Troubleshooting checklist (continued) – Warner Electric CB Series and Super CB Series Sizes 2, 4, 5, 6 and 8 User Manual

Page 9

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9

Warner Electric • 800-825-9050

P-258 • 819-0406

Troubleshooting Checklist (continued)

Possible Problem

Recommended Action

5. Lubrication has been added to the unit. Added

lubrication can result in a mis-match of lubricants which
can cause a soapy appearing viscous material around
the collar and springs. This can cause the springs to
slide on the hubs rather than grab them. The result is a
loss of accuracy.

A. Disassemble the unit. Using a clean rag, wipe excess material

from the hubs, springs and collar.

Do not use solvent

. (If the

added lubricant was due to spray from a chain lubricator, take
steps to guard against reoccurrence.) When re-assembling the
unit ensure that the spring differential is set properly. See
Section 5 “CB Spring Differential Adjustment.”

6. Output does not repeat stopping point

Clutch

Tc

t

lc

CB-2

1.65

0.003

0.0116

CB-4

4.40

0.004

0.0450

CB-5

6.88

0.004

0.1663

CB-6

8.75

0.005

1.221 (0.75 in. bore)

1.138 (1.0 in. bore)

CB-8

20

0.005

9.43 (.075 in. bore)

9.32 (1.0 in. bore)

8.15 (1.5 in. bore)

I = (t) (Tc + To) (3700) - Ic

RPM

How to determine maximum inertia load.

T x 3700 x t = WR

2

RPM

A. There is not enough RPM or inertia to fully disengage the

clutch spring and fully engage the brake spring. Add inertia

by either increasing RPM or adding weight to the output

(such as a sprocket or flywheel). To calculate the inertia

required, use the following table and formula:

Minimum Inertia Calculation

l = Minimum inertia required to fully activate the

Clutch/Brake - lb.in.

2

t = Time – Seconds

T

c

= Torque required to fully activate the

Clutch/Brake – in.lb.

T

o

= Frictional Torque – in.lb.

RPM = Revolutions per minute

lc = Inertia at the output side of the clutch – lb. in.

2

Example:

CB-5 in a system running at 300 RPM with 10 in. lb.

drag. What inertia is required to fully activate the

Clutch/Brake?

I = (0.004) (6.88 + 10) (3700) - 0.1663 = 0.6664 lb. in.

2

300

B. There is too much system friction (drag) to allow the output

inertia to fully disengage the clutch spring and engage the

brake spring. Reduce system friction. Also, see Section A

above on minimum inertia.

C. The control tang is broken on the brake spring. Replace the

spring, using the “Light Repair Kit.”

D. There is binding caused by rigid mounting. (See Section 2).

Unit must not be rigidly mounted. The plate must be

allowed to “float” axially.

E. There is insufficient anti-back holding capability, or the unit

has been reversed. See technical ratings for anti-back

holding capabilities and re-evaluate sizing.

Important: Do

not reverse unit!

F. An overhauling load is causing misregistration. Verify the

unit has anti-overrun feature.

G. The anti-overrun holding capacity has been exceeded. See

technical ratings and re-evaluate unit sizing.

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