• 6.1.5 Define and use the terms stress, strain, and the Young modulus.
• 6.1.6 Describe an experiment to determine the Young modulus of a metal in the form of a wire.
• Calculate stress, strain, and Young’s modulus from given data.
• Interpret stress-strain graphs.
• Understand the concepts of elastic limit, tensile stress, and tensile strain.

• Students will be able to define key vocabulary related to material properties (stress, strain, Young’s modulus, tensile, elastic limit) in English.
• Students will be able to explain the relationship between stress and strain for a material in English.
• Students will be able to describe the experimental procedure for determining Young’s modulus in English.
• Students will be able to discuss results and sources of error in the Young’s modulus experiment in English.

Familiarize yourself with these essential terms for understanding the mechanical properties of materials.

Use online flashcards to master these terms. Search for «Stress Strain Young’s Modulus Physics» on Quizlet.

Search for Stress & Strain Flashcards on Quizlet

Embedding Example (replace with actual Quizlet embed code):

<!-- <iframe src="https://quizlet.com/your-set-id/flashcards/embed" height="500" width="100%" style="border:0;"></iframe> -->

Study Tip: Focus not just on the definitions, but also on the units for each term and how they relate to each other.

Stress (σ): The force applied per unit cross-sectional area of a material. It measures the internal forces that molecules within a continuous material ***-exert-***әсер етеді/оказывают on each other. Formula: σ = F/A. Units: Pascals (Pa) or N/m².

Strain (ε): The measure of ***-deformation-***деформация/деформация of a material in response to stress. For tensile strain, it is the ratio of the extension (ΔL) to the original length (L₀). Formula: ε = ΔL/L₀. Strain is a dimensionless quantity (it has no units).

Young Modulus (E): A measure of the ***-stiffness-***қаттылық/жёсткость of a solid material. It is defined as the ratio of stress (σ) to strain (ε) in the elastic region of deformation. Formula: E = σ/ε. Units: Pascals (Pa) or N/m².

Elastic Limit: The maximum stress a material can ***-withstand-***шыдау/выдержать without permanent deformation. If stress exceeds the elastic limit, the material will not return to its original shape when the stress is removed.

Tensile: Relating to tension; caused by pulling forces that tend to ***-stretch-***созу/растягивать a material.

Content adapted from A-Level Physics resources (e.g., SaveMyExams, OpenStax).

When forces are applied to a solid object, the object can deform. The way a material responds to such forces is described by concepts like stress, strain, and the Young modulus. These are crucial for understanding the ***-mechanical properties-***механикалық қасиеттер/механические свойства of materials.

Stress (σ)

Tensile stress is defined as the force acting per unit cross-sectional area when a material is stretched by a force.

σ = F / A

Where:

• σ (sigma) = stress, measured in Pascals (Pa) or N/m²
• F = applied force, measured in Newtons (N)
• A = cross-sectional area, measured in square meters (m²)

Stress is a measure of how ***-concentrated-***шоғырланған/концентрирована a force is over an area. A larger force or a smaller area will result in higher stress.

Strain (ε)

Tensile strain is the fractional change in length of a material when it is stretched. It is a measure of how much the material deforms relative to its original size.

ε = ΔL / L₀

Where:

• ε (epsilon) = strain (dimensionless)
• ΔL = extension or change in length, measured in meters (m)
• L₀ = original length, measured in meters (m)

Since strain is a ratio of two lengths, it has no units.

Young Modulus (E)

The Young modulus (also known as the modulus of elasticity) is a measure of a material’s stiffness or ***-resistance-***қарсылық/сопротивление to elastic deformation under tensile or compressive stress. It is defined as the ratio of stress to strain, provided the elastic limit is not exceeded (i.e., within the region where Hooke’s Law is obeyed).

E = σ / ε

Substituting the formulas for stress and strain:

E = (F/A) / (ΔL/L₀) = (F * L₀) / (A * ΔL)

Where:

• E = Young modulus, measured in Pascals (Pa) or N/m² (same units as stress)

A material with a high Young modulus is stiff (e.g., steel), meaning it requires a large stress to produce a small strain. A material with a low Young modulus is more flexible (e.g., rubber).

The Stress-Strain Graph

A graph of stress against strain for a material shows its behavior under load.

(You should replace this with an actual diagram of a stress-strain graph, showing the elastic region, yield point, plastic region, and breaking point).

Key points on a typical stress-strain graph for a ductile metal include:

• Limit of Proportionality (P): The point up to which stress is directly proportional to strain (Hooke’s Law is obeyed). The graph is a straight line.
• Elastic Limit (E): The maximum stress that can be applied to a material without causing permanent deformation. If the stress is removed before this point, the material returns to its original length. Very close to P.
• Yield Point (Y): The point at which the material starts to undergo significant plastic (permanent) deformation with little or no increase in stress. Some materials show distinct upper and lower yield points.
• Ultimate Tensile Strength (UTS): The maximum stress the material can withstand before it starts to neck (narrow) and eventually fracture.
• Breaking Point / Fracture Point (B): The point at which the material breaks.

The ***-gradient-***градиент/градиент of the linear part of the stress-strain graph (up to the limit of proportionality) is equal to the Young modulus (E).

Experiment to Determine the Young Modulus of a Metal in the Form of a Wire

The Young modulus of a metal wire can be determined using an apparatus like Searle’s apparatus or a similar setup.

(Replace this with an actual diagram of the experimental setup, e.g., Searle’s apparatus or a simple setup with a clamped wire, weights, and a way to measure extension.)

The procedure involves:

1. Setup:
   • A long wire (test wire) is clamped firmly at one end. A reference wire of the same material and length may be used alongside to compensate for thermal expansion or sagging of the support.
   • A scale (often a micrometer screw gauge or a spirit level arrangement with a micrometer for measuring small extensions – as in Searle’s apparatus) is used to measure the extension of the wire.
   • A known load (weights) is applied to the free end of the test wire, usually via a hanger.
2. Measurements:
   • Original Length (L₀): Measure the length of the wire from the clamp to the point where the extension is measured, using a meter rule. This should be done carefully.
   • Diameter (d) of the wire: Measure the diameter of the wire at several points along its length and in different orientations using a micrometer screw gauge. Calculate the average diameter to determine the cross-sectional area (A = π(d/2)² = πd²/4). ***-Accuracy-***дәлдік/точность here is crucial as ‘d’ is squared.
   • Applied Force (F): Apply a series of known weights (masses ‘m’) to the wire. The force is F = mg, where g is the acceleration due to gravity.
   • Extension (ΔL): For each applied force, measure the corresponding extension of the wire. It’s good practice to load and unload the wire a few times initially to remove kinks. Take readings for increasing loads and then for decreasing loads to check for hysteresis or permanent deformation.
3. Calculations:
   • For each pair of F and ΔL values, calculate stress (σ = F/A) and strain (ε = ΔL/L₀).
   • Plot a graph of stress (y-axis) against strain (x-axis).
   • The gradient of the initial linear portion of this graph is the Young modulus (E).
     
     Alternatively, if you plot F against ΔL, the gradient is (E A / L₀). So, E = (gradient * L₀) / A.
4. Precautions and Sources of Error:
   • Ensure the wire is straight and free from kinks.
   • Avoid exceeding the elastic limit to prevent permanent deformation.
   • Measure the diameter carefully at multiple points. Parallax error should be avoided when reading scales.
   • Use a long wire to produce a larger, more measurable extension for a given load, reducing percentage uncertainty in ΔL.
   • The support must be rigid.

Questions on the Theory:

1. (Easy) What are the standard SI units for stress and Young’s Modulus?
   
   Answer

   Pascals (Pa) or Newtons per square meter (N/m²).
   
2. (Medium) Explain why strain is a dimensionless quantity.
   
   Answer

   Strain is calculated as the ratio of the change in length (extension, ΔL) to the original length (L₀). Since both ΔL and L₀ are measured in units of length (e.g., meters), the units cancel out in the division (m/m = 1). Therefore, strain is a dimensionless quantity.
   
3. (Medium) In an experiment to determine Young’s Modulus for a wire, why is it preferable to use a long wire and measure the diameter at multiple points?
   
   Answer

   Using a long wire is preferable because for a given stress (applied force), a longer wire will produce a larger, more easily measurable extension (ΔL). This reduces the percentage uncertainty in the measurement of ΔL.
   Measuring the diameter at multiple points and averaging is important because the wire’s diameter may not be perfectly uniform. Since the cross-sectional area (A = πd²/4) depends on the square of the diameter, any small error in ‘d’ is magnified. Averaging helps to get a more representative value for the diameter and thus a more accurate cross-sectional area, leading to a more accurate Young’s Modulus value.
   
4. (Hard — Critical Thinking) Two wires, A and B, are made of the same material and have the same original length. Wire A has twice the diameter of wire B. If both wires are subjected to the same tensile force (load), how will the extension of wire A compare to the extension of wire B? Justify your answer using the formula for Young’s Modulus.
   
   Answer

   Let L₀ be the original length and F be the tensile force, both same for A and B.
   Let E be the Young’s Modulus, same for both as they are of the same material.
   Diameter of A (d_A) = 2 * Diameter of B (d_B).
   Area A (A_A) = π(d_A/2)² = π(2d_B/2)² = πd_B².
   Area B (A_B) = π(d_B/2)² = πd_B²/4.
   So, A_A = 4 * A_B.

   Young’s Modulus E = (F * L₀) / (A * ΔL).
   Rearranging for extension ΔL: ΔL = (F * L₀) / (A * E).

   For wire A: ΔL_A = (F * L₀) / (A_A * E)
   For wire B: ΔL_B = (F * L₀) / (A_B * E)

   Since A_A = 4A_B, we can substitute this into the equation for ΔL_A:
   ΔL_A = (F * L₀) / (4A_B * E)
   ΔL_A = (1/4) * [(F * L₀) / (A_B * E)] ΔL_A = (1/4) * ΔL_B.

   Therefore, the extension of wire A will be one-fourth (1/4) the extension of wire B.
   Wire A is thicker, so it has a larger cross-sectional area. For the same force, this means lower stress on wire A. Since it’s the same material (same E), lower stress results in lower strain, and thus less extension.
   

Fill in the blanks with the appropriate terms:

1. The ratio of force to cross-sectional area is called _______________.
   
   Answer

   stress
2. _______________ is defined as the change in length per unit original length.
   
   Answer

   Strain
3. The stiffness of a material is quantified by its _______________.
   
   Answer

   Young Modulus
4. If a material returns to its original shape after a deforming force is removed, it was deformed within its _______________.
   
   Answer

   elastic limit
5. The units for Young Modulus are the same as for _______________.
   
   Answer

   stress (Pascals or N/m²)

Match the term with its description:

1. Stress (σ)
2. Strain (ε)
3. Young Modulus (E)
4. Elastic Limit

A) Ratio of extension to original length.
B) Maximum stress before permanent deformation occurs.
C) Force per unit cross-sectional area.
D) Ratio of stress to strain in the elastic region.

Watch this video to visualize stress, strain, Young’s Modulus, and the experiment:

Description: This video (example) explains the concepts of stress, strain, and Young’s Modulus, often including graphical representations and experimental considerations relevant to A-Level Physics.

Further Watching — Similar Topics:

• A-Level Physics — Young’s Modulus Experiment Explained (Focus on experimental setup)
• Crash Course Physics: The Strength of Materials (Engaging overview)
• Search for «Determining Young’s Modulus of a Wire Experiment» on YouTube for more practical demonstrations.

Work through these examples to apply your understanding.

Example 1: Calculating Stress, Strain, and Young’s Modulus

A metal wire of length 2.50 m and diameter 0.80 mm is stretched by a force of 60 N. The wire extends by 1.5 mm. Calculate:

a) The stress in the wire.

b) The strain in the wire.

c) The Young modulus of the metal.

Example 2: Finding Extension

A steel rod has a Young’s modulus of 2.0 × 10¹¹ Pa and a cross-sectional area of 0.50 cm². Its original length is 3.0 m. What tensile force would be required to stretch the rod by 0.30 mm?

While a direct Young’s Modulus experiment simulator for wires might be specific, you can explore related concepts of force, extension, and stiffness using PhET’s «Masses and Springs» simulation.

Access the simulator: PhET Masses and Springs

(If the direct embed doesn’t work, use the link above.)

Task: Investigate Hooke’s Law and Spring Constant (Analogous to Stiffness)

Focus on the «Lab» or «Stretch» sections.

1. Select a spring. Keep its «Spring Constant» (stiffness) fixed for the first part.
2. Apply different masses (which exert a force F=mg, where g ≈ 9.8 N/kg). Record the mass and the resulting displacement (extension, x or ΔL) from the equilibrium position. You can use the ruler provided.
3. Plot a graph of Force (y-axis) against Extension (x-axis). What does the gradient of this graph represent for the spring?
4. Repeat with a spring of a different «Spring Constant». How does the graph change? How does this relate to the concept of Young’s Modulus (stiffness) for a material?

Brief Questions & Answers:

1. What relationship did you observe between the applied force and the extension of the spring (within its elastic limit)?
   
   Brief Answer

   Within the elastic limit, the extension of the spring is directly proportional to the applied force (Hooke’s Law: F = kx).
   
2. What does the gradient of your Force vs. Extension graph represent?
   
   Brief Answer

   The gradient represents the spring constant (k) of the spring, which is a measure of its stiffness. Units: N/m.
   
3. How does a higher «Spring Constant» in the simulation affect the extension for a given force? How is this analogous to a material with a high Young’s Modulus?
   
   Brief Answer

   A higher spring constant means the spring is stiffer, so for a given force, the extension will be smaller. This is analogous to a material with a high Young’s Modulus, which is also stiffer and will show less strain (and thus less extension for a given length) for a given stress.
   

Note: This simulation demonstrates Hooke’s Law (F=kx). Young’s Modulus (E = (F L₀)/(A ΔL)) is a material property, while ‘k’ here is for the specific spring. However, the principle of stiffness (resistance to deformation) is analogous.

Activity: Planning the Perfect Young's Modulus Experiment

Platform Suggestion: Use a collaborative document (Google Docs, Microsoft OneDrive) or a shared digital whiteboard (Miro, Jamboard).

Task:

In your group, you are tasked with writing a detailed plan for an experiment to determine the Young's Modulus of an unknown metal wire. Your plan should be clear enough for another group of students to follow and achieve accurate results.

Consider and include the following in your plan:

1. Apparatus List: What specific equipment will you need? (e.g., type of wire, method for applying force, method for measuring length, diameter, extension). Be precise.
2. Diagram: Sketch a clear, labeled diagram of your proposed experimental setup.
3. Procedure: Write step-by-step instructions.
   • How will you measure L₀?
   • How will you measure the diameter 'd' accurately? Why is this measurement particularly critical?
   • How will you apply varying forces (F)?
   • How will you measure the extension (ΔL) accurately for each force?
   • What range of forces will you use and why? (Consider the elastic limit).
4. Data Collection: Design a table to record your measurements.
5. Analysis:
   • What calculations will you perform? (Stress, Strain, Area).
   • How will you determine Young's Modulus from your data? (e.g., graphical method). Sketch what your expected graph would look like.
6. Safety Precautions & Minimizing Errors: List at least 3 safety precautions and 3 specific ways to minimize experimental errors/uncertainties.

Online Tool Suggestion:

• Formative (goformative.com): The teacher could create a "Show Your Work" question where groups upload their plan, including sketches.
• LearningApps.org: While not for full plan submission, you could create a "Mind Map" or "Notebook" task for brainstorming initial ideas.

Groups can then present their plans or peer-review other groups' plans for completeness and accuracy.

Answer the following questions, showing all working where calculations are needed.

1. Analysis & Calculation: A nylon guitar string has a diameter of 1.00 mm and an original length of 0.750 m. When a tension of 150 N is applied, the string stretches by 3.75 mm.
   
   a) Calculate the tensile stress in the string.
   
   b) Calculate the tensile strain in the string.
   
   c) Determine the Young’s modulus for nylon.
   
   d) If the elastic limit for this nylon is reached at a strain of 0.006, what is the maximum stress the string can withstand before permanent deformation occurs?
   
   Answer

   Given:
   d = 1.00 mm = 1.00 × 10^-3 m => r = 0.50 × 10^-3 m
   L₀ = 0.750 m
   F = 150 N
   ΔL = 3.75 mm = 3.75 × 10^-3 m
   Elastic limit strain ε_el = 0.006

   a) Tensile Stress (σ):
   Area A = πr² = π(0.50 × 10^-3 m)² = π(0.25 × 10^-6 m²) ≈ 7.854 × 10^-7 m²
   σ = F/A = 150 N / (7.854 × 10^-7 m²) ≈ 1.9099 × 10⁸ Pa ≈ 191 MPa

   b) Tensile Strain (ε):
   ε = ΔL/L₀ = (3.75 × 10^-3 m) / 0.750 m = 0.005

   c) Young’s Modulus (E):
   E = σ/ε = (1.9099 × 10⁸ Pa) / 0.005 ≈ 3.8198 × 10¹⁰ Pa ≈ 38.2 GPa

   d) Maximum stress at elastic limit (σ_el):
   Assuming E is constant up to the elastic limit: σ_el = E * ε_el
   σ_el = (3.8198 × 10¹⁰ Pa) * 0.006 ≈ 2.2919 × 10⁸ Pa ≈ 229 MPa

2. Experimental Design & Analysis: An experiment to determine the Young’s modulus of a metal wire produced the following data for applied force (F) and measured extension (ΔL). The original length of the wire was 2.00 m and its diameter was 0.56 mm.

   Force F (N)	Extension ΔL (mm)
   0	0
   10	0.42
   20	0.83
   30	1.25
   40	1.67
   50	2.08

   a) Calculate the cross-sectional area of the wire in m².
   
   b) For each load, calculate the stress and strain.
   
   c) Plot a graph of stress (y-axis) against strain (x-axis).
   
   d) From your graph, determine the Young’s modulus for the metal.
   
   e) Identify one significant source of uncertainty in this experiment and suggest how its effect could be minimized.
   

   Answer

   Given: L₀ = 2.00 m, d = 0.56 mm = 0.56 × 10^-3 m

   a) Cross-sectional Area (A):
   Radius r = d/2 = 0.28 × 10^-3 m
   A = πr² = π(0.28 × 10^-3 m)² = π(0.0784 × 10^-6 m²) ≈ 2.463 × 10^-7 m²

   b) Stress and Strain Calculations:
   (ΔL should be converted to meters: ΔL(m) = ΔL(mm) × 10^-3)

   F (N)	ΔL (mm)	ΔL (m)	Stress σ = F/A (Pa)	Strain ε = ΔL/L0
   0	0	0	0	0
   10	0.42	0.00042	10 / (2.463×10-7) ≈ 4.060×107	0.00042 / 2.00 = 0.00021
   20	0.83	0.00083	20 / (2.463×10-7) ≈ 8.120×107	0.00083 / 2.00 = 0.000415
   30	1.25	0.00125	30 / (2.463×10-7) ≈ 1.218×108	0.00125 / 2.00 = 0.000625
   40	1.67	0.00167	40 / (2.463×10-7) ≈ 1.624×108	0.00167 / 2.00 = 0.000835
   50	2.08	0.00208	50 / (2.463×10-7) ≈ 2.030×108	0.00208 / 2.00 = 0.00104

   c) Plot graph of stress vs. strain:
   (Students would plot these values. The graph should be approximately linear.)

   

   d) Determine Young’s Modulus (E) from graph:
   E = gradient of the stress-strain graph.
   Using the first and last data points (excluding zero) for an approximate gradient:
   E ≈ (2.030×10⁸ Pa — 4.060×10⁷ Pa) / (0.00104 — 0.00021)
   E ≈ (1.624×10⁸ Pa) / (0.00083) ≈ 1.957 × 10¹¹ Pa ≈ 196 GPa.
   (A more accurate value would be obtained from the line of best fit.)

   e) Significant source of uncertainty and minimization:
   Source: Measurement of extension (ΔL). Small extensions can have high percentage uncertainty, especially if using a simple ruler or if there’s parallax error.
   Minimization: Use a more precise instrument like a vernier scale, micrometer attached to the wire setup (e.g., part of Searle’s apparatus), or an optical lever to magnify the extension. Ensure the scale is read at eye level to avoid parallax error. Taking readings for loading and unloading and averaging can also help reduce random errors. Using a longer wire also helps increase ΔL for a given load, reducing its percentage uncertainty.
   Another significant source is the diameter measurement (d), as A depends on d². Minimize by measuring ‘d’ at several points and orientations using a micrometer screw gauge and averaging.

3. Synthesis & Application: A civil engineer is choosing between two types of steel for a bridge construction. Steel Type X has a Young’s modulus of 200 GPa and an elastic limit stress of 250 MPa. Steel Type Y has a Young’s modulus of 210 GPa and an elastic limit stress of 230 MPa.
   
   a) Which steel type is stiffer? Explain.
   
   b) If a particular component in the bridge must not permanently deform under a working stress of 240 MPa, which steel type could be used? Explain your reasoning.
   
   c) If minimizing extension under typical loads is the primary concern (well below any elastic limit), which steel might be preferred, and why?
   
   Answer

   a) Which steel type is stiffer?
   Steel Type Y is stiffer because it has a higher Young’s modulus (210 GPa vs 200 GPa for Type X). A higher Young’s modulus indicates greater resistance to elastic deformation for a given stress.

   b) Which steel type for working stress of 240 MPa without permanent deformation?
   The working stress (240 MPa) must be below the elastic limit stress of the chosen steel to avoid permanent deformation.
   — Steel Type X: Elastic limit = 250 MPa. Since 240 MPa 230 MPa, Steel Type Y would undergo permanent deformation and could NOT be used.
   Therefore, only Steel Type X could be used.

   c) Minimizing extension under typical loads (primary concern):
   If minimizing extension (strain) under typical loads is the primary concern, Steel Type Y would be preferred.
   Strain (ε) = Stress (σ) / Young’s Modulus (E).
   For a given stress (σ), a higher Young’s Modulus (E) will result in a lower strain (ε), meaning less extension. Since Steel Type Y has a higher Young’s Modulus (210 GPa) compared to Steel Type X (200 GPa), it will stretch less under the same typical load (assuming the load creates stress well within the elastic limit of both).

4. Critical Evaluation: A student performing the Young’s Modulus experiment uses a very short, thick wire. They find it difficult to get reliable results. Explain two reasons why using a very short, thick wire might lead to less accurate results for Young’s Modulus compared to using a long, thin wire.
   
   Answer

   Using a very short, thick wire can lead to less accurate results for Young’s Modulus for these main reasons:

   1. Small, Difficult-to-Measure Extensions (ΔL):
   Young’s Modulus E = (F L₀) / (A ΔL). Rearranging for extension: ΔL = (F L₀) / (A E).
   If L₀ (original length) is very short, the resulting extension (ΔL) for a given force F will also be very small. Measuring these tiny extensions accurately is difficult and leads to a high percentage uncertainty in ΔL. Any small absolute error in measuring ΔL becomes a large percentage of the reading itself, propagating into a large uncertainty in the calculated Young’s Modulus.

   2. Large Cross-Sectional Area (A) reduces extension and requires large forces:
   If the wire is thick, its cross-sectional area (A) is large. From ΔL = (F L₀) / (A E), a larger ‘A’ also contributes to a smaller ΔL for a given force and length. To achieve a measurable extension with a thick wire, significantly larger forces (F) would be required. Applying very large forces can:
   a) Pose safety risks or exceed the capacity of standard lab equipment.
   b) Increase the risk of the support system deforming or slipping, which would be mistaken for wire extension.
   c) If the forces are not large enough to cause a clearly measurable extension, the problem of high percentage uncertainty in ΔL remains.
   Furthermore, any non-uniformity in a thick wire’s diameter has a larger absolute impact on the area calculation than for a thin wire, though the percentage error in diameter might be similar. The primary issue is the very small ΔL.

   A long, thin wire is preferred because:
   — A long L₀ yields a larger, more measurable ΔL, reducing percentage uncertainty in ΔL.
   — A thin wire (smaller A) yields a larger ΔL for a given force, making it easier to measure and requiring smaller, safer loads.

5. Material Comparison: Compare the expected stress-strain graphs for a brittle material (like glass or cast iron) and a ductile material (like copper or mild steel). Sketch both graphs on the same axes (or separate labeled axes) and highlight the key differences in terms of elastic region, plastic deformation, and fracture point. What do these differences imply about how these materials behave under tension?
   
   Answer

   
   
   (Replace with actual comparative graphs. Ductile: shows elastic region, yield point, plastic deformation, necking, fracture. Brittle: shows elastic region and then sudden fracture with little to no plastic deformation.)

   Key Differences:

   1. Elastic Region: Both materials will show an initial linear elastic region where stress is proportional to strain and they obey Hooke’s Law. The gradient here is the Young’s Modulus.
   2. Plastic Deformation:
      • Ductile Material: Exhibits significant plastic deformation after the elastic limit (or yield point) is exceeded. It can be stretched considerably, undergoing permanent changes in shape before fracturing. This region shows the material ‘yielding’ and then ‘strain hardening’ up to the Ultimate Tensile Strength (UTS), followed by ‘necking’ before fracture.
      • Brittle Material: Shows very little or no plastic deformation. It fractures suddenly soon after the elastic limit is reached. There is no significant yielding or necking.
   3. Fracture Point:
      • Ductile Material: Fractures after considerable plastic deformation. The strain at fracture is large.
      • Brittle Material: Fractures at a relatively low strain, often just beyond the elastic limit. The stress at fracture might be high, but the material doesn’t stretch much.
   4. Energy Absorption (Toughness):
      • Ductile Material: The area under the stress-strain graph represents the energy absorbed per unit volume before fracture (toughness). Ductile materials have a large area due to significant plastic deformation, meaning they can absorb a lot of energy.
      • Brittle Material: The area under the graph is much smaller, indicating lower toughness. They absorb less energy before fracturing.

   Implications of Behavior Under Tension:

   • Ductile Materials: Can be drawn into wires, bent, and shaped due to their ability to deform plastically. They give warning before failure by undergoing visible deformation (necking). This makes them suitable for applications where some ‘give’ is required or where sudden failure is catastrophic (e.g., structural steel, copper wiring).
   • Brittle Materials: Fail suddenly without warning once their elastic limit is exceeded. They are often strong in compression but weak under tension if there are any surface flaws (stress concentrators). They are not suitable for applications requiring significant deformation or impact resistance (e.g., glass windows are strong but shatter easily on impact; cast iron engine blocks are strong but can crack).

• Save My Exams — Properties of Materials: Deformation of Solids (Save My Exams — CIE A-Level) (Check for your specific exam board if different)
• PhysicsAndMathsTutor — Materials Notes & Questions: Materials Section (PhysicsAndMathsTutor)
• OpenStax University Physics Vol 1 — Chapter 12: Static Equilibrium and Elasticity: Stress, Strain, and Elastic Modulus
• HyperPhysics — Elasticity: Elasticity & Young’s Modulus (HyperPhysics)
• A-Level Physics Online — Young Modulus: Young Modulus Page

Consider the following questions to reflect on your learning:

• What is the physical meaning of a high Young’s modulus compared to a low Young’s modulus? Give an example of a material for each.
• Describe one key challenge or source of error in the experiment to determine Young’s modulus of a wire, and how you would try to minimize it.
• How do stress and strain differ from force and extension? Why are stress and strain more useful for characterizing material properties?
• What was the most interesting or surprising thing you learned in this lesson?

Jot down your thoughts or discuss them with a classmate.