1. Torsion Torsional Deformation of a circular shaft,
Torsion Formula , Power Transmission

2. Torsional Deformation of a circular shaft
Length BD
when

3. The Torsion Formula Angular strain is proptional to shear stress:
Mean:
highest shear stress: will be at farthest away from center
At the center point, there will be no angular strain and therefore no shear stress is developed.

4. The Torsion Formula J: Polar moment of area
Maximum shear stress due to torsion
Shear stress due to torsion at radius r

5. Polar Moment of Inertia
Polar Moment of Inertia (J)
Solid shaft
Tubular shaft

6. Solve it!
The solid circular shaft is subjected to an internal torque of T= 5kNm. Determine the shear stress developed at point A and B. Represent each state of stress on a volume of element.

7. Solve it!
The solid circular shaft is subjected to an internal torque of T= 5kNm. Determine the shear stress developed at point A and B. Represent each state of stress on a volume of element.

8. Example 1
The shaft shown in figure is supported by two bearings and is subjected to three torques. Determine the shear stress developed at points A and B, located at section a-a of the shaft.

9. Point A
R = m
J = Jtotal
T = internal torque
Point B
R = m
J = Jtotal
T = internal torque
T=4.25 – 3.0 = 1.25 kNm

10. Solve it!
Determine the maximum shear stress developed in the shaft at section a-a.

11. Maximum Torsional Shear
T:2100 Nm = 2100(10)3 Nmm
r: 40 mm
J: tubular
Polar Moment of Area
Max torsional shear when r = 40mm

12. Solve it!
The solid shaft has a diameter of 40 mm. Determine the absolute maximum shear stress in the shaft.

13. Maximum Torsional Shear
T:70 Nm = 70(10)3 Nmm
r: 20 mm
J: solid r =20 mm

14. Power Transmission Power P: power (1 Watt = 1 Nm/s) T: torque (Nm)
w: radian/s
Input n(rpm)
Input frequency of shaft rotation

15. Solve it !
The gear motor can develop 1.6kW when it turns at 450 rev/min. If the shaft has a diameter of 25mm, determine the maximum shear stress developed in the shaft.

16. Calculate the T
Maximum shear stress

17. Solve it !
The gear motor can develop 2.4kW when it turns at 150 rev/min. If the allowable shear stress for the shaft is tallow = 84 Mpa, determine the smallest of the shaft to nearest multiples of 5mm that can be used.

18. Solve it !
Calculate the T
Maximum shear stress
Calculate d

19. Angle Twist
Lecture 1

20. Sign Convention Right hand rule
+ve: direction of the thumb is away from the part
Lecture 1

21. Solve it!
The splined ends and gears attached to the A-36 steel shaft are subjected to the torques shown. Determine the angle of twist of end B with respect to end A. The shaft has a diameter of 40 mm.

22. Total angle of twist: 300Nm 400Nm 500Nm 200Nm 300Nm Tac= 300Nm 500Nm
Tcd=200Nm
300Nm
Tdb =400Nm
400Nm
Total angle of twist:

23. Solve it !
The two shafts are made of A-36 steel. Each has a diameter of 25mm, and they are supported by bearings at A, B and C, which allow free rotation. If the support at D is fixed, determine of the angle of twist of end B and A when the toques are applied to the assembly as shown.

24. Direction of FF
Equilibrium of shaft AGFB FF
Equilibrium at gear F and E
Direction of FE FE

25. Equilibrium of shaft DCE
Direction of FF
Equilibrium of shaft DCE
9ONm
3ONm
12ONm FE
3ONm
9ONm
9ONm

26. Length of arc 12 = length of 13
Angle of twist at end B = angle of twist at F  FB : there is no torque
Angle of twist at end A