This is a simple clamping force problem that I just want to be sure I am envisioning it correctly. If you have a plate with one bolt in it torqued to a value X, the clamping force will be Y. If you then add 3 more bolts torquing them all to the same value X, the total clamping force will still be Y since they are acting in parallel (in other words, it is distributing the clamping force along the plate rather than increasing the total clamping force). Is that correct?
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Clamping Force 1
- Thread starter Creigbm
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Where are you measuring the "clamping force"? If you are trying to separate the plate from another piece, you must extert 4Y (assuming an ideal system).
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
- Thread starter
- #4
The exact problem I am having is dealing with a clamp band. The band is is broken into two 180 degree sections with a bolt torqued to X on each side. I was trying to solve for the vertical clamping force since clamp band is tapered on the inside. My approach was as follows
sigma(hoop)* Clamp Band Cross-Sectional Area = Internal Pressure * Area the pressure is acting on
Therefore,
Clamping Force = Internal Pressure * Area the pressure is acting on
I then solved for the internal pressure, did some good ole trig to get the vertical component, and called it a day. My concern is if the Clamping Force I use on the left side of the equation is suppose to be double since there are two bolts. I have talked myself into say yes and no. If you draw the FBD it appears as if you need to take 2 * the clamping force since you have one on each side. But realistically, I don't think that is the case but cannot prove it. If this does not make sense, ignore the tapered part and solve it as a simple clamp band and determine the resulting clamping force if bolted together. Thanks for your help.
sigma(hoop)* Clamp Band Cross-Sectional Area = Internal Pressure * Area the pressure is acting on
Therefore,
Clamping Force = Internal Pressure * Area the pressure is acting on
I then solved for the internal pressure, did some good ole trig to get the vertical component, and called it a day. My concern is if the Clamping Force I use on the left side of the equation is suppose to be double since there are two bolts. I have talked myself into say yes and no. If you draw the FBD it appears as if you need to take 2 * the clamping force since you have one on each side. But realistically, I don't think that is the case but cannot prove it. If this does not make sense, ignore the tapered part and solve it as a simple clamp band and determine the resulting clamping force if bolted together. Thanks for your help.
If you are assuming the fasteners are not prestressed, then the resisting force each one exerts to withstand the internal pressure is 0.5 * Clamping Force (as defined above).
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
Regards,
Cory
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
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