The Stress Formula defines the internal restoring force per unit area developed within a body when an external deforming force is applied to it. Mathematically, stress is expressed as σ = F/A, where F is the applied force in Newtons and A is the cross-sectional area in square metres. This concept is a core part of NCERT Class 11 Physics, Chapter 9 (Mechanical Properties of Solids). It also appears frequently in JEE Main and JEE Advanced problems on elasticity. This article covers the stress formula, its derivation, types, a complete formula sheet, solved examples, and CBSE exam tips for 2025-26.
Key Stress Formulas at a Glance
Quick reference for the most important stress and elasticity formulas.
- Stress: \( \sigma = \dfrac{F}{A} \)
- Longitudinal (Normal) Stress: \( \sigma = \dfrac{F_{\perp}}{A} \)
- Shear (Tangential) Stress: \( \tau = \dfrac{F_{\parallel}}{A} \)
- Volumetric (Bulk) Stress: \( \sigma_v = \dfrac{\Delta F}{A} = \Delta P \)
- Young's Modulus: \( Y = \dfrac{\sigma}{\varepsilon} = \dfrac{F \cdot L}{A \cdot \Delta L} \)
- Strain: \( \varepsilon = \dfrac{\Delta L}{L} \)
- SI Unit of Stress: Pascal (Pa) = N/m²
What is the Stress Formula?
The Stress Formula quantifies the internal restoring force that arises per unit area inside a solid body when external forces try to deform it. When you stretch, compress, or twist a solid object, its molecules resist the deformation. This resistance, measured per unit area, is called stress.
According to NCERT Class 11 Physics (Chapter 9 — Mechanical Properties of Solids), stress is defined as the restoring force per unit area. The SI unit of stress is the Pascal (Pa), which equals one Newton per square metre (N/m²). Stress has the same dimensions as pressure: [ML²T²] — wait, the correct dimensional formula is [ML−¹T−²].
There are three primary types of stress you must know for CBSE and JEE exams:
- Longitudinal (Normal) Stress: Force acts perpendicular to the cross-sectional area. It can be tensile (stretching) or compressive (squeezing).
- Shear (Tangential) Stress: Force acts parallel to the surface area, causing layers to slide over each other.
- Volumetric (Bulk) Stress: Uniform force acts from all directions, changing the volume of the body. This equals the change in pressure applied.
Understanding the stress formula is essential for solving problems on elastic behaviour, Young's modulus, bulk modulus, and modulus of rigidity in both CBSE board exams and competitive entrance tests.
Stress Formula — Expression and Variables
The general stress formula is written as:
\[ \sigma = \frac{F}{A} \]
Where \( \sigma \) (sigma) represents stress, \( F \) is the magnitude of the applied (or restoring) force, and \( A \) is the cross-sectional area over which the force acts.
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( \sigma \) | Stress | Pascal (Pa) = N/m² |
| \( F \) | Applied or Restoring Force | Newton (N) |
| \( A \) | Cross-sectional Area | Square metre (m²) |
| \( \tau \) | Shear Stress | Pascal (Pa) |
| \( \Delta P \) | Bulk Stress (change in pressure) | Pascal (Pa) |
Derivation of the Stress Formula
Consider a cylindrical rod of cross-sectional area \( A \) subjected to an external tensile force \( F \) applied along its axis.
Step 1: By Newton's Third Law, the molecules inside the rod develop an equal and opposite restoring force \( F_r \) to resist deformation. At equilibrium, \( F_r = F \).
Step 2: Stress is defined as the restoring force per unit area of cross-section:
\[ \sigma = \frac{F_r}{A} = \frac{F}{A} \]
Step 3: Since force is in Newtons (N) and area is in m², the unit of stress becomes N/m² = Pascal (Pa).
Step 4: The dimensional formula of stress is derived as:
\[ [\sigma] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}] \]
This derivation confirms that stress has the same dimensional formula as pressure, which makes physical sense.
Complete Elasticity Formula Sheet
The table below covers all key formulas related to stress, strain, and elastic moduli. These are essential for CBSE Class 11 board exams and JEE preparation.
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| General Stress | \( \sigma = F/A \) | F = Force, A = Area | Pa (N/m²) | Class 11, Ch 9 |
| Tensile/Compressive Stress | \( \sigma = F_{\perp}/A \) | F⊥ = Normal Force | Pa | Class 11, Ch 9 |
| Shear Stress | \( \tau = F_{\parallel}/A \) | F∥ = Tangential Force | Pa | Class 11, Ch 9 |
| Bulk (Volumetric) Stress | \( \sigma_v = \Delta P \) | ΔP = Change in Pressure | Pa | Class 11, Ch 9 |
| Longitudinal Strain | \( \varepsilon = \Delta L / L \) | ΔL = Change in Length, L = Original Length | Dimensionless | Class 11, Ch 9 |
| Shear Strain | \( \gamma = x / h = \tan\theta \approx \theta \) | x = Lateral Displacement, h = Height | Dimensionless | Class 11, Ch 9 |
| Volumetric Strain | \( \varepsilon_v = \Delta V / V \) | ΔV = Change in Volume, V = Original Volume | Dimensionless | Class 11, Ch 9 |
| Young's Modulus | \( Y = \sigma / \varepsilon = FL / (A \Delta L) \) | Y = Young's Modulus | Pa | Class 11, Ch 9 |
| Bulk Modulus | \( B = -\Delta P / (\Delta V / V) \) | B = Bulk Modulus | Pa | Class 11, Ch 9 |
| Modulus of Rigidity | \( G = \tau / \gamma \) | G = Shear Modulus, γ = Shear Strain | Pa | Class 11, Ch 9 |
| Elastic Potential Energy Density | \( u = \frac{1}{2} \sigma \varepsilon = \frac{1}{2} Y \varepsilon^2 \) | u = Energy per unit volume | J/m³ | Class 11, Ch 9 |
| Hooke's Law | \( \sigma = Y \varepsilon \) (within elastic limit) | Y = Young's Modulus | Pa | Class 11, Ch 9 |
Stress Formula — Solved Examples
The following three examples progress from basic CBSE-level to JEE-level difficulty. Work through each carefully to build exam confidence.
Example 1 (Class 9-10 Level): Basic Stress Calculation
Problem: A steel wire has a cross-sectional area of 2 × 10−&sup6; m². A force of 40 N is applied along its length. Calculate the stress developed in the wire.
Given:
- Force, F = 40 N
- Cross-sectional area, A = 2 × 10−&sup6; m²
Step 1: Write the stress formula: \( \sigma = \dfrac{F}{A} \)
Step 2: Substitute the given values:
\[ \sigma = \frac{40}{2 \times 10^{-6}} \]
Step 3: Perform the division: \( \sigma = 20 \times 10^{6} \) Pa = 2 × 10&sup7; Pa
Answer
Stress = 2 × 10&sup7; Pa = 20 MPa
Example 2 (Class 11-12 Level): Finding Extension Using Stress and Young's Modulus
Problem: A copper wire of length 2.0 m and cross-sectional area 1.5 × 10−&sup6; m² is subjected to a tensile force of 300 N. If Young's modulus of copper is 1.2 × 10¹¹ Pa, find (a) the stress in the wire and (b) the elongation produced.
Given:
- Length, L = 2.0 m
- Area, A = 1.5 × 10−&sup6; m²
- Force, F = 300 N
- Young's Modulus, Y = 1.2 × 10¹¹ Pa
Step 1: Calculate stress using \( \sigma = F/A \):
\[ \sigma = \frac{300}{1.5 \times 10^{-6}} = 2 \times 10^{8} \text{ Pa} \]
Step 2: Use Young's modulus to find strain. Recall \( Y = \sigma / \varepsilon \), so:
\[ \varepsilon = \frac{\sigma}{Y} = \frac{2 \times 10^{8}}{1.2 \times 10^{11}} = 1.67 \times 10^{-3} \]
Step 3: Calculate elongation using \( \varepsilon = \Delta L / L \):
\[ \Delta L = \varepsilon \times L = 1.67 \times 10^{-3} \times 2.0 = 3.33 \times 10^{-3} \text{ m} \]
Answer
(a) Stress = 2 × 10&sup8; Pa (b) Elongation = 3.33 × 10−³ m = 3.33 mm
Example 3 (JEE/NEET Level): Stress in a Composite Wire System
Problem: Two wires — one of steel (Y = 2 × 10¹¹ Pa) and one of aluminium (Y = 7 × 10¹&sup0; Pa) — are joined end to end and suspended from a rigid support. The steel wire has length 1.5 m and diameter 1 mm. The aluminium wire has length 2.0 m and diameter 2 mm. A load of 100 N is hung from the lower end. Find the stress in each wire.
Given:
- Steel wire: L⊂s; = 1.5 m, d⊂s; = 1 mm = 1 × 10−³ m
- Aluminium wire: L⊂a; = 2.0 m, d⊂a; = 2 mm = 2 × 10−³ m
- Load, F = 100 N (same for both wires in series)
Step 1: Calculate cross-sectional areas. For a circular cross-section, \( A = \pi r^2 = \pi (d/2)^2 \):
\[ A_s = \pi \left(\frac{1 \times 10^{-3}}{2}\right)^2 = \pi \times 0.25 \times 10^{-6} \approx 7.85 \times 10^{-7} \text{ m}^2 \]
\[ A_a = \pi \left(\frac{2 \times 10^{-3}}{2}\right)^2 = \pi \times 1 \times 10^{-6} \approx 3.14 \times 10^{-6} \text{ m}^2 \]
Step 2: Since the wires are in series, both carry the same tension of 100 N. Apply the stress formula to each:
\[ \sigma_s = \frac{F}{A_s} = \frac{100}{7.85 \times 10^{-7}} \approx 1.27 \times 10^{8} \text{ Pa} \]
\[ \sigma_a = \frac{F}{A_a} = \frac{100}{3.14 \times 10^{-6}} \approx 3.18 \times 10^{7} \text{ Pa} \]
Step 3: Note that the steel wire experiences much higher stress due to its smaller area. This is a key JEE insight: stress depends on area, not material, when the same force acts.
Answer
Stress in steel wire ≈ 1.27 × 10&sup8; Pa | Stress in aluminium wire ≈ 3.18 × 10&sup7; Pa
CBSE Exam Tips 2025-26
- Always state units: Write Pa or N/m² alongside every stress answer. CBSE examiners deduct marks for missing units.
- Distinguish types of stress: Board papers often ask you to identify whether a given situation involves tensile, compressive, shear, or bulk stress. We recommend making a quick comparison table in your notes.
- Memorise the dimensional formula: \( [ML^{-1}T^{-2}] \) is frequently asked in 1-mark questions. Do not confuse it with the dimensional formula of force.
- Link stress to Hooke's Law: Questions in the 3-mark category often combine the stress formula with Young's modulus. Practise multi-step problems like Example 2 above.
- Use the correct area: For a wire with diameter d, the area is \( A = \pi d^2 / 4 \), not \( \pi d^2 \). This is a very common calculation error in 2025-26 board papers.
- Elastic limit awareness: CBSE sometimes asks whether a body follows Hooke's Law. Remember that the stress formula applies both within and beyond the elastic limit, but Young's modulus is constant only within it.
Common Mistakes to Avoid
Our experts have identified the following errors that students repeatedly make with the stress formula:
-
Mistake 1: Confusing stress with pressure.
Both have the same unit (Pa) and dimensional formula, but pressure is an external force per unit area acting on a surface, while stress is the internal restoring force per unit area. Always clarify this distinction in exam answers. -
Mistake 2: Using diameter instead of radius to compute area.
If a wire has diameter d, then \( A = \pi (d/2)^2 = \pi d^2 / 4 \). Using \( \pi d^2 \) directly gives an area four times too large, leading to a stress four times too small. -
Mistake 3: Treating stress as a scalar quantity.
Stress is actually a tensor (it has both magnitude and direction). For CBSE purposes, it is treated as a scalar with a sign (tensile = positive, compressive = negative). Do not ignore the sign in problems that ask for the nature of stress. -
Mistake 4: Forgetting to convert units.
Force must be in Newtons and area must be in m² before applying \( \sigma = F/A \). If area is given in cm² or mm², convert first: 1 cm² = 10−&sup4; m², 1 mm² = 10−&sup6; m². -
Mistake 5: Applying the formula beyond its valid range.
The simple formula \( \sigma = F/A \) assumes uniform stress distribution across the cross-section. In real engineering problems (stress concentrations near holes, notches), the distribution is non-uniform. For CBSE and JEE, assume uniform distribution unless stated otherwise.
JEE/NEET Application of Stress Formula
In our experience, JEE aspirants encounter the stress formula in at least 2–3 questions per paper on Mechanical Properties of Solids. Here are the most common application patterns:
Pattern 1: Stress-Strain Graph Interpretation
JEE questions frequently present a stress-strain graph and ask students to identify the elastic limit, yield point, ultimate stress, and breaking stress. The slope of the linear region equals Young's modulus: \( Y = \sigma / \varepsilon \). Recognising the proportional limit on the graph is a key skill tested in JEE Advanced.
Pattern 2: Thermal Stress
When a rod is fixed at both ends and its temperature changes by \( \Delta T \), it cannot expand freely. The thermal stress developed is:
\[ \sigma_{\text{thermal}} = Y \alpha \Delta T \]
where \( \alpha \) is the coefficient of linear thermal expansion and Y is Young's modulus. This formula combines the stress formula with thermodynamics. It appears in both JEE Main and JEE Advanced every few years.
Pattern 3: Energy Stored in a Stretched Wire
The elastic potential energy stored per unit volume in a wire under stress is:
\[ u = \frac{1}{2} \times \text{stress} \times \text{strain} = \frac{\sigma^2}{2Y} \]
JEE problems often ask for the total energy stored: \( U = u \times V \), where V is the volume of the wire. NEET also tests this in the context of biological materials like tendons and bones, where elastic energy storage is physiologically important.
In our experience, JEE aspirants who master the relationship between stress, strain, and elastic moduli can solve these problems in under two minutes. We recommend practising at least 20 JEE-level problems on this topic before the exam.
FAQs on Stress Formula
Explore More Physics Formulas
Now that you have mastered the Stress Formula, strengthen your understanding of related topics. Visit our Complete Physics Formulas Hub for the full list of Class 11 and Class 12 formulas. You may also find these articles helpful:
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- Brewster's Law Formula — study polarisation of light for JEE and CBSE boards.
For the official NCERT textbook content on Mechanical Properties of Solids, refer to the NCERT Class 11 Physics Chapter 9 on the official NCERT website.