⚙️ Solid Mechanics 🔥 Day 1

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Solid Mechanics — Strength of Materials

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Solid Mechanics — Complete Study Notes

10 chapters · GATE, ESE (IES), SSC JE, State PSC, RRB JE

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Solid Mechanics (Strength of Materials) is the highest-weightage subject in GATE Civil and ESE (IES), and appears heavily in SSC JE too. This chapter covers every topic — from material properties and elastic constants, through stress transformation and Mohr's circle, theories of failure, bending and shear stresses, deflection of beams, torsion, thick/thin cylinders, columns, retaining walls and shear centre — with all formulae, diagrams, solved examples and exam-pattern analysis.

Learning Objectives

After studying this chapter you will be able to:

Describe the mechanical properties of engineering materials and read the stress-strain curve for mild steel, identifying every key point
Apply the elastic-constant inter-relations ($G$, $K$, $E$, $\nu$) and generalised Hooke's law to 3D stress states
Draw SFD and BMD for standard beams and apply the $\dfrac{dV}{dx} = -w$, $\dfrac{dM}{dx} = V$ relationships
Construct Mohr's circle of stress/strain, compute principal stresses, and apply the five classical theories of failure
Compute beam deflection using double integration, Macaulay's method, moment-area, conjugate beam and unit-load methods
Apply the flexure and shear-stress formulae ($\sigma = \dfrac{My}{I}$, $\tau = \dfrac{VQ}{Ib}$) to standard and composite sections
Analyse thin and thick (Lamé) pressure vessels, and design columns using Euler's and Rankine's formulae
Locate the shear centre of standard thin-walled sections and compute principal moments of inertia

Exam Relevance

ESE★★★★★

GATE★★★★★

SSC JE★★★★★

State PSC★★★★☆

💡 Solid Mechanics is the single highest-weightage subject in GATE Civil — Mohr's circle, theories of failure, beam deflection and torsion numericals appear in almost every paper, often worth 8–12 marks combined.

Concept Map

Properties of Materials → Stress-Strain Curve

↓

Elastic Constants (E, G, K, ν) → Generalised Hooke's Law

↓

SFD/BMD → Bending & Shear Stresses → Deflection of Beams

↓

Principal Stress (Mohr's Circle) → Theories of Failure

Thick/Thin Cylinders · Torsion & Springs

↓

Columns (Euler/Rankine) → Retaining Walls (Earth Pressure)

↓

Shear Centre & Principal Moments of Inertia

Where This Fits

Prerequisite for: Theory of Structures (analysis methods build directly on SFD/BMD and deflection here), and both Steel Structures and RCC/PCC Design (which apply these stress and section-property formulae to real member design).

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Ch 1 · Properties of Materials

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Mechanical Properties

Property	Definition	Unit / Symbol	Exam Relevance
Elasticity	Ability to regain original shape on removal of load	—	Basis of Hooke's law
Plasticity	Ability to retain deformation after load removal	—	Plastic hinge concept in RCC
Ductility	Ability to undergo large plastic deformation before fracture	% elongation; % reduction in area	Mild steel ductile; cast iron brittle
Brittleness	Fracture with little or no plastic deformation	—	Concrete, cast iron, glass
Toughness	Energy absorbed per unit volume up to fracture (area under σ–ε curve)	J/m³ or N·m/m³	Charpy/Izod impact tests
Resilience	Energy stored per unit volume within elastic limit $= \dfrac{\sigma_y^2}{2E}$	J/m³	Springs, proof resilience
Hardness	Resistance to surface indentation/scratch	BHN, Vickers HV, Rockwell	Wear resistance
Creep	Slow, time-dependent plastic deformation under sustained stress below yield	Creep rate $\dot\varepsilon$	High-temp applications; PSC losses
Fatigue	Failure under repeated cyclic loading below UTS	Endurance limit $\sigma_e$	S-N curve; notch sensitivity
Malleability	Ability to be hammered into thin sheets	—	Gold > Silver > Copper > Iron

Key Points on the Stress–Strain Curve

Point / Zone	Description	Key Value (typical)
Proportional limit	Hooke's law holds exactly (σ ∝ ε)	Below yield point
Elastic limit	Max stress with full elastic recovery (slightly above proportional limit)	≈ Proportional limit for metals
Upper yield point	Sudden drop; dislocations break free from solute atoms (Lüder's bands)	~250 MPa (Fe415 rebar)
Lower yield point	Stable plateau of plastic flow	~240 MPa
Ultimate Tensile Stress (UTS)	Maximum engineering stress; necking begins	~415 MPa (Fe415)
Fracture point	Actual failure; true stress > engineering stress	Ductile: cup-cone fracture

Factor of Safety and Proof Stress

Factor of Safety (FOS) = Ultimate Stress / Permissible (Working) Stress
Proof Stress (0.2% offset): stress at which 0.2% permanent strain occurs — used for materials without a well-defined yield point (Al alloys, HYSD bars)
Resilience (Modulus) = $\dfrac{\sigma_y^2}{2E}$ — energy stored per unit volume at yield
Toughness $\approx$ area under complete $\sigma$–$\varepsilon$ curve $\approx \sigma_{avg} \times \varepsilon_{fracture}$

Material Comparison

Property	Mild Steel	HYSD Steel (Fe415)	Concrete (M25)	Cast Iron
E (GPa)	200–210	200	25 ($\approx 5000\sqrt{f_{ck}}$ MPa)	100–170
$\sigma_y$ (MPa)	250	415	25 ($f_{ck}$)	No yield
UTS (MPa)	400–500	485 min	—	140–350
Behaviour	Ductile, defined yield	Ductile, no clear yield	Brittle in tension	Brittle
Poisson's ratio ν	0.25–0.30	0.25–0.30	0.15–0.20	0.25

💡 GATE Tip: E for steel = 2×10⁵ MPa (200 GPa). E for concrete = $5000\sqrt{f_{ck}}$ MPa (IS 456). Poisson's ratio for steel ν = 0.3; for concrete ν = 0.15 to 0.2. These values appear in almost every numerical question.

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Ch 1 · Formulas

Resilience (Modulus)

$$ U_r = \frac{\sigma_y^{2}}{2E} $$

Energy stored per unit volume within the elastic limit

J/m³

Factor of Safety

$$ FOS = \frac{\sigma_{ultimate}}{\sigma_{permissible}} $$

Ratio of ultimate stress to permissible (working) stress

—

Elastic Modulus — Steel & Concrete

$$ E_{steel} = 2\times10^{5}\ \text{MPa},\;\; E_{concrete} = 5000\sqrt{f_{ck}}\ \text{MPa} $$

IS 456; the two most-used E values in numericals

MPa

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Ch 1 · Solved Example

GATE CE / ESE level

A mild steel bar ($E = 2\times10^5$ MPa) has a yield stress $\sigma_y = 250$ MPa. Compute its modulus of resilience.

Approach hint

Apply $U_r = \dfrac{\sigma_y^2}{2E}$ directly, keeping units in N/mm² (MPa) throughout.

$U_r = \dfrac{250^2}{2 \times 2\times10^5} = \dfrac{62500}{4\times10^5}$
$U_r = 0.15625$ N·mm/mm³ = 0.156 MPa (equivalently MJ/m³)

✓ Modulus of resilience ≈ 0.156 N·mm/mm³ — the maximum elastic strain energy the steel can absorb per unit volume before yielding

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Ch 1 · Diagrams

Click any diagram to zoom

Fig 1 — Stress–Strain Curve for Mild Steel: O→A elastic; A upper yield, B lower yield; B→C strain hardening; C→D necking; D fracture

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Ch 1 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

Toughness of a material is best represented by:

Toughness is the total energy absorbed per unit volume up to fracture — the area under the entire stress-strain curve, not just the elastic region ✓

Question 2

Proof stress (0.2% offset) is used for materials that:

Materials like Al alloys and HYSD reinforcement bars have a gradual, continuous stress-strain curve without a sharp yield point, so 0.2% proof stress is used as a practical substitute ✓

Question 3

On the standard mild-steel stress-strain curve, necking begins at:

Necking (localised cross-section reduction) begins right at the UTS point, after which engineering stress drops even though true stress keeps rising until fracture ✓

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Ch 2 · Stress–Strain & Elastic Constants

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Stress and Strain Definitions

Normal Stress: $\sigma = \dfrac{P}{A}$ (Pa = N/m²; MPa = N/mm²)
Shear Stress: $\tau = V/A$ (shear force / area)
Normal Strain: $\varepsilon = \dfrac{\delta}{L}$ (dimensionless; δ = deformation, L = original length)
Shear Strain: $\gamma = \dfrac{\tau}{G}$ (angular distortion in radians)
Volumetric Strain: $e = \Delta V/V = \epsilon_x + \epsilon_y + \epsilon_z$ (for triaxial loading)

Elastic Constants

Constant	Symbol	Definition	Typical Range (steel)
Young's Modulus	E	σ / ε (axial stress / axial strain)	200–210 GPa
Shear Modulus (Rigidity)	G	τ / γ (shear stress / shear strain)	80–82 GPa
Bulk Modulus	K	Hydrostatic stress / volumetric strain = σ / e	160–170 GPa
Poisson's Ratio	ν	Lateral strain / longitudinal strain (magnitude)	0.25–0.30

Inter-Relations Between Elastic Constants

$G = \dfrac{E}{2(1+\nu)}$
$K = \dfrac{E}{3(1-2\nu)}$
$E = \dfrac{9KG}{3K+G}$
$\nu = \dfrac{3K-2G}{2(3K+G)}$

Limits of ν:

$\nu = 0 \to K = \dfrac{E}{3}, G = \dfrac{E}{2}$ (cork — no lateral strain)
$\nu = 0.5 \to K \to \infty$ (incompressible material, rubber)
$\nu < 0$ → auxetic material (rare, expands laterally under tension)
Theoretical range: $-1 \le \nu \le 0.5$ (for isotropic elastic materials)

⭐ Key memory: G = E/[2(1+ν)] and K = E/[3(1−2ν)]. These two are the most tested inter-relations. For steel: $E \approx 2\times10^{5}$ MPa, $G \approx 0.8\times10^{5}$ MPa, $\nu \approx 0.25 \to G = 2\dfrac{\times10^{5}}{2\times1.25} = 0.8\times10^{5}$ ✓

Generalised Hooke's Law (3D)

$$ \epsilon_x = \frac{1}{E}[\sigma_x - \nu(\sigma_y + \sigma_z)] $$ $$ \epsilon_y = \frac{1}{E}[\sigma_y - \nu(\sigma_x + \sigma_z)] $$ $$ \epsilon_z = \frac{1}{E}[\sigma_z - \nu(\sigma_x + \sigma_y)] $$

$\gamma_{xy} = \dfrac{\tau_{xy}}{G}$; $\gamma_{yz} = \dfrac{\tau_{yz}}{G}$; $\gamma_{zx} = \dfrac{\tau_{zx}}{G}$
Volumetric strain: $e = (1-2\nu)\dfrac{\sigma_x+\sigma_y+\sigma_z}{E} = \dfrac{\sigma_{hydro}}{K}$

Thermal Stress

Free expansion: $\delta_T = \alpha \cdot \Delta T \cdot L$ (no stress if unconstrained)
Thermal stress (fully constrained bar): $\sigma_T = E \cdot \alpha \cdot \Delta T$ (compressive if temperature rises)
Partially constrained (support settles by δ < δ_T): $\sigma = \dfrac{E(\alpha\Delta T L - \delta)}{L}$

Composite Bars and Statically Indeterminate Axial Problems

Compatibility (same deformation): $\delta_1 = \delta_2 \to P_1 L/(A_1 E_1) = P_2 L/(A_2 E_2) \to \dfrac{P_1}{P_2} = A_1 \dfrac{E_1}{A_2 E_2}$
Equilibrium: $P_{external} = P_1 + P_2$ (if parallel)
Modular ratio: $m = \dfrac{E_{steel}}{E_{concrete}}$ (used in RCC transformed section); $m = \dfrac{280}{3\sigma_{cbc}}$ per IS 456 [WSM]

Plane Stress and Plane Strain

Condition	Plane Stress	Plane Strain
Definition	σ_z = τ_yz = τ_zx = 0; thin plates/discs	ε_z = γ_yz = γ_zx = 0; long dams, tunnels, retaining walls
σ_z	0 (given)	ν(σ_x+σ_y) — induced by constraint
ε_z	−ν(σ_x+σ_y)/E (non-zero)	0 (given by condition)
Effective E & ν	E, ν	E/(1−ν²), ν/(1−ν)
Examples	Thin plates, flanges under in-plane loads	Long dams, tunnels, wide beams

💡 GATE Tip: Plane stress → σ_z = 0 but ε_z ≠ 0. Plane strain → ε_z = 0 but σ_z ≠ 0. This is a classic source of confusion and is directly tested.

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Ch 2 · Formulas

Elastic Constant Inter-Relations

$$ G = \frac{E}{2(1+\nu)},\;\; K = \frac{E}{3(1-2\nu)} $$

The two most frequently tested relations

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E in Terms of K and G

$$ E = \frac{9KG}{3K+G} $$

Used when K and G are given directly

—

Thermal Stress (Fully Constrained)

$$ \sigma_T = E\alpha\Delta T $$

Compressive when temperature rises

MPa

Modular Ratio (RCC, WSM)

$$ m = \frac{280}{3\sigma_{cbc}} $$

Per IS 456 Working Stress Method

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Ch 2 · Solved Example

GATE CE / ESE level

A steel rail, 30 m long, is laid at 10°C with no expansion gap and fully restrained. If the rail temperature rises to 45°C, find the thermal stress induced. Take $E = 2\times10^5$ MPa, $\alpha = 12\times10^{-6}$/°C.

Approach hint

Apply $\sigma_T = E\alpha\Delta T$ directly — note the rail's length doesn't affect the stress for a fully constrained bar.

$\Delta T = 45 - 10 = 35°C$
$\sigma_T = 2\times10^5 \times 12\times10^{-6} \times 35$
$\sigma_T = 84$ MPa (compressive)

✓ Thermal stress ≈ 84 MPa, compressive — note this is independent of the rail's length since it's fully restrained

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Ch 2 · Diagrams

Click any diagram to zoom

Fig 2 — Fully constrained bar: thermal stress σ = EαΔT

Fig 3 — Composite bar: equal deformation compatibility

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Ch 2 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

For a Poisson's ratio ν = 0.5 (e.g. rubber), the bulk modulus K:

$K = \dfrac{E}{3(1-2\nu)}$; as ν → 0.5, the denominator → 0, so K → ∞, meaning the material is incompressible ✓

Question 2

In a plane STRESS condition (thin plate):

Plane stress sets σ_z = 0 (thin plate, free surface), but the resulting ε_z = −ν(σ_x+σ_y)/E is generally non-zero — the plate is free to thin/thicken ✓

Question 3

A fully constrained bar experiences a temperature RISE. The induced thermal stress is:

A temperature rise makes the bar want to expand; full restraint prevents this, inducing a compressive stress σ = EαΔT, independent of length ✓

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Ch 3 · Shear Force and Bending Moment Diagrams

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Sign Convention

ℹ️ Standard (Sagging Positive): Sagging BM is positive (+ve). Shear force: left portion has upward force → +ve SF.

Relationship between load, SF and BM:

$\dfrac{dV}{dx} = -w(x)$ [rate of change of SF = negative of distributed load]
$\dfrac{dM}{dx} = V(x)$ [rate of change of BM = SF]
$\dfrac{d^2M}{dx^2} = -w(x)$
Consequence: Where V = 0 → BM is maximum (or minimum)

Standard SFD/BMD Results

Beam Type	Loading	Max SF	Max BM	Location of Max BM
SS beam	Central point load P	P/2	PL/4	Midspan
SS beam	UDL w over full span	wL/2	wL²/8	Midspan
SS beam	Point load P at 'a' from A	Pb/L (at A), Pa/L (at B)	Pab/L	Under load
Cantilever	Point load P at free end	P	PL (hogging)	Fixed end
Cantilever	UDL w over full span	wL	wL²/2 (hogging)	Fixed end
Cantilever	UVL (0 at free, w at fixed)	wL/2	wL²/6	Fixed end
SS beam	UVL (0 at A, w at B)	wL/6 at A, wL/3 at B	$\dfrac{wL^{2}}{9\sqrt{3}}$ at $x=\dfrac{L}{\sqrt{3}}$	$x=\dfrac{L}{\sqrt3}$ from A

SFD/BMD Shapes for Standard Loads

Load type → SFD shape → BMD shape:

No load (span between loads) → Constant SF → Linear BM
Point load → Step (jump) in SF → Kink in BM
UDL → Linear SF → Parabolic (2nd degree) BM
UVL → Parabolic SF → Cubic (3rd degree) BM
Applied moment M₀ at a point → No change in SF → Step (jump) in BM

Point of Contraflexure

The point of contraflexure is where the bending moment changes sign (BM = 0, and the beam changes from sagging to hogging). It exists only in beams where the BMD crosses the zero axis.

For propped cantilever (fixed at A, roller at B), UDL w: $M_A = -\dfrac{wL^2}{8}$ (hogging), $R_A = \dfrac{3wL}{8}$, $R_B = \dfrac{5wL}{8}$
BM = 0 (at support B) and at x = 3L/4 from fixed end A → point of contraflexure

Relationship Between Load, SF, and BM — Key Rules

The SFD between two point loads is horizontal (constant)
Under a UDL, SFD is linear, BMD is parabolic
At a free end → V = 0 and M = 0 (unless a point load/moment is applied)
At a roller/pin → M = 0 (but V may be non-zero)
At a fixed support → V and M are both non-zero reactions
A couple (moment M₀) applied at a point → SFD unchanged, BMD jumps by M₀

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Ch 3 · Formulas

Load–Shear–Moment Relations

$$ \frac{dV}{dx} = -w(x),\;\; \frac{dM}{dx} = V(x) $$

Where V = 0, BM is maximum or minimum

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SS Beam, Central Point Load

$$ M_{max} = \frac{PL}{4} $$

At midspan; max SF = P/2

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SS Beam, Full-Span UDL

$$ M_{max} = \frac{wL^{2}}{8} $$

At midspan; max SF = wL/2

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Propped Cantilever (UDL), Hogging Moment

$$ M_A = -\frac{wL^{2}}{8} $$

$R_A = \dfrac{3wL}{8}$, $R_B = \dfrac{5wL}{8}$; contraflexure at $x=\dfrac{3L}{4}$ from A

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Ch 3 · Solved Example

GATE CE / ESE level

A simply supported beam of span 6 m carries a UDL of 20 kN/m over its full length. Find the maximum shear force and maximum bending moment.

Approach hint

Apply the standard SS-beam-with-UDL results directly: $V_{max}=\dfrac{wL}{2}$, $M_{max}=\dfrac{wL^2}{8}$.

$V_{max} = \dfrac{wL}{2} = 20\dfrac{\times6}{2} = 60$ kN
$M_{max} = \dfrac{wL^{2}}{8} = 20\dfrac{\times6^{2}}{8} = 90$ kN·m

✓ Max SF = 60 kN (at each support); Max BM = 90 kN·m (at midspan)

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Ch 3 · Diagrams

Click any diagram to zoom

Fig 4 — SS beam, UDL w: SFD linear (±wL/2), BMD parabolic (max = wL²/8 at midspan)

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Ch 3 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

Under a UDL, the shape of the SFD and BMD are respectively:

Under a UDL, SFD is linear (first-degree) and BMD is parabolic (second-degree) — each integration raises the degree by one ✓

Question 2

At a point where the shear force V = 0 (and doesn't jump), the bending moment is:

Since $\dfrac{dM}{dx} = V$, when V = 0 the slope of the BMD is zero, meaning BM is at a local maximum or minimum ✓

Question 3

An applied concentrated moment M₀ at a point on a beam causes:

A concentrated moment does not change the shear force but causes the BMD to jump instantaneously by M₀ at that point ✓

⭕

Ch 4 · Principal Stress–Strain & Theories of Failure

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Stress Transformation (2D)

On an inclined plane at angle θ to the x-face:

$\sigma_\theta = \dfrac{\sigma_x+\sigma_y}{2} + \dfrac{\sigma_x-\sigma_y}{2} \cdot \cos2\theta + \tau_{xy}\cdot\sin2\theta$
$\tau_\theta = -\dfrac{\sigma_x-\sigma_y}{2} \cdot \sin2\theta + \tau_{xy}\cdot\cos2\theta$

Principal stresses ($\tau=0$):

$$ \sigma_1, \sigma_2 = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\frac{(\sigma_x-\sigma_y)^2}{4} + \tau_{xy}^2} $$

Max shear stress: $\tau_{max} = \sqrt{(\sigma_x-\sigma_y)\dfrac{^2}{4} + \tau_{xy}^2} = \dfrac{\sigma_1-\sigma_2}{2}$
Angle of principal planes: $\tan(2\theta_p) = 2\dfrac{\tau_{xy}}{\sigma_x-\sigma_y}$
Principal planes are at 90° to each other; max shear planes at 45° to principal planes

Mohr's Circle Construction — Step by Step

Draw σ-axis (horizontal) and τ-axis (vertical)
Plot point X = (σ_x, +τ_xy) and point Y = (σ_y, −τ_xy)
Centre C = midpoint of XY = ((σ_x+σ_y)/2, 0)
Radius $R = |CX| = \sqrt{(\sigma_x-\sigma_y)\dfrac{^2}{4} + \tau_{xy}^2}$
Principal stresses: $\sigma_1 = \sigma_{avg}+R$, $\sigma_2 = \sigma_{avg}-R$ (on σ-axis)
Max shear stress: $\tau_{max} = R$ (at top/bottom of circle)
Angle 2θ_p on circle = angle from X to horizontal (right); → θ_p on actual element

Mohr's Circle of Strain

Replace σ with ε and τ with γ/2 in all formulae:

Centre: $\epsilon_{avg} = \dfrac{\epsilon_x+\epsilon_y}{2}$
Radius: $R_\epsilon = \sqrt{(\epsilon_x-\epsilon_y)\dfrac{^2}{4} + (\dfrac{\gamma_{xy}}{2})^2}$
Principal strains: $\epsilon_1, \epsilon_2 = \epsilon_{avg} \pm R_\epsilon$
Max shear strain: $\gamma_{max} = 2R_\epsilon = \epsilon_1-\epsilon_2$

Strain rosette (0°–45°–90° rosette):

$$ \epsilon_x = \epsilon_{0°},\;\; \epsilon_y = \epsilon_{90°},\;\; \gamma_{xy} = 2\epsilon_{45°} - \epsilon_{0°} - \epsilon_{90°} $$

Theories of Failure

Failure theories predict when a material transitions from safe to unsafe under multiaxial stress, by comparison with the uniaxial yield stress σ_y (or UTS σ_u).

Theory	Failure Criterion	Safe for	Best Suited for
Maximum Principal Stress (Rankine)	σ₁ ≤ σ_y	σ₁ < σ_y	Brittle materials (cast iron, concrete)
Maximum Shear Stress (Tresca / Guest)	τ_max ≤ σ_y/2 → (σ₁−σ₂) ≤ σ_y	(σ₁−σ₂) < σ_y	Ductile metals (conservative)
Max Principal Strain (St. Venant)	ε₁ ≤ σ_y/E → σ₁−ν(σ₂+σ₃) ≤ σ_y	σ₁−ν(σ₂+σ₃) < σ_y	Brittle (less common)
Max Strain Energy (Beltrami-Haigh)	σ₁²+σ₂²+σ₃²−2ν(σ₁σ₂+σ₂σ₃+σ₃σ₁) ≤ σ_y²	—	Obsolete; not used practically
Distortion Energy (von Mises)	$\sigma_e = \sqrt{\tfrac12((\sigma_1-\sigma_2)^2+(\sigma_2-\sigma_3)^2+(\sigma_3-\sigma_1)^2)} \leq \sigma_y$	σ_e < σ_y	Ductile metals (most accurate)
Max Strain Energy of Distortion (Hencky)	Same as von Mises	—	Same as von Mises

Failure Criteria for Biaxial Stress (σ₃ = 0)

Rankine (Max Principal Stress): $\sigma_1 = \sigma_y$
Tresca (Max Shear Stress): $\sigma_1-\sigma_2 = \sigma_y$ (if σ₁ and σ₂ have opposite signs); $\sigma_1 = \sigma_y$ (if same sign — Tresca = Rankine in same-sign quadrants)
von Mises (Distortion Energy): $\sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2 = \sigma_y^2$ (ellipse in σ₁−σ₂ space)

For pure shear ($\sigma_1 = -\sigma_2 = \tau$):

Tresca: $\tau_y = \dfrac{\sigma_y}{2}$
von Mises: $\tau_y = \dfrac{\sigma_y}{\sqrt3} \approx 0.577\sigma_y$
von Mises is more accurate (agrees with experiments); Tresca is conservative (safe side)

⭐ Key comparison: For pure shear, Tresca gives τ_y = 0.5σ_y and von Mises gives τ_y = 0.577σ_y. von Mises predicts ~15% more capacity — important for GATE numericals. Use Tresca for conservative design; von Mises for accurate prediction of ductile failure.

⭕

Ch 4 · Formulas

Principal Stresses

$$ \sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\frac{(\sigma_x-\sigma_y)^2}{4}+\tau_{xy}^2} $$

τ = 0 on principal planes

MPa

Maximum Shear Stress

$$ \tau_{max} = \frac{\sigma_1-\sigma_2}{2} $$

= Radius of Mohr's circle

MPa

Tresca Criterion (Pure Shear)

$$ \tau_y = \frac{\sigma_y}{2} $$

Conservative; used for ductile-metal design

MPa

von Mises Criterion (Pure Shear)

$$ \tau_y = \frac{\sigma_y}{\sqrt3} \approx 0.577\sigma_y $$

More accurate; matches ductile-metal test data

MPa

⭕

Ch 4 · Solved Example

Mohr's Circle — GATE CE level

A stressed element has $\sigma_x = 80$ MPa, $\sigma_y = 20$ MPa, $\tau_{xy} = 30$ MPa. Find the principal stresses, maximum shear stress and the angle of the principal plane.

Approach hint

Compute $\sigma_{avg}$ and R first, then apply $\sigma_{1,2} = \sigma_{avg} \pm R$.

$\sigma_{avg} = \dfrac{80+20}{2} = 50$ MPa
$R = \sqrt{(80-20)\dfrac{^2}{4} + 30^2} = \sqrt{900+900} = \sqrt{1800} = 42.43$ MPa
$\sigma_1 = 50+42.43 = 92.43$ MPa; $\sigma_2 = 50-42.43 = 7.57$ MPa
$\tau_{max} = 42.43$ MPa
$2\theta_p = \arctan(2\dfrac{\times30}{80-20}) = \arctan(1) = 45° \to \theta_p = 22.5°$

✓ $\sigma_1 = 92.43$ MPa, $\sigma_2 = 7.57$ MPa, $\tau_{max} = 42.43$ MPa, $\theta_p = 22.5°$. Checking von Mises: $\sigma_e = \sqrt{92.43^2-92.43\times7.57+7.57^2} \approx 89$ MPa < σ₁, confirming the element is well within a ductile material's safe envelope for typical σ_y

⭕

Ch 4 · Diagrams

Click any diagram to zoom

Fig 5 — Mohr's Circle: Centre = ((σ_x+σ_y)/2, 0), Radius R = τ_max = (σ₁−σ₂)/2; σ₁, σ₂ at ends of horizontal diameter

⭕

Ch 4 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

The radius of Mohr's circle of stress represents:

The radius R of Mohr's circle equals $\tau_{max} = \dfrac{\sigma_1-\sigma_2}{2}$, the maximum shear stress at the point ✓

Question 2

Which theory of failure is best suited for BRITTLE materials such as cast iron and concrete?

Rankine's Maximum Principal Stress theory matches the brittle-fracture behaviour of cast iron and concrete, which fail by cleavage at the max principal stress ✓

Question 3

For pure shear, comparing Tresca and von Mises yield predictions:

Tresca gives τ_y = 0.5σ_y while von Mises gives τ_y = 0.577σ_y — about 15% higher, making Tresca the conservative (safe-side) choice ✓

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Ch 5 · Deflection of Beams

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Elastic Curve Equation

Euler-Bernoulli beam equation:

$$ EI\frac{d^2y}{dx^2} = M(x) $$ $$ EI\frac{d^3y}{dx^3} = V(x) = \frac{dM}{dx} $$ $$ EI\frac{d^4y}{dx^4} = -w(x) $$

Boundary conditions:

Fixed end: y = 0, dy/dx = 0
Simply supported: $y=0, M=0\ (\dfrac{d^2y}{dx^2}=0)$
Free end: M = 0, V = 0 (unless load applied)
Internal hinge: M = 0 at hinge; y and dy/dx continuous

Standard Deflection Formulae

Beam	Loading	Max Deflection δ_max	Location
SS beam	Central point load P	PL³/(48EI)	Midspan
SS beam	UDL w, full span	5wL⁴/(384EI)	Midspan
SS beam	Load P at 'a' from A (a<b)	Pa(3L²−4a²)b/(48EIL) at midspan approx	Exact: $x=\dfrac{\sqrt{L^2-b^2}}{3}$ from A if a>b
Cantilever	Point load P at free end	PL³/(3EI)	Free end
Cantilever	UDL w, full span	wL⁴/(8EI)	Free end
Cantilever	UDL w over half-span from fixed end	7wL⁴/(384EI)	Free end
Propped cantilever	UDL w	wL⁴/(185EI)	~0.42L from fixed end
Fixed-fixed beam	Central P	PL³/(192EI)	Midspan
Fixed-fixed beam	UDL w	wL⁴/(384EI)	Midspan

Methods for Computing Deflections

Double Integration Method

$EI\cdot y'' = M(x)$ → integrate twice → apply BCs to find constants C₁, C₂
Slope: $\theta = \dfrac{dy}{dx}$ | Deflection: y (positive downward convention or per problem)

Macaulay's Method (Singularity Functions)

Use <x−a>ⁿ brackets — integrated as (x−a)ⁿ⁺¹/(n+1) only when (x−a) > 0
Very efficient for multiple point loads / partial UDLs
Example: SS beam, load P at x = a: $EIy'' = R_A\cdot x - P\langle x-a\rangle^1$; integrate with the same bracket rule for slope and deflection

Moment Area Method (Mohr's Theorems)

Theorem I: Change in slope between two points A and B = $(\dfrac{1}{EI}) \times$ Area of $\dfrac{M}{EI}$ diagram between A and B
Theorem II: Tangential deviation of B from tangent at A: $t_{BA} = (\dfrac{1}{EI}) \times$ (Area of $\dfrac{M}{EI}$ diagram between A and B) $\times \bar{x}_B$, where $\bar{x}_B$ = distance of centroid of M/EI diagram from B
Best used for: cantilevers, overhanging beams, non-prismatic beams

Conjugate Beam Method

In the conjugate beam: load = M/EI diagram of real beam
Shear in conjugate beam = slope in real beam; Moment in conjugate beam = deflection in real beam
Real free end → Conjugate fixed end; Real fixed end → Conjugate free end; Real pin/roller → Conjugate pin/roller

Virtual Work / Unit Load Method

$$ \delta_C = \int \frac{M\cdot\bar{m}}{EI}\,dx $$

M = real bending moment (under actual loads); $\bar m$ = virtual bending moment (due to unit load at C in desired direction)
Slope: $\theta_C = \int(M\cdot\bar{m}_\dfrac{\theta}{EI})dx$ ($\bar m_\theta$ due to unit couple at C)

💡 GATE Tip: For a SS beam with UDL, max deflection = 5wL⁴/(384EI). For cantilever with tip load: PL³/(3EI). Ratio of these two is frequently asked to compare. Memorise the "5 in 384" for UDL SS beam.

📏

Ch 5 · Formulas

Euler-Bernoulli Equation

$$ EI\frac{d^2y}{dx^2} = M(x) $$

Core relation for all deflection methods

—

SS Beam, UDL — Max Deflection

$$ \delta_{max} = \frac{5wL^4}{384EI} $$

"5 in 384" — memorise this exact fraction

—

Cantilever, Tip Load — Max Deflection

$$ \delta_{max} = \frac{PL^3}{3EI} $$

At the free end

—

Unit Load Method

$$ \delta_C = \int \frac{M\bar{m}}{EI}dx $$

m̄ = virtual BM due to unit load at C

—

📏

Ch 5 · Solved Example

GATE CE / ESE level

A cantilever beam of length 3 m ($EI = 4\times10^{13}$ N·mm²) carries a point load of 10 kN at its free end. Find the maximum deflection.

Approach hint

Apply $\delta_{max}=\dfrac{PL^3}{3EI}$ directly, keeping all units consistent in N and mm.

$P = 10{,}000$ N, $L = 3000$ mm, $EI = 4\times10^{13}$ N·mm²
$\delta_{max} = \dfrac{10{,}000 \times 3000^3}{3\times4\times10^{13}} = \dfrac{2.7\times10^{14}}{1.2\times10^{14}}$
$\delta_{max} = 2.25$ mm

✓ Maximum deflection at the free end ≈ 2.25 mm

📏

Ch 5 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

For a simply supported beam with a full-span UDL, the maximum deflection at midspan is:

The "5 in 384" formula 5wL⁴/(384EI) is the standard result for an SS beam under full-span UDL ✓

Question 2

Macaulay's method is particularly efficient for beams with:

Macaulay's singularity-function brackets allow a single continuous expression to handle multiple discontinuous loads, making it efficient for complex loading ✓

Question 3

In the Conjugate Beam Method, the bending moment in the conjugate beam corresponds to which quantity in the real beam?

In the conjugate beam method: shear in the conjugate beam = slope in the real beam; moment in the conjugate beam = deflection in the real beam ✓

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Ch 6 · Bending & Shear Stresses in Beams

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Flexure Formula (Bending Stress)

Euler-Bernoulli Bending Formula:

$$ \frac{\sigma}{y} = \frac{M}{I} = \frac{E}{R} $$

$\sigma = M\cdot \dfrac{y}{I}$ (bending stress at distance y from NA)
$\sigma_{max} = M\cdot \dfrac{y_{max}}{I} = \dfrac{M}{Z}$
M = bending moment at section; I = second moment of area about NA; y = distance from neutral axis (NA)
$Z = \dfrac{I}{y_{max}}$ = section modulus (m³ or mm³); R = radius of curvature of neutral axis; E = Young's modulus

⭐ Key: Bending stress is zero at NA and maximum at the extreme fibres. Sagging BM → tension at bottom, compression at top. Hogging BM → opposite.

Section Modulus and Moment of Resistance

Section	I (about NA)	y_max	Z = I/y
Rectangle (b × d)	bd³/12	d/2	bd²/6
Solid Circle (dia D)	πD⁴/64	D/2	πD³/32
Hollow Circle (D, d)	π(D⁴−d⁴)/64	D/2	π(D⁴−d⁴)/(32D)
Triangle (b, h)	bh³/36	2h/3 (from apex)	bh²/24
I-Section	BD³/12 − 2·[(B−b)d³/12]	D/2	I/(D/2)

Shear Stress in Beams

Shear stress at distance y from NA:

$$ \tau = \frac{V\cdot Q}{I\cdot b} $$

V = shear force at section; Q = A'·ȳ' = first moment of area of section above (or below) y about NA
I = moment of inertia of full section about NA; b = width of section at distance y

Maximum shear stress for standard sections:

Rectangle: $\tau_{max} = 1.5\times V/A$ (at NA) — parabolic distribution
Solid Circle: $\tau_{max} = \dfrac{4V}{3A}$ (at NA)
I-section: $\tau_{max} \approx V/A_{web}$ (approximately uniform in web)
Thin-walled hollow circle: $\tau_{max} = \dfrac{2V}{A}$

Shear Flow in Thin-Walled Sections

Shear flow: $q = V\cdot \dfrac{Q}{I}$ (force per unit length along the wall)
In open thin-walled sections (channels, angles): q varies along the wall; zero at free edges. Shear centre is the point through which V must pass for no twisting
In closed thin-walled sections (box, tube): $q = V\cdot \dfrac{Q}{I} + q_0$ (q₀ = statically indeterminate shear flow)

Composite Beams (Transformed Section)

Transform all materials to equivalent area of one material (usually concrete or steel): $A_{transformed} = A_{original} \times (\dfrac{E_{material}}{E_{reference}})$
Modular ratio $m = \dfrac{E_1}{E_2}$; Neutral axis: $\Sigma A_t\cdot\bar{y} = 0$ (centroid of transformed section)
Bending stress in material 1: $\sigma_1 = M\cdot \dfrac{y}{I_t}$; in material 2: $\sigma_2 = m\cdot M\cdot \dfrac{y}{I_t}$

Flitched Beams

A timber beam reinforced with steel plates on two faces. Both materials undergo same curvature (1/R), same strain at same y from NA:

Equal curvature: $\dfrac{\sigma_{steel}}{E_s} = \dfrac{\sigma_{timber}}{E_t}$ at same y
$\therefore \sigma_{steel} = m\times\sigma_{timber}$ (m = E_s/E_t ≈ 20 for steel-timber)
Moment of resistance: $M = M_{steel}+M_{timber}$

💡 GATE/ESE Tip: The relationship τ = VQ/(Ib) is the Jourawski formula. For an I-beam, the web carries ~80–90% of the shear force even though it has a much smaller area than the flanges. This is why web buckling (shear buckling) is critical in plate girders.

🏗

Ch 6 · Formulas

Flexure Formula

$$ \frac{\sigma}{y} = \frac{M}{I} = \frac{E}{R} $$

σ zero at NA, maximum at extreme fibre

—

Jourawski Shear Stress Formula

$$ \tau = \frac{VQ}{Ib} $$

Q = first moment of area above/below y

MPa

Max Shear Stress — Rectangle

$$ \tau_{max} = 1.5\frac{V}{A} $$

At NA; parabolic distribution

MPa

Max Shear Stress — Solid Circle

$$ \tau_{max} = \frac{4}{3}\frac{V}{A} $$

At NA

MPa

🏗

Ch 6 · Solved Example

GATE CE / ESE level

A rectangular beam (b = 100 mm, d = 200 mm) is subjected to a bending moment of 15 kN·m. Find the maximum bending stress.

Approach hint

Compute the section modulus $Z=\dfrac{bd^2}{6}$, then apply $\sigma_{max}=\dfrac{M}{Z}$.

$Z = \dfrac{bd^2}{6} = 100\dfrac{\times200^2}{6} = 666{,}667$ mm³
$\sigma_{max} = \dfrac{M}{Z} = 15\times10^6 / 666{,}667$
$\sigma_{max} = 22.5$ N/mm² (MPa)

✓ Maximum bending stress ≈ 22.5 MPa, occurring at the extreme top (compression) and bottom (tension) fibres

🏗

Ch 6 · Diagrams

Click any diagram to zoom

Fig 6 — Rectangular section: parabolic shear stress, τ_max = 1.5V/A at NA

Fig 7 — I-section: shear concentrated in web; τ_max ≈ V/A_web at NA

🏗

Ch 6 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

Bending stress in a beam is:

Bending stress $\sigma = \dfrac{My}{I}$ is proportional to y, so it's zero at the NA (y=0) and maximum at the extreme fibres ✓

Question 2

For a solid circular section, the maximum shear stress at the neutral axis is:

For a solid circular section, $\tau_{max} = \dfrac{4V}{3A}$, distinct from the rectangle's 1.5V/A ✓

Question 3

In an I-section beam, approximately what fraction of the total shear force is carried by the web?

Despite its smaller area, the web carries roughly 80–90% of the total shear force, which is why shear (web) buckling is a critical design check for plate girders ✓

🛢

Ch 7 · Thick & Thin Cylinders and Spheres

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Thin Cylinders (t/D < 1/20 or t/r < 1/10)

Internal pressure p, mean radius r, thickness t:

Hoop (circumferential) stress: $\sigma_H = \dfrac{pd}{2t} = \dfrac{pr}{t}$
Longitudinal stress: $\sigma_L = \dfrac{pd}{4t} = \dfrac{pr}{2t} = \dfrac{\sigma_H}{2}$

Volumetric strain of cylinder:

$e_v = (\dfrac{pd}{2tE})[\dfrac{5}{2}-2\nu] = (\dfrac{pd}{2tE})[\dfrac{5-4\nu}{2}]$
Hoop strain: $\epsilon_H = \dfrac{\sigma_H-\nu\sigma_L}{E} = (\dfrac{pd}{2tE})(1-\dfrac{\nu}{2})$
Longitudinal strain: $\epsilon_L = \dfrac{\sigma_L-\nu\sigma_H}{E} = (\dfrac{pd}{2tE})(\dfrac{1}{2}-\nu)$
Volumetric strain: $e_v = 2\epsilon_H+\epsilon_L = (\dfrac{pd}{2tE})(\dfrac{5}{2}-2\nu)$

Thin Spherical Shell

Both hoop and longitudinal stresses are equal:

$$ \sigma = \frac{pd}{4t} = \frac{pr}{2t} $$

Volumetric strain: $e_v = 3\epsilon = (\dfrac{3pd}{4tE})(1-\nu)$
Comparison: Sphere is more efficient — same stress in all directions; needs ~half the steel of a cylinder for same p and d

Thick Cylinders — Lamé's Equations

When t/d ≥ 1/20, stress variation through thickness is significant. Lamé's equations apply:

Radial stress: $\sigma_r = A-\dfrac{B}{r^2}$; Hoop stress: $\sigma_H = A+\dfrac{B}{r^2}$
For internal pressure $p_i$ only ($p_o=0$), $r_i$ = inner radius, $r_o$ = outer radius: $A = p_i \dfrac{r_i^2}{r_o^2-r_i^2}$, $B = p_i r_i^2 \dfrac{r_o^2}{r_o^2-r_i^2}$

Hoop stress (max at inner radius):

$$ \sigma_{H,max} = \frac{p_i(r_o^2+r_i^2)}{r_o^2-r_i^2} $$

Boundary conditions:

At $r=r_i$: $\sigma_r = -p_i$ (negative = compressive, pointing inward)
At $r=r_o$: $\sigma_r = 0$ (free outer surface, assuming no external pressure)

Compound Cylinders and Shrink-Fit

A compound cylinder (outer cylinder shrunk onto inner) introduces residual compressive hoop stress at the inner surface, allowing higher working pressure.

Junction pressure $p_j$ is found from the interference $\delta = r_j \times (e_{outer,inner}-e_{inner,outer})$
Net hoop stress = prestress (compressive) + operating pressure stress

💡 GATE Tip: For thin cylinder: Hoop stress = 2 × Longitudinal stress. This is the most frequent fact. For thick cylinder at inner surface: σ_H = p_i(r_o²+r_i²)/(r_o²−r_i²) — maximum — and σ_r = −p_i (compressive).

🛢

Ch 7 · Formulas

Thin Cylinder — Hoop Stress

$$ \sigma_H = \frac{pd}{2t} $$

= 2 × longitudinal stress

MPa

Thin Cylinder — Longitudinal Stress

$$ \sigma_L = \frac{pd}{4t} = \frac{\sigma_H}{2} $$

—

MPa

Thin Sphere — Stress

$$ \sigma = \frac{pd}{4t} $$

Equal in every direction; most efficient pressure-vessel shape

MPa

Thick Cylinder — Max Hoop Stress (Lamé)

$$ \sigma_{H,max} = \frac{p_i(r_o^2+r_i^2)}{r_o^2-r_i^2} $$

At inner radius; $\sigma_r=-p_i$ there

MPa

🛢

Ch 7 · Solved Example

GATE CE / ESE level

A thin cylindrical boiler shell has internal diameter 1.2 m, wall thickness 12 mm, and carries an internal pressure of 2 MPa. Find the hoop and longitudinal stresses.

Approach hint

Apply $\sigma_H = \dfrac{pd}{2t}$ and $\sigma_L = \dfrac{pd}{4t}$ directly with d and t in mm.

$d = 1200$ mm, $t = 12$ mm, $p = 2$ MPa
$\sigma_H = \dfrac{2\times1200}{2\times12} = \dfrac{2400}{24} = 100$ MPa
$\sigma_L = \dfrac{\sigma_H}{2} = 50$ MPa

✓ Hoop stress = 100 MPa; Longitudinal stress = 50 MPa

🛢

Ch 7 · Diagrams

Click any diagram to zoom

Fig 8 — Thin cylinder: Hoop stress σ_H = pd/(2t), Longitudinal stress σ_L = pd/(4t); σ_H = 2σ_L

Fig 9 — Thick cylinder (Lamé): Hoop stress σ_H is maximum at inner radius and decreases hyperbolically; Radial stress σ_r = −p_i at inner surface, 0 at outer

🛢

Ch 7 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

In a thin cylindrical shell under internal pressure, the relationship between hoop and longitudinal stress is:

σ_H = pd/2t and σ_L = pd/4t, so σ_H is exactly twice σ_L — the single most frequently tested fact for thin cylinders ✓

Question 2

Compared to a cylindrical shell of the same diameter and pressure, a thin spherical shell requires approximately:

The sphere's stress (pd/4t) equals the cylinder's lower longitudinal stress, so for the same maximum allowable stress it needs only about half the thickness/steel of a cylinder ✓

Question 3

In a thick cylinder under internal pressure only, the radial stress at the OUTER surface is:

With no external pressure applied, the outer surface is free, so σ_r = 0 there; σ_r = −p_i only at the inner surface ✓

🌀

Ch 8 · Torsion in Shafts & Springs

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Torsion Formula

$$ \frac{T}{J} = \frac{\tau}{r} = \frac{G\phi}{L} $$

T = applied torque (N·mm or N·m); J = polar moment of inertia of cross-section
τ = shear stress at radius r from axis; G = modulus of rigidity (shear modulus)
φ = angle of twist (radians); L = length of shaft

Maximum shear stress (at outer surface, r = R) and angle of twist:

$$ \tau_{max} = \frac{TR}{J} = \frac{T}{Z_p},\qquad \phi = \frac{TL}{GJ} $$

Polar Moment of Inertia J

Section	J	Z_p (Polar Section Modulus)
Solid circular shaft (dia D)	πD⁴/32	πD³/16
Hollow circular shaft (D, d)	π(D⁴−d⁴)/32	π(D⁴−d⁴)/(16D)
Thin-walled tube (mean radius r, thickness t)	2πr³t	2πr²t
Rectangle (non-circular, approx)	Uses St. Venant correction; J ≠ I_p exactly	—

Power Transmission

Power: $P = T\cdot\omega = T\times2\pi \dfrac{N}{60}$ (P in Watts, T in N·m, N = rpm, ω = rad/s)
$T = P\dfrac{\times60}{2\pi N} = 9550\times \dfrac{P(kW)}{N(rpm)}$ [in N·m]

Shafts in Series and Parallel

Series shafts (same torque T, different φ): $\phi_{total} = \phi_1+\phi_2 = T \dfrac{L_1}{G_1J_1} + T \dfrac{L_2}{G_2J_2}$
Parallel shafts (same φ, shared T): $T = T_1+T_2$; $\phi_1=\phi_2 \to \dfrac{T_1L_1}{G_1J_1} = \dfrac{T_2L_2}{G_2J_2}$
$\to T_1/T_2 = \dfrac{G_1J_1L_2}{G_2J_2L_1}$

Combined Bending and Torsion

For a circular shaft under BM = M and Torque = T simultaneously:

Equivalent bending moment (max normal stress, Rankine/Tresca): $M_e = \tfrac12[M+\sqrt{M^2+T^2}]$
Equivalent torque (max shear stress): $T_e = \sqrt{M^2+T^2}$ (both normal and shear stresses combined)
Design: $\sigma_{max} = M_e\dfrac{\times32}{\pi D^3} \le \sigma_{permissible}$ or $\tau_{max} = T_e\dfrac{\times16}{\pi D^3} \le \tau_{permissible}$

Close-Coiled Helical Springs

Shear stress in wire: $\tau = \dfrac{8WD}{\pi d^3}\times K_w$
Wahl's stress correction factor: $K_w = \dfrac{4C-1}{4C-4} + \dfrac{0.615}{C}$; Spring index: $C = \dfrac{D}{d}$ (D = mean coil dia, d = wire dia)
Axial deflection: $\delta = \dfrac{8WD^3n}{Gd^4} = \dfrac{WL}{GJ}$ (n = number of active coils)
Stiffness: $k = \dfrac{W}{\delta} = \dfrac{Gd^4}{8D^3n}$
Strain energy stored: $U = \dfrac{W^2D^3n}{Gd^4} = W\dfrac{\delta}{2} = \dfrac{\tau^2}{4G}\times Volume_{wire}$

Springs in Series and Parallel

Series springs (same load, deflections add): $\dfrac{1}{k_{eq}} = \dfrac{1}{k_1}+\dfrac{1}{k_2}+...$, so $\delta_{total}=\delta_1+\delta_2$
Parallel springs (same deflection, loads add): $k_{eq}=k_1+k_2+...$, so $W_{total}=W_1+W_2$

Open-Coiled and Closely Coiled Springs

Type	Helix Angle α	Loading	Stress type
Close-coiled	α ≈ 0 (small)	Axial load W	Pure torsion in wire
Open-coiled	α significant	Axial load W	Torsion + bending in wire
Close-coiled	α ≈ 0	Applied torque M	Pure bending in wire

💡 GATE Tip: For close-coiled springs under axial load, wire is in pure torsion. Key formula: δ = 8WD³n/(Gd⁴). Stiffness k ∝ Gd⁴/(D³n). Doubling D reduces k by 8×; doubling d increases k by 16×.

🌀

Ch 8 · Formulas

Torsion Equation

$$ \frac{T}{J} = \frac{\tau}{r} = \frac{G\phi}{L} $$

Core relation for all torsion problems

—

Power Transmitted by Shaft

$$ T = \frac{9550\,P(\text{kW})}{N(\text{rpm})} $$

T in N·m

N·m

Equivalent Torque (Combined Loading)

$$ T_e = \sqrt{M^2+T^2} $$

Used for max-shear-stress based design

N·m

Helical Spring Deflection

$$ \delta = \frac{8WD^3n}{Gd^4} $$

Stiffness $k = \dfrac{Gd^4}{8D^3n}$

🌀

Ch 8 · Solved Example

GATE CE / ESE level

A solid circular shaft of diameter 60 mm transmits 100 kW of power at 300 rpm. Find the maximum shear stress induced in the shaft.

Approach hint

First find torque T from the power formula, then apply $\tau_{max}=\dfrac{16T}{\pi D^3}$.

$T = 9550\dfrac{\times100}{300} = 3183.3$ N·m = $3.1833\times10^6$ N·mm
$\tau_{max} = \dfrac{16T}{\pi D^3} = 16\times3.1833\dfrac{\times10^6}{\pi\times60^3}$
$\tau_{max} = 5.093\dfrac{\times10^7}{6.786\times10^5} \approx 75.06$ MPa

✓ Maximum shear stress ≈ 75 MPa

🌀

Ch 8 · Diagrams

Click any diagram to zoom

Fig 10 — Circular shaft under torsion: shear stress zero at axis, maximum at outer surface; angle of twist φ = TL/GJ

🌀

Ch 8 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

In a circular shaft under torsion, the shear stress distribution is:

Since τ ∝ r (from T/J = τ/r), shear stress is zero at the axis and increases linearly to a maximum at the outer surface ✓

Question 2

If the wire diameter d of a close-coiled helical spring is doubled (all else constant), the stiffness k changes by a factor of:

Since $k \propto d^4$, doubling d increases k by $2^4 = 16\times$; doubling coil diameter D instead reduces k by $\dfrac{1}{2^3} = \dfrac{1}{8}$ ✓

Question 3

For springs connected in SERIES, which statement is correct?

In series, each spring carries the same load but deflects independently, so total deflection is the sum — equivalent to compliances (1/k) adding, exactly like resistors in parallel ✓

🏛

Ch 9 · Theory of Columns & Retaining Walls

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Classification of Columns

Type	Slenderness Ratio λ = L_eff/r	Failure Mode
Short column	λ < 40 (steel); λ < 12 (concrete)	Material crushing/yielding
Long / slender column	λ > 120	Elastic buckling (Euler)
Intermediate column	40 ≤ λ ≤ 120	Inelastic buckling (Johnson/Rankine)

Euler's Buckling Load

$$ P_{cr} = \frac{\pi^2 EI}{L_{eff}^2} $$

Effective length $L_{eff}$ for various end conditions:

Both ends pinned: $L_{eff}=L$
One fixed, one free: $L_{eff}=2L$
Both ends fixed: $L_{eff}=\dfrac{L}{2}$
One fixed, one pinned: $L_{eff}=\dfrac{L}{\sqrt2} \approx 0.707L$

Critical stress: $\sigma_{cr} = \dfrac{P_{cr}}{A} = \dfrac{\pi^2E}{\lambda^2}$ where $\lambda=\dfrac{L_{eff}}{r}$
r = radius of gyration = $\sqrt{\dfrac{I}{A}}$
Euler's formula valid only when $\sigma_{cr} <$ proportional limit ($\sigma_p$)

Rankine-Gordon Formula (Intermediate Columns)

$$ \frac{1}{P_{cr}} = \frac{1}{P_{crushing}} + \frac{1}{P_{Euler}} $$ $$ P_{cr} = \frac{\sigma_c\cdot A}{1+a(L_{eff}/r)^2} $$

$\sigma_c$ = crushing stress (yield or compressive strength); $a$ = Rankine's constant = $\dfrac{\sigma_c}{\pi^2E}$ [dimensionless]
For steel: $a=\dfrac{1}{7500}$ (approx); For cast iron: $a=\dfrac{1}{1600}$; For timber: $a=\dfrac{1}{3000}$

Eccentrically Loaded Columns (Secant Formula)

For a column with eccentric load P at eccentricity e:

$$ \sigma_{max} = \frac{P}{A}\left[1+\frac{ec}{k^2}\sec\left(\frac{L_{eff}}{2k}\sqrt{\frac{P}{AE}}\right)\right] $$

k = radius of gyration; c = distance from centroid to extreme fibre
Core of section (no tension): load must be within kern. Rectangular section: kern is a rhombus with diagonals b/3 and d/3. Circular section: kern is a circle of diameter D/4

Retaining Walls — Lateral Earth Pressure

Rankine's Theory

Active earth pressure coefficient: $K_a = \dfrac{1-\sin\phi}{1+\sin\phi} = \tan^2(45°-\dfrac{\phi}{2})$
Passive earth pressure coefficient: $K_p = \dfrac{1+\sin\phi}{1-\sin\phi} = \tan^2(45°+\dfrac{\phi}{2})$
$K_a\times K_p = 1$
Active pressure at depth z: $p_a = K_a\gamma z - 2c\sqrt{K_a}$ (c-φ soil); for cohesionless soil (c=0): $p_a = K_a\gamma z$
Total active force: $P_a = \tfrac12 K_a\gamma H^2$ (per unit length, acting at H/3 from base)
Total passive force: $P_p = \tfrac12 K_p\gamma H^2$ (acting at H/3 from base)

Coulomb's Theory (Wedge Theory)

$$ K_a\ (\text{Coulomb}) = \frac{\sin^2(\alpha+\phi)}{\sin^2\alpha\left(1+\sqrt{\dfrac{\sin(\phi+\delta)\sin(\phi-\beta)}{\sin(\alpha-\delta)\sin(\alpha+\beta)}}\right)^2} $$

α = inclination of back of wall to horizontal; β = inclination of backfill surface to horizontal
δ = wall friction angle (0 to 2φ/3); φ = angle of internal friction of soil
Rankine = Coulomb with α=90°, β=0°, δ=0° (smooth vertical wall, horizontal backfill)

Stability of Retaining Walls

Stability Check	Requirement	FOS
Overturning	FOS = M_restoring / M_overturning	≥ 1.5 (normal), ≥ 2.0 (earthquake)
Sliding	$FOS = \dfrac{\mu(\Sigma V)}{P_a}$ (+ passive resistance)	≥ 1.5
Bearing failure	q_max ≤ q_allowable	≥ 2–3
Eccentricity check	e = B/2 − a ≤ B/6 (for no tension)	—

💡 GATE Tip: L_eff memorisation: "Pin-Pin: Full | Fixed-Free: Double | Fixed-Fixed: Half | Fixed-Pin: Root-Half (0.707L)". Euler's formula is valid only for long columns; Rankine for intermediate. The effective length is the key input to both.

🏛

Ch 9 · Formulas

Euler's Critical Load

$$ P_{cr} = \frac{\pi^2 EI}{L_{eff}^2} $$

Valid only for long/slender columns

Rankine's Formula

$$ P_{cr} = \frac{\sigma_c A}{1+a(L_{eff}/r)^2} $$

Valid for short and intermediate columns too

Active Earth Pressure Coefficient

$$ K_a = \tan^2\left(45°-\frac{\phi}{2}\right) $$

$K_a \times K_p = 1$

—

Total Active Thrust

$$ P_a = \frac{1}{2}K_a\gamma H^2 $$

Acts at H/3 from the base

kN/m

🏛

Ch 9 · Solved Example

GATE CE / ESE level

A steel column ($E=2\times10^5$ MPa) is 4 m long with both ends pinned, and has $I = 8\times10^6$ mm⁴. Compute Euler's critical buckling load.

Approach hint

Pin-pin means $L_{eff}=L$; apply $P_{cr}=\dfrac{\pi^2EI}{L_{eff}^2}$ with consistent mm units.

$L_{eff} = L = 4000$ mm (pin-pin)
$P_{cr} = \pi^2\times2\times10^5\times8\times10^6 / 4000^2$
$P_{cr} = 1.579\dfrac{\times10^{13}}{1.6}\times10^7 \approx 986{,}960$ N ≈ 987 kN

✓ Euler's critical buckling load ≈ 987 kN

🏛

Ch 9 · Diagrams

Click any diagram to zoom

Fig 11 — Column end conditions and effective lengths: Pin-Pin L_eff=L; Fixed-Free L_eff=2L; Fixed-Fixed L_eff=L/2; Fixed-Pin L_eff=0.707L

🏛

Ch 9 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

For a column FIXED at both ends, the effective length is:

Fixed-fixed columns have the shortest effective length, L_eff = L/2, giving them the highest buckling capacity of the four standard end conditions ✓

Question 2

Euler's buckling formula is valid only when:

Euler's formula assumes purely elastic behaviour, so it's valid only if the resulting critical stress stays below the material's proportional limit — i.e., for long/slender columns ✓

Question 3

For cohesionless soil (c = 0), Rankine's active earth pressure at depth z is:

For cohesionless soil the cohesion term drops out, leaving the simple linear relation $p_a = K_a\gamma z$, increasing linearly with depth ✓

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Ch 10 · Shear Centre, Moment of Inertia & Principal Axes

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Moment of Inertia (Second Moment of Area)

$I_x = \int y^2\,dA$ (about x-axis / NA)
$I_y = \int x^2\,dA$ (about y-axis)
$I_{xy} = \int xy\,dA$ (product of inertia)
$J = I_x+I_y$ (polar moment of inertia = I_p)

Standard MI Values

Shape	I_xx (centroidal)	I_yy (centroidal)	J
Rectangle b×d	bd³/12	db³/12	bd(b²+d²)/12
Solid circle D	πD⁴/64	πD⁴/64	πD⁴/32
Triangle (base b, h)	bh³/36	bh³/48 (about median)	—
Semicircle (R)	(π/8 − 8/9π)R⁴ ≈ 0.11R⁴	πR⁴/8	—
Thin circular ring (R, t)	πR³t	πR³t	2πR³t

Parallel Axis Theorem

$$ I_x = I_{\bar x} + A\cdot d^2 $$

$I_{\bar x}$ = MI about centroidal axis parallel to x; A = area of section; d = distance between the two parallel axes

⚠ NEVER add d² alone — must be A·d². A common exam error!

Perpendicular Axis Theorem (2D)

For a planar lamina: $I_z = I_x+I_y = J$ (polar MI). This relates out-of-plane MI to in-plane MIs.

Principal Moments of Inertia and Principal Axes

Just as principal stresses are the extreme values of normal stress at a point, principal moments of inertia are the extreme (maximum and minimum) values of MI achieved by rotating the axes. The product of inertia I_xy = 0 about principal axes.

$$ I_{max}, I_{min} = \frac{I_x+I_y}{2} \pm \sqrt{\frac{(I_x-I_y)^2}{4}+I_{xy}^2} $$

Angle of principal axes: $\tan(2\theta_p) = -\dfrac{2I_{xy}}{I_x-I_y}$
Check: $I_{max}+I_{min} = I_x+I_y$ (sum invariant)
$I_{max}\times I_{min} - I_{xy}^2 = I_x\cdot I_y - I_{xy}^2$ (product invariant)

⭐ Notice the analogy: principal MI formulae are identical to principal stress formulae with σ replaced by I and τ replaced by I_xy. Mohr's circle construction works for both!

Mohr's Circle of MI

Construct exactly like Mohr's stress circle:

Plot point X = (I_x, +I_xy) and Y = (I_y, −I_xy)
Centre = ((I_x+I_y)/2, 0)
Radius $R = \sqrt{(I_x-I_y)\dfrac{^2}{4} + I_{xy}^2}$
I_max = centre + R, I_min = centre − R

Shear Centre (Flexural Centre)

The shear centre (SC) is the point in the cross-section through which the transverse shear force must pass to produce pure bending (no twisting). If load passes through SC → bending only. If load does not pass through SC → combined bending + torsion.

Shear centre location found by equating external moment of shear force about SC to sum of moment of shear flows in all elements about SC: $e = \Sigma Q_{flange}\times h/V$ (for open thin-walled sections)

Shear Centre Locations — Summary

Section	Shear Centre Location	Notes
Solid circular / hollow circular	At geometric centre	Symmetric section
I-section / H-section	At centroid (intersection of axes of symmetry)	Doubly symmetric
Channel (C-section)	Outside the web, on the axis of symmetry	$e = \dfrac{3b^2}{6b+h t_w/t_f}$
Angle section (equal legs)	At the corner of the angle	No shear in either leg passing through corner
T-section	At intersection of web and flange	On axis of symmetry
Z-section	At centroid	Point-symmetric section
Semi-circular thin ring	At 4R/π ≈ 1.27R from centre (outside)	On axis of symmetry

Unsymmetric Bending

When the bending moment is not in a plane of symmetry, or when I_xy ≠ 0 (oblique bending), the neutral axis is not perpendicular to the applied moment. The beam deflects in the direction perpendicular to the NA, which is not the direction of loading.

$$ \sigma = \frac{(M_x I_y - M_y I_{xy})y}{D} + \frac{(M_y I_x - M_x I_{xy})x}{D},\qquad D = I_x I_y - I_{xy}^2 $$

For oblique loading at angle α to principal x-axis: $\sigma = M\cos\alpha\cdot \dfrac{y}{I_x} - M\sin\alpha\cdot \dfrac{x}{I_y}$ (if I_xy = 0, i.e., x-y are principal axes)
Neutral axis angle β: $\tan\beta = (\dfrac{I_x}{I_y})\tan\alpha$
NA is NOT perpendicular to load unless $I_x=I_y$ (circular) or loading in plane of symmetry

⚠ Key ESE concept: For unsymmetric sections (angle, Z-section), the deflection is NOT in the direction of the applied load. The beam deflects in the direction of the principal axis closest to the load. Designers must account for this to prevent unexpected lateral deflection.

📍

Ch 10 · Formulas

Parallel Axis Theorem

$$ I_x = I_{\bar x} + Ad^2 $$

Must include A — a common exam trap

—

Principal Moments of Inertia

$$ I_{max,min} = \frac{I_x+I_y}{2} \pm \sqrt{\frac{(I_x-I_y)^2}{4}+I_{xy}^2} $$

Same structure as principal stress formula

—

Channel Section — Shear Centre Offset

$$ e = \frac{3b^2t_f}{6bt_f+ht_w} $$

Measured from the web centreline, outside the section

📍

Ch 10 · Solved Example

GATE CE / ESE level

A rectangular section 100 mm × 200 mm has its centroidal x-axis moved 150 mm away (parallel shift). If the centroidal $I_{\bar x} = \dfrac{bd^3}{12}$, find the moment of inertia about the new axis.

Approach hint

Compute $I_{\bar x}$ and A first, then apply the parallel axis theorem $I_x = I_{\bar x}+Ad^2$ — do not forget the A.

$I_{\bar x} = 100\dfrac{\times200^3}{12} = 6.667\times10^7$ mm⁴
$A = 100\times200 = 20{,}000$ mm²
$I_x = 6.667\times10^7 + 20{,}000\times150^2 = 6.667\times10^7 + 4.5\times10^8$
$I_x = 5.167\times10^8$ mm⁴

✓ $I_x \approx 5.167\times10^8$ mm⁴ — note the A·d² term dominates the total, showing why forgetting "A" is such a costly error

📍

Ch 10 · Diagrams

Click any diagram to zoom

Fig 12 — Channel section: shear centre SC lies outside the section, to the left of the web

Fig 13 — Shear centre: at corner for angle, at centroid for I-section & circle

📍

Ch 10 · Quick Quiz

3 questions · tap an option to reveal answer

Question 1

The Parallel Axis Theorem states:

The correct form always includes the area: $I_x = I_{\bar x} + Ad^2$; forgetting the "A" is one of the most common exam errors ✓

Question 2

For a channel (C) section, the shear centre is located:

The channel section's shear centre lies outside the physical section, offset from the web by $e = \dfrac{3b^2t_f}{6bt_f+ht_w}$ ✓

Question 3

If a transverse load is applied through the shear centre of a beam section, the beam experiences:

By definition, a load passing through the shear centre produces pure bending with no twisting; any offset from SC introduces torsion ✓

📚

Ch 11 · Quick Revision — All Formulae & Mnemonics

Cross-chapter recall sheet, mnemonics, exam-angle comparison, mini quiz and practice questions

○ Not Started

All Key Formulae — Solid Mechanics

Elastic Constants

$$ G=\frac{E}{2(1+\nu)},\; K=\frac{E}{3(1-2\nu)},\; E=\frac{9KG}{3K+G} $$

Steel: E=2×10⁵ MPa, G=0.8×10⁵ MPa, ν=0.25–0.30. Concrete: $E=5000\sqrt{f_{ck}}$, ν=0.15–0.20

—

Principal Stresses & Mohr's Circle

$$ \sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2}\pm\sqrt{\frac{(\sigma_x-\sigma_y)^2}{4}+\tau_{xy}^2} $$

$\tau_{max}=\dfrac{\sigma_1-\sigma_2}{2}$; $\tan2\theta_p=2\dfrac{\tau_{xy}}{\sigma_x-\sigma_y}$; max shear planes at 45° to principal planes

MPa

Theories of Failure (Biaxial, σ₃=0)

$$ \text{Tresca: }\sigma_1-\sigma_2=\sigma_y;\;\; \text{von Mises: }\sigma_1^2-\sigma_1\sigma_2+\sigma_2^2=\sigma_y^2 $$

$\tau_{yield}$: Tresca = $\dfrac{\sigma_y}{2}$; von Mises = $\dfrac{\sigma_y}{\sqrt3} \approx 0.577\sigma_y$ (von Mises less conservative)

—

Bending and Flexure

$$ \frac{\sigma}{y}=\frac{M}{I}=\frac{E}{R},\quad Z=\frac{I}{y_{max}} $$

Rectangle Z=bd²/6; Circle Z=πD³/32. τ=VQ/(Ib); Rect τ_max=1.5V/A; Circle τ_max=4V/3A

—

Deflection of Beams

$$ EIy''=M(x) $$

SS+central P: PL³/48EI. SS+UDL: 5wL⁴/384EI ("5 in 384"). Cantilever+P: PL³/3EI. Fixed-fixed+P: PL³/192EI

—

Thin Cylinders & Spheres

$$ \sigma_H=\frac{pd}{2t},\;\; \sigma_L=\frac{pd}{4t}=\frac{\sigma_H}{2} $$

Sphere: σ=pd/4t (equal all directions). Thick (Lamé): σ_H,max = p_i(r_o²+r_i²)/(r_o²−r_i²) at inner radius

MPa

Torsion and Springs

$$ \frac{T}{J}=\frac{\tau}{r}=\frac{G\phi}{L},\quad \tau_{max}=\frac{16T}{\pi D^3} $$

J: solid=πD⁴/32, hollow=π(D⁴−d⁴)/32. Spring: δ=8WD³n/Gd⁴, k=Gd⁴/8D³n. Series: 1/k_eq=Σ(1/k_i); Parallel: k_eq=Σk_i

—

Columns

$$ P_{cr}=\frac{\pi^2EI}{L_{eff}^2},\quad \sigma_{cr}=\frac{\pi^2E}{\lambda^2} $$

$L_{eff}$: Pin-Pin=L; Fixed-Free=2L; Fixed-Fixed=L/2; Fixed-Pin=$\dfrac{L}{\sqrt2}$. Rankine: $P_{cr}=\dfrac{\sigma_cA}{1+a\lambda^2}$

Earth Pressure

$$ K_a=\tan^2(45-\frac{\phi}{2}),\quad K_p=\tan^2(45+\frac{\phi}{2}) $$

$K_a K_p=1$. $P_a=\tfrac12 K_a\gamma H^2$ (at H/3 from base); $P_p=\tfrac12 K_p\gamma H^2$

kN/m

Shear Centre & MI

$$ I_{max,min}=\frac{I_x+I_y}{2}\pm\sqrt{\frac{(I_x-I_y)^2}{4}+I_{xy}^2} $$

Parallel axis: $I=I_{centroid}+Ad^2$ (must include A). SC at corner→angle; at centroid→I,circle,Z; outside→channel

—

Key Mnemonics

💡 Elastic constants: "G gives shape change (Shear); K keeps volume (Kompressibility); E for Everything axial"

Failure theories (increasing allowance): Rankine < Tresca < von Mises
"Rankine is Rigid (brittle); Tresca is Tough (ductile, safe); von Mises is the Most accurate"

Thin cylinder stresses: "Hoop is Half-done on top: pd/2t | Longitudinal is half of Hoop: pd/4t"

Effective lengths of columns:
"Fixed-Free = 2 (worst); Pin-Pin = 1; Fixed-Pin = $\sqrt{\dfrac{1}{2}}$; Fixed-Fixed = ½ (best)"
Or: 2L — L — 0.7L — 0.5L (decreasing → increasing strength)

Spring analogy: "Series springs are WEAK (less k); Parallel springs are STRONG (more k)"
"Springs in Series → like resistors in parallel for k"

Bending formula: $\dfrac{\sigma}{y} = \dfrac{M}{I} = \dfrac{E}{R} \to$ "σ over y = M over I = E over R → SIMON is very Emotional"

Shear centre: "When load passes through SC → Section bends cleanly; off-centre → it twists meanly"

Exam-Angle Comparison

Topic	GATE Focus	ESE Focus	SSC JE Focus
Elastic constants	Numerical: find G or K given E, ν	Derivation + numerical; 3D Hooke's law	Definitions; standard values
Mohr's circle	Principal stresses + angle; τ_max	Strain rosette; Mohr's circle of strain	Identify σ₁, σ₂; τ_max
Failure theories	Tresca vs von Mises; pure shear comparison	All 5 theories; plot on σ₁-σ₂ plane	Name and criterion only
Deflection	All 5 methods; propped cantilever	Virtual work; non-prismatic beams	Standard formulae only
Torsion	Solid/hollow shaft; power transmission	Combined bending+torsion; equivalent M and T	φ = TL/GJ formula
Columns	Euler load; effective length; Rankine	Eccentric load; secant formula; design	Types; Euler formula
Cylinders	Thin: σ_H = pd/2t; vol strain; sphere	Thick: Lamé; compound cylinder; shrink-fit	σ_H and σ_L only
Shear centre	Channel section e formula; angle at corner	Unsymmetric bending; principal axes of MI	Definition only

Mini Quiz

A material has E = 2×10⁵ MPa and ν = 0.25. Find its shear modulus G.

$G = \dfrac{E}{2(1+\nu)} = 2\dfrac{\times10^5}{2\times1.25} = 0.8\times10^5$ MPa ✓

A cantilever of length L carries a tip load P. How does its max deflection compare to that of an SS beam of the same length and EI under a central load P?

Cantilever: $\dfrac{PL^3}{3EI}$; SS beam: $\dfrac{PL^3}{48EI}$. Ratio = 48/3 = 16 — the cantilever deflects 16× more ✓

A thin cylinder (d = 1 m, t = 10 mm) carries internal pressure p = 1.5 MPa. Find the hoop stress.

$\sigma_H = \dfrac{pd}{2t} = 1.5\dfrac{\times1000}{2\times10} = \dfrac{1500}{20} = 75$ MPa ✓

A pin-ended column has Euler load $P_{cr}$. If both ends are changed to fixed, the new Euler load becomes:

$P_{cr}\dfrac{\propto1}{L_{eff}^2}$; $L_{eff}$ goes from L (pin-pin) to L/2 (fixed-fixed), so $P_{cr}$ increases by $(\dfrac{L}{L/2})^2=4\times$ ✓

Which correctly relates the parallel and perpendicular axis theorems?

Perpendicular axis theorem (planar lamina): $I_z=I_x+I_y=J$. Parallel axis theorem is the separate $I_x=I_{\bar x}+Ad^2$ relation ✓

Practice Questions

EasyState the three elastic-constant inter-relations and the typical values of E, G and ν for steel.

$G=\dfrac{E}{2(1+\nu)}$, $K=\dfrac{E}{3(1-2\nu)}$, $E=\dfrac{9KG}{3K+G}$. For steel: $E\approx2\times10^5$ MPa, $G\approx0.8\times10^5$ MPa, $\nu\approx0.25$–$0.30$.

EasyWhy is the "A" essential in the parallel axis theorem $I=I_{\bar x}+Ad^2$, and what happens if it is omitted?

The A·d² term represents the moment of inertia of the section's area treated as a point mass at distance d from the new axis — dropping A would leave dimensionally inconsistent units (d² alone is length², not length⁴) and drastically understate I, since for sections far from the centroidal axis the Ad² term typically dominates the total.

MediumA shaft transmits 50 kW at 500 rpm through a solid circular section, with allowable shear stress 60 MPa. Find the minimum required shaft diameter.

T = 9550×50/500 = 955 N·m = 955,000 N·mm. From $\tau_{max}=\dfrac{16T}{\pi D^3}$: $D^3 = 16\dfrac{\times955{,}000}{\pi\times60} = 81{,}056$ mm³, so D ≈ 43.3 mm.

MediumTwo identical helical springs, each of stiffness k, are first connected in series and then in parallel. Compare the equivalent stiffness in each case and explain which arrangement gives a stiffer (less flexible) system.

Series: $\dfrac{1}{k_{eq}}=\dfrac{1}{k}+\dfrac{1}{k}=\dfrac{2}{k} \to k_{eq}=\dfrac{k}{2}$ (half as stiff — more flexible). Parallel: $k_{eq}=k+k=2k$ (twice as stiff). Parallel gives a stiffer system since both springs share the load and deflect together, while series springs each carry the full load but add their deflections.

HardExplain, from the underlying stress state, why von Mises consistently predicts a higher allowable shear capacity than Tresca for the SAME uniaxial yield stress σ_y, and why this makes Tresca the conservative choice in design.

Tresca's criterion depends only on the difference between the maximum and minimum principal stresses (σ₁−σ₃), ignoring the intermediate principal stress σ₂ entirely. Von Mises accounts for the full triaxial stress state through a root-mean-square-like combination of all three principal stress differences, which for pure shear (σ₁=−σ₃=τ, σ₂=0) works out to allow $\tau_y=\dfrac{\sigma_y}{\sqrt3}\approx0.577\sigma_y$ versus Tresca's $\tau_y=\dfrac{\sigma_y}{2}=0.5\sigma_y$. Because Tresca's criterion is always the more restrictive lower bound (it "wraps inside" the von Mises ellipse on the σ₁-σ₂ plane), any state satisfying Tresca automatically satisfies von Mises too — making Tresca inherently safe-sided, at the cost of being ~15% more conservative (less material-efficient) than the more accurate von Mises prediction.

Property	Mild Steel	HYSD Steel (Fe415)	Concrete (M25)	Cast Iron
E (GPa)	200–210	200	25 (\(\approx 5000\sqrt{f_{ck}}\) MPa)	100–170
\(\sigma_y\) (MPa)	250	415	25 (\(f_{ck}\))	No yield
UTS (MPa)	400–500	485 min	—	140–350
Behaviour	Ductile, defined yield	Ductile, no clear yield	Brittle in tension	Brittle
Poisson's ratio ν	0.25–0.30	0.25–0.30	0.15–0.20	0.25

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Solid Mechanics — Strength of Materials

Learning Objectives

Exam Relevance

Concept Map

Where This Fits

Mechanical Properties

Key Points on the Stress–Strain Curve

Factor of Safety and Proof Stress

Material Comparison

Stress and Strain Definitions

Elastic Constants

Inter-Relations Between Elastic Constants

Generalised Hooke's Law (3D)

Thermal Stress

Composite Bars and Statically Indeterminate Axial Problems

Plane Stress and Plane Strain

Sign Convention

Standard SFD/BMD Results

SFD/BMD Shapes for Standard Loads

Point of Contraflexure

Relationship Between Load, SF, and BM — Key Rules

Stress Transformation (2D)

Mohr's Circle Construction — Step by Step

Mohr's Circle of Strain

Theories of Failure

Failure Criteria for Biaxial Stress (σ₃ = 0)

Elastic Curve Equation

Standard Deflection Formulae

Methods for Computing Deflections

Double Integration Method

Macaulay's Method (Singularity Functions)

Moment Area Method (Mohr's Theorems)

Conjugate Beam Method

Virtual Work / Unit Load Method

Flexure Formula (Bending Stress)

Section Modulus and Moment of Resistance

Shear Stress in Beams

Shear Flow in Thin-Walled Sections

Composite Beams (Transformed Section)

Flitched Beams

Thin Cylinders (t/D < 1/20 or t/r < 1/10)

Thin Spherical Shell

Thick Cylinders — Lamé's Equations

Compound Cylinders and Shrink-Fit

Torsion Formula

Polar Moment of Inertia J

Power Transmission

Shafts in Series and Parallel

Combined Bending and Torsion

Close-Coiled Helical Springs

Springs in Series and Parallel

Open-Coiled and Closely Coiled Springs

Classification of Columns

Euler's Buckling Load

Rankine-Gordon Formula (Intermediate Columns)

Eccentrically Loaded Columns (Secant Formula)

Retaining Walls — Lateral Earth Pressure

Rankine's Theory

Coulomb's Theory (Wedge Theory)

Stability of Retaining Walls

Moment of Inertia (Second Moment of Area)

Standard MI Values

Parallel Axis Theorem

Perpendicular Axis Theorem (2D)

Principal Moments of Inertia and Principal Axes

Mohr's Circle of MI

Shear Centre (Flexural Centre)

Shear Centre Locations — Summary

Unsymmetric Bending

All Key Formulae — Solid Mechanics

Key Mnemonics

Exam-Angle Comparison

Mini Quiz

Practice Questions

Related Topics & Continue Learning