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STRESSES IN SHAFTS DUE TO SHEAR FORCES AND BENDING MOMENTS Lecture #6 Course Name : DESIGN OF MACHINE ELEMENTS Course Number: MET 214 After generating shear and moment diagrams for a shaft, the stresses accompanying the internal shear forces and bending moments needs to be investigated so that the shaft can be designed (specify material type and dimensions) to accommodate the stresses and/or deflections that results from the transverse loads applied to the shaft. In order to develop relationships relating bending moments in shafts to the amount of stress the bending moments generate, it is essential to understand the geometry associated with bending. Shown in the figures below are a unloaded beam having a rectangular cross section and the same beam experiencing a transverse loading condition. A side view of the situation under analysis is depicted below. Note: F a Y Y Since Y Y C max Y FY max a C The force FY will produce a moment MY about the neutral axis. M M Y F YY Y Y C max aY 2 M Y Y a max C To determine the total moment, all of the individual moments must be added together. M ( x) MY max max C Y 2 a C Y 2 a Recall the definition of second moment of area or the moment of inertia of the cross section about its own centroidal axes (not to be confused with polar moment of area J or mass moment of inertia Jm ) IX Y M (z) 2 is area moment of inertia of cross section about centroidal X axis. a max C Ix M ( z ) Total internal resisting moment. Found from moment diagram. Rearranging the formula max M max C Ix Since shafts are frequently formed from solid circular cylinders, recall the definition of I for a circular cross section as shown below. Accordingly, for solid circular cylindrical shafts, M TC max I I d 64 4 1 J 2 where J= Polar area moment Since stress is proportional to distance max C M TC Y I Y C M TY I Also note that M can be expressed in terms of M T max I C I S C max S Section Modulus To determine the shaft diameter needed in a particular application, use the formula for bending moments and substitute the expression for I in terms of D and solve for the diameter D. MT MT MT max I C 4 max D D / 2 64 32 max D 3 Formula may be rearranged and solved for D as shown below D 3 32 M T max MT : Take greatest value from composite moment diagram σmax : determined from material choice for the shaft and factor of safety = If the shaft has a diameter of 50 mm, determine the absolute maximum bending stress in the shaft. Additional situations exist where it is of interest to combine stresses arising from different types of loading. As an example that will be encountered frequently in the design of machine elements, consider the shaft shown below which is supported by two bearings and carries two V belt sheaves. During steady state operation, the sheaves (pulleys) apply reaction torques to the shafts as indicated by the arrows shown on the figure. Due to tension in the belts, the belts also apply a transverse load to the shaft resulting in bending moments. Previously, torsional shear stresses and the stresses due to bending were treated independently. Due to the nature of the loading applied to the shaft in this example, it is necessary to determine the effects of both stresses acting in combination. The accompanying shear and moment diagrams are shown below. Note max MC I 2 M D D 4 M 32 D 3 1540 32 3 . 14 1 . 25 3 8030 psi 64 Recall a negative sign for bending moment indicates the upper fibers (i.e. fibers lying above the neutral axis) are in tension. To calculate the amount of shear stress due to torsion, note the amount of torque declared for the shaft as indicated in original figure. T=1100 in-lb xy TC J 2 T D D 4 16 T D 3 16 1100 3 . 14 1 . 25 3 2870 psi 32 These stress levels exist at the outside radius of the shaft as shown below. Stresses on element k appear below Note: the bending moment associated with the transverse loading leads to bending stresses which are tensile or compressive in nature as depicted above on element k. The torque applied to the shaft leads to a torsional shear stress on element k as depicted above. To assist in the visualization of the stresses existing on element k, envision a band aide applied to the shaft before the shaft experiences loading and then note how the shape of the band aid changes after the shaft is placed under load. The bending forces due to belt tension would have a tendency to stretch the band aid as the shaft bows in response to the transverse loads. See diagram on previous slide. The torque applied to the shaft would have a tendency to twist the band aid and thereby create torsional shear stresses. Since element k involves both tensile stresses (due to bending moments) and torsional shear stresses, the maximum amount of shear stress associated with the element must be found by utilizing equations associated with Mohr’s circle. Using equation 4-4 from the book by Mott and noting σY=0, for expression for τmax can be rewritten as follows 2 X 2 max 2 XY Substituting the definitions for σY and τXY results in the following expression for τmax. 32 M X 3 D XY 16 T D 3 2 max 32 M 16 T 3 3 2 D D max 16 M 3 D 2 16 M 3 D max 2 2 T T 2 2 2 The above equation for τmax does not include effects of varying loads and stress concentrations. If the shaft diameter must be determined for a particular application, the above equation may be rearranged to solve for D. As will be evidently shortly, the equation involving max on the previous slides is very similar to the equation adopted by the ASME in 1927 for transmission shaft design. Accordingly to the book titled “Machine Design for Mobile and Industrial Applications” by Krutz et al, although the code has been replaced by a newer and more sophisticated one, it is often still used due to its simplicity. As indicated in the book titled “Machine Design for Mobile and Industrial Applications” by Krutz et al, stresses due to the use of a keyway in the shaft and/or stresses due to press fitting the hub of a component onto a shaft must be accounted for when sizing a shaft accordingly to the 1927 standard. As suggested in the book, one way to handle each of these problems is to increase the factor of safety existing in the equation of the standard. Although other texts books assert that the 1927 standard is obsolete, it is believed that adopting the suggestions made in the text book by Krutz et al will help the student obtain insight concerning factors that must be considered in the design of a shaft. The use of a simple approach enables a context to be established for the material that is consistent with material the student has been exposed to previously, such as Mohr’s circle. Such considerations would be lacking in alternative approaches that emphasize detailed calculations that are beyond the comprehension of most technology students at this level. It is for these reasons the 1927 standard will be used to illustrate the basic issues that must be considered when integrating several different components onto a shaft including bearings. After an initial exposure to the material has been established, a more contemporary approach may be undertaken using any number of text books including the book by Mott. Accordingly, the 1927 standard will be used as a bridge to enable the student to approach more contemporary information by introducing the students to a context consistent with prior exposure. Shafts, Keys and Couplings The accepted code for the design of transmission shafting is the American Standards Association Standard B17C-1927. It refers to all shafts for transmitting power and includes the effect of the type of load. Common usage, however, classifies shafts in many ways. For Example, there are shafts for prime movers such as generator, engine, and turbine shafts; shafts for power transmission such as line shafts, countershafts, and jack shafts, as well as machine shafts and spindles. A member carrying rotating wheels, gears, etc., is generally referred to as an axle; for example, the rear axle of an automobile. An axle may be either stationary or rotating. Short shafts on machines are usually referred to as spindles. When one or more pieces of shafting joined by couplings receive motion from a prime mover, and transmit motion to machines, the term line shafting is applied. When a shaft is placed is placed between the driver and the line shaft, it is a counter shaft or jack shaft. A stub shaft or head shaft is connected directly to a motor or engine. In addition to various types of shafts indicated, shafts are usually classified as to form. The most common form of shafting is of a solid circular cross section and is called a solid shaft. There are also hollow shafts. Flexible shafts make possible the transmission of power around corners, and examples are found in dental tools, small power tools, and the common speedometer drive in automobiles. A flexible shaft is of necessity limited to relatively short lengths and the transmission of light loads. Shock and Fatigue factors: The code mentioned previously gives factors to be used for various manners in which the torsional and moment loads are applied. The below table lists the recommended factors . They are to be applied to the torque T and moment M so that equation becomes a 16 d 3 K t T 2 K m M 2 Values for Km and Kt Types of loading Km Kt Stationary shafts: Gradually applied Suddenly applied 1.0 1.5 to 2.0 1.0 1.5 to 2.0 1.5 1.5 to 2.0 2.0 to 3.0 1.0 1.0 to 1.5 1.5 to 3.0 Rotating shafts: Gradually applied or steady Suddenly applied, minor shocks Suddenly applied, heavy shocks It must be emphasized that the 1927 equation can be rewritten in several equivalent, but alternative ways. d 3 16 K t T 2 K m M 2 a Where = ℎ Ex: A solid circular shaft free of keyways is transmitting 50 hp and rotates at 200 rpm. Assuming that 48,000 psi material is used in the shaft with a safety factor of 6 and that the shaft is subject to steady torsional stresses only, what shaft diameter is required? Ex. A solid circular shaft for power transmission with keyways is subject to a maximum torque of 6000 in-lb. What is the required shaft diameter, if the allowable shear stress is 6000 psi and if there will be sudden applications of load with minor shock? Shear in straight members Beams generally support both shear and moment loadings. The shear V is the result of a transverse shear-stress distribution that acts over the beam’s cross-section. Due to the complementary property of shear, shear stresses will also act along longitudinal planes of the beam. For example, a typical element removed from the interior point on the cross section is subjected to both transverse and longitudinal shear stress as shown below. Shear Stress Distribution Shear stress distribution for a Rectangular cross section A situation that is encountered frequently in power transmission systems involves combining a torsional shear stress distribution with a shear stress distribution arising from a shear force. Such a situation exists at a bearing support that is supporting a section of shaft experiencing pure torsion. The shear stresses due to the shear forces arises from the bearing reactions existing at the bearing supports. A typical situation involving shear stress combinations is shown below. The situation shown on the previous slide involves a solid circular shaft experiencing torsional shear stresses and shear stresses due to a shear force which could be envisioned as resulting from a bearing reaction force applied to the shaft. As will be evident shortly, other situations exist which could lead to the same combination shown on the previous slide. Due to the nature of the distributions and how shear stresses combine, the two distributions shown on the previous slide can only be added when the distributions are either parallel or antiparallel. However, it is still possible to identify the maximum value of stress that exists in the combined distribution. The maximum value of stress existing with the combined distributions can be calculated from the following formula also shown on the previous slide. = + = 4 3 + Where = ℎ = ℎ To identify how the shear stress distributions described previously may be combined to establish the maximum and minimum shear stress level existing when both distributions are present on the same cutting plane, consider the example shown below involving a simplified representation of a gear transmitting a particular level of torque established by the force and the geometry of the situation. Ex: A 3 inch diameter gear transmits 5HP to a short 1 inch diameter shaft at 300 rpm. To determine F note the following. P Tn T 1050 in lbs 63000 T rF F 700 lbs To investigate the shear stress distribution existing with the above applied load, an equivalent representation may be utilized involving a shear force acting through the centroid perpendicular to the neutral axis of the shaft and a torque T as shown below. The shear stress distribution resulting from the equivalent shear force is shown below. where max 4V max 4 3 A 700 3 1 2 1190 psi 4 The shear stress distribution resulting from torsion is shown below. max TC J 2 T D D 4 16 T D 3 32 max 16 1050 5350 3 . 14 1 3 psi Shear stress levels associated with the two distributions shown above can be added only when the shear stresses from both distributions are parallel or anti-parallel. Since the orientation of the shear stress distribution due to torsion varies across the cutting plane, the shear stress levels associated with the equivalent representation can only be obtained along the horizontal axis in this particular instance. Fortunately, the stress levels obtain by combining the two distributions for points along the horizontal axis include the maximum and minimum values as illustrated below. It should be noted that the above example transformed a tangential force acting at the radius of a shaft to a force acting through the centroid of the shaft and a torque applied to the shaft. The equivalent force acting through the centroid give rise to shear stresses and bending stresses and the force is also responsible for creating deflections in the shaft assuming the location of the component is not at a bearing support. The effect of the force as it contributes to bending stresses will be accounted for in shear and moment diagrams and consequently, the force’s influence as it pertains to bending stresses is accounted for in the equation existing with the 1927 standard for shafting. However, it should be noted that if the equation existing with the 1927 standard is used to determine the shaft diameter, then the stresses due to shearing forces will be neglected. Given the disparity that typically exists between shear stresses due to shear forces and bending stresses due to transverse loads leading to bending moment, such an assumption is usually acceptable.