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Shafts are fundamental components in many mechanical systems, particularly in transmitting power from one part to another - like in engines, gearboxes, and turbines. When a shaft transmits torque, it experiences a twisting action known as torsion. This twisting induces internal forces that generate shear stresses within the shaft material. Understanding these stresses is crucial for designing shafts that are both strong and reliable, preventing failures during operation.
Torsion can be thought of as the action of twisting a shaft by applying torque- a twisting moment. Imagine gripping a cylindrical shaft with both hands and twisting it: the shaft undergoes deformation. This deformation causes internal shear stresses within the material as the shaft resists the applied torque.
The essential terms you'll need to know are:
Figure: Cross-section of a circular shaft showing the radius r, the applied torque T, and the direction of shear stress τ at the outer surface where it is maximum.
To quantify the shear stress inside a circular shaft under torque, let's consider a small ring element in the shaft cross-section at radius \( r \). The torsional moment applied must be balanced by the resistive shear forces acting on this ring.
The total torque is the sum of all infinitesimal shear forces multiplied by their lever arms (radius):
\[T = \int_A \tau \, r \, dA\]We observe from experiments and theory that shear stress varies linearly with radius in a circular shaft, i.e.,
\[\tau \propto r\]Assuming the maximum shear stress \( \tau_{\max} \) occurs at the outer radius \( R \), the shear stress at any radius \( r \) is:
\[\tau = \frac{r}{R} \tau_{\max}\]Substituting into the torque equilibrium, we get:
\[T = \int_A \left(\frac{r}{R} \tau_{\max}\right) r \, dA = \frac{\tau_{\max}}{R} \int_A r^2 \, dA\]The integral \( \int_A r^2 \, dA \) is the definition of the polar moment of inertia \( J \). Therefore,
\[T = \frac{\tau_{\max}}{R} J \quad \Rightarrow \quad \tau_{\max} = \frac{T R}{J}\]At any point inside the shaft at radius \( r \), the shear stress is:
\[\boxed{\tau = \frac{T r}{J}}\]The polar moment of inertia \( J \), also called the torsional constant, quantifies a shaft's resistance to twisting. It depends solely on the geometry of the shaft's cross section.
For a solid circular shaft of diameter \( d \):
\[J = \frac{\pi d^4}{32}\]For a hollow circular shaft with outer diameter \( d_o \) and inner diameter \( d_i \):
\[J = \frac{\pi}{32} \left(d_o^4 - d_i^4\right)\]This difference accounts for the hollow section's lack of material, which reduces its torsional rigidity.
| Type of Shaft | Polar Moment of Inertia \( J \) (m4) | Description |
|---|---|---|
| Solid Circular Shaft | \(\displaystyle \frac{\pi d^4}{32}\) | Used when shaft is fully solid, e.g., steel transmission shafts. |
| Hollow Circular Shaft | \(\displaystyle \frac{\pi}{32} (d_o^4 - d_i^4)\) | Used when shaft has a hollow centre for weight saving or hollow pipes. |
When a shaft is subjected to torque, it not only develops shear stress but also twists by a certain angle along its length. This angle of twist is a measure of the shaft's torsional deformation and is vital in mechanical design to ensure functional performance and prevent failure.
The angle of twist \( \theta \) (in radians) for a shaft of length \( L \) under torque \( T \) is given by:
\[\boxed{\theta = \frac{T L}{G J}}\]where:
Figure: Shaft fixed at one end twists by angle \( \theta \) at free end under applied torque \( T \), shaft length \( L \).
Step 1: Convert diameter to meters.
Given \( d = 40 \text{ mm} = 0.040 \text{ m} \).
Step 2: Calculate the polar moment of inertia \( J \) for the solid circular shaft:
\[ J = \frac{\pi d^4}{32} = \frac{3.1416 \times (0.040)^4}{32} = \frac{3.1416 \times 2.56 \times 10^{-7}}{32} \approx 2.513 \times 10^{-8} \, \text{m}^4 \]
Step 3: Identify the outer radius \( R = d/2 = 0.020 \text{ m} \).
Step 4: Calculate the maximum shear stress \( \tau_{\max} \) using:
\[ \tau_{\max} = \frac{T R}{J} = \frac{1200 \times 0.020}{2.513 \times 10^{-8}} = \frac{24}{2.513 \times 10^{-8}} = 9.55 \times 10^{8} \text{ Pa} = 95.5 \text{ MPa} \]Answer: The maximum shear stress in the shaft is approximately 95.5 MPa.
Step 1: Convert all dimensions to meters:
Step 2: Calculate the polar moment of inertia \( J \):
\[ J = \frac{\pi}{32} (d_o^4 - d_i^4) = \frac{3.1416}{32} \left(0.080^4 - 0.050^4\right) \] \[ d_o^4 = (0.080)^4 = 4.096 \times 10^{-5}, \quad d_i^4 = (0.050)^4 = 6.25 \times 10^{-6} \] \[ J = \frac{3.1416}{32} (4.096 \times 10^{-5} - 6.25 \times 10^{-6}) = \frac{3.1416}{32} \times 3.471 \times 10^{-5} \approx 3.41 \times 10^{-6} \, \text{m}^4 \]Step 3: Find the outer radius \( R = d_o/2 = 0.040 \text{ m} \).
Step 4: Calculate maximum shear stress:
\[ \tau_{\max} = \frac{T R}{J} = \frac{1500 \times 0.040}{3.41 \times 10^{-6}} = \frac{60}{3.41 \times 10^{-6}} \approx 17.6 \times 10^{6} \text{ Pa} = 17.6 \text{ MPa} \]Step 5: Calculate angle of twist \( \theta \):
\[ \theta = \frac{T L}{G J} = \frac{1500 \times 2}{80 \times 10^{9} \times 3.41 \times 10^{-6}} = \frac{3000}{272.8 \times 10^{3}} \approx 0.011 \text{ radians} \]Answer:
Step 1: Convert diameter to meters:
\( d = 60 \text{ mm} = 0.060 \text{ m} \)
Step 2: Calculate torque \( T \) using power formula:
\[ P = \frac{2 \pi N T}{60} \quad \Rightarrow \quad T = \frac{60 P}{2 \pi N} \] \[ P = 15 \text{ kW} = 15,000 \text{ W}, \quad N=1200 \text{ rpm} \] \[ T = \frac{60 \times 15000}{2 \pi \times 1200} = \frac{900000}{7539.82} \approx 119.4 \text{ Nm} \]Step 3: Calculate polar moment of inertia \( J \):
\[ J = \frac{\pi d^4}{32} = \frac{3.1416 \times (0.060)^4}{32} = \frac{3.1416 \times 1.296 \times 10^{-5}}{32} = 1.27 \times 10^{-6} \text{ m}^4 \]Step 4: Calculate outer radius \( R = 0.030 \text{ m} \) and maximum shear stress:
\[ \tau_{\max} = \frac{T R}{J} = \frac{119.4 \times 0.030}{1.27 \times 10^{-6}} = \frac{3.582}{1.27 \times 10^{-6}} \approx 2.82 \times 10^{6} \text{ Pa} = 2.82 \text{ MPa} \]Answer:
Step 1: Convert diameter to meters:
\( d = 50 \text{ mm} = 0.050 \text{ m} \)
Step 2: Calculate polar moment of inertia \( J \) and moment of inertia \( I \):
Torsion formula requires \( J \) for shear stress, bending requires \( I \) for bending stress.
\[ J = \frac{\pi d^4}{32} = \frac{3.1416 \times (0.050)^4}{32} = 3.07 \times 10^{-7} \, \text{m}^4 \] \[ I = \frac{\pi d^4}{64} = \frac{3.1416 \times (0.050)^4}{64} = 1.54 \times 10^{-7} \, \text{m}^4 \]Step 3: Calculate outer radius \( R = 0.025 \text{ m} \).
Step 4: Calculate shear stress due to torque \( \tau \):
\[ \tau = \frac{T R}{J} = \frac{600 \times 0.025}{3.07 \times 10^{-7}} = 4.88 \times 10^{7} \text{ Pa} = 48.8 \text{ MPa} \]Step 5: Calculate bending stress \( \sigma_b \) (normal stress on outer fiber):
\[ \sigma_b = \frac{M R}{I} = \frac{800 \times 0.025}{1.54 \times 10^{-7}} = 1.30 \times 10^{8} \text{ Pa} = 130 \text{ MPa} \]Step 6: Since bending stress is normal and torsion causes shear stress, resultant shear stress on the surface can be found by:
\[ \tau_{\text{resultant}} = \sqrt{\tau^2 + \left(\frac{\sigma_b}{2}\right)^2} \] where the bending stresses produce principal stresses and the maximum shear stress from combined loading is calculated by this von Mises criterion approximation. \[ \tau_{\text{resultant}} = \sqrt{(48.8)^2 + (130/2)^2} = \sqrt{2381 + 4225} = \sqrt{6606} = 81.3 \text{ MPa} \]Answer: The maximum resultant shear stress in the shaft is approximately 81.3 MPa.
Step 1: Convert diameter and length to meters:
Step 2: Calculate the volume of the shaft:
\[ V = \pi \left(\frac{d}{2}\right)^2 L = 3.1416 \times (0.025)^2 \times 2.5 = 3.1416 \times 0.000625 \times 2.5 = 0.00491 \, \text{m}^3 \]Step 3: Calculate mass from volume and density:
\[ m = \rho V = 7850 \times 0.00491 = 38.5 \, \text{kg} \]Step 4: Compute cost:
\[ \text{Cost} = m \times \text{cost per kg} = 38.5 \times 60 = INR 2310 \]Answer: The approximate cost of the steel shaft material is INR 2310.
When to use: In every numerical problem to ensure consistency of units and avoid errors.
When to use: When calculating polar moment of inertia for hollow shafts to avoid overestimating stiffness.
When to use: To quickly find the maximum shear stress in circular shafts without calculating distribution.
When to use: In problems involving rotating shafts transmitting power, to switch between torque and power.
When to use: In combined loading and complex torsion problems where direction matters for stress signs.
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