Question On Calculating The Maximum Torque That A Screwdriver Can Transmit

[Ben Jamin]

Hi i need some help on solving a question from a past paper. I am revising for an exam in applied engineering mechanics.

Question:
A screwdriver is designed so that the bit is 5mm wide and 1mm deep and is 25mm long, this is welded to a cylindrical bar 12mm in diameter and 100mm long. If the maximum twist allowable is 1 degree and the allowable shear stress within the bar is 40MPa and G=80Gpa. Determine the maximum torque that the screwdriver can transmit.

You are also given a table of coefficients for rectangular bars in torsion

 

[tiny-tim]

Hi Ben!

A screwdriver is designed so that the bit is 5mm wide and 1mm deep and is 25mm long, this is welded to a cylindrical bar 12mm in diameter and 100mm long. If the maximum twist allowable is 1 degree and the allowable shear stress within the bar is 40MPa and G=80Gpa. Determine the maximum torque that the screwdriver can transmit.

You are also given a table of coefficients for rectangular bars in torsion

Show us what you've tried, and where you're stuck, and then we'll know how to help!

 

[Ben Jamin]

Basically what I am finding difficult is working out how to include both the bar and the bit. Do i do the entire calculation taking it as one singular piece or do i calculate the torque required to revolve the bar by 1 degree and then do the same separately for the bit. Would i then just take the maximum allowable torque as the piece which could withstand the lower torque value.

I couldn't see how i could do it as one piece as the both the bar and the bit deform in different ways?

I calculated J(bar) and then worked out the Torque value from (angle of twist) =(TL)/(JG)

But this only includes the bar, i thought by incorporating (angle of twist) =(TL)/(C*ab^3*G)

For the rectangular bit would give me two torque values and i could choose the smaller one but the question does not provide G for the bit.

This is where I am lost?

 

[tiny-tim]

the torque on each section is the same (why?) …

so the total angle of twist, for any particular torque, τ, is the sum of the two individuals angles of twist for that torque, τ

 

[Ben Jamin]

Well the torque is transmitted through the bar i suppose. I guess you take the end of the bit to be fixed. That the screwdriver is being used in a traditional sense. I don't see how you calculate the torque in the bit

 

[tiny-tim]

I don't see how you calculate the torque in the bit

the torque is fixed

you need to calculate the angle of twist in the bit

 

[Ben Jamin]

Were not given the torque in the question.

So would i need to calculate the torque in the cylindrical bar using the angle of twist, J, the length L and G?

(angle of twist) =(TL)/(JG)

or, and,

(G (angle of twist))/L = T/J

They both give me the same result. (I don't understand why iv been give allowable shear stress in the bar if I am not using.)

You said i need to calculate the angle of twist in the bit. Iv been given a maximum allowable angle of twist. I need to find out the maximum torque which can be transmitted.

 

[nvn]

Ben Jamin: G is the same for both the cylindrical bar and the rectangular bit, and is given in post 1. Yes, you take the end of the bit as fixed.

(1) By the way, always leave a space between a numeric value and its following unit symbol. E.g., 5 mm, not 5mm.

(2) Always use correct spelling and capitalization of units. E.g., 80 GPa, not 80 Gpa.​

Do not worry about allowable shear stress yet. We can discuss it later.

First, as mentioned by tiny-tim in post 4, add the twist angle of the bar and bit together. I.e., add together the right-hand sides of your two equations in post 3, then set this summation equal to 1 deg, converted to radians. Then solve for torque T. Give it a try.

 

[Ben Jamin]

Thank you for your help i understand now. Basically the only piece of information that confused me was that G was equal for both the bar and the bit. I can calculate the final torques now and just in time for my exam.

Thanks
Ben