4. Torsion of Circular Shafts
Torsion is the twisting of an object caused by a moment acting about the object’s longitudinal axis. Torque is the moment that causes the twisting.
Consider a shaft transmitting power, as shown in Figure 6.
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\[\begin{align}\tag{13}
\text{The angle of twist, } \theta = \frac{TL}{GJ}
\end{align}\]
Where \(T\) is the applied torque, \(L\) is the length of the bar, \(G\) is the shear modulus, and \(J\) is the polar moment of inertia. The value of \(G\) can be obtained using an experiment. This \(G\) can be done by measuring the angle of twist after applying a known torque to a bar of known length and cross-section.
Figure 7. Polar moment.
\[\begin{align}\tag{14}
\text{The polar moment of inertia, } J = \frac{\pi}{2} \big( R^4 - r^4 \big)
\end{align}\]
For a solid circular bar, such as Figure 7 (iii), \(r = 0\) and \(J = \frac{\pi}{2} \big( R^4 \big)\) |
4.1. The angle of twist
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Example 1 |
A moment of \(1500\ Nm\) is acting on a solid cylinder shaft with diameter \(150\ mm\) and length \(2\ m\). The shaft is made in steel with a modulus of rigidity \(79\ GPa\). (a) Calculate the angular deflection of the shaft. |
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Solution:
\[\begin{aligned}
\text{The angle of twist, } \theta &= \frac{TL}{GJ} \\
\text{The polar moment of inertia, } J &= \frac{\pi}{2} \big( R^4 - r^4 \big)
\end{aligned}\]
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4.2. Torsional Stress and Strain
Torsion generates stresses and trains in the bar which can cause it to deform or fail. In the design of drive shafts used in various machinery, it is important to determine how much torsional stress and strain a given bar can withstand.
After torsional deformation, section \(ABCD\) of the circular bar shown in Figure 6(i), deforms to \(ABC'D'\) in Figure 6(ii). Since a circular bar is axisymmetric, \(AB\) remains horizontal to \(C'D'\) even after the deformation. However, a shear strain \(\gamma\) is induced, which corresponds to the angle \(\gamma = D'BD = Q'PQ\) in Figure 6(ii).
Here, similar to Equation 7, the shear strain can be calculated as follows.
From Equation 15, we can calculate the shear strain on the surface of the bar. As seen in Figure 7(ii), the shear strain increases linearly as you move from the center of the circular cross-section towards the circumference. The area at the surface of the bar experiences the maximum torque \(\tau_{max}\).
Similar to shear strain, shear stress increases linearly from the center of the circular cross-section. The maximum shear stress occurs at the outermost surface, while the minimum shear stress occurs at the center of the circular bar. Hollow bars are efficient at carrying torsional loads since the center part of the bar only resists a small fraction of the total torque load.
To calculate the shear stress, consider a small area \(dA\) in Figure 7(iii), located at distance \(\rho\) from the center of a bar. The force acting on this small area is \(=\tau \times dA\), where \(\tau\) is the shear stress. Since the system remains in equilibrium, the moments caused by the forces acting on all area elements within the cross-section sum up to the total torque \(T\). This can be represented as follows.
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Example 2 |
A solid circular shaft of diameter \(100\ mm\) is subjected to a torque of \(25\ kNm\). The angle of twist over a length of \(3\ m\) is observed to be \(0.09\ rad\). Find the modulus of rigidity of the material. |
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Solution:
\[\begin{aligned}
%T &=25 \mathrm{kNm}=25 \times 10^6 \mathrm{Nmm}\\
J &=\frac{\pi}{32} \times D^4 \\
\frac{T}{J} &=G \frac{\theta}{L}\\
G &=\frac{T L}{J \theta}\\
%&=\frac{25 \times 10^6 \times 3000}{(\pi / 32) \times 100^4 \times 0.09}\\
%&=84,882 \mathrm{~N} / \mathrm{mm}^2 =84.88 \mathrm{GPa}
\end{aligned}\]
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Example 3 |
A hollow circular shaft has an external diameter of \(120\ mm\) and the internal diameter is three-fourths of the external diameter. If the stress on a fibre is \(36\ MPa\) due to the applied torque, T, |
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Solution:
(i) Calculate the applied torque T.
\[\begin{aligned}
&\text { Internal diameter }=\frac{3}{4} \times 120=90 \mathrm{~mm} \\
&J=\frac{\pi}{32}\left((\text{external diameter})^4-(\text{internal diameter})^4\right) \\
&\tau=\frac{T}{J} r \\
&T=\frac{\tau J}{r} \\
%&\tau=36 \mathrm{~N} / \mathrm{mm}^2 \\
%&r=90 \mathrm{~mm} \\
%&T=\frac{36 \times}{90} \frac{\pi \times 10^4}{32} \times 14,175\\
%&=5,566,510 \mathrm{~N} \mathrm{~mm}=5.567 \mathrm{kNm}
\end{aligned}\]
(ii) Calculate the maximum shear stress.
\[\begin{aligned}
\tau_{\max } &=\frac{T}{J} R \\
%&=\frac{5,566,510 \times 120}{(\pi / 32) \times 10^4 \times 14,175}\\
%&=\frac{36 \times 120}{90}=48 \mathrm{~N} / \mathrm{mm}^2 \\
\end{aligned}\]
(iii) Calculate the angle of twist per unit length.
\[\begin{aligned}
\text { Angle of twist per unit length }&=\frac{T}{G J}\\
%&=\frac{5,566,510}{85,000 \times(\pi / 32) \times 10^4 \times 14,175} \\
%&=4.7 \times 10^{-6} \mathrm{rad} / \mathrm{mm} \\
%&=0.0047 \mathrm{rad} / \mathrm{m}=0.27^{\circ} / \mathrm{m}\\
\end{aligned}\]
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Example 4 |
A solid circular shaft has a diameter of \(80\ mm\). Find the maximum shear stress and the twist angle on a length of \(2\ m\) when the shaft is subjected to a torque of \(10\ kNm\). \(G = 85\ GPa\). |
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Solution:
From the torsion equation \(T / J=\tau / r, \tau\) will be maximum when \(r\) is equal to the radius of the shaft.
\[\begin{aligned}
\tau_{\max } &=\frac{T R}{J}\\
%&=\frac{16 T}{\pi D^3}=\frac{16 \times 10 \times 10^6}{\pi \times(80)^3}\\
%&=99.5 \mathrm{~N} / \mathrm{mm}^2
\end{aligned}\]
Also, \(T / J=G \theta / L\), where \(\theta\) is the angle of twist over a length \(L\) of the shaft.
\[\begin{aligned}
\theta &=\frac{T L}{G J}\\
%&=\frac{10 \times 10^6 \times 2000 \times 32}{85,000 \times \pi \times(80)^4}\\
%&=0.0585 \mathrm{rad} \text { or } 3.35^{\circ}
\end{aligned}\]
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Comments (7)
Ae we doing chapters 6, 7, 8 & 9?
No. That's for Solid II.
Kindly confirm for us the formula for volumetric strain
Hello, are the notes here all we will have for SSM2 or is there a possibility for addition?
Solid Mechanics II starts from chapter 6.
Could you confirm if in example 1 the shear force diagram is correct since from my shear force calculations the forces are positive meaning the diagram ought to be upwards
Yes, in example 1, the shear force diagram is correct. There are two point loads acting downwards, which tend to shear the bar in the negative direction.