K Factors-Example
Hex Cap Screw 1/2 - 13 zinc plated grade 5
Using Torque = K x d x F K = .20
d = 0.50 in.
F = (75% of the proof load) = 0.75 x 85,000 psi x 0.1419 sq.in = 9045 lbs.
T = (0.20) (0.5 in.) (9045 lbs.) T= 905 in.-lbs. = 75 ft.-lbs.
In the previous example, we considered K = 0.20. This is a common K factor for zinc-plated fasteners.
We must now wonder what may have happened if the person installing the fastener would have excessively lubricated the bolt and installed it with the same torque. In such conditions, K may drop to a value of 0.10.
If T = K x d x F, then we can also determine the preload, F, in the same way.
F = T / (K x d)
F = 905 in.-lbs. / (0.10 x 0.50 in.) = 18,100 lbs.
The value obtained for our preload, 18,100 lbs., would exceed the tensile strength of a 1/2 - 13 grade 5
bolt. More than likely the bolt in question would have failed. The exact opposite may have also happened resulting in a lower clamp load than may be required resulting in a field failure.
