SRSD 2093: Engineering Mechanics 2SRRI SECTION 19 ROOM 7, LEVEL 14, MENARA RAZAK PLAN OF THE LECTURE TORSION Today’s Objectives: Students will be able to: Determine the shear stresses in a circular shaft due to torsion Determine the angle of twist Analyze statically indeterminate torque-loaded members In-Class Activities: 1. Applications 2. Torsion Formula 3. Angle of Twist 4. Statically indeterminate torque-loaded members APPLICATIONS The tubular drive shaft for this truck was The shaft attached to the subjected to an excessive torque center of this wheel is resulting in failure caused by yielding of subjected to a torque, and the material. the maximum stress it creates must be resisted by the shaft to prevent failure. TORSIONAL DEFORMATION OF A CIRCULAR SHAFT • Torsion is a moment that tends to twist a member about its longitudinal axis. • Engineers need to understand the effects of torque to properly design axles or drive shafts used in vehicles and machinery. • Torque causes shear stresses to develop in shafts. • If the angle of rotation is small, the length of the shaft and its radius will remain unchanged. THE TORSION FORMULA • For solid shaft, shear stress varies from zero at shaft’s longitudinal axis to maximum value at its outer surface. • When material is linear-elastic, Hooke’s law applies. THE TORSION FORMULA • If the shaft has a solid circular cross section, • If a shaft has a tubular cross section, EXAMPLE 1 Determine the shear stress developed at points A and B, located at section a–a of the shaft. EXAMPLE 1: SOLUTION Free-body diagram The polar moment of inertia for the shaft is 4 Since point A is at ρ = c = 75 mm, Likewise for point B, at ρ =15 mm, we have POWER TRANSMISSION • 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 EXAMPLE 1 A 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 n = 175 rpm and the steel has an allowable shear stress of allow τallow =100 MPa, determine the required diameter of the shaft to the nearest mm. Solution: The torque on the shaft is Since As 2c = 21.84 mm, select a shaft having a diameter of 22 mm. ANGLE OF TWIST • Integrating over the entire length L of the shaft, we have Φ = angle of twist T(x) = internal torque J(x) = shaft’s polar moment of inertia G = shear modulus of elasticity for the material • Assume material is homogeneous, G is constant, thus • Sign convention is determined by right hand rule, EXAMPLE 1 The solid steel shaft has a diameter of 20 mm. If it is subjected to the two torques, determine the reactions at the fixed supports A and B. Solution: By inspection of the free-body diagram, Since the ends of the shaft are fixed, Using the sign convention, Solving Eqs. 1 and 2 yields TA = -345 Nm and TB = 645 Nm. The End.