[7] Torsion. [7.1] Torsion. [7.2] Statically Indeterminate Torsion. [7] Torsion Page 1 of 21

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1 [7] Torsion Page 1 of 21 [7] Torsion [7.1] Torsion [7.2] Statically Indeterminate Torsion

2 [7] Torsion Page 2 of 21 [7.1] Torsion SHEAR STRAIN DUE TO TORSION 1) A shaft with a circular cross section is often used to transmit power Q: What is the typical example of a torsionally loaded member? A: Drive shaft of a car 2) If a torque T is applied to a circular shaft, it tends to twist the shaft and causes angular deformation: shear strain 3) The shear strain developed in the shaft can be analyzed if the following assumptions are made: a) The circular cross section remains circle, and ends of shaft remain flat b) The radial lines on these ends remain straight c) The length of the shaft and its radius remain unchanged

3 [7] Torsion Page 3 of 21 4) From the geometry, the arc length BD can be defined as: BD d dx d dx 5) Assuming the linear variation of the shear strain along the radius of the shaft, the maximum shear strain max occurs at the outer surface of the shaft where c, then: d max [Shear Strain under Torsion] max dx c c

4 [7] Torsion Page 4 of 21 THE TORSION FORMULA 1) If a torque T is applied to a circular shaft, it develops a corresponding internal torque within the shaft: shear stress 2) The assumption of linear variation in shear strain along the radius of the shaft leads to a corresponding linear variation in shear stress as well 3) Using Hooke s law in shear: G, substituted into the shear strain equation yields the shear stress equation as: c max 4) The stress distribution over the entire cross section is equivalent to the resultant internal torque T at the cross section: T ( da) max da T max A c 2 da c A A

5 [7] Torsion Page 5 of ) Define polar moment of inertia, J da Tc max [Torsion Formula] J A where, max : Maximum shear stress developed in the shaft (occurs at the outer surface of the shaft) T : Resultant internal torque at the cross section J : Polar moment of inertia of the cross-sectional area c : Outer radius of the shaft POLAR MOMENT OF INERTIA 1) The polar moment of inertia of a solid circular cross section can be given by 2 integration definition as: J da A 2) The integration is evaluated as: J J A 2 da 2 c c 0 c c ( 2 d) 2 d 2 c [Polar Moment of Inertia for the Solid Cross Section] 3) Similarly, the polar moment of inertia of a tubular cross section can be given as: 0 2 J 4 ( c outer c 2 4 inner ) [Polar Moment of Inertia for the Tubular Cross Section]

6 [7] Torsion Page 6 of 21 CLASS EXAMPLE The copper shaft has an outer diameter of 1.5 in. and inner diameter of 1 in. If it is subjected to the applied torques as shown, determine the maximum shear stress developed in the shaft. The shaft is supported by two smooth journal bearings at A and B.

7 Name: Student ID: [7] Torsion Page 7 of 21 HOMEWORK R3-2 The shaft shown in the figure is supported by two bearings and is subjected to three torsional loadings. Determine the shear stress developed at points A and B, located at section aa of the shaft in the figure.

8 [7] Torsion Page 8 of 21 POWER TRANSMISSION Q: What is power? A: Power is defined as the work done per unit time 1) If during the time dt, an applied torque T causes the shaft to rotate d, the power required is defined as: T d P dt Define shaft s angular velocity: d (rad/s) dt P T [Power Transmission Formula] 2) If the frequency of the shaft s rotation f (H z = rev / s) is given instead of : 1 rev = 2 rad 2 f P 2 f T (Note: rpm = rev / min) SHAFT DESIGN 1) When the power transmitted by a shaft (P) and its frequency (f ) are known, the torque developed in the shaft can be determined as: T P [Torque Developed in the Shaft] 2 f 2) Knowing T and the allowable shear stress for the material ( allow ), you can determine the required size of the shaft s cross section as: J T [Required Size of the Shaft] c allow

9 [7] Torsion Page 9 of 21 CLASS EXAMPLE The 3 horsepower motor can turn at 330 rev/min. If the shaft has a diameter of 3/4 in., determine the maximum shear stress that will be developed in the shaft.

10 Name: Student ID: [7] Torsion Page 10 of 21 HOMEWORK R3-4 A solid steel shaft AB shown in the figure is to be used to transmit 5 horsepower from the motor M. If the shaft rotates at 175 rpm and the steel has an allowable shear stress of 14.5 ksi, determine the required diameter of the shaft to the nearest 1/8 inch.

11 [7] Torsion Page 11 of 21 [7.2] Statically Indeterminate Torsion ANGLE OF TWIST 1) The design of a shaft depends on restricting the amount of rotation or twist that may occur when the shaft is subjected to a torque 2) The angle of twist of one end of a shaft with respect to its other end can be analyzed for the shaft with a circular cross section 3) Recall that from geometry: d dx 4) Applying Hooke s law for shear: G, or G and torsion formula: Tc J d T ( x) dx J ( x) G 5) Integrating over the entire shaft, the angle of twist is given as: L T( x) dx [Angle of Twist Formula] J ( x) G 0

12 [7] Torsion Page 12 of 21 where, : Angle of twist of the shaft (rad) T (x): Internal resultant torque at a cross section at a position x J (x): Polar moment of inertia of a cross section at a position x G : Shear modulus of elasticity 6) If T (resultant internal torque at the cross section) is constant and the shaft is prismatic (constant cross section), and the material is homogeneous: TL [Simplified Angle of Twist Formula] JG 7) If the shaft is composite (consists of several discrete segments of shafts): TL [Composite Angle of Twist Formula] JG

13 [7] Torsion Page 13 of 21 8) The right-hand rule is applied for sign convention of torque and angle of twist 9) The angle of twist is often written in the form: A B Angle of twist of end A with respect to end B of the shaft

14 [7] Torsion Page 14 of 21 CLASS EXAMPLE The shaft is made of 2014-T6 aluminum (shear modulus 27 GPa) and consists of a solid section AB and a tube BD which is fixed to the wall at D. Determine the angle of twist at its end A relative to C when it is subjected to the torsional loading shown. Section AB has a diameter of 25 mm, and BD has an inner diameter of 40 mm and an outer diameter of 50 mm T6 Aluminum: G = 27 GPa

15 Name: Student ID: [7] Torsion Page 15 of 21 HOMEWORK The two solid steel shafts (shear modulus 80 GPa) shown in the figure are coupled together using a gear. 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. Determine the angle of twist of end A of shaft AB when the torque 45 Nm is applied.

16 [7] Torsion Page 16 of 21 CLASS EXAMPLE The shaft has a radius c and is subjected to a torque per unit length of t 0, which is distributed uniformly over the entire length L. If it is fixed at its far end A, determine the angle of twist of end B. The shear modulus is G.

17 Name: Student ID: [7] Torsion Page 17 of 21 HOMEWORK The 2-in.-diameter solid cast-iron post (shear modulus ksi) shown in the figure is buried 24 in. in soil. Assume that the torque (generated by a couple of 25-lb forces) is about to turn the post, and the soil exert a uniform torsional resistance of t (lbin./in.) along its 24-in. buried length. Determine the maximum shear stress in the post and the angle of twist at its top.

18 [7] Torsion Page 18 of 21 STATICALLY INDETERMINATE TORSION Q: Again, what does statically indeterminate mean? A: Equations of equilibrium (statics) are no longer sufficient to determine all the reaction torques (redundant supports: more unknowns than numbers of equations available) 1) It is required to have an extra condition in addition to the equations of equilibrium to solve the statically indeterminate problems: it is called compatibility (or kinematic) condition 2) The equations of equilibrium can provide a relationship as: x M 0: T T A T 0 (1) B 3) Since the both sides of the shaft are fixed supported, we know that there will be no total angular deformation on the shaft: [compatibility condition] 0 A / B 0 C / A B / C

19 [7] Torsion Page 19 of 21 TL 4) Applying the angle of twist formula: JG T L A JG AC T L B JG BC 0 (2) 5) Solving the equations (1) and (2), the reaction torques can be found as: T T A B L T L BC L T L AC Note: L L AC L CB

20 [7] Torsion Page 20 of 21 CLASS EXAMPLE The A-36 steel shaft has a diameter of 50 mm and is fixed at its ends A and B. If it is subjected to the 300 Nm couple, determine the maximum shear stress in regions AC and CB of the shaft.

21 Name: Student ID: [7] Torsion Page 21 of 21 HOMEWORK The solid steel shaft shown in the figure has a diameter of 20 mm. If it is subjected to the two torsional loadings, determine the reactions at the fixed supports A and B.