×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

Wheel skidding due to braking

Wheel skidding due to braking

Wheel skidding due to braking

(OP)
I would like to work out if my wheels are going to skid, but am a bit stuck

We have wheel diameter = 460
Speed = 50kph, 13.9m\s
Weight on wheel = 3600kg
Effective braking force = 300Nm (worked out from brake radius & brake clamp force Cf of pads)
CF wheel to rail = 0.5

Can any one help with this, do I need to provide further info

Regards Brendon

RE: Wheel skidding due to braking

(OP)
Also we are assuming the full effect of the brakes are applied in an instant

RE: Wheel skidding due to braking

300 N.m braking torque applied at radius 0.23 m (half the wheel diameter) gives 1304 N braking force at the wheel rim.

3600 kg on earth at normal surface conditions gives 3600 x 9.807 = 35,305 N

1304 / 35305 = 0.037

That is the minimum coefficient of friction necessary to avoid skidding. More than that (which is virtually certain, even on ice) and there is no skid.

The other information that was provided, is irrelevant.

Mind you, with such little braking force being applied, the vehicle is going to take a loooooong time to stop. If your numbers are wrong, I can't help it.

Not sure why that was so hard.

RE: Wheel skidding due to braking

The calc looks straightforward, but the key info is the last one - wheel to rail. 0.5 cf looks ok for dry, but wet or with leaves etc can fall to s low as 0.1 to.0.15 so makes your 0.03 may be not so close after all. Still very low though and the distance to stop will be quite long.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

RE: Wheel skidding due to braking

Any hills involved?

RE: Wheel skidding due to braking

I'm going to be a pedant...
kg is not weight. It is mass.
Nm is not force. It is torque or moment.

If I assume this is a 4 wheeled contraption, then we can extrapolate it to be 14,400kg.
Stopping distance becomes 268m. A long way.
It can run away on an incline greater than 2.1 deg.

The basic nature of the question and the confusion over elementary units leads me to believe either:
1) This is a student posting.
2) The original poster may be in over his head. I don't know that forum guidance is a good way to design 14 Ton carts.

RE: Wheel skidding due to braking

(OP)
Thanks Brian, you were spot on, It is simpler than I thought, FYI the brake torque info supplied was incorrect, but that's all sorted now

Re. Imcjoek's comments. hay, sorry for asking, and for mixing my P's & Q's.
I'm not designing by forum, there are hundreds of calcs in our design, just got a little stuck on that one and our guru was away that day, so I thought I could ask here..
But you are right, I should be more careful with my units, lesson learned, thanks

Cheers Brendon

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources