This document discusses centrifugal force, inertia force, and inertia torque. It provides equations to calculate the maximum stress on rotor blades due to centrifugal force. It also describes how inertia forces on aircraft propeller blades produce a twisting moment, and provides equations to calculate the normal force, twisting force, and twisting moment on blade elements. Finally, it reviews the equations of motion that relate the sum of forces and sum of torques to acceleration and angular acceleration of a rigid body in plane motion, and the relationship between torque and moment of inertia.

1. Tribhuwan University
Institute of Engineering
Thapathali Campus
A Presentation on
Centrifugal force, Inertia force and Inertia torque
Presenter:
Sushil Chapai (Champ)
071-BME-440

2. Introduction
To design the parts of a machine or mechanism for strength, it is necessary to
determine the forces and torques acting on the individual links. Each
component of a complete machine, however small, should be carefully
analysed for its role in transmitting force.

3. Centrifugal Force
In rotor, inertia force, the product of mass and acceleration, is known as
centrifugal force. In high speed rotors with blades, centrifugal forces tend to
separate the blades from the rotor. Given figure shows a simple type of bladed
rotor.
Using relations:
𝑑𝐹 = 𝑑𝑀 ∗ 𝐴 = 𝑑𝑀𝜔2
𝑅
𝑑𝐹 = 𝜌𝑏𝑡 𝑑𝑅 𝜔2
𝑅
∴ 𝐹 = 𝑏𝑡𝜌 𝜔2
𝑅=𝑅𝑖
𝑅=𝑅𝑜
𝑅𝑑𝑅
𝑆𝑜 𝑡ℎ𝑒 max 𝑠𝑡𝑟𝑒𝑠𝑠, 𝑆 =
𝐹
𝑏𝑡
= 𝜌𝜔2
𝑅=𝑅𝑖
𝑅=𝑅𝑜
𝑅𝑑𝑅

4. Similarly for rotor blades of fan, the fan has the form of a disk with
slots between blades. The element of inertia force dF is the same as
given by previous equation in which mass of the element is;
𝑑𝑀 = 𝜌𝑡𝑅𝑑∅𝑑𝑅
𝑑𝐹 = 𝜌𝑡𝜔2 𝑅2 𝑑𝑅𝑑∅
∴ 𝐹 = 𝜌𝑡𝜔2
∅=0
∅=2𝜋/𝑁
𝑅=𝑅𝑖
𝑅=𝑅𝑜
𝑅2 𝑑𝑅 𝑑∅
𝐹 =
2𝜋
𝑁
𝜌𝑡𝜔2
𝑅=𝑅𝑖
𝑅=𝑅𝑜
𝑅2 𝑑𝑅
𝑆𝑜 𝑡ℎ𝑒 max 𝑠𝑡𝑟𝑒𝑠𝑠, 𝑆 = 𝜌𝜔2/𝑅𝑖
𝑅=𝑅𝑖
𝑅=𝑅𝑜
𝑅2 𝑑𝑅

5. Also in aircraft propellers, the blades are set at a blade angle as shown
in figure. In such cases, inertia forces produce a twisting moment on
the blade. With reference to the figure the inertia force dF on an
element t dx dR is;
𝑑𝐹 = 𝜔2 𝑎 𝑑𝑀
This force has two components normal to the axis of shaft, Fn (known
as tensile force and parallel to the axis of shaft, Ft (which is responsible
for twisting moment dMt).
So, normal force is;
𝑑𝐹𝑛 =
𝑅
𝑎
𝑑𝐹 = 𝜔2 𝑅 𝑑𝑀
𝑑𝐹𝑛 = 𝜌𝑡𝜔2 𝑅 𝑑𝑥 𝑑𝑅
∴ 𝐹𝑛 = 𝜌𝑡𝜔2
𝑥=0
𝑥=𝑏
𝑅=𝑅𝑖
𝑅=𝑅𝑜
𝑅 𝑑𝑅 𝑑𝑥

6. And the twisting force is;
𝑑𝐹𝑡 =
𝑥 cos 𝛽
𝑎
𝑑𝐹 = 𝜔2
𝑥 cos 𝛽 𝑑𝑀
𝑑𝐹𝑡 = 𝜌𝑡𝜔2
𝑥 cos 𝛽 𝑑𝑥 𝑑𝑅
Also the twisting moment is given by;
𝑑𝑀𝑡 = 𝑥 sin 𝛽 𝑑𝐹
𝑑𝑀𝑡 = 𝜌𝑡 cos 𝛽 𝑥 sin 𝛽 𝜔2
𝑥2
𝑑𝑥 𝑑𝑅
∴ 𝑀𝑡 = 𝜌𝑡 cos 𝛽 𝑥 sin 𝛽 𝜔2
𝑥=0
𝑥=𝑏
𝑅=𝑅𝑖
𝑅=𝑅𝑜
𝑥2
𝑑𝑥 𝑑𝑅

7. Inertia Force and Inertia Torque
From the study of mechanics, it is known that the following equations of
motion apply for a rigid body in plane motion.
∑F=MA=R
∑Ƭ=Iα
In which ∑F is the vector sum, or the resultant R, of a system of forces
acting on the body in the plane of motion; M is the mass of the body;
and A is the acceleration of the centre of gravity of the body. ∑Ƭ is the
sum of the moments of the forces and torques about an axis through the
mass centre normal to the plane of motion; I is the moment of inertia of
the body about the same axis through the mass centre; and a is the
angular acceleration of the body in the plane of motion .

8. Now, if we consider e be the distance of CG from the axis of rotation in
case of rigid body in plane motion then,
Torque = Re
i.e, Iα = Re

9. Thank you!