# Mechanical Engineering Design Notes

 Design Contents

A 1.75 mm pitch x 12 mm diameter grade 8.8 bolt is used to fix a part. Assume that the steel part has a cross section area of 200 mm2 and is subjected to a cyclic load varying between 0 and 20000 N. It is assumed that the bolt threads give rise to a stress concentration factor of 3.

Determine: a) The FoS of bolt when there is no pre-load.
b) The minimum required pre-load, Fo to prevent loss of compression.
c) The FoS for the bolt when the pre-load, Fo = 22000 N.
d) The minimum force in the part when the pre-load is 22000 N.

Solution:
It should be noted that from BS 6104, Pt 1, 1981, Table 6, the ultimate tensile load of this size and type bolt is given as 67,400 N. The bolt core area is 84.3 mm2
The UTS is 800 MPa, we assume that the yield strength is 500 MPa and that the fatigue endurance strength is half the UTS, ie 400 MPa.

To determine the FoS the mean stress and the stress amplitude, multiplied by the stress concentration factor, will be plotted on a modified Goodman diagram. With no pre-load the entire load is carried by the bolt. For a load varying between zero and 20000 N, the mean is 10000 N and the amplitude is 10000 N.

The mean stress, Sm = 20000 /(2 x 84.3) = 118.6 MPa, which in this case also equals the amplitude of the stress.

On the modified Goodman diagram the tensile strength is marked on the horizontal axis and the fatigue endurance strength of the actual component (after the fatigue strength reduction factors have been applied) is marked on the vertical axis. A line is drawn connecting these two points

The mean stress in the bolt is plotted on the horizontal axis and the stress amplitude on the vertical axis.
The point A falls above and to the right of the line, outside the safe zone, so the design is unsafe, the FoS, based on the tensile strength, is given by the ratio of the lengths of the lines OB / OA from the modified Goodman diagram:

Or from the applet or algebraically:

1 / FoS = (mean stress / Suts) + (stress amplitude / Se)

1/FoS = (118.6 / 800) + (118.6 / 133)

FoS = 0.961

b) The minimum Fo required to prevent loss of compression.
For bodies of equal length and equal modulii, the spring constants are proportional to the csa. The csa of the bolt is 84.3 mm2.
When the part has zero compression:

Fo = kp Pmax / (kp + kb) =200 x 20000 / 284.3 = 14070 N

c) Find the FoS in the bolt when Fo = 22000 N

Fb average = Paverage kb /(kp + kb) + Fo

Fb mean = (10000 x 84.3 / 284.3) + 22000 = 24970 N

stressmean = 24970 / 84.3 = 296.1 MPa

Fb amplitude = kb Pa/(kp + kb) =84.3 x 10000 / 284.3 = 2965 N

stressamplitude = 2965 / 84.3 = 35.17 MPa

Plotting this on the diagram, shows point C is inside the 'safe zone', the bolt now has a FoS greater than 1, shown graphically by the ratio of the lengths OD / OC, From the applet or algebraically: 1 / FoS = (296.1 / 800) + (35.17 / 133) = 1.58

The effect of the pre-load is to greatly reduce the magnitude of the alternating force and stress in the bolt (which is normally much more critical) while increasing the mean bolt force and stress (normally less critical) resulting in the bottom diagram.

d) The minimum pre-load in the part when the bolt pre-load, Fo = 22000 N.

Fp min = kp Pmax / (kp + kb) - Fo

Fp min = (200 x 20000 / 284.3) - 22000 = -7930 N

David J Grieve, Revised: 15th February 2010, Original: 10th February 2004.