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

Torsional Deformation of a circular shaft Length BD when

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.

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

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

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.

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.

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.

Point A R = 0.075 m J = Jtotal T = internal torque Point B R = 0.015 m J = Jtotal T = internal torque T=4.25 – 3.0 = 1.25 kNm

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

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

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

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

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

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.

Calculate the T Maximum shear stress

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.

Solve it ! Calculate the T Maximum shear stress Calculate d

Angle Twist Lecture 1

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

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.

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

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.

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

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

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