The latch of a door is constructed from stamped steel 350 BH

The latch of a door is constructed from stamped steel (350 BHN) which rubs against a softer bronze locking bar(140 BHN). If the latch experiences a normal force in contact with the locking bar of 30 N, and the sliding distance is 1, 5 cm per operation (only in contact during closing), How many cycles will it take for the locking bar to lose 2 mm^3 of material? Assume unlubed operation. If the latch is 2 mm thick and is rubbing contact with the bar for half of the bar perimeter, what depth of groove will the loss of 2 mm^3 produce?

Solution

To evaluate, one uses an equation due to Archard, saying

V = K * P.L/H where V = volume removed, P applied laod, L sliding distance, H = hardness

For Bronze the wear rate constant is approx 10^(-9) ( see Ashby Materials Engineering)

The equation gives the amount of material removed per cycle,

L = 1.5 c, = 1.5/100 m, P = 30 N, H =140

Putting in the figures, volume of material (in M^3) per cycle = K* 30*(1.5/100)/140

= 3.2 * 10 ^(-12) m^3 per cycle

A) To lose 2 mm^3 material ( = 2* 10 ^(-9) m^3)) would need ( 2/3.2) *1000 cycles or approx 625 cycles.

b) if the latch is in contact for half the distance of 1.5 cm = 7.5 mm, and the width is 2mm, the depth for a volume of 2mm^3 is 0.1333 mm