Highway Defect

EXAMINE WHY A MANHOLE COVER JUMPED OUT OF ITS CASTING

THE TIDDLY WINK EFFECT

Review of the Event

A vehicle was driving along a main road when it struck an open sewer manhole, lost control and crashed into a tree. The road had been re-surfaced a few months prior. The sewer manhole cover had been raised to match the new roadway surface with a casting riser. After the re-surfacing the manhole cover began banging whenever a vehicle drove over it. A few weeks prior to the crash a neighbor had seen the manhole cover pop out of the casting and roll away. The neighbor found the cover and replaced it . As cars drove over the manhole cover it banged until one night the cover either broke and fell into the manhole, rolled away or someone took it.

Picture

View video of tire rolling over properly seated manhole cover at 30 mph.
Tire rolling over properly seated manhole  (opens in YouTube window)
 
View video of tire rolling over improperly seated manhole cover at 30 mph
Tire rolling over unseated manhole cover  (opens in YouTube window)
 
Graph of Kinetic Energy vs. Potential Energy
When the Kinetic Energy exceeds the Potential Energy of the manhole cover, the cover will jump/slide out of its casting.

 

Kinetic Energy vs Potential Energy of Improperly Seated Manhole Cover (opens in YouTube window)

 After the MH cover has been given the initial rotation because of the tire rolling over it, it is free to continue to rotate.What is the limit of rotation and what is the height that one end of the MH cover can attain as it raises up from the casting?
Procedure: Convert linear velocity to angular velocity:  Vf2 = Vo2 + 2 × g × h   Where  h = Height that the MH cover must reach to clear the casting.  ωf2 × r2 = ωo2 × r2 + 2 × α × r × hWhere  ωf= Final angular velocity of the MH Cover = 0 (Vertical Direction) ωo = Initial angular velocity of rotation of the MH cover in rad/sec.
r= Radius of the MH cover in inches, and;ωf = ωo + α × t;   α = ωf × tIf we estimate that the tire rolls over a section of the manhole equal to the radius then the equation reduces to:ω=2 × h ÷ r ÷ tWhere tis the time for the tire to roll over a portion (chord), of the MH cover equivalent to a function of the radius.

Resulting graph:

Picture

 

After the MH cover has been given the initial rotation because of the tire rolling over it, it is free to continue to rotate. What is the limit of rotation and what is the height that one end of the MH cover can attain as it raises up from the casting? Procedure: Convert linear velocity to angular velocity:

Vf2 = Vo2 + 2 × g × h

Where h = Height that the MH cover must reach to clear the casting.

ωf2 × r2 = ωo2 × r2 + 2 × α × r × h

Where;
ωf = Final angular velocity of the MH Cover = 0 (Vertical Direction)
ωo= Initial angular velocity of rotation of the MH cover in rad/sec.
r = Radius of the MH cover in inches and;
ω= ω+ α × t;   α = ω× t

If we estimate that the tire rolls over a section of the manhole equal to the radius then, the equation reduces to:

ω=2 × h ÷ r ÷ t

Where t is the time for the tire to roll over the MH cover.

Last update: 08/24/2012