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Torsion is the twisting of a structural member caused by an applied torque or twisting moment. Imagine holding a cylindrical rod fixed at one end and twisting the other end with your hand - the rod experiences torsion. This twisting action induces internal stresses and deformations within the member.
In civil engineering, torsion is a critical consideration in shafts, beams, and columns that are subjected to twisting forces. For example, drive shafts in machinery, bridge components, and structural beams may experience torsion during operation or due to external loads such as wind or seismic forces.
Understanding torsion helps engineers design safer and more efficient structures by ensuring that materials can withstand twisting without failure. In this chapter, we use metric units (meters, newtons, pascals) consistent with Indian engineering standards.
When a torque \( T \) is applied to a circular shaft, it causes the shaft to twist about its longitudinal axis. This twisting action produces shear stresses within the material, which vary across the shaft's cross-section.
Let's define the key terms:
The shear stress at any radius \( \rho \) from the center is given by the torsion formula:
This formula tells us that shear stress increases linearly from zero at the center (\( \rho = 0 \)) to a maximum value at the outer surface (\( \rho = c \), where \( c \) is the outer radius).
Why does shear stress vary linearly? Because the twisting effect is zero at the shaft's center (no radius to cause rotation) and maximum at the outer surface, where the material experiences the greatest rotational displacement.
The angle of twist \( \theta \) measures how much the shaft twists over its length \( L \) due to the applied torque \( T \). It is expressed in radians and is given by:
The shear modulus \( G \) is a material property indicating stiffness against shear deformation. Higher \( G \) means less twist for the same torque.
The polar moment of inertia \( J \) quantifies the shaft's resistance to twisting based on its cross-sectional shape. For circular shafts, \( J \) depends strongly on the diameter.
For a solid circular shaft of diameter \( d \):
For a hollow circular shaft with outer diameter \( d_o \) and inner diameter \( d_i \):
Why is the polar moment of inertia important? It determines how resistant a shaft is to twisting. Larger \( J \) means less twist and lower shear stress for the same torque. Notice how \( J \) depends on the diameter raised to the fourth power - even small increases in diameter greatly increase torsional resistance.
Step 1: Convert diameter to meters: \( d = 50 \text{ mm} = 0.05 \text{ m} \).
Step 2: Calculate the polar moment of inertia \( J \) for the solid shaft:
\[ J = \frac{\pi d^4}{32} = \frac{3.1416 \times (0.05)^4}{32} = \frac{3.1416 \times 6.25 \times 10^{-7}}{32} = 6.136 \times 10^{-8} \text{ m}^4 \]
Step 3: Calculate the outer radius \( c = \frac{d}{2} = 0.025 \text{ m} \).
Step 4: Calculate maximum shear stress \( \tau_{max} \):
\[ \tau_{max} = \frac{T c}{J} = \frac{200 \times 0.025}{6.136 \times 10^{-8}} = 8.15 \times 10^{7} \text{ Pa} = 81.5 \text{ MPa} \]
Answer: The maximum shear stress in the shaft is 81.5 MPa.
Step 1: Convert diameters to meters:
\( d_o = 0.08 \text{ m}, \quad d_i = 0.06 \text{ m} \)
Step 2: Calculate polar moment of inertia \( J \):
\[ J = \frac{\pi}{32} (d_o^4 - d_i^4) = \frac{3.1416}{32} \left( (0.08)^4 - (0.06)^4 \right) \]
\( (0.08)^4 = 4.096 \times 10^{-5}, \quad (0.06)^4 = 1.296 \times 10^{-5} \)
\[ J = \frac{3.1416}{32} (4.096 \times 10^{-5} - 1.296 \times 10^{-5}) = \frac{3.1416}{32} \times 2.8 \times 10^{-5} = 2.75 \times 10^{-6} \text{ m}^4 \]
Step 3: Use angle of twist formula:
\[ \theta = \frac{T L}{G J} = \frac{500 \times 2}{80 \times 10^{9} \times 2.75 \times 10^{-6}} = \frac{1000}{220000} = 0.004545 \text{ radians} \]
Step 4: Convert radians to degrees:
\( \theta = 0.004545 \times \frac{180}{\pi} = 0.260^\circ \)
Answer: The angle of twist is approximately 0.26 degrees.
Step 1: Use maximum shear stress formula:
\[ \tau_{max} = \frac{T c}{J} \]
For a solid shaft, \( J = \frac{\pi d^4}{32} \) and \( c = \frac{d}{2} \), so:
\[ \tau_{max} = \frac{T \times \frac{d}{2}}{\frac{\pi d^4}{32}} = \frac{16 T}{\pi d^3} \]
Step 2: Rearranging for \( d \):
\[ d^3 = \frac{16 T}{\pi \tau_{max}} \]
Step 3: Substitute values:
\( T = 1000 \text{ Nm}, \quad \tau_{max} = 60 \times 10^{6} \text{ Pa} \)
\[ d^3 = \frac{16 \times 1000}{3.1416 \times 60 \times 10^{6}} = \frac{16000}{188495559} = 8.49 \times 10^{-5} \text{ m}^3 \]
Step 4: Calculate \( d \):
\( d = \sqrt[3]{8.49 \times 10^{-5}} = 0.0444 \text{ m} = 44.4 \text{ mm} \)
Answer: The minimum shaft diameter required is approximately 44.4 mm.
Step 1: Convert diameter to meters: \( d = 0.04 \text{ m} \), radius \( c = 0.02 \text{ m} \).
Step 2: Calculate polar moment of inertia \( J \):
\[ J = \frac{\pi d^4}{32} = \frac{3.1416 \times (0.04)^4}{32} = 8.042 \times 10^{-7} \text{ m}^4 \]
Step 3: Calculate moment of inertia \( I \) for bending:
\[ I = \frac{\pi d^4}{64} = \frac{3.1416 \times (0.04)^4}{64} = 4.021 \times 10^{-7} \text{ m}^4 \]
Step 4: Calculate maximum shear stress due to torsion:
\[ \tau_{max} = \frac{T c}{J} = \frac{300 \times 0.02}{8.042 \times 10^{-7}} = 7.46 \times 10^{6} \text{ Pa} = 7.46 \text{ MPa} \]
Step 5: Calculate maximum bending stress:
\[ \sigma_{max} = \frac{M c}{I} = \frac{500 \times 0.02}{4.021 \times 10^{-7}} = 24.87 \times 10^{6} \text{ Pa} = 24.87 \text{ MPa} \]
Step 6: Combine stresses using maximum shear stress theory:
Maximum shear stress due to combined loading is:
\[ \tau_{combined} = \sqrt{\left(\frac{\sigma_{max}}{2}\right)^2 + \tau_{max}^2} = \sqrt{\left(\frac{24.87}{2}\right)^2 + 7.46^2} = \sqrt{12.435^2 + 7.46^2} = 14.5 \text{ MPa} \]
Answer: Maximum shear stress is 14.5 MPa, maximum normal stress is 24.87 MPa.
Step 1: For thin-walled rectangular sections, the torsional constant \( J_t \) is approximated by:
\[ J_t = \frac{4 A_m^2 t}{P} \]
where:
Step 2: Calculate median dimensions:
\( b_m = 100 - 5 = 95 \text{ mm} = 0.095 \text{ m} \)
\( h_m = 50 - 5 = 45 \text{ mm} = 0.045 \text{ m} \)
Step 3: Calculate median area \( A_m \):
\( A_m = b_m \times h_m = 0.095 \times 0.045 = 0.004275 \text{ m}^2 \)
Step 4: Calculate median perimeter \( P \):
\( P = 2 (b_m + h_m) = 2 (0.095 + 0.045) = 0.28 \text{ m} \)
Step 5: Calculate torsional constant \( J_t \):
\[ J_t = \frac{4 \times (0.004275)^2 \times 0.005}{0.28} = \frac{4 \times 1.827 \times 10^{-5} \times 0.005}{0.28} = 1.305 \times 10^{-6} \text{ m}^4 \]
Step 6: Calculate maximum shear stress:
For thin-walled sections, maximum shear stress is approximately:
\[ \tau_{max} = \frac{T}{2 t A_m} = \frac{400}{2 \times 0.005 \times 0.004275} = \frac{400}{4.275 \times 10^{-5}} = 9.35 \times 10^{6} \text{ Pa} = 9.35 \text{ MPa} \]
Answer: The estimated maximum shear stress is 9.35 MPa.
When to use: When calculating shear stress distribution in torsion problems.
When to use: When computing \( J \) for different shaft geometries.
When to use: At the start of every problem.
When to use: When calculating deformation under torsion.
When to use: When shafts experience multiple types of loads.
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