## Number of bolts required for a flanged coupling?

Hi everyone.

This question is giving me a real headache (literally). It should be simple, but I’ve spend probably the last 4-5 hours trying to figure it out. The left side of my brain is redundant.

1. The problem statement, all variables and given/known data

The following specifications are used for the design of a flanged coupling between two coaxial shafts:

Speed: 650rpm
Power transmitted: 550 kW
Bolt diameter: 12mm
Pitch-circle diameter: 200mm
Material: Mild steel
Factor of safety: 4

Determine the number of bolts required, assuming the bolts are equally loaded.

2. Relevant equations

Ultimate Shear Strength of mild steel = 360 N/mm^2

Shear stress (MPa) = Force (Newtons) / Area (mm^2)

Factor of safety = Ultimate Shear Strength / Shear Stress

3. The attempt at a solution

I cannot get anything down at all. And I also cannot understand why they have given me Speed and Power. Is there maybe some way to work out the force for the shear stress equation?

Also for your information, this is question 27.11 from Val Ivanoff’s Engineering Mechanics.

Any help would be much appreciated!

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hi jacoja06! welcome to pf!
 Quote by jacoja06 … I also cannot understand why they have given me Speed and Power. Is there maybe some way to work out the force for the shear stress equation?
go back to basic definitions …
power = energy/time = work/time (same thing)

= force x distance/time

= force x distance x angle/time

 Thanks for the reply and welcome! :) So using that formula; P = F x distance/time 550kw = F x (650rpm x ((2 x pi)/60)) F = 484.81N?

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## Number of bolts required for a flanged coupling?

 Quote by jacoja06 550kw = F x (650rpm x ((2 x pi)/60))

 Oh, sorry... I didn't put the time in seconds in my last post. I was supposed to put; 550kw = F x (650rpm x ((2 x pi)/60)/60) So; 550kw = F x (68.0678rad/60) F = 484.81N Where would the radius come into the equation?
 Recognitions: Homework Help Hint: Torque = Force * Distance
 Recognitions: Homework Help You've also got an extra factor of 60 in your RPM conversion. Remember 1 min = 60 sec
 Recognitions: Homework Help To solve the larger problem, how many bolts, you are trying to transmit a certain torque from one shaft to another at the coupling. You are given a standard size bolt to use, along with the material characteristics. How is the torque going to be transmitted? What type of stress will be set up in the bolts?
 Thanks, steamking. I think he distance part is confusing me the most at the moment. There isn't any specified distance I can use to find the force.

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 Quote by jacoja06 Thanks, steamking. I think he distance part is confusing me the most at the moment. There isn't any specified distance I can use to find the force.
The pitch-circle is the circle concentric with the shaft where the bolt holes are located.

 total torque to be transmitted => T=9550P/N (N-m) => 9550*550/650 = 8080N-m force on the coupling= 8080/0.250 =40400 N assuming its a mild steel of yield strength 240MPa and shear strength of 120 MPa,with a factor of safety of 4 the design shear strength is 30Mpa we have area of each bolt = Pi*12*12/4 =113.04mm2 total load that a single bolt can take = 113*30 = 3390N so you will need 40400/3390 = 12 bolts. I hope this clarifies your question and i stand to be corrected if I'm wrong.
 speed and power helps you to find the torque..... you can use the equation I specified above, or you can also use this Power =(2*Pi*N*T)/60 also, we design machines with factor of safety as a function of yield strength rather than UTS. :)
 Sorry!! the force on the coupling would be 8080/0.125 (consider the radius... not diameter. silly error :) ) follow the same methodology
 Recognitions: Gold Member I have received nasty Warnings in the past from PF about revealing the entire solution to the OP without giving them an adequate chance to solve it themselves, and have had my replies removed from the thread. How come no one is receiving Notifications in this thread?

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