Unit 3: Stress and strain - Subjective Questions

BTY730 — Biomechanics • Practice Questions with Detailed Answers

20 questions

Define stress and strain. State their SI units and explain why strain is a dimensionless quantity.

Explain the relationship between applied forces and deformations in a loaded body. How does the concept of equilibrium relate to internal forces?

Describe the three basic loading configurations (tension, compression, and shear) with neat diagrams and give one biomechanical example of each.

State and explain Hooke's Law. Derive the relationship between stress and strain, and define the modulus of elasticity.

Describe the uniaxial tension test. Explain the procedure, the type of specimen used, and the quantities measured during the test.

Distinguish between a load-elongation diagram and a stress-strain diagram. Why is the stress-strain diagram preferred for characterizing material behavior?

Load-Elongation Diagram:

Plot of applied load ( $F$ ) on the y-axis versus elongation ( $\Delta L$ ) on the x-axis.
Directly obtained from raw test data.
Depends on specimen dimensions (cross-sectional area and length).

Stress-Strain Diagram:

Plot of stress ( $\sigma = F/A_0$ ) versus strain ( $\epsilon = \Delta L/L_0$ ).
Obtained by normalizing load and elongation by geometry.
Independent of specimen dimensions — a true material property.

Key differences:

Feature	Load-Elongation	Stress-Strain
Axes	Load vs Elongation	Stress vs Strain
Geometry dependence	Yes	No
Represents	Structural response	Material behavior
Units	N vs mm	Pa vs dimensionless

Why stress-strain is preferred:

It removes the effect of specimen size and shape.
The curve becomes a fundamental material characteristic, allowing comparison between different specimens and materials.
Design parameters like modulus, yield strength, and ultimate strength are directly read from it.

With a neat sketch, explain the salient points on the stress-strain diagram of a ductile material (mild steel). Label the proportional limit, elastic limit, yield point, ultimate strength, and fracture point.

Define the following material properties as obtained from the stress-strain diagram: elasticity, plasticity, ductility, brittleness, and toughness.

Distinguish between ductile and brittle materials based on their stress-strain behavior. Give two examples of each relevant to biomechanics.

Ductile Materials:

Exhibit large plastic deformation before fracture.
Show a distinct yield point and pronounced necking.
Give warning before failure (large strain).
High percentage elongation (> 5%).
Examples: Mild steel, copper, and biologically, cortical bone in some conditions and tendon/ligament (show large elongation).

Brittle Materials:

Exhibit little or no plastic deformation before fracture.
No distinct yield point; fracture occurs near the elastic limit.
Fail suddenly without warning.
Low percentage elongation.
Examples: Cast iron, glass, ceramics, and biologically, bone under high strain rates or tooth enamel.

Comparison Table:

Feature	Ductile	Brittle
Plastic deformation	Large	Very small
Yield point	Distinct	Absent/unclear
Warning before failure	Yes	No
Necking	Present	Absent
Energy absorbed	High	Low

Note: Many biological materials are viscoelastic, so their ductile/brittle behavior can depend on the rate of loading.

Explain the concept of an idealized model for material behavior. Describe the rigid, linear elastic, elastic-perfectly plastic, and elastic-plastic with strain hardening models with sketches.

Define Young's modulus (E), shear modulus (G), and bulk modulus (K). State the relationships among them and Poisson's ratio.

Derive the expression for the elongation ( $\Delta L$ ) of a prismatic bar of length $L$ , cross-sectional area $A$ , subjected to an axial load $P$ , made of a material with Young's modulus $E$ .

Explain the terms proportional limit, elastic limit, and yield strength. How do they differ from one another?

These three terms describe transition points on the stress-strain curve but represent different concepts.

Proportional Limit:

The maximum stress up to which stress is directly proportional to strain (Hooke's law valid).
The stress-strain curve is a straight line up to this point.

Elastic Limit:

The maximum stress the material can withstand without any permanent (residual) deformation.
On unloading before this point, the specimen returns fully to its original shape.
Usually slightly higher than the proportional limit.

Yield Strength:

The stress at which the material begins to deform plastically (permanent deformation begins).
For materials without a clear yield point, it is defined using the 0.2% offset method (offset yield strength).

Key Differences:

Term	Definition	Curve feature
Proportional limit	Linearity ends	End of straight line
Elastic limit	Permanent deformation begins	End of full recovery
Yield strength	Plastic flow begins	Yield/offset point

Order (typically): Proportional limit $\leq$ Elastic limit $\leq$ Yield strength.

Describe the 0.2% offset method for determining the yield strength of materials that do not exhibit a well-defined yield point. Illustrate with a diagram.

Define modulus of resilience and modulus of toughness. Derive/explain how they are obtained from the stress-strain diagram.

Both quantities represent energy absorbed per unit volume by a material, obtained from areas under the stress-strain curve.

Modulus of Resilience ( $U_r$ ):

The energy absorbed per unit volume in stretching the material up to the elastic (proportional) limit.
It is the area under the elastic portion of the stress-strain curve.

$U_r = \int_0^{\epsilon_p} \sigma \, d\epsilon$

For a linear elastic material up to proportional limit stress $\sigma_p$ :

$U_r = \frac{1}{2}\sigma_p \epsilon_p = \frac{\sigma_p^2}{2E}$

Represents the ability to absorb energy without permanent deformation (elastic energy).

Modulus of Toughness ( $U_t$ ):

The total energy absorbed per unit volume up to fracture.
It is the total area under the entire stress-strain curve (elastic + plastic).

$U_t = \int_0^{\epsilon_f} \sigma \, d\epsilon$

Represents the material's ability to absorb energy before breaking, combining strength and ductility.

Comparison:

Property	Region	Meaning
Resilience	Up to elastic limit	Elastic energy storage
Toughness	Up to fracture	Total energy to break

Units: Both are expressed in $J/m^3$ (energy per unit volume).

Distinguish between engineering (nominal) stress-strain and true stress-strain. Why does the true stress continue to rise while engineering stress falls after necking?

A steel rod of diameter $20\,mm$ and length $2\,m$ is subjected to an axial tensile load of $50\,kN$ . If $E = 200\,GPa$ , calculate the stress, strain, and elongation of the rod.

Explain the phenomenon of necking in a ductile material during a tension test. What is meant by ultimate strength and breaking strength?

Define Poisson's ratio. Explain its physical significance and typical values. How is lateral strain related to axial strain?

Compare and contrast the mechanical behavior of biological materials (like bone, tendon, and cartilage) with engineering materials (like steel). Highlight the concepts of anisotropy and viscoelasticity.

Engineering Materials (e.g., steel):

Generally homogeneous and isotropic (same properties in all directions).
Behavior is largely elastic-plastic and time-independent (rate-independent within normal ranges).
Well-defined, reproducible stress-strain curves.

Biological Materials (bone, tendon, cartilage):

Heterogeneous and anisotropic — properties depend on direction and location.
Exhibit viscoelastic behavior — response depends on the rate and duration of loading.
Capable of remodeling/adaptation (living tissue).

Anisotropy:

A material is anisotropic if its mechanical properties vary with direction.
Example: Bone is stronger along its longitudinal axis (direction of osteons) than transversely.

Viscoelasticity:

A combination of elastic (spring-like) and viscous (dashpot-like) behavior.
Characteristic phenomena:
- Creep: progressive deformation under constant load.
- Stress relaxation: decrease in stress under constant strain.
- Hysteresis: energy loss between loading and unloading curves.
- Strain-rate dependence: stiffer/stronger at higher loading rates.

Comparison Table:

Property	Steel	Biological Tissue
Isotropy	Isotropic	Anisotropic
Time dependence	Minimal	Viscoelastic
Homogeneity	Homogeneous	Heterogeneous
Self-repair	No	Yes (living)

Conclusion: Biological materials require more complex models (viscoelastic, anisotropic) than the simple linear-elastic models used for many engineering materials.

Question 1

Define stress and strain. State their SI units and explain why strain is a dimensionless quantity.

Tap card to reveal answer

Answer 1

Stress is defined as the internal resistance offered by a body per unit area against an externally applied force (deforming force). Mathematically:

$\sigma = \frac{F}{A}$

where $F$ is the applied force and $A$ is the cross-sectional area.

SI unit of stress: pascal (Pa) or $N/m^2$ . In biomechanics, MPa is common.

Strain is the measure of deformation of a body relative to its original dimension. For axial (normal) strain:

$\epsilon = \frac{\Delta L}{L_0}$

where $\Delta L$ is the change in length and $L_0$ is the original length.

Strain is dimensionless because it is the ratio of two lengths (change in length divided by original length). The units of length cancel out, so strain has no units (sometimes expressed as mm/mm or as a percentage).

Tap card to flip back

Question 2

Explain the relationship between applied forces and deformations in a loaded body. How does the concept of equilibrium relate to internal forces?

Tap card to reveal answer

Answer 2

When an external force is applied to a body, it tends to change the shape or size of the body, producing a deformation.

Key points:

Applied forces are the external loads (tension, compression, shear) acting on a body.
Deformation is the resulting change in dimensions (elongation, shortening, or angular distortion).
The magnitude of deformation depends on the material stiffness, geometry, and the magnitude/direction of the applied force.

Equilibrium and internal forces:

When a body is in static equilibrium, the net external force is zero: $\sum F = 0$ .
To resist the external load, the material develops internal resisting forces distributed over its cross-section.
By using the method of sections, we can expose these internal forces and relate them to stress:

$\sigma = \frac{F_{internal}}{A}$

The internal force must balance the external force for the body to remain in equilibrium, which is the basis for computing stress at any section.

Tap card to flip back

Question 3

Describe the three basic loading configurations (tension, compression, and shear) with neat diagrams and give one biomechanical example of each.

Tap card to reveal answer

Answer 3

The three basic loading configurations describe how forces act on a structural member.

1. Tension (Axial Tensile Loading)

Forces act away from each other along the axis, tending to stretch/elongate the member.
Stress: $\sigma = \dfrac{F}{A}$ (positive/tensile)
Biomechanical example: Tendons and ligaments under muscular pull.

<---[=====]---> (pulling apart)

2. Compression (Axial Compressive Loading)

Forces act toward each other, tending to shorten/crush the member.
Stress is compressive (negative).
Biomechanical example: Vertebral column and femur bearing body weight.

--->[=====]<--- (pushing together)

3. Shear Loading

Forces act parallel to the cross-section but in opposite directions, causing sliding/angular distortion.
Shear stress: $\tau = \dfrac{V}{A}$
Biomechanical example: Shear forces at joints, e.g., anterior shear at the knee.

--->
[=====]
<---

Each configuration produces a distinct stress and deformation pattern that must be considered in the analysis of biological tissues.

Tap card to flip back

Question 4

State and explain Hooke's Law. Derive the relationship between stress and strain, and define the modulus of elasticity.

Tap card to reveal answer

Answer 4

Hooke's Law states that within the elastic limit, the stress developed in a material is directly proportional to the strain produced.

Statement:

$\sigma \propto \epsilon$

Removing the proportionality by introducing a constant:

$\sigma = E \cdot \epsilon$

where:

$\sigma$ = normal stress
$\epsilon$ = normal strain
$E$ = modulus of elasticity (Young's modulus)

Derivation / Definition of $E$ :

Rearranging Hooke's law:

$E = \frac{\sigma}{\epsilon}$

Substituting $\sigma = \dfrac{F}{A}$ and $\epsilon = \dfrac{\Delta L}{L_0}$ :

$E = \frac{F/A}{\Delta L / L_0} = \frac{F \cdot L_0}{A \cdot \Delta L}$

Modulus of Elasticity ( $E$ ):

It is the slope of the linear elastic portion of the stress-strain curve.
It represents the stiffness of a material — a higher $E$ means a stiffer material.
Unit: pascal (Pa) or GPa.

Note: Hooke's law is valid only up to the proportional limit; beyond it, stress and strain are no longer linearly related.

Tap card to flip back

Question 5

Describe the uniaxial tension test. Explain the procedure, the type of specimen used, and the quantities measured during the test.

Tap card to reveal answer

Answer 5

The uniaxial tension test is a standard mechanical test used to determine the material properties of a specimen by subjecting it to a controlled tensile (pulling) load along a single axis.

Specimen:

A standardized specimen (often dog-bone/dumbbell shaped) with a reduced central gauge section of known cross-sectional area $A_0$ and gauge length $L_0$ .
The enlarged ends are gripped by the testing machine.

Procedure:

The specimen is clamped in the grips of a universal testing machine (UTM).
An axial tensile load is applied gradually and steadily.
The applied load ( $F$ ) is measured by a load cell.
The corresponding elongation ( $\Delta L$ ) is measured by an extensometer across the gauge length.
Loading continues until the specimen fractures.

Quantities measured/computed:

Load vs elongation data → load-elongation diagram.
Engineering stress: $\sigma = \dfrac{F}{A_0}$
Engineering strain: $\epsilon = \dfrac{\Delta L}{L_0}$
These give the stress-strain diagram, from which properties like $E$ , yield strength, ultimate strength, and ductility are obtained.

Purpose: To characterize the material's elastic behavior, strength, and ductility for design and analysis.

Tap card to flip back

Question 6

Distinguish between a load-elongation diagram and a stress-strain diagram. Why is the stress-strain diagram preferred for characterizing material behavior?

Tap card to reveal answer

Answer 6

Load-Elongation Diagram:

Plot of applied load ( $F$ ) on the y-axis versus elongation ( $\Delta L$ ) on the x-axis.
Directly obtained from raw test data.
Depends on specimen dimensions (cross-sectional area and length).

Stress-Strain Diagram:

Plot of stress ( $\sigma = F/A_0$ ) versus strain ( $\epsilon = \Delta L/L_0$ ).
Obtained by normalizing load and elongation by geometry.
Independent of specimen dimensions — a true material property.

Key differences:

Feature	Load-Elongation	Stress-Strain
Axes	Load vs Elongation	Stress vs Strain
Geometry dependence	Yes	No
Represents	Structural response	Material behavior
Units	N vs mm	Pa vs dimensionless

Why stress-strain is preferred:

It removes the effect of specimen size and shape.
The curve becomes a fundamental material characteristic, allowing comparison between different specimens and materials.
Design parameters like modulus, yield strength, and ultimate strength are directly read from it.

Tap card to flip back

Question 7

Tap card to reveal answer

Answer 7

The stress-strain diagram of a ductile material (e.g., mild steel) shows several important points:

Stress
^ U (Ultimate)
| /\
| / _ F (Fracture)
| Y /
| /
| /P,E
| /
| /
|/____> Strain

Salient points:

Proportional Limit (P): The point up to which stress is directly proportional to strain (Hooke's law valid). The curve is a straight line up to here.
Elastic Limit (E): The maximum stress the material can sustain without permanent deformation. On unloading before this point, the specimen returns to original shape.
Yield Point (Y): The stress at which the material begins to deform plastically (permanent deformation). Often shows an upper and lower yield point in mild steel.
Ultimate Strength (U): The maximum stress the material can withstand. Beyond this, necking begins.
Fracture/Breaking Point (F): The point at which the specimen finally breaks/ruptures. The stress here is called the breaking/rupture stress.

Regions:

Elastic region: O to E (recoverable deformation).
Plastic region: E to F (permanent deformation).

Tap card to flip back

Question 8

Define the following material properties as obtained from the stress-strain diagram: elasticity, plasticity, ductility, brittleness, and toughness.

Tap card to reveal answer

Answer 8

The following properties are derived from analysis of the stress-strain diagram:

Elasticity: The ability of a material to return to its original shape after removal of the applied load. Reflected by the elastic (linear) portion of the curve.
Plasticity: The ability of a material to undergo permanent deformation without fracture when stressed beyond the elastic limit. Represented by the region beyond the yield point.
Ductility: The ability of a material to be drawn into wires or undergo large plastic deformation before fracture. Measured by percentage elongation or percentage reduction in area. Ductile materials (e.g., copper, mild steel) show a large plastic region.
Brittleness: The tendency of a material to fracture with little or no plastic deformation. Brittle materials (e.g., cast iron, glass, bone under high strain rate) fail suddenly with a small strain.
Toughness: The ability of a material to absorb energy up to fracture. It is measured by the total area under the stress-strain curve:

$U_t = \int_0^{\epsilon_f} \sigma \, d\epsilon$

A tough material has both good strength and good ductility.

Tap card to flip back

Question 9

Distinguish between ductile and brittle materials based on their stress-strain behavior. Give two examples of each relevant to biomechanics.

Tap card to reveal answer

Answer 9

Ductile Materials:

Exhibit large plastic deformation before fracture.
Show a distinct yield point and pronounced necking.
Give warning before failure (large strain).
High percentage elongation (> 5%).
Examples: Mild steel, copper, and biologically, cortical bone in some conditions and tendon/ligament (show large elongation).

Brittle Materials:

Exhibit little or no plastic deformation before fracture.
No distinct yield point; fracture occurs near the elastic limit.
Fail suddenly without warning.
Low percentage elongation.
Examples: Cast iron, glass, ceramics, and biologically, bone under high strain rates or tooth enamel.

Comparison Table:

Feature	Ductile	Brittle
Plastic deformation	Large	Very small
Yield point	Distinct	Absent/unclear
Warning before failure	Yes	No
Necking	Present	Absent
Energy absorbed	High	Low

Note: Many biological materials are viscoelastic, so their ductile/brittle behavior can depend on the rate of loading.

Tap card to flip back

Question 10

Tap card to reveal answer

Answer 10

Real material behavior is complex, so engineers use idealized (simplified) models to make analysis tractable. Each model approximates the true stress-strain response.

1. Rigid Model:

Assumes the material undergoes no deformation under load.
Stress-strain curve is a vertical line.
Idealization for very stiff bodies.

2. Linear Elastic Model:

Stress is linearly proportional to strain ( $\sigma = E\epsilon$ ) with full recovery on unloading.
Curve is a straight line through origin.

σ | /
| /
| /
|/__ ε

3. Elastic–Perfectly Plastic Model:

Linear elastic up to the yield stress $\sigma_Y$ , then flows at constant stress (horizontal line) with increasing strain.

σ |
| /
| /
|/__ ε

4. Elastic–Plastic with Strain Hardening:

Linear elastic region followed by a plastic region with a positive slope, i.e., stress increases with strain after yielding (work hardening).

σ | _/
| /
| /
|/__ ε

Purpose of idealization:

Simplifies mathematical modeling and design.
Allows closed-form solutions.
The chosen model depends on the required accuracy and the material's actual response.

Tap card to flip back

Question 11

Define Young's modulus (E), shear modulus (G), and bulk modulus (K). State the relationships among them and Poisson's ratio.

Tap card to reveal answer

Answer 11

These are the elastic constants describing a material's response to different loading types.

1. Young's Modulus ( $E$ ):

Ratio of normal stress to normal strain in the elastic region.
$E = \frac{\sigma}{\epsilon}$
Measures axial stiffness.

2. Shear Modulus / Modulus of Rigidity ( $G$ ):

Ratio of shear stress to shear strain.
$G = \frac{\tau}{\gamma}$
Measures resistance to angular/shape distortion.

3. Bulk Modulus ( $K$ ):

Ratio of volumetric (hydrostatic) stress to volumetric strain.
$K = \frac{\sigma_v}{\Delta V / V}$
Measures resistance to uniform compression (volume change).

4. Poisson's Ratio ( $\nu$ ):

Ratio of lateral strain to longitudinal strain.
$\nu = -\frac{\epsilon_{lateral}}{\epsilon_{longitudinal}}$

Relationships among elastic constants:

$E = 2G(1 + \nu)$

$E = 3K(1 - 2\nu)$

$G = \frac{3KE}{9K - E}$

These relations allow one constant to be found if the others are known.

Tap card to flip back

Question 12

Tap card to reveal answer

Answer 12

Given:

Prismatic bar of length $L$
Cross-sectional area $A$
Axial load $P$
Young's modulus $E$

Derivation:

The normal stress in the bar:

$\sigma = \frac{P}{A}$

The normal strain (deformation per unit length):

$\epsilon = \frac{\Delta L}{L}$

From Hooke's law (within elastic limit):

$\sigma = E \cdot \epsilon$

Substituting the expressions:

$\frac{P}{A} = E \cdot \frac{\Delta L}{L}$

Rearranging to solve for $\Delta L$ :

$\boxed{\Delta L = \frac{P L}{A E}}$

Interpretation:

Elongation is directly proportional to the load $P$ and length $L$ .
Elongation is inversely proportional to the cross-sectional area $A$ and modulus $E$ .
$AE$ is termed the axial rigidity of the bar.

Note: For a bar with varying section or load, integrate over the length:

$\Delta L = \int_0^L \frac{P(x)}{A(x) E} \, dx$

Tap card to flip back

Question 13

Explain the terms proportional limit, elastic limit, and yield strength. How do they differ from one another?

Tap card to reveal answer

Answer 13

These three terms describe transition points on the stress-strain curve but represent different concepts.

Proportional Limit:

The maximum stress up to which stress is directly proportional to strain (Hooke's law valid).
The stress-strain curve is a straight line up to this point.

Elastic Limit:

The maximum stress the material can withstand without any permanent (residual) deformation.
On unloading before this point, the specimen returns fully to its original shape.
Usually slightly higher than the proportional limit.

Yield Strength:

The stress at which the material begins to deform plastically (permanent deformation begins).
For materials without a clear yield point, it is defined using the 0.2% offset method (offset yield strength).

Key Differences:

Term	Definition	Curve feature
Proportional limit	Linearity ends	End of straight line
Elastic limit	Permanent deformation begins	End of full recovery
Yield strength	Plastic flow begins	Yield/offset point

Order (typically): Proportional limit $\leq$ Elastic limit $\leq$ Yield strength.

Tap card to flip back

Question 14

Describe the 0.2% offset method for determining the yield strength of materials that do not exhibit a well-defined yield point. Illustrate with a diagram.

Tap card to reveal answer

Answer 14

Many materials (e.g., aluminum, high-strength alloys, and many biological tissues) do not show a distinct yield point. For these, the 0.2% offset method (proof stress method) is used.

Procedure:

Plot the stress-strain curve from test data.
Locate a strain value of 0.002 (i.e., 0.2%) on the strain axis.
From this point, draw a line parallel to the initial linear (elastic) portion of the curve.
The point where this offset line intersects the stress-strain curve defines the offset yield strength ( $\sigma_{0.2\%}$ ).

Stress
^ _ curve
| /|
| / | <- intersection = yield strength
| / |
| / /
| / / <- offset line (parallel)
| / /
| / /
|//____> Strain
0.002

Significance:

Provides a consistent, reproducible measure of yield strength.
The 0.2% permanent strain is an accepted small amount of plastic deformation.
Widely used in material specifications and design codes.

Tap card to flip back

Question 15

Define modulus of resilience and modulus of toughness. Derive/explain how they are obtained from the stress-strain diagram.

Tap card to reveal answer

Answer 15

Both quantities represent energy absorbed per unit volume by a material, obtained from areas under the stress-strain curve.

Modulus of Resilience ( $U_r$ ):

The energy absorbed per unit volume in stretching the material up to the elastic (proportional) limit.
It is the area under the elastic portion of the stress-strain curve.

$U_r = \int_0^{\epsilon_p} \sigma \, d\epsilon$

For a linear elastic material up to proportional limit stress $\sigma_p$ :

$U_r = \frac{1}{2}\sigma_p \epsilon_p = \frac{\sigma_p^2}{2E}$

Represents the ability to absorb energy without permanent deformation (elastic energy).

Modulus of Toughness ( $U_t$ ):

The total energy absorbed per unit volume up to fracture.
It is the total area under the entire stress-strain curve (elastic + plastic).

$U_t = \int_0^{\epsilon_f} \sigma \, d\epsilon$

Represents the material's ability to absorb energy before breaking, combining strength and ductility.

Comparison:

Property	Region	Meaning
Resilience	Up to elastic limit	Elastic energy storage
Toughness	Up to fracture	Total energy to break

Units: Both are expressed in $J/m^3$ (energy per unit volume).

Tap card to flip back

Question 16

Distinguish between engineering (nominal) stress-strain and true stress-strain. Why does the true stress continue to rise while engineering stress falls after necking?

Tap card to reveal answer

Answer 16

Engineering (Nominal) Stress and Strain:

Based on the original cross-sectional area ( $A_0$ ) and original length ( $L_0$ ).

$\sigma_{eng} = \frac{F}{A_0}, \quad \epsilon_{eng} = \frac{\Delta L}{L_0}$

True Stress and Strain:

Based on the instantaneous (actual) cross-sectional area ( $A$ ) and instantaneous length.

$\sigma_{true} = \frac{F}{A}, \quad \epsilon_{true} = \ln\left(\frac{L}{L_0}\right)$

Relationships (before necking, constant volume):

$\sigma_{true} = \sigma_{eng}(1 + \epsilon_{eng})$
$\epsilon_{true} = \ln(1 + \epsilon_{eng})$

Why true stress rises while engineering stress falls after necking:

After the ultimate point, necking occurs — the cross-sectional area reduces sharply at a localized region.
Engineering stress uses the constant original area $A_0$ , so as load $F$ drops, $\sigma_{eng} = F/A_0$ decreases.
True stress uses the actual reduced area $A$ , which decreases faster than the load, so $\sigma_{true} = F/A$ continues to increase.
Thus, the true stress-strain curve keeps rising up to fracture, giving a more accurate picture of the material's real behavior.

Tap card to flip back

Question 18

Explain the phenomenon of necking in a ductile material during a tension test. What is meant by ultimate strength and breaking strength?

Tap card to reveal answer

Answer 18

Necking:

Necking is the localized reduction in cross-sectional area that occurs in a ductile specimen after the ultimate strength is reached.
Beyond the ultimate point, deformation is no longer uniform; it concentrates in one region, forming a 'neck'.
As the area rapidly reduces in this region, the load-bearing capacity drops, and fracture eventually occurs at the neck.

Ultimate Strength (Ultimate Tensile Strength, UTS):

The maximum stress the material can withstand, corresponding to the peak of the engineering stress-strain curve.

$\sigma_u = \frac{P_{max}}{A_0}$

Represents the maximum load-carrying capacity of the material.

Breaking / Rupture Strength:

The stress at the point of fracture (final failure).

$\sigma_b = \frac{P_{fracture}}{A_0}$

On the engineering stress-strain curve, the breaking strength appears lower than the ultimate strength (because of necking, but $A_0$ is used).
On the true stress-strain curve, the breaking stress is actually higher due to the reduced actual area.

Summary: Necking marks the onset of instability leading to failure; UTS is the peak stress and breaking strength is the stress at rupture.

Tap card to flip back

Question 19

Define Poisson's ratio. Explain its physical significance and typical values. How is lateral strain related to axial strain?

Tap card to reveal answer

Answer 19

Poisson's Ratio ( $\nu$ ):

It is defined as the negative ratio of lateral (transverse) strain to axial (longitudinal) strain within the elastic limit.

$\nu = -\frac{\epsilon_{lateral}}{\epsilon_{axial}}$

Physical Significance:

When a material is stretched axially, it tends to contract laterally (and vice versa for compression).
Poisson's ratio quantifies this coupling between longitudinal and transverse deformations.
It is a dimensionless material property.

Relationship between lateral and axial strain:

$\epsilon_{lateral} = -\nu \cdot \epsilon_{axial}$

The negative sign indicates that lateral strain is opposite in sign to axial strain (elongation causes lateral contraction).

Typical Values:

Most metals: $\nu \approx 0.25 - 0.35$ (steel $\approx 0.3$ ).
Rubber: $\nu \approx 0.5$ (nearly incompressible).
Cork: $\nu \approx 0$ (little lateral change).
Bone: $\nu \approx 0.3$ .

Bounds: Theoretically, $-1 \leq \nu \leq 0.5$ . A value of $0.5$ indicates an incompressible material (constant volume).

Tap card to flip back

Question 20

Tap card to reveal answer

Answer 20

Engineering Materials (e.g., steel):

Generally homogeneous and isotropic (same properties in all directions).
Behavior is largely elastic-plastic and time-independent (rate-independent within normal ranges).
Well-defined, reproducible stress-strain curves.

Biological Materials (bone, tendon, cartilage):

Heterogeneous and anisotropic — properties depend on direction and location.
Exhibit viscoelastic behavior — response depends on the rate and duration of loading.
Capable of remodeling/adaptation (living tissue).

Anisotropy:

A material is anisotropic if its mechanical properties vary with direction.
Example: Bone is stronger along its longitudinal axis (direction of osteons) than transversely.

Viscoelasticity:

A combination of elastic (spring-like) and viscous (dashpot-like) behavior.
Characteristic phenomena:
- Creep: progressive deformation under constant load.
- Stress relaxation: decrease in stress under constant strain.
- Hysteresis: energy loss between loading and unloading curves.
- Strain-rate dependence: stiffer/stronger at higher loading rates.

Comparison Table:

Property	Steel	Biological Tissue
Isotropy	Isotropic	Anisotropic
Time dependence	Minimal	Viscoelastic
Homogeneity	Homogeneous	Heterogeneous
Self-repair	No	Yes (living)

Conclusion: Biological materials require more complex models (viscoelastic, anisotropic) than the simple linear-elastic models used for many engineering materials.

Tap card to flip back

1 / 20

Unit2 Unit4

Unit 3: Stress and strain - Subjective Questions

Did this save you a night before the exam?