RADIAL LOAD: HOW IT IS CALCULATED, EXERCISES SOLVED - PHYSICAL - 2024

The radial load is the force exerted perpendicular to the axis of symmetry of an object whose line of action passing through the axis. For example, a belt on a pulley imposes a radial load on the bearing or bearing of the pulley shaft.

In Figure 1 the yellow arrows represent radial forces or loads on the shafts due to the tension of the belt passing through the pulleys.

Figure 1. Radial load on pulley shafts. Source: self made.

The unit of measurement for radial load in the international or SI system is the Newton (N). But other units of force are also often used to measure it, such as the kilogram-force (Kg-f) and the pound-force (lb-f).

How is it calculated?

To calculate the value of the radial load on the elements of a structure, the following steps must be followed:

- Make the diagram of forces on each element.

- Apply the equations that guarantee translational equilibrium; that is, that the sum of all the forces is zero.

- Consider the equation of torques or moments so that rotational equilibrium is fulfilled. In this case the sum of all torques must be zero.

- Calculate the forces to be able to identify the radial loads that act on each of the elements.

Solved exercises

-Exercise 1

The following figure shows a pulley through which a tensioned pulley passes with tension T. The pulley is mounted on a shaft that is supported by two bearings. The center of one of them is at a distance L ₁ from the center of the pulley. At the other end is the other bearing, at distance L ₂.

Figure 2. Pulley through which a tensioned belt passes. Source: self made.

Determine the radial load on each of the journal bearings, assuming the shaft and pulley weight are significantly less than the applied stress.

Take as value for the belt tension 100 kg-f and for the distances L ₁ = 1 m and L ₂ = 2 m.

Solution

First, a diagram of the forces acting on the shaft is made.

Figure 3. Force diagram of exercise 1.

The pulley tension is T, but the radial load on the shaft at the pulley position is 2T. The weight of the shaft and pulley is not taken into account because the problem statement tells us that it is considerably less than the tension applied to the belt.

The radial reaction of the supports on the shaft is caused by the radial forces or loads T1 and T2. The distances L1 and L2 from the supports to the center of the pulley are also indicated in the diagram.

The coordinate system is also displayed. The total torque or moment on the axis will be calculated taking as the center the origin of the coordinate system and will be positive in the Z direction.

Equilibrium conditions

Now the equilibrium conditions are established: sum of forces equal to zero and sum of torques equal to zero.

From the second equation the radial reaction on the axis in support 2 (T ₂) is obtained, substituting in the first and solving for the radial reaction on the axis in support 1 (T ₁) is obtained.

T ₁ = (2/3) T = 66.6 kg-f

And the radial load on the shaft in the position of the support 2 is:

T ₂ = (4/3) T = 133.3 kg-f.

Exercise 2

The following figure shows a system made up of three pulleys A, B, C all of the same radius R. The pulleys are connected by a belt that has a tension T.

Shafts A, B, C go through lubricated bearings. The separation between the centers of axes A and B is 4 times the radius R. Similarly, the separation between axes B and C is also 4R.

Determine the radial load on the axes of pulleys A and B, assuming that the belt tension is 600N.

Figure 4. Pulley system. Exercise 2. (Own elaboration)

Solution

We begin by drawing a diagram of the forces that act on pulley A and on B. On the first we have the two tensions T ₁ and T ₂, as well as the force F _A that the bearing exerts on the axis A of the pulley.

Similarly, on pulley B there are the tensions T ₃, T ₄ and the force F _B that the bearing exerts on its axis. The radial load on the pulley shaft A is the force F _A and the radial load on the force F B is the _B.

Figure 5. Force diagram, exercise 2. (Own elaboration)

Since axes A, B, C form an isorectangle triangle, angle ABC is 45 °.

All the tensions T ₁, T ₂, T ₃, T ₄ shown in the figure have the same modulus T, which is the belt tension.

Balance condition for pulley A

Now we write the equilibrium condition for pulley A, which is nothing other than the sum of all the forces acting on pulley A must be zero.

Separating the X and Y components of the forces and adding (vectorially) the following pair of scalar equations is obtained:

F _{A X} -T = 0; F _{A Y} - T = 0

These equations lead to the following equality: F _AX = F _AY = T.

Therefore the radial load has magnitude given by:

F _A = (T² + T²) ^1/2 = 2 ^1/2 ∙ T = 1.41 ∙ T = 848.5 N. with direction of 45 °.

Balance condition for pulley B

Similarly, we write the equilibrium condition for pulley B. For component X we have: F _{B X} + T + T ∙ Cos45 ° = 0

Y for component Y: F _{B Y} + T ∙ Sen45 ° = 0

Thus:

F _BX = - T (1 + 2 ^-1/2) and F _BY = -T ∙ 2 ^-1/2

That is, the magnitude of the radial load on pulley B is:

F _B = ((1 + 2 ^-1/2) ² + 2 ^-1) ^1/2 ∙ T = 1.85 ∙ T = 1108.66 N and its direction is 135 °.

References

1. Beer F, Johnston E, DeWolf J, Mazurek, D. Mechanics of materials. Fifth edition. 2010. Mc Graw Hill. 1-130.
2. Gere J, Goodno, B. Mechanics of materials. Eighth edition. Cengage Learning. 4-220.
3. Giancoli, D. 2006. Physics: Principles with Applications. 6 ^th Ed. Prentice Hall. 238-242.
4. Hibbeler R. Mechanics of materials. Eighth edition. Prentice Hall. 2011. 3-60.
5. Valera Negrete, J. 2005. Notes on General Physics. UNAM. 87-98.