Define normal stress and shear stress. Give SI units.

Normal stress acts perpendicular to a section; shear stress (or ) acts tangent to the section. Both have SI unit or (often in SOM).

What is the difference between engineering stress and true stress?

Engineering stress uses original area:. True stress uses instantaneous area:. After necking, true stress keeps rising while engineering stress falls.

Define longitudinal and lateral strain. How are they related by Poisson’s ratio?

Longitudinal strain; lateral strain. Poisson’s ratio (negative sign because lateral strain is opposite in sense for uniaxial tension).

State Hooke’s law in uniaxial form and the range of validity.

within the elastic (proportional) limit. Beyond that, stress–strain is nonlinear; unloading may leave permanent set.

What is volumetric strain? Express it for a cube under hydrostatic pressure.

Volumetric strain. Under hydrostatic pressure,, where is the bulk modulus.

Define modulus of elasticity, modulus of rigidity, and bulk modulus.

(direct stress/strain); (shear); (volumetric). All have units of stress.

What is factor of safety? How is allowable stress obtained?

Factor of safety (FoS). Allowable (working) stress (brittle) or (ductile design).

Back to Strength of Materials (SOM)

Study notes

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Qwestrum Engineering360 · Mechanical Engineering · Strength of Materials (SOM)

Stress and Strain Basics

Stress is internal force intensity

\sigma=P/A

(normal) or

\tau=V/A

(shear). Strain is relative deformation

\varepsilon=\delta/L

\gamma=\delta_s/L

. The engineering stress–strain curve (RK Bansal / GATE SOM) defines elastic limit, yield, UTS, and fracture; true stress–strain accounts for instantaneous area.

Exam tip: keep units consistent — N with mm² gives MPa directly (1 MPa = 1 N/mm²).

Key formulas & points

Skim these first — then read the full notes below.

Normal stress: $\sigma=\dfrac{P}{A}$ — tensile (positive) or compressive; SI unit $\mathrm{N/m^2}=\mathrm{Pa}$ (often MPa).
Shear stress: $\tau=\dfrac{V}{A}$ on a plane parallel to the force.
Normal strain: $\varepsilon=\dfrac{\delta}{L}=\dfrac{\Delta L}{L}$ (dimensionless); shear strain $\gamma\approx\tan\theta$ (or $\gamma\approx\theta$ in radians) for small angles.
Hooke’s region: $\sigma=E\varepsilon$ until proportional limit; $E$ = Young’s modulus.
Engineering vs true: $\sigma_{\mathrm{eng}}=P/A_0$ (original area), $\sigma_{\mathrm{true}}=P/A_{\mathrm{inst}}$ (instantaneous area), $\varepsilon_{\mathrm{true}}=\ln(L/L_0)$ .
Poisson’s ratio: $\nu=-\varepsilon_{\mathrm{lat}}/\varepsilon_{\mathrm{long}}$ (typically $0.25$ – $0.33$ for metals).
Factor of safety: $\mathrm{FoS}=\sigma_{\mathrm{allow}}/\sigma_{\mathrm{working}}$ (or based on yield/UTS as specified).

Topic details

Definition and physical meaning

Stress is the intensity of internal forces that develop within a body to resist external loads. At a cut section, resolve the resultant into a normal component

P

(perpendicular to the section) and a shear (tangential) component

V

Normal (direct) stress

\sigma=\frac{P}{A}

- Tension: fibres elongate; conventionally

\sigma>0

.
- Compression: fibres shorten;

\sigma<0

(or reported as compressive magnitude).

Shear stress

\tau=\frac{V}{A}

where

A

is the area of the plane parallel to

V

Strain measures deformation relative to original size:

\varepsilon=\frac{\delta}{L}=\frac{L-L_0}{L_0}\qquad\text{(normal)}

\gamma=\frac{\delta_s}{L}\approx\theta\quad(\theta\text{ in radians, small)}

Physical parameters

Symbol	Meaning	SI unit
$P$ , $V$	Axial / shear force	$\mathrm{N}$
$A$	Cross-sectional area	$\mathrm{m^2}$
$\sigma$ , $\tau$	Normal / shear stress	$\mathrm{Pa}$ ( $\mathrm{N/m^2}$ )
$\delta$ , $L$	Elongation / length	$\mathrm{m}$
$\varepsilon$ , $\gamma$	Normal / shear strain	— (dimensionless)
$E$	Young’s modulus	$\mathrm{Pa}$
$\nu$	Poisson’s ratio	—

Practical unit:

1\,\mathrm{MPa}=10^6\,\mathrm{Pa}=1\,\mathrm{N/mm^2}

Fig 10.1 — Normal stress σ acts perpendicular to the cut area; shear stress τ acts parallel. Bending σ is linear in y; torsion τ is zero at the centre and maximum at the outer fibre.

Schematic diagram for study — aligned with standard B.Tech / GATE syllabus.

Core assumptions (state these in exams)

1. Continuum — material is continuous; stress/strain defined at a point as averages over a small area/volume.
2. Homogeneous and isotropic unless anisotropy is stated (composites, wood).
3. Uniform stress on the section for simple axial/shear formulas (Saint-Venant: valid away from load application points).
4. Small deformations — geometry based on undeformed dimensions (engineering stress/strain).
5. Plane sections remain plane for elementary bar theory.
6. Quasi-static loading — inertia neglected; no wave propagation.
7. Temperature constant unless thermal strain is included separately.

If the bar is tapered or stepped,

\sigma=P/A(x)

varies; integrate for elongation rather than using a single

A

Engineering stress–strain curve

A standard tensile test plots engineering stress

\sigma=P/A_0

versus engineering strain

\varepsilon=\delta/L_0

Key points on the curve (mild steel)
1. Proportional limit — end of linear

\sigma

–

\varepsilon

; Hooke’s law holds.
2. Elastic limit — unloading returns to zero permanent set (nearly coincides with proportional limit for many metals).
3. Yield point (upper/lower for mild steel) — large strain at nearly constant stress;

\sigma_y=P_y/A_0

.
4. Ultimate tensile strength (UTS) — maximum engineering stress;

\sigma_u=P_{\max}/A_0

.
5. Fracture / breaking point — specimen separates; engineering stress falls after necking because

A_{0}

is fixed while

P

drops.

Ductile vs brittle
- Ductile (mild steel, Al): large plastic strain, clear yield, necking.
- Brittle (cast iron, concrete in tension): little plasticity; fracture near elastic limit.

True stress and true strain

\sigma_{\mathrm{true}}=\frac{P}{A_{\mathrm{inst}}}\qquad\varepsilon_{\mathrm{true}}=\int_{L_0}^{L}\frac{dL}{L}=\ln\left(\frac{L}{L_0}\right)

Before necking (volume constancy

A L = A_{0} L_{0}

\sigma_{\mathrm{true}}=\sigma_{\mathrm{eng}}(1+\varepsilon_{\mathrm{eng}})

Fig 1.1 — Tension test curve, mild steel (ductile)

O → A: Proportional limit — σ = Eε, slope = E (Young's modulus).
B: Elastic limit — max stress with full recovery on unloading.
B → C: Yield plateau — upper/lower yield (mild steel).
C → D: Strain hardening — stress rises to ultimate σᵤ.
D: Ultimate tensile strength — necking begins.
E: Fracture point — specimen separates.

Schematic diagram for study — aligned with standard B.Tech / GATE syllabus.

Types of loading and stress states

Axial (tensile/compressive): uniform

\sigma=P/A

on a transverse section (ideal bar).

Shear (single/double): rivet or pin;

\tau=P/A

\frac{P}{(2A)}

for double shear.

Bearing / crushing: contact pressure

\sigma_b=P/(d\cdot t)

for a pin in a plate of thickness

t

Bending: normal stress varies linearly over depth (flexure formula — separate module).

Torsion: shear stress varies with radius in circular shafts.

Combined loading: use superposition in the elastic range; principal stresses via Mohr’s circle when both

\sigma

and

\tau

act.

Volumetric strain (small strain, isotropic):

\varepsilon_v=\varepsilon_x+\varepsilon_y+\varepsilon_z

For uniaxial stress with Poisson effect:

\varepsilon_v=\varepsilon(1-2\nu)

Bulk modulus relates hydrostatic pressure to volumetric strain:

K=-p/\varepsilon_v

Poisson’s ratio and lateral strain

When a bar is stretched longitudinally, it contracts laterally:

\nu=-\frac{\varepsilon_{\mathrm{lateral}}}{\varepsilon_{\mathrm{longitudinal}}}

For uniaxial tension

\sigma_x

(others free):

\varepsilon_x=\frac{\sigma_x}{E},\qquad\varepsilon_y=\varepsilon_z=-\nu\frac{\sigma_x}{E}

Limits: theoretically

-1<\nu<0.5

for isotropic stable solids; metals

\nu\approx 0.25

–

0.33

; rubber

\approx 0.5

(nearly incompressible).

Exam note: If

\nu=0.5

and plastic incompressible flow is assumed,

\varepsilon_v=0

Step-by-step problem approach

1. Identify load type: axial, shear, bearing, or combined.
2. Draw FBD; find internal

P

V

on the critical section.
3. Compute area carefully —

A=\pi d^2/4

(not

\pi r^2

with

d

), hollow, or net area after holes.
4.

\sigma=P/A

\tau=V/A

; convert to MPa.
5. For deformation:

\delta=PL/(AE)

in the elastic range (Hooke).
6. On stress–strain questions: locate the described point (yield, UTS, fracture).
7. Check units: N and mm² → MPa directly; N and m² → Pa, then ÷

10^6

for MPa.
8. State assumptions (uniform stress, small strain, elastic).

Common mistakes in exams

• Using radius in

A=\pi d^2/4

, or diameter in

A=\pi r^2

, or forgetting the

/4

.
• Confusing engineering stress with true stress after necking.
• Mixing N/mm² and N/m² (factor

10^6

).
• Applying

\sigma=P/A

at a stress concentration without a factor

K_{t}

.
• Taking double-shear area as

A

instead of

2A

.
• Reporting strain with units (strain is dimensionless).
• Using ultimate stress as allowable without dividing by FoS.

Calculator

Axial stress

\sigma = \frac{P}{A}

Axial load PNArea Amm²

Result

100N/mm² (MPa)

σ = P/A = 50000 / 500 = 100 N/mm²

Worked examples

Try the problem first — open the solution when you are ready to check.

Axial stress and strain in a steel rod

Problem

A steel rod has diameter

d=20\,\mathrm{mm}

, length

L=1.5\,\mathrm{m}

, and carries axial tension

P=40\,\mathrm{kN}

. Take

E=200\,\mathrm{GPa}

. Find stress

\sigma

, strain

\varepsilon

, and elongation

\delta

Solution

Formulas (Indian textbook notation)

$A=\frac{\pi d^2}{4}=\frac{\pi(20)^2}{4}=314.16\,\mathrm{mm^2}$

Formulas (Indian textbook notation)

$\sigma=\frac{P}{A}=\frac{40\times 10^3}{314.16}=127.3\,\mathrm{N/mm^2}=127.3\,\mathrm{MPa}$

Formulas (Indian textbook notation)

$\varepsilon=\frac{\sigma}{E}=\frac{127.3}{200\times 10^3}=6.365\times 10^{-4}$

Formulas (Indian textbook notation)

$\delta=\varepsilon L=(6.365\times 10^{-4})(1500)=0.955\,\mathrm{mm}$

Alternatively:

\delta=\dfrac{PL}{AE}=\dfrac{(40\times 10^3)(1500)}{314.16\times(200\times 10^3)}=0.955\,\mathrm{mm}

Single vs double shear in a pin

Problem

A pin of diameter

d=12\,\mathrm{mm}

connects a clevis joint transmitting

P=18\,\mathrm{kN}

. Find shear stress for (a) single shear and (b) double shear.

Reading the stress–strain curve

Problem

A mild-steel specimen has

A_0=100\,\mathrm{mm^2}

. Yield load

P_y=25\,\mathrm{kN}

, maximum load

P_{\max}=40\,\mathrm{kN}

, fracture load

P_f=32\,\mathrm{kN}

. Gauge length

L_0=50\,\mathrm{mm}

; elongation at fracture

\delta_f=12\,\mathrm{mm}

. Find

\sigma_y

\sigma_u

, and fracture strain.

Practice questions

Most-asked interview and GATE questions for this topic — expand any item for a model answer.

1
Define normal stress and shear stress. Give SI units.
Model answer
Normal stress $\sigma=P/A$ acts perpendicular to a section; shear stress $\tau=V/A$ (or $\frac{F}{A}$ ) acts tangent to the section. Both have SI unit $\mathrm{N/m^2}$ or $\mathrm{Pa}$ (often $\mathrm{MPa}$ in SOM).
2
What is the difference between engineering stress and true stress?
Model answer
Engineering stress uses original area $A_{0}$ : $\sigma_{\mathrm{eng}}=P/A_0$ . True stress uses instantaneous area $A$ : $\sigma_{\mathrm{true}}=P/A$ . After necking, true stress keeps rising while engineering stress falls.
3
Define longitudinal and lateral strain. How are they related by Poisson’s ratio?
Model answer
Longitudinal strain $\varepsilon_l=\delta L/L$ ; lateral strain $\varepsilon_{\mathrm{lat}}=\delta d/d$ . Poisson’s ratio $\nu=-\varepsilon_{\mathrm{lat}}/\varepsilon_l$ (negative sign because lateral strain is opposite in sense for uniaxial tension).
4
State Hooke’s law in uniaxial form and the range of validity.
Model answer
$\sigma=E\varepsilon$ within the elastic (proportional) limit. Beyond that, stress–strain is nonlinear; unloading may leave permanent set.
5
What is volumetric strain? Express it for a cube under hydrostatic pressure.
Model answer
Volumetric strain $\varepsilon_v=\Delta V/V=\varepsilon_x+\varepsilon_y+\varepsilon_z$ . Under hydrostatic pressure $p$ , $\varepsilon_v=-p/K$ , where $K$ is the bulk modulus.
6
Define modulus of elasticity, modulus of rigidity, and bulk modulus.
Model answer
$E=\sigma/\varepsilon$ (direct stress/strain); $G=\tau/\gamma$ (shear); $K=-p/(\Delta V/V)$ (volumetric). All have units of stress ( $\mathrm{Pa}$ ).
7
Write the relation connecting $E$ , $G$ , and $\nu$ .
Model answer
$E=2G(1+\nu)$ . Also $E=3K(1-2\nu)$ and $E = \frac{9KG}{(3K+G)}$ .
8
What is factor of safety? How is allowable stress obtained?
Model answer
Factor of safety (FoS) $=\sigma_{\mathrm{failure}}/\sigma_{\mathrm{allowable}}$ . Allowable (working) stress $=\sigma_{\mathrm{ultimate}}/\mathrm{FoS}$ (brittle) or $\sigma_{\mathrm{yield}}/\mathrm{FoS}$ (ductile design).
9
Differentiate ductile and brittle materials using the stress–strain curve.
Model answer
Ductile materials show large plastic strain before fracture (clear yield, necking). Brittle materials fail with little plastic deformation; fracture stress is close to ultimate with small elongation.
10
What is resilience and toughness?
Model answer
Resilience: ability to absorb energy in the elastic range (area under $\sigma$ – $\varepsilon$ up to elastic limit). Toughness: total energy to fracture (entire area under the curve to rupture).
11
Explain proof resilience and modulus of resilience.
Model answer
Proof resilience is maximum elastic strain energy stored in a body. Modulus of resilience is proof resilience per unit volume: $U_r=\sigma_y^2/(2E)$ for linear elastic behaviour up to yield.
12
What is complementary shear stress? Why must it exist?
Model answer
On mutually perpendicular planes, shear stresses are equal in magnitude and form a couple pair. Equilibrium of moments on a differential element requires complementary shear; otherwise the element would spin.
13
Define principal stresses. What is the shear stress on a principal plane?
Model answer
Principal stresses are the maximum and minimum normal stresses on planes where shear stress is zero. On a principal plane, $\tau=0$ by definition.
14
State the maximum shear stress in terms of principal stresses for plane stress.
Model answer
$\tau_{\max}=(\sigma_1-\sigma_2)/2$ in the plane. If $\sigma_1$ and $\sigma_2$ have opposite signs, this is also the absolute maximum; if both same sign, compare with $\sigma_1/2$ or $\sigma_2/2$ (out-of-plane).
15
What is thermal stress? When does it develop?
Model answer
Thermal strain $\varepsilon_T=\alpha\Delta T$ . If free expansion is fully/partially constrained, stress develops: for full constraint $\sigma=E\alpha\Delta T$ (compressive on heating for fixed ends). No thermal stress if free to expand.

Exams & GATE

1
Textbook: RK Bansal Ch. 1–2 / Strength of Materials.
2
Always state whether stress is engineering or true.
3
Prefer consistent units: N with mm² gives MPa directly ( $1\,\mathrm{MPa}=1\,\mathrm{N/mm^2}$ ). If you mix m², convert carefully ( $1\,\mathrm{m^2}=10^6\,\mathrm{mm^2}$ ).
4
GATE favourites: stress–strain curve point identification, Poisson effect on volume, and elastic vs plastic regions.

📖 Standard books (India)

Strength of Materials — RK Bansal
Read: Ch. 1–2
SOM — beams, torsion, columns, and deflection

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After exams and interviews, see how engineers actually built careers — milestones and decisions from people in the field.

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📚Key formulas & points

Topic details

Definition and physical meaning

Core assumptions (state these in exams)

Engineering stress–strain curve

Types of loading and stress states

Poisson’s ratio and lateral strain

Step-by-step problem approach

Common mistakes in exams

Axial stress

Worked examples

Axial stress and strain in a steel rod

Single vs double shear in a pin

Reading the stress–strain curve

Practice questions

📝Exams & GATE

📖 Standard books (India)

Explore related topics

See real mechanical engineering careers

Key formulas & points

Exams & GATE