lecture 4 - torsion
DESCRIPTION
Lecture 4 - TorsionTRANSCRIPT
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Dr/Ahmed Fathi Mohamed SalihUniversity Of BahriCivil Engineering DepartmentMechanics Of MaterialTorsion
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3 - *
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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
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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
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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
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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.
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Tubular shaft
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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
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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.
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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
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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.
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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,
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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
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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.
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Solution for Example 6From free body diagram, the internal torque is T.
Thus, at x,For angle of twist,
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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
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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
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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