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In mechanical engineering, power transmission refers to the process of transferring mechanical power from a source (like an engine or motor) to a machine or device that performs work (such as a conveyor belt, pump, or machine tool). Efficient power transmission is critical for the smooth functioning of industrial machinery, vehicles, and countless mechanical systems.
Power transmission involves components that carry torque and rotational motion from one part to another. Common methods include shafts, belt drives, and gear systems. These components are designed to handle forces and transmit motion with minimal losses.
This chapter covers the foundational mechanics concepts related to power transmission, including how torque and angular velocity relate to power, how stresses develop in shafts due to torsion, and how belt drives function and transfer power.
In all analyses, metric units are used: torque in Newton-meters (Nm), power in Watts (W), rotational speed in revolutions per minute (rpm), and lengths in meters (m). Mastery of these concepts is essential for design, analysis, and problem-solving in power transmission systems.
Let's begin by understanding the relationship between torque, angular velocity, and power.
Torque (T) is a measure of the rotational force applied to an object causing it to rotate about an axis. It is expressed in Newton-meters (Nm). Imagine trying to open a door: the force exerted on the door handle times the distance from the hinge is torque.
Angular velocity (\(\omega\)) is the rate of rotation, measured in radians per second (rad/s). It tells us how fast something is spinning.
Power (P) is the rate at which work is done or energy is transferred. For rotating systems, power is related to torque and angular velocity by
Because many motors and machines express speed in revolutions per minute (rpm), we convert rpm to rad/s using
\[\omega = \frac{2\pi N}{60}\]where \(N\) is the speed in rpm.
Combining both, power transmitted by a rotating shaft is
\[P = T \times \frac{2\pi N}{60} = \frac{2\pi N T}{60}\]This formula allows calculation of power when the torque and rotational speed are known.
A rotating shaft transmits torque by twisting about its longitudinal axis. This twisting action induces torsional shear stresses within the shaft's material.
Consider a circular shaft of diameter \(d\) subjected to a torque \(T\). The resistance of the shaft to twisting depends on its cross-sectional geometry quantified by the polar moment of inertia (\(J\)). For a solid circular shaft,
\[J = \frac{\pi d^{4}}{32}\]Torsion causes shear stress \(\tau\) distributed over the shaft's cross-section. The maximum shear stress occurs at the outer surface (radius \(r = d/2\)) and is given by
\[\tau = \frac{T r}{J} = \frac{T d}{2 J}\]This formula highlights that the shear stress increases with torque and radius but decreases with a larger polar moment of inertia.
Additionally, the shaft undergoes a twist angle \(\theta\) over its length \(L\) given by
\[\theta = \frac{T L}{G J}\]where \(G\) is the modulus of rigidity (shear modulus) of the shaft material, indicating its resistance to shear deformation.
Why is understanding torsion important? Because designers need to ensure shafts can withstand the twisting stresses without failure or excessive deformation that could misalign machinery or cause breakdowns.
Belt drives are widely used to transmit power between rotating shafts that are some distance apart. A belt loops over two or more pulleys, transmitting motion and power through frictional forces.
The belt has two tension sides:
The power transmitted by the belt is
\[P = (T_1 - T_2) \times v\]where \(v\) is the belt velocity (m/s), calculated by
\[v = \frac{\pi d N}{60}\]with \(d\) as the pulley diameter (m) and \(N\) as its rotational speed (rpm).
The ratio of tensions relates to the coefficient of friction \(\mu\) and the angle of lap \(\alpha\) (the contact angle between belt and pulley, in radians):
\[\frac{T_1}{T_2} = e^{\mu \alpha}\]This exponential relation shows how tension difference-and thus power capacity-depends on frictional grip and belt wrap angle.
Slip and creep are important factors reducing efficiency. Slip denotes the difference between belt speed and pulley surface speed causing power loss. Creep refers to the small relative movement of belt with respect to pulley surface due to belt elasticity.
Proper tensioning and choosing correct materials minimize these effects for efficient power transmission.
Step 1: Note down given data:
Step 2: Use power formula
\[ P = \frac{2 \pi N T}{60} \]
Step 3: Substitute values
\[ P = \frac{2 \times 3.1416 \times 1200 \times 150}{60} = 18849.56 \text{ Watts} \]
Step 4: Convert power to kilowatts
\[ P = \frac{18849.56}{1000} = 18.85 \text{ kW} \]
Answer: The power transmitted by the shaft is approximately 18.85 kW.
Step 1: Given data:
Step 2: Calculate polar moment of inertia \(J\):
\[ J = \frac{\pi d^4}{32} = \frac{3.1416 \times (0.04)^4}{32} = 2.01 \times 10^{-7} \text{ m}^4 \]
Step 3: Calculate radius \(r = \frac{d}{2} = 0.02 \text{ m}\)
Step 4: Use shear stress formula
\[ \tau = \frac{T r}{J} = \frac{500 \times 0.02}{2.01 \times 10^{-7}} = 49.75 \times 10^{6} \text{ Pa} = 49.75 \text{ MPa} \]
Answer: The maximum shear stress in the shaft is approximately 49.75 MPa.
Step 1: Given data:
Step 2: Calculate design shear stress:
\[ \tau_{design} = \frac{\tau_{allow}}{n} = \frac{40}{2} = 20 \text{ MPa} = 20 \times 10^6 \text{ Pa} \]
Step 3: Find torque \(T\) transmitted using power formula:
\[ P = \frac{2 \pi N T}{60} \Rightarrow T = \frac{60 P}{2 \pi N} = \frac{60 \times 20000}{2 \pi \times 1000} = 191 \text{ Nm} \]
Step 4: Use shear stress formula to solve for diameter \(d\):
\[ \tau = \frac{16 T}{\pi d^3} \Rightarrow d^3 = \frac{16 T}{\pi \tau} \]
Rearranged for \(d\):
\[ d = \left( \frac{16 T}{\pi \tau} \right)^{1/3} \]
Substitute values:
\[ d = \left( \frac{16 \times 191}{3.1416 \times 20 \times 10^6} \right)^{1/3} = \left(4.864 \times 10^{-5}\right)^{1/3} = 0.0367 \text{ m} = 36.7 \text{ mm} \]
Answer: The shaft diameter should be approximately 37 mm to safely transmit the required power with the given factor of safety.
Step 1: Convert angle of lap to radians:
\[ \alpha = 160^{\circ} = \frac{160 \pi}{180} \approx 2.7925 \text{ radians} \]
Step 2: Calculate belt velocity \(v\):
\[ v = \frac{\pi d N}{60} = \frac{3.1416 \times 0.5 \times 600}{60} = 15.707 \text{ m/s} \]
Step 3: Calculate tension difference:
\[ P = (T_1 - T_2) \times v \Rightarrow T_1 - T_2 = \frac{P}{v} = \frac{5000}{15.707} = 318.3 \text{ N} \]
Step 4: Use tension ratio formula:
\[ \frac{T_1}{T_2} = e^{\mu \alpha} = e^{0.25 \times 2.7925} = e^{0.6981} = 2.01 \]
Step 5: Let \(T_2 = T\), then \(T_1 = 2.01 T\):
\[ T_1 - T_2 = 2.01 T - T = 1.01 T = 318.3 \]
Step 6: Solve for \(T_2\):
\[ T = \frac{318.3}{1.01} = 315.1 \text{ N} \]
Thus, \(T_1 = 2.01 \times 315.1 = 633.3 \text{ N}\)
Answer: The slack side tension \(T_2\) is approximately 315 N and the tight side tension \(T_1\) is approximately 633 N.
Step 1: Calculate velocity of belt based on driver pulley:
\[ v = \frac{\pi d N}{60} = \frac{3.1416 \times 0.6 \times 900}{60} = 28.27 \text{ m/s} \]
Step 2: Calculate ideal speed of driven pulley (without slip):
\[ N_2 = \frac{v \times 60}{\pi d_2} = \frac{28.27 \times 60}{3.1416 \times 0.4} = 1350 \text{ rpm} \]
Step 3: Calculate actual speed considering slip of 3%:
\[ N_{2(actual)} = N_2 \times (1 - 0.03) = 1350 \times 0.97 = 1309.5 \text{ rpm} \]
Step 4: Calculate power loss due to slip:
Power loss = \( \text{slip} \times \text{input power} = 0.03 \times 10,000 = 300 \text{ W} \)
Answer: The actual speed of the driven pulley is approximately 1310 rpm, and power loss due to slip is 300 W.
When to use: Calculations involving angular velocity or power transmitted through rotating shafts.
When to use: All problems involving shaft dimensions and stress calculations.
When to use: Quickly finding belt tensions and analyzing belt drive problems.
When to use: Design and dimensioning exam questions.
When to use: All mechanical analysis problems involving power transmission.
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