Torsion of a Square Bar

[BrianC]

To solve this problem you must calculate the torsional deflection of a square steel bar. The problem is solved using superposition, which I understand. However, when solving for the torsional deflection due to the bar/load farthest away from the wall, a length of L = 12 inch is used in the solution. I believe should be L = 18 inch, which will result in a larger deflection.

Can anyone confirm or reject this potential mistake in the solution?

 

[Oughtsix]

I agree. I believe the solution is wrong and the twist for the further bar should be calculated using 18".

 

[gaidox]

i dont have sms md.

is this cantilever with free end loaded on L=12 with Ltotal=18?

 

[BrianC]

This problem has a square bar with two torque loads axially along the bar. I believe the problem can be solved by two methods.

1) Superposition where the length of the bar is measured from the wall, and the torque loads are evaluated individually.

2) Or by using a torque diagram, where the bar length is detemined by the spacing of the torques (bar broken down into sections).

 

[MechGale]

I solved it using the 2nd method, with the bar as two sections, so solve for twist at 6" and at 12", then add them together. The mistake I made was actually using the twist equation for a round bar (angle = TL/JG) instead of for a square bar. That's one I don't think I'll make again though, so glad I made that simple mistake before the exam.

 

[Oughtsix]

My strength book has an example similar to this but the method is not explained. You are all correct using superposition: T@6" x 6" and T@18" x 12". I'm not getting the concept. It makes perfect sense in loaded beams, but I'm having trouble applying it here. The torque at 18" does act on the entire bar and I intuitively want to sum the deflections over the length acted and not in sections. How are you all picturing this concept?