← Engineering Mechanics, SOM & Structural Analysis · AAI Manager (Civil)

Chapter 2 of 13

Strength of Materials

In the AAI Manager (Civil) syllabus under Engineering Mechanics, SOM & Structural Analysis · 3 parts

Simple Stress & Strain · Elastic Constants · Principal Stresses, Principal Planes & Mohr's Circle

CIVILGYAN TEST SERIES · civilgyantests.com — free study notes for civil engineering exams

📑 Contents (25 sections)

Part 1 of 3

Simple Stress & Strain

Last reviewed 16 Sept 2026 · 10 min read

Load, stress and strain

When external forces act on a body, the material develops internal resisting forces. The internal resisting force per unit area is stress.

σ = \frac{P}{A}

where $P$ is the axial load and $A$ the cross-sectional area normal to it. The SI unit is N/m² (pascal); in engineering practice N/mm² = MPa is used. $1 N/mm^{2} = 1 MPa = 1 0^{6} Pa$ .

Strain is the deformation per unit original dimension. It has no unit.

ε = \frac{δ L}{L}

DefinitionTypes of stress

Normal (direct) stress — acts perpendicular to the section. Tensile when it lengthens the member, compressive when it shortens it.
Shear stress $τ = Q / A$ — acts parallel (tangential) to the section, as in a rivet or bolt resisting sliding of plates.
Bearing (crushing) stress $= P / (d t)$ — contact pressure between a bolt/rivet and the plate hole, on the projected area.
Simple stress — stress caused by a single direct load, uniformly spread over the section (this chapter).

DefinitionTypes of strain

Longitudinal (linear) strain $ε = δ L / L$ — along the load.
Lateral strain $ε_{l a t} = δ d / d$ — perpendicular to the load, of opposite sign.
Shear strain $γ$ — change in a right angle (radians).
Volumetric strain $ε_{v} = δ V / V$ — change in volume per unit volume (see Elastic Constants).

Hooke's law and modulus of elasticity

Within the limit of proportionality, stress is proportional to strain:

σ = E ε, τ = G γ

$E$ is Young's modulus (modulus of elasticity) and $G$ is the modulus of rigidity. They have the units of stress. A higher $E$ means a stiffer material.

Material	Typical $E$ (GPa)
Structural steel	200 (IS 800 takes $2 \times 1 0^{5}$ N/mm²)
Cast iron	100 – 150
Copper	about 110
Aluminium alloys	about 70
Timber (along grain)	8 – 14
Concrete	$5000 f_{c k}$ N/mm² as per IS 456 (short-term)
Rubber	a few MPa only

Stress–strain curve of mild steel

A tension test on a mild-steel specimen gives the classic curve below. Each point is a favourite exam question.

FigureTensile stress–strain curve for mild steel

A: limit of proportionality (elastic limit just beyond it) · B/C: upper and lower yield points · D: ultimate stress · E: breaking point (nominal stress drops because of necking).

O–A, proportional limit — straight line; Hooke's law holds.
Elastic limit — highest stress up to which the material recovers fully on unloading. For mild steel it is very close to A.
Yield point (B, C) — the material elongates with little or no increase in load. The upper yield point depends on test conditions; the lower yield point is taken as the yield strength $f_{y}$ .
Strain hardening (C–D) — the material gains strength with further strain.
Ultimate stress (D) — maximum load ÷ original area. Necking starts here.
Breaking point (E) — fracture. Nominal stress falls because the load is divided by the original area, although the true stress (load ÷ actual area) keeps rising.

DefinitionProof stress

Materials without a clear yield point — high-strength deformed (HYSD) bars, aluminium, copper — are given a 0.2% proof stress: the stress at which a line parallel to the initial slope, drawn from 0.2% strain, cuts the curve. IS 456 uses 0.2% proof stress as the characteristic strength of cold-worked bars.

Ductile and brittle behaviour

Ductile (mild steel, copper, aluminium)	Brittle (cast iron, concrete, glass)
Large plastic deformation before fracture	Little or no plastic deformation
Percentage elongation usually > 5%	Elongation < 5%
Cup-and-cone fracture in tension	Flat, granular fracture
Clear warning before failure	Sudden failure
Strong in tension and compression alike	Much stronger in compression than in tension

Percentage elongation $= \frac{L _{f} - L _{0}}{L _{0}} \times 100$ and percentage reduction in area $= \frac{A _{0} - A _{f}}{A _{0}} \times 100$ measure ductility.

RememberMechanical properties in one line each

Ductility — can be drawn into wire (large plastic strain in tension).
Malleability — can be hammered into sheets (plastic strain in compression).
Toughness — energy absorbed up to fracture (area under the whole curve).
Resilience — energy stored up to the elastic limit (recoverable).
Hardness — resistance to indentation or scratching.
Creep — slow increase of strain under constant load with time.
Fatigue — failure under repeated or reversed loading below the static strength.

Working stress and factor of safety

A member is not allowed to reach its failure stress. The permissible (working) stress is

σ_{w or k in g} = \frac{failure stress}{factor of safety}

For ductile materials the failure stress is usually taken as the yield stress; for brittle materials, the ultimate stress. A factor of safety covers uncertainty in loads, material strength, workmanship and analysis.

Deformation of axially loaded bars

Uniform bar

δ L = \frac{P L}{A E}

The quantity $A E$ is the axial rigidity and $A E / L$ the axial stiffness (load per unit elongation).

Bars of several segments (stepped bars)

Find the internal force in each segment from a free-body diagram, then add the elongations:

δ L = \sum \frac{P _{i} L _{i}}{A _{i} E _{i}}

Principle of superposition — when several loads act, the total deformation is the algebraic sum of the deformations caused by each load acting alone (valid within the elastic range and small deformations).

Tapering circular bar

For a bar whose diameter varies linearly from $d_{1}$ to $d_{2}$ over length $L$ :

δ L = \frac{4 P L}{π E d _{1} d _{2}}

Exam TipQuick check

If $d_{1} = d_{2} = d$ the formula becomes $4 P L / (π E d^{2}) = P L / A E$ — the uniform-bar result. Also note the tapered bar does not stretch the same as a uniform bar of mean diameter; it stretches as a bar of diameter $d_{1} d_{2}$ .

Tapering rectangular bar

Width varying linearly from $b_{1}$ to $b_{2}$ , constant thickness $t$ :

δ L = \frac{P L}{E t ( b _{1} - b _{2} )} ln \frac{b _{1}}{b _{2}}

Bar hanging under its own weight

For a prismatic bar of length $L$ , density $ρ$ (unit weight $γ = ρ g$ ):

δ L = \frac{ρ g L ^{2}}{2 E} = \frac{W L}{2 A E}

where $W$ is the total weight. The bar elongates half as much as if its whole weight hung at the free end. The stress is zero at the free end and maximum, $ρ g L$ , at the support.

For a conical bar hanging from its base: $δ L = \frac{ρ g L ^{2}}{6 E}$ .

Bar of uniform strength

To keep the stress constant at $σ$ everywhere in a vertical bar carrying a load and its own weight, the area must increase towards the support:

A_{x} = A_{0} e^{ρ g x / σ}

where $A_{0}$ is the area at the loaded (lower) end and $x$ is measured upward from it.

Part 2 of 3

Elastic Constants

Last reviewed 16 Sept 2026 · 6 min read

The four elastic constants

For a homogeneous, isotropic material loaded within the elastic range, four constants describe how it deforms:

Constant	Definition	Symbol
Young's modulus	normal stress ÷ linear strain	$E = σ / ε$
Modulus of rigidity (shear modulus)	shear stress ÷ shear strain	$G$ or $C$ or $N$ $= τ / γ$
Bulk modulus	direct stress (equal in all directions) ÷ volumetric strain	$K = σ / ε_{v}$
Poisson's ratio	lateral strain ÷ linear strain (magnitude)	$μ$ or $ν$ $= - ε_{l a t} / ε$

Only two of them are independent for an isotropic material. Knowing any two, the other two follow from the relations derived below.

RememberNumber of independent elastic constants

Isotropic material: 2
Orthotropic material (e.g. timber, laminated composites): 9
General anisotropic material: 21

Poisson's ratio

When a bar is stretched it becomes longer and thinner. The ratio of the lateral contraction strain to the axial extension strain is constant within the elastic range:

μ = \frac{lateral strain}{longitudinal strain} = \frac{δ d / d}{δ L / L}

Material	Typical Poisson's ratio
Steel	0.25 – 0.30 (0.3 commonly used)
Aluminium	about 0.33
Cast iron	0.21 – 0.26
Concrete	0.15 – 0.20 (IS 456 uses 0.2 for elastic analysis)
Rubber	nearly 0.5
Cork	nearly 0
Perfectly plastic material (volume constant)	0.5

Exam TipWhy cork is used as a bottle stopper

Cork has $μ \approx 0$ : pushing it into the neck does not make it bulge sideways, so it goes in and stays sealed. Rubber, with $μ \approx 0.5$ , bulges and grips.

Limits of Poisson's ratio

From $E = 3 K (1 - 2 μ)$ , since $E$ and $K$ are positive, $1 - 2 μ > 0$ → $μ < 0.5$ . From $E = 2 G (1 + μ)$ , $μ > - 1$ . So theoretically

- 1 < μ < 0.5

For ordinary engineering materials $0 \leq μ \leq 0.5$ . At $μ = 0.5$ the material is incompressible ( $K \to \infty$ , no volume change).

Strains in three dimensions (generalised Hooke's law)

When normal stresses $σ_{x}$ , $σ_{y}$ , $σ_{z}$ act together (tension positive), each produces its own strain along its direction and lateral contraction in the other two:

ε_{x} = \frac{σ _{x}}{E} - μ \frac{σ _{y}}{E} - μ \frac{σ _{z}}{E}

ε_{y} = \frac{σ _{y}}{E} - μ \frac{σ _{z}}{E} - μ \frac{σ _{x}}{E}

ε_{z} = \frac{σ _{z}}{E} - μ \frac{σ _{x}}{E} - μ \frac{σ _{y}}{E}

Volumetric strain

Volumetric strain is the sum of the three linear strains (for small strains):

ε_{v} = \frac{δ V}{V} = ε_{x} + ε_{y} + ε_{z} = \frac{( σ _{x} + σ _{y} + σ _{z} ) ( 1 - 2 μ )}{E}

FormulaVolumetric strain — standard cases

Bar under axial stress $σ$ : $ε_{v} = \frac{σ}{E} (1 - 2 μ)$
Rectangular block: $ε_{v} = \frac{δ L}{L} + \frac{δ b}{b} + \frac{δ t}{t}$
Cylindrical rod: $ε_{v} = \frac{δ L}{L} + 2 \frac{δ d}{d}$
Sphere: $ε_{v} = 3 \frac{δ d}{d}$
Equal stress $σ$ in all three directions: $ε_{v} = \frac{3 σ ( 1 - 2 μ )}{E}$

Relation between E, K and μ

Apply equal tensile stress $σ$ on all faces of a cube. Each linear strain is

ε = \frac{σ}{E} (1 - 2 μ)

so $ε_{v} = 3 ε = \frac{3 σ ( 1 - 2 μ )}{E}$ . By definition $K = σ / ε_{v}$ , therefore

E = 3 K (1 - 2 μ)

Part 3 of 3

Principal Stresses, Principal Planes & Mohr's Circle

Last reviewed 16 Sept 2026 · 7 min read

Why stresses on inclined planes matter

The stress on a plane depends on the orientation of that plane. A bar in simple tension has no shear stress on its cross-section, yet it has shear on planes inclined to the axis — which is why ductile bars can fail by shear along 45° lines. Design against failure therefore needs the largest normal and shear stresses at a point, whatever planes they act on.

DefinitionTerms

Principal planes — planes on which the shear stress is zero.
Principal stresses — the normal stresses on principal planes; they are the maximum and minimum normal stresses at the point.
Planes of maximum shear — inclined at 45° to the principal planes.
Obliquity — angle between the resultant stress on a plane and the normal to that plane.

Sign convention used here: tensile normal stress positive; shear stress positive when it tends to rotate the element clockwise on the $x$ -face as drawn in most Indian textbooks. The plane angle $θ$ is measured from the plane on which $σ_{x}$ acts (equivalently, the normal of the inclined plane is at $θ$ from the $x$ -axis). Examiners accept any consistent convention; the magnitudes do not change.

Case 1 — uniaxial stress

A bar under direct stress $σ$ . On a plane whose normal makes angle $θ$ with the axis:

σ_{n} = σ cos^{2} θ = \frac{σ}{2} (1 + cos 2 θ), τ = \frac{σ}{2} sin 2 θ

$σ_{n}$ is maximum ( $= σ$ ) at $θ = 0$ — the cross-section.
$τ$ is maximum ( $= σ /2$ ) at $θ = 45°$ , where $σ_{n} = σ /2$ as well.

Exam TipExam favourite

In simple tension or compression, maximum shear stress $= σ /2$ on planes at 45°. That explains the 45° shear failure of a short cast-iron cylinder in compression and the cup-and-cone fracture of mild steel.

Case 2 — two perpendicular normal stresses (biaxial)

$σ_{x}$ and $σ_{y}$ act without shear:

σ_{n} = \frac{σ _{x} + σ _{y}}{2} + \frac{σ _{x} - σ _{y}}{2} cos 2 θ, τ = \frac{σ _{x} - σ _{y}}{2} sin 2 θ

Principal stresses are $σ_{x}$ and $σ_{y}$ themselves.
$τ_{ma x} = \frac{σ _{x} - σ _{y}}{2}$ at 45°.
If $σ_{x} = σ_{y}$ (equal biaxial, like a thin sphere), $τ = 0$ on every plane.

Case 3 — pure shear

Only $τ_{x y}$ acts. Then $σ_{n} = τ_{x y} sin 2 θ$ and the principal stresses are $\pm τ_{x y}$ on planes at 45°. A shaft in torsion is in this state.

Case 4 — general two-dimensional stress

With $σ_{x}$ , $σ_{y}$ and $τ_{x y}$ :

FormulaStresses on an inclined plane

σ_{n} = \frac{σ _{x} + σ _{y}}{2} + \frac{σ _{x} - σ _{y}}{2} cos 2 θ + τ_{x y} sin 2 θ

τ_{t} = \frac{σ _{x} - σ _{y}}{2} sin 2 θ - τ_{x y} cos 2 θ

The sum of normal stresses on any two perpendicular planes is constant: $σ_{n} + σ_{n}^{'} = σ_{x} + σ_{y}$ (first stress invariant).

Principal planes

Setting $τ_{t} = 0$ :

tan 2 θ_{p} = \frac{2 τ _{x y}}{σ _{x} - σ _{y}}

This gives two values of $θ_{p}$ , 90° apart — the two principal planes.

Principal stresses

FormulaPrincipal stresses and maximum shear

σ_{1, 2} = \frac{σ _{x} + σ _{y}}{2} \pm (\frac{σ _{x} - σ _{y}}{2})^{2} + τ_{x y}^{2}

τ_{ma x} = (\frac{σ _{x} - σ _{y}}{2})^{2} + τ_{x y}^{2} = \frac{σ _{1} - σ _{2}}{2}

On the planes of maximum shear the normal stress is $\frac{σ _{x} + σ _{y}}{2}$ (not zero, unless $σ_{x} = - σ_{y}$ ). Maximum-shear planes are at 45° to the principal planes.

Common MistakeIn-plane vs absolute maximum shear

$(σ_{1} - σ_{2}) /2$ is the maximum in-plane shear. In a real 3-D body the third principal stress (often $σ_{3} = 0$ on a free surface) matters: if $σ_{1}$ and $σ_{2}$ have the same sign, the absolute maximum shear is $σ_{1} /2$ (taking $σ_{1}$ as the larger magnitude). Thin pressure vessels are the classic case.

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