Mechanical Properties of Solids

1. Elasticity & Plasticity

• Elasticity : ability of a deformed body to return to its original shape and size when the deforming forces are removed.
  e.g. rubber band, catapult (slingshot).
• Plasticity : inability of a deformed body to return to its original shape and size when the forces are removed.
  e.g. putty / clay, crushed can.
• Note : Steel wire is more elastic than a rubber band (steel needs a much larger force for the same strain, i.e. larger modulus of elasticity).

2. Stress (σ)

• Stress = internal restoring force per unit area of cross-section between layers of the material.

• Cut the rod (in equilibrium) at any section : each part is in equilibrium ⇒ internal tension T = F.

3. Strain (φ)

• Strain is a ratio ⇒ unitless & dimensionless.

4. Hooke's Law – Modulus of Elasticity

• Within the limit of proportionality, stress ∝ strain :

• E = modulus of elasticity → depends on the material and temperature (not on the size of the body).
• Unit of E : N/m² or Pa (same as stress, since strain has no unit).
• Like a spring : mass m → extension x ; mass 2m → extension 2x.

Effect of temperature :

• As temperature increases, Young's modulus decreases (material becomes easier to stretch).

5. Longitudinal Stress & Strain

• Only change in length is considered (wire or rod).

• Assumption : area of cross-section A remains the same.

Stress σ = FA ,   Strain φ = Δll ,   Stress = Y (Strain)

• Y → Young's modulus (modulus of elasticity for change in length).

6. Wire as a Spring

• Wire : F = (AY/l) Δl ;  Spring : F = k x  ⇒ a wire behaves like a spring of constant

• For the same material : k ∝ A/l ∝ r²/l.

7. Series & Parallel Combination

Series (wires joined end to end) :

Parallel (wires side by side) :

(use k = AY/l for each wire)

8. Elastic Energy Stored in a Wire

• Wire and spring behave alike, so energy stored = work done in stretching :

U = ½ Y (Al)(Δl)²l² = ½ Y (Vol.)(Strain)²

(Volume of wire = A l)

• Energy per unit volume (energy density) — using Stress = Y (Strain) :

• Also U = ½ F Δl (= ½ × load × extension).

9. Breaking Stress & Breaking Force

• Breaking stress σ_b = minimum stress at which the wire breaks (= maximum stress at which it will not break).
• σ_b depends only on the material and temperature ⇒ same for all steel wires (at the same temp.), whatever their thickness.

• Thicker wire (bigger A) needs a bigger force to break : F₁ = σ_bA₁ , F₂ = σ_bA₂.

10. Solved Example – Work in Stretching

11. Variable Tension, Stress & Strain

• If the rod is accelerating or has weight, tension T varies along its length : T = T(x).
• Find T by Newton's laws on a part of the rod ; then at that section

Total elongation :

• Element of length dx stretches by δx :

Strain = δxdx = TAY  ⇒  δx = T dxAY

Energy stored :

dU = (Stress)²2Y(Vol.) = (T/A)²2Y(A dx)

12. Solved Example – Hanging Rod

(1) Stress & strain at x :

• Tension at x = weight of the part below = mlx · g

T = mgxl ,  Stress = mgxAl ,  Strain = mgxAYl

(2) Total elongation :

Δl = ∫₀^l T dxAY = mgAYl∫₀^l x dx = mgAYl · l²2

(3) Elastic energy stored :

U = ∫₀^l T²2AYdx = m²g²2AYl²∫₀^l x² dx = m²g²2AYl² · l³3

(Δl is half of what the full weight mg would produce if hung at the end.)

13. Rod Pulled by Two Unequal Forces

• Rod of mass m, length l on a smooth floor ; F₁ (left) and F₂ (right), F₂ > F₁.

Acceleration : a = F₂ − F₁m

Left part (length x) : T − F₁ = m₁a = mlx · F₂ − F₁m

• T varies linearly from F₁ to F₂ ; Stress = T/A, Strain = T/AY.

Total elongation :

Δl = 1AY∫₀^l [F₁ + F₂ − F₁lx] dx

    = 1AY[F₁l + F₂ − F₁l · l²2]

    = lAY[F₁ + F₂ − F₁2]

• Both tensile / both compressive → use F₁ + F₂.
• One tensile, one compressive → use the difference (tensile − compressive).

14. Cases

(Ex 5 : front half stretched, back half compressed — net change in length is zero.)

15. Composite Rod (Equilibrium)

• F_net = −10F + F + 2F + 3F + 4F = 0 ⇒ acc. = 0.
• Going left → right : T₁ = 10F, T₂ = 10F − F = 9F, T₃ = 9F − 2F = 7F, T₄ = 7F − 3F = 4F.
• If all parts have the same A and Y : Δl = Σ T_il/AY = (10 + 9 + 7 + 4)FlAY = 30Fl/AY.

16. Poisson Ratio (μ)

• Stretching a wire increases its length but decreases its radius.

• Transverse strain = strain along radius ; longitudinal strain = strain along length.
• Range : −1 to +0.5 (theoretical). For most materials 0 < μ < 0.5.
• Negative μ ⇒ material expands sideways when stretched (positive transverse strain with positive longitudinal strain).

17. Stress – Strain Graph

• Wire of length l loaded with mass m : Stress = mg/A , Strain = Δl/l. Increase the load and plot.

18. Brittle & Ductile Materials

• Brittle → very small plastic zone (B to D close) ; breaks soon after the elastic limit (e.g. glass, cast iron).
• Ductile → large plastic zone (B to D far) ; can be drawn into wires (e.g. copper, mild steel).

19. Volume Stress, Strain & Bulk Modulus (β)

• Volume stress = change in pressure (ΔP) — equal force per area on all faces.
• Volume strain = change in volumeoriginal volume = ΔVV

• Negative sign : pressure increase ⇒ volume decreases ; β is positive.

For a gas :

• Isothermal (PV = const) : dPdV = −PV ⇒ β = P
• Adiabatic (PV^γ = const) : dPdV = −γPV ⇒ β = γP

Gases are the most compressible, solids the least.

20. Change in Volume & Density with Depth

• At depth h : ΔP = ρgh (ρ = 1000 kg/m³ for water).
• As we go down, water is compressed ⇒ density increases.
• Mass constant : ρV = const ⇒ Δρρ + ΔVV = 0

21. Shear Stress, Strain & Modulus of Rigidity (η)

• Shear stress : sliding (tangential) stress between layers of the material. Force F_∥ acts parallel to the top face (area A) ; the bottom face is held fixed.

• η → modulus of rigidity (shear modulus) ; φ in radian (small angle).

22. Solved Example – Oblique Section

23. Relation between Y, η, β and μ

(Young's modulus & modulus of rigidity)

(Young's modulus & bulk modulus)

(Y, η and β — obtained by eliminating μ from the two relations above)

• β = bulk modulus, μ = Poisson ratio, η = modulus of rigidity, Y = Young's modulus.
• β > 0 ⇒ 1 − 2μ > 0 ⇒ μ < 0.5 ; η > 0 ⇒ 1 + μ > 0 ⇒ μ > −1. This gives the range −1 < μ < 0.5.

24. Formula Summary