## Bolt Pre - Load Example

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.

a) FoS with zero pre-load.
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.

The amplitude of the stress multiplied by the stress concentration factor (3) is 355.9 MPa.

When this is plotted on the modified Goodman diagram, the point A falls above and to the right of the line, outside the safe zone, the design is unsafe, the FoS, based on the yield strength, is given by the ratio of the lengths of the lines OB / OA or algebraically from the modified Goodman diagram:

1 / FoS = (mean stress / Sy) + (stress amplitude x stress conc. / Se)

1/FoS = (118.6 / 500) + (3 x 118.6 / 400)

FoS = 0.89

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, approximately = 1.2 or algebraically: 1 / FoS = (296.1 / 500) + (3 x 35.17 / 400) = 1.17

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