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When we apply forces to solid objects-say, stretching a metal wire or compressing a wooden beam-they change their shape or size slightly. This change is known as deformation. Understanding how solids deform under various loads is essential for Mechanical Engineering, ensuring structures are safe, efficient, and durable.
Two key quantities help describe deformation: the internal forces per unit area (called stress) and the relative change in dimension (called strain). The relationship between stress and strain lays the foundation for designing all mechanical components.
Hooke's law, a cornerstone in the study of material behavior, establishes a simple yet powerful connection between stress and strain - valid as long as the material is not permanently deformed. This principle is extensively tested in competitive exams, capturing both conceptual understanding and numerical problem-solving.
Hooke's lawstress (\(\sigma\)) produced is directly proportional to the corresponding strain (\(\varepsilon\)). Mathematically,
This proportionality applies only up to the limit of proportionality, beyond which the relationship becomes nonlinear. The material returns to its original shape only if the load does not exceed the elastic limit. Beyond this, permanent deformation (plastic behavior) occurs.
Before delving deeper into Hooke's law, it's essential to precisely define stress and strain.
Stress is the internal resistance provided by a material per unit area when subjected to an external force. For axial loading (force applied along the length), normal stress is given by
Stress has units of Pascal (Pa), where 1 Pa = 1 N/m².
Strain is a measure of relative deformation - the ratio of change in length to the original length:
Since strain is a ratio of lengths, it is dimensionless and usually expressed as a small decimal or in microstrain (1 microstrain = \(10^{-6}\)).
For common materials, typical stress and strain values within elastic range are:
The constant of proportionality in Hooke's law is called the elastic modulus or Young's modulus (\(E\)). It quantifies the stiffness of a material - higher \(E\) means the material resists deformation more strongly.
Rearranging Hooke's law:
Young's modulus depends only on the material, not on its shape or size.
A steel rod of length 2 m and diameter 20 mm is stretched by a force of 30 kN, resulting in an elongation of 1 mm. Calculate the normal stress and strain in the rod.
Step 1: Calculate the cross-sectional area \(A\).
Diameter, \(d = 20\) mm = 0.02 m
Area, \(A = \pi d^2 / 4 = \pi (0.02)^2 / 4 = \pi \times 0.0004 / 4 = 0.000314\) m² (approximately)
Step 2: Calculate the normal stress \(\sigma\).
Force, \(F = 30\,000\) N
\[ \sigma = \frac{F}{A} = \frac{30\,000}{0.000314} = 95\,541\, \text{Pa} = 95.54\, \text{MPa} \]
Step 3: Calculate the normal strain \(\varepsilon\).
Original length, \(L = 2\) m
Elongation, \(\Delta L = 1\) mm = 0.001 m
\[ \varepsilon = \frac{\Delta L}{L} = \frac{0.001}{2} = 0.0005 \]
Answer: Normal stress = 95.54 MPa, strain = 0.0005 (dimensionless)
A copper wire 1.5 m long and 10 mm diameter is stretched such that the strain in the wire is 0.0003. Given the Young's modulus for copper is 1.1 x 1011 Pa, find the force applied.
Step 1: Calculate cross-sectional area \(A\).
Diameter \(d = 10\) mm = 0.01 m
\[ A = \frac{\pi d^2}{4} = \frac{\pi \times (0.01)^2}{4} = 7.85 \times 10^{-5} \text{ m}^2 \]
Step 2: Calculate the stress using Hooke's law \(\sigma = E \varepsilon\).
\[ \sigma = 1.1 \times 10^{11} \times 0.0003 = 3.3 \times 10^{7} \text{ Pa} = 33 \text{ MPa} \]
Step 3: Calculate the force \(F = \sigma \times A\).
\[ F = 3.3 \times 10^{7} \times 7.85 \times 10^{-5} = 2,590 \text{ N} \]
Answer: Force applied = 2,590 N
A circular steel shaft of diameter 50 mm and length 1.2 m is subjected to a tensile force of 20 kN and a torsional moment simultaneously. If the shear strain due to torsion is given as 0.002 and Young's modulus is 2 x 1011 Pa, find the total axial strain using Hooke's law.
Step 1: Calculate axial tensile stress \(\sigma\).
Diameter \(d = 50\) mm = 0.05 m
Cross-sectional area, \[ A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.05)^2 = 0.0019635 \text{ m}^2 \]
Force \(F = 20\,000\) N
\[ \sigma = \frac{F}{A} = \frac{20\,000}{0.0019635} = 10.19 \times 10^{6} \text{ Pa} = 10.19 \text{ MPa} \]
Step 2: Calculate axial strain due to tensile force using Hooke's law.
\[ \varepsilon_{axial} = \frac{\sigma}{E} = \frac{10.19 \times 10^{6}}{2 \times 10^{11}} = 5.095 \times 10^{-5} \]
Step 3: Note that shear strain from torsion is given as 0.002.
Axial strain is generally independent of shear strain; thus, total axial strain remains the same as the tensile strain if no coupling effects are considered:
\[ \varepsilon_{total} = \varepsilon_{axial} = 5.095 \times 10^{-5} \]
Note: For combined loading including torsion, Hooke's law applies separately for axial and shear components; they do not add directly for axial strain.
Answer: Axial strain = \(5.095 \times 10^{-5}\)
A steel rod of length 3 m and diameter 25 mm is used in construction. The allowable tensile stress is 140 MPa, and the Young's modulus is 2 x 1011 Pa. If the cost of steel is Rs.60 per kg and the density is 7850 kg/m³, estimate the cost of the rod.
Step 1: Calculate the volume \(V\) of the rod.
Diameter, \(d = 25\) mm = 0.025 m
Cross-sectional area, \[ A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.025)^2 = 4.91 \times 10^{-4} \text{ m}^2 \]
Length = 3 m
\[ V = A \times L = 4.91 \times 10^{-4} \times 3 = 0.001473 \text{ m}^3 \]
Step 2: Calculate mass \(m\) of the rod.
Density \(\rho = 7850\) kg/m³
\[ m = \rho V = 7850 \times 0.001473 = 11.56 \text{ kg} \]
Step 3: Calculate cost.
Cost/kg = Rs.60
\[ \text{Cost} = 11.56 \times 60 = Rs.693.60 \]
Answer: The cost of the steel rod is approximately Rs.693.60.
In a tensile test, a steel specimen with original length 1 m and cross-sectional area 50 mm² elongates by 0.5 mm when subjected to an axial load of 20 kN. Determine the Young's modulus of the steel specimen.
Step 1: Convert units where necessary.
Cross-sectional area \(A = 50\) mm² = \(50 \times 10^{-6}\) m² = \(5 \times 10^{-5}\) m²
Elongation \(\Delta L = 0.5\) mm = 0.0005 m
Step 2: Calculate stress \(\sigma = \frac{F}{A}\).
Force \(F = 20\,000\) N
\[ \sigma = \frac{20\,000}{5 \times 10^{-5}} = 4 \times 10^{8} \text{ Pa} = 400 \text{ MPa} \]
Step 3: Calculate strain \(\varepsilon = \frac{\Delta L}{L}\).
Original length \(L = 1\) m
\[ \varepsilon = \frac{0.0005}{1} = 0.0005 \]
Step 4: Calculate Young's modulus \(E = \frac{\sigma}{\varepsilon}\).
\[ E = \frac{4 \times 10^{8}}{0.0005} = 8 \times 10^{11} \text{ Pa} \]
Answer: Young's modulus \(E = 8 \times 10^{11} \text{ Pa}\)
Note: This value is unusually high for steel (typically ~2 x 1011 Pa), indicating possible experimental error or assumptions in problem statement.
When to use: When solving problems to avoid incorrect stress-strain assumptions.
When to use: During every numerical problem to maintain accuracy.
When to use: Useful for saving time in exams.
When to use: While interpreting word problems and verifying answers.
When to use: Always check final answers for logical consistency.
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