Pure Torsion MCQ Quiz - Objective Question with Answer for Pure Torsion - Download Free PDF

Concept:

We use the torsion equation to analyze the relationship between torsional rigidity and angle of twist in a shaft under torque.

Given:

• Torsion equation: \(\frac{T}{J} = \frac{G \cdot \theta}{L}\)
• Torsional rigidity: \(G J\)
• Angle of twist: \(\theta\)

Step 1: Rearrange the torsion equation

To express angle of twist explicitly:

\(\theta = \frac{T L}{G J}\)

Step 2: Analyze the relationship

The equation shows:

\(\theta \propto \frac{1}{G \cdot J}\)

This means:

• When torsional rigidity (\(G J\)) increases, angle of twist (\(\theta\)) decreases
• When torsional rigidity decreases, angle of twist increases

Step 3: Physical interpretation

Torsional rigidity represents a shaft's resistance to twisting:

• Higher \(G J\) (stiffer shaft) → Less twist under same torque
• Lower \(G J\) (more flexible shaft) → More twist under same torque

Increasing torsional rigidity decreases the angle of twist in a shaft under torsion.

Explanation:

Torsional Stiffness of a Shaft

• Torsional stiffness is a measure of a shaft's resistance to twisting under the application of torque.
• It is an essential parameter in the design of mechanical systems where rotational motion and torque transmission are involved, such as in drive shafts, axles, and various rotating machinery components.

Key Factors Affecting Torsional Stiffness:

• Several factors influence the torsional stiffness of a shaft, including the material properties, the geometry of the shaft, and the boundary conditions.
• However, for a given material and shaft geometry, the primary factor that affects the torsional stiffness is the Modulus of Rigidity (also known as Shear Modulus).

The torsional stiffness (k) of a shaft can be mathematically expressed using the following formula:

k = (G × J) / L

Where:

• G is the Modulus of Rigidity (Shear Modulus) of the material.
• J is the Polar Moment of Inertia of the shaft cross-section.
• L is the length of the shaft.

Importance of Modulus of Rigidity (G):

• The Modulus of Rigidity (G) is a material property that measures the material's ability to resist shear deformation. It is a fundamental parameter in defining the torsional stiffness of a shaft.
• For a given shaft geometry, the torsional stiffness is directly proportional to the Modulus of Rigidity. A higher Modulus of Rigidity results in greater torsional stiffness, meaning the shaft will experience less angular deformation under an applied torque.

To understand why the Modulus of Rigidity is the primary factor, let's delve into the relationship between the torque (T) applied to a shaft and the resulting angle of twist (θ). The relationship is given by:

θ = (T × L) / (G × J)

This equation shows that for a given torque (T) and shaft length (L), the angle of twist (θ) is inversely proportional to the Modulus of Rigidity (G) and the Polar Moment of Inertia (J). Therefore, increasing the Modulus of Rigidity reduces the angle of twist, indicating higher torsional stiffness.

Concept:

The torsional stiffness of a shaft depends on its polar moment of inertia (\( J \)), which is a function of the shaft diameter.

The torsional stiffness (\( K \)) is given by:

\( K = \frac{G J}{L} \)

Since the shafts are of the same material (\( G \) is constant) and same length (\( L \) is constant), the stiffness is proportional to the polar moment of inertia:

\( K \propto J \)

Calculation:

The polar moment of inertia for a solid circular shaft is:

\( J = \frac{\pi d^4}{32} \)

Let the diameter of the smaller shaft be \( d \) and the larger shaft be \( 2d \).

For the smaller shaft:

\( J_{\text{small}} = \frac{\pi d^4}{32} \)

For the larger shaft:

\( J_{\text{large}} = \frac{\pi (2d)^4}{32} = \frac{\pi \times 16d^4}{32} = 16 J_{\text{small}} \)

Since \( K \propto J \), the torsional stiffness of the larger shaft is:

\( K_{\text{large}} = 16 K_{\text{small}} \)

The torsional stiffness of the larger shaft is 16 times that of the smaller shaft.

Explanation:

Shear stress, \(\tau~=~\frac{16T}{\pi D^3}\)

Where, T = Torque, D = Diameter of the shaft

Given that:

T = 2.355 KN-m, \(\tau\) = 120 MPa

∴ 120 = \(\frac{16~\times~2.355~\times~1000~\times~1000}{\pi ~\times~D^3}\)

⇒ D = 46.41 mm

Explanation:
In a beam subjected to transverse loading, different types of stresses develop across its cross-section.

Bending Stress: Bending stress is zero at the neutral axis but varies linearly with the distance from the neutral axis, reaching its maximum at the top and bottom fibers of the beam.

Shear Stress: Shear stress is maximum at the neutral axis and decreases towards the top and bottom surfaces of the beam.

Therefore, along the neutral axis of the central cross-section of a beam:

The bending stress is present but is zero since the neutral axis is the line of zero bending stress.The shear stress is at its maximum value along the neutral axis.

Hence, along the neutral axis, there is both bending stress (zero in value at that axis) and significant shearing stress present.

Explanation:

Two types of shear stresses are generated due to the action of pure torsion on a shaft.

(i) Circumferential Stress:

• Due to the application of torsion, one layer of the shaft cross-section tends to move relative to the other. Hence circumferential shear stress is generated.
• It can be expressed as:

 \(\frac{T}{J} =\;\frac{τ }{r} = \frac{{G\theta }}{L}\) 

Where, T = Torsion, J = Polar Section Modulus, τ = Circumferential Shear Stress, r = Radius of the shaft, G = Modulus of Rigidity of the shaft, θ = Angle of rotation, L = length of the shaft

(ii) It is also important to realize that the shear stresses acting on the cross-sectional planes are accompanied by shear stresses of the same magnitude in the longitudinal planes following the principle of complementary shear stresses.

Explanation:

Let the strength of the solid and hollow shaft is T_S and T_H 

\({{\bf{T}}_{\bf{S}}} = \frac{\pi }{{16}}{{\bf{d}}^3}{τ _{per}}\), 

\({{\bf{T}}_{\bf{H}}} = \frac{\pi }{{16}}{{\bf{D}}^3}\left( {1 - {{\bf{k}}^4}} \right){τ _{per}}\)

where τ_per = permissible shear strength, k = Ratio of inner diameter (din) to outer diameter (D_o) i.e. \(\left( \frac{d_{in}}{D_o} \right)\)

The ratio of the strength of the hollow shaft to that of the solid shaft when the outer diameter of both the shafts are the same provided RPM, Material and length are the same is given by

\(\frac{{{{\bf{T}}_{\bf{H}}}}}{{{{\bf{T}}_{\bf{S}}}}} = 1 - {{\bf{k}}^4}\)

Here the outer diameter of both the shaft is the same 

D_o = D, d_in = \(\frac{{\rm{D}}}{2}\)

∴ k = \(\frac{{\rm{1}}}{2}\)

\(\frac{{{{\bf{T}}_{\bf{H}}}}}{{{{\bf{T}}_{\bf{S}}}}} = 1 - {\left( {\frac{1}{2}} \right)^4} = \frac{{15}}{{16}}\)

\(\frac{{{{\bf{T}}_{\bf{S}}}}}{{{{\bf{T}}_{\bf{H}}}}}\; = \;\frac{{16}}{{15}}\)

Alternate Method 

This problem can also be solved by using \({M \over I} = {σ \over y}\) 

Where strength can be correlated with M (MOMENT RESISTED BY THE SECTION)\(I_{Solid} = {\pi × D^4 \over 64} \ and\ I_{Hollow} = {\pi [D^4 - ({D/2})]^2\over 64}\)

\(Y_{Solid} = D/2\ and\ Y_{Hollow} = D/2\)

Putting the above values in M/I = σ/y

We get M = (I/Y) × σ (Assuming both are having same stress), we can say that strength depends on I/Y only

Putting the value of I and Y from the above equation, we get

\({{Strength \ of \ Solid} \over {Strength \ of \ hollow\ shaft} } = {I\over Y}\)

\({{{\pi \times D^4\over 64 } \times {D \over 2} } \over {{\pi [D^4 - {(D/2})^2]\over 64 } \times {D \over 2} }} = {1 \over (15/16)} = {16 \over 15}\)

Concept:

Torsion formula, \(\frac{T}{J} = \frac{\tau }{r} = \frac{{Gθ }}{L}\)

Angle of twist, \(θ = \frac{{TL}}{{GJ}}\)

Rate of twist is given by, \(\frac{θ }{L} = \frac{T}{{GJ}}\)

Where, J = Polar moment of inertia

G = Modulus of rigidity

θ = Angle of twist in radian 

L = Length of the shaft

R = Radius of shaft

Series combination of shaft: 

Series combination of the shaft is done to transmit torque from one shaft to another shaft. For series, connection couplings are used. Torque in all shaft connections in series will be equal.

T1 = T2 = T

Total angle of twist from A to C

θAC = θAB + θBC

\({θ _{AC}} = \frac{{T{L_1}}}{{{G_1}{J_1}}} + \frac{{T{L_2}}}{{{G_2}{J_2}}}\)

Calculation:

This is the case of series combination means if we have to apply torque M_t at B then torque developed at point A is also M_t.

ΣMt = 0, 

M_tA = M_tB

So in the given shaft, Twist (Mt), Modulus of rigidity, and Polar moment of inertia are constant.

So the ratio of the rate of twist (θ/L) at points A to B is 1.

Concept:

Strain energy: It is the energy stored in a body on account of deformations produced.

Strain energy due to various forces –  

(1) Axial force: \(\frac{{{P^2}L}}{{2AE}}\)

(2) Bending Moment:\(\frac{{{M^2}L}}{{2EI}}\)

(3) Torsional Moment: \(\frac{{{T^2}L}}{{2GJ}}\)

Now \(U = \frac{{{T^2}L}}{{2GJ}} \ and \ V = \pi r^2l\)

we have \(\tau = \frac {Tr}{J} \ and \ J = \frac{\pi}{32}d^4\)

Taking U/V and eliminating T from the above equations, we get 

Torsional strain energy per unit volume in cylindrical shaft = \(\frac{{{\tau ^2}}}{{4G}}\)

Note: Torsional strain energy per unit volume in rectangular shaft = \(\frac{{{\tau ^2}}}{{2G}}\)

Where , τ = Maximum shear stress produced

G = Modulus of Rigidity.

Explanation:

Distribution of shear stresses in circular Shafts subjected to torsion is given by,

\(\frac{\tau }{r} = \frac{T}{J} = \frac{{G\theta }}{L}\)

\(\tau = \frac{{Tr}}{J}\)

where, T = twisting moment, r = Radius of the shaft and J = polar moment of inertia

This states that the shearing stress varies directly as the distance ‘r' from the axis of the shaft.

The maximum shear stress occurs on the outer surface of the shaft where r = R.

For the solid shaft, at the centre, the stress is zero.

Stress is never zero in the hollow shaft although it is minimum at the inner surface of the shaft.

The correct answer is "The length and the radius of the shaft remain fixed"

Solution:

Following assumptions are made for torsional force in a circular shaft:

1.Material of the shaft is homogenous throughout the length of the shaft.

2.Shaft is straight and of uniform circular cross section over its length.

3.Radial lines remain radial during torsion.

4.Torsion is constant along the length of the shaft.

5. Cross section of the shaft remains plane.  

Concept:

From the torsional equation of the shaft

\(\mathop \tau \nolimits_{\max } = \frac{{16T}}{{\pi \mathop D\nolimits^3 }}\)

\(T = \frac{{\mathop \tau \nolimits_{\max } \pi \mathop D\nolimits^3 }}{{16}}\)

Calculation:

Given:

D = 4 cm

\(\mathop \tau \nolimits_{\max } = 20 ~\frac{kN}{cm^2}\)

\(T = \frac{{\mathop \tau \nolimits_{\max } \pi \mathop D\nolimits^3 }}{{16}}\)

\(T =\frac{20 \;\times\; \pi\; \times \;4^3}{16}=80\pi\) kN-cm

Concept:

The torsion equation is given by:

\(\frac{T}{J} = \frac{τ }{R} = \frac{{G\theta }}{L}\)

Where,

T is the torque applied, J = polar moment of inertia, τ = shear strength of material, G = modulus of rigidity, θ = angular deflection of shaft (in radians).

\(\frac{T}{J} = \frac{τ }{r}\)

\(T = \frac{\tau J}{r} =\frac{\tau}{r} \times \frac{\pi d^4}{32}=\frac{\pi \tau d^3}{16}\)

Calculation:

τ = 48 N/cm2, d = 3 cm

\(T =\frac{\pi \tau d^3}{16}=\frac{\pi \times 48 \times3^3}{16}=81\pi~Ncm\)

Explanation:

Torsional equation for the shaft is given by,

\( \frac{T}{J} = \frac{\tau }{r} = \frac{{G\theta }}{L}\)

Torque per radian twist is known as torsional stiffness (k)

\(k=\frac{T}{\theta}=\frac{GJ}{L}\)

The parameter GJ is called torsional rigidity of the shaft. 

Torsional rigidity is also defined as torque per unit angular twist per unit length

\(GJ=\frac{T}{\theta/L}\)

Explanation:

As the cross-section area is the same,

\(\frac{\pi }{4}\left( {d_1^2 - d_2^2} \right) = \frac{\pi }{4}d_s^2\)

\(\therefore d_1^2 - d_2^2 = d_s^2\)

\(\frac{τ }{r} = \frac{T}{J} = \frac{{G\theta }}{L}\)

\(T=\frac{τJ }{r}\)

Now torque ∝ polar moment

\(\therefore \frac{{{T_h}}}{{{T_s}}} = \frac{{\frac{\pi }{{32}}\left( {d_1^4 - d_2^4} \right)}}{{\frac{\pi }{{32}}\left( {d_s^4} \right)}} = \frac{{d_1^4 - d_2^4}}{{d_s^4}}= \frac{{(d_1^2 - d_2^2)}{(d_1^2 + d_2^2)}}{{d_s^4}}\)

\(\frac{{{T_h}}}{{{T_s}}} = \frac{{d_1^2 + d_2^2}}{{d_s^2}}\)