A solid circular shaft of diameter d and length L is fixed at one end and free at the other end. A torque T is applied at the free end. The shear modulus of the material is G. The angle of twist at three free ends is
Shaft Angle of Twist Calculation
This problem involves determining the angle of twist for a solid circular shaft subjected to a torque at its free end. To solve this, we need to apply the fundamental principles of torsion in shafts, specifically the formula for the angle of twist and the polar moment of inertia for a solid circular cross-section.
Torsion in a Solid Circular Shaft
Torsion is the twisting deformation of a shaft caused by an applied torque. For a uniform circular shaft, the angle of twist (\(\theta\)) along its length is directly proportional to the applied torque (\(\text{T}\)) and the length of the shaft (\(\text{L}\)), and inversely proportional to the material's shear modulus (\(\text{G}\)) and the cross-sectional polar moment of inertia (\(\text{J}\)).
The general formula that relates these parameters for the angle of twist is:
$$ \theta = \frac{TL}{GJ} $$Here's what each term represents:
- T: The externally applied torque (in N·m or lb·ft).
- L: The length of the shaft (in meters or feet).
- G: The shear modulus (or modulus of rigidity) of the shaft material (in Pa or psi). This property describes the material's resistance to shear deformation.
- J: The polar moment of inertia of the shaft's cross-section (in m\(^4\) or in\(^4\)). This geometric property indicates how the area of the cross-section is distributed around the axis of rotation, affecting its resistance to twisting.
Polar Moment of Inertia for a Solid Circular Shaft
For a solid circular shaft, the polar moment of inertia (\(\text{J}\)) depends solely on its diameter (\(\text{d}\)). It quantifies the shaft's resistance to torsional deformation based on its shape and size. The formula for the polar moment of inertia of a solid circular cross-section is:
$$ J = \frac{\pi d^4}{32} $$Where \(\text{d}\) is the diameter of the solid circular shaft.
Step-by-Step Angle of Twist Calculation
Let's apply the formulas to find the angle of twist at the free end of the given shaft.
Given Information:
- Shaft type: Solid circular shaft
- Diameter of shaft: \(\text{d}\)
- Length of shaft: \(\text{L}\)
- Applied torque: \(\text{T}\)
- Shear modulus of material: \(\text{G}\)
Step 1: Write down the fundamental angle of twist formula.
$$ \theta = \frac{TL}{GJ} $$Step 2: Substitute the formula for the polar moment of inertia (\(\text{J}\)) for a solid circular shaft into the angle of twist equation.
We know that for a solid circular shaft, \( J = \frac{\pi d^4}{32} \). Substituting this value into the angle of twist formula:
$$ \theta = \frac{TL}{G \left( \frac{\pi d^4}{32} \right)} $$Step 3: Simplify the expression to get the final formula for the angle of twist.
To simplify the equation, we can bring the denominator of the fraction in \(\text{J}\) to the numerator of the main expression:
$$ \theta = \frac{TL \times 32}{G \times \pi d^4} $$ $$ \theta = \frac{32TL}{\pi d^4 G} $$Final Angle of Twist Result
Based on our derivation, the angle of twist at the free end of the solid circular shaft is:
$$ \theta = \frac{32TL}{\pi d^4 G} $$Comparing with the Given Options
Let's compare our derived formula with the options provided in the question:
| Option | Formula |
|---|---|
| 1 | \(\dfrac{{16{\rm{TL}}}}{{{\rm{\pi }}{{\rm{d}}^4}{\rm{G}}}}\) |
| 2 | \(\dfrac{{32{\rm{TL}}}}{{{\rm{\pi }}{{\rm{d}}^4}{\rm{G}}}}\) |
| 3 | \(\dfrac{{64{\rm{TL}}}}{{{\rm{\pi }}{{\rm{d}}^4}{\rm{G}}}}\) |
| 4 | \(\dfrac{{128{\rm{TL}}}}{{{\rm{\pi }}{{\rm{d}}^4}{\rm{G}}}}\) |
Our calculated formula, \(\dfrac{{32{\rm{TL}}}}{{{\rm{\pi }}{{\rm{d}}^4}{\rm{G}}}}\), perfectly matches Option 2.
Important Questions from Equation of Torsion
- If two aluminum bar have a different length (L1 = 2L2) and diameter (d1 = 2d2) with an identical angle of a twist then, find torque value for bar 1, If bar 2 torque value is 50 N-m
Which of the following assumptions are True for torsion theory for axisymmetric sections?
A tubular shaft, having an inner diameter of 30 mm and an outer diameter of 40 mm, is to be used to transmit 80 kW of power. The speed of rotation of the shaft so that the shear stress will not exceed 50 MPa is
A circular solid shaft of span L = 5 m is fixed at one end and free at the other end. A torque T = 100 kN.m is applied at the free end. The shear modulus and polar moment of inertia of the section are denoted as G and J, respectively. The torsional rigidity GJ is 50,000 kN.m2 /rad. The following are reported for this shaft:
Statement i) The rotation at the free end is 0.01 rad
Statement ii) The torsional strain energy is 1.0 kN.m
With reference to the above statements, which of the following is true?
The magnitude of shear stress induced in a shaft due to applied torque varies from:
