Dr/Ahmed Fathi Mohamed SalihUniversity Of BahriCivil Engineering DepartmentMechanics Of MaterialTorsion

3 - *

Torque is a moment that twists a member about its longitudinal axis.If the angle of rotation is small, the length of the shaft and its radius will remain unchangedIntroduction

Torsional Loads on Circular ShaftsInterested in stresses and strains of circular shafts subjected to twisting couples or torquesGenerator creates an equal and opposite torque T on the shaft (action-reaction principle)Shaft transmits the torque to the generatorTurbine exerts torque T on the shaft

When material is linear-elastic, Hookes law applies. A linear variation in shear strain leads to a corresponding linear variation in shear stress along any radial line on the cross section.3. The Torsion Formula= maximum shear stress in the shaft= shear stress= resultant internal torque= polar moment of inertia of cross-sectional area= outer radius of the shaft= intermediate distance

5.2 THE TORSION FORMULASolid shaftJ can be determined using area element in the form of a differential ring or annulus having thickness d and circumference 2 .For this ring, dA = 2 dJ is a geometric property of the circular area and is always positive. Common units used for its measurement are mm4 and m4.

Tubular shaft

Integrating over the entire length L of the shaft, we have

Assume material is homogeneous, G is constant, thus

*Used when several different torque, cross section and shear modulus variedSign convention is determined by right hand rule, Angle of Twist = angle of twistT(x) = internal torqueJ(x) = shafts polar moment of inertia G = shear modulus of elasticity for the material

Example 4The shaft is supported by two bearings and is subjected to three torques. Determine the shear stress developed at points A and B, located at 75 mm as outer diam and 15 mm as inner diam respectively.

Solution for Example 4From the free-body diagram of the left segment,

The polar moment of inertia for the shaft is

Since point A is at = c = 75 mm,

Likewise for point B, at =15 mm, we have

Example 5The 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.

Solution for Example 5From free body diagram,

Angle of twist at C is

Since the gears at the end of the shaft are in mesh,

Solution for Example 5Since the angle of twist of end A with respect to end B of shaft AB caused by the torque 45 Nm,

The rotation of end A is therefore

Example 6The tapered shaft is made of a material having a shear modulus G. Determine the angle of twist of its end B when subjected to the torque.

Solution for Example 6From free body diagram, the internal torque is T.

Thus, at x,For angle of twist,

Power is defined as the work performed per unit of time.For a rotating shaft with a torque, the power is

Since , the ,power equation is

For shaft design, the design or geometric parameter is Power Transmission

Example 7A solid steel shaft AB is to be used to transmit 3750 W from the motor M to which it is attached. If the shaft rotates at =175 rpm and the steel has an allowable shear stress of allow allow =100 MPa, determine the required diameter of the shaft to the nearest mm.4. Power Transmission

Solution for Example 7The torque on the shaft is

Since,

As 2c = 21.84 mm, select a shaft having a diameter of 22 mm Power Transmission