Learning topic

Power Transmission by Shafts

Relationship between shaft power, torque, and rotational speed using P = Tω and T ≈ 9550P/n, followed by torsional strength and stiffness checks.

0 practice tasks · 0 subtopics

This topic explains the relationship between transmitted mechanical power, shaft torque, and rotational speed. It covers P = Tω and T ≈ 9550P/n, consistent units, conversion from drive parameters to design torque, and the subsequent shaft checks for torsional shear stress and angle of twist.

In a mechanical drive, a shaft transmits energy between a motor, gearbox, coupling, and driven machine. For shaft strength calculations, power and rotational speed must first be converted into torque.

Mechanical power in rotational motion is:

$$P=T\omega.$$

If rotational speed is given in revolutions per minute:

$$\omega=\frac{2\pi n}{60},\qquad T=\frac{60P}{2\pi n}.$$

P — power, W;
T — torque, N·m;
ω — angular speed, rad/s;
n — rotational speed, rpm.

In the common engineering form, for P in kW and n in rpm: $T\approx9550P/n$ N·m.

Physical meaning

For the same transmitted power, reducing rotational speed increases torque. This is why an idealized reduction gearbox that lowers output speed increases the torque available at its output shaft.

From drive parameters to shaft calculation

After determining $T$, use it in the torsion relations. For a circular shaft, determine the polar section properties, maximum torsional shear stress, and angle of twist.

For a straight solid or hollow circular shaft under elastic torsion:

$$\tau(\rho)=\frac{T\rho}{J_p},\qquad \tau_{\max}=\frac{T}{W_p}.$$

If $T$, $G$, and $J_p$ are constant over a segment, the angle of twist is:

$$\varphi=\frac{TL}{GJ_p}.$$

$T$ — torque;
$\rho$ — radial distance from the shaft axis;
$J_p$ — polar second moment of area;
$W_p$ — polar section modulus;
$G$ — shear modulus;
$L$ — segment length;
$\varphi$ — angle of twist, rad.

For a stepped shaft, the total twist is the algebraic sum $\varphi=\sum_i T_iL_i/(G_iJ_{p,i})$.

Example

A shaft transmits $P=15\ \text{kW}$ at $n=1500\ \text{rpm}$. Then:

$$T\approx\frac{9550\cdot15}{1500}=95.5\ \text{N·m}.$$

This is the basic torque for the subsequent shaft calculation. A real drive may additionally require allowances for transmission efficiency, load nonuniformity, starting conditions, and dynamic effects.

Units and a common error

In the formula containing the coefficient 9550, power $P$ must be entered in kW and rotational speed $n$ in rpm; torque $T$ is obtained in N·m. When using $P=T\omega$, all quantities must be expressed in a consistent SI unit system.