7.5 Analyze introductory belt and journal-bearing friction | University of Alberta ENGG 130: Engineering Mechanics: Statics

University of Alberta ENGG 130: Engineering Mechanics: Statics

Friction

ENGG 130 Engineering Mechanics: Statics — Study topic 7.5

Belt friction and journal-bearing friction both resist relative motion, but they are modeled differently. For a belt wrapped around a fixed drum, the limiting tension ratio depends on the friction coefficient and the wrap angle. For a journal bearing, tangential friction at the journal resists impending rotation and creates a moment about the shaft centre. The examples use idealized introductory models. In each problem, define the isolated body, identify the impending motion, apply the appropriate friction relation, and check equilibrium with consistent units.

What you will learn

Distinguish the introductory friction models for a belt on a fixed drum and a journal bearing.
Use the belt friction relation to determine the limiting tension ratio at impending slip.
Find limiting journal friction and its moment using equilibrium and the friction model.
Choose friction directions from the stated impending motion and check force and moment equilibrium.

Belt friction: the limiting tension ratio

Consider a flexible belt wrapped around a fixed, rough drum. Isolate the part of the belt in contact with the drum. The belt tensions act at the two ends of this segment; the drum applies distributed normal contact and friction forces along the wrap. Friction opposes the belt’s impending motion relative to the drum.

At impending slip, the belt-friction relation connects the larger, tight-side tension to the smaller, slack-side tension. The wrap angle must be expressed in radians, and the coefficient of friction is dimensionless. Since the relation gives a ratio, an additional given tension or other equilibrium information is needed to find actual tension values.

Determine which side is tight from the stated impending direction. Place that tension in the numerator. If the assumed impending direction reverses, the tight and slack sides can exchange roles. If the belt is not at impending slip, do not automatically impose the limiting ratio.

\frac{T_{\mathrm{tight}}}{T_{\mathrm{slack}}}=e^{\mu\theta}

Use the limiting relation at impending slip, not as an equality for every belt condition.
Convert the wrap angle to radians.
The tight-side tension is the numerator.

Journal-bearing friction in planar statics

A journal is the cylindrical part of a shaft supported by a bearing. In the introductory model, represent the contact by a normal force and a tangential friction force. The normal force acts perpendicular to the contact surface; friction acts along it and opposes the journal’s impending motion. At impending motion, the friction-force magnitude is the coefficient of friction times the normal-force magnitude.

For a journal of radius

r

, tangential friction creates a moment about the shaft centre. Its perpendicular distance from the centre is

r

, so the friction-moment magnitude is the friction force multiplied by the radius. Find the normal force from equilibrium of the isolated body; it equals the applied load only in a suitably simple case, such as one vertical load balanced by one vertical normal force.

Choose a positive moment direction and keep it throughout. To determine the friction moment’s sign, use the force location and direction: the moment about a point is positive for counterclockwise turning and negative for clockwise turning. An applied couple that holds the journal in equilibrium must oppose the friction moment. Include any other external forces required to balance the contact forces.

F_f=\mu N,\qquad |M_f|=F_f r

At impending sliding, the friction magnitude is the coefficient of friction times the normal force.
The friction moment magnitude is the friction force times the journal radius.
A force and a moment have different units: newtons and newton-metres.

A consistent solution and checking method

First define the isolated body: for a belt problem, consider the belt segment in contact with the drum; for a journal problem, isolate the shaft or journal specified. Draw and label forces, reactions, and any applied couple. Identify the likely impending motion so the friction direction follows from the contact motion rather than guesswork.

Choose axes and a moment sign convention. For a belt, convert the wrap angle to radians and use the belt relation together with the given tension information. For a journal, apply force equilibrium to find the normal reaction, then use the friction model and the radius to find the limiting friction force and moment. If a couple is required to prevent impending rotation, its moment must balance the friction moment.

Use consistent units when substituting. Convert journal radii to metres when using newtons and reporting newton-metres. The belt tension ratio is dimensionless; a journal moment is force multiplied by length. Finish by checking horizontal and vertical force balances and moment balance. A negative signed result means the actual direction is opposite the assumed direction.

\sum F_x=0,\qquad \sum F_y=0,\qquad \sum M_O=0

State whether the body is at impending slip or impending rotation.
Find reactions from equilibrium before using them in a friction calculation.
Check force and moment balance separately, with consistent signs and units.

When to use the limiting model

The belt relation applies to a flexible belt in contact with a fixed drum through a specified wrap angle when the belt is at impending slip. The journal model uses a specified friction coefficient and an idealized contact force. Confirm that the stated contact and motion condition match the model before applying it.

At impending motion, use limiting friction. If a problem supplies a particular friction force rather than asking for the limiting condition, use the supplied force in equilibrium instead of replacing it automatically with the limiting value.

Match the friction model to the stated contact and motion condition.
A friction coefficient is dimensionless; friction force and friction moment have units.

Worked example

Find the holding tension for a belt

A belt wraps around a fixed drum through

150^\circ

. The coefficient of friction is

0.25

. The belt is about to slip, and the tight-side tension is

800\,\mathrm{N}

. Find the slack-side tension.

Convert the contact angle
The belt relation uses radians, so convert the given wrap angle before substituting.
$\theta=150^\circ\left(\frac{\pi}{180^\circ}\right)=2.618\,\mathrm{rad}$
Apply the limiting relation
The tight side is specified, so its tension is the larger value in the ratio. Rearrange the relation to find the slack-side tension.
$T_{\mathrm{slack}}=\frac{800\,\mathrm{N}}{e^{(0.25)(2.618)}}\approx415.9\,\mathrm{N}$
Check the ratio
The ratio of the two tensions matches the limiting ratio for the given friction coefficient and wrap angle.
$\frac{800\,\mathrm{N}}{415.9\,\mathrm{N}}\approx1.924=e^{(0.25)(2.618)}$

Answer: The slack-side tension is approximately

416\,\mathrm{N}

.

Check: The tight-side tension is larger, as specified, and the tension ratio is dimensionless. It equals the limiting ratio for impending slip.

Worked example

Find the largest tight-side belt tension

A belt wraps around a fixed drum through

120^\circ

. The coefficient of friction is

0.30

. The slack-side tension is

300\,\mathrm{N}

. Find the largest tight-side tension at impending slip.

Convert the wrap angle
Convert degrees to radians before using the belt relation.
$\theta=120^\circ\left(\frac{\pi}{180^\circ}\right)=2.094\,\mathrm{rad}$
Calculate the limiting ratio
At impending slip, the tight-to-slack tension ratio is the exponential of the friction coefficient times the wrap angle.
$e^{\mu\theta}=e^{(0.30)(2.094)}\approx1.874$
Find the tight-side tension
Multiply the given slack-side tension by the limiting ratio.
$T_{\mathrm{tight}}=(300\,\mathrm{N})(1.874)\approx562.2\,\mathrm{N}$

Answer: The largest tight-side tension at impending slip is approximately

562\,\mathrm{N}

.

Check: The calculated tight-to-slack ratio is approximately

562.2\,\mathrm{N}/300\,\mathrm{N}=1.874

, matching the limiting ratio.

Worked example

Journal-bearing friction force and resisting moment

A horizontal shaft is supported by a journal bearing at its lower contact. A downward load of

1.20\,\mathrm{kN}

acts through the shaft centre. The journal radius is

25\,\mathrm{mm}

and the friction coefficient is

0.08

. A horizontal force through the shaft centre balances the bearing friction force. Find the limiting friction force and resisting moment. If the shaft is about to rotate clockwise, find the applied couple required for equilibrium.

Find the normal force
The vertical forces are the upward contact normal and the downward load. Vertical equilibrium gives the normal force. The horizontal force through the centre must balance friction.
$\sum F_y=0:\quad N-1.20\,\mathrm{kN}=0,\quad N=1200\,\mathrm{N}$
Calculate limiting friction
At impending rotation, apply the limiting friction relation using the normal force.
$F_f=\mu N=(0.08)(1200\,\mathrm{N})=96\,\mathrm{N}$
Calculate the friction moment
Convert the radius to metres. At the bottom contact, the position from the centre is downward and friction acts to the right. This force tends to turn the shaft counterclockwise, so its moment is positive under a counterclockwise-positive convention.
$M_f=F_f r=(96\,\mathrm{N})(0.025\,\mathrm{m})=2.40\,\mathrm{N\cdot m}$
Balance forces and moments
Clockwise impending rotation is opposed by the counterclockwise friction moment. Therefore the required applied couple is clockwise. The horizontal force $H$ balances friction, while the normal force balances the load. Taking moments about the shaft centre, the normal force and both centre-applied forces have zero moment arm.
$\sum F_x=96\,\mathrm{N}-96\,\mathrm{N}=0,\quad \sum F_y=1200\,\mathrm{N}-1200\,\mathrm{N}=0,\quad \sum M_O=2.40\,\mathrm{N\cdot m}-2.40\,\mathrm{N\cdot m}=0$

Answer: The limiting friction force is

96\,\mathrm{N}

and the resisting moment is

2.40\,\mathrm{N\cdot m}

counterclockwise. The required applied couple is

2.40\,\mathrm{N\cdot m}

clockwise.

Check: The contact normal balances the downward load, and the centre-applied horizontal force balances friction. The upward normal acts through the centre line, so it has zero moment about the centre; the lower-contact friction acts right and produces a counterclockwise moment. The clockwise applied couple balances that moment.

Common mistakes and how to avoid them

Using degrees directly in the exponential belt relation.

Correction: Convert the wrap angle to radians before evaluating the relation.

Putting slack-side tension in the numerator.

Correction: At impending slip, place the larger tight-side tension in the numerator.

Assuming the bearing normal always equals one listed load.

Correction: Use force equilibrium to find the normal resultant from all forces on the isolated body.

Treating the journal friction force as a moment.

Correction: First find the friction force in newtons, then multiply by the radius in metres to find the moment in newton-metres.

Choosing the friction direction without considering impending motion.

Correction: Friction opposes the stated or assumed relative motion; keep that direction consistent in the diagram and equations.

Lesson summary

At impending slip on a fixed drum, the belt’s tight-to-slack tension ratio is $e^{\mu\theta}$ , with $\theta$ in radians.
For an introductory journal-bearing model at impending motion, use $F_f=\mu N$ and a friction moment magnitude of $F_f r$ .
State the impending direction, use equilibrium to determine reactions, and verify force and moment balance with consistent units.

Check your understanding

Question 1

A belt has

\mu=0.20

and wraps around a drum through

90^\circ

. At impending slip, what is the tight-to-slack tension ratio?

$e^{0.20(\pi/2)}\approx1.37$
$e^{0.20(90)}$
$0.20(\pi/2)\approx0.31$
$e^{0.20(2\pi)}\approx3.51$

Show answer and explanation

e^{0.20(\pi/2)}\approx1.37

Convert

90^\circ

to

\pi/2

radians, then evaluate

e^{\mu\theta}

. The ratio is dimensionless.

Question 2

A journal bearing supports a

500\,\mathrm{N}

downward load and has

\mu=0.10

. With no other vertical forces, what is its limiting friction force?

$5\,\mathrm{N}$
$50\,\mathrm{N}$
$500\,\mathrm{N}$
$5000\,\mathrm{N}$

Show answer and explanation

50\,\mathrm{N}

Vertical equilibrium gives a normal force of

500\,\mathrm{N}

. Thus

F_f=(0.10)(500\,\mathrm{N})=50\,\mathrm{N}

.

Key terms

Impending slip: The limiting condition just before relative sliding begins.
Wrap angle: The angle through which a belt contacts a drum; use radians in the belt relation.
Journal bearing: A bearing that supports a shaft at its cylindrical journal surface.
Limiting friction: The maximum friction force in the introductory model just before relative motion begins.

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7.5 · Analyze introductory belt and journal-bearing friction

University of Alberta ENGG 130: Engineering Mechanics: Statics

ENGG 130 Engineering Mechanics: Statics — Study topic 7.5

What you will learn

Belt friction: the limiting tension ratio

Journal-bearing friction in planar statics

A consistent solution and checking method

When to use the limiting model

Find the holding tension for a belt

Find the largest tight-side belt tension

Journal-bearing friction force and resisting moment

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through ENGG 130

About this lesson and its review