SoE - Design - Stresses in valve stem due to valve bounce

1. Introduction
This takes a simple approach based on data from a Ford Fiesta engine. The worst case scenario is examined - where valve bounce is occurring and the valve closing is driven totally by the valve spring and the cam has no moderating contact.

2. Assumptions

• Valve spring stiffness is about 6N/mm
• Valve lift is 6.5mm
• Valve stem diameter is 5mm
• Valve stem length is 70mm
• Mass of valve is 33g
• Mass of spring is 25.5g
• Mass of both colletts 1.1g
• Mass of cap 8.2g
• Mass of tappet and spacer 35.5g
• Total reciprocating mass of valve etc. is 80g

3. Calculations
As an approximation it is assumed that the average valve closing force exerted by the spring is 30N.
NB: It should be noted that typical maximum valve accelerations are considerably higher than the value calculated below. Hence the valve springs have significant compression when fitted and generate a significantly higher force than used in the example below. This means that the stress calculated is low and may well be exceeded in a real engine should valve bounce occur. The high stresses explain why failure of the valve train can happen rapidly when valve bounce occurs.

Using: Force = mass x acceleration, this gives the acceleration of the closing valve to be 30/0.08 = 375m/s2
Using: v2 = u2 + 2as,

v2 = 0 + 2 x 375 x 0.0065 = 4.875 gives: v=2.2m/s Convert this to a drop height problem, determine the height of drop needed to give rise to a velocity of 2.2m/s
Again use: v2 = u2 + 2as, 4.875 = 0 + 2 x 9.81 x s,
So equivalent drop height is: s = 0.2485m
The strain energy in a bar subject to tension or compression is given by: U = (sigma)2AL/2E = w2L/2AE
where W is the load, A the cross section area of the rod stem and L its length.
It is assumed that the valve assembly of weight P falls from a height h (0.2485m in this case) and is stopped when the seat on the head of the valve hits the valve seat insert in the cylinder head and this results in a slight elastic stretching of the stem, by a small amount delta. The loss of potential energy through P falling freely through the height h, is equated to the work done by the equivalent static force PE that would produce the same static deflection. P(h + delta) = PEdelta/2 = U, delta is very small compared to h and can be ignored on the left hand side of the above equation. Also PE =A E delta/L, so
Ph = A E (delta)2/2L and substituting values gives:
0.08 x 9.81 x 0.2485 = 3.14159 X (0.005)2 x 200E9 (delta)2/( 4 x 2 x 0.07)
(delta)2 = 6.953E-9m, so delta = 83.3E-6m
substitute this value of delta in: PE =A E delta/L, gives
PE = 3.14159 x (0.005)2 x 200E9/(4 x 0.07) = 4678N
Substitute this value of PE into sigma = PE/A gives sigma = 4678 x 4/(3.14159 x (0.005)2
sigma = 238MPa.

This is a very rough calculation but it gives some indication of the high stresses that can arise when valve bounce occurs. Although this value will be well below the yield stress of a medium carbon steel, which is sometimes used for valve stems, the semi-circular grooves in the stem to retain the colletts will give rise to stress raisers. FEA on another page indicates this stress concentration factor can be up to about 2.5. This will cause local stresses that may well be too high for a medium carbon steel.

David J Grieve, 14th February 2007.