Shaft Design on Torsional Rigidity Basis MCQ Quiz - Objective Question with Answer for Shaft Design on Torsional Rigidity Basis - Download Free PDF

Explanation:

Line shaft:

• A shaft directly coupled to a power source is referred to as a Line Shaft. This type of shaft is used to transmit power from the primary source to various machine components or systems in mechanical operations. The term specifically denotes the direct connection between the power source and the shaft, ensuring efficient power transmission without intermediate mechanisms.
• The Line Shaft operates by being physically connected to the power source, such as an engine or motor. This connection allows the rotational motion generated by the power source to be directly transmitted to the shaft. The shaft then distributes this mechanical energy to other connected components, such as pulleys, gears, or belts, which perform specific tasks or operations. Its direct coupling eliminates the need for secondary power transmission systems, enhancing efficiency and minimizing energy losses.

Applications: Line Shafts are commonly used in industries that require direct and robust power transmission systems, such as:

• Textile manufacturing units for operating multiple machines simultaneously.
• Conveyor systems in material handling processes.
• Machining operations in workshops and factories.
• Power transmission setups in agricultural equipment.

Important InformationCounter Shaft

• A Counter Shaft is used as an intermediate shaft in power transmission systems. It typically connects the main power source to other shafts or components, allowing for speed or torque adjustments.

Jack Shaft:

• A Jack Shaft is another type of intermediate shaft used in power transmission systems. It often serves as a connecting shaft between two main shafts or between the power source and the final driven component. Similar to the Counter Shaft, its function involves intermediate transmission rather than direct coupling to the power source.

Flexible Shaft

• A Flexible Shaft is designed to transmit rotary motion between components that are not perfectly aligned or have relative motion between them. It is commonly used in applications requiring flexibility in power transmission, such as handheld tools or equipment.

Explanation:

The polar sectional modulus (Z_p) of a shaft having a hollow cross-section is higher in comparison to the shaft having a solid cross-section.

Polar section modulus of a solid shaft (Z_ps) \(= \frac {J}{R}= \frac{\pi d^3}{16}\)

and,

Polar section modulus of hollow shaft (Zph) \(= \frac {J}{R}= \frac{\pi {d_o}^3 (1-k^4)}{16}\)

\(k= \frac {d_i}{d_o} = \frac {d}{D} < 1\)

where, d = Diameter of the solid shaft, d_i = Inner diameter of the hollow shaft, d_o = Outer-diameter of the hollow shaft.

Power/Torque transmission capacity:

Torque Transmission (T) \( = Z_p \times \tau_(permissible)\)

\( \frac {T_s}{T_h} = \frac {​​​​\frac{\pi\times d^3}{16}}{ \frac{\pi {d_o}^3 (1-k^4)}{16}} = \frac{d^3}{d_o^3}\times \frac{1}{(1-k^4)}\)

\( \frac {T_s}{T_h} = \frac{d^3}{d_o^3}\times \frac{1}{(1-k^2)}\times \frac{1}{(1+k^2)}\)

For shafts having the same cross-sectional area:

​\(A_s = A_h\)

\(\frac{\pi\times d^2}{4} = \frac{\pi\times {d_o}^2}{4}\times (1-k^2)\)

\(\frac{d}{d_o} = \sqrt{(1-k^2)}\)​​

On solving,

\( \frac {T_s}{T_h} = \frac{d^3}{d_o^3}\times \frac{1}{(1-k^2)}\times \frac{1}{(1+k^2)}\)

\( \frac {T_s}{T_h}= (1-k^2)^{3/2}\times \frac{1}{(1-k^2)}\times \frac{1}{(1+k^2)}\)

\( \frac {T_s}{T_h} = \frac{(1-k^2)^{1/2}}{(1+k^2)}\)

\(T_s < T_h\)

Conclusion:

T has a greater value for shafts having a hollow cross-section as compared to the shafts having a solid cross-section, So hollow shafts are stronger and more rigid than the solid shaft.

Concept:

The Torque equation is given as, 

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

where T is torque, J is the polar moment of inertia, τ is shear stress, R is the radius of the shaft, G is Modulus of rigidity, θ is angle of twist and L is length of shaft.

Calculation:

Given:

G = 100 GPa = 100 × 10 9 Pa, L = 120 mm = 0.120 m, T = 592.6 Nm, \(θ = 20° =\frac{\pi}{180} \times 20~rad\)

\(\frac{T }{J} = \frac{{Gθ }}{L}\)

The polar moment of inertia is given as, \(J=\frac{\pi}{32}D^4\)

\(\therefore ~J = \frac{TL}{G \theta}\)

\(\therefore ~\frac{\pi}{32}D^4 = \frac{TL}{G \theta}\)

\(\therefore ~\frac{\pi}{32} \times D^4 = \frac{592.6 \times 0.120}{100 \times 10^9 \times \frac{\pi}{180} \times 20}\)

D = 0.012 m = 12 mm

Explanation:

Torsion equation is

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

\(\begin{array}{l} \tau = \frac{T}{Z}\\ \end{array}\)

Here Z = Sectional modulus of the shaft

\(z = \frac{{\pi \mathop d\nolimits^3 }}{{16}}\)

Torque transited by shaft

\(T = \frac{\pi }{{16}}\tau {d^3}\)

Weight of shaft (ω) = Volume × density \( = \frac{\pi }{4}{d^2} \times l \times \rho \)

Required ratio is:

\(\begin{array}{l} \frac{T}{\omega } = \frac{{\frac{\pi }{{16}}\tau {d^3}}}{{\frac{\pi }{4}{d^2}l\rho }}\\ \Rightarrow \frac{T}{\omega } \propto d \end{array}\)

Concept:

As we know that for shaft under pure torsion

\(τ = \frac{16T}{\pi d^3}\)

As T is constant

∴ \(τ\ \ \ \alpha\ \ \ \frac{1}{{{d^3}}}\) 

\({τ_1\overτ_2} =\frac{d_2^3}{{{d_1^3}}}\)

Calculation:

Given:

τ₁ = 16, d₁ = d, \(d_2 = \frac{d}{2}\)

Then, \({τ_1\overτ_2} =\frac{d_2^3}{{{d_1^3}}}\Rightarrow {16\overτ_2} =\frac{(d/2)^3}{{{d^3}}}\)

τ₂ = 16 × 8 = 128 MPa

Concept:

The Torque equation is given as, 

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

where T is torque, J is the polar moment of inertia, τ is shear stress, R is the radius of the shaft, G is Modulus of rigidity, θ is angle of twist and L is length of shaft.

Calculation:

Given:

G = 100 GPa = 100 × 10 9 Pa, L = 120 mm = 0.120 m, T = 592.6 Nm, \(θ = 20° =\frac{\pi}{180} \times 20~rad\)

\(\frac{T }{J} = \frac{{Gθ }}{L}\)

The polar moment of inertia is given as, \(J=\frac{\pi}{32}D^4\)

\(\therefore ~J = \frac{TL}{G \theta}\)

\(\therefore ~\frac{\pi}{32}D^4 = \frac{TL}{G \theta}\)

\(\therefore ~\frac{\pi}{32} \times D^4 = \frac{592.6 \times 0.120}{100 \times 10^9 \times \frac{\pi}{180} \times 20}\)

D = 0.012 m = 12 mm

Explanation:

The polar sectional modulus (Z_p) of a shaft having a hollow cross-section is higher in comparison to the shaft having a solid cross-section.

Polar section modulus of a solid shaft (Z_ps) \(= \frac {J}{R}= \frac{\pi d^3}{16}\)

and,

Polar section modulus of hollow shaft (Zph) \(= \frac {J}{R}= \frac{\pi {d_o}^3 (1-k^4)}{16}\)

\(k= \frac {d_i}{d_o} = \frac {d}{D} < 1\)

where, d = Diameter of the solid shaft, d_i = Inner diameter of the hollow shaft, d_o = Outer-diameter of the hollow shaft.

Power/Torque transmission capacity:

Torque Transmission (T) \( = Z_p \times \tau_(permissible)\)

\( \frac {T_s}{T_h} = \frac {​​​​\frac{\pi\times d^3}{16}}{ \frac{\pi {d_o}^3 (1-k^4)}{16}} = \frac{d^3}{d_o^3}\times \frac{1}{(1-k^4)}\)

\( \frac {T_s}{T_h} = \frac{d^3}{d_o^3}\times \frac{1}{(1-k^2)}\times \frac{1}{(1+k^2)}\)

For shafts having the same cross-sectional area:

​\(A_s = A_h\)

\(\frac{\pi\times d^2}{4} = \frac{\pi\times {d_o}^2}{4}\times (1-k^2)\)

\(\frac{d}{d_o} = \sqrt{(1-k^2)}\)​​

On solving,

\( \frac {T_s}{T_h} = \frac{d^3}{d_o^3}\times \frac{1}{(1-k^2)}\times \frac{1}{(1+k^2)}\)

\( \frac {T_s}{T_h}= (1-k^2)^{3/2}\times \frac{1}{(1-k^2)}\times \frac{1}{(1+k^2)}\)

\( \frac {T_s}{T_h} = \frac{(1-k^2)^{1/2}}{(1+k^2)}\)

\(T_s < T_h\)

Conclusion:

T has a greater value for shafts having a hollow cross-section as compared to the shafts having a solid cross-section, So hollow shafts are stronger and more rigid than the solid shaft.

Explanation:

Line shaft:

• A shaft directly coupled to a power source is referred to as a Line Shaft. This type of shaft is used to transmit power from the primary source to various machine components or systems in mechanical operations. The term specifically denotes the direct connection between the power source and the shaft, ensuring efficient power transmission without intermediate mechanisms.
• The Line Shaft operates by being physically connected to the power source, such as an engine or motor. This connection allows the rotational motion generated by the power source to be directly transmitted to the shaft. The shaft then distributes this mechanical energy to other connected components, such as pulleys, gears, or belts, which perform specific tasks or operations. Its direct coupling eliminates the need for secondary power transmission systems, enhancing efficiency and minimizing energy losses.

Applications: Line Shafts are commonly used in industries that require direct and robust power transmission systems, such as:

• Textile manufacturing units for operating multiple machines simultaneously.
• Conveyor systems in material handling processes.
• Machining operations in workshops and factories.
• Power transmission setups in agricultural equipment.

Important InformationCounter Shaft

• A Counter Shaft is used as an intermediate shaft in power transmission systems. It typically connects the main power source to other shafts or components, allowing for speed or torque adjustments.

Jack Shaft:

• A Jack Shaft is another type of intermediate shaft used in power transmission systems. It often serves as a connecting shaft between two main shafts or between the power source and the final driven component. Similar to the Counter Shaft, its function involves intermediate transmission rather than direct coupling to the power source.

Flexible Shaft

• A Flexible Shaft is designed to transmit rotary motion between components that are not perfectly aligned or have relative motion between them. It is commonly used in applications requiring flexibility in power transmission, such as handheld tools or equipment.

Explanation:

Torsion equation is

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

\(\begin{array}{l} \tau = \frac{T}{Z}\\ \end{array}\)

Here Z = Sectional modulus of the shaft

\(z = \frac{{\pi \mathop d\nolimits^3 }}{{16}}\)

Torque transited by shaft

\(T = \frac{\pi }{{16}}\tau {d^3}\)

Weight of shaft (ω) = Volume × density \( = \frac{\pi }{4}{d^2} \times l \times \rho \)

Required ratio is:

\(\begin{array}{l} \frac{T}{\omega } = \frac{{\frac{\pi }{{16}}\tau {d^3}}}{{\frac{\pi }{4}{d^2}l\rho }}\\ \Rightarrow \frac{T}{\omega } \propto d \end{array}\)

Concept:

The Torque equation is given as, 

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

where T is torque, J is the polar moment of inertia, τ is shear stress, R is the radius of the shaft, G is Modulus of rigidity, θ is angle of twist and L is length of shaft.

Calculation:

Given:

G = 100 GPa = 100 × 10 9 Pa, L = 120 mm = 0.120 m, T = 592.6 Nm, \(θ = 20° =\frac{\pi}{180} \times 20~rad\)

\(\frac{T }{J} = \frac{{Gθ }}{L}\)

The polar moment of inertia is given as, \(J=\frac{\pi}{32}D^4\)

\(\therefore ~J = \frac{TL}{G \theta}\)

\(\therefore ~\frac{\pi}{32}D^4 = \frac{TL}{G \theta}\)

\(\therefore ~\frac{\pi}{32} \times D^4 = \frac{592.6 \times 0.120}{100 \times 10^9 \times \frac{\pi}{180} \times 20}\)

D = 0.012 m = 12 mm

Explanation:

The polar sectional modulus (Z_p) of a shaft having a hollow cross-section is higher in comparison to the shaft having a solid cross-section.

Polar section modulus of a solid shaft (Z_ps) \(= \frac {J}{R}= \frac{\pi d^3}{16}\)

and,

Polar section modulus of hollow shaft (Zph) \(= \frac {J}{R}= \frac{\pi {d_o}^3 (1-k^4)}{16}\)

\(k= \frac {d_i}{d_o} = \frac {d}{D} < 1\)

where, d = Diameter of the solid shaft, d_i = Inner diameter of the hollow shaft, d_o = Outer-diameter of the hollow shaft.

Power/Torque transmission capacity:

Torque Transmission (T) \( = Z_p \times \tau_(permissible)\)

\( \frac {T_s}{T_h} = \frac {​​​​\frac{\pi\times d^3}{16}}{ \frac{\pi {d_o}^3 (1-k^4)}{16}} = \frac{d^3}{d_o^3}\times \frac{1}{(1-k^4)}\)

\( \frac {T_s}{T_h} = \frac{d^3}{d_o^3}\times \frac{1}{(1-k^2)}\times \frac{1}{(1+k^2)}\)

For shafts having the same cross-sectional area:

​\(A_s = A_h\)

\(\frac{\pi\times d^2}{4} = \frac{\pi\times {d_o}^2}{4}\times (1-k^2)\)

\(\frac{d}{d_o} = \sqrt{(1-k^2)}\)​​

On solving,

\( \frac {T_s}{T_h} = \frac{d^3}{d_o^3}\times \frac{1}{(1-k^2)}\times \frac{1}{(1+k^2)}\)

\( \frac {T_s}{T_h}= (1-k^2)^{3/2}\times \frac{1}{(1-k^2)}\times \frac{1}{(1+k^2)}\)

\( \frac {T_s}{T_h} = \frac{(1-k^2)^{1/2}}{(1+k^2)}\)

\(T_s < T_h\)

Conclusion:

T has a greater value for shafts having a hollow cross-section as compared to the shafts having a solid cross-section, So hollow shafts are stronger and more rigid than the solid shaft.

Explanation:

From the torsion equation

\(\frac{T}{J} = \frac{{\mathop τ \nolimits_{\max } }}{R} = \frac{{Gθ }}{L}\)

Given:

G = 40 GPa, τ_max = 110 MPa, θ = 0.3 radians, L = 0.5 m;

Now,

Considering strength and rigidity criterion together

\(\begin{array}{l} \frac{{\mathop τ \nolimits_{\max } }}{R} = \frac{{Gθ }}{L}\\ R = \frac{{\mathop τ \nolimits_{\max } L}}{{Gθ }}\\ D = \frac{{2\mathop τ \nolimits_{\max } L}}{{Gθ }} = \frac{{2 \times 0.5 \times 110 \times \mathop {10}\nolimits^6 }}{{40 \times \mathop {10}\nolimits^9 \times 0.3}} = 9.167 \; mm \end{array}\)

\(T = \frac{{\mathop τ \nolimits_{\max } J}}{R} = \frac{{\mathop τ \nolimits_{\max } \pi \mathop D\nolimits^3 }}{{16}} = 110 \times \mathop {10}\nolimits^6 \times \frac{\pi }{{16}} \times \mathop {9.167}\nolimits^3 \times \mathop {10}\nolimits^{ - 9} = 16.64 \; Nm\)

Concept:

As we know that for shaft under pure torsion

\(τ = \frac{16T}{\pi d^3}\)

As T is constant

∴ \(τ\ \ \ \alpha\ \ \ \frac{1}{{{d^3}}}\) 

\({τ_1\overτ_2} =\frac{d_2^3}{{{d_1^3}}}\)

Calculation:

Given:

τ₁ = 16, d₁ = d, \(d_2 = \frac{d}{2}\)

Then, \({τ_1\overτ_2} =\frac{d_2^3}{{{d_1^3}}}\Rightarrow {16\overτ_2} =\frac{(d/2)^3}{{{d^3}}}\)

τ₂ = 16 × 8 = 128 MPa

Given:

L = 0.5 m, P = 80 kW, N = 300 rpm

θ = 0.25°, τ = 36 MPa, G = 85 GPa

\(P = \frac{{2\pi NT}}{{60}} \Rightarrow T = \frac{{P \times 60}}{{2\pi N}}\)

\(T = \frac{{80 \times {{10}^3} \times 60}}{{2\pi \times 300}} = 2546.5\;N.m\)

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

\(\frac{{G\theta }}{L} = \frac{\tau }{r} \Rightarrow \frac{{G\theta }}{L} = \frac{\tau }{{\frac{D}{2}}}\)

\(D = \frac{{2\tau L}}{{G\theta }} = \frac{{2 \times 36 \times {{10}^6} \times 0.5}}{{85 \times {{10}^9} \times 0.00436}} = 0.09714\;m\)

D = 97.14 mm

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

\(\frac{{\pi \left( {{D^4} - {d^4}} \right)}}{{32}} = \frac{{2546.5 \times 0.5}}{{85 \times {{10}^9} \times 0.00436}} = 3.436 \times {10^{ - 6}}\)

D⁴ – d⁴ = 3.49 × 10^-5

(0.09714)⁴ – d⁴ = 3.49 × 10^-5

Explanation:

Line shaft:

• A shaft directly coupled to a power source is referred to as a Line Shaft. This type of shaft is used to transmit power from the primary source to various machine components or systems in mechanical operations. The term specifically denotes the direct connection between the power source and the shaft, ensuring efficient power transmission without intermediate mechanisms.
• The Line Shaft operates by being physically connected to the power source, such as an engine or motor. This connection allows the rotational motion generated by the power source to be directly transmitted to the shaft. The shaft then distributes this mechanical energy to other connected components, such as pulleys, gears, or belts, which perform specific tasks or operations. Its direct coupling eliminates the need for secondary power transmission systems, enhancing efficiency and minimizing energy losses.

Applications: Line Shafts are commonly used in industries that require direct and robust power transmission systems, such as:

• Textile manufacturing units for operating multiple machines simultaneously.
• Conveyor systems in material handling processes.
• Machining operations in workshops and factories.
• Power transmission setups in agricultural equipment.

Important InformationCounter Shaft

• A Counter Shaft is used as an intermediate shaft in power transmission systems. It typically connects the main power source to other shafts or components, allowing for speed or torque adjustments.

Jack Shaft:

• A Jack Shaft is another type of intermediate shaft used in power transmission systems. It often serves as a connecting shaft between two main shafts or between the power source and the final driven component. Similar to the Counter Shaft, its function involves intermediate transmission rather than direct coupling to the power source.

Flexible Shaft

• A Flexible Shaft is designed to transmit rotary motion between components that are not perfectly aligned or have relative motion between them. It is commonly used in applications requiring flexibility in power transmission, such as handheld tools or equipment.