Mechanics of materials torque on solid shaft

1. Jul 12, 2015

MrJoseBravo

1. The problem statement, all variables and given/known data
http://prntscr.com/7rtrep
http://prntscr.com/7rts95
2. Relevant equations
theta = TL/GJ
t(max)= Tc/J

3. The attempt at a solution
The two questions i have are rather similar so i put them up at the same time, i hope thats okay.
For the first one ( the first link) i started by

angle of twist = 0 , thus TbcLbc + TcdLcd + TdaLda = 0 (1)

so then i have to find some way to relate those torques to the reactions at ends A and B. I assumed Tb to be positive (in tension) and worked my way up the shaft drawing free body diagrams.

my results : (2)
Tb = -Tbc
Tcd = -Tb + 800

then i plugged equations 2 into 1 and solved for Tb.
my result was 645N-m which doesnt make sense to me since the net torque experienced by the member should be 300N-m.
I may have some conceptual information messed up in my head, any insight would be wonderful!

for the second link, id only like to know if im thinking about it correctly.
I can find the reactions at the fixed supports, and create equations that relate them to the torques along different segments of the bar. Would i then do this

tmax=Tac(c)/J

?

thank you in advance for any insight!

2. Jul 12, 2015

CWatters

If it's not accelerating wouldn't the net torque (sum of all four torques) be zero?

3. Jul 12, 2015

SteamKing

Staff Emeritus
Even though these two problems are very similar, PF prefers that you post one problem per HW thread. It's easier for people replying to your question to keep answers straight.

I think that the problem you are having in understanding these two questions relates to the figures used to illustrate the two shafts. In problem 7rtrep, it's not really clear, but both ends of the shaft at A and B are fixed, so there are two unknown torque reactions to be determined at those points, which will keep the shaft in static equilibrium. I didn't realize that both ends of the shaft were fixed in this problem until I looked at the illustration in the other problem.

4. Jul 12, 2015

MrJoseBravo

This does make sense. What i meant by the net torque being 300N-m is the torque applied.

So if the torque experienced at fixed end B is 645N-m, and the applied torque is 300, then it follows that the torque at A must be something such that the net torque equals zero.
Thus 345 N-m.
is this logic sound ?

5. Jul 12, 2015

Engineer at UIC

This is a statically indeterminate problem. It's fixed at both ends. You need a compatibility equation that sets the torsional deformation of all three sections to zero.

6. Jul 12, 2015

MrJoseBravo

sorry about that i will keep it in mind for next time!
I do think that these illustrations are hard to understand but i did gather that there are two unknown torques since its fixed on both ends. I guess (if my reply above is sound) that i wasnt able to bridge the last part of the problem, that the net torque equals zero.
I likely wasnt clear but i did equilibrium and used the compatibility equations giving me the results above. ( i used the angle of twist formulas for all three segments)
then the GJ terms in the denominators cancelled out and i was left with equation (1) above.

7. Jul 12, 2015

haruspex

Yes, but how did you get 645?
I don't understand your equations in the OP, since you don't define any of the variables.
Without ever having studied this area, it seems to me that the share each end gets for an applied torque is inversely proportional to its distance from that torque. So if a rod length L has a torque T applied at distance X from and A then end A's share is T(1/X)/(1/X+1/(L-X))=T(1-X/L). Applying that to each applied torque here and summng gives 345Nm at A.

8. Jul 12, 2015

CWatters

What happens if you calculate Ta the same way you did Tb . Then see if they sum to zero?