7.6 journal bearings - civil engineering...7.6 journal bearings example 1, page 1 of 3 f 140 n 80 mm...

7.6 Journal Bearings

7.6 Journal Bearings Example 1, page 1 of 3

F

140 N

80 mm

1. The pulley has a radius of 80 mm and has

negligible weight. If the pulley fits loosely on a

12-mm-diameter fixed shaft, and the coefficient of

static friction is s = 0.2, determine the minimum

force F required to start the pulley rotating

clockwise.

1 The angle of friction and the radius of the circle

of friction are

s = tan-1 s

= tan-10.2

= 11.3099°

rf = r sin s

= (12/2 mm) sin 11.3099°

= 1.1767 mm (1)


140 N

2

Pulley

P1

Fixed shaft

Contact

point

For clarity, the difference in size between the

radius of the shaft and the inner radius of the

pulley has been exaggerated in the diagram.

Suppose F = 140 N

F

140 N

Pulley

P1

New contact

point

P2

Pulley rotates clockwise.

Position of pulley if rope

forces were equal Rotated position of pulley caused by

increase in force F.

3


Normal force N and friction force f

from shaft acting on pulley.

4

NR

f

Line of action of

resultant, R, is tangent

to circle of friction.

Frictional force f

opposes clockwise

rotation of pulley

Pulley

P2

Free-body diagram of pulley5

RyR = Rx2 + Ry

2

Rx rf

O

F

140 N

80 mm

80 mm

Fx = 0: F + 140 N Rx = 0

Fy = 0: Ry = 0

MO = 0: (140 N)(80 mm) F(80 mm)

+ Rx2 + Ry

2 (rf) = 0

1.1767 mm, by Eq. 1

Solving gives

Rx = 284 N

Ry = 0

F = 144.2 N Ans.

7 Equilibrium equations

++

+

Pulley

P2

rf

O

For convenience, express

R in terms of horizontal

and vertical components

(rather than in terms of N

and f).

6


135 N

140 N

80 mm

2. The pulley has a radius of 80 mm and has negligible

weight. If the pulley fits loosely on a 12-mm-diameter

fixed shaft, and the pulley rotates with constant angular

velocity counterclockwise, determine the coefficient of

kinetic friction, µk.

1 The angle of friction, k, and radius of the circle, rf ,

can be expressed in terms of k:

k = tan-1k

rf = r sin k

= (12/2 mm) sin ( tan-1k)

= 6 sin ( tan-1k) (1)


135 N

2

Pulley

P1

Fixed shaft

Contact

point




135 N

140 N

Pulley

P1

New contact

point

Pulley rotates counterclockwise.

Position of pulley if belt

forces were equal Rotated position of pulley caused by

increase of belt force to 140 N.

3

Suppose that the

140 N force is

replaced by a 135

N force.

P2




4

Line of action of resultant, R, is

tangent to circle of friction.

Frictional force opposes

counterclockwise

rotation of pulley.

P2R

f

N

rf


P2

Ry

rf

Rx

R = Rx2 + Ry

2

140 N

80 mm

80 mm

135 N

Free-body diagram of pulley5

Fx = 0: 135 N + 140 N Rx = 0

Fy = 0: Ry = 0

MO = 0: (140 N)(80 mm) (135 N)(80 mm)

Rx2 + Ry

2 (rf) = 0

Solving gives

Rx = 275 N

Ry = 0

rf = 1.455 mm


++

+

8 Substituting the value of rf in Eq. 1 gives

rf = 6 sin (tan-1k) (Eq. 1 repeated)

Thus

sin-1(1.455/6) = tan-1k

k = tan [sin-1(1.455/6)]

= 0.25 Ans.

1.455 mm

6 For convenience, express R in

terms of horizontal and vertical

components (rather than in

terms of N and f).

O


130 mm

1

F

100 N

3. The force from the belt causes the 130-mm-radius

pulley to rotate counterclockwise with constant angular

velocity. The pulley fits loosely on a fixed shaft of

24-mm diameter. Determine the value of the belt force F

if the coefficient of kinetic friction is µk = 0.3. Assume

that no slipping occurs between the belt and pulley.

Calculate the angle of friction and the radius of

the circle of friction.

k = tan-1 k

= tan-1 0.3

= 16.6992°

rf = r sin k

= (24/2 mm) sin 16.6992°

= 3.4482 mm (1)


Suppose F = 100 N

100 N

Pulley A

P1

Fixed shaft

Contact point

100 N

Pulley A

P1

New contact point

P2

Pulley rotates counterclockwise.

F

For clarity, the difference in size between the radius

of the shaft and the inner radius of the pulley has

been exaggerated in the diagram.

2 Position of pulley if belt force were equal

3 Rotated position of pulley caused by a

decrease in value of force F.


Pulley A

P2

rf

f

NR


counterclockwise rotation of

pulley.

Line of action of resultant, R, is tangent to circle of

friction.


from shaft action on pulley

4


100 N

Pulley A

P2

F

rf

R = Rx2 + Ry

2

130 mm

130 mm Rx

Ry

O


Fx = 0: Rx 100 N = 0

Fy = 0: Ry F = 0

MO = 0: (100 N)(130 mm) F(130 mm)

Rx2 + Ry

2 (rf) = 0

Solving gives

Rx = 100 N

Ry = 96.3 N

F = 96.3 N Ans.

+

++

3.4482 mm, by Eq. 1




terms of N and f).


8 in.

4 in.

20 lb 40.5 lb

4. The stepped pulley has radii of 4 in. and 8 in. and fits

loosely on a 0.5-in.-diameter fixed shaft. If the given loads

cause the pulley to rotate clockwise with a constant angular

velocity, determine the normal and frictional forces from

the shaft acting on the pulley. Also find the coefficient of

kinetic friction, µk .

1 The angle of friction, k, and radius of friction circle,

rf , can be expressed in terms of µk:

k = tan-1µk

rf = r sin k

= (0.5/2 in) sin (tan-1µk)

= 0.25 sin (tan-1µk) (1)


P1

Contact pointFixed shaft

Pulley

40.5 lb20 lb

Replace by 40 lb

P2

New contact

point


For clarity, the difference in size between the radius

of the shaft and the inner radius of the pulley has

been exaggerated in the diagram.

2 Position of pulley if belt force were

slightly altered to produce equal moments

(20 lb 8 in. = 40 lb 4 in.) about the

center of the pulley.

3 Rotated position of pulley caused by an increase of

belt force on the right to 40.5 lb.

8 in. 4 in.

P1

40.5 lb20 lb

8 in. 4 in.




4

Frictional force f opposes the

clockwise rotation of the pulley.



P2

N

R

f

rf


O

R = Rx2 + Ry

2

Equilibrium equations

Fx = 0: Rx = 0

Fy = 0: Ry 20 lb 40.5 lb = 0

MO = 0: (20 lb)(8 in.) (40.5 lb)(4 in.)

+ Rx2 + Ry

2 rf = 0

Solving gives

Rx = 0

Ry = 60.5 lb

rf = 0.0331 in.

5 Free-body diagram of pulley

7

+

++

6 For convenience,

express R in terms of

horizontal and

vertical components

(rather than in terms

of N and f).

P2

rf

40.5 lb20 lb

8 in. 4 in.

Rx

Ry


9

R = Rx2 + Ry

2 = 0 + (60.5 lb)2 = 60.5 lb

(line of action passes through P2 and is tangent to the

friction circle)

8 Substituting the value of rf just computed into Eq. 1

gives

rf = 0.25 sin (tan-1µk) (Eq.1 repeated)

Thus,

sin-1 (0.0331/0.25) = tan-1µk

or

µk = tan sin-1 (0.0331/0.25)

= 0.134 Ans.

The normal force N and friction force f can be

found by resolving the resultant R into components:

f = R cos

= (60.5 lb) cos 82.392°

= 8.01 lb Ans.

N = R sin

= (60.5 lb) sin 82.392°

= 60.0 lb Ans.

0.0331 in.

O

rf = 0.0331 in.

rpulley

= 0.25 in.

N (line of action passes

through P2 and center of

circle, O)

f

P2

= cos-1(0.0331 in./0.25 in.)

= 82.392°

Circle of friction

drawn to scale

relative to the

0.25-in. radius

of the hole in the

pulley


P

A

B

C

P150 mm150 mm

5. The couple forces P are intended to rotate the shaft and wind

the cable around the drum, thus raising the 5-kg mass. The

shaft has a diameter of 30 mm and fits loosely in the journal

bearing B. If the combined mass of the shaft handle A and

drum C is 15 kg, and the coefficient of static friction is µs = 0.2,

determine the minimum value of P required to initiate upward

motion of the 5-kg mass. Assume the drum C is attached

rigidly to the shaft.

Diameter of drum = 200 mm

1 The angle of friction and the radius of circle of

friction are

s = tan-1µs

= tan-10.2

= 11.3099°

rf = r sin 11.3099°

= (30 mm/2) sin 11.3099°

= 2.9394 mm (1)

5 kg


P1 P1

P2

Fixed support bearing at B

Shaft (free to rotate

in support bearing)Contact

point

New contact

point

Rotation of shaft counterclockwise


radius of the shaft and the radius of the support

bearing has been exaggerated in the diagram.

2 Position of the shaft, if external

forces were balanced so that no

friction force exists at point P1. That

is, no tendency exists to either raise

or lower the weight.

3 Rotated position of shaft caused by

increasing the forces P on the handle (The

shaft tends to "climb" the wall of the bearing

until it reaches a point where slipping

occurs).


P1

P2

N

R

f

rf

Frictional force opposes the

counterclockwise rotation of the shaft.

Line of action of

resultant, R, is tangent

to circle of friction

4 Normal force N and friction force f

from bearing acting on the shaft.


R = Rx2 + Ry

2

Rx

Ry

P

ShaftDrum C

O

P

150 mm 150 mm

Weight of handle and drum = (15 kg)(9.81 m/s2

= 147.15 N

200/2 mm = 100 mm

rf

Free body of shaft, handle A, and drum5


Fx = 0: Rx = 0

Fy = 0: Ry + P P 147.15 N 49.05 N = 0

MO = 0: (2 P)(150 mm) (49.05 N)(100 mm)

(Rx2 + Ry

2) (rf) = 0

Solving gives

Rx = 0

Ry = 196.2 N

P = 18.3 N Ans.

7

+

++

2.9394 mm, by Eq. 1

Weight of block

= (5 kg)(9.81 m/s2

= 49.05 N




terms of N and f).


10 in. 10 in.

A B

100 lb W

6. Pulleys A and B are identical. Each has a radius of 10 in.,

weighs 10 lb and fits loosely on a 0.5-in.-diameter fixed shaft.

Determine the maximum value of the weight W that may be

supported without causing the pulleys to rotate. Also calculate

the value of the tension in the cord between the pulleys. The

coefficient of static friction is µs = 0.2 .

1 The angle of friction and the radius of the circle

of friction are

s = tan-1µs

= tan-10.2

= 11.3099°

rf = r sin s

= (0.5 in./2) sin 11.3099°

= 0.0490 in. (1)


Pulley A

100 lb

Suppose

cord tension

is T = 100 lbP1

Fixed

shaft

Contact

point

Pulley A

100 lb

T > 100 lb

P1

New contact

point P2

Pulley rotates clockwise (We are to find the

maximum value of the weight W, so

impending motion of the cord is to the right).




Position of pulley A if cord

forces were equalPosition of pulley A caused by

increase in value of force T

2 3


Pulley A

P2

rf

NR

f

Frictional force

opposes clockwise

rotation of pulley.




from shaft acting on pulley A


Pulley A

P2

rf

Ry

Rx

R = Rx2 + Ry

2

100 lb

T

10 in.

10 in.

10 lb

(Weight of the pulley)

O

5 Free-body diagram of pulley A Equilibrium equations

Fx = 0: T Rx = 0

Fy = 0: Ry 100 lb 10 lb = 0

MO = 0: (100 lb)(10 in.) T(10 in.)

(Rx2 + Ry

2)(rf) = 0

Solving simultaneously gives

Rx = 100.7309 lb

Ry = 110.0000 lb

T = 100.7309 lb Ans.

7

+

++

0.0490 in., by Eq. 1

6 For convenience, express

R in terms of horizontal

and vertical components

(rather than in terms of N

and f).


Pulley B

T = 100.7309 lb

Q1

Fixed

shaft

Contact point

Suppose W = T = 100.7309 lb

Pulley B

T = 100.7309 lb

Q1

New contact

point

W > 100.7309 lb

Q2


8 Position of pulley B if cord forces were equal 9 Rotated position of pulley B, caused by

increase in value of weight W


Pulley B

Q2

N

S

f

Frictional force

opposes clockwise

rotation of pulley.

Line of action of resultant, S, is


rf


from shaft acting on pulley B


rf

Pulley B

Sx

Sy

Q2

10 in.

10 in.

O

T = 100.7309 lb

W 10 lb

(Weight of the pulley)

S = Sx2 + Sy

2

11 Free-body diagram of pulley B Write Equilibrium equations

Fx = 0: Sx 100.7309 lb = 0

Fy = 0: Sy W 10 lb = 0

MO = 0: (100.7309 lb)(10 in.) W(10 in.)

+ Sx2 + Sy

2 (rf) = 0

Solving Simultaneously gives

Sx = 100.7 lb

Sy = 111.5 lb

W = 101.5 lb Ans.

12

+

++

0.0490 in., by Eq. 1


100 kg

P

7. The 100-kg cart has four 200-mm-diameter wheels,

25-mm-diameter axles, and center of mass G. Each wheel has

a mass of 10 kg. The coefficient of kinetic friction is 0.06.

Determine the horizontal force P required to move the cart at

constant speed. Assume that the axles do not rotate, the wheels

fit loosely on the axles, and rolling resistance between the

wheels and the plane is negligible.

The angle of friction and the radius of the circle of

friction are

k= tan-1µs = tan-10.06 = 3.4336°

rf = r sin k

= (25/2 mm) sin 3.4336°

= 0.7486 mm (1)

1

G

100 mm

400 mm 400 mm


Weight of 100-kg cart = 981 N

2G12G2

2F1 2F2

2 Free-body diagram of cart

Factor of "2"

because 2 rear

wheels.

Forces G1 and G2 normal to the

plane and F1 and F2 parallel to the

plane act on the four wheels.

3

P

G

100 mm

400 mm 400 mm


Fx = 0: P 2F1 2F2 = 0 (1)

Fy = 0: 2G1 + 2G2 981 N 2(98.1 N) 2(98.1 N) = 0 (2)

MA= 0: 2(98.1 N)(400 mm) 2(98.1 N)(400 mm)

+ (2G2)(400 mm) (2G1)(400 mm) P(100 mm) = 0 (3)

++

+

A

Rolling resistance is negligible (The

coefficient of rolling resistance is

zero), so the force from the plane

acting on the wheel is concentrated at

a single point the point where the

wheel is tangent to the plane.

Weight of 2

10-kg wheels

= 2(98.1 N)


Wheel

P1

Contact point

Axle (does not

rotate)

Wheel P1

Clockwise rotation of wheel

P2New contact point

4 Relative position of axle and wheel when

cart is not moving (No friction force from

the plane acts on the wheel).

5 Relative position of axle and wheel when

cart is moving (As the friction force from

the plane increases from zero, it causes

the wheel to rotate about the contact point

P1 initially. But this rotation causes the

contact point to shift up and to the left on

the axle. As the force increases further

and the wheel rotates more, the contact

point moves further left and up until

slipping occurs).

Friction force from plane

acting on wheel


P2

N

R

f

rf


clockwise rotation

Line of action of

resultant, R, is

tangent to circle of

friction.

Normal force N and friction

force f from axle acting on wheel.6


Ry1

Rx1

R = (Rx1)2 + (R

y1)2

200/2 mm = 100 mm

7 Free-body diagram of a rear wheel Equilibrium equations:

Fx = 0: Rx1 F1 = 0 (4)

Fy = 0: G1 98.1 N Ry1

= 0 (5)

MO = 0: (F1)(100 mm) + (Rx1)

2 + (Ry1

)2 (rf) = 0 (6)

A free-body diagram of a front wheel of the cart would appear

similar to the free-body diagram of the rear wheel. Thus the

equilibrium equations for a front wheel are identical to those

above except the subscript "1" is replaced by "2":

Fx = 0: Rx2 F2 = 0 (7)

Fy = 0: G2 98.1 N Ry2

= 0 (8)

MO = 0: (F2)(100 mm) + (Rx2)

2 + (Ry2

)2 (rf) = 0 (9)

Substituting the value rf = 0.7486 mm and then solving Eqs. 1-9

gives

Rx1 = F1 = 1.83 N Rx2 = F2 = 1.84 N G1 = 342.89 N

G2 = 343.81 N Ry1

= 244.79 N Ry2

= 245.71 N

P = 7.34 N Ans.

8

+

++

P2

rf

G1

F1

Weight of wheel = 98.1 N

O


8. The 100-kg cart has four 300-mm-diameter wheels,

30-mm-diameter axles, and center of mass G.

Determine the angle for which the cart will roll down

the incline at constant speed. Also determine the

reactions from the incline acting on each wheel. The

coefficient of kinetic friction is 0.2. Assume that

rolling resistance and the weight of the wheels are

negligible, the wheels fit loosely on the axles, and the

axles do not rotate.

1 The angle of friction and the radius of the

circle of friction are

k = tan-1µs = tan-10.08 = 4.5739°

rf = r sin k

= (30/2 mm) sin 4.5739°

= 1.1962 mm (1)

G200 mm

500 mm

500 mm


x

y

Weight of 100 kg mass = 981 N

Factor of 2 to account

for 2 front wheels

2 Free-body diagram of cart.

Rolling resistance is negligible (The

coefficient of rolling resistance is

zero), so the force from the plane

acting on the wheel is concentrated

at a single point the point where

the wheel is tangent to the plane.


Fx = 0: 2F1 + 2F2 (981 N) sin = 0 (2)

Fy = 0: 2G1 + 2G2 (981 N) cos = 0 (3)

MA = 0: [(981 N) sin ](200 mm) [(981 N) cos ](500 mm)

+ (2G2)(500 mm + 500 mm) = 0 (4)

3

+

+

G

200 mm

500 mm

500 mm

A

+

2G1

2G2

2F1

2F2


P1

Contact point

between axle

and wheel

Axle (does not

rotate)

P1

P2

New contact

pointCounterclockwise

rotation of wheel

4 Relative position of wheel and axle when cart is not

moving (We suppose that the cart is held in place by

a force that is parallel to the plane and is equal and

opposite to the component of the weight down the

plane). No friction force from the plane acts on the

wheel.

5

Wheel

Relative position of wheel and axle when

cart is moving (As the friction force from the

plane increases from zero, it causes the

wheel to rotate about the contact point P1

initially. But this rotation causes the contact

point to shift up and to the right on the axle.

As the force increases further and the wheel

rotates more, the contact point moves further

right and up until slipping occurs).

Friction force from plane

acting on wheel


P2

N

R

f

Frictional force

opposes the

counterclockwise

rotation of the

wheel.

R is the resultant of the normal

force, N, and the frictional force, f,

from the axle acting on the wheel.

The line of action of R is tangent to

the circle of friction and passes

through the contact point P2.

6 Force from axle acting on wheel.

Circle of friction

(radius rf)

P2

Rrf

G1

F1

8 Free-body diagram of front wheel

rf

We can save some work if we

use the fact that the wheel is a

two-force member. Thus the

force R from the axle must

have the same line of action as

the resultant of the forces F1

and G1 from the inclined plane.

That is, both lines of action

pass through P2 and the point

of contact, A, with the inclined

plane. It follows from

geometry that F1 and G1 must

be related by the angle :

tan F1/G1 (5)

7

9 A

Resultant of

forces F1 and

G1 from the

plane acting

on the wheel


10 From geometry,

= sin-1(rf /150)

1.1962 mm, by Eq. 1

= 0.4569°

Using this result in Eq. 5 gives

tan = F1/G1 (Eq. 5 repeated)

or

tan 0.4569° = F1/G1 (6)

Because the geometry is the same for both wheels,

tan 0.4569° = F2/G2 (7)

Solving Eqs. 2-4, 6 and 7 gives

F1 = 1.96 N, G1 = 246 N

F2 = 1.95 N, G2 = 244 N

= 0.457° Ans.

P2

Rrf

300/2 mm = 150 mm

A

7.6 journal bearings - civil engineering...7.6 journal bearings example 1, page 1 of 3 f 140 n 80 mm...

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