🔗Statics and Strength of Materials Unit 10 Review

Power transmission through shafts is the process of moving mechanical energy from a source (like a motor) to a load (like a machine) via rotational motion. Understanding the relationship between power, torque, and shaft deflection is essential for sizing shafts correctly and preventing mechanical failure.

Power Transmission in Shafts

Fundamentals of Power Transmission

A circular shaft under torsion develops a shear stress distribution that varies linearly from zero at the center to a maximum at the outer surface. This non-uniform distribution is why the outer fibers of a shaft are the most critical for design.

The angle of twist a shaft experiences depends on four things:

The applied torque (more torque = more twist)
The shaft's length (longer shaft = more twist)
The shear modulus of the material, $G$ (stiffer material = less twist)
The polar moment of inertia of the cross-section, $J$ (larger cross-section = less twist)

These relationships are captured in the angle of twist formula covered below.

Calculating Power Transmission

The core relationship is straightforward: power equals torque times angular velocity.

$P = T \cdot \omega$

where $P$ is power in watts, $T$ is torque in N·m, and $\omega$ is angular velocity in rad/s.

Since rotational speed is usually given in RPM rather than rad/s, you'll need to convert:

Convert RPM to angular velocity: $\omega = \frac{2\pi N}{60}$ where $N$ is the rotational speed in RPM.
Find torque from power and speed: Rearranging $P = T \cdot \omega$ gives: $T = \frac{P}{\omega} = \frac{60P}{2\pi N}$
Check the shaft's capacity. Once you know the torque, verify that the shaft's material and geometry can handle the resulting torsional stress and deflection without failure.

Always keep your units consistent. If power is in watts, torque comes out in N·m and angular velocity in rad/s. If power is given in horsepower, convert first (1 hp = 745.7 W).

Torsional Deflection of Shafts

Torsional deflection is the angular deformation (twisting) a shaft undergoes when torque is applied. The angle of twist is given by:

$\phi = \frac{TL}{GJ}$

where $\phi$ is the angle of twist in radians, $T$ is the applied torque, $L$ is the shaft length, $G$ is the shear modulus, and $J$ is the polar moment of inertia.

This equation tells you that twist increases with torque and length, and decreases with stiffer materials and larger cross-sections.

Excessive torsional deflection is a real design concern. Even if stresses remain below the yield point, too much twist can cause:

Misalignment of connected components (gears, couplings)
Vibration and noise in the system
Reduced efficiency and premature wear

Shaft Stiffness and Rigidity

Torsional Stiffness

Torsional stiffness measures how much a shaft resists twisting per unit of applied torque. It's defined as:

$k_t = \frac{T}{\phi} = \frac{GJ}{L}$

where $k_t$ is in N·m/rad. A higher value means less twist for the same torque.

You'll need the polar moment of inertia for the cross-section:

Solid circular shaft: $J = \frac{\pi d^4}{32}$
Hollow circular shaft: $J = \frac{\pi (D^4 - d^4)}{32}$ where $D$ is the outer diameter and $d$ is the inner diameter.

Notice that $J$ depends on diameter to the fourth power. Even a small increase in diameter significantly increases stiffness.

Torsional Rigidity

Torsional rigidity is the product $GJ$ , and it represents the shaft's inherent resistance to torsion independent of length. This is the property you're comparing when choosing between shaft designs.

Two practical comparisons:

A steel shaft ( $G = 80 \text{ GPa}$ ) vs. an aluminum shaft ( $G = 26 \text{ GPa}$ ) of the same diameter: the steel shaft has roughly 3 times the torsional rigidity, so it twists about one-third as much under the same torque.
A solid shaft vs. a hollow shaft with the same outer diameter and material: the solid shaft has higher rigidity because removing the inner core reduces $J$ . However, hollow shafts offer a much better strength-to-weight ratio, which is why they're common in practice.

To increase torsional rigidity, you can either increase the shaft diameter or select a material with a higher shear modulus.

Power, Torque, and Rotational Speed

Relationship between Power, Torque, and Rotational Speed

The equation $P = T \cdot \omega$ connects all three variables. The key takeaway for design:

For a fixed power output, increasing rotational speed decreases the required torque, and vice versa. This is why high-speed shafts can be smaller in diameter than low-speed shafts transmitting the same power.
Torque is the rotational equivalent of linear force. It's what the prime mover (motor, engine) supplies to drive the shaft.
Rotational speed can be modified using gears, belts, or chain drives to match what the driven machinery needs.

Gear Ratios and Speed Changes

Gear ratios and pulley ratios let you trade speed for torque (or vice versa) within a power transmission system.

Gear systems:

The gear ratio equals the number of teeth on the driven gear divided by the number of teeth on the driving gear:

$\text{Gear Ratio} = \frac{N_{\text{driven}}}{N_{\text{driving}}}$

A ratio greater than 1 means speed reduction and torque increase.
A ratio less than 1 means speed increase and torque reduction.
Example: A driving gear with 20 teeth meshing with a driven gear of 60 teeth gives a 3:1 ratio. The output shaft spins at one-third the input speed but delivers three times the torque (minus losses).

Belt-driven systems:

The speed ratio equals the diameter of the driven pulley divided by the diameter of the driving pulley:

$\text{Speed Ratio} = \frac{D_{\text{driven}}}{D_{\text{driving}}}$

Example: A 100 mm driving pulley paired with a 200 mm driven pulley gives a 2:1 ratio, cutting the output speed in half while doubling the torque.

In both cases, power is conserved (minus efficiency losses). If speed goes down, torque goes up proportionally.

System Efficiency

Real power transmission systems always lose some energy to friction, slippage, and heat. Efficiency is defined as:

$\eta = \frac{P_{\text{out}}}{P_{\text{in}}}$

For example, a gearbox rated at 95% efficiency converts 95% of input power to useful output, with the remaining 5% lost as heat from friction between gear teeth, bearing losses, and oil churning.

When sizing a motor or shaft, account for these losses. If you need 10 kW at the output and your gearbox is 95% efficient, the input must be:

$P_{\text{in}} = \frac{10}{0.95} \approx 10.53 \text{ kW}$

Selecting low-friction bearings, proper lubrication, and well-matched components all help maximize system efficiency.

🔗Statics and Strength of Materials Unit 10 Review