Chapter Objectives. Copyright 2011 Pearson Education South Asia Pte Ltd

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1 Chapter Objectives To determine the torsional deformation of a perfectly elastic circular shaft. To determine the support reactions when these reactions cannot be determined solely from the moment equilibrium equation. To determine the maximum power that can be transmitted by a shaft.

2 In-class Activities 1. Reading Quiz 2. Applications 3. Torsion Formula 4. Angle of Twist 5. Statically indeterminate torque-loaded members 6. Solid non-circular shafts 7. Stress concentration 8. Concept Quiz

3 READING QUIZ 1) Given the angle of rotation is small and the materials remain linear elastic, which statement below is incorrect for the torsional behavior of a long straight circular shaft. a) Section-shape remains unchanged b) Straight member remains straight c) Plane section remains plane d) End of member may wrap

4 READING QUIZ (cont) 2) The unit for a shaft s polar moment of inertia J is: a) kpa b) m 4 c) m 2 d) m 3

5 READING QUIZ (cont) 3) Which one of the statements below is incorrect? Inelastic torsion of a circular shaft leads to a) non-linear distribution of shear stress b) plane section remaining plane c) change of sectional shape d) wrapping at the end of member

6 APPLICATIONS

7 APPLICATIONS (cont)

8 TORSION FORMULA Assumptions: Linear and elastic deformation Plane section remains plane and undistorted

9 TORSION FORMULA (cont) Linear distribution of stress: c max Torsion shear relationship: T T A c max max Tc J A da A 2 da Similarily, T J c max da

10 TORSION FORMULA (cont) Polar moment of inertia For solid shaft: For tubular shaft: c J d d da J C C C A c o c i J

11 EXAMPLE 1 The pipe shown in Fig. 5 12a has an inner diameter of 80 mm and an outer diameter of 100 mm. If its end is tightened against the support at A using a torque wrench at B, determine the shear stress developed in the material at the inner and outer walls along the central portion of the pipe when the 80-N forces are applied to the wrench.

12 EXAMPLE 1 (cont) Solutions The only unknown at the section is the internal torque T 80 M y 0; T T 40 N m The polar moment of inertia for the pipe s cross-sectional area is 0 J m For any point lying on the outside surface of the pipe, c m Tc J MPa (Ans)

13 EXAMPLE 1 (cont) Solutions And for any point located on the inside surface, i Tc i J The resultant internal torque is equal but opposite c i MPa 0.04 m (Ans)

14 ANGLE OF TWIST dx d d x x For constant torque and cross-sectional area: T J G TL JG dx

15 ANGLE OF TWIST (cont) Sign convention for both torque and angle of twist positive if (right hand) thumb directs outward from the shaft

16 EXAMPLE 2 The two solid steel shafts are coupled together using the meshed gears. Determine the angle of twist of end A of shaft AB when the torque 45 Nm is applied. Take G to be 80 GPa. Shaft AB is free to rotate within bearings E and F, whereas shaft DC is fixed at D. Each shaft has a diameter of 20 mm.

17 EXAMPLE 2 (cont) Solutions From free body diagram, F 45/ N Angle of twist at C is Nm T D Since the gears at the end of the shaft are in mesh, B C TL DC JG x rad rad

18 EXAMPLE 2 (cont) Solutions Since the angle of twist of end A with respect to end B of shaft AB caused by the torque 45 Nm, T L JG 45 2 AB AB A/ B rad The rotation of end A is therefore A B A/ B rad (Ans)

19 STATICALLY INDETERMINATE TORQUE- LOADED MEMBERS Procedure for analysis: use both equilibrium and compatibility equations Equilibrium Draw a free-body diagram of the shaft in order to identify all the torques that act on it. Then write the equations of moment equilibrium about the axis of the shaft. Please refer to the website for the animation: Statically Indeterminate Torque-Loaded Members Compatibility To write the compatibility equation, investigate the way the shaft will twist when subjected to the external loads, and give consideration as to how the supports constrain the shaft when it is twisted.

20 STATICALLY INDETERMINATE TORQUE- LOADED MEMBERS (cont) Express the compatibility condition in terms of the rotational displacements caused by the reactive torques, and then use a torque-displacement relation, such as Φ = TL/JG, to relate the unknown torques to the unknown displacements. Solve the equilibrium and compatibility equations for the unknown reactive torques. If any of the magnitudes have a negative numerical value, it indicates that this torque acts in the opposite sense of direction to that indicated on the free-body diagram.

21 EXAMPLE 3 The solid steel shaft shown in Fig. 5 23a has a diameter of 20 mm. If it is subjected to the two torques, determine the reactions at the fixed supports A and B.

22 EXAMPLE 3 (cont) Solution It is seen that the problem is statically indeterminate since there is only one available equation of equilibrium and there are 2 unknowns M T b x T A 0 (1) Since the ends of the shaft are fixed, the angle of twist of one end of the shaft with respect to the other must be zero. A / B 0

23 EXAMPLE 3 (cont) Solution Using the sign convention established, T T 0.3 TB JG T 645 Nm B B JG (Ans) B JG 0 Using Eq. 1, T A 345 N m The negative sign indicates that acts in the opposite direction of that shown in Fig. 5 23b.