Finding maximum shearing stress in connected shafts

  • Thread starter deus_ex_86
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  • #1

Homework Statement


Shaft AB is fixed at point A, and has a flange attached at B. Shaft CD is fixed at D, and has a flange attached at C. The flanges at B and C are bolted together, but with bolts that are small enough to allow one shaft to rotate by 1.5[itex]^{\circ}[/itex] before the two begin to rotate together. Shaft AB has a diameter of 1.25 in. and a length of 2 ft., while shaft CD has a diameter of 1.5 in., and a length of 3 ft. The modulus of rigidity for both shafts is 11.2 [itex]\times[/itex] 10[itex]^{6}[/itex] psi. Find the maximum shearing stress in shafts AB and CD if a torque of 420 kip-ft (!) is applied to the flange at B.

To summarize:
L[itex]_{AB}[/itex] = 2 ft. = 24 in.
L[itex]_{CD}[/itex] = 3 ft. = 36 in.
[itex]\phi_{AB}[/itex] = 1.25 in.
[itex]\phi_{CD}[/itex] = 1.5 in.
[itex]\theta_{C}[/itex] = [itex]\theta_{B}[/itex] - 1.5[itex]^{\circ}[/itex]
G = 11 [itex]\times[/itex] 10[itex]^{6}[/itex] psi.
T = 420 kip-ft = 5.04 [itex]\times[/itex] 10[itex]^{6}[/itex] lb-in.

Homework Equations


[itex]\tau_{max,AB}[/itex] = [itex]\frac{2T}{c^{3}\pi}[/itex]


The Attempt at a Solution


For shaft AB, I just plugged in the numbers, since the torque was being applied directly to that member:

[itex]\tau_{max,AB}[/itex] = [itex]\frac{2 * 5040000 psi}{0.625^{4}*\pi}[/itex], which I won't even calculate, because I can tell you right now it isn't right.

That 420 kip-ft seems to me to be comically overlarge for an apparatus of these dimensions. It causes such a massive angle of twist between the two flanges that it would snap like a pretzel rod. That said, I had a plan of attack: Find the rotation of the flange at C by the relationship between the two angles, from there find the torsion in shaft CD, and use that to find the maximum shearing stress in the shaft. It's pretty much the same as gear problems where the angles are multiples of one another, except here the angles are related through a sum. With numbers this absurdly large, though, I have to wonder how I would even check the answer against the solutions manual! Am I at least on the right track in terms of my approach?
 

Answers and Replies

  • #2
I hate to "bump" a thread, but I'm out of options. Literally no one I've talked to can work this out! I desparately need help. This problem is out of Beer's Mechanics of Materials, Sixth Edition, Chapter 3, Problem 55, if that helps.
 
  • #3
nvn
Science Advisor
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deus_ex_86: Your equation posted in section 3 of post 1 is an incorrect concept. Why do you think 100 % of the full, applied torque goes to shaft AB? All applied torque goes to shaft AB only until flange B rotates 1.5 deg. Think about it.

The given problem statement appears to have a typographic mistake. Assume a torque of 4.20 kip*ft is applied to flange B, not 420 kip*ft. Assume the allowable shear stress of the shaft material is Ssa = 416 MPa (although you do not need this value). Use all data as given in the problem statement, except change Tb = 420 kip*ft to Tb = 4.20 kip*ft. Now solve the problem. Show your work. Here are a few hints, below, to get you started.

Hint 1: Tb = Tb1 + Tb2. Hint 2: Ta1 = Tb1 = pi*(d_ab^4)*G*theta_b1/(32*L_ab), where theta_b1 = 1.5 deg = 0.026 180 rad. Therefore, Ta1 = Tb1 = pi*[(1.25 in)^4](11 200 ksi)(0.026 180 rad)/[32(24 in)] = 2.9283 kip*in. Hint 3: Tb2 = Tb - Tb1 = (50.40 kip*in) - (2.9283 kip*in) = 47.4717 kip*in. Hint 4: The final answers are tau_ab = 59.595 ksi (410.89 MPa), tau_cd = 41.567 ksi. Do these answers match the solutions manual?
 

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