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In mechanical engineering, understanding how materials respond to applied forces is essential for safe and efficient design. Elastic properties describe a material's ability to return to its original shape and size after the forces causing deformation are removed. This means the material temporarily deforms but recovers completely without permanent damage.
When a solid material, such as steel or aluminum, is subjected to forces, it deforms. This deformation can be broadly classified into two types: elastic deformation and plastic deformation. Elastic deformation is reversible; the material regains its original shape once the load is removed. Plastic deformation is permanent; the material undergoes irreversible changes.
Elastic properties are critical in the design of structures and mechanical components to ensure that stresses and strains stay within limits where materials behave elastically, preventing failure or permanent deformation.
Let's begin by understanding the basic quantities used to describe deformation: stress and strain.
Stress is the internal force per unit area developed within a material when an external force acts upon it. It helps quantify how intensely the load is acting inside the material.
Mathematically, normal stress (denoted by \( \sigma \)) is defined as the force applied perpendicular to a cross-sectional area divided by that area:
Stress can be tensile (pulling apart, positive stress) or compressive (pushing together, negative stress). Besides normal stress, there is also shear stress, which acts tangentially to the surface.
Strain measures the deformation of a material as the ratio of change in length to the original length. It is a dimensionless quantity (no units) since it is a ratio.
If a specimen of length \( L \) changes length by \( \Delta L \) under load, the normal strain \( \varepsilon \) is:
Illustration: A rod fixed at one end is loaded axially at the other end, causing elongation. The original length is \( L \), and the deformed length after loading is \( L + \Delta L \). The axial force applied is \( F \) acting over a cross-sectional area \( A \).
Within the limits of elastic deformation, stress and strain are related in a simple and direct manner. This relationship is called Hooke's Law, named after the English scientist Robert Hooke.
Hooke's Law states:
"The stress applied to a material is directly proportional to the strain produced, as long as the material remains within its elastic limit."
This proportionality can be written as:
Young's modulus (E) is a fundamental elastic constant that indicates how stiff a material is. The higher the \( E \), the less it deforms under a given load.
Graph Explanation: The stress-strain curve is a straight line up to the proportional limit, indicating direct proportionality (Hooke's law holds). Beyond this point, the curve deviates as elastic behavior ends.
The stress-strain diagram for a material under tension shows how stress varies with strain throughout the deformation process.
This typical curve for a ductile metal under tensile test includes:
Step 1: Convert all dimensions to SI units.
Diameter, \( d = 20\, \text{mm} = 0.02\, \text{m} \)
Length, \( L = 2\, \text{m} \)
Force, \( F = 10\, \text{kN} = 10\,000\, \text{N} \)
Step 2: Calculate cross-sectional area of rod (circular):
\[ A = \frac{\pi}{4} d^2 = \frac{3.1416}{4} \times (0.02)^2 = 3.1416 \times 10^{-4} \text{ m}^2 \]
Step 3: Calculate normal stress:
\[ \sigma = \frac{F}{A} = \frac{10000}{3.1416 \times 10^{-4}} \approx 3.183 \times 10^{7} \text{ Pa} = 31.83 \text{ MPa} \]
Step 4: Calculate strain using Hooke's Law:
\[ \varepsilon = \frac{\sigma}{E} = \frac{3.183 \times 10^{7}}{2.0 \times 10^{11}} = 1.591 \times 10^{-4} \]
Answer: Normal stress = 31.83 MPa, Strain = \(1.59 \times 10^{-4}\) (dimensionless)
Step 1: Convert all units to SI:
Diameter, \( d = 10\, \text{mm} = 0.01\, \text{m} \)
Length, \( L = 1.5\, \text{m} \)
Force, \( F = 5\, \text{kN} = 5000\, \text{N} \)
Elongation, \( \Delta L = 0.5\, \text{mm} = 0.0005\, \text{m} \)
Step 2: Calculate the cross-sectional area:
\[ A = \frac{\pi}{4} d^2 = \frac{3.1416}{4} \times (0.01)^2 = 7.854 \times 10^{-5} \text{ m}^2 \]
Step 3: Calculate normal stress:
\[ \sigma = \frac{F}{A} = \frac{5000}{7.854 \times 10^{-5}} \approx 6.368 \times 10^{7} \text{ Pa} \]
Step 4: Calculate strain:
\[ \varepsilon = \frac{\Delta L}{L} = \frac{0.0005}{1.5} = 3.333 \times 10^{-4} \]
Step 5: Calculate Young's modulus \( E \):
\[ E = \frac{\sigma}{\varepsilon} = \frac{6.368 \times 10^{7}}{3.333 \times 10^{-4}} = 1.911 \times 10^{11} \text{ Pa} \]
Answer: Young's modulus of the material is approximately \(1.91 \times 10^{11}\) Pa (or 191 GPa).
Step 1: Convert units:
Diameter, \( d = 25\, \text{mm} = 0.025\, \text{m} \)
Length, \( L = 3\, \text{m} \)
Force, \( F = 8\, \text{kN} = 8000\, \text{N} \)
Step 2: Cross-sectional area of the rod:
\[ A = \frac{\pi}{4} d^2 = \frac{3.1416}{4} \times (0.025)^2 = 4.9087 \times 10^{-4} \text{ m}^2 \]
Step 3: Calculate stress:
\[ \sigma = \frac{F}{A} = \frac{8000}{4.9087 \times 10^{-4}} = 1.629 \times 10^{7} \text{ Pa} \]
Step 4: Use Hooke's law to find strain:
\[ \varepsilon = \frac{\sigma}{E} = \frac{1.629 \times 10^{7}}{7.0 \times 10^{10}} = 2.327 \times 10^{-4} \]
Step 5: Calculate elongation:
\[ \Delta L = \varepsilon \times L = 2.327 \times 10^{-4} \times 3 = 6.981 \times 10^{-4} \text{ m} = 0.698 \text{ mm} \]
Answer: Elongation of the aluminium rod is approximately 0.698 mm.
Step 1: Convert units and calculate area:
Diameter, \( d=40\, \text{mm}=0.04\, \text{m} \)
Cross-sectional area \( A = \frac{\pi}{4} d^2 = \frac{3.1416}{4} \times (0.04)^2 = 1.2566 \times 10^{-3} \, m^2 \)
Step 2: Calculate normal axial stress:
\[ \sigma = \frac{F}{A}=\frac{20000}{1.2566 \times 10^{-3}} = 1.5915 \times 10^{7} \text{ Pa} = 15.92 \text{ MPa} \]
Step 3: Calculate shear stress due to torque:
Polar moment of inertia for circular shaft:
\[ J = \frac{\pi}{32} d^4 = \frac{3.1416}{32} \times (0.04)^4 = 8.042 \times 10^{-7} \, m^4 \]
Shear stress at outer surface:
\[ \tau = \frac{T \cdot c}{J} = \frac{500 \times 0.02}{8.042 \times 10^{-7}} = 1.243 \times 10^{7} \text{ Pa} = 12.43 \text{ MPa} \]
Note: \( c = \frac{d}{2} = 0.02 m \)
Step 4: Calculate axial strain:
\[ \varepsilon = \frac{\sigma}{E} = \frac{1.5915 \times 10^{7}}{2.0 \times 10^{11}} = 7.96 \times 10^{-5} \]
Step 5: Calculate shear strain:
\[ \gamma = \frac{\tau}{G} = \frac{1.243 \times 10^{7}}{8.0 \times 10^{10}} = 1.554 \times 10^{-4} \]
Step 6: Calculate equivalent (von Mises) stress for combined loading:
\[ \sigma_{eq} = \sqrt{\sigma^2 + 3 \tau^2} = \sqrt{(15.92)^2 + 3 \times (12.43)^2} \text{ MPa} \]
\[ = \sqrt{253.6 + 3 \times 154.6} = \sqrt{253.6 + 463.8} = \sqrt{717.4} = 26.78 \text{ MPa} \]
Answer:
Step 1: Identify given values:
\( \sigma_x = +40\, \text{MPa}, \; \sigma_y = -10\, \text{MPa}, \; \tau_{xy} = 15\, \text{MPa} \)
Step 2: Calculate average normal stress:
\[ \sigma_{avg} = \frac{\sigma_x + \sigma_y}{2} = \frac{40 + (-10)}{2} = 15\, \text{MPa} \]
Step 3: Calculate radius of Mohr's Circle:
\[ R = \sqrt{\left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2} = \sqrt{\left(\frac{40 - (-10)}{2}\right)^2 + (15)^2} \]
\[ = \sqrt{(25)^2 + 225} = \sqrt{625 + 225} = \sqrt{850} = 29.15\, \text{MPa} \]
Step 4: Calculate principal stresses:
\[ \sigma_1 = \sigma_{avg} + R = 15 + 29.15 = 44.15\, \text{MPa} \]
\[ \sigma_2 = \sigma_{avg} - R = 15 - 29.15 = -14.15\, \text{MPa} \]
Step 5: Maximum shear stress is equal to radius \( R \):
\[ \tau_{max} = R = 29.15\, \text{MPa} \]
Answer:
When to use: At the start of every problem to avoid unit mismatch errors.
When to use: When dealing with stress-strain data or estimating strain from stress.
When to use: Always during strain calculation and interpreting results.
When to use: When given normal and shear stresses at a point in problems involving plane stress.
When to use: For shafts, beams, or members under complex real-world loading.
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