I have a scenario I would like to get double checked. I have a shaft spinning at 1800 rpm driven by a 500 hp electric motor. I am bolting another drive shaft to it with (qty 8) 3\8 dia shoulder bolts on a 4-1/4" bolt circle. The bolts have a shear strength of 10.5 ksi. The stresses on the bolts are purely shear as all the other forces and mitigated by other means. I am assuming a uniform load on all 8 bolts and I am ignoring starting and just focusing on the running state. I came up with a safety factor of 9, but I seldom venture into this realm and you like a sage guide to confirm the path.
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shear strength of drive shaft bolts
- Thread starter Guanadon
- Start date
Hi
I usually work in metric however if your 10.5 ksi is 10500 lbs/in^2 then according to my calculations the shear stress in each bolt is 9324 lb/in^2 meaning the safety margin is 1.126 per bolt.
I think you need more bolts because starting the motor or in an overload condition which you've said your ignoring, the stress good be 3 to 4 times the stress due to running.
In addition, the chances of getting 8 bolts to equally share the load would be less than winning a lottery, I would assume 4 bolts.
The material the bolts pass through should also be checked for bearing stress based on 4 bolts.
I usually work in metric however if your 10.5 ksi is 10500 lbs/in^2 then according to my calculations the shear stress in each bolt is 9324 lb/in^2 meaning the safety margin is 1.126 per bolt.
I think you need more bolts because starting the motor or in an overload condition which you've said your ignoring, the stress good be 3 to 4 times the stress due to running.
In addition, the chances of getting 8 bolts to equally share the load would be less than winning a lottery, I would assume 4 bolts.
The material the bolts pass through should also be checked for bearing stress based on 4 bolts.
- Thread starter
- #3
I agree with the 9324 lb/in^2 but my understanding was that I would then multiply by the distance from the center of the shaft which is .1771 Ft to reach 1651.1 ft/lbs of torque required. My initial post has an excel attachment where I show my work.
I really appreciate the help.
I really appreciate the help.
Hi Guanadon
The torque is based on the motor power, then once you calculate the motor torque you divide it by the radius of your bolt circle and then the next step is to divide it by the number of bolts which then gives you the force per bolt.
The bolt area is then divided into the load per bolt which gives the bolt shear stress, the shear stress figure obtained is compared with the allowable stress which in this case is 10.5 ksi.
I think your mixing stress and torque figures up, if you multiply lbs/in^2 by inches, then the units become lbs/in but torque is measured in lbs-in and not lbs/in
The torque is based on the motor power, then once you calculate the motor torque you divide it by the radius of your bolt circle and then the next step is to divide it by the number of bolts which then gives you the force per bolt.
The bolt area is then divided into the load per bolt which gives the bolt shear stress, the shear stress figure obtained is compared with the allowable stress which in this case is 10.5 ksi.
I think your mixing stress and torque figures up, if you multiply lbs/in^2 by inches, then the units become lbs/in but torque is measured in lbs-in and not lbs/in
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