When we use HSFG bolt and consider friction, the number of bolts required are 3 but at the same time when we use HSFG bolt and do not consider friction, the number of bolts required are 2. I don't understand this. Shouldn't the number of bolts required increase if friction is not there?
When slip is allowed, the plate slides and bearing action is developed. Hence HSFG bolts are designed as Ordinary Black Bolts. We require high Fo value for no slip condition so we require more bolts.
The thing is you are considering friction between two plates will resist external load. Go to assumptions of bolted connections. We neglect the effect of friction developed between plates due to self weight. Actually the thing is we are developing high frictional resistance between two plates against external applied load by means of applying high Fo value by tightening the bolt. That is the principle behind HSFG bolts
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At 5:26 (length of joint) lj> 15 d
Lj
When we use HSFG bolt and consider friction, the number of bolts required are 3 but at the same time when we use HSFG bolt and do not consider friction, the number of bolts required are 2. I don't understand this. Shouldn't the number of bolts required increase if friction is not there?
somebody should answer this.
Maybe more bolts account for more friction as slip is not allowed. But in case of lesser bolts slip is more likely to occur.
When slip is allowed, the plate slides and bearing action is developed. Hence HSFG bolts are designed as Ordinary Black Bolts.
We require high Fo value for no slip condition so we require more bolts.
The thing is you are considering friction between two plates will resist external load. Go to assumptions of bolted connections. We neglect the effect of friction developed between plates due to self weight. Actually the thing is we are developing high frictional resistance between two plates against external applied load by means of applying high Fo value by tightening the bolt. That is the principle behind HSFG bolts