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39 Pages

### NEW 375 Lecture10(2)

Course: ENGIN 375, Spring 2008
School: Cincinnati
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Word Count: 1169

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DO Introduction WHAT A JAR OF RELISH, A DOORKNOB AND A CROSS WRENCH HAVE IN COMMON? TORQUE = T = F x d =COUPLE T F F TORQUE = T = F x d =COUPLE F T d F TORQUE = T = F x d =COUPLE d T F F Torsion: Objectives Determine the effects of applying a torsional loading to a long straight member such as a shaft or tube. Determine the stress distribution within the member. Determine the angle of twist when...

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DO Introduction WHAT A JAR OF RELISH, A DOORKNOB AND A CROSS WRENCH HAVE IN COMMON? TORQUE = T = F x d =COUPLE T F F TORQUE = T = F x d =COUPLE F T d F TORQUE = T = F x d =COUPLE d T F F Torsion: Objectives Determine the effects of applying a torsional loading to a long straight member such as a shaft or tube. Determine the stress distribution within the member. Determine the angle of twist when the member behaves in a linear-elastic manner. What is torque? Torque is a moment that tends to twist a member about its longitudinal axis. T T (right hand rule) Engineering Applications The most common application is provided by transmission shafts, which are used to transmit power from one point to another, as from a steam turbine to an electric generator, or from a motor to a machine tool, or from the engine to the rear axle of an automobile. Example of a Transmission Shaft Preliminary Discussion of the Stresses in a Shaft Shaft AB Subjected to Torque FBD of Section BC of Shaft = We must have shearing forces dF perpendicular to the axis of the shaft, i.e., on the plane of shaft, which portion AC exerts on BC when the shaft is twisted. INTERNAL TORSIONAL MOMENT Line of action of the differential force dF is in the plane of the cross-section and is perpendicular to the radius r dF = t dA r 90 deg. T dT r dF r tdA Note that the sum of the moments of the differential shearing forces [ dF = t dA ] about the axis of the shaft is equal in magnitude to the torque T. If we knew the distribution of shear stress (t), then we could evaluate the integral. As shown later, t is not distributed uniformly. t c a t h d Note that shear stress cannot take place only in the plane of the cross-section (plane abcd) t c a t h d t Note that shear stress cannot take place only in the plane of the cross-section (plane abcd) t t c a t h t t THEOREM OF RECIPROCITY d Note that shear stress cannot take place only in the plane of the cross-section (plane abcd) t Therefore, shear stresses exist on the faces perpendicular to the axis of the member, AND equal shear stresses on the faces parallel to the axis of the member. The following figure demonstrates that shearing stresses occur on longitudinal planes as well as on planes perpendicular to the axis of the shaft. Because of the longitudinal shearing stresses, there will be a sliding tendency as shown here. While sliding will not actually occur in the shaft made of a homogeneous and cohesive material, the tendency for sliding will exist, showing that stresses occur on longitudinal planes as well as on planes perpendicular to the axis of the shaft. If we were to mark a small rectangle on the surface of an elastic circular shaft (a pencil eraser, for example), and subject the shaft to torsion, the rectangular element would deform into a rhombus. Behavior of Circular Shafts When a circular shaft is subjected to torsion, every cross section remains plane and undistorted. While the various cross sections along the shaft rotate through different amounts, each cross section rotates as a solid rigid slab. As long as the shaft is circular, solid or hollow, this behavior remains true. This behavior is because of axisymmetric (that is, the shaft remains the same when it is viewed from a fixed position and rotated about its axis through an arbitrary angle) nature of a circular shaft. Behavior of Non-Circular Members For non-circular cross sections, the cross-sections warped are and do NOT remain plane. Distribution of Shearing Strains Consider deformations at radius r for applied torque T T Segment shown below These two sides do not distort because these are two circles on the shaft that do not warp. g Before Applying Torque After Applying Torque g For small values of shearing strain g, the arc length AA' = L g Also, AA' = r f Therefore: Lg = r f rf g L Shear strain equation g The shearing strain g at a given point of a shaft in torsion is proportional to the angle of twist f. The shearing strain is also proportional to the distance r from the axis of the shaft to the point under consideration. The shearing strain in a circular shaft varies linearly with the distance from the axis of the shaft. Therefore, the shearing strain is maximum on the surface of the shaft, where r = the radius c. f f g max r max c L L r and g g max c STRESSES IN ELASTIC RANGE Consider the case when the torque T is such that all shearing stresses in the shaft remain below the yield strength. Hooke's law will apply and there will be no permanent deformation. tG g t = shear stress G= modulus of rigidity g = shearing strain t Slope=G g Multiplying both sides of the previous equation relating shearing strain variation on the cross-section by the shear modulus G. r r G g G g max G g max c c From : t G g r t t max c NOTE: The shear stress distribution is linear with respect to radius. The Shearing Stress Distribution in Solid and Hollow Shafts REVISIT THE FOLLOWING RELATIONSHIP dF = t dA r Line of action of the differential force dF is in the plane of the cross-section and is perpendicular to the radius r T dT r dF r tdA Elastic Torsion Formulas T r t dA r r t max dA c t max 2 T r dA c r 2 dA is the polar moment of inertia J of the The integral cross section with respect to its center. t max T J c Tc t max J r r Tc From t t max c c J Tr Then t J POLAR MOMENT OF INERTIA FOR CIRCULAR CROSSSECTIONS Solid circular cross-section J r dA r 2 2rdr 2 A A c 3 c 1 4c 2 r dr r 0 2 0 1 4 J c 2 co ci 1 4 J co ci4 2 Hollow circular cross-section Elastic Torsion Formula Tr t J NOTE THE SIMILARITY BETWEEN THE FORM OF THE SHEAR STRESS EQUATION FOR TORSION AND THE FORM OF THE NORMAL STRESS EQUATION FOR AXIAL LOAD. P P 1 A A t max Tc J Tc 45 J While the element "a"on the left is in pure shear, the element "c" on the right is subjected to normal stresses of the same magnitude, Tc/J, two of the normal stresses being tensile and two compressive. In a torsion test ductile materials, which generally fail in shear, will break along a plane perpendicular to the axis of the specimen. In a torsion test brittle materials, which are weaker in tension than in shear, will break along surfaces forming a 45 degree angle with that axis. Demonstration of a Brittle Material Subjected to Torque Demonstration of a Brittle Material Subjected to Torque Example 1 a) Calculate T that causes tmax=45 MPa. b) Calculate tmax caused by the same T as part (a) if a solid cylinder shaft of the same cross-sectional area of that in part (a) is used. 30 mm 45 mm T Example 1 (Cont.) Part (a ) Jt max Tc t max T J c 6 45 10 4 4 45 30 6 10 T 2 45 6 T 5.17 10 N mm 5.17 kN m Example 1 (Cont.) Part (b) A r (45 30 ); r 33.54mm Tc t max J 6 5.17 10 33.54 2 t max 87.2 N / mm 4 33.54 2 t max 87.2 MPa 2 2 2
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