# Stress, Strain & Tensile Strength ## Tensile Stress - Applied forces don't just cause motion – they also alter the physical dimensions of objects - When opposite forces act along the axis of a material, the material experiences internal resistance - **Tensile stress** quantifies this internal resistance per unit area

Tensile stress $\sigma$ is the force per unit cross-sectional area: $\sigma = \frac{F}{A}$

- Where: - $\sigma$ = tensile stress (Pa or N m$^{-2}$) - $F$ = applied force (N) - $A$ = original cross-sectional area (m$^{2}$) - Units are pascals (Pa), identical to pressure units since both represent force per area - Stress represents how intensely force is distributed across a material's cross-section - A thin wire experiences higher stress than a thick rod under the same load ## Tensile Strain - Strain describes the fractional change in a material's dimensions - Unlike stress, strain has no units – it's a pure ratio

Tensile strain $\varepsilon$ is the extension per unit original length: $\varepsilon = \frac{x}{L}$

- Where: - $\varepsilon$ = tensile strain (no units, dimensionless) - $x$ = extension from original length (m) - $L$ = original unstretched length (m) - Strain expresses deformation as a proportion rather than an absolute measurement - A 2 m cable extending by 4 mm has the same strain as a 1 m cable extending by 2 mm - Sometimes expressed as a percentage: multiply the ratio by 100 - Example: A 0.5 m wire extending 0.01 m has strain = (0.01 ÷ 0.5) = 0.02 or 2% ## Ultimate Tensile Strength - Every material has a maximum stress it can withstand before fracturing - This critical value is called the **ultimate tensile strength**

Ultimate tensile strength is the maximum stress a material tolerates before breaking

- Measured in pascals (Pa), typically megapascals (MPa) for solid materials - Different materials have vastly different tensile strengths based on their atomic structure ### Tensile Strength Values | Material | Tensile Strength (MPa) | |----------|------------------------| | Concrete | 2–5 | | Rubber | 16 | | Human skin | 20 | | Glass | 33 | | Human hair | 200 | | Steel | 840 | | Diamond | 2800 | The chart below illustrates the dramatic variation in tensile strength across different materials:

- Brittle materials like glass fracture near their tensile strength with little warning - Ductile materials like steel can deform significantly before reaching this limit - Biological materials like skin and hair have surprisingly high tensile strengths for their composition ## Worked Example: Calculating Stress and Strain **A titanium wire has length 3.20 m and diameter 0.35 mm. A tensile force of 75 N is applied. Calculate:** a) The tensile stress in the wire b) The tensile strain if the wire extends to 3.22 m ### Answer: **Part (a) – Tensile stress** Step 1: List known quantities - Force, $F$ = 75 N - Diameter, $d$ = 0.35 mm = 0.35 × 10$^{-3}$ m - Radius, $r$ = $d$/2 = 0.175 × 10$^{-3}$ m Step 2: Calculate cross-sectional area - Wire has circular cross-section: $A = \pi r^2$ - $A = \pi \times (0.175 \times 10^{-3})^2$ - $A = 9.621 \times 10^{-8}$ m$^2$ Step 3: Apply stress equation - $\sigma = \frac{F}{A}$ - $\sigma = \frac{75}{9.621 \times 10^{-8}}$ - $\sigma = 779.5 \times 10^6$ Pa - $\sigma = 780$ MPa (to 2 s.f.) **Part (b) – Tensile strain** Step 1: Identify original and final lengths - Original length, $L$ = 3.20 m - Final length = 3.22 m - Extension, $x$ = 3.22 – 3.20 = 0.02 m Step 2: Apply strain equation - $\varepsilon = \frac{x}{L}$ - $\varepsilon = \frac{0.02}{3.20}$ - $\varepsilon = 6.25 \times 10^{-3}$ - $\varepsilon = 6.3 \times 10^{-3}$ (to 2 s.f.)

Always use the original cross-sectional area and original length when calculating stress and strain

## Key Points - Stress concentrates force over area – thinner materials experience higher stress - Strain measures proportional deformation – identical strains produce different extensions in different lengths - Cross-sectional area must be calculated from diameter measurements (often given in mm, requiring conversion) - Extension equals final length minus original length - Units matter: convert all measurements to SI base units (metres for length, square metres for area) - Since strain is dimensionless, length units must match but need not be metres