-bearing surface. A very deep, narrow beam could buckle (bend sideways); the flanges help prevent that too.
3. How high can a mountain be?
The rock at the base of a mountain of height \(h\) carries the weight above it, so the stress there is about \(h\rho g\). If this exceeds the elastic limit of rock (about \(30\times10^{7}\ \text{N m}^{-2}\)), the base rock starts to flow and the mountain sinks. With \(\rho \approx 3\times10^{3}\ \text{kg m}^{-3}\):
Symbols and assumptions: h = height (m); ρ = rock density (kg/m³); g = gravitational acceleration (m/s²); σ_limit = limiting stress (Pa). The model treats the mountain as a uniform column; it is an order-of-magnitude estimate.
A steel cable of radius 1.5 cm supports a chairlift at a ski area. If the stress must not exceed \(10^{8}\ \text{N m}^{-2}\), what is the maximum load it can support?
Show solution
- \(A = \pi(1.5\times10^{-2})^2 = 7.07\times10^{-4}\ \text{m}^2\).
- \(F_\text{max} = \sigma A = 10^{8}\times7.07\times10^{-4} = 7.1\times10^{4}\ \text{N}\).
About 7.1 × 10⁴ N (≈ 7 tonnes-weight)
A wooden plank (b = 10 cm, d = 5 cm, l = 2 m) is laid flat across a stream and sags 12 mm under a person. The same plank is turned on its edge (so b = 5 cm, d = 10 cm). What is the new sag?
Show solution
- \(\delta \propto \dfrac{1}{b\,d^{3}}\). Flat: \(b d^{3} = 10\times5^{3} = 1250\). On edge: \(5\times10^{3} = 5000\).
- \(\delta_\text{new} = 12\times\dfrac{1250}{5000} = 3\ \text{mm}\).
3 mm: four times stiffer, from the same plank. (This is why floor joists stand on edge.)
Try it
Problem patterns and shortcuts
Almost every numerical in this chapter is one formula, \(\Delta L = \dfrac{FL}{AY}\), dressed up in a different situation. Learn to spot which quantity is held fixed, and the answer usually follows as a ratio in one line.
Pattern 1: Wire under its own weight
The bottom element carries nothing; the top element carries the whole weight. Tension grows linearly, so the average tension is \(Mg/2\):
\[ \Delta l = \frac{MgL}{2AY} = \frac{\rho g L^{2}}{2Y} \]
With an extra load \(m\) at the bottom: \(\Delta l = \dfrac{(m + M/2)gL}{AY}\). The wire breaks first at the top, and the longest wire that can hang is \(L_\text{max} = \sigma_\text{break}/\rho g\), independent of area.
Pattern 2: Same material, different sizes
\(\Delta L \propto \dfrac{F L}{r^{2}}\) (or \(FL/d^{2}\)). Write the ratio and cancel. Common case: "length doubled, radius doubled, same load" → \(\Delta L\) halves.
Pattern 3: Same volume (wire drawn or melted and recast)
\(A = V/L\), so \(\Delta L = \dfrac{F L^{2}}{VY} \propto L^{2} \propto \dfrac{1}{A^{2}}\). For equal extension, \(F \propto A^{2}\). So areas A and 3A need forces F and 9F. Stretching a wire to n times its length (same volume) makes a given load stretch it \(n^{2}\) times more.
Pattern 4: A wire is a spring, \(k = YA/L\)
Series (end to end, same force): \(\dfrac{1}{k} = \dfrac{1}{k_1}+\dfrac{1}{k_2}\), and the extensions add. For equal areas, \(Y_\text{eq} = \dfrac{L_1+L_2}{L_1/Y_1 + L_2/Y_2}\); for equal lengths as well, \(Y_\text{eq} = \dfrac{2Y_1Y_2}{Y_1+Y_2}\).
Parallel (side by side, same extension): \(k = k_1 + k_2\), and the load is shared in the ratio \(k_1 : k_2\). For equal lengths, \(Y_\text{eq} = \dfrac{Y_1A_1 + Y_2A_2}{A_1 + A_2}\).
A mass on a wire oscillates with \(T = 2\pi\sqrt{\dfrac{mL}{YA}}\).
Pattern 5: Find the tension first
- Atwood pulley (masses \(m_1, m_2\)): \(T = \dfrac{2m_1m_2 g}{m_1+m_2}\), not \(m_1 g\) or \(m_2 g\).
- Lift accelerating upward at \(a\): \(T = m(g+a)\); downward: \(m(g-a)\); free fall: 0, so no extension.
- Rod on a smooth floor pulled by \(F\) at one end: tension falls linearly to 0 at the free end, so \(\Delta l = \dfrac{FL}{2AY}\). Pulled by \(F_1\) and \(F_2\) at the two ends: \(\Delta l = \dfrac{(F_1+F_2)L}{2AY}\).
- Two masses hung one below the other: the upper wire carries both.
- Load released suddenly: maximum extension is \(2\times\) the static extension (energy argument).
Pattern 6: Graph slopes
Stress–strain slope \(= Y\). Strain–stress slope \(= 1/Y\). Load–extension slope \(= YA/L = k\). Extension–load slope \(= L/YA\). Always check which quantity is on which axis before deciding whether "steeper" means "stiffer".
Pattern 7: Strength, units and quick ratios
- Breaking load \(\propto r^{2}\), independent of length; breaking stress depends only on the material.
- Stress, pressure, every modulus and energy density all have unit Pa (J m⁻³ = N m⁻²) and dimensions \([\mathrm{ML^{-1}T^{-2}}]\). Strain and Poisson's ratio are dimensionless; compressibility is Pa⁻¹.
- Volume strain = 3 × linear strain (sphere or cube).
Masses of 3 kg and 5 kg hang from the two ends of a steel wire passing over a light frictionless pulley. The total length of the wire is 2 m and its cross-section is 1 mm². Find the extension of the wire. \((Y = 2\times10^{11}\ \text{Pa},\ g = 10)\)
Show solution
- The masses accelerate, so the tension is not 30 N or 50 N: \(T = \dfrac{2m_1m_2g}{m_1+m_2} = \dfrac{2\times3\times5\times10}{8} = 37.5\ \text{N}\).
- The same tension acts along the whole 2 m: \(\Delta l = \dfrac{TL}{AY} = \dfrac{37.5\times2}{10^{-6}\times2\times10^{11}} = 3.75\times10^{-4}\ \text{m}\).
0.375 mm
A steel wire and a copper wire are each 1 m long with cross-section 1 mm². \((Y_s = 2.0\times10^{11},\ Y_c = 1.1\times10^{11}\ \text{Pa})\) (a) They hang side by side and support a rigid bar carrying 310 N, with both wires stretching equally. Find the extension and the load on each. (b) They are instead joined end to end and carry 310 N. Find the total extension.
Show solution
- \(k_s = \dfrac{YA}{L} = 2.0\times10^{5}\ \text{N m}^{-1}\), \(k_c = 1.1\times10^{5}\ \text{N m}^{-1}\).
- (a) Parallel: \(k = 3.1\times10^{5}\), \(\Delta l = 310/3.1\times10^{5} = 1.0\ \text{mm}\). Loads: steel \(k_s\Delta l = 200\ \text{N}\), copper \(110\ \text{N}\). The stiffer wire takes more of the load.
- (b) Series: same 310 N in each. \(\Delta l = \dfrac{310}{2\times10^{5}} + \dfrac{310}{1.1\times10^{5}} = 1.55 + 2.82 = 4.37\ \text{mm}\).
(a) 1.0 mm; 200 N and 110 N (b) ≈ 4.4 mm
A uniform steel wire of mass 0.5 kg, length 2 m and cross-section 1 mm² hangs from the ceiling with a 2 kg mass at its lower end. Find its extension. \((Y = 2\times10^{11}\ \text{Pa},\ g = 9.8)\) Also find the longest steel wire that could hang without breaking under its own weight, if the breaking stress is \(4\times10^{8}\) Pa and \(\rho = 7800\ \text{kg m}^{-3}\) \((g = 10)\).
Show solution
- Effective load \(= m + M/2 = 2 + 0.25 = 2.25\ \text{kg}\).
- \(\Delta l = \dfrac{2.25\times9.8\times2}{10^{-6}\times2\times10^{11}} = 2.2\times10^{-4}\ \text{m}\).
- The top carries \(\rho A L g\), so the stress there is \(\rho L g\). Setting \(\rho L g = \sigma_b\): \(L_\text{max} = \dfrac{4\times10^{8}}{7800\times10} \approx 5.1\times10^{3}\ \text{m}\).
0.22 mm; about 5 km (independent of the wire's thickness)
A: A rod lying on a smooth horizontal floor and pulled by a force \(F\) at one end stretches by \(FL/2AY\), half of what it would if fixed at the other end. R: In the accelerating rod, the tension decreases linearly from \(F\) at the pulled end to zero at the free end.
Show solution
- The rod accelerates with \(a = F/m\). The part beyond a point at distance \(x\) from the free end has mass \(mx/L\), so the tension there is \(T = (mx/L)a = Fx/L\). This is linear, which makes R true.
- The average tension is \(F/2\), so \(\Delta l = \dfrac{(F/2)L}{AY}\). A is true.
- R directly explains A.
Both true; R is the correct explanation of A.
Night-before checklist
Read this the night before the test. If any line feels shaky, click back to its section.
- More elastic = larger modulus. Steel is more elastic than rubber. A perfectly rigid body has \(Y = G = B = \infty\).
- Stress \(= F/A\) is an internal restoring force per area. Strain is a ratio with no unit and no dimensions. Stress and all moduli: Pa, \([\mathrm{ML^{-1}T^{-2}}]\).
- Tensile/compressive → length; shear → shape (volume same); hydraulic → volume (shape same).
- Curve letters: A proportional limit · B yield point/elastic limit \((\sigma_y)\) · D ultimate strength \((\sigma_u)\) · E fracture. Unloading beyond B is parallel to OA and leaves a permanent set.
- D and E far apart → ductile; close → brittle. Rubber and aorta are elastomers: large elastic range, nonlinear over most of that range, no well-defined plastic region.
- \(Y\) depends only on the material (and temperature), not on \(L\), \(A\) or the load.
- \(\Delta L = FL/AY \propto L/d^{2}\). Square the diameter. Halve the diameter before using \(\pi r^{2}\); 1 mm² = \(10^{-6}\) m².
- Same volume: \(\Delta L \propto L^{2}\); equal extension needs \(F \propto A^{2}\) (A and 3A → F and 9F).
- Own weight: \(\Delta l = MgL/2AY = \rho g L^{2}/2Y\) (do not forget the ½).
- Breaking load \(\propto r^{2}\), independent of length; breaking stress is a material property.
- Wire = spring with \(k = YA/L\). Series: add \(1/k\); parallel: add \(k\).
- Energy stored \(= \tfrac12 F\Delta l\) (= \(Mgl/2\), not \(Mgl\)); per unit volume \(= \tfrac12\) stress × strain.
- Sphere under pressure: \(\Delta V/V = p/B\), \(\Delta r/r = p/3B\). Fluids have \(G = 0\); gases are the most compressible.
- Thermal stress \(= Y\alpha\Delta T\) and force \(= YA\alpha\Delta T\): no \(L\).
- Beam sag \(\delta = Wl^{3}/4bd^{3}Y\): doubling the depth makes it 1/8. That is why girders are deep I-sections. Max mountain height ≈ 10 km.
Formula sheet
| Quantity | Formula | Unit |
|---|---|---|
| Stress | \(F/A\) | Pa |
| Longitudinal / shear / volume strain | \(\Delta L/L\) · \(\Delta x/L = \theta\) · \(\Delta V/V\) | none |
| Young's modulus | \(Y = \dfrac{FL}{A\Delta L} = \dfrac{MgL}{\pi r^{2}\Delta L}\) | Pa |
| Shear modulus | \(G = \dfrac{F}{A\theta} = \dfrac{FL}{A\Delta x}\) | Pa |
| Bulk modulus | \(B = -\dfrac{V\Delta p}{\Delta V}\); \(k = 1/B\) | Pa; Pa⁻¹ |
| Poisson's ratio | \(\sigma = -\dfrac{\Delta d/d}{\Delta L/L}\); \(\Delta V/V = (1-2\sigma)\Delta L/L\) | none |
| Moduli relations (aside) | \(Y = 2G(1+\sigma) = 3B(1-2\sigma)\) | – |
| Elastic energy | \(U = \tfrac12 F\Delta l\); \(u = \tfrac12\,\text{stress}\times\text{strain} = \dfrac{\text{stress}^{2}}{2Y}\) | J; J m⁻³ |
| Wire as spring | \(k = YA/L\) | N m⁻¹ |
| Own weight | \(\Delta l = \dfrac{MgL}{2AY} = \dfrac{\rho gL^{2}}{2Y}\) | m |
| Thermal stress | \(Y\alpha\Delta T\); \(F = YA\alpha\Delta T\) | Pa; N |
| Beam sag | \(\delta = \dfrac{Wl^{3}}{4bd^{3}Y}\) | m |
| Mountain height | \(h_\text{max} = \sigma_\text{limit}/\rho g\) | m |