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Torsion is a type of mechanical loading that involves twisting a structural member about its longitudinal axis. Imagine holding a cylindrical shaft - such as the axle of a motor or the drive shaft in a car - and applying a torque at one end while the other end is fixed. The shaft experiences internal stresses and deforms, twisting relative to its fixed end. This twisting action is known as torsion.
Understanding torsion is crucial in mechanical engineering because many practical components (like shafts, axles, and couplings) transmit power by rotating under torque. The ability to predict the internal stresses and the resulting angular deformation ensures that these components can be designed for safety and reliability.
Common scenarios requiring torsion analysis include:
In this chapter, you will learn how torsion causes shear stresses inside circular shafts, how to calculate the angle of twist, and how power transmission is related to applied torque and rotational speed. We will build from basic principles towards complex situations like combined bending and torsion, and special cases such as thin-walled tubes.
When a torque \( T \) is applied to a circular shaft, it induces shear stresses within the shaft material. These stresses vary across the radius of the shaft's cross-section, being zero at the center and maximum at the outer surface.
Why shear stress? Because the applied torque tries to rotate one cross-section relative to another, causing internal layers of the shaft material to slide over each other in a shear manner.
The shear stress at any radius \( r \) (measured from the center of the shaft) can be determined using the torsion formula:
The parameter \( J \), called the polar moment of inertia, reflects the resistance of the shaft cross section to torsion. A larger \( J \) means the shaft can carry more torque for the same amount of shear stress.
For a solid circular shaft of diameter \( d \), \( J \) is given by:
The distribution of shear stress across the shaft radius is linear. This means it is zero at the center (\(r=0\)) and reaches its maximum value at the outer radius \( r = \frac{d}{2} \). It is important to note that the maximum shear stress, \( \tau_{\max} \), occurs at the outer surface and is:
Under torsion, not only do internal stresses develop, but the shaft also undergoes a deformation called angle of twist. The angle of twist \( \theta \) is the angular displacement of one end relative to the other due to the applied torque. It is measured in radians.
Physically, twisting causes the shaft material to shear, and materials resist shear through their modulus of rigidity, denoted by \( G \). Modulus of rigidity (or shear modulus) measures a material's resistance to shear deformation.
For a shaft of length \( L \) subjected to a torque \( T \), the angle of twist is given by the formula:
The angle of twist \( \theta \) is directly proportional to applied torque and shaft length and inversely proportional to the shaft's stiffness parameters \( G \) and \( J \). This means a longer or more flexible (smaller \( G \)) shaft twists more under the same torque.
Step 1: Convert all units to SI base units
Diameter \( d = 40 \text{ mm} = 0.04 \text{ m} \)
Torque \( T = 200 \text{ N·m} \)
Step 2: Calculate the polar moment of inertia \( J \) for a solid circular shaft:
\[ J = \frac{\pi d^{4}}{32} = \frac{\pi (0.04)^{4}}{32} = \frac{\pi \times 2.56 \times 10^{-7}}{32} \approx 2.51 \times 10^{-8} \text{ m}^4 \]
Step 3: Calculate the maximum shear stress using:
\[ \tau_{\max} = \frac{T c}{J} \]
where \( c = \frac{d}{2} = 0.02 \text{ m} \)
Therefore,
\[ \tau_{\max} = \frac{200 \times 0.02}{2.51 \times 10^{-8}} = \frac{4}{2.51 \times 10^{-8}} = 1.59 \times 10^{8} \text{ Pa} = 159 \text{ MPa} \]
Answer: Maximum shear stress induced in the shaft is approximately 159 MPa.
Step 1: Convert units to SI:
Diameter \( d = 30 \text{ mm} = 0.03 \text{ m} \)
Length \( L = 2 \text{ m} \)
Torque \( T = 150 \text{ N·m} \)
Modulus of rigidity \( G = 80 \times 10^{9} \text{ Pa} \)
Step 2: Calculate polar moment of inertia \( J \):
\[ J = \frac{\pi d^{4}}{32} = \frac{\pi (0.03)^{4}}{32} = \frac{\pi \times 8.1 \times 10^{-8}}{32} \approx 7.95 \times 10^{-9} \text{ m}^4 \]
Step 3: Use angle of twist formula:
\[ \theta = \frac{T L}{G J} = \frac{150 \times 2}{80 \times 10^{9} \times 7.95 \times 10^{-9}} = \frac{300}{636} = 0.4717 \text{ radians} \]
Convert radians to degrees:
\[ \theta = 0.4717 \times \frac{180}{\pi} \approx 27.03^{\circ} \]
Answer: The shaft twists through approximately 27 degrees under the applied torque.
Step 1: Given torque \( T = 250 \text{ N·m} \), speed \( N = 1200 \text{ rpm} \).
Step 2: Convert rotational speed to angular velocity:
\[ \omega = \frac{2 \pi N}{60} = \frac{2 \pi \times 1200}{60} = 125.66 \text{ rad/s} \]
Step 3: Use power transmission formula:
\[ P = T \omega = 250 \times 125.66 = 31,415 \text{ W} = 31.415 \text{ kW} \]
Answer: The power transmitted by the shaft is approximately 31.4 kW.
Step 1: Given diameter \( d = 0.05 \text{ m} \), bending moment \( M = 500 \text{ N·m} \), torque \( T = 300 \text{ N·m} \).
Step 2: Calculate polar moment of inertia \( J \) and moment of inertia \( I \) (for bending):
\[ J = \frac{\pi d^{4}}{32} = \frac{\pi (0.05)^4}{32} = 3.07 \times 10^{-7} \text{ m}^4 \]
\[ I = \frac{\pi d^{4}}{64} = \frac{3.07 \times 10^{-7}}{2} = 1.535 \times 10^{-7} \text{ m}^4 \]
Step 3: Calculate maximum bending stress \( \sigma_b \):
Maximum bending stress occurs at outer surface at radius \( c = d/2 = 0.025 \text{ m} \).
\[ \sigma_b = \frac{M c}{I} = \frac{500 \times 0.025}{1.535 \times 10^{-7}} = 8.14 \times 10^{7} \text{ Pa} = 81.4 \text{ MPa} \]
Step 4: Calculate maximum shear stress \( \tau_{max} \) due to torsion:
\[ \tau_{max} = \frac{T c}{J} = \frac{300 \times 0.025}{3.07 \times 10^{-7}} = 24.4 \times 10^{6} \text{ Pa} = 24.4 \text{ MPa} \]
Step 5: Calculate principal stresses at the surface combining bending and torsion stresses:
The bending stress is normal (tensile or compressive) stress, shear stress is perpendicular to it, so principal stresses are:
\[ \sigma_{1,2} = \frac{\sigma_b}{2} \pm \sqrt{\left(\frac{\sigma_b}{2}\right)^2 + \tau_{max}^2} \]
Calculate intermediate terms:
\[ \frac{\sigma_b}{2} = \frac{81.4}{2} = 40.7 \text{ MPa} \]
\[ \sqrt{(40.7)^2 + (24.4)^2} = \sqrt{1656.5 + 595.4} = \sqrt{2252} = 47.47 \text{ MPa} \]
Thus,
\[ \sigma_1 = 40.7 + 47.47 = 88.17 \text{ MPa} \]
\[ \sigma_2 = 40.7 - 47.47 = -6.77 \text{ MPa} \]
Answer:
Step 1: Convert diameters to meters:
\( d_o = 0.06 \) m, \( d_i = 0.05 \) m, \( L = 1.5 \) m.
Step 2: Calculate polar moment of inertia \( J \) for hollow shaft:
\[ J = \frac{\pi}{32} (d_o^4 - d_i^4) = \frac{\pi}{32} (0.06^4 - 0.05^4) \]
Calculate each term:
\( 0.06^4 = 0.00001296 \), \( 0.05^4 = 0.00000625 \)
\[ J = \frac{\pi}{32} (0.00001296 - 0.00000625) = \frac{\pi}{32} \times 0.00000671 = 6.59 \times 10^{-7} \text{ m}^4 \]
Step 3: Calculate shear stress \( \tau \) on the outer surface:
Using formula \( \tau = \frac{T c}{J} \), where \( c = d_o/2 = 0.03 \text{ m} \)
\[ \tau = \frac{180 \times 0.03}{6.59 \times 10^{-7}} = 8.19 \times 10^{6} \text{ Pa} = 8.19 \text{ MPa} \]
Step 4: Calculate the angle of twist:
\[ \theta = \frac{T L}{G J} = \frac{180 \times 1.5}{80 \times 10^9 \times 6.59 \times 10^{-7}} = \frac{270}{52720} = 0.00512 \text{ radians} \]
Convert to degrees:
\[ \theta = 0.00512 \times \frac{180}{\pi} = 0.293^{\circ} \]
Answer: Shear stress in the tube wall is approximately 8.19 MPa and angle of twist is about 0.293 degrees.
When to use: Applying torsion shear stress formula on circular shafts.
When to use: Throughout all torsion problems to avoid unit mismatch.
When to use: Power transmission problems involving shafts rotating at rpm.
When to use: Calculating angular deformation under torsion.
When to use: Evaluating design stress limits and safety.
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