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Chapter 2 of 9

Principal Stress and Theory of Failure

In the RSMSSB JE Civil (Degree) syllabus under Strength of Materials · 2 parts

Principal Stresses, Principal Planes & Mohr's Circle · Theories of Failure

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📑 Contents (19 sections)

Part 1 of 2

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.

Part 2 of 2

Theories of Failure

Last reviewed 16 Sept 2026 · 7 min read

Why theories of failure are needed

The strength of a material is measured in a simple tension test, which gives a single number — the yield stress $f_{y}$ (or ultimate stress for brittle materials). Real members are under combined stresses: a shaft under bending and torsion, a pressure vessel under hoop and longitudinal stress. A theory of failure is a rule that says which quantity (stress, strain or energy) governs failure, so that the combined state can be compared with the one-dimensional test value.

Notation below: principal stresses $σ_{1} \geq σ_{2} \geq σ_{3}$ ; yield stress in simple tension $f_{y}$ ; Poisson's ratio $μ$ . For plane stress, $σ_{3} = 0$ . Design values use $f_{y} / FOS$ in place of $f_{y}$ .

1. Maximum principal stress theory (Rankine)

Failure occurs when the maximum principal stress reaches the yield (or ultimate) stress in simple tension, or the minimum principal stress reaches the compressive strength.

σ_{1} = f_{y t} or ∣ σ_{3} ∣ = f_{y c}

Suitable for brittle materials (cast iron, concrete, glass) which fail by tension.
Unsafe for ductile materials under shear: in pure shear it predicts failure at $τ = f_{y}$ , whereas ductile materials yield at about $0.5 f_{y}$ to $0.577 f_{y}$ .
Boundary in the $σ_{1}$ – $σ_{2}$ plane: a square.

2. Maximum shear stress theory (Tresca, Guest)

Failure occurs when the maximum shear stress reaches the maximum shear stress at yield in simple tension, $f_{y} /2$ .

τ_{ma x} = \frac{σ _{ma x} - σ _{min}}{2} = \frac{f _{y}}{2} \Rightarrow σ_{1} - σ_{3} = f_{y}

For plane stress:

If $σ_{1}$ and $σ_{2}$ have opposite signs: $σ_{1} - σ_{2} = f_{y}$ .
If they have the same sign: the larger magnitude $= f_{y}$ (because $σ_{3} = 0$ enters).
Suitable for ductile materials; gives slightly conservative (safe) results.
In pure shear: yield at $τ = 0.5 f_{y}$ .
Boundary: a hexagon inscribed in the von Mises ellipse.
Basis of the equivalent twisting moment $T_{e} = M^{2} + T^{2}$ for shafts.

3. Maximum principal strain theory (St Venant)

Failure occurs when the maximum principal strain reaches the strain at yield in simple tension, $f_{y} / E$ .

σ_{1} - μ (σ_{2} + σ_{3}) = f_{y}

Gives reasonable results for some brittle materials; not reliable for ductile metals.
Predicts that a body under equal all-round tension can carry more than $f_{y}$ — and under hydrostatic compression it would predict failure, which does not happen.
Boundary: a rhombus (parallelogram) in the $σ_{1}$ – $σ_{2}$ plane.
Pure shear: yield at $τ = f_{y} / (1 + μ)$ .

4. Total strain energy theory (Haigh)

Failure occurs when the total strain energy per unit volume reaches the strain energy per unit volume at yield in simple tension, $f_{y}^{2} /2 E$ .

σ_{1}^{2} + σ_{2}^{2} + σ_{3}^{2} - 2 μ (σ_{1} σ_{2} + σ_{2} σ_{3} + σ_{3} σ_{1}) = f_{y}^{2}

Boundary: an ellipse.
Weakness: under hydrostatic pressure a material stores large strain energy but does not yield, so the theory predicts failure where none occurs.
Pure shear: yield at $τ = f_{y} / 2 (1 + μ)$ .

5. Shear strain energy (distortion energy) theory — von Mises, Hencky

Strain energy is split into a part that changes volume and a part that changes shape (distortion). Only distortion causes yielding.

Failure occurs when the distortion energy per unit volume reaches the distortion energy at yield in simple tension.

Formulavon Mises criterion

\frac{1}{2} [(σ_{1} - σ_{2})^{2} + (σ_{2} - σ_{3})^{2} + (σ_{3} - σ_{1})^{2}] = f_{y}^{2}

Plane stress ( $σ_{3} = 0$ ): $σ_{1}^{2} + σ_{2}^{2} - σ_{1} σ_{2} = f_{y}^{2}$

In terms of $σ$ and $τ$ (beam or shaft): $σ^{2} + 3 τ^{2} = f_{y}$

Best agreement with experiments for ductile materials.
Pure shear: yield at $τ = f_{y} / 3 = 0.577 f_{y}$ .
Boundary: an ellipse with axes at 45°, circumscribing the Tresca hexagon.
No failure under hydrostatic stress — consistent with experiments.
The octahedral shear stress theory gives the identical criterion.

Code ProvisionIS 800:2007 — combined stresses in welds

Where a weld carries both normal stress $f_{a}$ and shear $q$ , IS 800 checks the equivalent stress $f_{e} = f_{a}^{2} + 3 q^{2}$ — the von Mises (distortion energy) form. Objective questions test this expression rather than a clause number.

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