Mechanics of Materials: Torsion

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SUMMARY

The discussion focuses on calculating the torque distribution in a composite shaft consisting of a steel tube bonded to a brass core under applied torque T_D. The fraction of the torque resisted by the steel at point C, denoted as T_c, is expressed by the formula T_c = T_D [G_s J_s / (G_b J_b + G_s J_s)] [L_ad / L]. Participants emphasize the importance of using free body diagrams to analyze the system and the need to consider the angle of twist, which is zero at the fixed ends. The conversation clarifies the equilibrium of torques and the necessity of understanding material properties when dealing with bonded components.

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  • Understanding of torsion mechanics in materials
  • Familiarity with the concepts of shear modulus (G) and polar moment of inertia (J)
  • Knowledge of free body diagram construction
  • Basic principles of equilibrium in static systems
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  • Learn about the properties of materials, specifically shear modulus for steel and brass
  • Explore the application of free body diagrams in solving mechanics problems
  • Investigate the effects of fixed supports on torque distribution in shafts
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steak313
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Homework Statement


A shaft fixed on both ends is shown below. It is made of a steel tube, which is bonded to a brass core. A torque T_D is applied at location shown. The tube is fixed at both ends.

[PLAIN]http://img202.imageshack.us/img202/2940/problem2eb.jpg

Show that the fraction of the torque, T_D, resisted by the steel at C labeled as T_c is given by the following expression: T_c = T_D \left[\frac{G_s J_s}{G_b J_b + G_s J_s}\right]\left[\frac{L_{ad}}{L}\right]

Homework Equations



\phi = \frac{TL}{JG}

The Attempt at a Solution



So I am thinking I should start with a free body diagram. The shaft is fixed which makes me think that the angle of twist will be zero allowing me to use this equation to solve for torques. If I am not mistaken the torque as C will be T_D in the opposite direction.

At point C I have \frac{T_c L}{G_sJ_s}= \frac{T_dL_{ad}}{G_bJ_b} + \frac{T_d L_{ad}}{G_sJ_s}

Obviously this isn't correct; it seems like they may have simply combined the modulus of rigidiy and polar moments for the part of the shaft AD. Any input, am I on the correct track? Or did I possibly miss something about combining these components when a shaft is bonded to a core of different material?
 
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steak313 said:
So I am thinking I should start with a free body diagram.
Yes, which free body diagram?
The shaft is fixed which makes me think that the angle of twist will be zero
Where will it be 0?
allowing me to use this equation to solve for torques. If I am not mistaken the torque as C will be T_D in the opposite direction.
. No. If the torque at C is equal and opposite to T_D, that implies that the torque at A is 0, which seems illogical.
Or did I possibly miss something about combining these components when a shaft is bonded to a core of different material?
Hint: The angle of twist at D is the same immediately to the left and right of D. What's the torque immediately to the left of D? What's the torque immediately to the right of D? You must draw free body diagrams of sections cut through these points and about one end to find out. And don't forget the equilibrium equation for sum of torques.
 
Thank you very much for your reply! Earlier some students and I worked the problem together and it is much clearer now. Thanks!
 

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